CBSE Class 12 Physics Short Answer Question Bank Worksheet Set 01

CBSE Class 12 Physics Short Answer Question Bank (1).The Short Answer questions in the worksheets have been specifically designed by best Physics teachers so that the students can practice them to clear their Short Answer concepts and get better marks in class 12 Physics tests and examinations. Students can free download these Short Answer Question Bank worksheets in pdf and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the Short Answer questions chapter and other subjects too. Use them for better understanding of the subjects.

1. Name the physical quantity whose S.I. unit is J/C. Is it a scalar or a vector quantity?

Ans: (i)electric potential (P.d), (ii) scalar quantity

2. How does a torque affect the dipole in an electric field?

Ans: Torque to align the dipole along the electric field

3. What is the work done in moving a charge 10 nC Between two point on equipotential surface?

Ans: No work is done to move charge on equipotential surface.

Electric Charge and Electric Field

Question 1. Name the physical quantity whose S.I. unit is J/C. Is it a scalar or a vector quantity?
Answer: The physical quantity with the S.I. unit of Joules per Coulomb (J/C) is the electric potential (or potential difference). It is a scalar quantity.
In simple words: Electric potential represents the amount of work needed to move a unit charge, and since it only has size without direction, it is a scalar.

Exam Tip: Always specify both the name of the physical quantity and its scalar/vector status clearly. Remember that \( 1\text{ J/C} = 1\text{ V} \).

Question 2. How does a torque affect the dipole in an electric field?
Answer: The torque acting on an electric dipole in an external electric field works to rotate and align the dipole parallel to the direction of the electric field.
In simple words: The twisting force aligns the positive and negative ends of the dipole along the direction of the electric field lines.

Exam Tip: Be sure to write the formula \( \vec{\tau} = \vec{p} \times \vec{E} \) to support your answer, showing that the torque is zero when the angle \( \theta = 0^\circ \).

Question 3. What is the work done in moving a charge 10 nC Between two point on equipotential surface?
Answer: The work done is zero. Since the electric potential is identical at all points on an equipotential surface, the potential difference between the two points is zero. Consequently, no work is required to move a charge across it.
In simple words: Because the voltage is exactly the same everywhere on this surface, there is no change in electrical pressure to fight against, so moving a charge takes zero effort.

Exam Tip: Use the formula \( W = q \cdot \Delta V \) to show that since \( \Delta V = 0 \), the work done \( W \) must be zero.

Question 4. Calculate the capacity of sphere of radius of 10km.
Answer: The capacitance \( C \) of a spherical conductor of radius \( r \) is calculated using the formula:
\( C = 4\pi\varepsilon_0 r \)
Given \( r = 10\text{ km} = 10^4\text{ m} \):
\( C = \frac{10^4}{9 \times 10^9} \approx 1.1 \times 10^{-6}\text{ F} = 1.1\text{ }\mu\text{F} \)
Therefore, the capacity of the sphere is \( 1.1\text{ }\mu\text{F} \).
In simple words: To find the capacitance, we multiply the sphere's radius by a constant. A 10-kilometer sphere can hold 1.1 microfarads of electrical capacity.

Exam Tip: Convert kilometers to meters first. Don't forget that \( \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2\text{/C}^2 \).

Question 5. Is the force acting between two point electric charges q1 and q2, kept at some distance apart in air, attractive or repulsive, when (i) q1q2 > 0 (ii) q1q2 < 0 ?
Answer: (i) When \( q_1 q_2 > 0 \), both charges are of the same sign (either both positive or both negative). Thus, the force acting between them is repulsive.
(ii) When \( q_1 q_2 < 0 \), the charges are of opposite signs (one is positive, and the other is negative). Hence, the force acting between them is attractive.
In simple words: If multiplying the two charges gives a positive result, they are like charges and push each other away. If it gives a negative result, they are unlike charges and pull together.

Exam Tip: Always clearly label sections (i) and (ii) in your response to keep it organized and easy for the examiner to grade.

