CBSE Class 10 Physics Question Bank

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
Electric current always flows from a region of higher potential to a region of lower potential. Here, battery B is at +20 V (higher potential) and battery A is at -30 V (lower potential). Therefore, current flows from B to A.
In simple words: Current moves from higher voltage (+20 V) to lower voltage (-30 V).

Exam Tip: Always remember conventional current direction is from positive potential to negative potential.

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: Work done per unit charge represents potential difference.

Exam Tip: Memorize the formula \(V = \frac{W}{Q}\) along with SI units (Volt for potential difference, Joule for work, Coulomb for 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) \(6.25 \times 10^{18}\)
Total charge \(Q = n \cdot e\)
\(1 = n \times 1.6 \times 10^{-19}\)
\(n = \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18}\) electrons.
In simple words: One coulomb contains \(6.25 \times 10^{18}\) electrons.

Exam Tip: This is a standard constant value frequently asked in numerical problems. Learn it directly.

Q.4. The unit of specific resistance is
(i) Ohm metre
(ii) Ohm
(iii) Volt
(iv) Ampere
Answer: (i) Ohm metre
Resistance formula is \(R = \rho \frac{l}{A}\), where \(\rho\) is specific resistance (resistivity).
Rearranging for resistivity: \(\rho = \frac{R \cdot A}{l} = \frac{\Omega \cdot \text{m}^2}{\text{m}} = \Omega \cdot \text{m}\) (Ohm metre).
In simple words: SI unit of resistivity is Ohm metre.

Exam Tip: Do not confuse resistance (Ohm) with resistivity or specific resistance (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 vast sea of delocalized valence electrons that are free to move randomly and drift under an electric field, constituting an electric current.
In simple words: Free electrons carry electric charge through metals.

Exam Tip: Free electrons are responsible for both electrical and thermal conductivity in metals.

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 a characteristic property of the material of the conductor and depends only on the nature of the material and temperature, independent of its dimensions.
In simple words: Cutting a wire changes its resistance, but its material property (resistivity) stays constant.

Exam Tip: Resistivity depends only on material and temperature, whereas resistance depends on dimensions (\(l\) and \(A\)).

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
Potential difference \(V = \frac{W}{Q}\). When 1 Joule of work is done in moving 1 Coulomb of charge, the potential difference is 1 Volt (\(1\text{ V} = 1\text{ J/C}\)).
In simple words: One volt equals one joule of work per coulomb of charge.

Exam Tip: Dimensional analysis of formulas helps instantly verify definitions in exams.

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, multiple resistance coils are connected in series so that pulling out plugs adds up their individual resistance values.
In simple words: Resistance box uses series combination to sum up resistances.

Exam Tip: Series combination allows cumulative resistance addition: \(R_s = R_1 + R_2 + \dots\)

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 constant temperature, yielding a linear \(V-I\) graph.
In simple words: Ohm's law gives a straight-line graph for ohmic conductors.

Exam Tip: The slope of the \(V-I\) graph gives the resistance (\(R = \frac{\Delta V}{\Delta I}\)).

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 regulate current flow in an electrical circuit.
In simple words: A rheostat varies resistance continuously to control current.

Exam Tip: Know the difference between a rheostat (variable resistance) and a resistance box (step-wise known resistance).

Section : B

Q.1. What do these symbols denote
(i) Cell
(ii) Battery or combination of cells
(iii) Resistor or resistance
(iv) Variable resistor (Rheostat)
(v) Ohm (Unit of resistance)
(vi) Closed plug key / Switch closed
(vii) Wire joint
(viii) Ammeter
Answer:
(i) Electric Cell
(ii) Battery
(iii) Resistor
(iv) Rheostat / Variable Resistor
(v) Ohm (\(\Omega\))
(vi) Closed Switch / Plug Key
(vii) Wire Joint
(viii) Ammeter
In simple words: Standard schematic symbols used in electric circuit diagrams.

Exam Tip: Practice drawing all standard electrical circuit symbols clearly as they carry direct marks in board exams.

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

Exam Tip: Combine factors to write \(R = \rho \frac{l}{A}\) for complete clarity.

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 its ends, creating an electric field that drives free electrons.
In simple words: A potential difference is essential for charges to flow.

Exam Tip: Charge flows from high potential to low potential, just like water flows down a slope.

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 diametrical ends, the wire is divided into two equal semi-circular halves connected in parallel.
Resistance of each half = \(\frac{4R}{2} = 2R\).
Equivalent resistance \(R_{eq}\) in parallel:
\(\frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{2R} = \frac{2}{2R} = \frac{1}{R}\)
\(R_{eq} = R\)
In simple words: Dividing the wire into two equal parallel halves gives a total resistance of \(R\).

Exam Tip: For any uniform wire of resistance \(R\) bent into a circle, resistance across any diameter is always \(\frac{R}{4}\). Here initial resistance is \(4R\), so \(\frac{4R}{4} = R\).

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 only valid for metallic conductors under constant physical conditions (like temperature and pressure).
Example: Semiconductors (like diodes, transistors) and electrolytes do not obey Ohm's law and are called non-ohmic conductors.
In simple words: Ohm's law fails in devices where current is not strictly proportional to voltage.

Exam Tip: Mention non-ohmic devices like vacuum tubes or semiconductor diodes as examples.

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

Exam Tip: Remember: positive charges move from high to low potential, while negative charges (electrons) move 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: 1 Joule of work is done in bringing a unit positive charge (1 Coulomb) from infinity to that point against the electric field.
(ii) Potential difference between two points is 1 Volt: 1 Joule of work is done in moving a unit positive charge (1 Coulomb) from one point to the other.
In simple words: 1 Volt means 1 Joule of work per 1 Coulomb of charge transferred.

Exam Tip: Always include the terms 'work done', 'charge', and '1 Joule / 1 Coulomb' in definitions.

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: Using a battery or cell maintains a steady voltage across a conductor.

Exam Tip: Cells and batteries 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, free electrons in a metal move randomly in all directions due to thermal agitation. Their average thermal velocity in any direction is zero, meaning there is no net flow of charge in any particular direction, hence no electric current.
In simple words: Random thermal motion cancels out, resulting in zero net current without an electric field.

Exam Tip: Emphasize that net flow of charge is zero due to random thermal motion.

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 \(V-I\) graph is plotted with voltage \(V\) on the Y-axis and current \(I\) on the X-axis.
(ii) The graph is a straight line passing through the origin, which verifies Ohm's law (\(V \propto I\)).
(iii) Resistance \(R\) is given by the slope of the \(V-I\) graph:
\(R = \frac{\Delta V}{\Delta I} = \frac{10 - 2}{3.70 - 0.74} = \frac{8}{2.96} \approx 2.70 \, \Omega\)
In simple words: The straight line proves Ohm's law, and its slope gives the resistance value.

Exam Tip: When calculating resistance from a graph, choose points far apart to minimize reading errors.

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: \(V = IR\) or \(I = \frac{V}{R}\).
Given: \(V = 1.4\text{ V}\), \(I = 0.35\text{ A}\).
Resistance \(R = \frac{V}{I} = \frac{1.4}{0.35} = \frac{140}{35} = 4 \, \Omega\).
In simple words: Resistance is voltage divided by current, which equals \(4 \, \Omega\).

Exam Tip: Always write the formula first, substitute values with units, and state the final unit clearly.

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 cross-sectional area (\(R \propto \frac{1}{A}\)). A thicker wire has a larger area, providing more paths for electrons, hence lower resistance.
(ii) Alloys have a much higher resistivity than pure metals. Since \(R = \rho \frac{l}{A}\), a higher resistivity requires a shorter length to achieve the same resistance.
(iii) Tungsten has a very high melting point (\(3380^\circ\text{C}\)) and high resistivity, so it glows brightly without melting when heated to incandescence.
In simple words: (i) Thicker wire offers easier flow. (ii) Alloys have higher resistivity requiring less length. (iii) Tungsten withstands very high temperatures without melting.

Exam Tip: Questions on reasons are standard board favorites; always link them directly to \(R = \rho \frac{l}{A}\).

Module 10P (04 & 05)

Section : A

Q.1. SI unit of electric energy is
(i) Kwh
(ii) Joules
(iii) J sec
(iv) \(\text{NC}^{-1}\)
Answer: (ii) Joules (or kWh for commercial unit, but SI unit of energy is Joule).
Energy is the capacity to do work, and its SI unit is the Joule (\(\text{J}\)).
In simple words: Joule is the standard SI unit of energy.

Exam Tip: Joule is the SI unit for all forms of energy, whereas kilowatt-hour (kWh) is the commercial 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 because of its high melting point and high resistivity.
In simple words: Tungsten resists high heat without melting.

Exam Tip: High melting point and high resistivity are key properties of bulb filaments.

Q.3. Electric heaters are made from wire of
(i) Copper
(ii) Nichrome
(iii) Constantan
(iv) Manganin
Answer: (ii) Nichrome
Nichrome has high resistivity and does not oxidize easily at high temperatures.
In simple words: Nichrome produces immense heat efficiently.

Exam Tip: Heating devices use alloys like Nichrome due to high resistance and high-temperature durability.

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

Exam Tip: In series use \(P = I^2R\); in parallel use \(P = \frac{V^2}{R}\).

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 energy is dissipated entirely as heat in a pure resistor.
In simple words: Electrical work done in a resistor converts into heat energy.

Exam Tip: Joule's heating formula is \(H = I^2Rt\).

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: (ii) More heat is produced in thin wire
A thin wire has a smaller cross-sectional area, hence higher resistance than a thick wire. Since \(H = I^2Rt\), higher resistance produces more heat for the same current.
In simple words: Thinner wire has higher resistance, generating more heat.

Exam Tip: Resistance is inversely proportional to area (\(R \propto \frac{1}{A}\)).

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

Exam Tip: Always check units carefully in multiple-choice questions.

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 \(n\) equal resistors of resistance \(R\) in parallel, equivalent resistance \(R_{eq} = \frac{R}{n}\).
\(R_{eq} = \frac{2}{10} = 0.2 \, \Omega\).
In simple words: Ten 2-ohm resistors in parallel give 0.2 ohm.

Exam Tip: Shortcut for identical resistors in parallel: \(R_{eq} = \frac{R}{n}\).

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 \(3 \, \Omega\) and \(6 \, \Omega\) resistors in parallel, and then connect this parallel combination in series with the \(2 \, \Omega\) resistor.
Parallel combination of 3 and 6 ohm:
\(R_p = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 \, \Omega\)
Total resistance in series with 2 ohm resistor:
\(R_{total} = 2 + R_p = 2 + 2 = 4 \, \Omega\).
In simple words: Connect 3 and 6 ohms in parallel, then add 2 ohms in series.

Exam Tip: Trial and error with parallel/series combinations is key to solving such circuit puzzles.

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

Exam Tip: Series = Higher resistance; Parallel = Lower resistance.

Q.3. A one metre long wire is bent at \(180^\circ\) in the middle and two halves one twisted together. What will be the effect on resistance?
Answer:
When the wire is bent in half and twisted, its length becomes \(l' = \frac{l}{2}\) and its cross-sectional area becomes twice (\(A' = 2A\)).
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 area reduces resistance to one-fourth.

Exam Tip: Always analyze how length and area change simultaneously when folding wires.

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) which have very high electrical resistance. According to Joule's Law (\(H = I^2Rt\)), heat produced is directly proportional to resistance, hence a large amount of heat is generated.
In simple words: High resistance of the alloy coil produces immense heat when current passes through.

Exam Tip: Mention Joule's heating formula \(H = I^2Rt\) and high resistance of alloys.

Q.5. What is the meaning of mark 100W, 220V on an electric bulb?
Answer:
It means that when the bulb is connected across a 220 V supply mains, it consumes electrical power at the rate of 100 Watts (i.e., it consumes 100 Joules of electrical energy per second).
In simple words: The bulb uses 100W of power when operated at 220V.

Exam Tip: This rating helps calculate the operating current (\(I = P/V\)) and resistance (\(R = V^2/P\)).

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 equal sides of \(1 \, \Omega\) each. When current enters at vertex A and leaves at vertex B (adjacent vertex), the hexagon splits into two paths in parallel between A and B:
Path 1 (direct side AB): \(1 \, \Omega\)
Path 2 (remaining 5 sides in series): \(1 + 1 + 1 + 1 + 1 = 5 \, \Omega\)
These two paths are in parallel:
\(R_{eq} = \frac{1 \times 5}{1 + 5} = \frac{5}{6} \, \Omega\).
In simple words: Equivalent resistance is the parallel combination of 1 ohm and 5 ohms, which gives \(\frac{5}{6} \, \Omega\).

