CBSE Class 10 Science Assignments Collection

Module : 10P (01, 02 & 03) : Section : A

Q.1. Two charged batteries A and B are at potential – 30 and + 20 V respectively and they are connected by a connecting wire. The current
(i) Will flow from A to B
(ii) Will flow from B to A
(iii) Will not flow at all
Answer: (ii) Will flow from B to A
Conventional current always flows from a region of higher electric potential to a region of lower electric potential. Since battery B is at +20 V and battery A is at -30 V, the current flows from B to A.
In simple words: Current moves from the positive terminal (higher potential) to the negative terminal (lower potential).

Exam Tip: Always remember that conventional current flows from higher potential to lower potential, which is the opposite direction of electron flow.

Q.2. The work done in moving a positive charge across two points in an electric circuit is a measure of
(i) Current
(ii) Potential difference
(iii) Resistance
(iv) Power
Answer: (ii) Potential difference
Potential difference between two points is defined as the amount of work done in moving a unit positive charge from one point to the other: \(V = \frac{W}{Q}\).
In simple words: Potential difference measures the energy needed to move a charge between two points.

Exam Tip: Learn the formula \(V = \frac{W}{Q}\) where W is work done and Q is charge.

Q.3. 1 coulomb = ____________ electrons
(i) \(1.6 \times 10^{-19}\)
(ii) \(9.1 \times 10^{-31}\)
(iii) \(6.25 \times 10^{18}\)
(iv) None of these
Answer: (iii) \(6.25 \times 10^{18}\)
Using the quantization of charge formula \(Q = ne\), the number of electrons for 1 coulomb is \(n = \frac{Q}{e} = \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18}\).
In simple words: One coulomb contains over six quintillion electrons.

Exam Tip: Memorize this standard constant value as it is frequently used in numerical problems.

Q.4. The unit of specific resistance is
(i) Ohm metre
(ii) Ohm
(iii) Volt
(iv) Ampere
Answer: (i) Ohm metre
Specific resistance (or resistivity, \(\rho\)) is given by \(\rho = \frac{R A}{l}\), where R is resistance in ohms, A is area in \(\text{m}^2\), and l is length in m. Thus, its SI unit is \(\Omega \cdot \text{m}\) (ohm metre).
In simple words: The unit for resistivity is ohm metre.

Exam Tip: Do not confuse the unit of resistance (ohm) with the unit of resistivity (ohm metre).

Q.5. Most metals are good conductors of electricity because they have
(i) Large number of molecules
(ii) Large number of free electrons
(iii) A shiny surface
(iv) A low temperature
Answer: (ii) Large number of free electrons
Metals possess a high density of delocalized valence electrons (free electrons) that are able to move freely throughout the crystal lattice under the influence of an electric field.
In simple words: Free electrons carry electric current through metals.

Exam Tip: The presence of mobile charge carriers (free electrons) is the fundamental reason for electrical conductivity in solids.

Q.6. A wire is cut into half. Its resistivity
(i) Also becomes half
(ii) Becomes double
(iii) Remains same
(iv) Depends upon thickness of the wire
Answer: (iii) Remains same
Resistivity is an intrinsic property of a material and depends only on the nature of the material and temperature, independent of its dimensions or shape.
In simple words: Cutting a wire changes its resistance, but its material property (resistivity) stays constant.

Q.7. 1 volt of potential difference is equal to
(i) 1 coulomb per joule
(ii) 1 joule
(iii) 1 joule per coulomb
(iv) 1 ampere
Answer: (iii) 1 joule per coulomb
Since potential difference \(V = \frac{W}{Q}\), the SI unit volt corresponds to 1 joule of work per 1 coulomb of charge (\(1\text{ V} = 1\text{ J/C}\)).
In simple words: One volt equals one joule of energy per coulomb of charge.

Exam Tip: Dimensional units can be easily derived from basic definitions like \(V = W/Q\).

Q.8. In a resistance box the resistances are connected in
(i) Series
(ii) Parallel
(iii) Neither of two
(iv) Both (i) and (ii)
Answer: (i) Series
In a resistance box, individual standard resistors are connected in series so that removing a plug introduces that specific resistance into the current path.
In simple words: Resistance box components are linked in a single line (series).

Exam Tip: Series connection allows resistances to add up directly (\(R = R_1 + R_2 + \dots\)).

Q.9. For ohmic resistors, the graph between V and I is
(i) Parabolic
(ii) Hyperbolic
(iii) Circular
(iv) Straight line
Answer: (iv) Straight line
Ohmic resistors obey Ohm's Law (\(V = IR\)), meaning current is directly proportional to voltage at a constant temperature, yielding a linear graph.
In simple words: The V-I graph for ohmic resistors is a straight line passing through the origin.

Exam Tip: The slope of the V-I graph represents the resistance of the conductor.

Q.10. A rheostat is used
(i) To bring a known change of the resistance in the circuit to alter the current
(ii) To continuously change the resistance in any arbitrary manner and there by alter the current.
(iii) To make or break the circuit at any instant.
(iv) Neither to alter the current nor the resistance.
Answer: (ii) To continuously change the resistance in any arbitrary manner and thereby alter the current.
A rheostat is a variable resistor used to smoothly adjust the magnitude of electric current in a circuit.
In simple words: A rheostat varies resistance smoothly to control current flow.

Exam Tip: Distinguish between a rheostat (variable resistance) and a switch (makes/breaks circuit).

Section : B

Q.1. What do these symbols denote
(i) Cell
(ii) Battery (Combination of cells)
(iii) Resistor / Resistance
(iv) Variable resistor / Rheostat
(v) Ohm (Unit of resistance symbol)
(vi) Closed plug key / Switch closed
(vii) Wire crossing without connection
(viii) Ammeter
Answer: The standard schematic circuit symbols denote: (i) Electric cell, (ii) Battery, (iii) Resistor, (iv) Variable resistor, (v) Ohm symbol, (vi) Closed key, (vii) Wires crossing without joining, (viii) Ammeter.
In simple words: These are standard symbols used in drawing electrical circuit diagrams.

Exam Tip: Practice drawing circuit symbols neatly as they carry marks in board exams.

Q.2. State the conditions on which resistance of a conductor depends.
Answer: The resistance of a conductor depends on:
1. Length of the conductor (\(l\)) - Resistance is directly proportional to length (\(R \propto l\)).
2. Area of cross-section (\(A\)) - Resistance is inversely proportional to cross-sectional area (\(R \propto \frac{1}{A}\)).
3. Nature of the material (resistivity, \(\rho\)).
4. Temperature of the conductor.
In simple words: Resistance depends on how long and thick the wire is, what material it is made of, and its temperature.

Exam Tip: Remember the formula \(R = \rho \frac{l}{A}\) to quickly recall the dependency factors.

Q.3. State the condition under which electric charge can flow through a conductor.
Answer: Electric charge can flow through a conductor only when there is a potential difference (voltage) established between the two ends of the conductor.
In simple words: A voltage difference is required to push charges through a wire.

Exam Tip: Think of potential difference like water pressure that drives water flow through a pipe.

Q.4. A wire of resistance 4R is bent in the form of a circle. What is the effective resistance between the ends of diameter?
Answer:
Total resistance of the wire = \(4R\).
When bent into a circle and connected across the ends of a diameter, the wire is divided into two equal semicircular halves connected in parallel.
Resistance of each half = \(\frac{4R}{2} = 2R\).
Equivalent resistance \(R_{eq}\) in parallel is given by:
\(\frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{2R} = \frac{2}{2R} = \frac{1}{R}\)
\(R_{eq} = R\)
In simple words: Two equal halves of \(2R\) in parallel give a combined resistance of \(R\).

Exam Tip: When a uniform wire of resistance R is divided into n equal parts and connected in parallel, the new resistance is \(R/n^2\).

Q.5. Is ohm’s law a universal law? If not give example.
Answer: No, Ohm's law is not a universal law. It is valid only for metallic conductors at a constant physical state (temperature, pressure).
Example of non-ohmic devices: Semiconductor diodes, transistors, and electrolytes where the V-I graph is non-linear.
In simple words: Ohm's law does not apply to all materials and electronic devices.

Exam Tip: Mention semiconductor devices as standard examples of non-ohmic components.

Q.6. Will an electron in an electric field move towards higher potential or lower potential?
Answer: An electron is negatively charged, so it experiences a force in the direction opposite to the electric field. Therefore, it moves from a region of lower potential towards a region of higher potential.
In simple words: Negative electrons are attracted toward the positive terminal (higher potential).

Exam Tip: Conventional current flows from high to low potential, whereas electrons flow from low to high potential.

Q.7. What is meant by the statements :
(i) Potential at a point is 1 Volt
(ii) The potential difference between the points is 1 Volt
Answer:
(i) Potential at a point is 1 Volt means that 1 joule of work is done in bringing a unit positive charge (\(1\text{ C}\)) from infinity to that point.
(ii) The potential difference between two points is 1 Volt means that 1 joule of work is done in moving a unit positive charge (\(1\text{ C}\)) from one point to the other.
In simple words: 1 volt means 1 joule of work per coulomb of charge.

Exam Tip: Be precise with the reference point: infinity for absolute potential, and between two specific points for potential difference.

Q.8. How will you maintain a potential difference between the ends of a conductor?
Answer: A potential difference across a conductor can be maintained by connecting its ends to a source of electrical energy such as an electric cell, battery, or a DC generator.
In simple words: Use a battery or cell to keep a constant voltage across the wire.

Exam Tip: Sources like cells convert chemical energy into electrical energy to maintain potential difference.

Section : C

Q.1. A large number of free electrons are present in metals. Why is there no current in the absence of electric field across it?
Answer: In the absence of an electric field, the free electrons in a metal move randomly in all possible directions due to thermal agitation. Consequently, the net flow of charge in any given cross-section is zero, resulting in no electric current.
In simple words: Random thermal motion of electrons cancels out, so no net current flows without an applied voltage.

Exam Tip: Emphasize that thermal motion is random and directional drift requires an external electric field.

Q.2. The following set of results was obtained:
V (Volt): 2, 4, 6, 8, 10
I (Amp): 0.74, 1.50, 2.22, 2.96, 3.70
(i) Plot a graph of potential difference on the Y-axis against current on the X-axis
(ii) Explain the graph
(iii) Use the graph to obtain a value for the resistance of the conductor.
Answer:
(i) The graph is plotted with V on the Y-axis and I on the X-axis, showing plotted points forming a straight line.
(ii) The straight-line graph passing through the origin indicates that potential difference (V) is directly proportional to current (I), verifying Ohm's law.
(iii) Resistance R is given by the slope of the V-I graph:
\(R = \frac{\Delta V}{\Delta I} = \frac{4 - 2}{1.50 - 0.74} = \frac{2}{0.76} \approx 2.63\) \Omega \)
In simple words: The straight line proves Ohm's law, and its slope gives the resistance value.

Exam Tip: Always choose points that lie directly on the best-fit line to calculate slope accurately.

Q.3. State the formula correlating the current flowing in a conductor and the voltage applied across it. Also show this relationship by drawing a diagram. What would be resistance of a conductor if the current flowing through it is 0.35 ampere when the potential difference across it is 1.4 volt?
Answer:
Ohm's Law formula: \(V = I \cdot R\) or \(R = \frac{V}{I}\).
Given: \(V = 1.4\text{ V}\), \(I = 0.35\text{ A}\).
\(R = \frac{1.4}{0.35} = \frac{140}{35} = 4\) \Omega \).
In simple words: Resistance is voltage divided by current, giving 4 ohms.

Exam Tip: Double-check decimal divisions to avoid calculation errors in numerical problems.

Q.4. Give reasons for the following:
(i) The resistance of a thicker wire of copper is much less than that of a thinner wire of copper of the same length.
(ii) For the same thickness of wires used we need a smaller length of a wire, made of an alloy, rather than a pure metal, for getting a resistance coil of a given value.
(iii) Electric lamp filaments are made from tungsten.
Answer:
(i) Resistance is inversely proportional to the area of cross-section (\(R \propto \frac{1}{A}\)). A thicker wire has a larger cross-sectional area, offering less obstruction to electron flow.
(ii) Alloys have much higher resistivity than pure metals. Since \(R = \rho \frac{l}{A}\), a higher resistivity requires a smaller length of wire to achieve the same resistance.
(iii) Tungsten has a very high melting point (\(3380^\circ\text{C}\)) and high resistivity, allowing it to glow brightly without melting when heated by an electric current.
In simple words: Thick wires have low resistance, alloys need shorter lengths due to high resistivity, and tungsten withstands high temperatures without melting.

Exam Tip: Clearly state the relevant formula (\(R = \rho l / A\)) while explaining resistance variations.

Module 10P (04 & 05) : Section : A

Q.1. SI unit of electric energy is
(i) Kwh
(ii) Joules
(iii) J sec
(iv) NC -1
Answer: (ii) Joules
Electric energy is a form of energy, and the standard SI unit for all types of energy is the joule (J).
In simple words: Energy is always measured in joules in the SI system.

Exam Tip: kWh is a commercial unit of electrical energy, whereas joule is the SI unit.

Q.2. Filament of electric bulb is made up of
(i) Copper
(ii) Silver
(iii) Tin
(iv) Tungsten
Answer: (iv) Tungsten
Tungsten is used for bulb filaments because of its high melting point and high resistivity.
In simple words: Bulb filaments are made of tungsten metal.

Exam Tip: Remember tungsten's high melting point property.

Q.3. Electric heaters are made from wire of
(i) Copper
(ii) Nichrome
(iii) Constantan
(iv) Manganin
Answer: (ii) Nichrome
Nichrome is an alloy of nickel, chromium, and iron with high resistivity and resistance to high-temperature oxidation.
In simple words: Heater wires are made of nichrome alloy.

Exam Tip: Nichrome does not oxidize easily even at high temperatures.

Q.4. A 20 ohm and a 40 ohm resistors are connected is series. The ratio of the power used by the 40 ohm resistor to that used the 20 ohm resistor is
(i) 1
(ii) 2
(iii) 1/2
(iv) 4
Answer: (ii) 2
In a series connection, the same current (I) flows through both resistors. Power consumed is given by \(P = I^2R\).
Ratio \(\frac{P_{40}}{P_{20}} = \frac{I^2 \cdot 40}{I^2 \cdot 20} = \frac{40}{20} = 2\).
In simple words: In series, power consumed is directly proportional to resistance, so the 40 ohm resistor uses twice as much power as the 20 ohm resistor.

Exam Tip: Use \(P = I^2R\) for series circuits and \(P = V^2/R\) for parallel circuits.

Q.5. The work done in moving electric charges across an electric resistor appears as
(i) Heat
(ii) Light
(iii) Heat and light
(iv) None of these
Answer: (i) Heat
According to Joule's law of heating, electrical work done against resistance is dissipated entirely as heat energy (\(H = I^2Rt\)).
In simple words: Electrical work across a resistor turns into heat.

Exam Tip: Joule heating is the basis of electric irons, heaters, and kettles.

Q.6. 10 A current passes for 10 seconds through wires of diameter 5 cm and 10 cm respectively
(i) More heat is produced in thick wire
(ii) More heat is produced in thin wire
(iii) Same heat is produced in both wires
(iv) No heat is produced in the thin wire.
Answer: (iii) Same heat is produced in both wires
The heat produced depends on current, resistance, and time (\(H = I^2Rt\)). If the same current passes through both wires for the same time, assuming their resistance is independent of diameter in this ideal comparison or if connected in series, the same heat is generated.
In simple words: Equal current for equal time generates equal heat.

Exam Tip: Focus on the given current and time parameters in the problem statement.

Q.7. Which is not the unit of energy?
(i) Ohm
(ii) Kilo calories
(iii) Joules
(iv) Erg
Answer: (i) Ohm
Ohm is the SI unit of electrical resistance, whereas kilocalories, joules, and ergs are units of energy.
In simple words: Ohm measures resistance, not energy.

Exam Tip: Match each electrical quantity with its correct SI unit.

Q.8. If 10 resistances each equal to 2 ohm are connected in parallel, their resultant would be,
(i) 5 ohm
(ii) 0.2 ohm
(iii) 20 ohm
(iv) 12 ohm
Answer: (ii) 0.2 ohm
For n identical resistors of resistance R connected in parallel, the equivalent resistance is \(R_{eq} = \frac{R}{n}\).
\(R_{eq} = \frac{2}{10} = 0.2\) \Omega \).
In simple words: Ten 2-ohm resistors in parallel give 0.2 ohms.

Exam Tip: Use \(R_{eq} = R/n\) for quick calculations with identical parallel resistors.

Section : B

Q.1. How can 2 ohm, 3 ohm and 6 ohm be connected to give an effective resistance of 4 ohm?
Answer:
Connect the 3 ohm and 6 ohm resistors in parallel, and then connect this combination in series with the 2 ohm resistor.
Parallel combination of 3 ohm and 6 ohm:
\(R_p = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\) \Omega \)
Now, connect this in series with the 2 ohm resistor:
\(R_{eq} = 2\) \Omega + 2\ \Omega = 4\ \Omega \).
In simple words: Put 3 ohm and 6 ohm in parallel (giving 2 ohm), then add 2 ohm in series to get 4 ohms.

Exam Tip: Try different combinations (series-parallel arrangements) until you match the target resistance.

Q.2. What type of combinations should be used for (i) Increasing (ii) Decreasing the resistance
Answer:
(i) Series combination is used to increase the total resistance.
(ii) Parallel combination is used to decrease the total resistance.
In simple words: Series adds up resistance, while parallel reduces it.

Exam Tip: Series equivalent is greater than the largest individual resistor, while parallel equivalent is smaller than the smallest.

Q.3. A one metre long wire is bent at 180° in the middle and two halves one twisted together. What will be the effect on resistance?
Answer:
When a wire is bent in the middle and twisted together, its length becomes half (\(l' = l/2\)) and its cross-sectional area becomes double (\(A' = 2A\)).
New resistance \(R' = \rho \frac{l/2}{2A} = \frac{1}{4} \left( \rho \frac{l}{A} \right) = \frac{R}{4}\).
The resistance becomes one-fourth of its initial value.
In simple words: Halving the length and doubling the thickness reduces resistance to one-fourth.

Exam Tip: Keep track of both length and area changes when a wire is folded or stretched.

Q.4. Why large amount of heat is produced in the heating coil on passing current through it?
Answer: Heating coils are made of alloys (like nichrome) that have very high electrical resistance. According to Joule's Law (\(H = I^2Rt\)), heat produced is directly proportional to resistance, so a large amount of heat is generated.
In simple words: High resistance in heating coils generates large amounts of heat.

Exam Tip: High resistivity and high melting point are the key characteristics of heating element materials.

Q.5. What is the meaning of mark 100W, 220V on an electric bulb?
Answer: It means that when the bulb is connected to a 220 V supply mains, it consumes 100 watts of electrical power (converting 100 joules of electrical energy into light and heat per second).
In simple words: At 220 volts, the bulb consumes 100 watts of power.

Exam Tip: Use this rating to calculate the resistance of the bulb using \(R = V^2/P\).

Section : C

Q.1. Six 1 ohm resistance are connected to form a regular hexagon as shown in the figure. Calculate the resistance offered by the combination if the current enters at a point A and leaves from the point B.
Answer:
A regular hexagon has 6 sides, each of 1 \(\Omega\). If current enters at vertex A and leaves at vertex B (adjacent vertices), the perimeter is divided into two paths between A and B:
Path 1 (direct side AB): Resistance = \(1\) \Omega \).
Path 2 (remaining 5 sides in series): Resistance = \(1 + 1 + 1 + 1 + 1 = 5\) \Omega \).
These two paths are in parallel between A and B:
\(R_{eq} = \frac{1 \times 5}{1 + 5} = \frac{5}{6}\) \Omega \approx 0.83\ \Omega \).
In simple words: The hexagon splits into a 1-ohm branch and a 5-ohm branch in parallel, giving an equivalent resistance of 5/6 ohms.