Question 6. You are given three capacitors of value 2µF, 3µF, 6µF. How will you connect them to a resultant capacity of 4µF?
Answer: To obtain an equivalent capacitance of \( 4\text{ }\mu\text{F} \), connect the \( 3\text{ }\mu\text{F} \) and \( 6\text{ }\mu\text{F} \) capacitors in series, and then connect this combination in parallel with the \( 2\text{ }\mu\text{F} \) capacitor.
The series capacitance of \( 3\text{ }\mu\text{F} \) and \( 6\text{ }\mu\text{F} \) is:
\( C_s = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\text{ }\mu\text{F} \)
Adding this in parallel to the remaining \( 2\text{ }\mu\text{F} \) capacitor yields:
\( C_p = C_s + 2\text{ }\mu\text{F} = 2\text{ }\mu\text{F} + 2\text{ }\mu\text{F} = 4\text{ }\mu\text{F} \).
In simple words: Put the 3 and 6 microfarad capacitors in series to make them behave like a single 2 microfarad capacitor. Then add the other 2 microfarad capacitor in parallel to get 4 microfarads in total.

Exam Tip: Show the step-by-step calculations for both the series and parallel parts of the circuit to secure full marks.

Question 7. An uncharged insulated conductor A is brought near a charge insulated conductor B what happens to charge and potential of B?
Answer: (i) The net charge on conductor B remains exactly the same.
(ii) The electric potential of conductor B decreases. This happens because conductor B induces an opposite charge on the closer side of the uncharged conductor A, which acts to lower the potential of conductor B.
In simple words: Conductor B keeps its original charge, but its voltage drops because it induces an opposite electrical charge on conductor A nearby, drawing some electric potential away.

Exam Tip: Emphasize that electrostatic induction causes this reduction in potential without any actual transfer of physical charge between the objects.

Question 8. Find the ratio of potential difference that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitance in the ratio 1:3 so that the energy stored in the two cases is same.
Answer: Let the capacitances be \( C_1 = x \) and \( C_2 = 3x \).
For the parallel combination:
\( C_p = C_1 + C_2 = x + 3x = 4x \)
For the series combination:
\( C_s = \frac{C_1 C_2}{C_1 + C_2} = \frac{x \cdot 3x}{x + 3x} = \frac{3x}{4} \)
Since the stored energy is the same in both configurations:
\( U_p = U_s \)
\( \implies \frac{1}{2} C_p V_p^2 = \frac{1}{2} C_s V_s^2 \)
\( \implies \frac{V_p^2}{V_s^2} = \frac{C_s}{C_p} = \frac{3x/4}{4x} = \frac{3}{16} \)
\( \implies \frac{V_p}{V_s} = \frac{\sqrt{3}}{4} \)
The required ratio of potential differences is \( \sqrt{3} : 4 \).
In simple words: By equating the energy stored in both setups and substituting the values of series and parallel equivalent capacitances, we find the ratio of voltages is square root of 3 over 4.

Exam Tip: Be sure to write down the formula for energy stored in a capacitor (\( U = \frac{1}{2} C V^2 \)) at the beginning of your calculation.

Question 9. (i) Can two equipotential surfaces intersect each other? Give reasons.
(ii) Two charges −q and +q are located at points A (0, 0, −a) and B (0, 0, +a) respectively. How much work is done in moving a test charge from point P (7, 0, 0) to Q (−3, 0, 0)?
Answer: (i) No, two equipotential surfaces can never intersect. If they did, there would be two different values of electric potential at the point of intersection, leading to two different directions of the electric field at that single point, which is physically impossible.
(ii) Zero work is done. The points P (7, 0, 0) and Q (-3, 0, 0) both lie on the x-axis, which is the equatorial plane (perpendicular bisector) of the dipole. The electric potential is zero at every point on this equatorial plane, meaning \( V_P = V_Q = 0 \). Hence, the potential difference \( \Delta V = 0 \), so the work done \( W = q \cdot \Delta V = 0 \).
In simple words: (i) Electric field lines cannot point in two directions at once, so voltage lines can never cross. (ii) No work is needed because the two starting and ending points are along a center line where positive and negative voltages completely balance out.

Exam Tip: State clearly that the electric field is perpendicular to equipotential surfaces. This principle is key to answering part (i) successfully.