Exam Tip: Identify series and parallel branches 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:
Given: Number of electrons \(n = 6 \times 10^{17}\)
Time \(t = 1 \text{ minute} = 60 \text{ seconds}\)
Charge \(Q = n \cdot e = 6 \times 10^{17} \times 1.6 \times 10^{-19} = 0.096 \text{ C}\)
Current \(I = \frac{Q}{t} = \frac{0.096}{60} = 0.0016 \text{ A} = 1.6 \text{ mA}\).
In simple words: Current is total charge divided by time, equal to 1.6 mA.

Exam Tip: Always convert time into SI units (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 resistors be \(R_1 = 19 \, \Omega\) and \(R_2 = 20 \, \Omega\). (Note: usually textbook problems use 30 and 20 or similar, let's use exact given values).
Equivalent resistance \(R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{19 \times 20}{19 + 20} = \frac{380}{39} \, \Omega \approx 9.74 \, \Omega\).
Total voltage \(V = I \cdot R_p = 6 \times \frac{380}{39} = \frac{760}{13} \text{ V} \approx 58.46 \text{ V}\).
Current in \(R_1\) (\(I_1\)) = \(\frac{V}{R_1} = \frac{760/13}{19} = \frac{40}{13} \text{ A} \approx 3.08 \text{ A}\).
Current in \(R_2\) (\(I_2\)) = \(\frac{V}{R_2} = \frac{760/13}{20} = \frac{38}{13} \text{ A} \approx 2.92 \text{ A}\).
In simple words: Find total voltage and divide by individual resistances to find branch currents.

Exam Tip: In parallel circuits, voltage across each branch is identical and equals total voltage.

Q.4. Calculate the resistance between A and B of the circuit shown
Answer:
Examining the standard bridge/ladder network of \(5 \, \Omega\) resistors: Due to symmetry, the bridge is balanced, and the middle resistor carries no current, so it can be removed.
Simplifying the series and parallel branches yields an equivalent resistance of \(5 \, \Omega\) between A and B.
In simple words: Use bridge symmetry to simplify and calculate equivalent resistance as \(5 \, \Omega\).

Exam Tip: Look for Wheatstone bridge symmetry in complex resistor networks.

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 resistance \(R_s = R_1 + R_2 + R_3 = 2 + 3 + 6 = 9 \, \Omega\).
(ii) Parallel resistance \(\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1 \implies R_p = 1 \, \Omega\).
Ratio \(\frac{R_s}{R_p} = \frac{9}{1} = 9 : 1\).
In simple words: Series resistance is 9 ohms, parallel is 1 ohm, giving a ratio of 9:1.

Exam Tip: Always show formulas for both series and parallel additions clearly.

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 \(t = 3 \text{ hours} = 3 \times 3600 = 10800 \text{ seconds}\)
Energy consumed = \(P \times t = 2.2 \text{ kW} \times 3 \text{ h} = 6.6 \text{ kWh}\) (or \(2200 \times 10800 = 2.376 \times 10^7 \text{ J}\)).
Current drawn \(I = \frac{P}{V} = \frac{2200 \text{ W}}{220 \text{ V}} = 10 \text{ A}\).
In simple words: Energy is power times time (6.6 kWh), and current is power divided by voltage (10 A).

Exam Tip: Keep power in kW and time in hours for commercial units (kWh), or Watts and seconds for Joules.

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 \(t = 5 \text{ hrs/day} \times 10 \text{ days} = 50 \text{ hours}\)
Total energy consumed = \(P \times t = 0.2 \text{ kW} \times 50 \text{ h} = 10 \text{ kWh}\) (units)
Cost = Total energy \(\times\) rate = \(10 \text{ units} \times 0.40 \text{ Rs/unit} = \text{Rs. } 4.00\).
In simple words: Total energy is 10 kWh, costing Rs. 4.00 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 \(V = 220 \text{ V}\)
Current \(I = \frac{V}{R} = \frac{220}{660} = \frac{1}{3} \text{ A} \approx 0.33 \text{ A}\)
Time \(t = 11 \text{ minutes} = 11 \times 60 = 660 \text{ seconds}\)
Heat produced \(H = I^2Rt = \left(\frac{1}{3}\right)^2 \times 660 \times 660\) (or use \(\frac{V^2}{R}t\))
\(H = \frac{220^2}{660} \times 660 = 220^2 \times 660\) ... wait, using \(H = \frac{V^2}{R}t\):
\(H = \frac{220 \times 220}{660} \times 660 = 48400 \text{ Joules}\) (Wait, time was 11 minutes = 660 s):
\(H = \frac{V^2}{R}t = \frac{220^2}{660} \times 660 = 48400 \text{ J}\) (Incorrect substitution, let's recalculate correctly):
\(H = I^2Rt = \left(\frac{1}{3}\right)^2 \times 660 \times 660 = \frac{1}{9} \times 435600 = 48400 \text{ J}\).
In simple words: Current is 0.33 A, and heat produced is 48,400 Joules over 11 minutes.

Exam Tip: Use \(H = \frac{V^2}{R}t\) to avoid fractions and calculation errors.

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 much higher resistivity than pure metals and do not oxidize (burn) readily at high temperatures.
In simple words: Alloys generate more heat and withstand high temperatures without burning.

Exam Tip: High resistivity and high melting point without oxidation are the key reasons.

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:
Materials (iii) and (v) with resistivity around \(10^{-4}\) to \(10^{-5} \, \Omega\cdot\text{m}\) are typical alloys (like Nichrome) suitable for heating elements.
In simple words: Intermediate resistivity values characteristic of alloys are chosen for heating elements.

Exam Tip: Good conductors have very low resistivity (\(10^{-8}\)), insulators have very high resistivity (\(10^{12}\)), and heating elements use alloys with moderate resistivity (\(10^{-5}\)).

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, offering lower resistance (\(R \propto \frac{1}{A}\)).
In simple words: Thicker wires offer less resistance, allowing more current to flow.

Exam Tip: Relate current inversely to resistance (\(I = V/R\)).

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. It remains constant because resistivity is an intrinsic property of the material, independent of its dimensions.
In simple words: Stretching a wire changes its resistance, but resistivity stays exactly the same.

Exam Tip: A classic trick question in exams - remember resistivity depends only on material and temperature.

Q.3. Nichrome and copper wires of same length and same radius are connected in series current 'l' is passed through them. Why does the nichrome wire get heated first?
Answer:
Since nichrome has much higher resistivity than copper, its resistance is much higher. According to Joule's law (\(H = I^2Rt\)), for the same current in series, more heat is produced in the nichrome wire.
In simple words: Nichrome has higher resistance, so it generates more heat than copper.

Exam Tip: In series connection, current remains constant, so heat produced is directly proportional to resistance (\(H \propto R\)).

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 will remain unchanged (\(\rho\)) because resistivity depends only on the material nature and temperature, not on stretching or dimensions.
In simple words: Resistivity is independent of stretching and remains \(\rho\).

Exam Tip: Be careful: resistance becomes 9 times, but resistivity remains constant.

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 wire is divided into two equal halves of \(10 \, \Omega\) each, connected in parallel across a diameter.
\(R_{eq} = \frac{10 \times 10}{10 + 10} = 5 \, \Omega\).
In simple words: Two 10-ohm halves in parallel give an equivalent resistance of 5 ohms.

Exam Tip: Resistance across any diameter is always \(\frac{R}{4}\) of the total resistance (\(20 / 4 = 5 \, \Omega\)).

Q.6. Name a substance whose resistance decreases with temperature.
Answer:
Semiconductors (like silicon, germanium) and carbon / electrolytes show a decrease in resistance (increase in conductivity) with an increase in temperature.
In simple words: Semiconductors have negative temperature coefficient of resistance.

Exam Tip: Metals have positive temperature coefficient (resistance increases with temperature), 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 thin tungsten wires with high resistance. Since \(H = I^2Rt\), less heat is generated in low-resistance cables.
In simple words: Cables have low resistance, minimizing heat loss during transmission.

Exam Tip: Low resistance transmission lines prevent wastage of electrical energy as heat.

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

Exam Tip: Remember rule of thumb: Lower power rating = Higher resistance for same rated voltage.

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 across a voltage source, an extremely large current flows through it, causing overheating and burning out the instrument.
In simple words: Ammeter's very low resistance causes a short circuit and burnout if connected in parallel.

Exam Tip: Ammeter must always be connected in series, and voltmeter 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 (\(R = \frac{\Delta V}{\Delta I}\)). A steeper slope indicates higher resistance. Since resistance of metals increases with temperature (\(R \propto T\)), the line with the higher temperature has a greater slope and resistance.
In simple words: Higher slope means higher resistance and hence higher temperature for metallic conductors.

Exam Tip: Slope = Resistance; higher temperature = higher resistance 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:
Equivalent resistance in series is higher than in parallel (\(R_s > R_p\)). Therefore, the slope of the \(V-I\) graph for series must be steeper than for parallel. The correctly labeled graph is the one where the series line has a larger slope.
Ohm's law is applicable only when physical conditions like temperature remain constant.
In simple words: Series has higher resistance (steeper slope) than parallel. Ohm's law requires constant temperature.

Exam Tip: Steeper slope on a \(V-I\) graph always signifies higher total resistance.

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 cross-sectional area (\(R = \rho \frac{l}{A}\)). Unless the ratio \(\frac{l}{A}\) is identical for both, resistances will differ.
In simple words: Different lengths and thicknesses result in different resistances.

Exam Tip: Resistance depends on dimensions, not just the material.

Q.5. Can you produce high voltage in your body without getting a shock?
Answer:
Yes, high static voltages can be generated on our body through friction (e.g., rubbing shoes on a carpet on a dry day), but no shock is felt because the current (charge flow rate) is extremely small due to high resistance or very brief discharge time.
In simple words: High static voltage without current flow does not cause an electric shock.

Exam Tip: It is current, not voltage alone, that causes electric shock and physiological damage.

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, running computers, hospitals, and electrical appliances improving modern life quality.
(ii) Unfavourable images: Electrical shocks, short-circuit fires, electrocution, power wastage, and environmental pollution from thermal power generation.
(iii) Common factor: Electric current and flow of electrons are fundamental to all these phenomena.
In simple words: Electricity is a powerful servant but a dangerous master; both benefits and hazards stem from electron flow.

Exam Tip: Value-based questions test balanced awareness of technology utility and safety precautions.

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 fields are represented using directed magnetic field lines.

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 poles (North-North or South-South) repel each other, while unlike poles attract.
In simple words: Similar magnetic poles repel each other.

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 Maxwell's corkscrew rule, but Right hand thumb rule is standard).
In simple words: Right hand thumb rule gives the direction of magnetic field around a current-carrying wire.

Exam Tip: Thumb points along current, curled fingers show magnetic field lines direction.

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
Field line density is directly proportional to magnetic field strength.
In simple words: Closer field lines indicate a stronger magnetic field.

Exam Tip: Strongest magnetic field is near the poles where lines are most crowded.

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
Magnetic field lines around a straight current-carrying conductor are concentric circles centered on the wire.
In simple words: The magnetic field forms concentric circles around a straight wire.

Exam Tip: Plane of magnetic field circles is 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
Magnetic field strength increases with current and decreases with distance from the wire.
In simple words: Field strength depends on both current magnitude and distance from the conductor.

Exam Tip: \(B \propto I\) and \(B \propto \frac{1}{r}\).

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 permeability and low retentivity, making it ideal for temporary electromagnets.
In simple words: Soft iron magnetizes and demagnetizes easily, perfect for electromagnets.

Exam Tip: Soft iron for electromagnets (temporary); 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
All these materials (carbon steel, chromium steel, alnico) retain their magnetism strongly and are used for permanent magnets.
In simple words: Special steel alloys and alnico maintain permanent magnetism well.

Exam Tip: Alnico is an alloy of Aluminum, Nickel, and Cobalt used for strong permanent magnets.

Section: B

Q.1. What are the factors on which magnetic field due to a solenoid depends?
Answer:
1. Magnitude of electric current flowing through the solenoid.
2. Number of turns per unit length in the solenoid.
3. Nature of the core material placed inside the solenoid (e.g., soft iron core increases field strength).
In simple words: Field strength depends on current, number of turns, and core material.

Exam Tip: More turns and higher current produce a stronger magnetic field inside a solenoid.

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 straight wire.
(ii) Circular loops around the wire that merge into nearly straight parallel lines at the center of the coil.
(iii) Parallel straight lines inside the solenoid, similar to a bar magnet.
In simple words: Diagrams show concentric circles for straight wire, loops for circular coil, and uniform parallel lines for solenoids.

Exam Tip: Always draw arrowheads indicating magnetic field direction (North to South outside, South to North inside).

Q.3. State various uses of electromagnets.
Answer:
1. Lifting heavy iron loads and scrap metal in factories using cranes.
2. Used in electric bells, relays, loudspeakers, and telephones.
3. Used in medical equipment like MRI scanners and lifting iron splinters from eyes.
In simple words: Electromagnets are used in cranes, electric bells, and medical devices.

Exam Tip: Electromagnets can be switched on and off, making them versatile for control applications.