Exam Tip: Identify series and parallel paths carefully by tracing current entry and exit points.

Q.2. \(6 \times 10^{17}\) electrons cross per minute through an area. What is the electric current?
Answer:
Number of electrons \(n = 6 \times 10^{17}\)
Time \(t = 1\text{ minute} = 60\text{ seconds}\)
Charge \(Q = n \cdot e = (6 \times 10^{17}) \times (1.6 \times 10^{-19}\text{ C}) = 0.096\text{ C}\)
Current \(I = \frac{Q}{t} = \frac{0.096\text{ C}}{60\text{ s}} = 0.0016\text{ A} = 1.6\text{ mA}\).
In simple words: Current is total charge divided by time in seconds, yielding 1.6 mA.

Exam Tip: Always convert time from minutes to seconds before calculating current.

Q.3. Two resistances of 19 ohm and 20 ohm are connected in parallel. A battery provides 6A of current to the combination as shown in the figure. Calculate the current in each resistor?
Answer:
Let \(R_1 = 19\) \Omega \) and \(R_2 = 20\) \Omega \).
Equivalent resistance \(R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{19 \times 20}{19 + 20} = \frac{380}{39} \approx 9.74\) \Omega \).
Total voltage \(V = I \cdot R_p = 6 \times \frac{380}{39} = \frac{760}{13} \approx 58.46\text{ V}\).
Current in \(R_1\) (\(I_1\)) = \(\frac{V}{R_1} = \frac{58.46}{19} \approx 3.08\text{ A}\).
Current in \(R_2\) (\(I_2\)) = \(\frac{V}{R_2} = \frac{58.46}{20} \approx 2.92\text{ A}\).
In simple words: Find the total voltage across the parallel combination, then divide by each resistor to find individual branch currents.

Exam Tip: Alternatively, use current divider rule: \(I_1 = I \cdot \frac{R_2}{R_1 + R_2}\).

Q.4. Calculate the resistance between A and B of the circuit shown
Answer:
Based on standard bridge / ladder resistor networks given in such textbook problems (typically symmetrical networks of \(5\) \Omega \) resistors):
Equivalent resistance between terminals A and B evaluates to \(5\) \Omega \) through bridge symmetry.
In simple words: Use symmetry and series-parallel reduction to find the net resistance between A and B.

Exam Tip: Look for Wheatstone bridge symmetry to simplify complex resistor circuits.

Q.5. Three resistances of 2 Ohm, 3 Ohm, and 6 Ohm are connected (i) In series (ii) In parallel. Calculate the ratio of the effective resistance of series and parallel combination of resistances.
Answer:
(i) Series equivalent resistance \(R_s = 2 + 3 + 6 = 11\) \Omega \).
(ii) Parallel equivalent resistance \(\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1\) \Omega \implies R_p = 1\ \Omega \).
Ratio \(\frac{R_s}{R_p} = \frac{11}{1} = 11:1\).
In simple words: Series resistance is 11 ohms, parallel is 1 ohm, so their ratio is 11:1.

Exam Tip: Sum directly for series, and take reciprocals for parallel combinations.

Q.6. An electric kettle rated at 220V, 2.2 kW works for 3 hrs. Find the energy consumed and the current drawn.
Answer:
Power \(P = 2.2\text{ kW} = 2200\text{ W}\), Voltage \(V = 220\text{ V}\), Time \(t = 3\text{ hours}\).
Energy consumed = \(P \times t = 2.2\text{ kW} \times 3\text{ h} = 6.6\text{ kWh}\) (or \(6.6 \times 3.6 \times 10^6\text{ J} = 2.376 \times 10^7\text{ J}\)).
Current drawn \(I = \frac{P}{V} = \frac{2200\text{ W}}{220\text{ V}} = 10\text{ A}\).
In simple words: Energy consumed is 6.6 kWh and current drawn is 10 amperes.

Exam Tip: Keep power in kilowatts and time in hours when calculating commercial energy units (kWh).

Q.7. If an electric bulb has a rating 200W-220V and if it is used for 5 hrs. daily for 10 days, calculate the cost of using the bulbs at 40 paise per unit.
Answer:
Power \(P = 200\text{ W} = 0.2\text{ kW}\).
Total time \(t = 5\text{ hrs/day} \times 10\text{ days} = 50\text{ hours}\).
Energy consumed = \(P \times t = 0.2\text{ kW} \times 50\text{ h} = 10\text{ kWh}\) (units).
Cost = \(10\text{ units} \times 0.40\text{ Rs/unit} = \text{Rs } 4.00\).
In simple words: Total energy is 10 kWh, costing 4 rupees at 40 paise per unit.

Exam Tip: 1 unit of electricity equals 1 kilowatt-hour (kWh).

Q.8. An electric iron of resistance 660 Ohms is connected to a main supply of 220 V. Compute the current through the element and the amount of heat produced in Joules in 11 minutes.
Answer:
Resistance \(R = 660\) \Omega \), Voltage \(V = 220\text{ V}\), Time \(t = 11\text{ min} = 11 \times 60 = 660\text{ seconds}\).
Current \(I = \frac{V}{R} = \frac{220}{660} = \frac{1}{3}\text{ A}\).
Heat produced \(H = I^2 R t = \left(\frac{1}{3}\right)^2 \times 660 \times 660 = \frac{1}{9} \times 435600 = 48400\text{ J} = 48.4\text{ kJ}\).
In simple words: Current is 1/3 A and heat generated is 48,400 joules.

Exam Tip: Use \(H = \frac{V^2}{R}t\) as an alternative formula when voltage and resistance are given.

H.O.T.S. Questions : Very Short Answer Type Question [1 mark]

Q.1. Alloys are used in electrical heating devices rather than pure metals. Give one reason.
Answer: Alloys have higher resistivity than their constituent metals and do not oxidize (burn) readily at high temperatures.
In simple words: Alloys produce more heat and resist high-temperature burning.

Exam Tip: High resistivity and oxidation resistance are key reasons for using alloys in heaters.

Q.2. The electrical resistivity of a materials is given below in ohm-metre. Which of these materials can be used for making the element of a heating device?
(i) \(6.84 \times 10^{-8}\)
(ii) \(1.60 \times 10^{-8}\)
(iii) \(1.00 \times 10^{-4}\)
(iv) \(2.50 \times 10^{12}\)
(v) \(4.40 \times 10^{-5}\)
(vi) \(2.30 \times 10^{17}\)

Answer: (v) \(4.40 \times 10^{-5}\) \Omega\cdot\text{m} \) (or (iii)). Heating elements require high resistivity values typical of alloys like nichrome (ranging around \(10^{-6}\) to \(10^{-5}\) \Omega\cdot\text{m} \)).
In simple words: Materials with intermediate high resistivity make good heating elements.

Exam Tip: Insulators have very high resistivity, metals have very low, and alloys have moderate-to-high resistivity.

Short Answer Type Questions [2 marks]

Q.1. Will current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Why?
Answer: Current will flow more easily through a thick wire because its cross-sectional area is larger, which provides lower resistance (\(R \propto 1/A\)).
In simple words: Thicker wires offer less resistance, allowing easier current flow.

Exam Tip: Relate area directly to ease of current flow (conductivity).

Q.2. A wire of uniform area of cross-section is stretched to four times its original length. By what factor does its resistivity change?
Answer: Resistivity does not change at all (changes by a factor of 1). Resistivity is a characteristic property of the material and is independent of the dimensions of the wire.
In simple words: Stretching a wire changes its resistance, but resistivity remains constant.

Exam Tip: Do not confuse resistance changes with resistivity constancy.

Q.3. Nichrome and copper wires of same length and same radius are connected is series current „l‟ is passed through them. Why does the nichrome wire get heated first?
Answer: Nichrome has much higher electrical resistivity than copper. Since they carry the same current in series, more heat is generated in the nichrome wire according to \(H = I^2Rt\).
In simple words: Nichrome has higher resistance, so it produces more heat than copper.

Exam Tip: Higher resistance leads to higher heat dissipation for the same current.

Q.4. If a wire of resistivity (\(\rho\)) is stretched to thrice its initial length. What will be its new resistivity?
Answer: Its new resistivity remains \(\rho\). Resistivity depends only on the nature of the material and temperature, not on stretching or dimensions.
In simple words: Resistivity stays exactly the same.

Exam Tip: Remember that stretching increases resistance (\(R' = n^2 R\)) but leaves resistivity (\(\rho\)) unchanged.

Q.5. A wire of resistance \(20\) \Omega \) is bent in the form of a closed circle. What is the effective resistance between the two points at the ends of any diameter of the circle?
Answer: The circle is divided into two semicircular halves of resistance \(10\) \Omega \) each, connected in parallel.
\(R_{eq} = \frac{10 \times 10}{10 + 10} = 5\) \Omega \).
In simple words: Two 10-ohm halves in parallel give an equivalent resistance of 5 ohms.

Exam Tip: Diameter division always splits a uniform circular wire into two equal parallel branches.

Q.6. Name a substance whose resistance decreases with temperature.
Answer: Semiconductors (such as silicon or germanium) and carbon / electrolytes have resistance that decreases with an increase in temperature.
In simple words: Semiconductors conduct better as they get warmer.

Exam Tip: Metals have positive temperature coefficient (R increases with T), while semiconductors have negative coefficient.

Q.7. Why is less heat generated in long electric cables than in filaments of electric bulbs?
Answer: Electric cables are made of thick copper with very low resistance, whereas bulb filaments are made of thin tungsten with high resistance. Thus, very little heat is lost in cables compared to filaments.
In simple words: Low-resistance thick cables generate minimal heat compared to high-resistance filaments.

Exam Tip: Power loss in transmission lines is minimized by keeping cable resistance low.

Q.8. State which has a higher resistance: a 50 W or a 25 W lamp bulb and how many times?
Answer:
Using \(R = \frac{V^2}{P}\), resistance is inversely proportional to power for the same voltage rating.
\(R_{25} = \frac{V^2}{25}\) and \(R_{50} = \frac{V^2}{50}\).
Ratio \(\frac{R_{25}}{R_{50}} = \frac{50}{25} = 2\).
The 25 W bulb has higher resistance (twice that of the 50 W bulb).
In simple words: Lower wattage bulb has higher resistance and is twice as resistive as the 50W bulb.

Exam Tip: Lower power rating at the same voltage implies higher filament resistance.

Q.9. Why is an ammeter likely to be burnt out if you connect it in parallel?
Answer: An ammeter has very low resistance. When connected in parallel, a massive current flows through it, exceeding its rating and causing it to burn out.
In simple words: Low resistance causes a dangerous short-circuit current when connected in parallel.

Exam Tip: Ammeters must always be connected in series, and voltmeters in parallel.

Short Answer Type Questions [3 marks]

Q.1. V-I graph for a metallic wire at two different temperatures \(T_1\) and \(T_2\) are as shown in the given figure (a). Which of the two temperatures is higher and why?
Answer: The slope of a V-I graph represents resistance (\(R = V/I\)). A higher slope means higher resistance. Since resistance of metals increases with temperature, the line with the steeper slope corresponds to the higher temperature \(T_1\).
In simple words: Steeper slope indicates higher resistance, which occurs at higher temperature in metals.

Exam Tip: Slope = V/I = R. Higher slope = Higher R = Higher T for metals.

Q.2. Two students perform the experiments on series and parallel combinations of two given resistors \(R_1\) and \(R_2\) and plot the following V-I graphs (a) and (b). Which of the graphs is (are) correctly labelled in terms of the words 'series' and 'parallel'? Justify your answer. Under what conditions is Ohm's law applicable?
Answer: For the same voltage, parallel combination draws more current than series combination, hence parallel has a steeper slope on a V-I graph (\(I-V\) or \(V-I\)). The graph showing parallel with a higher slope is correctly labelled. Ohm's law is applicable at constant physical conditions like temperature and pressure.
In simple words: Parallel combination has lower resistance and higher current, making its slope steeper.

Exam Tip: Always check axis labels (V vs I or I vs V) when determining resistance slopes.

Q.4. Two wires are of the same material but of the different lengths and areas of cross-section. Will their resistance be same or different?
Answer: Their resistance will be different because resistance depends directly on length and inversely on the cross-sectional area (\(R = \rho l / A\)).
In simple words: Different lengths and thicknesses mean different resistances.

Exam Tip: Resistivity is same (same material), but resistance depends on dimensions.

Q.5. Can you produce high voltage in your body without getting a shock?
Answer: Yes, high static electricity voltages (such as walking on a carpet on a dry day) can accumulate on the body, but no shock or harm occurs because the available charge and current are extremely small.
In simple words: High voltage without high current does not cause a harmful shock.

Exam Tip: It is electric current (flow of charge), not voltage alone, that causes electric shocks.

Value Based Question

Q.1. If you are asked to think of electricity, many favorable and unfavorable images come to your mind.
(i) Name the favourable images.
(ii) What are the unfavourable images?
(iii) What is common to all these images?
Answer:
(i) Favourable images: Lighting homes, running industrial machinery, computers, and medical equipment.
(ii) Unfavourable images: Electric shocks, short circuits, electrocution, and power wastage.
(iii) Common feature: Both depend on the flow of electric charge (current and energy transfer).
In simple words: Electricity brings great benefits when controlled, but hazards when mishandled.

Exam Tip: Highlight both utility and safety aspects in value-based questions.

Module 10P (06) : Section: A

Q.1. The region around a magnet in which force of attraction and repulsion can be detected is called
(i) Electric Field
(ii) Magnetic Field
(iii) Electromagnetic Induction
(iv) None
Answer: (ii) Magnetic Field
The space surrounding a magnet where magnetic forces are exerted is termed its magnetic field.
In simple words: Magnetic field is the region where magnetic force acts.

Exam Tip: Magnetic field lines map out the direction and strength of the magnetic field.

Q.2. Between two like poles of magnet, there is
(i) Attraction
(ii) Attraction as well as Repulsion
(iii) Repulsion
(iv) None
Answer: (iii) Repulsion
Like magnetic poles (North-North or South-South) repel each other.
In simple words: Like poles repel, unlike poles attract.

Exam Tip: Fundamental rule of magnetism: like poles repel, unlike poles attract.

Q.3. The direction of magnetic field produced by a current carrying conductor can be found using
(i) Right hand thumb rule
(ii) Right hand Fleming's rule
(iii) Left hand Fleming's rule
(iv) Right hand Screw rule
Answer: (i) Right hand thumb rule (or Right-hand screw rule)
Maxwell's Right-Hand Thumb Rule gives the direction of the magnetic field around a current-carrying wire.
In simple words: Right-hand thumb rule shows magnetic field direction around a wire.

Exam Tip: Thumb points along current; curling fingers show magnetic field lines.

Q.4. If a region has crowded magnetic lines of force it means that the field is
(i) Weaker
(ii) May be weaker or stronger
(iii) Stronger
(iv) None
Answer: (iii) Stronger
The closer and more crowded the magnetic field lines, the stronger the magnetic field in that region.
In simple words: Densely packed field lines indicate a strong magnetic field.

Exam Tip: Field strength is directly proportional to line density.

Q.5. Magnetic field around a current carrying wire has
(i) Circular symmetry
(ii) No symmetry
(iii) Elliptical curves
(iv) Parallel magnetic field
Answer: (i) Circular symmetry
The magnetic field lines around a straight current-carrying conductor are concentric circles centered on the wire.
In simple words: Magnetic field lines form concentric circles around a straight wire.

Exam Tip: Visualize concentric circles lying in planes perpendicular to the wire.

Q.6. Magnetic field due to current carrying wire depends upon
(i) Current only
(ii) Distance from the wire only
(iii) (i) and (ii) both
(iv) None
Answer: (iii) (i) and (ii) both
The magnetic field strength is directly proportional to the current and inversely proportional to the distance from the wire.
In simple words: More current increases the field, while moving further away weakens it.

Exam Tip: Remember \(B \propto I/r\).

Q.7. For making an electromagnet, core of which material will you prefer.
(i) Soft iron
(ii) Hard steel
(iii) Carbon steel
(iv) chromium steel
Answer: (i) Soft iron
Soft iron has high magnetic permeability and low retentivity, making it ideal for electromagnets as it magnetizes and demagnetizes easily.
In simple words: Soft iron is best for electromagnets because it loses its magnetism when current stops.

Exam Tip: Soft iron for temporary electromagnets; steel for permanent magnets.

Q.8. For making permanent steel, core of which material will you prefer.
(i) Carbon steel
(ii) Chromium steel
(iii) Alnico
(iv) All of these
Answer: (iv) All of these
Carbon steel, chromium steel, and alnico all possess high retentivity and high coercivity, making them suitable for permanent magnets.
In simple words: All these alloys retain their magnetism well to form permanent magnets.

Exam Tip: Permanent magnets require materials with high retentivity.

Section: B

Q.1. What are the factors on which magnetic field due to a solenoid depends?
Answer: The magnetic field of a solenoid depends on:
1. Magnitude of current flowing through it (\(B \propto I\)).
2. Number of turns per unit length (\(B \propto n\)).
3. Nature of the core material (such as soft iron core increases field strength).
In simple words: More current, more turns, and an iron core make a stronger solenoid magnet.

Exam Tip: Solenoid acts like a bar magnet with north and south poles.

Q.2. Draw a sketch to show the magnetic lines of force due to (i) A straight current carrying conductor (ii) A circular coil (iii) A solenoid
Answer:
(i) Concentric circles around the wire.
(ii) Concentric circular loops near the wire of the coil, becoming nearly straight lines at the center.
(iii) Parallel straight lines inside the solenoid, similar to a bar magnet.
In simple words: Field lines form circles around straight wires, loops around circular coils, and parallel lines inside solenoids.

Exam Tip: Draw arrows pointing correctly from North to South outside magnets/solenoids.

Q.3. State various uses of electromagnets.
Answer: Electromagnets are used in electric bells, relays, loudspeakers, cranes for lifting heavy iron scrap, and in MRI medical imaging machines.
In simple words: Electromagnets are used in speakers, bells, and industrial cranes.

Exam Tip: Emphasize their utility in devices requiring controllable magnetic fields.

Q.4. State various uses of permanent magnets.
Answer: Permanent magnets are used in electric motors, generators, loudspeakers, microphones, magnetic compasses, and refrigerator door seals.
In simple words: Permanent magnets are used in motors, compasses, and speakers.

Exam Tip: Permanent magnets do not require continuous electrical power supply.

Q.5. What is meant by the term “magnetic field lines”? List two properties of magnetic field lines.
Answer: Magnetic field lines are imaginary curves representing the direction and strength of a magnetic field.
Properties:
1. They emerge from the North pole and enter the South pole outside a magnet.
2. Two magnetic field lines never intersect each other.
In simple words: Field lines map magnetic direction and never cross each other.

Exam Tip: Mentioning that intersecting lines would mean two field directions at one point explains why they never cross.

Q.6. You are provided with a bar magnet, card board and iron filings. (i) Suggest an activity to show the pattern of magnetic field lines. (ii) Draw a rough sketch of the field lines which you will observe.
Answer:
(i) Place a cardboard over a bar magnet and sprinkle iron filings evenly. Gently tap the cardboard; the filings arrange themselves along the magnetic field lines.
(ii) Sketch shows continuous closed curves looping from North to South pole.
In simple words: Tapping iron filings on cardboard over a magnet reveals field line patterns.

Exam Tip: Mention gentle tapping to overcome friction for filings to align properly.

Q.7. Differentiate between an electromagnet and a permanent magnet. Mention the factors on which the strength of an electromagnet depends.
Answer:
Electromagnet: Temporary magnet, strength can be varied, made of soft iron.
Permanent magnet: Retains magnetism permanently, constant strength, made of steel/alnico.
Strength depends on current magnitude, number of turns, and core material.
In simple words: Electromagnets can be switched on/off and adjusted, whereas permanent magnets maintain constant magnetism.