Question 10. A spherical Gaussian surface encloses a charge of 8.85 × 10-10C.
(i) Calculate the electric flux passing through the surface.
(ii) How would the flux change if the radius of the Gaussian surface is doubled and why?
Answer: (i) According to Gauss's Law, the total electric flux \( \Phi \) is given by:
\( \Phi = \frac{q}{\varepsilon_0} \)
Given \( q = 8.85 \times 10^{-10}\text{ C} \) and \( \varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2} \):
\( \Phi = \frac{8.85 \times 10^{-10}}{8.85 \times 10^{-12}} = 100\text{ N m}^2\text{/C} \)
(ii) The electric flux remains unchanged if the radius of the Gaussian surface is doubled. This is because the electric flux depends only on the total charge enclosed within the surface, which does not change.
In simple words: (i) The total number of electric field lines passing out of the sphere is 100. (ii) Making the sphere twice as big doesn't change this number because the total charge trapped inside is still the same.

Exam Tip: Remember that Gauss's Law is independent of the size and shape of the closed boundary. Clearly state this fact to gain full marks.

Question 11. Three charges –q, +Q and –q are placed at equal distance on a straight line. If the potential energy of the system of three charges is zero, find the ratio of Q:q
Answer: Let the three charges \( -q \), \( +Q \), and \( -q \) be placed at equal distances \( r \) along a line (at points A, B, and C respectively). The distance between the two end charges is \( AC = 2r \).
The total electrostatic potential energy \( U \) of the system is:
\( U = \frac{1}{4\pi\varepsilon_0} \left[ \frac{(-q)(Q)}{r} + \frac{(Q)(-q)}{r} + \frac{(-q)(-q)}{2r} \right] \)
Since the potential energy of the system is zero:
\( \frac{1}{4\pi\varepsilon_0} \left[ \frac{-qQ}{r} - \frac{qQ}{r} + \frac{q^2}{2r} \right] = 0 \)
\( \implies -2 \frac{qQ}{r} + \frac{q^2}{2r} = 0 \)
\( \implies 2 \frac{qQ}{r} = \frac{q^2}{2r} \)
\( \implies 2Q = \frac{q}{2} \)
\( \implies \frac{Q}{q} = \frac{1}{4} \)
Thus, the ratio of \( Q : q \) is \( 1 : 4 \).
In simple words: We calculate and add up the mutual potential energy of all three pairs of charges. Setting this sum to zero shows that charge Q must be one-fourth the size of charge q.

Exam Tip: Be careful with the distance between the two end charges; it is \( 2r \), not \( r \). This is a common place where students make errors.

Current Electricity

Question 1. How does the drift velocity of electrons in a metallic conductor vary with increase in temperature?
Answer: As the temperature of a metallic conductor increases, the thermal vibrations of its lattice ions increase. This causes more frequent collisions of free electrons, which decreases the relaxation time (\( \tau \)). Since drift velocity is given by \( v_d = \frac{e E \tau}{m} \), the drift velocity decreases with rising temperature.
In simple words: Heating a metal makes its atoms shake faster. This gets in the way of flowing electrons, causing them to crash more often and slow down.

Exam Tip: Mention the relation \( v_d \propto \tau \) explicitly to show the direct mathematical link between relaxation time and drift speed.

Question 2. The colours of four bands are yellow, violet, brown and gold. What is the resistance with tolerance limit.
Answer: According to the standard resistor color code:
- Yellow represents 4
- Violet represents 7
- Brown represents a multiplier of \( 10^1 = 10 \)
- Gold represents a tolerance of \( \pm 5\% \)
Combining these values, the resistance is:
\( R = 47 \times 10^1\text{ }\Omega \pm 5\% = 470\text{ }\Omega \pm 5\% \).
In simple words: The colored stripes tell us the resistance is 470 ohms, with an expected margin of error of plus or minus 5 percent.

Exam Tip: Use the standard sentence 'B B ROY of Great Britain had a Very Good Wife' to recall color codes quickly under exam pressure.

Question 3. How does the relaxation time of electron in the conductor change when temperature of the conductor decreases ?
Answer: When the temperature of a conductor decreases, the thermal vibrations of its metal ions are reduced. As a result, the free electrons experience fewer collisions, which increases the relaxation time (\( \tau \)).
In simple words: Cooling a conductor slows down the vibration of its atoms, making it easier for electrons to slide past with fewer crashes, so the time between collisions increases.

Exam Tip: State the relation \( \tau \propto \frac{1}{T} \) to explain that relaxation time is inversely proportional to temperature.