Q.4. State various uses of permanent magnets.
Answer:
1. Used in electric meters (galvanometers, voltmeters, ammeters).
2. Used in loudspeakers, microphones, and headphones.
3. Used in electric motors and generators, compasses, and refrigerator door seals.
In simple words: Permanent magnets are used in meters, speakers, motors, and compasses.

Exam Tip: Permanent magnets do not require a continuous power supply to maintain their magnetic field.

Q.5. What is meant by the term "magnetic field lines"? List two properties of magnetic field lines.
Answer:
Magnetic field lines are continuous 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 the magnet.
2. Two magnetic field lines never intersect each other.
In simple words: Field lines map out magnetic force; they never cross each other.

Exam Tip: If lines intersected, it would mean two directions of magnetic field at the same point, which is impossible.

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 curves originating from North pole and terminating at South pole.
In simple words: Iron filings align along magnetic field lines when tapped on a cardboard over a magnet.

Exam Tip: Mention gentle tapping to overcome friction so filings can 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, strong magnetic field, strength can be varied, uses soft iron core.
Permanent Magnet: Permanent magnetism, relatively weaker field, strength cannot be altered easily, made of steel/alnico.
Factors affecting electromagnet strength: Current, number of turns, and core material.
In simple words: Electromagnets can be turned on/off and adjusted; permanent magnets retain magnetism constantly.

Exam Tip: Tabular comparison works best for differences in board exams.

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 says 'decreases as distance increases').
Answer:
Place a magnetic compass closer to a current-carrying wire, and the deflection is large. When moved farther away, the deflection decreases, proving that magnetic field strength decreases with increasing distance.
In simple words: Compass deflection drops as distance from the wire increases, showing a weaker magnetic field.

Exam Tip: Compass needle deflection directly indicates 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 solenoid, magnetic field lines are parallel straight lines, indicating a uniform magnetic field.
In simple words: The magnetic field inside a solenoid is uniform and parallel to its axis.

Exam Tip: Uniform magnetic field inside a solenoid is a key concept in electromagnetism.

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.
Moving charges (electric current) produce magnetic fields, whereas stationary charges produce only electric fields.
In simple words: Stationary charges do not produce magnetic fields.

Exam Tip: Current (moving charge) creates magnetism; static charge creates electrostatics.

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, so it requires a thicker fuse wire with a higher current-carrying capacity before melting.
In simple words: Higher current circuits need thicker fuse wires.

Exam Tip: Thicker fuse wire allows more current to pass safely without melting prematurely.

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 so that turning off the switch disconnects the appliance from high voltage safely.
In simple words: Switches must always be on the live wire for safety.

Exam Tip: Connecting a switch to the neutral wire 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 appliance receives the same voltage and can be operated independently.
In simple words: Parallel connection ensures independent operation and full voltage for all appliances.

Exam Tip: If appliances were in series, turning off one would turn off all others.

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 transmitted over long distances across the country with minimum power loss using step-up transformers.
In simple words: AC can be transmitted over long distances with very little energy loss.

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

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 (mechanical rotation from current) and electric generators/dynamos (current generation from mechanical rotation) involve coils rotating in magnetic fields.
In simple words: Both motors and generators feature rotating coils in magnetic fields.

Exam Tip: Motor converts electrical energy to mechanical; Generator converts mechanical energy to electrical.

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 energy 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 electrical appliances of high power rating are connected to a single socket, drawing current beyond the safe limit of the wires, causing overheating.
In simple words: Drawing excessive current beyond safe wire capacity is called overloading.

Exam Tip: Overloading can damage wiring insulation 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 force is found using Fleming's Left-Hand Rule.
In simple words: Fleming's Left-Hand Rule gives the direction of magnetic force on a current-carrying wire.

Exam Tip: Thumb = Force, Forefinger = Magnetic Field, Middle finger = Current (FBI: Force, B-field, Current).

Q.3. Why is electricity transmitted over long distances at very high voltage?
Answer:
Transmitting at high voltage reduces the current proportionally for the same power, which drastically minimizes heat energy losses (\(H = I^2Rt\)) along transmission cables.
In simple words: High voltage lowers current, reducing heat energy losses during transmission.

Exam Tip: Lower current means significantly less \(I^2R\) power dissipation in power lines.

Q.4. What is the difference between a heating wire and a fuse wire?
Answer:
Heating wire: Made of nichrome, has high resistance and high melting point, designed to produce steady heat.
Fuse wire: Made of tin-lead alloy, has low melting point and moderate resistance, designed to melt quickly and break the circuit when excess current flows.
In simple words: Heating wire produces continuous heat; fuse wire melts to break circuits during overloads.

Exam Tip: Key distinction: Fuse wire must have a low melting point; heating wire must have a high melting point.

Q.5. A bird sitting on 11,000V wire is not hurt while a person touching 220 V wire dies. Why?
Answer:
The bird sits only on a single high-voltage wire with both feet at the same potential, so no potential difference exists across its body and no current flows. A person standing on the ground touching a 220V wire completes a circuit to the earth, creating a large potential difference and dangerous current flow through their body.
In simple words: No potential difference means no current flows through the bird; touching earth completes a fatal circuit for humans.

Exam Tip: Electric shock requires a closed circuit and potential difference for current to flow.

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 independently.
2. If one appliance fails or is switched off, other appliances continue to work uninterrupted.
In simple words: Parallel connection provides full voltage to each appliance and allows independent switching.

Exam Tip: Contrast with series circuits where failure of one component breaks the entire circuit.

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 an extremely high current to flow.
Fuse protection: The high current instantly overheats and melts the fuse wire, breaking the circuit and protecting appliances.
In simple words: Short circuit causes dangerous high current; the fuse melts to cut off power.

Exam Tip: Short circuits result from damaged insulation or faulty appliance wiring.

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:
Force acts on a current-carrying conductor in a magnetic field, given by \(F = BIL \sin\theta\).
- Maximum: When the conductor is perpendicular to the magnetic field (\(\theta = 90^\circ\)).
- Minimum: When the conductor is parallel to the magnetic field (\(\theta = 0^\circ \text{ or } 180^\circ\)).
In simple words: Force is maximum when wire is perpendicular to the magnetic field, and zero when parallel.

Exam Tip: Remember \(\sin(90^\circ) = 1\) (maximum) and \(\sin(0^\circ) = 0\) (minimum).

Section : C

Q.1. Define electromagnetic induction. State the factors on which the strength of induced current depends.
Answer:
Electromagnetic induction is the phenomenon of producing an induced electric current in a circuit by changing the magnetic flux linked with it.
Factors:
1. Relative speed between the magnet and the 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; speed, turns, and field strength determine its magnitude.

Exam Tip: Discovered by Michael Faraday; forms the working principle of generators and transformers.

Q.2. Describe the hazards and safety measures in the use of electricity in our houses.
Answer:
Hazards: Electric shocks, electrical fires from short circuits, and appliance burnout.
Safety measures: Proper earthing of metallic appliances, using fused circuits and MCBs, wearing insulated shoes, and using quality ISI-certified wiring.
In simple words: Electricity hazards include shocks and fires; safety measures include earthing, fuses, and insulated wiring.

Exam Tip: Earthing provides a safe low-resistance path for leakage current into the ground.

Q.3. Why is it necessary to connect an earth wire to electric appliances having metallic covers?
Answer:
If the live wire touches the metallic body of an appliance due to a fault, the user touching it could get a severe shock. The earth wire conducts any leakage current safely to the ground, preventing electrocution.
In simple words: Earth wire safely drains leakage current to the ground, preventing electric shocks.

Exam Tip: Earth wire is green-colored and connected to the metal casing of heavy appliances like refrigerators and irons.

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 magnetic field and middle finger in the direction of current, then the thumb points in the direction of force.
Direction of force: Current is East, B-field is North. Applying left-hand rule, force acts vertically upwards (out of the page).
In simple words: Left-hand rule relates magnetic field, current, and force directions. Force here is vertically upwards.

Exam Tip: Always use your left hand for motors/force and right hand for generators/induction.

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 (e.g., from cells/batteries).
Alternating Current (AC): Periodically reverses its direction of flow at regular intervals (e.g., 50 Hz in India).
Advantage of AC: AC can be transmitted over very long distances with minimal power loss using transformers.
In simple words: DC flows one way; AC reverses direction periodically and transmits over long distances easily.

Exam Tip: AC frequency in India is 50 Hz, meaning it changes direction 100 times every second.

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 vector (both direction and momentum) changes due to magnetic force, though its kinetic energy and speed remain constant.
In simple words: The proton's direction of motion and momentum change in a magnetic field.

Exam Tip: Magnetic force does zero work on a charged particle, so kinetic energy remains unchanged.

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

Exam Tip: Perpendicular orientation yields maximum force (\(F = BIL\)).

Q.3. An alternating electric current has a frequency of 50 Hz. How many times does it change its direction in one second?
Answer:
It changes its direction \(50 \times 2 = 100\) times in one second (since each complete cycle of AC consists of two directional reversals).
In simple words: At 50 Hz, AC changes direction 100 times per second.

Exam Tip: Frequency is cycles per second, and each cycle has two directional reversals.

Q.4. Why is soft-iron not used for making a permanent magnet?
Answer:
Soft iron has very low retentivity and low coercivity, meaning it loses its magnetism easily as soon as the external magnetizing field is removed.
In simple words: Soft iron cannot retain permanent magnetism.

Exam Tip: Soft iron is ideal for temporary magnets (electromagnets), while steel is used for permanent magnets.

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: Current (motion of positive charge) is West, Force is North, so the magnetic field must be directed vertically upwards (out of the page).
In simple words: The magnetic field acts vertically upwards.

Exam Tip: Direction of current for positive charges is the same as the direction of motion of the particle.

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, magnetic fields produced by individual circular turns add up constructively along the axis, creating a strong resultant field, unlike a single straight wire.
In simple words: Magnetic fields from numerous turns in a solenoid reinforce each other to create a strong field.

Exam Tip: More turns per unit length = stronger magnetic field inside the solenoid.

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) When pushed into the coil, magnetic flux changes, inducing a current and causing the galvanometer to deflect.
(ii) When held stationary, magnetic flux is constant, so no induced current flows and the galvanometer shows zero deflection.
In simple words: (i) Motion induces current. (ii) Stationary magnet induces zero current.

Exam Tip: Electromagnetic induction requires 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 direction of electric current, which reverses the direction of the magnetic field and thus reverses the compass needle deflection.
In simple words: Observation (i) is wrong because reversing current reverses magnetic field direction and compass deflection.

Exam Tip: Current direction determines magnetic field polarity (Right-Hand Thumb Rule).

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) Graph (a) represents Direct Current (DC); Graph (b) represents Alternating Current (AC).
(ii) Source of DC: Cell or battery; Source of AC: AC generator / mains supply.
(iii) Frequency of AC in India is 50 Hz.
In simple words: (a) is DC from batteries, (b) is AC from generators at 50 Hz in India.

Exam Tip: DC graph is a straight line parallel to time axis; AC graph is a sinusoidal wave.

Value Based Question

Q.1. The use of magnetic therapy for pain relief has become increasingly popular in the last few years...
(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 applied to body parts improve blood circulation and stimulate nerve healing.
(ii) Side effects: Generally no major side effects when used moderately, but unverified claims require caution.
(iii) Precautions: People with pacemakers or metallic implants must avoid strong magnetic therapy.
In simple words: Magnetic therapy uses magnets for pain relief, but pacemaker wearers must avoid it.

Exam Tip: Scientific validation is important; medical devices like pacemakers can malfunction near strong magnets.

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).
In simple words: Solar cells are semiconductor devices made of silicon or germanium.

Exam Tip: Silicon (Si) is the most widely used semiconductor in photovoltaic solar 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 frequencies/energies than ultraviolet rays.
In simple words: Infrared rays carry less energy than ultraviolet rays.

Exam Tip: Energy is inversely proportional to wavelength in 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 photovoltaic effect.
In simple words: Solar cells convert light energy directly into electrical energy.

Exam Tip: Solar heaters and cookers convert light 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 mirrors converge solar radiation onto a single focal point to generate high temperatures.
In simple words: Concave reflectors focus sunlight to concentrate heat.

Exam Tip: Concave mirrors are converging reflectors used in solar cookers and furnaces.

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

Exam Tip: VIBGYOR order: Wavelength increases from Violet to Red.

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 is thermal radiation perceived as heat.
In simple words: Infrared rays provide the heating sensation from sunlight.

Exam Tip: Infrared rays lie just beyond the red end of the visible spectrum.

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 the globe.
In simple words: Solar energy is the most abundant renewable energy source.

Exam Tip: Solar energy is inexhaustible and clean.

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

Exam Tip: Wind farms require regions with consistent high wind speeds.

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 (anaerobic digestion) takes place in the absence of oxygen (air) in a slurry medium containing water.
In simple words: Anaerobic bacteria break down organic waste in water without oxygen to form biogas.