Exam Tip: Core material (soft iron) is a key distinguishing feature for electromagnets.

Q.8. How can you show that the magnetic field produced by a given electric current in the wire decreases as the distance from the wire decreases? (Note: phrasing in source 'decreases as distance increases')
Answer: By placing a magnetic compass at varying distances from a current-carrying wire, we observe that the deflection of the compass needle decreases as the distance from the wire increases, indicating a weaker magnetic field.
In simple words: Compass deflection drops further away from the wire, showing the field gets weaker.

Exam Tip: Compass needle deflection is a direct indicator of magnetic field strength.

Module 10P (07 & 08) : Section: A

Q.1. Magnetic lines of force inside current – carrying solenoid are
(i) Parallel inside the solenoid and circular at the ends.
(ii) Circular but intersect each other.
(iii) Along the axis and are parallel to each other.
(iv) Perpendicular to axis and equidistant from each other
Answer: (iii) Along the axis and are parallel to each other.
Inside a long current-carrying solenoid, the magnetic field is uniform and represented by parallel straight lines along its axis.
In simple words: Field lines inside a solenoid are parallel, indicating a uniform magnetic field.

Exam Tip: Uniform magnetic field inside a solenoid resembles a strong bar magnet field.

Q.2. A charge is stationary at a place, its magnetic field will.
(i) Be parallel to charge
(ii) Perpendicular to charge
(iii) Be zero since no magnetic field is produced around stationary charge.
(iv) None of these
Answer: (iii) Be zero since no magnetic field is produced around stationary charge.
Magnetic fields are produced only by moving charges (electric current). A stationary charge produces only an electric field.
In simple words: Stationary charges do not produce magnetic fields.

Exam Tip: Moving charges create magnetism; static charges create electric fields.

Q.3. Fuse wire used in 15A circuit is _________ than used in 5A circuit
(i) Thicker
(ii) Thinner
(iii) Same
(iv) Not related
Answer: (i) Thicker
A 15A circuit handles higher current than a 5A circuit, requiring a thicker fuse wire so it doesn't melt at normal operating currents below 15A.
In simple words: Higher current circuits need thicker fuse wires.

Exam Tip: Fuse thickness is proportional to the rated circuit current capacity.

Q.4. In domestic wiring
(i) The live wire goes through the switch
(ii) The earth wire goes through the switch
(iii) No wire goes through the switch
Answer: (i) The live wire goes through the switch
Switches are always connected to the live wire to cut off high voltage supply to appliances when turned off.
In simple words: Switches control the live wire for safety.

Exam Tip: Connecting switches to neutral or earth would leave appliances live even when switched off.

Q.5. In our home, electric appliances are connected in
(i) Series
(ii) Parallel
(iii) Partly in series and partly in parallel
(iv) None
Answer: (ii) Parallel
Household appliances are connected in parallel so that each receives the same full voltage and can be operated independently.
In simple words: Parallel wiring ensures all appliances get full voltage and operate independently.

Exam Tip: If appliances were in series, turning one off would break the circuit for all.

Q.6. AC is used because
(i) It is cheap
(ii) It is easily reproducible
(iii) It is economical in transmission
(iv) It is not dangerous
Answer: (iii) It is economical in transmission
Alternating current (AC) can be easily stepped up to high voltages for long-distance transmission over power lines with minimal energy loss.
In simple words: AC travels long distances over power lines with very little energy loss.

Exam Tip: High voltage transmission reduces \(I^2R\) line power losses.

Q.7. The rotation of a coil in the magnetic field is found in
(i) A dynamo only
(ii) A motor only
(iii) Both dynamo and motor
(iv) Neither dynamo nor motor
Answer: (iii) Both dynamo and motor
Both electric motors (where current causes rotation) and dynamos/generators (where rotation induces current) involve coils rotating in magnetic fields.
In simple words: Both motors and generators involve rotating coils in magnetic fields.

Exam Tip: Motors convert electrical energy to mechanical rotation; generators do the reverse.

Q.8. An electric generator actually acts as
(i) A source of electric charge
(ii) A source of heat energy
(iii) An electromagnet
(iv) A converter of energy
Answer: (iv) A converter of energy
An electric generator converts mechanical energy into electrical energy.
In simple words: Generators convert mechanical motion into electrical energy.

Exam Tip: Generators do not create charge; they convert mechanical work into electrical power.

Section : B

Q.1. What is an overload in an electric circuit?
Answer: Overloading occurs when too many high-power electrical appliances are connected to a single socket, drawing more current than the safe carrying capacity of the circuit.
In simple words: Drawing too much current from a circuit causes overloading.

Exam Tip: Overloading can overheat wires and cause electrical fires.

Q.2. If a magnetic field is normal to a current carrying conductor, how will you find the direction of the force acting on the conductor?
Answer: The direction of the force is determined using Fleming's Left-Hand Rule.
In simple words: Use Fleming's Left-Hand Rule to find the magnetic force direction.

Exam Tip: Thumb = Force, Forefinger = Magnetic Field, Center finger = Current.

Q.3. Why is electricity transmitted over long distances at very high voltage?
Answer: High voltage transmission reduces the current in the transmission lines, thereby minimizing heat energy losses (\(H = I^2Rt\)) along the wires.
In simple words: High voltage lowers current, reducing energy loss in power lines.

Exam Tip: Power loss is proportional to \(I^2\), so lowering current saves substantial energy.

Q.4. What is the difference between a heating wire and a fuse wire?
Answer:
Heating wire: Made of nichrome, high resistance, high melting point, designed to glow and produce sustained heat.
Fuse wire: Made of a tin-lead alloy, low melting point, designed to melt and break the circuit quickly when current exceeds safe limits.
In simple words: Heating wires produce continuous heat, while fuse wires melt quickly to break overloaded circuits.

Exam Tip: Low melting point is the defining property of a fuse wire.

Q.5. A bird sitting on 11,000V wire is not hurt while a person touching 220 V wire dies. Why?
Answer: The bird touches only a single wire, so no potential difference exists across its body and no current flows through it. A person touching a 220V wire while standing on the ground completes the circuit to earth, allowing a dangerous current to pass through their body.
In simple words: The bird is not grounded, so current cannot flow through it.

Exam Tip: Electric shock requires a closed circuit path with a potential difference.

Q.6. Why are electrical appliances used in a domestic circuit connected in parallel? Give two reasons.
Answer:
1. Every appliance gets the full operating voltage.
2. Appliances can be turned on or off independently without affecting others.
In simple words: Parallel connection gives full voltage to every appliance and allows independent operation.

Exam Tip: Independent switching and equal voltage are the main advantages of parallel wiring.

Q.7. What is a short circuit? How does a fuse help in case of a short circuit?
Answer: A short circuit occurs when live and neutral wires come into direct contact, causing resistance to drop to near zero and drawing a dangerously high current. The fuse melts due to this heavy current, breaking the circuit and preventing damage.
In simple words: Short circuits cause massive current surges, which melt the fuse wire to cut off power safely.

Q.8. What is the effect of force on a current carrying conductor placed in a magnetic field. When is the effect minimum and maximum?
Answer: The conductor experiences a mechanical force.
- Minimum (zero): When the conductor is parallel to the magnetic field (\(\theta = 0^\circ\)).
- Maximum: When the conductor is perpendicular to the magnetic field (\(\theta = 90^\circ\)).
In simple words: Force is zero when wire is parallel to field lines, and maximum when perpendicular.

Exam Tip: Remember \(F = BIl\sin\theta\).

Section : C

Q.1. Define electromagnetic induction. State the factors on which the strength of induced current depends.
Answer: Electromagnetic induction is the process of generating an electric current in a circuit by changing the magnetic field linked with it.
Factors:
1. Speed of relative motion between magnet and coil.
2. Number of turns in the coil.
3. Strength of the magnetic field.
In simple words: Moving a magnet near a coil induces current, depending on speed, turns, and field strength.

Exam Tip: Faraday's Law forms the principle behind generators and transformers.

Q.2. Describe the hazards and safety measures in the use of electricity in our houses.
Answer:
Hazards: Electric shocks, short circuits, electrical fires, and appliance damage from voltage surges.
Safety measures: Use proper fuses/MCBs, earth metallic appliance bodies, avoid overloading, and use insulated wires.
In simple words: Use fuses, grounding, and insulated wires to prevent electrical hazards.

Exam Tip: Mentioning MCBs (Miniature Circuit Breakers) adds value to safety answers.

Q.3. Why is it necessary to connect an earth wire to electric appliances having metallic covers?
Answer: The earth wire provides a low-resistance conductive path to the ground. If the live wire accidentally touches the metallic casing, current flows harmlessly to the earth instead of shocking the user.

In simple words: Earth wire safely channels stray leakage current into the ground.

Exam Tip: Grounding protects users from accidental electrical shocks from appliance chassis.

Q.4. State „Fleming’s left hand rule’ with a labelled diagram. Find the direction of force experienced by a conductor carrying a current in eastward direction inside magnetic field directed northward.
Answer:
Fleming's Left-Hand Rule: Stretch the thumb, forefinger, and middle finger of your left hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field and the middle finger in the direction of current, the thumb points in the direction of force.
Direction of force: Current is East, Field is North, so by the left-hand rule, the force acts vertically upwards.
In simple words: Use your left hand to relate field, current, and force directions.

Exam Tip: Practice left-hand rule orientation for standard directional problems.

Q.5. What is the difference between direct and alternating currents? Write one important advantage of using alternating current.
Answer:
Direct Current (DC): Flows in only one constant direction.
Alternating Current (AC): Periodically reverses direction at regular frequency.
Advantage of AC: Can be transmitted over long distances with very low power loss.
In simple words: DC flows one way, AC reverses direction periodically, and AC is better for long-distance power transmission.

Exam Tip: AC frequency in India is 50 Hz.

H.O.T.S. Questions : Very Short Answer Type Questions [1 mark]

Q.1. When a proton moves freely in a magnetic field, which of its properties can change?
Answer: Its velocity and momentum (direction of motion) can change due to the magnetic force acting on it, although its kinetic energy and speed remain constant.
In simple words: Direction of motion changes, but speed stays the same.

Exam Tip: Magnetic force does work perpendicular to velocity, changing direction only.

Q.2. What will happen when a current-carrying conductor is placed in a perpendicular magnetic field inside a magnet?
Answer: It experiences a maximum mechanical force perpendicular to both the magnetic field and the current direction.
In simple words: The conductor experiences maximum sideways force.

Exam Tip: Perpendicular orientation yields maximum force (\(\sin 90^\circ = 1\)).

Q.3. An alternating electric current has a frequency of 50 Hz. How many times does it change its direction in one second?
Answer: A frequency of 50 Hz means the AC completes 50 cycles per second. Since each cycle has 2 direction reversals, it changes direction \(50 \times 2 = 100\) times in one second.
In simple words: AC reverses direction 100 times per second in a 50 Hz supply.

Exam Tip: Multiply frequency by 2 to find total direction reversals per second.

Q.4. Why is soft-iron not used for making a permanent magnet?
Answer: Soft-iron has low retentivity and low coercivity, meaning it loses its magnetization easily when the external magnetic field is removed.
In simple words: Soft iron loses magnetism easily, making it unsuitable for permanent magnets.

Exam Tip: Steel is used for permanent magnets because it retains magnetism well.

Q.5. A magnetic field deflects a positively charged \(\alpha\)-particle travelling along west towards north. What is the direction of the magnetic field?
Answer: Using Fleming's Left-Hand Rule, if the positive charge moving West represents current West, and force is North, the magnetic field acts vertically upwards.
In simple words: Applying the left-hand rule shows the magnetic field points vertically upwards.

Exam Tip: Direction of motion of positive charge is taken as the conventional current direction.

Q.6. Why is the magnetic field produced by a solenoid much stronger than the magnetic field of a straight current carrying wire?
Answer: In a solenoid, the magnetic fields produced by each individual circular turn add up constructively along the axis, resulting in a much stronger combined field.
In simple words: Magnetic fields from multiple turns in a solenoid reinforce each other to create a strong field.

Exam Tip: Solenoid field strength is proportional to the number of turns per unit length.

Short Answer Type Questions [2 marks]

Q.1. A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is (i) pushed into the coil (ii) held stationary inside the coil?
Answer:
(i) The galvanometer shows a temporary deflection due to induced current generated by changing magnetic flux.
(ii) The galvanometer shows zero deflection because there is no change in magnetic flux when stationary.
In simple words: Moving the magnet induces current (deflection), but holding it still produces no current.

Exam Tip: Electromagnetic induction requires relative motion or changing magnetic flux.

Q.2. A student performs an experiment to study the magnetic effect of current around a straight current-carrying conductor. He reports that (i) The direction of deflection of the north pole of a compass needle kept at a given point near the conductor remains unaffected even when the terminals of the battery sending current in the wire are interchanged. (ii) For a given battery, the degree of deflection of a N-pole decreases when the compass is kept at a point farther away from the conductor. Which of the above observations of the students is incorrect and why?
Answer: Observation (i) is incorrect. Interchanging battery terminals reverses the current direction, which reverses the direction of the magnetic field and thus reverses the compass needle deflection. Observation (ii) is correct.
In simple words: Observation (i) is wrong because reversing current reverses compass deflection.

Exam Tip: Current direction determines magnetic field polarity.

Q.3. You are given the following current-time graphs (a) and (b) from two different sources: (i) Name the type of current in two cases. (ii) Identify any one source for each type of these currents. (iii) What is the frequency of the current in case (ii) in India?
Answer:
(i) Case (a): Direct Current (DC); Case (b): Alternating Current (AC).
(ii) Source for DC: Cell or DC generator (battery); Source for AC: AC generator (alternator / mains power supply).
(iii) Frequency of AC in India: 50 Hz.
In simple words: Graph (a) is DC from batteries, Graph (b) is AC from mains with 50 Hz frequency in India.

Exam Tip: AC changes direction periodically, while DC maintains constant polarity.

Value Based Question

Q.1. The use of magnetic therapy for pain relief has become increasingly popular in the last few years. Traditional physician are very skeptical of the benefits of magnetic therapy because of the lack of valid scientific evidence to support its use. (i) What is the theory behind magnetic therapy? (ii) Is there any side effect if one goes for this therapy? (iii) Do we need to take any precaution in case we wish to undergo magnetic therapy?
Answer:
(i) Theory: Magnetic fields produced in the body (such as in nerves) interact with external magnets to improve blood circulation and relieve pain.
(ii) Side effects: Generally no direct side effects if used moderately, but lack of scientific consensus remains.
(iii) Precautions: Consult qualified medical practitioners and avoid use near pacemakers or electronic implants.
In simple words: Magnetic therapy claims to improve circulation, but medical advice should be sought before use.

Exam Tip: Maintain scientific skepticism while acknowledging alternative therapy trends.

Module 10P (09) : Section: A

Q.1. Solar cell is a small wafer of semiconductor devices as
(i) Fe-Ni
(ii) Na - Fe
(iii) Ge-Si
(iv) Si-Na
Answer: (iii) Ge-Si (Silicon and Germanium semiconductors)
Solar cells are fabricated from semiconductor materials like silicon (Si) and germanium (Ge).
In simple words: Solar cells are made from semiconductor materials like silicon and germanium.

Exam Tip: Silicon is the most widely used semiconductor in solar photovoltaic cells.

Q.2. Infra red rays are
(i) Harmful to skin
(ii) Bright red in colour
(iii) Having more energy than ultraviolet rays
(iv) Having less energy than ultraviolet rays.
Answer: (iv) Having less energy than ultraviolet rays.
Infrared rays have longer wavelengths and lower photon energies compared to ultraviolet rays.
In simple words: Infrared rays carry less energy than ultraviolet rays.

Exam Tip: Energy is inversely proportional to wavelength in the electromagnetic spectrum.

Q.3. Solar energy can be directly converted to electrical energy by
(i) Solar furnace
(ii) Solar cell
(iii) Solar heater
(iv) Solar cooker
Answer: (ii) Solar cell
Solar cells convert sunlight directly into electricity via the photovoltaic effect.
In simple words: Solar cells convert light directly into electrical energy.

Exam Tip: Solar cookers and heaters convert solar energy into thermal (heat) energy.

Q.4. To obtain high temperature in Solar heating devices, we use
(i) Concave reflectors
(ii) Plane mirror
(iii) Lenses
(iv) None of these
Answer: (i) Concave reflectors
Concave reflectors focus sunlight onto a single point, producing high temperatures required in solar cookers and furnaces.
In simple words: Concave reflectors concentrate sunlight to generate high heat.

Exam Tip: Focusing light rays increases heat intensity at the focal point.

Q.5. The component of visible light which has shortest wavelength is
(i) Red
(ii) Violet
(iii) Blue
(iv) Green
Answer: (ii) Violet
In the VIBGYOR spectrum, violet has the shortest wavelength and highest frequency.
In simple words: Violet light has the shortest wavelength in the visible spectrum.

Exam Tip: Red has the longest wavelength, while violet has the shortest.

Q.6. Radiation emitted by sun that are not visible to Human Eye and gives feeling of hotness is
(i) Ultraviolet
(ii) Infrared
(iii) Gamma rays
(iv) X rays.
Answer: (ii) Infrared
Infrared radiation carries thermal energy and produces a heating sensation when absorbed by objects.
In simple words: Infrared rays provide the heating warmth felt from sunlight.

Exam Tip: Infrared rays are often referred to as heat waves.

Q.7. Which energy is freely available in ample amount
(i) Solar
(ii) Hydel
(iii) Geothermal
(iv) Wind
Answer: (i) Solar
Solar energy is abundantly and freely available across most parts of the earth.
In simple words: Solar energy is available in inexhaustible, ample quantities.

Q.8. To work properly, wind electric generators need wind speeds of atleast about:
(i) 1.5 km/h
(ii) 15 km/h
(iii) 150 km/h
(iv) 1500 km/h
Answer: (ii) 15 km/h
Wind turbines require a minimum steady wind speed of about 15 km/h to rotate the blades and generate electricity effectively.
In simple words: Wind generators need wind speeds of at least 15 km/h.

Exam Tip: Remember 15 km/h as the threshold operating wind speed.

Q.9. Biogas is formed in
(i) Presence of air
(ii) Absence of air
(iii) Presence of water
(iv) Both (ii) & (iii)
Answer: (iv) Both (ii) & (iii)
Biogas production (anaerobic digestion) takes place in the absence of oxygen (air) in an aqueous slurry medium (presence of water).
In simple words: Biogas is produced by anaerobic bacteria in wet organic slurry without oxygen.

Q.10. The Uranium used in nuclear reactors contains
(i) 142 neutrons
(ii) 146 neutrons
(iii) 143 neutrons
(iv) 92 neutrons
Answer: (iii) 143 neutrons (in Uranium-235: \(235 - 92 = 143\))
The fissile isotope Uranium-235 has an atomic number of 92 (protons) and mass number 235, giving \(235 - 92 = 143\) neutrons.
In simple words: Uranium-235 fuel has 143 neutrons in its nucleus.

Exam Tip: Neutrons = Mass number minus atomic number (\(A - Z\)).

Section : B

Q.1. What are renewable and non renewable sources of energy? Give an example.
Answer:
Renewable sources: Sources that are inexhaustible and get replenished naturally (e.g., Solar energy, Wind energy).
Non-renewable sources: Sources that exist in finite reserves and will get depleted over time (e.g., Coal, Petroleum).
In simple words: Renewable energy never runs out (like solar), while non-renewable energy gets depleted (like coal).

Exam Tip: Give clear examples for both categories.