Question 4. Two wire of equal length one copper and manganin have same resistance , which wire is thicker?
Answer: The resistance of a wire is given by \( R = \rho \frac{l}{A} \). For wires of equal length (\( l \)) and equal resistance (\( R \)), the cross-sectional area is directly proportional to the resistivity (\( A \propto \rho \)). Since manganin has a higher resistivity than copper (\( \rho_{\text{manganin}} > \rho_{\text{copper}} \)), the manganin wire must have a larger cross-sectional area. Therefore, the manganin wire is thicker.
In simple words: Manganin naturally resists electricity more than copper. To make their total resistance equal, the manganin wire must be made wider to let current pass through more easily.

Exam Tip: Always state the resistance formula first and explain the proportionality of the variables to build a strong logical argument.

Question 5. In the given graph of voltage vs current for a semiconductor,Identify the negative resistance region.
Answer: In the voltage-current characteristic curve, the negative resistance region is represented by the segment BC. In this region, an increase in voltage causes a decrease in current, yielding a negative dynamic resistance (\( r = \frac{\Delta V}{\Delta I} < 0 \)).
In simple words: The BC section is the negative resistance area. In this part, as voltage goes up, the current actually drops instead of rising.

Exam Tip: Clearly point out that negative resistance is defined as a negative slope on a voltage-current characteristic graph.

Question 6. Establish a relation between current and drift velocity.
Answer: Let \( l \) be the length of a conductor and \( A \) be its cross-sectional area. If \( n \) is the number density of free electrons, the total number of free electrons in the conductor is \( N = n \cdot A \cdot l \).
The total charge \( q \) of these electrons is:
\( q = N \cdot e = n A l e \)
The time \( t \) taken by the electrons to cross the length of the conductor with drift velocity \( v_d \) is:
\( t = \frac{l}{v_d} \)
The electric current \( I \) flowing through the conductor is:
\( I = \frac{q}{t} = \frac{n A l e}{l / v_d} = n e A v_d \)
Thus, the relation between current and drift velocity is \( I = n e A v_d \).
In simple words: Current is the total moving charge divided by time. By expressing the total charge of free electrons and their transit time in terms of drift speed, we show that current equals the density, charge, area, and drift velocity multiplied together.

Exam Tip: Define each variable clearly at the beginning of your derivation, and write down the final expression with its standard units.

Question 7. Define resistivity of a conductor. Draw the variation of resistivity versus temperature for (i) Nichrome (ii) Silicon
Answer: The resistivity (\( \rho \)) of a material is defined as the resistance of a conductor made of that material with a unit length and a unit cross-sectional area.
(i) For Nichrome (an alloy), resistivity is high even at low temperatures and increases almost linearly with temperature.
(ii) For Silicon (a semiconductor), resistivity decreases exponentially as the temperature increases due to the liberation of more charge carriers.

In simple words: Resistivity is a material's natural opposition to electrical flow. Heating up a metal alloy like nichrome increases its resistance, but heating up a semiconductor like silicon makes its resistance go down.

Exam Tip: When drawing the Nichrome graph, make sure the line does not touch the origin, as alloys have a high residual resistivity even at absolute zero.

Question 8. What happens to the drift velocity (vd) of electrons and to the resistance R if length of a conductor is doubled (keeping potential difference unchanged)? Justify.
Answer: (i) The drift velocity is halved. Since \( v_d = \frac{e V \tau}{m l} \) and the potential difference \( V \) remains constant, drift velocity is inversely proportional to length (\( v_d \propto \frac{1}{l} \)).
(ii) If the length of the conductor is doubled by stretching it, its volume remains constant. This means when the length is doubled (\( l' = 2l \)), its cross-sectional area is halved (\( A' = \frac{A}{2} \)). The new resistance becomes:
\( R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4 R \).
Thus, the resistance increases to four times its original value.
In simple words: Because the voltage stays the same, doubling the length reduces the internal electrical pull, halving the speed of electrons. Stretching the wire also makes it thinner, which combines to make the overall resistance four times greater.

Exam Tip: Pay attention to whether the length is doubled by "stretching" (which changes the area) or by "adding" another piece of wire (where area stays constant).