Exam Tip: Methane is the major component of biogas (about 75%).

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: Mass number 235 - Atomic number 92 = 143 neutrons).
In simple words: Uranium-235 isotope used in nuclear reactors has 143 neutrons.

Exam Tip: \(\text{Neutrons} = \text{Mass Number} - \text{Atomic Number}\).

Section : B

Q.1. What are renewable and non renewable sources of energy? Give an example.
Answer:
Renewable sources: Sources that are inexhaustible and can be replenished naturally in a short time (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 (solar), while non-renewable energy gets exhausted (coal).

Exam Tip: Emphasize sustainability and replenishment time in definitions.

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

Exam Tip: Burning fossil fuels releases greenhouse gases like carbon dioxide.

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 energy generated by harnessing the kinetic energy of falling water.
Principle: Potential energy of stored water in dams is converted into kinetic energy of flowing water, which rotates turbines coupled to generators producing electricity.
(ii) Advantages: Renewable, clean energy with no air pollution.
Disadvantages: Submerges large land areas, displaces populations, and disrupts aquatic ecosystems.
In simple words: Hydroelectricity uses falling water to turn turbines. It is clean and renewable but displaces communities and ecosystems.

Exam Tip: Dams convert gravitational potential energy of water into 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:
(a) Glass sheet: Traps heat inside via greenhouse effect (allows incoming short-wave radiation, blocks outgoing infrared radiation).
(b) Plane mirror: Reflects and redirects additional sunlight into the cooker box.
(c) Black painted vessels: Absorb maximum heat radiation efficiently.
In simple words: Glass traps heat, mirror reflects extra light, and black paint absorbs maximum heat.

Exam Tip: Black surfaces are good absorbers and poor reflectors 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 solar energy directly into electrical energy.
Examples of materials: Silicon, Germanium, Gallium arsenide.
Uses: Powering artificial satellites, calculators, street lights, and remote water pumps.
In simple words: Solar cells generate electricity from sunlight and power satellites and calculators.

Exam Tip: Solar panels consist of multiple solar cells connected in series/parallel.

Q.6. What are the limitations of solar heating devices?
Answer:
1. They cannot be used effectively on cloudy days or at night.
2. They have limited heating capacity and cannot reach very high temperatures compared to electrical or gas heaters.
In simple words: Solar devices depend entirely on sunlight and fail on cloudy days.

Exam Tip: Intermittency of sunlight is the main drawback of solar energy.

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

Exam Tip: Photovoltaic cells operate on the principle of the photoelectric effect.

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:
Wind energy rotates wind turbine blades, driving a generator to produce electricity.
Advantages: Renewable, pollution-free.
Limitations: Requires high wind speeds (>15 km/h), requires large land areas for wind farms.
In simple words: Wind turns turbines to generate clean electricity, but requires windy sites and large space.

Exam Tip: Wind energy is completely eco-friendly but intermittent.

Q.9. What is biogas? Name the major component of biogas.
Answer:
Biogas is a gaseous fuel produced by the anaerobic decomposition of organic wastes (like animal dung and plant matter).
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: Leftover slurry from biogas plants makes excellent organic manure.

Q.10. Describe the construction and working of a biogas plant with the help of a labeled diagram.
Answer:
Construction: Consists of a mixing tank, digester tank (sealed dome), and outlet tank.
Working: Animal dung and water slurry is fed into the digester. Anaerobic bacteria decompose the waste in the absence of oxygen, producing biogas which collects in the dome and is supplied through pipes.
In simple words: Slurry decomposes anaerobically in a digester tank to produce combustible biogas.

Exam Tip: Label the mixing tank, digester, gas holder dome, and outlet slurry tank clearly in diagrams.

Q.11. Define nuclear fission and nuclear fusion with an example.
Answer:
Nuclear fission: The splitting of a heavy nucleus into smaller nuclei with release of immense energy.
Example: Bombarding Uranium-235 with neutrons yielding Barium and Krypton.
Nuclear fusion: The combining of two lighter nuclei to form a heavier nucleus with release of massive energy.
Example: Fusion of hydrogen isotopes to form helium in the Sun.
In simple words: Fission splits heavy nuclei (uranium); fusion combines light nuclei (hydrogen).

Exam Tip: Nuclear fusion powers the Sun and hydrogen bombs; fission powers nuclear reactors and atom 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 is a converging mirror that focuses parallel sun rays onto a small focal area to produce high temperatures.
In simple words: Concave mirrors concentrate sunlight to generate high temperatures for cooking.

Exam Tip: Concave reflectors concentrate solar heat efficiently.

Q.2. Name two types of energy which do not relate to the sun.
Answer:
1. Geothermal energy
2. Nuclear energy (fission/fusion)
In simple words: Geothermal and nuclear energies do not originate from solar radiation.

Exam Tip: Geothermal taps Earth's internal heat; nuclear energy taps atomic binding energy.

Q.3. Biomass is a renewable source of energy. Why?
Answer:
Because plant and animal matter can be continuously regrown and replenished through agriculture and biological cycles.
In simple words: Biomass replenishes naturally through plant growth and animal waste.

Exam Tip: Renewable means capable of natural regeneration over short time spans.

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 harmful soot or unburnt hydrocarbon residues.
2. It produces very low emissions of sulfur dioxide, carbon monoxide, and greenhouse gases compared to petrol or diesel.
In simple words: CNG burns cleanly with minimal toxic emissions.

Exam Tip: CNG stands for Compressed Natural Gas, primarily composed of methane.

Q.2. Mention any two ways by which water can be used to produce hydro-electricity?
Answer:
1. Constructing dams across rivers to create large water reservoirs with high potential energy.
2. Utilizing run-of-river hydroelectric systems where flowing river current turns turbines.
In simple words: Hydroelectricity is generated using river dams or run-of-river water flow.

Exam Tip: Both methods convert kinetic/potential energy of water into electrical energy via turbines.

Q.3. In many application, solar cells are connected to rechargeable batteries. Why is it so?
Answer:
Solar cells generate electricity only during daytime when sunlight is available. Connecting them to rechargeable batteries stores surplus energy for use at night or on cloudy days.
In simple words: Batteries store solar electricity for continuous power supply at night or during cloudy weather.

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: 1 gram of cooking gas produces 50 kJ of heat energy upon burning.

Exam Tip: Calorific value is expressed in kJ/g or kJ/kg.

Q.5. A nuclear reaction is represented by the following equation:
\({}^{235}_{92}\text{U} + {}^1_0\text{n} \longrightarrow {}^{139}_{56}\text{Ba} + {}^{94}_{36}\text{Kr} + xc + \text{Energy}\)
(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 \(x = 3\).
(iii) Nuclear power reactor.
(iv) Atom bomb.
In simple words: This equation shows nuclear fission of uranium producing neutrons and energy, used in reactors and atom bombs.

Exam Tip: Balance mass numbers and atomic numbers on both sides of nuclear equations.

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) Extremely high temperature and high pressure.
(iv) Hydrogen bomb.
In simple words: This shows nuclear fusion of deuterium, requiring extreme heat and pressure, utilized in hydrogen bombs.

Exam Tip: Nuclear fusion requires millions of degrees of temperature to overcome electrostatic repulsion between nuclei.

Value Based Question

Q.1. There are many dangers in the use of nuclear power plants...
(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, nuclear waste disposal hazards, and risk of catastrophic meltdowns.
(ii) Precautionary conduct: Strict regulatory safety standards, secure underground waste storage, and international treaties ensuring peaceful, safe nuclear management.
In simple words: Nuclear plants risk radiation leaks and waste hazards; nations must maintain strict safety and secure waste disposal.

Exam Tip: Nuclear energy is clean regarding carbon emissions but hazardous regarding radioactive waste.

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 real and diminished images when the object is placed beyond \(2F\).
In simple words: Concave mirrors can form real and diminished images.

Exam Tip: Convex mirrors always form virtual, erect, 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 (\(R = 2f\)).
In simple words: Focal length is half the radius of curvature for spherical mirrors.

Exam Tip: Formula applies to both concave and convex mirrors under paraxial approximation.

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
Convex mirrors always form virtual, erect, and diminished images regardless of object position.
In simple words: Convex mirrors always produce virtual and erect images.

Exam Tip: This makes convex mirrors ideal for rear-view mirrors in vehicles due to a wider field of view.

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 A, H, I, M, O, T, U, V, W, X, Y) do not show noticeable lateral inversion.
In simple words: Symmetrical letters like 'O' look identical in a mirror.

Exam Tip: Letters lacking lateral inversion possess vertical line 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
Plane mirrors always form virtual and erect images, so magnification \(m = +\frac{h_i}{h_o} = +1\) (always positive).
In simple words: Plane mirrors always produce erect virtual images, so magnification is positive (+1).

Exam Tip: Positive sign of magnification always denotes an erect 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
Mirror formula (\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)) holds true for spherical mirrors as well as plane mirrors (treating a plane mirror as a spherical mirror of infinite radius).
In simple words: Mirror formula applies universally to both plane and spherical mirrors.

Exam Tip: For plane mirrors, \(f = \infty\), leading to \(v = -u\).

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
Positive sign indicates virtual and erect image; magnitude \(\frac{2}{3} (< 1)\) indicates a diminished image. A convex mirror always forms virtual and diminished images.
In simple words: Positive and less than 1 magnification means a convex mirror.

Exam Tip: Magnification < 1 means diminished; positive means virtual/erect.

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 and becomes progressively smaller.
In simple words: Moving an object away from a convex mirror pushes its image closer to the focus.

Exam Tip: For convex mirrors, image always stays between pole and focus behind the mirror.

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 (\(R = \infty\), \(f = \infty\)).
In simple words: A plane mirror has an infinite radius of curvature.

Exam Tip: Focal length of a plane mirror is infinity.

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
Because a convex mirror is diverging, incident parallel rays appear to diverge from its principal focus behind the mirror.
In simple words: Parallel rays reflecting off a convex mirror diverge and appear to originate from its focus.

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

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 normally (\(\theta = 0^\circ\)). The angle of incidence is zero, so by the laws of reflection, the angle of reflection is also zero, causing the ray to retrace its path.
In simple words: Rays hitting the mirror perpendicularly (along radius) reflect back along the same path.

Exam Tip: Normal incidence means angle of incidence \(i = 0^\circ\) and reflection \(r = 0^\circ\).

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 either:
1. Between pole and focus (to get a virtual, erect, and magnified image).
2. Between focus and centre of curvature (to get a real, inverted, and magnified image).
In simple words: Use a concave mirror with the object placed between pole and focus or between focus and centre of curvature.

Exam Tip: Concave mirrors are the only mirrors that produce enlarged real and virtual images.

Q.3. Which property of a concave mirror can be used to determine its focal length?
Answer:
The property that rays of light parallel to the principal axis converge at the principal focus after reflection can be used to determine focal length by focusing a distant object (like the sun).
In simple words: Converging parallel rays to a sharp focus allows measurement of focal length.

Exam Tip: Distance between the pole and the sharp screen image of a distant object gives the focal length.

Q.4. What kind of mirror is used in vehicles to see the traffic following it?
Answer:
Convex mirror, because it always produces erect, diminished images and provides a much wider field of view than plane or concave mirrors.
In simple words: Convex mirrors are used as rear-view mirrors because they cover a wide field of view.

Exam Tip: Always remember: Rear-view = Convex mirror; Headlights/Shaving = Concave mirror.

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 (head): Convex mirror (produces diminished image).
- Middle part (body): Concave mirror (produces magnified image).
- Bottom part (legs): Plane mirror (produces normal-sized image).
In simple words: Convex at top for small head, concave in middle for fat body, plane at bottom for normal legs.

Exam Tip: Combine mirror characteristics based on image size and type.

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 outward due to surface tension.
In simple words: A fallen mercury drop forms a spherical bulging surface acting as a convex mirror.

Exam Tip: Surface tension gives liquid drops a curved, convex outer boundary.

Section - C

Q.1. Distinguish between real and virtual image.
Answer:
Real Image:
1. Formed by actual intersection of light rays.
2. Can be caught on a screen.
3. Always inverted.
Virtual Image:
1. Formed when rays appear to diverge from a point.
2. Cannot be caught on a screen.
3. Always erect.
In simple words: Real images can be caught on a screen and are inverted; virtual images cannot be caught on a screen and are erect.

Exam Tip: Use a clear comparative table for differences in exams.

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 Pole and Focus: Forms a virtual, erect, and magnified image behind the mirror.
(ii) Object between Centre of Curvature and Focus: Forms a real, inverted, and magnified image beyond the centre of curvature.
In simple words: (i) Virtual enlarged behind mirror. (ii) Real inverted magnified beyond C.

Exam Tip: Label arrows, pole (P), focus (F), and centre of curvature (C) accurately in all ray diagrams.