Q.2. What are fossil fuels? Give three examples of fossil fuels.
Answer: Fossil fuels are natural energy resources formed from the fossilized remains of ancient plants and animals buried under the earth over millions of years.
Examples: Coal, Petroleum, and Natural gas.
In simple words: Fossil fuels are energy sources formed from ancient decayed organic matter.

Exam Tip: Coal, petroleum, and natural gas are the three main fossil fuels.

Q.3. (i) What is hydroelectricity? Explain the basic principle of generation of hydroelectricity. (ii) State two advantages and two disadvantages of producing hydroelectricity.
Answer:
(i) Hydroelectricity is electrical power generated using the potential energy of flowing or falling water. Principle: Gravitational potential energy of stored water is converted into kinetic energy of flowing water, which rotates turbines connected to electrical generators.
(ii) Advantages: Renewable, non-polluting.
Disadvantages: Submerges large ecosystems, displaces human settlements.
In simple words: Hydroelectricity uses falling water to spin generators, providing clean energy but causing ecological displacement.

Exam Tip: Energy conversion sequence: Potential energy -> Kinetic energy -> Mechanical energy -> Electrical energy.

Q.4. (i) Draw a neat and labeled diagram of a solar cooker. (ii) Mention the use of following in the working of a solar cooker. (a) Glass sheet (b) Plane mirror (c) Black Painted vessels
Answer:
(ii) (a) Glass sheet: Traps heat inside via the greenhouse effect, preventing infrared radiation from escaping.
(b) Plane mirror: Reflects and redirects additional sunlight into the cooker box.
(c) Black Painted vessels: Absorb maximum heat radiation from sunlight efficiently.
In simple words: Mirrors reflect light in, glass traps the heat, and black paint absorbs it.

Exam Tip: Black surfaces are best absorbers of heat radiation.

Q.5. What is a solar cell? Give 2 examples and two uses of solar cell.
Answer: A solar cell is a device that converts sunlight directly into electrical energy.
Examples of semiconductor materials: Silicon, Germanium.
Uses: Powering artificial satellites, calculators, street lights, and remote solar power systems.
In simple words: Solar cells turn sunlight into electricity and power satellites and calculators.

Exam Tip: Mention silicon as the primary material and space satellites as a classic application.

Q.6. What are the limitations of solar heating devices?
Answer:
1. They do not work on cloudy days or at night.
2. They have limited efficiency and cannot produce very high temperatures instantly.
In simple words: Solar devices depend on sunlight and fail on cloudy days or at night.

Exam Tip: Intermittent availability of sunlight is the main drawback of solar devices.

Q.7. A solar cell transforms energy of one form to another. What are these forms?
Answer: Solar cells convert solar energy (light energy) into electrical energy.
In simple words: Light energy is converted into electrical energy.

Q.8. Explain how wind energy can be used to generate electricity. Also mention two advantages and two limitations of using wind energy for generating electricity.

Answer: High-speed wind rotates the blades of a windmill, which turns a turbine connected to an electric generator to produce electricity.
Advantages: Renewable, pollution-free.
Limitations: Requires minimum wind speed (15 km/h), requires large land areas for wind farms.
In simple words: Wind turns turbines to make electricity; it is clean but needs steady wind and large open spaces.

Exam Tip: Wind energy is pollution-free but site-dependent.

Q.9. What is biogas? Name the major component of biogas.
Answer: Biogas is a gaseous fuel produced by the anaerobic decomposition of biomass (organic wastes like animal dung).
Major component: Methane (\(\text{CH}_4\), about 75%).
In simple words: Biogas is a clean fuel produced from organic waste, with methane as its main component.

Exam Tip: Methane is the flammable gas responsible for biogas energy content.

Q.10. Describe the construction and working of a biogas plant with the help of a labeled diagram.
Answer: A biogas plant consists of a mixing tank, a sealed underground digester tank, a gas holder dome, and an outlet tank. Organic waste slurry is fed into the digester where anaerobic bacteria break it down in the absence of oxygen to produce biogas, which collects in the dome.
In simple words: Organic waste ferments in an underground tank without air to produce combustible biogas.

Exam Tip: Mention the absence of oxygen (anaerobic conditions) in your description.

Q.11. Define nuclear fission and nuclear fusion with an example.
Answer:
Nuclear fission: Splitting of a heavy nucleus into lighter nuclei with release of energy (e.g., Uranium-235 splitting under neutron bombardment).
Nuclear fusion: Combining of two light nuclei to form a heavier nucleus with release of immense energy (e.g., Hydrogen isotopes fusing to form helium in the Sun).
In simple words: Fission splits heavy atoms, while fusion combines light atoms, both releasing huge amounts of energy.

Exam Tip: Fission powers nuclear reactors; fusion powers the Sun and hydrogen bombs.

H.O.T.S. Questions : Very Short Answer Type Question [1 mark]

Q.1. What kind of mirror – concave, convex or plane-would be suited for use in a solar cooker? Why?
Answer: Concave mirror, because it converges sunlight to a focal point, producing the high temperatures required for cooking.
In simple words: Concave mirrors concentrate sunlight to generate high heat.

Exam Tip: Converging property of concave mirrors is essential for solar heating.

Q.2. Name two types of energy which do not relate to the sun.
Answer: Geothermal energy and Nuclear energy.
In simple words: Geothermal and nuclear energy do not rely on solar radiation.

Exam Tip: Geothermal relies on earth's internal heat; nuclear relies on atomic binding energy.

Q.3. Biomass is a renewable source of energy. Why?
Answer: Biomass is derived from plants and animals that continuously grow and reproduce, ensuring a continuous, renewable supply of organic matter.
In simple words: Biomass is renewable because plants and organic matter grow back continuously.

Exam Tip: Biological cycles ensure continuous replenishment of biomass.

Short Answer Type Question [2 marks]

Q.1. Why is CNG considered an environmental friendly fuel? State two reasons for it?
Answer:
1. It burns completely without leaving any smoke or soot residue.
2. It produces significantly lower emissions of greenhouse gases and sulfur dioxide compared to petrol or diesel.
In simple words: CNG burns cleanly with minimal pollution and harmful emissions.

Q.2. Mention any two ways by which water can be used to produce hydro-electricity?
Answer:
1. Building dams across rivers to create high-head water reservoirs.
2. Utilizing run-of-river water flow or tidal flow to turn turbines.
In simple words: Using dam reservoirs or river currents to spin hydroelectric generators.

Exam Tip: Dams store gravitational potential energy for reliable electricity generation.

Q.3. In many application, solar cells are connected to rechargeable batteries. Why is it so?
Answer: Solar cells generate electricity only when sunlight is available. Connecting them to rechargeable batteries stores the excess energy for use at night or on cloudy days.
In simple words: Batteries store solar electricity for use when the sun isn't shining.

Exam Tip: Energy storage overcomes the intermittent nature of solar power.

Q.4. “The calorific value of cooking gas is 50KJ/g.” What does it mean?
Answer: It means that complete combustion of 1 gram of cooking gas releases 50 kilojoules of heat energy.
In simple words: Burning 1 gram of the gas produces 50 kJ of heat.

Exam Tip: Calorific value measures the energy density of a fuel in kJ/g or kJ/kg.

Q.5. A nuclear reaction is represented by the following equation:
\(^{232}_{92}\text{U} + ^1_0\text{n} \longrightarrow ^{139}_{56}\text{Ba} + ^{94}_{36}\text{Kr} + xc + \text{Energy}\)
(i) Name the process represented by this equation.
(ii) Identify the particle c and the number x.
(iii) Name one installation where the above reaction is utilized.
(iv) What type of bomb is based on similar type of reaction?
Answer:
(i) Nuclear fission.
(ii) Particle c is a neutron (\(^1_0\text{n}\)), and number \(x = 3\) (three neutrons are released).
(iii) Nuclear power reactor.
(iv) Atomic bomb (Fission bomb).
In simple words: This is a nuclear fission reaction producing neutrons, used in nuclear power plants and atomic bombs.

Exam Tip: Balance mass numbers and atomic numbers on both sides to find missing particles and coefficients.

Q.6. A nuclear reaction is represented by the following equation:
\(^2_1\text{H} + ^2_1\text{H} \longrightarrow ^3_2\text{He} + xc + \text{Energy}\)
(i) Name the process represented by this equation.
(ii) Identify the particle c and the number x.
(iii) State two conditions under which such a reaction takes place.
(iv) What type of bomb is based on similar type of reaction?
Answer:
(i) Nuclear fusion.
(ii) Particle c is a neutron (\(^1_0\text{n}\)), and number \(x = 1\).
(iii) Conditions: Extremely high temperature (millions of degrees) and high pressure.
(iv) Hydrogen bomb (Thermonuclear bomb).
In simple words: This is a nuclear fusion reaction requiring extreme heat and pressure, similar to the reaction in hydrogen bombs.

Exam Tip: Nuclear fusion requires extreme temperatures to overcome electrostatic repulsion between nuclei.

Value Based Question

Q.1. There are many dangers in the use of nuclear power plants. There have been a number of situations in which these dangers have become real disasters, giving birth to safety and regulatory agencies.
(i) Name three main dangers of nuclear power plants.
(ii) How should the nuclear nations conduct themselves to avert these dangers?
Answer:
(i) Dangers: Accidental radiation leaks, improper nuclear waste disposal, and risk of core meltdown.
(ii) Precautions: Implement rigorous safety standards, secure containment structures, and practice international cooperation for waste management.
In simple words: Nuclear plants pose radiation and waste disposal risks, requiring strict safety measures and proper waste management.

Exam Tip: Highlight both environmental safety and regulatory compliance.

Module 10P (11 & 12) : Section – A

Q.1. A real and diminished image of an object can be obtained by,
(i) Plane Mirror
(ii) Concave Mirror
(iii) Convex Mirror
(iv) Both Concave & Convex Mirrors
Answer: (ii) Concave Mirror
A concave mirror forms a real and diminished image when the object is placed beyond the center of curvature (C).
In simple words: Concave mirrors can form real, smaller images when objects are placed far away.

Exam Tip: Convex mirrors always form virtual and diminished images.

Q.2. The relation between f and R in case of a concave mirror is,
(i) \(f = 2R\)
(ii) \(f = \frac{2}{R}\)
(iii) \(f = \frac{R}{2}\)
(iv) \(f = \frac{1}{R}\)
Answer: (iii) \(f = \frac{R}{2}\)
The focal length of a spherical mirror is half of its radius of curvature (\(f = R/2\)).
In simple words: Focal length is always half the radius of curvature.

Exam Tip: Remember \(R = 2f\) or \(f = R/2\).

Q.3. Image formed by reflection from a convex mirror is,
(i) Real only
(ii) Virtual only
(iii) Both real & virtual
(iv) Neither real nor virtual
Answer: (ii) Virtual only
A convex mirror always forms virtual, erect, and diminished images regardless of object position.
In simple words: Convex mirrors always form upright, virtual images.

Exam Tip: Convex mirrors never form real images.

Q.4. Which of the following letter doesn’t show lateral inversion,
(i) N
(ii) Q
(iii) P
(iv) O
Answer: (iv) O
Letters with vertical line symmetry (like O, H, I, M, T, U, V, W, X) do not show noticeable lateral inversion.
In simple words: Symmetrical letters like 'O' look identical in a mirror.

Exam Tip: Look for letters with vertical axes of symmetry.

Q.5. Linear magnification of a plane mirror is,
(i) Always positive
(ii) Always negative
(iii) Both +ive and -ive
(iv) Depends upon position of the object
Answer: (i) Always positive
A plane mirror always forms a virtual and erect image of the same size, so magnification \(m = +1\) (always positive).
In simple words: Plane mirror magnification is always +1.

Exam Tip: Positive sign indicates an erect (upright) image.

Q.6. Mirror formula is applicable to,
(i) Plane mirrors only
(ii) Spherical mirror only
(iii) Both plane and spherical mirror
(iv) Neither plane nor spherical
Answer: (iii) Both plane and spherical mirror
The mirror formula (\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)) applies to both spherical mirrors and plane mirrors (taking plane mirror as a sphere of infinite radius).
In simple words: Mirror formula works for both plane and spherical mirrors.

Exam Tip: Plane mirrors can be treated as a special case of spherical mirrors with infinite focal length.

Q.7. Magnification of mirror is \(+\frac{2}{3}\). The type of mirror is,
(i) Concave
(ii) Plane
(iii) Convex
(iv) Combination of all
Answer: (iii) Convex
A positive magnification less than 1 indicates a virtual, erect, and diminished image, which is characteristic of a convex mirror.
In simple words: Positive and less than 1 magnification means a convex mirror.

Exam Tip: Positive m = virtual/erect; Magnitude < 1 = diminished (convex mirror).

Q.8. When an object placed in front of a convex mirror is moved away from it, the image –
(i) Also moves away from focus
(ii) Moves closer to focus
(iii) Becomes larger and larger
(iv) Becomes real and inverted
Answer: (ii) Moves closer to focus
As an object moves away from a convex mirror, its virtual image shifts closer to the focus behind the mirror and becomes smaller.
In simple words: Moving the object away pushes the image closer to the focal point.

Exam Tip: For convex mirrors, image distance is always less than object distance and bounded by the focus.

Q.9. Plane mirror is a part of sphere of radius
(i) Zero
(ii) Infinity
(iii) Nil, since its not a part of sphere
(iv) Any value depending upon its size
Answer: (ii) Infinity
A plane mirror is considered a spherical mirror with an infinite radius of curvature (\(R = \infty\)).
In simple words: A plane mirror has an infinite radius of curvature.

Exam Tip: Infinite radius means infinite focal length (\(f = \infty\)).

Q.10. A ray of light traveling parallel to the principal axis after reflection from a convex mirror,
(i) Always passes through its focus
(ii) Always passes through its centre of curvature
(iii) Appears to come from its focus behind the mirror
(iv) Retraces its path
Answer: (iii) Appears to come from its focus behind the mirror
A convex mirror diverges incident rays parallel to the principal axis so that they appear to diverge from its principal focus behind the mirror.
In simple words: Parallel rays diverge from the focal point behind a convex mirror.

Exam Tip: Convex mirrors have a virtual focus located behind the mirror.

Section – B

Q.1. Why a ray of light passing through the centre of curvature of a concave mirror retrace its path after reflection?
Answer: A ray passing through the centre of curvature strikes the spherical mirror surface normally (\(90^\circ\) angle of incidence), so the angle of reflection is also zero (\(\angle i = \angle r = 0^\circ\)), causing it to retrace its path.
In simple words: Rays hitting the mirror perpendicularly reflect right back along the same path.

Exam Tip: Normal incidence means light strikes along the radius of curvature.

Q.2. One wants to see an enlarged image of an object in a mirror. What kind of mirror should one use and where should the object be placed?
Answer: Use a concave mirror. The object should be placed between the focus (F) and the pole (P) of the mirror to obtain an enlarged, virtual, and erect image.
In simple words: Use a concave mirror and place the object between the focus and pole for an enlarged image.

Q.3. Which property of a concave mirror can be used to determine its focal length?
Answer: Rays parallel to the principal axis converge at the principal focus after reflection. Measuring the distance between the pole of the mirror and the sharp image of a distant object gives the focal length.
In simple words: The distance where parallel rays converge gives the focal length.

Exam Tip: Distant object method is a standard experimental way to find focal length.

Q.4. What kind of mirror is used in vehicles to see the traffic following it?
Answer: Convex mirror, because it always forms a virtual, erect, and diminished image, providing a wide field of view.
In simple words: Convex mirrors are used as rear-view mirrors because they provide a wide field of view.

Exam Tip: Wide field of view and upright images make convex mirrors ideal for driving safety.

Q.5. A man standing in front of a special mirror finds his image having a very small head, a fat body and legs of normal size. What are the shapes of the three parts of the mirror?
Answer:
- Top part (small head): Convex mirror (forms diminished image).
- Middle part (fat body): Concave mirror (forms enlarged image).
- Bottom part (normal legs): Plane mirror (forms same-size image).
In simple words: Convex, concave, and plane mirror sections create distorted amusement park mirror reflections.

Exam Tip: Relate image characteristics (diminished, magnified, normal) to mirror types.

Q.6. What type of mirror is formed when a mercury drop falls on the earth?
Answer: A convex mirror, because its upper surface bulges outwards due to surface tension.
In simple words: A spherical mercury droplet acts as a convex mirror.

Exam Tip: Droplets bulge outward, creating a convex reflective surface.

Section – C

Q.1. Distinguish between real and virtual image.
Answer:
Real Image: Can be caught on a screen, formed by actual intersection of light rays, always inverted.
Virtual Image: Cannot be caught on a screen, formed by diverging rays appearing to meet, always erect.
In simple words: Real images can be projected on screens, while virtual images can only be seen in mirrors.

Exam Tip: Screen projection is the definitive test to distinguish real from virtual images.

Q.2. Draw ray diagrams to show the formation of images when the object is placed in front of a concave mirror. (i) Between its pole and focus point. (ii) Between its centre of curvature and focus point.
Answer:
(i) Object between P and F: Image is virtual, erect, and enlarged behind the mirror.
(ii) Object between C and F: Image is real, inverted, and enlarged beyond C.
In simple words: Ray diagrams show virtual magnified images close to the mirror and real inverted images further away.

Exam Tip: Always draw arrowheads on incident and reflected rays in ray diagrams.

Q.3. How will you distinguish between a plane mirror, a concave mirror and a convex mirror without touching them?
Answer: Look at your face in the mirror:
- Image of same size: Plane mirror.
- Image magnified (enlarged): Concave mirror (when close).
- Image diminished (smaller): Convex mirror.
In simple words: Compare image size (same, larger, smaller) to identify the mirror type without touching it.

Exam Tip: Size comparison provides a quick way to identify mirror types.

Q.4. Define : (i) Radius of curvature. (ii) Principal focus
Answer:
(i) Radius of curvature: The radius of the hollow sphere of which the spherical mirror is a part.
(ii) Principal focus: A point on the principal axis where rays parallel to the axis converge (or appear to diverge) after reflection.
In simple words: Radius of curvature is the sphere's radius, and focus is where parallel light rays converge.

Exam Tip: Use precise scientific terminology when writing definitions.

Q.5. Show that image distance is same as object distance when an extended object is placed in front of the plane mirror.
Answer: By laws of reflection, the perpendicular distance of an object in front of a plane mirror equals the perpendicular distance of its virtual image behind the mirror (\(u = v\)).
In simple words: Images appear as far behind a plane mirror as the object is in front of it.

Exam Tip: Use congruent triangle geometry to prove object distance equals image distance in plane mirrors.

Section – D

Q.1. A short linear object of 2 cm. lies on the axis of a concave mirror of 15 cm. focal length at a distance of 30 cm. from the mirror. What is the size of the image?
Answer:
Given: \(h_1 = 2\text{ cm}\), \(f = -15\text{ cm}\), \(u = -30\text{ cm}\).
Using mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
\(\frac{1}{-15} = \frac{1}{v} + \frac{1}{-30}\)
\(\frac{1}{v} = -\frac{1}{15} + \frac{1}{30} = \frac{-2 + 1}{30} = -\frac{1}{30} \implies v = -30\text{ cm}\).
Magnification \(m = -\frac{v}{u} = \frac{h_2}{h_1}\)
\(-\frac{-30}{-30} = \frac{h_2}{2} \implies -1 = \frac{h_2}{2} \implies h_2 = -2\text{ cm}\).
Size of the image is 2 cm (real and inverted).
In simple words: Using the mirror and magnification formulas, the image size is calculated to be 2 cm.

Exam Tip: Always apply sign conventions correctly for u, v, and f.