Q.3. How will you distinguish between a plane mirror, a concave mirror and a convex mirror without touching them?
Answer:
Bring your face close to each mirror:
1. Plane mirror: Forms an erect image of the same size as your face at all distances.
2. Concave mirror: Forms an enlarged and erect image when close, but inverts as you move further away.
3. Convex mirror: Forms a diminished and erect image at all distances.
In simple words: Observe image size and orientation as you move closer or further away.

Exam Tip: Image size variation with distance is the easiest way to identify mirrors optically.

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 (concave) or appear to diverge from (convex) after reflection.
In simple words: (i) Radius of the sphere forming the mirror. (ii) Point where parallel rays converge or diverge.

Exam Tip: Precise definitions carry full credit in theory questions.

Q.5. Show that image distance is same as object distance when an extended object is placed in front of the plane mirror.
Answer:
Using laws of reflection for a plane mirror, angle of incidence equals angle of reflection. By congruent triangle geometry formed by incident and reflected rays, object distance \(u\) equals image distance \(v\) (\(v = -u\)).
In simple words: Geometric reflection proves object distance equals image distance in plane mirrors.

Exam Tip: Plane mirror image characteristics: Virtual, erect, same size, \(v = u\), laterally inverted.

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: Object size \(h_o = 2 \text{ cm}\)
Focal length \(f = -15 \text{ cm}\)
Object distance \(u = -30 \text{ cm}\) (at \(2f\))
Using mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
\(\frac{1}{-15} = \frac{1}{v} + \frac{1}{-30} \implies \frac{1}{v} = -\frac{1}{15} + \frac{1}{30} = -\frac{1}{30} \implies v = -30 \text{ cm}\)
Magnification \(m = -\frac{v}{u} = \frac{h_i}{h_o}\)
\(-\frac{-30}{-30} = \frac{h_i}{2} \implies -1 = \frac{h_i}{2} \implies h_i = -2 \text{ cm}\).
Image size is 2 cm (inverted).
In simple words: Object at \(2f\) forms a real, inverted image of the same size (\(2\text{ cm}\)) at \(2f\).

Exam Tip: Recognizing that object at \(2f\) gives an identical sized image saves calculation time.

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:
Given: Radius \(R = 68 \text{ cm} \implies f = +\frac{68}{2} = +34 \text{ cm}\)
Image distance \(v = +22 \text{ cm}\) (behind mirror)
Using mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
\(\frac{1}{34} = \frac{1}{22} + \frac{1}{u} \implies \frac{1}{u} = \frac{1}{34} - \frac{1}{22} = \frac{11 - 17}{374} = \frac{-6}{374}\)
\(u = -\frac{374}{6} \approx -62.33 \text{ cm}\)
Magnification \(m = -\frac{v}{u} = -\frac{22}{-374/6} = \frac{132}{374} \approx +0.35\) (virtual, erect, diminished).
In simple words: Object is at 62.33 cm in front; image is virtual, erect, and diminished.

Exam Tip: Always double check sign conventions: \(u\) is always negative, convex mirror \(f\) and \(v\) are positive.

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: \(h_o = 4 \text{ cm}\), \(u = -10 \text{ cm}\), \(f = -20 \text{ cm}\)
Using mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
\(\frac{1}{-20} = \frac{1}{v} + \frac{1}{-10} \implies \frac{1}{v} = -\frac{1}{20} + \frac{1}{10} = +\frac{1}{20} \implies v = +20 \text{ cm}\)
Magnification \(m = -\frac{v}{u} = -\frac{20}{-10} = +2\)
Image size \(h_i = m \cdot h_o = 2 \times 4 = 8 \text{ cm}\).
Nature: Virtual, erect, and magnified (8 cm high) located 20 cm behind the mirror.
In simple words: Object placed between pole and focus of a concave mirror forms a virtual, magnified, and erect image behind the mirror.

Exam Tip: When object is within focal length of a concave mirror, image is always virtual and behind the mirror.

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 \(f = \infty\) in mirror formula:
\(\frac{1}{\infty} = \frac{1}{v} + \frac{1}{u} \implies 0 = \frac{1}{v} + \frac{1}{u} \implies v = -u\).
This proves image distance equals object distance, confirming the formula applies to plane mirrors.
In simple words: Setting focal length to infinity in the mirror formula yields \(v = -u\) for plane mirrors.

Exam Tip: Plane mirrors are a special case of spherical mirrors with infinite radius of curvature.

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 diminished (\(m = -\frac{1}{2}\)).
Let focal length be \(f\). Object is at \(u = -(f + 10)\).
Magnification \(m = -\frac{v}{u} = -\frac{1}{2} \implies v = \frac{u}{2} = -\frac{f + 10}{2}\).
Using mirror formula \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\):
\(\frac{1}{-f} = \frac{1}{-(f+10)/2} + \frac{1}{-(f+10)}\)
Solving yields focal length \(f = 10 \text{ cm}\).
In simple words: Solving mirror formula with given magnification yields a focal length of 10 cm.

Exam Tip: Set up simultaneous equations for \(u, v, f\) using magnification and mirror formula.

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 \(f = -20 \text{ cm}\). Magnification \(m = \pm 2\).
Case 1 (Real image): \(m = -2\)
\(-\frac{v}{u} = -2 \implies v = 2u\)
\(\frac{1}{-20} = \frac{1}{2u} + \frac{1}{u} = \frac{3}{2u} \implies 2u = -60 \implies u = -30 \text{ cm}\).
Case 2 (Virtual image): \(m = +2\)
\(-\frac{v}{u} = +2 \implies v = -2u\)
\(\frac{1}{-20} = \frac{1}{-2u} + \frac{1}{u} = \frac{1}{2u} \implies 2u = -20 \implies u = -10 \text{ cm}\).
Object positions are at \(30 \text{ cm}\) and \(10 \text{ cm}\) in front of the mirror.
In simple words: Double magnification can occur for both real (at 30 cm) and virtual (at 10 cm) images.

Exam Tip: Always consider both positive and cases for magnification (\(m = \pm 2\)) when nature of image is unspecified.

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.
In simple words: An object appears blue because it reflects blue light and absorbs other colors.

Exam Tip: Color of an object depends on the wavelengths of light it reflects.

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 angle of incidence (\(i\)) and angle of reflection (\(r\)) are \(0^\circ\).
In simple words: Normal incidence means both incidence and reflection angles are zero.

Exam Tip: Normal incidence means the ray strikes perpendicular to the mirror surface.

Q.3. A concave mirror is placed in water. Will there be any change in focal length?
Answer:
No, focal length of a mirror depends only on its radius of curvature (\(f = R/2\)) and is independent of the surrounding medium.
In simple words: Focal length of mirrors does not change in water.

Exam Tip: Unlike lenses (whose refraction depends on refractive index), mirrors rely on reflection, which is medium-independent.

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 produces erect, highly diminished images and provides a wide field of view.
In simple words: Convex mirror gives a wide view and fits tall objects into a small image.

Exam Tip: Convex mirrors are ideal for wide-angle viewing.

Q.5. What type of image is formed on a cinema screen?
Answer:
Real and inverted image (projected upside down by the projector lens, but viewed upright due to film orientation).
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 and the principal focus of the concave mirror to obtain an erect, magnified, and virtual image.
In simple words: Teeth should be placed within the focal length to see a magnified virtual image.

Exam Tip: Dentists use this setup for an enlarged view of teeth cavities.

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 \(2\theta\) (twice the angle of rotation of the mirror).
In simple words: Rotating a plane mirror by \(\theta\) shifts the reflected ray by \(2\theta\).

Exam Tip: Standard optical theorem: mirror rotation by \(\theta\) rotates 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) Concave mirror (object placed between F and C).
(ii) Concave mirror (object placed between P and F).
In simple words: Concave mirrors produce both real and virtual enlarged images depending on object distance.

Exam Tip: Concave mirrors are versatile and form all types of images (real/virtual, magnified/diminished).

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

Exam Tip: Convex mirrors only form virtual diminished images; concave form real diminished images beyond C.

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

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

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 a 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 = bends towards normal; Denser to rarer = bends away from 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 (\(f = +15 \text{ cm}\)).
In simple words: Positive focal length signifies convex lens and convex mirror.

Exam Tip: Convex lens and convex mirror have positive focal lengths; concave lens and concave mirror have negative 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 only when the object is placed within its focal length (\(u < f\)). Since object distance is 18 cm and it forms a virtual image, focal length must be greater than 18 cm (\(f > 18 \text{ cm}\)).
In simple words: For a virtual image in a convex lens, object must be within focal length (\(f > 18\text{ cm}\)).

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

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
Depending on object position, a convex lens can form real images (beyond F) or virtual images (within F).
In simple words: Convex lenses can form both real and virtual images.

Exam Tip: Convex lens is converging and forms real images mostly, except when object is very close (\(u < f\)).

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 form virtual and erect images.

Exam Tip: Concave lens behaves like a convex mirror in terms of image nature (always virtual, erect, 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 \(P = \frac{1}{f(\text{in meters})} \implies f = \frac{1}{+0.5} = +2 \text{ m}\). Positive power indicates a convex lens.
In simple words: Positive power indicates a convex lens with a 2-meter focal length.

Exam Tip: Power of convex lens is positive; power of concave lens is negative.

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, parallel light converges at its focus, meaning focal length \(f_{mirror} = 10 \text{ cm}\).
Radius of curvature \(R = 2 \times 10 = 20 \text{ cm}\).
Given that focal length of the convex lens equals the radius of curvature of the mirror, lens focal length \(f_{lens} = 20 \text{ cm}\).
Therefore, the convex lens converges parallel light at its principal focus, which is 20 cm away from the optical center.
In simple words: The lens converges the parallel beam at its focal length of 20 cm.

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

Q.2. Show by ray diagram the formation of images by a convex lens in any two cases.
Answer:
Case 1: Object at \(2F\) - forms a real, inverted image of the same size at \(2F\) on the other side.
Case 2: Object between optical center and focus - forms a virtual, erect, and magnified image on the same side.
In simple words: Diagrams show real image at \(2F\) and virtual magnified image when object is close to lens.

Exam Tip: Use arrows for light rays and label optical center (O), focus (F), and \(2F\).

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:
To the fish, the bird will appear to be flying at a distance greater than its actual height due to refraction of light going from rarer medium (air) to denser medium (water).
In simple words: Refraction makes the bird appear higher than it actually is to the fish.

Exam Tip: Objects in rarer medium appear raised when viewed from denser medium.

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 cm (iii) The magnification
Answer:
(i) The lens is a convex lens (only convex lenses form real images).
(ii) Given \(u = -20 \text{ cm}\) (at \(f\)), \(f = +20 \text{ cm}\).
Using lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
\(\frac{1}{20} = \frac{1}{v} - \frac{1}{-20} \implies \frac{1}{v} = \frac{1}{20} - \frac{1}{20} = 0 \implies v = \infty\).
(iii) Magnification \(m = \frac{v}{u} = \infty\) (infinitely large, real, and inverted image at infinity).
In simple words: Object at focus of convex lens forms a real image at infinity with infinite magnification.

Exam Tip: Object at focus (\(u = f\)) always 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. Computation of power of this lens.
Answer:
Myopia (short-sightedness): A defect where distant objects appear blurred because light focuses in front of the retina.
Correction: Corrected using a concave lens.
Given: Focal length \(f = -50 \text{ cm} = -0.5 \text{ m}\).
Power \(P = \frac{1}{f(\text{in meters})} = \frac{1}{-0.5} = -2.0 \text{ D}\).
In simple words: Myopia is short-sightedness corrected by a concave lens of -2.0 D power.

Exam Tip: Negative focal length and negative power always indicate a concave (remedial) lens for myopia.

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: \(h_o = 3 \text{ cm}\), \(u = -15 \text{ cm}\), \(f = +10 \text{ cm}\)
Using lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
\(\frac{1}{10} = \frac{1}{v} - \frac{1}{-15} \implies \frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = +\frac{1}{30} \implies v = +30 \text{ cm}\)
Magnification \(m = \frac{v}{u} = \frac{+30}{-15} = -2\)
Size \(h_i = m \cdot h_o = -2 \times 3 = -6 \text{ cm}\).
Nature: Real, inverted, and magnified (6 cm high).
In simple words: Image is formed at 30 cm, is real, inverted, and 6 cm in size.

Exam Tip: Negative sign of image height indicates a real and inverted image.

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:
Letters will appear much nearer (higher apparent shift) below the plastic cube because it has a higher refractive index than glass, causing greater refraction and upward shift of the image.
In simple words: Higher refractive index causes greater bending of light, making letters appear closer under the plastic cube.

Exam Tip: Apparent depth is inversely proportional to refractive index (\(\text{Apparent depth} = \frac{\text{Real depth}}{\mu}\)).

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 centre and converges parallel light rays to a single point. A concave lens is thinner at the centre and spreads out (diverges) parallel light rays.
In simple words: Convex lenses converge light rays together, while concave lenses spread them apart.

Exam Tip: Convex = Converging; Concave = Diverging.

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:
A thin lens behaves like a parallel glass slab at its optical centre, where emergent rays are parallel to incident rays with negligible lateral displacement.
In simple words: Light passing through the optical centre goes straight without bending.

Exam Tip: Optical centre is a key reference point in lens ray tracing where deviation is zero.

Q.4. The focal length of a convex lens is f. How does the size of 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 \(2f\) towards \(f\), the real image formed on the other side moves away from the lens and progressively increases in size (magnification increases from 1 to infinity).
In simple words: Moving the object closer to the focus makes the real image progressively larger.

Exam Tip: At \(2f\), image equals object size; as object approaches \(f\), image grows larger towards infinity.

H.O.T.S. Questions

Very Short Answer Type Question [1 mark]

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 in D (refractive index 1.33 - lowest optical density).
(ii) Slowest in C (refractive index 1.83 - highest optical density).
In simple words: Light travels fastest in the medium with the lowest refractive index (D) and slowest in the highest (C).

Exam Tip: Speed of light is inversely proportional to refractive index (\(v = c/n\)).

Q.2. When light undergoes refraction at the surface of separation of two media, what happens to its wavelength?
Answer:
When light enters a denser medium, its speed decreases, and since frequency remains constant, its wavelength decreases accordingly (\(\lambda = v/f\)).
In simple words: Wavelength changes during refraction as light speed changes, while frequency remains constant.

Exam Tip: Frequency never changes when light crosses a boundary between media.

Q.3. When does a ray of light neither refract nor deviate when passing through a glass slab?
Answer:
When light falls normally (at an angle of incidence \(i = 0^\circ\)) to the surface of the glass slab.
In simple words: Normal incidence results in zero deviation or refraction.

Exam Tip: Zero angle of incidence means ray passes straight through without bending.

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 (indicated by positive power sign).
In simple words: Positive power means it is a convex lens.

Exam Tip: Reading glasses correct hypermetropia (long-sightedness) using 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:
Sodium-vapour lamps emit monochromatic yellow light, so you will see only a single yellow band, not a spectrum of seven colors.
In simple words: Monochromatic sodium light produces only a yellow band through a prism.

Exam Tip: Dispersion into VIBGYOR requires polychromatic white light, not monochromatic light.

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, but the intensity (brightness) of the image will be reduced because fewer light rays pass through.
In simple words: The lens still forms a complete image, but it will be dimmer.

Exam Tip: Every part of a lens forms a complete image; covering part of it only dims the brightness.

Short Answer Type Question [2 marks]

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:
- Bird looking at fish: Fish appears deeper than it actually is.
- Fish looking at bird: Bird appears raised (closer/further away? higher up) than it actually is.
In simple words: Refraction makes underwater objects appear deeper and objects in air appear raised.

Exam Tip: Apparent depth rules depend on which medium the observer is in relative to the object.

Q.7. What happens to the speed of light when it undergoes refraction? Explain why?
Answer:
The speed of light changes (decreases in denser medium, increases in rarer medium) when undergoing refraction because optical density of the medium alters wave propagation speed.
In simple words: Light slows down when entering a denser medium due to optical interaction.

Exam Tip: Refractive index is defined as the ratio of speed of light in vacuum to speed in medium (\(n = c/v\)).

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) Refraction at the parallel air-glass and glass-air interfaces produces equal and opposite bending, making the emergent ray parallel to the incident ray.
(ii) The sideward shift (lateral displacement) occurs because the ray travels a finite distance obliquely inside the slab before emerging.
In simple words: Parallel faces cause equal opposite bending (parallel emergent ray) with a sideways lateral shift.

Exam Tip: Angle of incidence equals angle of emergence for a rectangular glass slab.

Value Based Question

Q.1. A spherical mirror in which reflecting surface is towards the centre... 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 focus).
(ii) Convex mirror has a virtual focus (reflected rays appear to diverge from it).
(iii) Staying focused implies concentrating our attention and energy towards our goals without getting distracted.
In simple words: Concave focus is real; convex focus is virtual. 'Staying focused' means maintaining concentration on goals.

Exam Tip: Real focus can be caught on a screen; virtual focus cannot.

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 an object can be seen clearly without strain by a normal human eye is 25 cm.
In simple words: Least distance of distinct vision is 25 cm.

Exam Tip: Also known as near point of the human eye.

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 adjust the curvature and focal length of the eye lens to focus on objects at varying distances (accommodation).
In simple words: Ciliary muscles change the shape of the eye lens to focus properly.

Exam Tip: Ciliary muscles relax for distant vision and contract for near vision.

Q.3. Persistence 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 stays on the retina for about \(\frac{1}{16}\text{th}\) of a second.
In simple words: Persistence of vision lasts for roughly 1/16th of a second.

Exam Tip: This principle is utilized in cinematography and moving pictures.

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 photoreceptors on the retina to the brain.
In simple words: Optic nerves carry visual signals from the retina to the brain.

Exam Tip: Blind spot has no photoreceptor cells, so no image is formed there.

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 transplantation is performed for corneal blindness.
In simple words: Cornea is the transparent tissue transplanted during eye donation.

Exam Tip: Cornea is avascular (has no blood vessels), making tissue rejection rare in transplants.

Q.6. Accommodation 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 human eye can focus clearly on objects ranging from 25 cm up to infinity.
In simple words: Normal eye range spans from 25 cm to infinity.

Exam Tip: Near point is 25 cm; far point is infinity for a normal eye.

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 VIBGYOR colors due to varying wavelengths and refractive indices.
In simple words: Splitting of white light into component colors is called dispersion.

Exam Tip: Violet deviates the most, and red deviates the least during 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 goggles detect infrared thermal radiation emitted or reflected by objects.
In simple words: Night vision devices detect infrared radiation.

Exam Tip: Infrared rays are invisible thermal radiation.

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 minimum refractive index in glass, hence it deviates the least.
In simple words: Red light undergoes the least deviation through a prism.

Exam Tip: Wavelength is inversely proportional to deviation angle.

Q.10. Which one of the following colour has maximum frequency
(i) Red
(ii) Violet
(iii) Green
(iv) Yellow
Answer: (ii) Violet
Violet light has the shortest wavelength and highest frequency in the visible spectrum.
In simple words: Violet light has the highest frequency.

Exam Tip: Frequency \(\nu = c/\lambda\); shorter wavelength means higher 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:
For a short-sighted (myopic) person, far point \(d = 1.5 \text{ m}\).
Focal length of remedial lens \(f = -d = -1.5 \text{ m}\).
Power \(P = \frac{1}{f(\text{in meters})} = \frac{1}{-1.5} = -0.67 \text{ D}\).
Nature: Concave lens.
In simple words: Remedial lens is concave with -1.5 m focal length and -0.67 D power.

Exam Tip: For myopia correction, focal length equals negative of the myopic person's far point.

Q.2. Complete the following figure.
Answer:
Parallel rays passing through a concave lens diverge such that their backward extensions pass through the principal focus.
In simple words: Rays diverging from a concave lens appear to come from the focus.

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 colors have different wavelengths. Since refractive index of glass varies with wavelength (Cauchy's dispersion), each color travels at a different speed and bends by a different angle.
In simple words: Different wavelengths bend by different angles due to varying refractive indices.

Exam Tip: Shorter wavelengths (violet) bend more than longer wavelengths (red).

Q.4. What is colour blindness?
Answer:
Colour blindness is a genetic visual defect where a person cannot distinguish between certain colors (typically red and green) due to defective cone cells in the retina.
In simple words: Inability to distinguish certain colors due to defective retinal cone cells.

Exam Tip: Rods respond to light intensity; cones respond to colors.

Q.5. What are Rods and Cones? State briefly the functions of Rod & Cones.
Answer:
Rods: Photoreceptor cells sensitive to dim light (intensity), responsible for night vision.
Cones: Photoreceptor cells sensitive to bright light and colors, responsible for color vision.
In simple words: Rods handle dim light vision, while cones handle color vision.

Exam Tip: Remember: Rods for Rods (dim light), Cones for Colors.

Q.6. What is the reason behind twinkling of stars? Explain.
Answer:
Twinkling of stars is caused by atmospheric refraction of starlight. As starlight passes through varying layers of the Earth's atmosphere with fluctuating density and temperature, light rays bend continuously, causing apparent position and brightness to flicker.
In simple words: Changing atmospheric density refracts starlight continuously, making them twinkle.

Exam Tip: Planets do not twinkle because they act as extended sources of light 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 of human eye:
- Cornea: Transparent outer membrane that refracts light.
- Pupil: Aperture regulating light entry.
- Iris: Colored diaphragm controlling pupil size.
- Crystalline Lens: Focuses image on retina.
- Retina: Light-sensitive screen containing rods and cones.
- Optic Nerve: Transmits electrical signals to brain.
In simple words: Cornea refracts, iris adjusts light, lens focuses, and retina captures the image.

Exam Tip: Practice labeling 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: Short-sightedness where distant objects cannot be seen clearly because light focuses in front of the retina.
Correction: A concave lens diverges incoming parallel rays so they appear to come from the person's far point, allowing the eye lens to focus them correctly on the retina.
In simple words: Myopia focuses images in front of the retina; a concave lens diverges rays to correct it.

Exam Tip: Always show focal point falling in front of the retina for uncorrected myopia.

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: Long-sightedness where nearby objects appear blurred because light focuses behind the retina.
Correction: Corrected using a converging convex lens that converges rays before entering the eye so they focus sharply on the retina.
In simple words: Hypermetropia focuses images behind the retina; a convex lens converges rays to correct it.

Exam Tip: Corrected with convex lens (positive power); myopia corrected with concave lens (negative power).

Q.4. Why is the colour of the clear sky blue?
Answer:
According to Rayleigh scattering, molecules in the atmosphere scatter shorter wavelengths of light (blue) much more efficiently than longer wavelengths (red), making the sky appear blue.
In simple words: Atmospheric molecules scatter blue light more than other colors, giving us a blue sky.
Exam Tip: Rayleigh scattering intensity is inversely proportional to the fourth power of wavelength (\(I \propto 1/\lambda^4\)).

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 alter the curvature and thickness of the eye lens.
In simple words: Ciliary muscles alter lens curvature to change focal length.

Exam Tip: Accommodation relies entirely on ciliary muscle flexibility.

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) Myopia (short-sightedness).
(ii) Hypermetropia (long-sightedness).
In simple words: (i) High converging power causes myopia. (ii) Large focal length causes hypermetropia.

Exam Tip: High power = short focal length = myopia; Low power = long focal length = hypermetropia.

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 (air molecules) scatter blue light.
(ii) Very large particles (dust, water droplets) scatter all wavelengths equally, making scattered light appear white (clouds).
In simple words: Small particles scatter blue light; large particles scatter all colors to appear white.

Exam Tip: Mie scattering applies to large particles (white clouds); Rayleigh scattering applies to small molecules (blue sky).

Q.4. Which property of the eye is used by cinematography to show motion pictures?
Answer:
Persistence of vision (\(\frac{1}{16}\text{th}\) of a second).
In simple words: Persistence of vision creates the illusion of smooth motion in movies.

Exam Tip: Successive still frames flashed faster than 16 frames per second appear as 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 very long distances.
In simple words: Red light scatters the least and travels the farthest through fog and smoke.

Exam Tip: Longest wavelength means minimum scattering (\(I \propto 1/\lambda^4\)).

Q.6. What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
Image distance in the eye remains constant because the distance between the eye lens and retina is fixed. Ciliary muscles adjust the focal length instead.
In simple words: Image distance stays constant; the eye adjusts its focal length for varying object distances.

Exam Tip: Unlike cameras where lens-to-film distance changes, human eye changes focal length via accommodation.

Short Answer Type Question [2 marks]

Q.1. Why do clouds look white?
Answer:
Clouds consist of water droplets and dust particles that are much larger than the wavelength of light. They scatter all wavelengths of white sunlight equally, producing white light.
In simple words: Large water droplets in clouds scatter all colors equally, making them appear white.

Exam Tip: Equal scattering of all colors results in white light (Mie scattering).

Q.2. Why is a normal eye not able to see clearly the objects closer than 25 cm?
Answer:
Because ciliary muscles cannot contract beyond their maximum limit to increase the curvature of the eye lens sufficiently to focus objects closer than 25 cm.
In simple words: Ciliary muscles reach their maximum limit of accommodation at 25 cm.

Exam Tip: 25 cm is the standard least distance of distinct vision for a normal healthy eye.

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

Exam Tip: Rayleigh scattering explains the blue sky; Mie scattering explains white clouds.

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

Exam Tip: Rainbow formation involves refraction, internal reflection, and dispersion combined.

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).
(ii) Bifocal lenses are required: upper part with concave lens for distant vision, lower part with convex lens for reading/near vision.
In simple words: Defect requires bifocal lenses to correct both near and distant vision limits.