Q.2. The image behind a convex mirror (\(R = 68\text{ cm}\)) is located at 22 cm. from the mirror. Where is the object located? What is the magnification produced? Write the nature of image formed.
Answer:
Radius \(R = +68\text{ cm}\), so focal length \(f = \frac{R}{2} = +34\text{ cm}\).
Image distance \(v = +22\text{ cm}\) (behind the convex mirror).
Using mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
\(\frac{1}{34} = \frac{1}{22} + \frac{1}{u}\)
\(\frac{1}{u} = \frac{1}{34} - \frac{1}{22} = \frac{11 - 17}{374} = -\frac{6}{374} \implies u = -\frac{374}{6} \approx -62.33\text{ cm}\).
Magnification \(m = -\frac{v}{u} = -\frac{22}{-62.33} \approx +0.35\).
Nature: Virtual, erect, and diminished.
In simple words: Object distance is 62.33 cm in front, forming a virtual, erect, and diminished image.

Exam Tip: Focal length and image distance for convex mirrors are positive (behind the mirror).

Q.3. Find the position, nature and size of the image of an object 4 cm high placed at a distance 10 cm from a concave mirror of focal length 20 cm.
Answer:
Given: \(h_1 = 4\text{ cm}\), \(u = -10\text{ cm}\), \(f = -20\text{ cm}\).
Using mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
\(\frac{1}{-20} = \frac{1}{v} + \frac{1}{-10}\)
\(\frac{1}{v} = -\frac{1}{20} + \frac{1}{10} = \frac{-1 + 2}{20} = +\frac{1}{20} \implies v = +20\text{ cm}\).
Magnification \(m = -\frac{v}{u} = -\frac{20}{-10} = +2\).
Image size \(h_2 = m \cdot h_1 = 2 \times 4 = 8\text{ cm}\).
Nature: Virtual, erect, and magnified (enlarged).
In simple words: The image forms 20 cm behind the mirror, is virtual, erect, and 8 cm high.

Exam Tip: When object is between F and P of a concave mirror, the image is virtual and magnified.

Q.4. Show that the mirror formula for spherical mirror also holds good for a plane mirror too.
Answer: For a plane mirror, focal length \(f = \infty\). Substituting \(f = \infty\) into the mirror formula: \(\frac{1}{\infty} = \frac{1}{v} + \frac{1}{u} \implies 0 = \frac{1}{v} + \frac{1}{u} \implies v = -u\). This shows object distance equals image distance, proving the mirror formula holds for plane mirrors.
In simple words: Setting focal length to infinity in the mirror formula gives \(v = -u\), confirming plane mirror properties.

Exam Tip: Treat plane mirrors as spherical mirrors with infinite radius and focal length.

Q.5. An object is placed at 10 cm from focus farther away from the mirror. If the image formed is half the size of the object and real in nature, find the focal length of the mirror.
Answer:
Real image and half size means magnification \(m = -0.5 = -\frac{1}{2}\).
\(m = -\frac{v}{u} = -\frac{1}{2} \implies v = \frac{u}{2}\).
Object distance \(u = f - 10\).
Using mirror formula and solving yields focal length \(f = -10\text{ cm}\) (or matching standard mirror parameters).
In simple words: Using magnification and object position equations, the focal length is calculated.

Q.6. An image double the size of the object is formed on placing an object in front of a mirror of focal length 20 cm. Find the position (s) of the object.
Answer:
Given \(f = -20\text{ cm}\).
Case I (Real image): \(m = -2\).
\(m = -\frac{v}{u} = -2 \implies v = 2u\).
Using mirror formula: \(\frac{1}{-20} = \frac{1}{2u} + \frac{1}{u} = \frac{3}{2u} \implies 2u = -60 \implies u = -30\text{ cm}\).
Case II (Virtual image): \(m = +2\).
\(v = -2u\).
\(\frac{1}{-20} = \frac{1}{-2u} + \frac{1}{u} = \frac{1}{2u} \implies 2u = -20 \implies u = -10\text{ cm}\).
Object positions are at \(30\text{ cm}\) or \(10\text{ cm}\) in front of the mirror.
In simple words: Solving for both real and virtual image cases gives object distances of 30 cm and 10 cm.

Exam Tip: Always consider both real (inverted) and virtual (erect) possibilities when magnification magnitude is given without sign.

H.O.T.S. Questions : Very Short Answer Type Question [1 mark]

Q.1. Which colour of the light reflected by an object will make it appear blue?
Answer: Blue colour light.
In simple words: An object appears blue because it reflects blue light.

Exam Tip: Objects appear the colour of the light they reflect.

Q.2. What are the values of (i) the angle of incidence and (ii) the angle of reflection for the normal incidence on a plane mirror surface?
Answer: Both the angle of incidence and the angle of reflection are \(0^\circ\).
In simple words: Normal incidence means angle of incidence and reflection are both zero.

Exam Tip: Angles are measured with respect to the normal, not the mirror surface.

Q.3. A concave mirror is placed in water. Will there be any change in focal length?
Answer: No change. The focal length of a mirror depends only on its radius of curvature and is independent of the surrounding medium.
In simple words: Placing a mirror in water does not change its focal length.

Exam Tip: Refraction depends on medium (lenses), but reflection depends on geometry (mirrors).

Q.4. A person wants to see the full image of tall building in a small mirror. Which type of mirror should be used by him?
Answer: Convex mirror, because it provides a wide field of view and forms diminished images.
In simple words: Convex mirrors fit large views into small mirror reflections.

Exam Tip: Diminished images enable wide-area viewing.

Q.5. What type of image is formed on a cinema screen?
Answer: Real and inverted image.
In simple words: Cinema screens display real images.

Exam Tip: Only real images can be projected onto screens.

Q.6. To use a concave mirror as a dentist’s mirror, what should be the position of the teeth?
Answer: Between the pole (P) and the principal focus (F) of the concave mirror.
In simple words: Teeth must be placed between the focus and pole to see an enlarged image.

Exam Tip: This position gives an erect and magnified virtual image.

Q.7. If a plane mirror is rotated by an angle \(\theta\), by how will the angle between the incident and the reflected ray change?
Answer: The reflected ray rotates by an angle of \(2\theta\).
In simple words: Rotating the mirror by theta rotates the reflected ray by twice that angle (\(2\theta\)).

Exam Tip: Standard optical law: rotating a plane mirror by \(\theta\) shifts the reflected ray by \(2\theta\).

Short Answer Type Question [2 marks]

Q.1. Name the kind of mirror used to obtain. (i) a real and enlarged image (ii) virtual and enlarged image.
Answer:
(i) Real and enlarged: Concave mirror (object between C and F).
(ii) Virtual and enlarged: Concave mirror (object between F and P).
In simple words: Concave mirrors produce both real and virtual enlarged images depending on object distance.

Exam Tip: Concave mirrors are versatile and produce various image types.

Q.2. Name the kind of mirror used to obtain. (i) virtual and diminished image (ii) real and diminished image.
Answer:
(i) Virtual and diminished: Convex mirror.
(ii) Real and diminished: Concave mirror.
In simple words: Convex mirrors give virtual diminished images, while concave mirrors give real diminished images.

Exam Tip: Convex mirrors never form real images.

Q.3. What happens to the speed of light when it undergoes reflection? Why?
Answer: The speed of light remains unchanged during reflection because light stays in the same optical medium.
In simple words: Light speed stays constant upon reflection since it does not change medium.

Exam Tip: Speed changes only during refraction (changing medium), not reflection.

Short Answer Type Question [3 marks]

Q.1. In the figure (a), AB is the object, A' B' is its image and MM' is the position of the mirror. Complete the ray diagram and find the position of centre of curvature and focus of the mirror. Also find the nature of the mirror
Answer: The mirror is a concave mirror. By drawing converging rays from object tip A to image tip A', the intersection with the principal axis gives the center of curvature C, and focal point F lies midway between pole and C.
In simple words: Tracing rays between object and image reveals a concave mirror with its focus and center of curvature.

Exam Tip: Ray tracing geometry helps locate optical centers and focal points.

Module 10P (13 & 14) : Section : A

Q.1. A ray of light travelling in air falls obliquely on the surface of a calm pond. It will
(i) Go into the water without deviating from its path.
(ii) Deviate away from the normal.
(iii) Deviate towards the normal.
(iv) Turn back on its original path.
Answer: (iii) Deviate towards the normal.
When light travels from an optically rarer medium (air) to a denser medium (water), it bends towards the normal.
In simple words: Light bends towards the normal when entering water from air.

Exam Tip: Rarer to denser medium bends light towards the normal.

Q.2. A thin lens and a spherical mirror have a focal length of + 15 cm each.
(i) Both are convex.
(ii) The lens is convex and the mirror is concave.
(iii) The lens is concave and the mirror is convex.
(iv) Both are concave.
Answer: (i) Both are convex.
A convex lens and a convex mirror both have positive focal lengths according to sign conventions (\(f > 0\)).
In simple words: Positive focal length indicates convex optical components.

Exam Tip: Convex lens and convex mirror both possess positive focal lengths.

Q.3. A convex lens forms a virtual image when an object is placed at a distance of 18 cm from it. The focal length will be
(i) Greater than 36 cm
(ii) Greater than 18 cm
(iii) Less than 36 cm
(iv) Less than 18 cm
Answer: (ii) Greater than 18 cm
A convex lens forms a virtual, magnified image when the object is placed within its focal length (\(u < f\)). Therefore, \(f > 18\text{ cm}\).
In simple words: For a virtual image, the object must be within the focal length, meaning focal length is greater than 18 cm.

Exam Tip: Virtual image with convex lens requires \(u < f\).

Q.4. An object is placed before a convex lens. The image formed.
(i) Is always real
(ii) May be real or virtual
(iii) Less than virtual
(iv) Is always erect
Answer: (ii) May be real or virtual
A convex lens forms real images when the object is beyond F, and virtual images when the object is within F.
In simple words: Convex lenses can form both real and virtual images.

Exam Tip: Concave lenses form virtual images only; convex lenses form both.

Q.5. An object is placed before a concave lens. The image formed
(i) Is always erect
(ii) May be erect or inverted
(iii) Is always inverted
(iv) Is always real
Answer: (i) Is always erect
A concave lens always forms virtual, erect, and diminished images for all object positions.
In simple words: Concave lenses always produce upright (erect) virtual images.

Exam Tip: Concave lens images are always virtual, erect, and diminished.

Q.6. A lens has a power of + 0.5 D. It is
(i) A concave lens of focal length 5 m
(ii) A convex lens of focal length 5 cm
(iii) A convex lens of focal length 2 m
(iv) A concave lens of focal length 2 m
Answer: (iii) A convex lens of focal length 2 m
Power \(P = +\frac{1}{f(\text{in m})} \implies f = \frac{1}{0.5} = +2\text{ m}\). Positive power indicates a convex lens.
In simple words: Positive power means a convex lens with a 2-meter focal length.

Exam Tip: Positive power = convex lens; Negative power = concave lens.

Section : B

Q.1. Focal length of a convex lens is same as radius of curvature of a concave mirror. If the mirror converges the light parallel to the principal axis at a distance of 10 cm from the pole where does the same light beam converge by the lens?
Answer:
For the concave mirror, focal length \(f_m = 10\text{ cm}\), so its radius of curvature \(R_m = 2 \times 10 = 20\text{ cm}\).
Given that the focal length of the convex lens equals this radius of curvature: \(f_l = 20\text{ cm}\).
Therefore, parallel light incident on the convex lens converges at its principal focus, which is at a distance of \(20\text{ cm}\).
In simple words: The lens focal length is 20 cm, so parallel rays converge at 20 cm.

Exam Tip: Read multi-step conditional problems carefully to link mirror and lens parameters.

Q.2. Show by ray diagram the formation of images by a convex lens in any two cases.
Answer:
Case 1: Object at infinity -> Image forms at focus F, real, inverted, and point-sized.
Case 2: Object beyond \(2F\) -> Image forms between F and \(2F\), real, inverted, and diminished.
In simple words: Ray diagrams show convex lenses forming real inverted images at various positions.

Exam Tip: Draw principal rays (parallel to axis and through optical center) for clear diagrams.

Q.3. A bird flies down vertically towards a water surface. To a fish inside the water, vertically below the bird, how will the bird appears to be?
Answer: Due to refraction of light going from a rarer medium (air) to a denser medium (water), the bird will appear to be at a higher altitude than its actual position to the fish.
In simple words: Refraction makes the bird appear higher than it actually is to the fish.

Q.4. A lens forms a real image of an object when placed at 20 cm from the lens. Find (i) Type of lens (ii) The image distance if \(f = 20\text{ cm}\) (iii) The magnification
Answer:
(i) Type of lens: Convex lens (only lens that forms real images).
(ii) Image distance: If object is at \(u = -20\text{ cm}\) (at focus \(f = +20\text{ cm}\)), applying lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \implies \frac{1}{20} = \frac{1}{v} - \frac{1}{-20} \implies \frac{1}{v} = 0 \implies v = \infty\).
(iii) Magnification: \(m = \frac{v}{u} = \infty\).
In simple words: Object at the focus of a convex lens forms a real image at infinity with infinite magnification.

Exam Tip: Object at focus (\(u = f\)) forms an image at infinity.

Q.5. Explain Myopia with the help of suitable ray diagrams. How can this defect of vision be corrected? A boy uses spectacles of focal length – 50 cm. Name the defect of vision he is suffering from. Characteristics / power calculation:
Answer:
Myopia (near-sightedness): A defect where distant objects appear blurry because images form in front of the retina. Corrected using a concave lens.
Power calculation: Given \(f = -50\text{ cm} = -0.5\text{ m}\).
Power \(P = \frac{1}{f(\text{in m})} = \frac{1}{-0.5} = -2.0\text{ D}\).
In simple words: Myopia is near-sightedness corrected with a -2.0D concave lens.

Exam Tip: Negative focal length and power always indicate a concave correcting lens.

Q.6. An object of size 3 cm is placed at a distance of 15 cm from a convex lens of focal length 10 cm. Calculate the distance and size of the image so formed. What will be the nature of the image?
Answer:
Given: \(h_1 = 3\text{ cm}\), \(u = -15\text{ cm}\), \(f = +10\text{ cm}\).
Using lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
\(\frac{1}{10} = \frac{1}{v} - \frac{1}{-15}\)
\(\frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30} \implies v = +30\text{ cm}\).
Magnification \(m = \frac{v}{u} = \frac{30}{-15} = -2\).
Size \(h_2 = m \cdot h_1 = (-2) \times 3 = -6\text{ cm}\).
Nature: Real, inverted, and enlarged (size 6 cm).
In simple words: Image forms 30 cm on the other side, real, inverted, and 6 cm high.

Exam Tip: Lens formula has a minus sign (\(1/f = 1/v - 1/u\)), whereas mirror formula has a plus sign.

Section : C

Q.1. Two cubes of equal size are placed on a newspaper. One cube is made of glass and the other is made of a transparent plastic whose refractive index is higher than that of glass. When viewed from above, below which cube will the letters appear nearer? Show the image formation in any one case.
Answer: The letters will appear nearer below the plastic cube because it has a higher refractive index, resulting in greater bending of light and a larger apparent shift (greater upward displacement).
In simple words: Higher refractive index causes greater upward shift, making letters appear closer under the plastic cube.

Exam Tip: Apparent depth decreases more when refractive index is higher.

Q.2. Why is a convex lens called a converging lens and a concave lens, a diverging lens?
Answer: A convex lens is thicker at the center and bends parallel light rays inward to converge at a point. A concave lens is thinner at the center and spreads parallel light rays outward, causing them to diverge.
In simple words: Convex lenses converge light rays together, while concave lenses spread them apart.

Q.3. How do you conclude that a ray of light incident towards the optical center of a thin lens is transmitted almost without any deviation?
Answer: Because a thin lens behaves like a parallel glass slab at its optical center, the emergent ray is parallel to the incident ray with negligible lateral displacement.
In simple words: Rays passing through the optical center of a thin lens pass straight through without bending.

Exam Tip: Optical center is the undeviated ray path point in lenses.

Q.4. The focal length of a convex lens is f. How does the size the image placed in front of it change as the object is brought progressively closer to the focus from a distance which is just greater than 2f?
Answer: As the object is brought closer from \(2f\) to \(f\), the real image shifts further away from the lens and its size increases progressively, becoming infinitely large at the focus.
In simple words: Bringing the object closer to the focus makes the real image progressively larger.

Q.1. The refractive index of media A, B, C & D are 1.44, 1.47, 1.83, 1.33 respectively. In which of these does the light travel (i) fastest (ii) slowest?
Answer:
(i) Fastest: Medium D (lowest refractive index 1.33, since speed is inversely proportional to refractive index).
(ii) Slowest: Medium C (highest refractive index 1.83).
In simple words: Light travels fastest in the medium with the lowest refractive index and slowest in the highest.

Exam Tip: Speed \(v = c/n\); lower n means higher speed.

Q.2. When light undergoes refraction at the surface of separation of two media, what happens to its wavelength?
Answer: The wavelength of light changes (decreases in a denser medium and increases in a rarer medium), while its frequency remains constant.
In simple words: Wavelength changes during refraction, but frequency stays the same.

Exam Tip: Frequency is a source property and never changes across media boundaries.

Q.3. When does a ray of light neither refract nor deviate when passing through a glass slab?
Answer: When the ray of light is incident normally (\(0^\circ\) angle of incidence) to the surface of the glass slab.
In simple words: Light striking straight-on (normally) passes through without bending.

Exam Tip: Normal incidence results in zero deviation (\(\angle i = \angle r = 0^\circ\)).

Q.4. The power of a lens used in a reading glass of a person is +1D. What is the nature of the lens?
Answer: Convex lens (converging lens), indicated by the positive sign of power.
In simple words: Positive power indicates a convex lens.

Exam Tip: Reading glasses for hypermetropia use positive power convex lenses.

Q.5. Light from a sodium-vapour street lamp is passed through a prism. What colours of light do you expect to see if the transmitted light is allowed to fall on a piece of white paper?
Answer: Only yellow light, because sodium-vapour lamps emit monochromatic (single-wavelength) yellow light, not a full spectrum of white light.
In simple words: Sodium lamps emit single-colour yellow light, not a rainbow spectrum.

Exam Tip: Monochromatic sources do not split into constituent colours.

Q.6. If one-half of a convex lens is painted black, will it form the complete image of an object.
Answer: Yes, it will form the complete image of the object, though the intensity (brightness) of the image will be reduced because fewer light rays pass through.
In simple words: The image will still be complete, just dimmer.

Q.1. When a bird looks at a fish in water does it appear raised or deeper than it actually is? When a fish looks at the bird does it appear nearer or further away?
Answer:
- Fish viewed by bird: Appears deeper than it actually is.
- Bird viewed by fish: Appears further away (higher) than it actually is.
In simple words: Refraction makes objects in denser media appear shallower, and objects in rarer media appear higher.

Exam Tip: Apparent depth/height depends on whether you view from rarer to denser or vice versa.

Q.2. Copy the figure (a) and complete the formation of the image. What is the type of lens used in the figure? What is the type of the image formed?
Answer: Type of lens: Concave lens. Type of image: Virtual, erect, and diminished.
In simple words: The ray diagram illustrates a concave lens forming a virtual, upright, and diminished image.

Exam Tip: Concave lens ray diagrams always produce virtual, erect, and diminished images.

Q.3. Complete the following ray diagram in Figure (I-a) in which AB is the object and A'B' is the image. Locate the lens and mark the focus of the lens by the letter F. What type of lens is this?
Answer: The lens is a convex lens. By tracing lines from object to image, the optical center and focal points F are located.
In simple words: Tracing the ray intersection points identifies a convex lens and its focal points.

Exam Tip: Ray intersection points define optical component positions.

Q.4. In Figure (I-b), state the position of object, position of image, nature of image when (i) convex lens is used as burning glass (ii) convex lens is used in observing biological specimens.
Answer:
(i) Burning glass: Object is at focus (F), image forms at infinity, real and extremely magnified.
(ii) Observing specimens (magnifying glass): Object is between optical center and focus (P and F), image is virtual, erect, and magnified behind the object.
In simple words: At focus it concentrates light for burning, while inside the focus it acts as a magnifying glass.

Q.5. Which lens assist in reading a book? Which lens assist in reading from a blackboard in a class?
Answer:
- Reading a book (hypermetropia correction): Convex lens.
- Reading from a blackboard (myopia correction): Concave lens.
In simple words: Convex lenses help read close-up books, while concave lenses help read distant blackboards.

Exam Tip: Convex for near vision (reading); Concave for distant vision (blackboard).

Q.6. Can one burn a piece of paper by just using a convex lens instead of a matchstick or any direct flame? Support your answer with the help of an appropriate ray diagram.
Answer: Yes. A convex lens converges parallel sunlight rays to a sharp focal point on the paper, concentrating solar heat energy enough to raise its ignition temperature and burn it.
In simple words: Concentrating sunlight at the focus of a convex lens generates enough heat to burn paper.

Exam Tip: This demonstrates the converging and heat-concentrating property of convex lenses.

Q.7. What happens to the speed of light when it undergoes refraction? Explain why?
Answer: The speed of light changes when passing from one medium to another. It slows down when entering a optically denser medium and speeds up when entering a rarer medium due to electromagnetic interactions with optical media atoms.
In simple words: Light slows down in denser media and speeds up in rarer media.

Exam Tip: Refractive index \(n = c/v\) relates light speed in vacuum to speed in the medium.

Short Answer Type Question [3 marks]

Q.1. In case of refraction of light through a rectangular glass slab, explain why (i) the emergent ray is parallel to the direction of the incident ray (ii) the emergent ray is shifted sideward slightly?
Answer:
(i) The extent of bending at the air-glass interface equals the opposite bending at the glass-air interface, making the emergent ray parallel to the incident ray.
(ii) It is shifted sideward (lateral displacement) because the refraction angles at both parallel boundaries shift the ray path sideways.
In simple words: Equal and opposite bending makes emergent and incident rays parallel, with a slight sideways shift.

Exam Tip: Parallel faces of a glass slab ensure lateral displacement without angular deviation.

Value Based Question

Q.1. A spherical mirror in which reflecting surface is towards the centre of the sphere of which the mirror is a part, is called concave mirror. The rays of light incident on concave mirror in a direction parallel to principal axis, actually meet at a single point „F’ on the principal axis of the mirror. This point is called principal focus of the mirror. Read the above passage and answer the following question:
(i) Is principal focus of a concave mirror, a real point or a virtual point?
(ii) What happens in case of a convex mirror?
(iii) Our teachers and parents advise us to stay focused. What does it imply?
Answer:
(i) Real point (reflected rays actually converge at F).
(ii) In a convex mirror, parallel rays appear to diverge from a virtual focus behind the mirror.
(iii) Staying focused implies concentrating our attention and efforts on our goals without getting distracted.
In simple words: Concave focus is real, convex focus is virtual, and in life, staying focused means concentrating on your goals.

Exam Tip: Connect scientific principles metaphorically to moral values in value-based questions.

Module 10P (15, 16 & 17) : Section : A

Q.1. The least distance of distinct vision for normal eye is
(i) Infinity
(ii) 25 m
(iii) 25 cm
(iv) 25 m
Answer: (iii) 25 cm
The nearest point at which a normal human eye can see objects clearly without strain is 25 cm.
In simple words: The standard least distance of distinct vision is 25 cm.

Exam Tip: Memorize 25 cm as the standard near point for normal human vision.

Q.2. The change in the focal length of eye lens is brought by.
(i) ciliary muscles
(ii) pupil
(iii) optical nerves
(iv) cornea
Answer: (i) ciliary muscles
Ciliary muscles alter the curvature and focal length of the eye lens to focus on near and distant objects (accommodation).
In simple words: Ciliary muscles adjust the eye lens shape to focus properly.

Q.3. Persistance of vision of eye is
(i) 1/10 sec.
(ii) 1/16 sec.
(iii) 1/100 sec.
(iv) cornea
Answer: (ii) 1/16 sec.
The impression of an image persists on the retina for about 1/16th of a second after the object is removed.
In simple words: Visual impressions last on the retina for 1/16th of a second.

Exam Tip: Persistence of vision is the principle behind cinematography (movies).

Q.4. Electric pulses from retina to brain is conveyed via
(i) ciliary muscles
(ii) blind spot
(iii) pupil
(iv) optical nerves
Answer: (iv) optical nerves
Optic nerves transmit electrical signals generated by retinal photoreceptors to the brain for visual processing.
In simple words: Optic nerves carry visual signals from the eye to the brain.

Exam Tip: Optic nerve acts as the communication link between eye and brain.

Q.5. Part of eye which is removed from eyes of dead body for eye donation is
(i) cornea
(ii) Retina
(iii) Iris
(iv) Pupil
Answer: (i) cornea
Cornea grafting is performed in eye donation because the cornea is avascular and can be successfully transplanted.
In simple words: The cornea is the eye part transplanted during eye donation.

Exam Tip: Corneal opacity is corrected via corneal transplantation from donors.

Q.6. Accomodation of normal eye is from
(i) 5 cm to 15 cm
(ii) 1m to 3m
(iii) 15 cm to 1 m
(iv) 25 cm to infinity
Answer: (iv) 25 cm to infinity
The normal eye can comfortably focus on objects ranging from 25 cm (near point) to infinity (far point).
In simple words: Normal vision ranges clearly from 25 cm up to infinity.

Exam Tip: Near point = 25 cm; Far point = infinity.

Q.7. The splitting of white light into different colours on passing through a prism is called
(i) Refraction
(ii) Reflection
(iii) Dispersion
(iv) Deviation
Answer: (iii) Dispersion
Dispersion is the splitting of white light into its constituent seven colours (VIBGYOR) due to wavelength-dependent refraction.
In simple words: Splitting white light into colours through a prism is called dispersion.

Exam Tip: Rainbow formation is a natural example of dispersion.

Q.8. Night vision device utilizes
(i) Infra – red rays
(ii) Ultra-violet rays
(iii) X - rays
(iv) \(\gamma\) – rays
Answer: (i) Infra – red rays
Night vision equipment detects infrared radiation emitted as heat by objects and living beings.
In simple words: Night vision devices detect infrared thermal radiation.

Exam Tip: Infrared imaging detects heat signatures in the dark.

Q.9. When white light falls on the glass prism, the least deviated colour is
(i) violet
(ii) orange
(iii) green
(iv) red
Answer: (iv) red
Red light has the longest wavelength and lowest refractive index in glass, experiencing the least deviation.
In simple words: Red light bends the least when passing through a prism.

Exam Tip: Red deviates least; violet deviates most.

Q.10. Which one of the following colour has maximum frequency
(i) Red
(ii) Violet
(iii) Green
(iv) Yellow
Answer: (ii) Violet
Frequency is inversely proportional to wavelength. Since violet has the shortest wavelength, it has the maximum frequency.
In simple words: Violet light has the highest frequency among visible colours.

Exam Tip: Violet = shortest wavelength, highest frequency; Red = longest wavelength, lowest frequency.

Section : B

Q.1. The far point of a shortsighted person is 1.5 m. Find the focal length, Power and nature of the remedial lens.
Answer:
Short-sightedness (myopia) is corrected by a concave lens whose focal length equals the negative of the person's far point: \(f = -1.5\text{ m}\).
Power \(P = \frac{1}{f(\text{in m})} = \frac{1}{-1.5} = -0.67\text{ D}\).
Nature: Concave lens (diverging lens).
In simple words: Focal length is -1.5 m with a power of -0.67 D using a concave lens.

Exam Tip: For myopia correction, lens focal length \(f = -d\), where d is the far point.

Q.2. Complete the following figure.
Answer: The ray diagram showing a concave lens diverging parallel rays so that they appear to diverge from the principal focus is completed.
In simple words: Complete the diagram by showing rays diverging away from the focal point through a concave lens.

Exam Tip: Show dashed lines extending backward to the virtual focus for concave lenses.

Q.3. Why do different colours deviate differently in a glass prism?
Answer: Different colours have different wavelengths. Since the refractive index of glass varies with wavelength (shorter wavelengths have higher refractive index), each colour bends by a different angle.
In simple words: Different colours travel at different speeds in glass, causing them to bend by varying amounts.

Exam Tip: Cauchy's dispersion relation links refractive index to wavelength.

Q.4. What is colour blindness?
Answer: Colour blindness is a genetic visual defect in which a person is unable to distinguish between certain colours (commonly red and green) due to defective cone cells in the retina.
In simple words: Colour blindness is an inability to distinguish certain colours due to defective cone cells.

Exam Tip: Cone cells are responsible for colour vision, while rod cells handle dim-light vision.

Q.5. What are Rods and Cones? State briefly the functions of Rod & Cones.
Answer:
Rods: Photoreceptor cells sensitive to dim light (night vision, black and white).
Cones: Photoreceptor cells sensitive to bright light and responsible for colour vision.
In simple words: Rods help us see in dim light, and cones provide colour vision in bright light.

Exam Tip: Remember: Rods for light intensity; Cones for colour perception.

Q.6. What is the reason behind twinkling of stars? Explain.
Answer: Twinkling of stars is caused by atmospheric refraction. Starlight passes through varying layers of the earth's atmosphere with fluctuating temperatures and densities, causing the apparent position and brightness of the star to constantly shift.
In simple words: Starlight bends through shifting layers of the atmosphere, making stars twinkle.

Exam Tip: Planets do not twinkle because they act as extended sources, averaging out fluctuations.

Section : C

Q.1. Draw a labeled diagram of the human eye and give the basic details about its various parts.
Answer: Key parts include:
- Cornea: Outer transparent protective membrane.
- Iris: Coloured diaphragm controlling pupil size.
- Pupil: Aperture regulating light entry.
- Eye lens: Crystalline lens focusing images.
- Retina: Screen containing photoreceptor cells (rods and cones).
In simple words: The eye focuses light through the cornea and lens onto the light-sensitive retina.

Exam Tip: Practice labelling cornea, iris, pupil, lens, retina, and optic nerve clearly.

Q.2. What is myopia? Draw ray diagrams showing the formation of image of an object at infinity and at its far point by a myopic eye. Explain how this defect is corrected by using a concave lens.
Answer: Myopia is near-sightedness where distant objects appear blurred because light focuses in front of the retina. It is corrected using a concave (diverging) lens that diverges incoming rays so they focus properly on the retina.
In simple words: Myopia is corrected using a concave lens to push the focus back onto the retina.

Exam Tip: Myopia = near-sightedness = corrected with concave lens.

Q.3. What is hypermetropia? Draw ray diagrams showing the formation of image by a hypermetropic eye. For (i) a nearby object (ii) an object at the near point of the eye. How is this defect corrected by using a convex lens?
Answer: Hypermetropia is far-sightedness where nearby objects appear blurred because light focuses behind the retina. It is corrected using a convex (converging) lens that converges light rays before they enter the eye, shifting the focus onto the retina.
In simple words: Hypermetropia is far-sightedness corrected with a convex lens.

Exam Tip: Hypermetropia = far-sightedness = corrected with convex lens.

Q.4. Why is the colour of the clear sky blue?
Answer: Molecules of air and fine particles scatter shorter wavelengths of light (blue and violet) much more strongly than longer wavelengths (red) according to Rayleigh scattering, making the sky appear blue.
In simple words: Air molecules scatter blue light in all directions more than other colours, making the sky look blue.

Exam Tip: Rayleigh scattering states that scattering is inversely proportional to the fourth power of wavelength (\(\propto 1/\lambda^4\)).

Module 10P (18) Revision for Annual Examinations : H.O.T.S. Questions : Very Short Answer Type Questions [1 mark]

Q.1. How does the focal length of the eye lens change?
Answer: By the action of ciliary muscles, which contract or relax to change the curvature and thickness of the crystalline eye lens.
In simple words: Ciliary muscles change the shape of the eye lens to alter its focal length.

Exam Tip: Ciliary muscle contraction makes the lens thicker (shorter focal length for near objects).

Q.2. Name the defect of vision due to (i) Power of the eye is too great (ii) Focal length of the eye lens is too great.
Answer:
(i) Power too great: Myopia (near-sightedness).
(ii) Focal length too great: Hypermetropia (far-sightedness).
In simple words: High eye power causes myopia, while excessively long focal length causes hypermetropia.

Exam Tip: High power means converging power is too strong, focusing light in front of the retina.

Q.3. Which colour light is produced due to the scattering of sunlight by (i) very small-sized (ii) very large-sized constituent particles of the atmosphere?
Answer:
(i) Very small particles: Blue light (Rayleigh scattering).
(ii) Very large particles (like water droplets/dust): White light or all colours scattered equally (Mie scattering).
In simple words: Small particles scatter blue light, while large particles scatter all colours equally as white light.

Q.4. Which property of the eye is used by cinematography to show motion pictures?
Answer: Persistence of vision.
In simple words: Movies use persistence of vision to create smooth motion from separate image frames.

Exam Tip: Persistence of vision blends consecutive still frames into continuous motion.

Q.5. Why the signal lights are red in colour?
Answer: Red light has the longest wavelength in the visible spectrum and undergoes the least scattering by fog, smoke, or dust, allowing it to be seen from long distances.
In simple words: Red light scatters the least, making it visible from afar for danger signals.

Exam Tip: Longest wavelength means minimum scattering.

Q.6. What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer: The image distance in the eye remains constant because the distance between the eye lens and the retina is fixed. Accommodation adjusts the focal length instead.
In simple words: Image distance in the eye stays fixed; the lens changes shape to focus instead.

Q.1. Why do clouds look white?
Answer: Clouds consist of large water droplets and ice particles that scatter all wavelengths of white sunlight equally (non-selective scattering), making clouds appear white.
In simple words: Large water droplets scatter all sunlight colours equally, making clouds look white.

Exam Tip: Large particles scatter all wavelengths equally, unlike tiny molecules that scatter blue.

Q.2. Why is a normal eye not able to see clearly the objects closer than 25 cm?
Answer: Because the ciliary muscles cannot increase the curvature of the eye lens beyond a certain maximum limit to focus light rays coming from objects closer than 25 cm.
In simple words: Ciliary muscles cannot strain enough to focus on objects closer than 25 cm.

Exam Tip: 25 cm is the physiological near point limit for normal human accommodation.

Q.3. What is the relation between the wavelength of light and the size of the particles causing scattering?
Answer:
- When particle size is much smaller than the wavelength, scattering is inversely proportional to the fourth power of wavelength (\(\propto 1/\lambda^4\), Rayleigh scattering).
- When particle size is much larger, all wavelengths are scattered equally.
In simple words: Smaller particles scatter shorter wavelengths, while larger particles scatter all wavelengths equally.

Exam Tip: Particle size relative to wavelength determines the type of scattering.

Q.4. When and where do we see a rainbow?
Answer: A rainbow is seen in the sky opposite to the direction of the Sun during or immediately after rain, caused by the dispersion, refraction, and internal reflection of sunlight by water droplets.
In simple words: Rainbows appear opposite to the sun when sunlight passes through raindrops in the sky.

Exam Tip: You always need the sun behind you and rain in front of you to see a rainbow.

Q.5. A person is able to see objects clearly only when objects are lying at distances between 50 cm and 300 cm from his eye. (i) What kind of defect of vision is he suffering from? (ii) What kind of lenses will be required to increase his range of vision from 25 cm to infinity? Explain briefly.
Answer:
(i) He is suffering from both Myopia (far point limited to 300 cm) and Presbyopia / Hypermetropia (near point shifted to 50 cm instead of 25 cm).
(ii) Bifocal lenses are required: concave lens in the upper portion for distance vision, and convex lens in the lower portion for near vision.
In simple words: He has both near and distant vision defects, requiring bifocal lenses.

Exam Tip: Bifocals correct both nearsightedness and farsightedness in aging eyes.

Short Answer Type Questions [3 marks]

Q.1. The alongside figure (a) shows a beam of white light falling on a glass prism gets split up into seven colours marked 1 to 7, as shown in the diagram. A student makes the following statement about the spectrum observed on the screen. (i) The colours at position marked 3 and 5 are similar to the colour of the sky and the core of a hard boiled egg respectively. Is the above statement made by the student correct or incorrect? Justify in which two positions correspond closely to the colour of – (a) A solution of potassium permagnate. (b) ‘danger’ or ‘stop’ signal lights.
Answer:
(i) The statement is correct (Position 3 is blue corresponding to the sky, Position 5 is yellow/white corresponding to egg yolk).
(a) Potassium permanganate solution: Purple / violet (Position 7).
(b) Danger or stop signal lights: Red (Position 1).
In simple words: Position 3 corresponds to blue sky, Position 5 to egg yolk, Position 7 to purple permanganate, and Position 1 to red signals.

Exam Tip: Spectrum order from top/bottom: Red, Orange, Yellow, Green, Blue, Indigo, Violet (VIBGYOR).

Q.2. A student finds the writing on the blackboard as blurred and unclear when sitting on the last desk in the classroom. He, however sees it clearly when sitting on the front desk at an approximate distance of 2 m from the blackboard. Draw ray diagrams to illustrate the formation of image of the blackboard writing by his eye lens when he is seated at the (i) last desk (ii) front desk. Name the kind of lens that would help him to see clearly even he is seated at the last desk. Draw a ray diagram to illustrate how this lens help him to see clearly?
Answer: The student is suffering from myopia (near-sightedness) because he can see close objects (front desk) but not distant objects (last desk). He needs a concave lens to correct this vision defect.
In simple words: He has myopia and needs a concave lens to see the distant blackboard clearly.

Exam Tip: Inability to see distant objects clearly indicates myopia, corrected by concave lenses.

Value Based Question

Q.1. Dolly and Ritu are two friends studying together in VI grade. They prefer to sit together. Dolly is uncomfortable reading the blackboard when they are sitting on the last bench. Dolly is depressed fearing that she may turn blind one day. Ritu explains to her that the problem may be due to some minor eye defect. Dolly accompanied by Ritu visits the Doctor who prescribed specs of suitable power for her. Dolly is now all smiles and thanks Ritu. Read the above passage and answer the following questions. (i) Name the eye defect Dolly is suffering from. (ii) What could be the cause of this defect? (iii) The far point of Dolly is 50cm. What is the power of the lens she should use to read from the black board? (iv) What values are displayed by Ritu?
Answer:
(i) Myopia (near-sightedness).
(ii) Cause: Excessive curvature of the eye lens or elongation of the eyeball.
(iii) Given far point \(d = 50\text{ cm} = 0.5\text{ m}\). Focal length \(f = -0.5\text{ m}\). Power \(P = \frac{1}{-0.5} = -2.0\text{ D}\).
(iv) Values displayed by Ritu: Empathy, scientific awareness, helpfulness, and scientific temperament.
In simple words: Dolly has myopia caused by an elongated eyeball, corrected with a -2.0D lens, and Ritu showed empathy and helpfulness.

Exam Tip: Combine optical calculations with social values in value-based case questions.

REVISION WORKSHEET : (Light, Electricity and Experiments on Light and Electricity.)

Q.1. Which of the following correctly describe OHM’s law?
i) \(V \propto I\)
ii) \(I \propto 1/R\)
iii) \(V \propto R\)
iv) None of the above.
Answer: (i) \(V \propto I\) (or \(V = IR\))
Ohm's law states that voltage is directly proportional to current at constant temperature.
In simple words: Voltage is proportional to current in Ohm's law.

Exam Tip: Ohm's law fundamental relation: \(V \propto I\) or \(V = IR\).

Q.2. What are the two quantities between which you have to draw a graph to study ohm’s law?
i) V and I
ii) I and R
iii) V and R
iv) Potential difference and current.
Answer: (i) V and I (Potential difference and current)
To verify Ohm's law experimentally, potential difference (V) is plotted against current (I).
In simple words: V and I are plotted on graph axes for Ohm's law verification.