Exam Tip: Bifocals are commonly prescribed for age-related vision defects combining myopia and hypermetropia.

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...
(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) Position 3 (blue, sky) and Position 5 (yellow/white, egg yolk) - statement is correct.
(a) Potassium permanganate solution is purple/violet (corresponds to position 7 or violet end).
(b) 'Danger' or 'stop' signal lights are red (corresponds to position 1).
In simple words: Position 3 is blue (sky), position 5 is yellow/white. Potassium permanganate is violet, and danger signals are red.

Exam Tip: VIBGYOR order: 1 is Red, 7 is Violet.

Q.2. A student finds the writing on the blackboard as blurred and unclear when sitting on the last desk in the classroom... 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 (short-sightedness) because he cannot see distant objects (blackboard from last desk) clearly.
Remedial lens: Concave lens.
Ray diagram: Diverges rays from distant blackboard so they appear to originate from his far point, enabling his eye lens to focus them sharply on the retina.
In simple words: Myopia prevents seeing distant blackboards; a concave lens corrects it.

Exam Tip: Short-sightedness = Myopia = Corrected with Concave lens.

Value Based Question

Q.1. Dolly and Ritu are two friends studying together in VI grade...
(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 (short-sightedness).
(ii) Cause: Excessive curvature of eye lens or elongation of eyeball.
(iii) Far point \(d = 50 \text{ cm} = 0.5 \text{ m}\). Focal length \(f = -50 \text{ cm} = -0.5 \text{ m}\).
Power \(P = \frac{1}{f(\text{in meters})} = \frac{1}{-0.5} = -2.0 \text{ D}\).
(iv) Values displayed by Ritu: Empathy, helpfulness, scientific temperament, and care for friends.
In simple words: Dolly has myopia, corrected with a -2.0 D concave lens. Ritu showed empathy and helpfulness.

Exam Tip: Value-based questions reward identifying both scientific calculations and positive human values.

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 \(V = IR\)).
In simple words: Ohm's law states that voltage is proportional to current (\(V \propto I\)).

Exam Tip: Always remember constant temperature condition for Ohm's law.

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).
In simple words: Voltage (V) and Current (I) are plotted to verify Ohm's law.

Exam Tip: Plot V on Y-axis and I on X-axis to get resistance as slope.

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
In simple words: A rheostat varies circuit current and voltage in Ohm's law experiments.

Exam Tip: Rheostat acts as a variable resistor.

Q.4. For a particular conducting wire what remains constant
i) V
ii) I
iii) R
iv) All the above.
Answer: iii) R
Resistance of a specific conducting wire remains constant at a given temperature.
In simple words: Resistance of a given metallic wire stays constant.

Exam Tip: V and I change when voltage supply is varied, but resistance R is fixed by wire dimensions.

Q.5. Which of the following correctly represents ohm's law experiment?
i) ... ii) ... iii) ... iv) ...
Answer: i) Straight line graph between V and I passing through origin.
In simple words: A straight line graph of V versus I proves Ohm's law.

Exam Tip: Linear proportionality always yields a straight line through the 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).
In simple words: Resistance depends on length, cross-sectional area, material, and temperature.

Exam Tip: Comprehensive formula: \(R = \rho \frac{l}{A}\).

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) \(R = \rho l/A\)
In simple words: Resistance formula is \(R = \rho \frac{l}{A}\).

Exam Tip: Rearranging gives resistivity \(\rho = \frac{RA}{l}\).

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.
Since \(R = \rho \frac{l}{A}\), if both length and area are doubled simultaneously, the ratio \(\frac{l}{A}\) remains unchanged, so resistances are equal.
In simple words: Doubling both length and area keeps resistance unchanged.

Exam Tip: Proportionality scaling: \(R_A = \rho \frac{2l}{2A} = \rho \frac{l}{A} = R_B\).

Q.9. The unit (s) of Resistivity is/are:
i) \(1/\Omega^{-1} \text{ m}^{-1}\)
ii) Volt meter ampere\(^{-1}\)
iii) \(\Omega \text{ m}\)
iv) All the above
Answer: iv) All the above
\(\Omega\cdot\text{m}\) is equivalent to Volt meter per Ampere (\(\frac{V}{I} \cdot \frac{A}{m} \dots\)).
In simple words: All listed units correctly represent resistivity.

Exam Tip: Dimensional consistency confirms equivalent units.

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 iii is also correct for metals, but ii is a primary dimensional factor).
In simple words: Resistance is inversely proportional to cross-sectional area.

Exam Tip: \(R \propto l\), \(R \propto 1/A\).

Q.11. What is the current in the following Resistance in the given diagram? (\(V = 24 \text{ V}\), \(R = 60 \, \Omega\))
i) 2.5 A
ii) 0.4 A
iii) 1440 A
iv) None of above
Answer: ii) 0.4 A
Current \(I = \frac{V}{R} = \frac{24}{60} = \frac{2}{5} = 0.4 \text{ A}\).
In simple words: Current equals 24 V divided by 60 ohms, which is 0.4 A.

Exam Tip: Direct application of Ohm's law (\(I = V/R\)).

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 length is doubled (\(l' = 2l\)) with constant volume, area becomes half (\(A' = A/2\)).
New resistance \(R' = \rho \frac{2l}{A/2} = 4 \left(\rho \frac{l}{A}\right) = 4R = 4 \times 10 = 40 \, \Omega\) (Wait! Original formula: when length is doubled by stretching, resistance increases by \(n^2\), so \(2^2 \times 10 = 40 \, \Omega\)).
Correct answer: iv) \(40 \, \Omega\).
In simple words: Stretching a wire to double its length multiplies resistance by \(n^2\) (4 times), yielding \(40 \, \Omega\).

Exam Tip: For stretched wires, remember shortcut: \(R' = n^2 R\).

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
In simple words: Lenses bend light via refraction.

Exam Tip: Lenses are transparent refracting media bounded by spherical surfaces.

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) \(\sin i / \sin r = \text{constant}\)
In simple words: Snell's law states that the ratio of sine of angle of incidence to sine of angle of refraction is constant.

Exam Tip: Constant represents refractive index (\(n_{21} = \frac{\sin i}{\sin r}\)).

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: ii) or i depending on notation, typically refractive index of glass with respect to air is written as \({}^{air}n_{glass}\) or \(n_g\).
In simple words: Refractive index of glass relative to air represents glass-air interface.

Exam Tip: Understand relative versus absolute refractive indices.

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) and iv are correct standard terms.
In simple words: P is optical center, LL' is aperture, and PF is focal length.

Exam Tip: Memorize all lens diagram labels and terminology.

Q.17. Give the nature of the image in the given figure.
i) Real, Erect, Enlarged.
ii) Virtual, Erect, Diminished
iii) Virtual, Erect, Enlarged
iv) Real, inverted, Enlarged.
Answer: iv) Real, inverted, Enlarged.
In simple words: Object placed between F and 2F of a convex lens produces a real, inverted, and enlarged image.

Exam Tip: Real images formed by lenses are always inverted.

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 project a magnified real image, object is placed between F and 2F.
In simple words: Place object between F and 2F to obtain an enlarged projected image.

Exam Tip: Projectors use this principle to cast magnified images on screens.

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 lens can form virtual images, erect images when used as magnifier, and corrects long-sightedness, not short-sightedness).
In simple words: All statements i, ii, iii are incorrect for convex lenses.

Exam Tip: Convex lens corrects hypermetropia (long-sightedness); concave lens corrects myopia (short-sightedness).

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) and iv are correct (Lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\) and magnification \(M = \frac{h_2}{h_1} = \frac{v}{u}\), so iii is missing a negative sign for mirrors but correct for lenses).
In simple words: Lens formula and magnification definitions are correctly stated in (i) and (iv).

Exam Tip: Note the minus sign in lens formula (\(1/v - 1/u\)) versus plus sign in mirror formula (\(1/v + 1/u\)).

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 (\(2F\)).
In simple words: Placing an object at \(2F\) in front of a convex lens produces an identical-sized real image at \(2F\) on the other side.

Exam Tip: Magnification at \(2F\) is exactly \(-1\).

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 virtual and erect images for any real object position.
In simple words: Concave lenses always produce virtual and erect images.

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

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 (yielding \(5 \, \Omega\)), and leave the third disconnected (or connect two in series and one in parallel... wait: two \(10 \, \Omega\) in parallel give \(5 \, \Omega\)).
In simple words: Connect two 10-ohm resistors in parallel to get 5 ohms.

Exam Tip: Parallel combination of two identical resistors halves their 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 even when dimensions and material remain constant.
In simple words: Change the wire's temperature to alter its resistance.

Exam Tip: Resistance varies directly with temperature for metallic conductors.

Q.25. Why do electricians wear rubber gloves or shoes while working with electricity?
Answer:
Rubber is an insulator with very high electrical resistance, preventing electric current from passing through the electrician's body to the ground, thus avoiding electric shocks.
In simple words: Rubber is an insulator that prevents electric shocks.

Exam Tip: Insulators protect against completion of electrical circuits through the human body.

Q.26. Give the final resistance of the given figure. (\(R = 1 \, \Omega\) network)
Answer:
Evaluating the bridge/series-parallel network of \(1 \, \Omega\) resistors yields an equivalent resistance of \(1 \, \Omega\).
In simple words: Equivalent resistance of the given symmetric network is \(1 \, \Omega\).

Exam Tip: Simplify complex circuits step-by-step by identifying series and parallel branches.

Q.27. In the given figure what is the potential difference of the battery? (\(I = 0.5 \text{ A}\), resistor values shown in circuit)
Answer:
Calculate total equivalent resistance \(R_{eq}\) and multiply by total current \(I = 0.5 \text{ A}\) using \(V = I \cdot R_{eq}\).
In simple words: Multiply total circuit resistance by current to get battery potential difference.

Exam Tip: Ohm's law applies to entire circuits using total equivalent resistance and total current.

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 simple words: Voltage remains identical across all branches in a parallel circuit.

Exam Tip: Parallel = Equal voltage; Series = Equal current.

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 cooled below \(4.2\text{ K}\).
In simple words: Mercury was the first metal in which superconductivity was discovered.

Exam Tip: Superconductivity is the zero electrical resistance exhibited by certain materials at very low temperatures.

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:
Given: Diameter \(d = 0.5 \text{ mm} = 5 \times 10^{-4} \text{ m} \implies \text{Radius } r = 2.5 \times 10^{-4} \text{ m}\)
Area \(A = \pi r^2 = \pi (2.5 \times 10^{-4})^2 = 1.96 \times 10^{-7} \text{ m}^2\)
Resistivity \(\rho = 1.6 \times 10^{-6} \, \Omega\cdot\text{m}\), Resistance \(R = 10 \, \Omega\)
Using \(R = \rho \frac{l}{A} \implies l = \frac{R \cdot A}{\rho} = \frac{10 \times 1.96 \times 10^{-7}}{1.6 \times 10^{-6}} \approx 1.23 \text{ meters}\).
In simple words: Substitute values into \(R = \rho \frac{l}{A}\) to find length \(l \approx 1.23 \text{ m}\).

Exam Tip: Always convert all units to SI (meters, ohms) 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:
Paints and sketch pens work on subtractive color mixing (pigments absorb light). Combining all pigment colors absorbs all wavelengths of incident light, reflecting none, resulting in black.
In simple words: Mixing pigment colors absorbs all light, appearing black, unlike mixing light rays which produces white.

Exam Tip: Pigment mixing is subtractive; light mixing is additive.

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).
In simple words: Total internal reflection occurs when light travels from denser to rarer medium and angle of refraction reaches \(90^\circ\).

Exam Tip: Conditions for TIR: Light must travel from denser to rarer medium, and angle of incidence must exceed critical angle.

Q.33. If I have two lenses of powers \(P_1 = -3D\) and \(P_2 = +5D\). 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 (zero net power), total power \(P_{net} = 0\).
\(P_1 + P_2 + P_3 = 0\)
\(-3 + 5 + P_3 = 0 \implies +2 + P_3 = 0 \implies P_3 = -2 \text{ D}\).
In simple words: Net power must be zero, so the third lens must have a power of -2D.

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 used to correct myopia (short-sightedness).
In simple words: Negative power means a concave lens 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 slab thickness, angle of incidence, and refractive index.
In simple words: Thicker slabs, higher incidence angles, and greater refractive index all increase lateral displacement.

Exam Tip: More refraction and thicker glass result in greater sideways shift.

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?
Answer:
Speed in glass \(v = \frac{c}{\mu} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \text{ m/s}\).
Frequency \(f = \frac{c}{\lambda_{air}} = \frac{3 \times 10^8}{500 \times 10^{-9}} = 6 \times 10^{14} \text{ Hz}\) (remains constant in glass).
Wavelength in glass \(\lambda_{glass} = \frac{\lambda_{air}}{\mu} = \frac{500 \text{ nm}}{1.5} = 333.33 \text{ nm}\).
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 never changes when light enters a new medium; speed and wavelength decrease by factor \(\mu\).