Exam Tip: V is typically on the Y-axis and I on the X-axis.

Q.3. In the experiment of ohm’s law which apparatus /device did you use to change the voltage?
i) Voltmeter
ii) Variable Resistance box
iii) Rheostat
iv) Ammeter.
Answer: (iii) Rheostat
A rheostat is used to vary resistance and thereby adjust current and voltage in an Ohm's law circuit.
In simple words: A rheostat is used to adjust voltage and current in circuit experiments.

Exam Tip: Rheostat controls current; voltmeter measures voltage; ammeter measures current.

Q.4. For a particular conducting wire what remains constant
i) V
ii) I
iii) R
iv) All the above.
Answer: (iii) R
At a constant temperature, the resistance (R) of a specific conducting wire remains constant regardless of applied V or resulting I.
In simple words: Resistance of a specific wire stays constant at fixed temperature.

Exam Tip: Resistance is an intrinsic property of the conductor at constant temperature.

Q.5. Which of the following correctly represents ohm’s law experiment?
i) Linear V-I graph passing through origin.
Answer: (i) Linear V-I graph passing through origin.
The correct graphical representation of Ohm's law is a straight line passing through the origin (V on Y-axis, I on X-axis).
In simple words: A straight line through the origin correctly represents Ohm's law.

Exam Tip: The graph must pass through (0,0) origin.

Q.6. On what factors does resistance of a conductor depend.
i) length of conductor.
ii) Area of cross-section.
iii) All the above.
Answer: (iii) All the above. (Along with material and temperature)
Resistance depends on length, cross-sectional area, material, and temperature.
In simple words: Resistance depends on length, thickness, and material.

Exam Tip: Recall \(R = \rho l/A\) to check all dependency factors.

Q.7. Which of the following equation is/are correct?
i) \(R = \rho l/A\)
ii) \(\rho = R l/A\)
iii) \(R = \rho A/l\)
iv) \(\rho = \sqrt{R} A/l\)
Answer: (i) \(R = \rho l/A\)
Resistance formula is \(R = \rho \frac{l}{A}\).
In simple words: Resistance equals resistivity times length divided by area.

Exam Tip: Memorize the standard resistance formula.

Q.8. I have two wires of same substance, A and B, The length of A is twice the length of B. The area of A is twice the area of B. Which of the following is /are correct?
i) Resistance of A is twice that of B.
ii) Resistance of A is four times that of B.
iii) Both the resistance of A&B are same.
iv) None of the above.
Answer: (iii) Both the resistance of A & B are same.
For wire A: \(R_A = \rho \frac{2l}{2A} = \rho \frac{l}{A} = R_B\). Since length and area both double, their ratio cancels out.
In simple words: Doubling both length and area leaves the resistance unchanged.

Exam Tip: If length and area scale by the same factor, resistance remains unchanged.

Q.9. The unit (s) of Resistivity is/are:
i) \(\Omega^{-1}\text{ m}^{-1}\)
ii) Volt meter ampere-1
iii) \(\Omega\text{ m}\)
iv) All the above
Answer: (iii) \(\Omega\text{ m}\)
The SI unit of electrical resistivity is ohm metre (\(\Omega\cdot\text{m}\)).
In simple words: Resistivity is measured in ohm metres.

Exam Tip: Do not confuse ohm (resistance) with ohm-metre (resistivity).

Q.10. Which of the following is correct?
i) \(R \propto I\)
ii) \(R \propto 1/A\)
iii) \(R \propto T\)
iv) \(R \propto A\)
Answer: (ii) \(R \propto 1/A\) (and also \(R \propto l\), \(R \propto T\) for metals)
Resistance is inversely proportional to cross-sectional area.
In simple words: Resistance decreases as cross-sectional area increases.

Q.11. What is the current in the following Resistance in the given diagram? (\(24\text{ V}, 60\) \Omega \))
i) 2.5 A
ii) 0.4 A
iii) 1440 A
iv) None of above
Answer: (ii) 0.4 A
Using Ohm's law: \(I = \frac{V}{R} = \frac{24\text{ V}}{60\) \Omega} = 0.4\text{ A} \).
In simple words: Voltage divided by resistance gives 0.4 amperes.

Exam Tip: Apply \(I = V/R\) directly.

Q.12. A piece of wire is redrawn by pulling it until its length is doubled, but volume of material remains constant. If the original Resistance was \(10\) \Omega \). What is the new resistance?
i) \(100\) \Omega \)
ii) \(20\) \Omega \)
iii) \(5\) \Omega \)
iv) \(40\) \Omega \)
Answer: (i) \(100\) \Omega \)
When a wire is stretched to n times its length keeping volume constant, new resistance \(R' = n^2 R\). Here \(n = 2\), so \(R' = 2^2 \times 10 = 4 \times 10 = 40\) \Omega \) (Wait, check: if length doubles and volume is constant, area halves, so \(R' = \rho \frac{2l}{A/2} = 4 \left(\rho \frac{l}{A}\right) = 4 \times 10 = 40\ \Omega\). Option (iv) is 40 \(\Omega\)).
In simple words: Doubling the length while keeping volume constant multiplies resistance by 4, giving 40 ohms.

Exam Tip: Formula for stretched wire: \(R' = n^2 R\).

Q.13. When light rays converge or diverge while passing through a lens, which law is followed
i) Reflection
ii) Diffraction
iii) Refraction
iv) Dispersion
Answer: (iii) Refraction
Light bending through lenses occurs via refraction.
In simple words: Lenses bend light through refraction.

Exam Tip: Lenses refract light; mirrors reflect light.

Q.14. What does snell’s law represent?
i) \(\angle i = \angle r\)
ii) \(\angle i / \angle r = \text{constant}\)
iii) \(\sin r / \sin i = \text{constant}\)
iv) \(\sin i / \sin r = \text{constant}\)
Answer: (iv) \(\sin i / \sin r = \text{constant}\)
Snell's Law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant (refractive index).
In simple words: Snell's law states that \(\sin i / \sin r\) is constant.

Exam Tip: Always put \(\sin i\) in the numerator and \(\sin r\) in the denominator.

Q.15. Which of the following represents refractive index of glass-air interface?
i) \(^{\text{air}}n_{\text{glass}}\)
ii) \(^{\text{glass}}n_{\text{air}}\)
iii) \(^{\text{air}}\mu_{\text{glass}}\)
iv) \(^{\text{glass}}\mu_{\text{air}}\)
Answer: (i) \(^{\text{air}}n_{\text{glass}}\) (or relative refractive index notation)
Refractive index of glass with respect to air is denoted as \(^{\text{air}}n_{\text{glass}}\) or \(n_{ga}\).
In simple words: Refractive index of glass relative to air is written as \(^{\text{air}}n_{\text{glass}}\).

Exam Tip: The medium light enters is written second in subscript notation.

Q.16. In the following figure what are the parameters?
i) \(P = \text{aperture}, LL' = \text{focal plane}, F = \text{focus}\)
ii) \(LL' = \text{aperture}, P = \text{optical center}, PF = \text{focal length}\)
iii) \(LL' = \text{aperture}, F = \text{focus}, PF = \text{focal length}\)
iv) \(PF = \text{focal length}, P = \text{optical center}, FF' \text{ principal axis}\)

Answer: (ii) \(LL' = \text{aperture}, P = \text{optical center}, PF = \text{focal length}\) (and standard lens parameters).
In simple words: LL' represents lens aperture, P optical center, and PF focal length.

Exam Tip: Learn standard lens diagram terminology.

Q.17. Give the nature of the image in the given figure.
Answer: Virtual, Erect, and Enlarged (based on standard magnifying glass ray diagram configuration for convex lens).
In simple words: The diagram shows a virtual, upright, and magnified image.

Exam Tip: Object within focal length of a convex lens produces virtual, erect, and magnified images.

Q.18. If you wish to use a convex lens as a projector then where should you keep the object so as to get a projected image
i) At focus F
ii) Between F & 2F
iii) At 2F
iv) Between F & O
Answer: (ii) Between F & 2F
To obtain a real, inverted, and enlarged image (projector use), the object must be placed between F and \(2F\).
In simple words: Place the object between F and 2F to get an enlarged real projection image.

Exam Tip: Projectors require real, inverted, and magnified images.

Q.19. Which of the following is /are correct for a convex lens?
i) The image formed is always real.
ii) The image is always inverted.
iii) It can be used to correct short sightedness.
iv) None of above.
Answer: (iv) None of above.
Convex lenses can form virtual images (when object is within F), can form erect images, and correct far-sightedness (hypermetropia), not near-sightedness.
In simple words: None of the first three statements are universally true for convex lenses.

Exam Tip: Check all lens properties carefully before selecting "None of above".

Q.20. Which of the following is/are correct?
i) \(1/f = 1/v - 1/u\)
ii) \(f = v-u\)
iii) \(M = v/u\)
iv) \(M = h_2/h_1\)
Answer: (i) \(1/f = 1/v - 1/u\) and (iv) \(M = h_2/h_1\) (Multiple correct options; lens formula and magnification ratio).
In simple words: Lens formula and height magnification ratio are correct formulas.

Exam Tip: Lens magnification is \(+v/u\), while mirror magnification is \(-v/u\).

Q.21. Where should an object be placed, so that a real and inverted image of the same size is obtained by a convex lens?
i) At the focus
ii) At twice the focal length
iii) At infinity
iv) Between optical center of lens & focus
Answer: (ii) At twice the focal length (\(2F\))
When an object is placed at \(2F\) in front of a convex lens, a real, inverted image of exactly the same size is formed at \(2F\) on the other side.
In simple words: Placing the object at 2F produces an equal-sized real image at 2F.

Exam Tip: At \(u = 2f\), magnification magnitude \(|m| = 1\).

Q.22. An object is placed at a distance of 50 cm from a concave lens of focal length 20cm. What is the nature of the image?
i) Real, inverted
ii) Virtual, erect
iii) Real, erect
iv) Virtual, inverted
Answer: (ii) Virtual, erect
A concave lens always forms a virtual, erect, and diminished image for any real object position.
In simple words: Concave lenses always form virtual and erect images.

Exam Tip: Concave lens images are always virtual and erect.

Q.23. Suppose you have three resistances of \(10\) \Omega \) each. You need a final resistance of \(5\) \Omega \). How can you do so?
Answer: Connect two \(10\) \Omega \) resistors in parallel (giving equivalent \(5\) \Omega \)), and disconnect or leave out the third resistor.
In simple words: Connect two 10-ohm resistors in parallel to get 5 ohms.

Exam Tip: Two identical resistors in parallel halve their individual resistance.

Q.24. You want to change the resistance of a wire without changing its material, length or area of cross-section. Then what should you do?
Answer: Change the temperature of the wire, because resistance depends on temperature (\(R = R_0(1 + \alpha\Delta T)\)).
In simple words: Change the temperature of the wire to alter its resistance.

Q.25. Why do electricians wear rubber gloves or shoes while working with electricity?
Answer: Rubber is an excellent electrical insulator. Wearing rubber gloves or shoes prevents electric current from passing through the electrician's body to the ground, protecting them from shocks.
In simple words: Rubber insulates against electricity, preventing dangerous shocks.

Exam Tip: Insulators like rubber and plastic prevent closed-circuit pathways through the body.

Q.26. Give the final resistance of the given figure. (\(R = 1\) \Omega \) network)
Answer: Evaluating the standard bridge or combination network shown in the figure yields the equivalent resistance (typically matching balanced bridge or simple series-parallel reduction, e.g., \(1\) \Omega \)).
In simple words: Use series and parallel formulas to calculate the total network resistance.

Exam Tip: Break complex resistor networks into simpler series and parallel blocks.

Q.27. In the given figure what is the potential difference of the battery?
Answer: Using the given circuit parameters (\(I = 0.5\text{ A}\) and given resistors like \(7.2\) \Omega \) and \(12\) \Omega \)):
Calculate equivalent resistance and apply \(V = I \cdot R_{eq}\) to find the battery voltage.
In simple words: Multiply total circuit resistance by current to find battery voltage.

Exam Tip: Find total resistance first before applying Ohm's law for total voltage.

Q.28 Which of the following is true?
i) Current distribution is equal in all resistances of a parallel circuit.
ii) Potential difference across the two terminals of a parallel circuit remains same for all resistances.
iii) Both the above.
iv) None of the above.
Answer: (ii) Potential difference across the two terminals of a parallel circuit remains same for all resistances.
In a parallel circuit, potential difference is identical across all parallel branches, while current divides inversely according to resistance.
In simple words: Voltage is the same across all branches in a parallel circuit.

Exam Tip: Same voltage in parallel; same current in series.

Q.29. In the discovery of superconductivity by H.K. Onnes. Which metal was used?
i) Iron
ii) Copper
iii) Silver
iv) Mercury
v) Gallium
Answer: (iv) Mercury
Heike Kamerlingh Onnes discovered superconductivity in solid mercury when cooled below 4.2 K.
In simple words: Mercury was the first metal in which superconductivity was discovered.

Exam Tip: Remember mercury and liquid helium temperatures for Onnes' discovery.

Q.30 A copper wire has diameter of 0.5 mm and resistivity of \(1.6 \times 10^{-6}\) \Omega\cdot\text{m} \). How much wire is required to make a \(10\) \Omega \) coil.
Answer:
Diameter \(d = 0.5\text{ mm} = 5 \times 10^{-4}\text{ m}\), radius \(r = 2.5 \times 10^{-4}\text{ m}\).
Area \(A = \pi r^2 = \pi (2.5 \times 10^{-4})^2 = 6.25\pi \times 10^{-8}\text{ m}^2\).
Resistance \(R = 10\) \Omega \), Resistivity \(\rho = 1.6 \times 10^{-6}\) \Omega\cdot\text{m} \).
Using \(R = \rho \frac{l}{A} \implies l = \frac{R \cdot A}{\rho} = \frac{10 \times (6.25\pi \times 10^{-8})}{1.6 \times 10^{-6}} \approx 1.23\text{ meters}\).
In simple words: Using the resistance formula and cross-sectional area, the required wire length is calculated.

Exam Tip: Convert all units to SI (meters) before calculating.

Q.31. We know that white light is made up of VIBGYOR colours. But when we combine all the VIBGYOR colours using sketch pens etc, we get black colour. Why?
Answer: Sketch pens use pigments/inks that act as subtractive colour mixers. Combining pigment colours absorbs all wavelengths of light and reflects none, resulting in black.
In simple words: Mixing pigment inks absorbs all light, appearing black, unlike mixing light colours which makes white.

Exam Tip: Distinguish between additive mixing (lights) and subtractive mixing (pigments/inks).

Q.32. When can you have Total Internal Reflection
i) Light passing from glass to air \(\angle i = 90^\circ\)
ii) Light passing from air to glass \(\angle i = 90^\circ\)
iii) Light passing from glass to air \(\angle r = 90^\circ\)
iv) Light passing from air to glass \(\angle r = 90^\circ\)
Answer: (iii) Light passing from glass to air \(\angle r = 90^\circ\) (when angle of incidence exceeds critical angle, resulting in angle of refraction / reflection conditions).
Total internal reflection occurs when light travels from a denser to a rarer medium and the angle of incidence exceeds the critical angle (where \(\angle r = 90^\circ\)).
In simple words: Total internal reflection happens when light goes from dense to rare medium at angles greater than the critical angle.

Exam Tip: Conditions: denser to rarer medium, and \(i > i_c\).

Q.33. If I have two lenses of powers \(P_1 = -3\text{ D}\) and \(P_2 = +5\text{ D}\). What should be the power of a third lens so that the combination acts like a plane piece of glass?
Answer:
For the combination to act like a plane piece of glass, net power \(P_{\text{net}} = 0\).
\(P_{\text{net}} = P_1 + P_2 + P_3 = 0\)
\(-3 + 5 + P_3 = 0 \implies 2 + P_3 = 0 \implies P_3 = -2\text{ D}\).
In simple words: The third lens must have a power of -2D so the combined power totals zero.

Exam Tip: Powers of lenses in contact add algebraically (\(P = P_1 + P_2 + P_3\)).

Q.34. My spectacles have a lens of power – 5.0D. What kind of lens do I have? What is my eyes suffering from?
i) Convex, Myopia.
ii) Concave, Hypermetropia.
iii) Convex, Hypermetropia.
iv) Concave, Myopia.
Answer: (iv) Concave, Myopia.
Negative power indicates a concave lens, which is prescribed to correct myopia (near-sightedness).
In simple words: Negative power means a concave lens for correcting myopia.

Exam Tip: Negative power = concave lens = myopia correction.

Q.35. Lateral displacement through a glass slab increases with
i) thickness of medium.
ii) angle of incidence.
iii) increase of refractive index.
iv) all the above.
Answer: (iv) all the above.
Lateral displacement is directly proportional to thickness, angle of incidence, and refractive index of the slab.
In simple words: Thicker slabs, higher angles, and greater refractive index all increase lateral displacement.

Exam Tip: All three factors increase sideways ray shifting.

Q.36. Light of wavelength 500 nm in air, enters a glass plate of \(\mu = 1.5\). What is the speed, wavelength and frequency of light in the glass. Assume frequency to be constant in both media.
Answer:
Speed in glass \(v = \frac{c}{\mu} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8\text{ m/s}\).
Frequency \(\nu = \frac{c}{\lambda_{\text{air}}} = \frac{3 \times 10^8}{500 \times 10^{-9}} = 6 \times 10^{14}\text{ Hz}\).
Wavelength in glass \(\lambda_{\text{glass}} = \frac{\lambda_{\text{air}}}{\mu} = \frac{500\text{ nm}}{1.5} = 333.33\text{ nm}\).
In simple words: Speed is \(2 \times 10^8\text{ m/s}\), frequency is \(6 \times 10^{14}\text{ Hz}\), and wavelength is 333.33 nm.

Exam Tip: Frequency remains unchanged when light enters a new medium.

Q.37. What is common among Infra Red, Ultraviolet and Visible light?
i) They all have same velocity.
ii) They al are electromagnetic in nature.
iii) They all have same wavelength.
iv) They all have same frequency.
Answer: (ii) They all are electromagnetic in nature. (They also travel at the same speed in a vacuum).
Infrared, ultraviolet, and visible light are all part of the electromagnetic spectrum.
In simple words: All of them are electromagnetic waves travelling at the speed of light in vacuum.

Exam Tip: All EM waves travel at \(c = 3 \times 10^8\text{ m/s}\) in a vacuum.

Q.38. The working of an optical fibre is due to
i) Refraction.
ii) Reflection
iii) Total Internal Reflection
iv) None of the above.
Answer: (iii) Total Internal Reflection
Optical fibers transmit light signals over long distances via continuous total internal reflection along the core.
In simple words: Optical fibres work on the principle of total internal reflection.

Exam Tip: Total internal reflection prevents light leakage in fiber optic cables.

Q.39. How does current flow through a metal?
i) Due to flow of electrons.
ii) Due to flow of negative ions.
iii) Due to flow of protons.
iv) Due to flow of positive ions.
Answer: (i) Due to flow of electrons.
Electric current in metals is carried by the drift of free electrons.
In simple words: Free electrons moving through metals carry electric current.

Exam Tip: Electrons carry current in solids, while ions carry current in liquids/electrolytes.

Q.40. Which of the following is /are not correct?
i) \(V = IR\)
ii) \(P = I^2R\)
iii) \(P = V^2/R\)
iv) \(W = Irt\)
Answer: (iv) \(W = Irt\) (Correct formula for work / heat is \(H = I^2Rt\), so \(Irt\) is incorrect).
In simple words: Formula (iv) is incorrect because work/heat requires \(I^2Rt\), not \(Irt\).

Exam Tip: Verify electrical power and energy formulas carefully (\(H = I^2Rt\)).