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

Exam Tip: All EM spectrum waves travel at \(c = 3 \times 10^8 \text{ m/s}\) in 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
In simple words: Optical fibres guide light via total internal reflection.

Exam Tip: Core has higher refractive index than cladding to ensure total internal reflection.

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.
In simple words: Electric current in metals is carried by free electrons.

Exam Tip: Electrons drift from negative to positive terminal under electric fields.

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) \(W = Irt\) is incorrect (correct formula is \(W = VIt = I^2Rt\)).
In simple words: Formula (iv) is incorrect; all others are standard electrical formulas.

Exam Tip: Memorize power and work formulas: \(P = VI = I^2R = V^2/R\).

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 iv is also numerically equal since watt second = joule).
In simple words: 1 kWh equals \(3.6 \times 10^6 \text{ Joules}\).

Exam Tip: \(1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}\).

Q.42. A current of 4 A flows through a 12 V car headlight for 10 minutes. How much energy transfer takes place?
Answer:
Given: \(I = 4 \text{ A}\), \(V = 12 \text{ V}\), \(t = 10 \text{ minutes} = 600 \text{ seconds}\)
Energy \(E = V \cdot I \cdot t = 12 \text{ V} \times 4 \text{ A} \times 600 \text{ s} = 28,800 \text{ Joules}\) (\(28.8 \text{ kJ}\)).
In simple words: Energy transferred is voltage times current times time, equal to 28,800 Joules.

Exam Tip: Always convert minutes to seconds before calculating energy in Joules.

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 \(P_{total} = P_1 + P_2 = 100 + 60 = 160 \text{ W}\).
Supply voltage \(V = 220 \text{ V}\).
Total current \(I = \frac{P_{total}}{V} = \frac{160 \text{ W}}{220 \text{ V}} = \frac{8}{11} \text{ A} \approx 0.73 \text{ A}\).
In simple words: Total power is 160W, drawing 0.73A current at 220V.

Exam Tip: In parallel, powers add up directly to find total power consumption.

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:
Given: Charge \(Q = 10^{-5} \text{ C}\), Work \(W = 2 \times 10^3 \text{ J}\)
Potential \(V = \frac{W}{Q} = \frac{2 \times 10^3}{10^{-5}} = 2 \times 10^8 \text{ Volts}\).
In simple words: Potential is work divided by charge, equal to \(2 \times 10^8 \text{ V}\).

Exam Tip: Formula \(V = W/Q\) defines electric potential.

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 in meters).
In simple words: Power of a lens is the reciprocal of its focal length in meters.

Exam Tip: SI unit of power is Dioptre (\(\text{D} = \text{m}^{-1}\)).

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: \(h_o = 4 \text{ cm}\), \(u = -27 \text{ cm}\), \(f = +18 \text{ cm}\)
Using lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
\(\frac{1}{18} = \frac{1}{v} - \frac{1}{-27} \implies \frac{1}{v} = \frac{1}{18} - \frac{1}{27} = \frac{3 - 2}{54} = +\frac{1}{54} \implies v = +54 \text{ cm}\)
Magnification \(m = \frac{v}{u} = \frac{54}{-27} = -2\)
Image size \(h_i = -2 \times 4 = -8 \text{ cm}\).
Nature: Real, inverted, and magnified (8 cm high) at 54 cm on the other side of the lens.
In simple words: Real, inverted, and magnified image is formed at 54 cm.

Exam Tip: Sign conventions for lenses: \(f\) is positive for convex, \(u\) is always negative.

Q.47. Why do diamonds sparkle?
Answer:
Diamonds sparkle due to total internal reflection. A diamond has a very high refractive index (2.42) and a very small critical angle (\(24.4^\circ\)). Once light enters, it undergoes multiple total internal reflections until it emerges from specific facets, causing intense brilliance.
In simple words: High refractive index and small critical angle cause multiple total internal reflections, making diamonds sparkle.

Exam Tip: Small critical angle ensures most entering light gets trapped and internally reflected.

Q.48. Why do stars twinkle?
Answer:
Starlight undergoes continuous atmospheric refraction through varying layers of fluctuating density and temperature in the Earth's atmosphere, causing apparent position and brightness to flicker.
In simple words: Atmospheric refraction of starlight causes twinkling.

Exam Tip: Planets do not twinkle because they are closer and appear as extended sources.

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 \(E_{total} = 5 \times 1.5 \text{ V} = 7.5 \text{ V}\)
Total internal resistance \(r_{total} = 0.2 + 0.3 + 0.4 + 0.5 + 1.2 = 2.6 \, \Omega\)
External resistance \(R = 10 \, \Omega\)
Total resistance \(R_{total} = R + r_{total} = 10 + 2.6 = 12.6 \, \Omega\)
Current \(I = \frac{E_{total}}{R_{total}} = \frac{7.5}{12.6} \approx 0.595 \text{ A}\).
In simple words: Total voltage is 7.5V and total resistance is 12.6 ohms, yielding 0.595A current.

Exam Tip: In series, internal resistances add up along with external resistance.

Q.50 In q. no. 49 if the cells are connected in parallel along with the \(10 \, \Omega\) resistance, then what will be the current?
Answer:
Note: Connecting identical cells in parallel maintains voltage at 1.5 V, but mixing cells with unequal internal resistances in parallel is complex. Assuming equivalent ideal parallel EMF of 1.5 V and parallel internal resistance calculation, current is evaluated using \(I = \frac{E}{R + r_{eq}}\).
In simple words: Parallel connection maintains 1.5 V total voltage and lowers internal resistance.

Exam Tip: Parallel cells share the load, increasing current capacity for identical cells.

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.
\(\text{Pb(NO}_3)_2(aq) + 2\text{KI}(aq) \longrightarrow \text{PbI}_2(\downarrow) + 2\text{KNO}_3(aq)\)
(ii) Effervescence of hydrogen gas is observed.
\(\text{Zn}(s) + 2\text{HCl}(dil) \longrightarrow \text{ZnCl}_2(aq) + \text{H}_2(\uparrow)\)
(iii) Slaked lime is formed with evolution of a large amount of heat (exothermic).
\(\text{CaO}(s) + \text{H}_2\text{O}(l) \longrightarrow \text{Ca(OH)}_2(aq) + \text{Heat}\)
(iv) Lime water turns milky due to formation of insoluble calcium carbonate.
\(\text{Ca(OH)}_2(aq) + \text{CO}_2(g) \longrightarrow \text{CaCO}_3(\downarrow) + \text{H}_2\text{O}(l)\)
In simple words: Chemical reactions produce precipitates, gases, or temperature changes with balanced equations.

Exam Tip: Always balance chemical equations and mention physical state symbols (\(s, l, g, aq\)).

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) Reaction of zinc with dilute sulfuric acid (exothermic temperature rise):
\(\text{Zn} + \text{H}_2\text{SO}_4 \longrightarrow \text{ZnSO}_4 + \text{H}_2 \uparrow + \text{Heat}\)
(ii) Precipitation of barium sulphate when barium chloride reacts with sodium sulphate:
\(\text{BaCl}_2(aq) + \text{Na}_2\text{SO}_4(aq) \longrightarrow \text{BaSO}_4(\downarrow) + 2\text{NaCl}(aq)\)
In simple words: (i) Zinc and acid release heat. (ii) Barium chloride and sodium sulphate form a white precipitate.

Exam Tip: Characteristics of chemical reactions include evolution of gas, change in color, change in temperature, and formation of precipitate.

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 number of atoms of each element on both sides.

Exam Tip: Balance metal atoms first, then non-metals, oxygen, and hydrogen last.

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:
\(\text{FeSO}_4(aq) + 2\text{NaOH}(aq) \longrightarrow \text{Fe(OH)}_2(\downarrow) + \text{Na}_2\text{SO}_4(aq)\)
In simple words: Ferrous sulphate and sodium hydroxide react to form a green ferrous hydroxide precipitate and sodium sulphate.

Exam Tip: Ferrous hydroxide precipitate is dirty green in color.

Q.5. What information does the following chemical equation convey?
(i) \(\text{CO}(g) + 2\text{H}_2(g) \xrightarrow[ZnO+CrO_3]{340^\circ\text{C}} \text{CH}_3\text{OH}(l)\)
(ii) \(\text{AgNO}_3(aq) + \text{NaCl}(aq) \longrightarrow \text{AgCl}(s) + \text{NaNO}_3(aq)\)
Answer:
(i) Carbon monoxide reacts with hydrogen gas at \(340^\circ\text{C}\) in the presence of \(\text{ZnO-CrO}_3\) catalyst to form liquid methanol. Catalyst, temperature, and physical states are specified.
(ii) Aqueous silver nitrate reacts with aqueous sodium chloride to form a white precipitate of silver chloride and aqueous sodium nitrate.
In simple words: Equations convey reactants, products, physical states, catalysts, and reaction conditions.

Exam Tip: Mentioning temperature, pressure, and catalyst above the arrow is essential in chemical 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.
- Endothermic: (ii) Electrolysis of water, (iii) Photosynthesis, (v) Decomposition of calcium carbonate.
In simple words: Exothermic reactions release heat; endothermic reactions absorb heat or energy.

Exam Tip: Respiration is exothermic because it releases energy for bodily functions.

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 decomposition into silver metal and chlorine gas.
\(2\text{AgCl}(s) \xrightarrow{\text{Sunlight}} 2\text{Ag}(s) + \text{Cl}_2(g)\)
Use: Used in black-and-white photography.
In simple words: Sunlight decomposes silver chloride into silver and chlorine; used in photography.

Exam Tip: This is a photochemical decomposition reaction stored in dark colored bottles.

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 out from hydrated copper sulphate crystals upon heating.
In simple words: Blue crystals turn white on heating as water of crystallization evaporates into droplets.

Exam Tip: Adding water to anhydrous copper sulphate turns it blue again, testing for moisture.

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 (\(\text{FeSO}_4\cdot 7\text{H}_2\text{O}\)) or ferrous sulphate crystals.
(ii) Thermal decomposition reaction.
(iii) \(2\text{FeSO}_4(s) \xrightarrow{\Delta} \text{Fe}_2\text{O}_3(s) + \text{SO}_2(g) + \text{SO}_3(g)\)
In simple words: Green ferrous sulphate decomposes thermally into black ferric oxide and sulphur gases.

Exam Tip: Recognize the characteristic suffocating odor of burning sulphur dioxide gas.

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 (\(\text{NO}_2\)).
(iii) \(2\text{Pb(NO}_3)_2(s) \xrightarrow{\Delta} 2\text{PbO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)\)
In simple words: Heating lead nitrate yields yellow lead oxide and brown nitrogen dioxide fumes.

Exam Tip: Lead oxide is yellow when hot and cold; brown fumes of \(\text{NO}_2\) are characteristic.

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: \(\text{HCl}\) (loses hydrogen / gets chlorinated); Reduced: \(\text{MnO}_2\) (loses oxygen).
(ii) Oxidized: \(\text{H}_2\text{S}\) (loses hydrogen); Reduced: \(\text{SO}_2\) (loses oxygen).
(iii) Oxidized: \(\text{PbS}\) (gains oxygen); Reduced: \(\text{H}_2\text{O}_2\) (loses oxygen / gains hydrogen).
In simple words: Oxidation involves loss of hydrogen or gain of oxygen; reduction involves loss of oxygen or gain of hydrogen.

Exam Tip: Remember OIL RIG: Oxidation Is Loss (of electrons/hydrogen), Reduction Is Gain.

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 common antioxidants added to fatty foods to prevent spoilage.

Exam Tip: Antioxidants prevent oxidation of fats, which causes unpleasant smell and taste (rancidity).

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: \(\text{H}_2\text{S}\) (loses hydrogen).
(ii) Reduced: \(\text{Cl}_2\) (gains hydrogen).
In simple words: \(\text{H}_2\text{S}\) is oxidized to sulphur, and \(\text{Cl}_2\) is reduced to HCl.

Exam Tip: Addition of electronegative element or removal of hydrogen is oxidation.

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 is Silver (\(\text{Ag}\)). Chloride is Silver Chloride (\(\text{AgCl}\)).
Type of reaction: Photochemical decomposition reaction.
In simple words: Metal X is silver, and its chloride decomposes in sunlight into grey silver.

Exam Tip: Silver is widely used in coinage metals and photography.

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 is Silver Nitrate (\(\text{AgNO}_3\)).
Yellow precipitate B is Silver Bromide (\(\text{AgBr}\)).
Reaction: \(\text{AgNO}_3(aq) + \text{NaBr}(aq) \longrightarrow \text{AgBr}(\downarrow) + \text{NaNO}_3(aq)\).
In simple words: A is silver nitrate, B is silver bromide (photosensitive yellow precipitate).

Exam Tip: Both \(\text{AgCl}\) and \(\text{AgBr}\) are photosensitive salts used in photography.

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