Q.41. 1 kilowatt hour is equal to
i) \(3.6 \times 10^6\text{ Joules/second}\)
ii) \(3.6 \times 10^6\text{ Joules}\)
iii) \(3.6 \times 10^6\text{ watt}\)
iv) \(3.6 \times 10^6\text{ watt second}\)
Answer: (ii) \(3.6 \times 10^6\text{ Joules}\) (and also equivalent to watt-second).
\(1\text{ kWh} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ J}\).
In simple words: One kilowatt-hour equals 3.6 million joules of energy.

Exam Tip: Memorize this conversion factor for commercial electrical energy calculations.

Q.42. A current of 4 A flows through a 12 V car headlight for 10 minutes. How much energy transfer takes place?
Answer:
Current \(I = 4\text{ A}\), Voltage \(V = 12\text{ V}\), Time \(t = 10\text{ min} = 600\text{ seconds}\).
Energy transfer \(W = V \cdot I \cdot t = 12\text{ V} \times 4\text{ A} \times 600\text{ s} = 28,800\text{ J} = 28.8\text{ kJ}\).
In simple words: Energy transfer is voltage multiplied by current and time, giving 28,800 joules.

Exam Tip: Use \(E = VIt\) or \(E = Pt\) to calculate electrical energy transfer.

Q.43. Two lamps, one rated 100W, 220 V and the other is 60 W, 220 V are connected in parallel to 220-V supply. What current is drawn from supply?
Answer:
Total power \(P_{\text{total}} = P_1 + P_2 = 100\text{ W} + 60\text{ W} = 160\text{ W}\).
Voltage \(V = 220\text{ V}\).
Total current \(I = \frac{P_{\text{total}}}{V} = \frac{160\text{ W}}{220\text{ V}} = \frac{8}{11}\text{ A} \approx 0.73\text{ A}\).
In simple words: Add both powers to get total power, then divide by voltage to find total current.

Exam Tip: In parallel circuits, total power is the sum of individual appliance powers.

Q.44. When a particle carrying a charge of \(10^{-5}\) coulomb is brought from infinity to a point P, \(2 \times 10^3\) Joules of work is done. What is the potential at the point P?
Answer:
Work \(W = 2 \times 10^3\text{ J}\), Charge \(Q = 10^{-5}\text{ C}\).
Potential \(V = \frac{W}{Q} = \frac{2 \times 10^3}{10^{-5}} = 2 \times 10^8\text{ V}\).
In simple words: Potential is work done divided by charge, yielding \(2 \times 10^8\text{ volts}\).

Exam Tip: Apply \(V = W/Q\) directly with correct exponents.

Q.45. Which of the following gives relation between Power and Focal length?
i) \(P = F^2\)
ii) \(P = F\)
iii) \(P = 1/F\)
iv) \(P = 1/F^2\)
Answer: (iii) \(P = 1/F\) (where F is focal length in meters).
Power of a lens is defined as the reciprocal of its focal length in meters (\(P = 1/f\)).
In simple words: Lens power is the reciprocal of focal length.

Exam Tip: Focal length must be in meters for power in dioptres.

Q.46. An object 4 cm high is placed at a distance of 27 cm in front of a convex lens of focal length 18 cm. Find the position and nature of the image formed.
Answer:
Given: \(h_1 = 4\text{ cm}\), \(u = -27\text{ cm}\), \(f = +18\text{ cm}\).
Using lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
\(\frac{1}{18} = \frac{1}{v} - \frac{1}{-27}\)
\(\frac{1}{v} = \frac{1}{18} - \frac{1}{27} = \frac{3 - 2}{54} = \frac{1}{54} \implies v = +54\text{ cm}\).
Magnification \(m = \frac{v}{u} = \frac{54}{-27} = -2\).
Nature: Real, inverted, and magnified (\(8\text{ cm}\) high).
In simple words: Image forms at +54 cm, real, inverted, and enlarged.

Exam Tip: Double-check signs when transposing terms in the lens formula.

Q.47. Why do diamonds sparkle?
Answer: Diamonds sparkle due to total internal reflection. A diamond has a very high refractive index (2.42), making its critical angle very small (\(24.4^\circ\)). Light entering the cut facets undergoes multiple total internal reflections before emerging out of specific faces.
In simple words: High refractive index and multiple total internal reflections inside cut facets make diamonds sparkle.

Exam Tip: Mention high refractive index and small critical angle as key factors.

Q.48. Why do stars twinkle?
Answer: Stars twinkle because of atmospheric refraction. Starlight passes through shifting atmospheric layers of varying densities and temperatures, causing the light rays to bend randomly and rapidly change apparent brightness and position.
In simple words: Atmospheric refraction bends starlight irregularly, making stars appear to twinkle.

Exam Tip: Point-sized stars twinkle; extended source planets do not.

Q.49. Five dry cells each of 1.5 V have internal resistance of 0.2, 0.3, 0.4, 0.5 and \(1.2\) \Omega \). When connected in series, what current will the five cells furnish through \(10\) \Omega \) resistance?
Answer:
Total EMF \(E_{\text{total}} = 5 \times 1.5\text{ V} = 7.5\text{ V}\).
Total internal resistance \(r_{\text{total}} = 0.2 + 0.3 + 0.4 + 0.5 + 1.2 = 2.6\) \Omega \).
External resistance \(R = 10\) \Omega \).
Total resistance \(R_{\text{total}} = R + r_{\text{total}} = 10 + 2.6 = 12.6\) \Omega \).
Current \(I = \frac{E_{\text{total}}}{R_{\text{total}}} = \frac{7.5}{12.6} \approx 0.6\text{ A}\).
In simple words: Total voltage divided by total series resistance (including internal resistances) gives 0.6 amperes.

Exam Tip: Add internal resistances directly when cells are connected in series.

Q.50 In q. no. 49 if the cells are connected in parallel a long with the \(10\) \Omega \) resistance, then what will be the current?
Answer: When identical cells are connected in parallel, the total voltage remains equal to that of a single cell (\(V = 1.5\text{ V}\)).
Current through the \(10\) \Omega \) resistor \(I = \frac{1.5\text{ V}}{10\) \Omega} = 0.15\text{ A} \) (ignoring small combined internal resistance in parallel).
In simple words: Parallel cells provide the voltage of one cell (1.5V), driving a current of 0.15 A through the 10 ohm resistor.

Exam Tip: Parallel cells maintain voltage equal to a single cell while increasing current capacity.

ASSIGNMENTS : Module : 1 & 2

Q.1. What happens when (give equation also).
(i) Potassium iodide solution is added to lead nitrate solution.
(ii) Dilute hydrochloric acid is added to zinc granules.
(iii) Water is added to quick lime.
(iv) Carbon-di-oxide gas passed through lime water.
Answer:
(i) A yellow precipitate of lead iodide is formed.
\(\text{Pb(NO}_3)_2(\text{aq}) + 2\text{KI}(\text{aq}) \longrightarrow \text{PbI}_2(\text{s})\ (\text{yellow ppt}) + 2\text{KNO}_3(\text{aq})\)
(ii) Effervescence of hydrogen gas is observed along with formation of zinc chloride.
\(\text{Zn}(\text{s}) + 2\text{HCl}(\text{aq}) \longrightarrow \text{ZnCl}_2(\text{aq}) + \text{H}_2(\text{g})\)
(iii) Slaked lime is formed with a vigorous release of heat (exothermic reaction).
\(\text{CaO}(\text{s}) + \text{H}_2\text{O}(\text{l}) \longrightarrow \text{Ca(OH)}_2(\text{aq}) + \text{Heat}\)
(iv) Lime water turns milky due to formation of insoluble calcium carbonate precipitate.
\(\text{Ca(OH)}_2(\text{aq}) + \text{CO}_2(\text{g}) \longrightarrow \text{CaCO}_3(\text{s})\ (\text{milky}) + \text{H}_2\text{O}(\text{l})\)
In simple words: These reactions produce yellow precipitates, hydrogen gas, heat, and milky calcium carbonate.

Exam Tip: Balance all chemical equations and state the colour of precipitates formed.

Q.2. (i) Give one chemical reaction characterised by the change in temperature. (ii) Give one chemical reaction characterised by the formation of precipitate.
Answer:
(i) Action of dilute hydrochloric acid on zinc granules (exothermic, temperature rises):
\(\text{Zn} + 2\text{HCl} \longrightarrow \text{ZnCl}_2 + \text{H}_2\)
(ii) Reaction of barium chloride with sodium sulphate (forms white barium sulphate precipitate):
\(\text{BaCl}_2 + \text{Na}_2\text{SO}_4 \longrightarrow \text{BaSO}_4(\text{s}) + 2\text{NaCl}\)
In simple words: Zinc with acid produces heat, while barium chloride with sodium sulphate forms a white precipitate.

Exam Tip: Temperature change and precipitation are primary characteristics of chemical reactions.

Q.3. Balance the following equations
(i) \(\text{N}_2 + \text{H}_2 \longrightarrow \text{NH}_3\)
(ii) \(\text{Al(OH)}_3 + \text{H}_2\text{SO}_4 \longrightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O}\)
(iii) \(\text{H}_2\text{O}_2 \longrightarrow \text{H}_2\text{O} + \text{O}_2\)
(iv) \(\text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2 \longrightarrow \text{CO}_2 + \text{H}_2\text{O}\)
(v) \(\text{Pb(NO}_3)_2 \longrightarrow \text{PbO} + \text{NO}_2 + \text{O}_2\)

Answer:
(i) \(\text{N}_2 + 3\text{H}_2 \longrightarrow 2\text{NH}_3\)
(ii) \(2\text{Al(OH)}_3 + 3\text{H}_2\text{SO}_4 \longrightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O}\)
(iii) \(2\text{H}_2\text{O}_2 \longrightarrow 2\text{H}_2\text{O} + \text{O}_2\)
(iv) \(\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \longrightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}\)
(v) \(2\text{Pb(NO}_3)_2 \longrightarrow 2\text{PbO} + 4\text{NO}_2 + \text{O}_2\)
In simple words: Balanced chemical equations ensure equal numbers of each atom on both sides.

Exam Tip: Balance complex molecules like nitrates and sulphates first.

Q.4. Convey the following in the form of a balanced chemical equation. “An aqueous solution of ferrous sulphate reacts with an aqueous of sodium hydroxide to form a precipitate of ferrous hydroxide and sodium sulphate remains in solution.”
Answer:
\(\text{FeSO}_4(\text{aq}) + 2\text{NaOH}(\text{aq}) \longrightarrow \text{Fe(OH)}_2(\text{s})\ (\text{dirty green ppt}) + \text{Na}_2\text{SO}_4(\text{aq})\)
In simple words: Ferrous sulphate reacts with sodium hydroxide to form a dirty green ferrous hydroxide precipitate.

Exam Tip: Note the characteristic dirty green colour of ferrous hydroxide precipitate.

Q.5. What information does the following chemical equation convey?
(i) \(\text{CO}(\text{g}) + 2\text{H}_2(\text{g}) \xrightarrow[ZnO + CrO_3]{340^\circ\text{C}} \text{CH}_3\text{OH}(\text{l})\)
(ii) \(\text{AgNO}_3(\text{aq}) + \text{NaCl}(\text{aq}) \longrightarrow \text{AgCl}(\text{s}) + \text{NaNO}_3(\text{aq})\)

Answer:
(i) Carbon monoxide reacts with hydrogen gas at \(340^\circ\text{C}\) in the presence of a ZnO and \(\text{CrO}_3\) catalyst to form liquid methanol.
(ii) Aqueous silver nitrate reacts with aqueous sodium chloride to form a white precipitate of silver chloride and aqueous sodium nitrate.
In simple words: These equations specify reactants, products, physical states, catalysts, and reaction conditions.

Exam Tip: Mention temperature, pressure, catalyst, and physical states when interpreting equations.

Q.6. Which of the following reactions are endothermic and which are exothermic.
(i) Burning of natural gas
(ii) Electrolysis of water
(iii) Photosynthesis
(iv) Respiration
(v) Decomposition of calcium carbonate
Answer:
- Exothermic: (i) Burning of natural gas, (iv) Respiration (release energy/heat).
- Endothermic: (ii) Electrolysis of water, (iii) Photosynthesis, (v) Decomposition of calcium carbonate (absorb energy/heat).
In simple words: Combustion and respiration release heat (exothermic), while electrolysis, photosynthesis, and decomposition absorb energy (endothermic).

Exam Tip: Exothermic reactions release heat; endothermic reactions require energy input.

Module : 3

Q.1. What happens when AgCl is exposed to sunlight? Write a balanced chemical equation for this reaction. Also give one use of such reaction.
Answer: White silver chloride turns grey in sunlight due to photochemical decomposition into silver metal and chlorine gas.
\(2\text{AgCl}(\text{s}) \xrightarrow{\text{Sunlight}} 2\text{Ag}(\text{s}) + \text{Cl}_2(\text{g})\)
Use: Used in black-and-white photography.
In simple words: Sunlight decomposes silver chloride into silver and chlorine, which is used in photography.

Exam Tip: Photochemical decomposition reactions require light energy.

Q.2. Crystals of copper sulphate are heated in a test tube for some time. What is the colour of copper sulphate - (i) before heating and (ii) after heating? What is the source of liquid droplets seen on the upper side of the test tube during the heating process?
Answer:
(i) Before heating: Blue (hydrated copper sulphate, \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\)).
(ii) After heating: White / greyish-white (anhydrous copper sulphate, \(\text{CuSO}_4\)).
Source of droplets: Water of crystallization driven off from the hydrated crystals during heating.
In simple words: Blue crystals turn white on heating as water of crystallization evaporates into droplets.

Q.3. When a green iron salt is heated strongly its colour finally changes to black and odour of burning sulphur is given out. (i) Name the iron salt. (ii) Name the type of reaction. (iii) Write the chemical equation for the reaction involved.
Answer:
(i) Ferrous sulphate heptahydrate / crystals (\(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\)).
(ii) Thermal decomposition reaction.
(iii) \(2\text{FeSO}_4(\text{s}) \xrightarrow{\Delta} \text{Fe}_2\text{O}_3(\text{s})\ (\text{black/red-brown}) + \text{SO}_2(\text{g}) + \text{SO}_3(\text{g})\)
In simple words: Heating green ferrous sulphate causes thermal decomposition into ferric oxide, sulphur dioxide, and sulphur trioxide.

Exam Tip: Smell of burning sulphur is due to \(\text{SO}_2\) and \(\text{SO}_3\) gases.

Q.4. A colourless salt, when heated produce a yellow residue and brown fumes. (i) Name the lead salt. (ii) Name the brown fumes and give its formula. (iii) Write a chemical equation for the reaction involved.
Answer:
(i) Lead nitrate (\(\text{Pb(NO}_3)_2\)).
(ii) Nitrogen dioxide gas (\(\text{NO}_2\)).
(iii) \(2\text{Pb(NO}_3)_2(\text{s}) \xrightarrow{\Delta} 2\text{PbO}(\text{s})\ (\text{yellow}) + 4\text{NO}_2(\text{g})\ (\text{brown fumes}) + \text{O}_2(\text{g})\)
In simple words: Heating lead nitrate produces yellow lead monoxide and brown nitrogen dioxide fumes.

Exam Tip: Brown fumes of \(\text{NO}_2\) are characteristic of nitrate thermal decomposition.

Q.5. Name the substance oxidized and substance reduced in the following reactions.
(i) \(\text{MnO}_2 + 4\text{HCl} \longrightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}\)
(ii) \(\text{SO}_2 + 2\text{H}_2\text{S} \longrightarrow 2\text{H}_2\text{O} + 3\text{S}\)
(iii) \(\text{PbS} + 4\text{H}_2\text{O}_2 \longrightarrow \text{PbSO}_4 + 4\text{H}_2\text{O}\)

Answer:
(i) Oxidized: HCl (loses hydrogen / gets chlorinated to \(\text{Cl}_2\)); Reduced: \(\text{MnO}_2\) (loses oxygen to form \(\text{MnCl}_2\)).
(ii) Oxidized: \(\text{H}_2\text{S}\) (loses hydrogen to form S); Reduced: \(\text{SO}_2\) (loses oxygen to form S).
(iii) Oxidized: PbS (gains oxygen to form \(\text{PbSO}_4\)); Reduced: \(\text{H}_2\text{O}_2\) (loses oxygen / gets reduced to \(\text{H}_2\text{O}\)).
In simple words: Substances losing hydrogen or gaining oxygen are oxidized, while those gaining hydrogen or losing oxygen are reduced.

Exam Tip: Oxidation = gain of oxygen / loss of hydrogen; Reduction = loss of oxygen / gain of hydrogen.

Q.6. Name two antioxidants which are usually added to food containing fat and oil to prevent rancidity.
Answer: BHA (Butylated Hydroxyanisole) and BHT (Butylated Hydroxytoluene).
In simple words: BHA and BHT are commonly added antioxidants that prevent food oils from turning rancid.

Exam Tip: Antioxidants slow down oxidation reactions in fatty foods.

H.O.T.S. Questions

Q.1. \(\text{H}_2\text{S} + \text{Cl}_2 \longrightarrow \text{S} + 2\text{HCl}\)
(i) Name the species undergoing oxidation.
(ii) Name the species undergoing reduction.
Answer:
(i) Oxidized species: \(\text{H}_2\text{S}\) (loses hydrogen to form sulfur).
(ii) Reduced species: \(\text{Cl}_2\) (gains hydrogen to form HCl).
In simple words: \(\text{H}_2\text{S}\) is oxidized and \(\text{Cl}_2\) is reduced.

Exam Tip: Removal of hydrogen is oxidation; addition of hydrogen is reduction.

Q.2. Chloride of a metal (X) (which is used to make coins) when exposed to sunlight, turns grey from white. Name the type of reaction and identify X.
Answer:
Metal X: Silver (Ag).
Chloride: Silver chloride (AgCl).
Type of reaction: Photochemical decomposition reaction.
In simple words: Metal X is silver, and the reaction is photochemical decomposition.

Exam Tip: Silver is widely used in making coins and photographic silver salts.

Q.3. Sodium bromide is added to an aqueous solution of a metal nitrate ‘A’. A yellow precipitate of ‘B’ is obtained which is used in photography. Compound ‘B’ is photosensitive and decomposes to its constituents, accompanied by the evolution of a reddish brown gas. Write down the balanced equation and identify A and B.
Answer:
Metal nitrate A: Silver nitrate (\(\text{AgNO}_3\)).
Yellow precipitate B: Silver bromide (\(\text{AgBr}\)).
Equation: \(\text{AgNO}_3(\text{aq}) + \text{NaBr}(\text{aq}) \longrightarrow \text{AgBr}(\text{s})\ (\text{yellow ppt}) + \text{NaNO}_3(\text{aq})\).
In simple words: A is silver nitrate and B is silver bromide, which forms a yellow precipitate used in photography.

Exam Tip: Silver bromide is photosensitive and decomposes in light.

Q.4. Metal ‘A’ is found in the earth crust and on exposure to moist air, it form a reddish brown flaky substance. When a container made up of metal ‘A’ is used to store a blue coloured solution of ‘B’, the blue colour changes to pale green and reddish brown metal ‘C’ is formed. Identify A, B, C and write a balanced equation for the reaction. Also in name the type of reaction.
Answer:
Metal A: Iron (Fe).
Solution B: Copper sulphate (\(\text{CuSO}_4\)).
Metal C: Copper (Cu).
Equation: \(\text{Fe}(\text{s}) + \text{CuSO}_4(\text{aq}) \longrightarrow \text{FeSO}_4(\text{aq})\ (\text{pale green}) + \text{Cu}(\text{s})\ (\text{reddish brown})\).
Type of reaction: Displacement reaction.
In simple words: Iron displaces copper from copper sulphate in a single displacement reaction.

Exam Tip: More reactive metal displaces less reactive metal from its salt solution.

Value Based Question