Question 9. Draw a plot showing the variation of terminal voltage (V) vs the current (I) drawn from the cell. Using this plot, how does one determine the internal resistance of the cell ?
Answer: The terminal potential difference \( V \) and current \( I \) of a discharging cell are related by the equation:
\( V = E - I r \)
where \( E \) is the emf and \( r \) is the internal resistance. A plot of \( V \) versus \( I \) is a downward straight line:

The internal resistance \( r \) is determined by calculating the absolute value of the slope of this graph:
\( \text{Slope} = \frac{\Delta V}{\Delta I} = -r \implies r = -(\text{Slope}) \).
In simple words: The voltage drops linearly as you draw more current. The steepness of this drop (the slope of the line) is equal to the cell's internal resistance.

Exam Tip: Clearly label the y-intercept as the EMF \( E \), as this point represents the open-circuit voltage when no current is drawn.

Question 10. Find the value of the unknown resistance X and the current drawn by the circuit from the battery if no current flows through the galvanometer. Assume the resistance per unit length of the wire is 0.01Ω cm-1.
Answer: Since no current flows through the galvanometer, the Wheatstone bridge is balanced. Using the balance condition:
\( \frac{X}{2} = \frac{120}{80} \)
\( \implies X = 2 \times \frac{120}{80} = 3\text{ }\Omega \).
Thus, the value of the unknown resistance \( X \) is \( 3\text{ }\Omega \).
In simple words: When a bridge circuit is balanced, the ratio of resistances on the left side equals the ratio of resistances on the right side. This relation shows the unknown resistor is 3 ohms.

Exam Tip: State the balanced Wheatstone bridge principle formula \( \frac{P}{Q} = \frac{R}{S} \) to show the base of your calculation.

Moving Charge and Magnetism (Magnetic Effect of Current)

Question 1. Where is the magnetic field at a current element (i) minimum and (ii) maximum?
Answer: According to the Biot-Savart law, the magnetic field is given by \( d B = \frac{\mu_0}{4\pi} \frac{I \cdot dl \cdot \sin\theta}{r^2} \).
(i) The magnetic field is minimum (zero) along the axis of the current element, where \( \theta = 0^\circ \) or \( \theta = 180^\circ \), because \( \sin\theta = 0 \).
(ii) The magnetic field is maximum in a plane perpendicular to the current element and passing through it, where \( \theta = 90^\circ \), because \( \sin 90^\circ = 1 \) is maximum.
In simple words: The magnetic field is completely zero along the line of the wire itself, and it reaches its strongest point directly outward at a right angle from the wire.

Exam Tip: Relate the maximum and minimum values directly to the angle \( \theta \) in the Biot-Savart equation to show a robust physics explanation.

Question 2. Consider the circuit shown, where APB and AQB are semicircles. What will be the magnetic field at the centre C of the circular loop?
Answer: The net magnetic field at the center \( C \) of the circular loop is zero. This occurs because the total current divides equally into the two identical semicircular paths APB and AQB, flowing in opposite directions. The magnetic fields produced by these two currents are equal in magnitude but opposite in direction, canceling each other out completely.
In simple words: Because the path splits evenly, the electrical current goes in opposite directions around the two halves. The magnetic fields they create fight against each other and completely cancel out at the exact center.

Exam Tip: State clearly that the symmetry of the parallel pathways ensures equal current distribution, which is essential for the fields to cancel to zero.

Question 3. What will be the path of a charged particle moving in a uniform magnetic field at any arbitrary angle?
Answer: The path of a charged particle moving at an arbitrary angle to a uniform magnetic field is helical. The component of velocity perpendicular to the field causes the particle to move in a circle, while the component parallel to the field moves it forward in a straight line, combining into a helix.
In simple words: The particle moves in a spiral path because one part of its speed keeps it turning in circles, while the other part pushes it straight along the magnetic field line.

Exam Tip: Explain both components of velocity (parallel and perpendicular) to justify why the combined motion is a helix.

Question 4. In a certain arrangement, a proton does not get deflected while passing through a magnetic field region. State the condition under which it is possible.
Answer: A proton will pass undeflected if the magnetic force acting on it is zero. According to \( F = q v B \sin\theta \), the magnetic force is zero when the velocity vector \( \vec{v} \) of the proton is parallel (\( \theta = 0^\circ \)) or antiparallel (\( \theta = 180^\circ \)) to the magnetic field \( \vec{B} \).
In simple words: The magnetic field cannot push the proton if the proton is moving along the exact same line as the magnetic field, either forward or backward.

Exam Tip: Write down the magnetic Lorentz force equation to support your explanation with solid math.

Question 5. A galvanometer gives full scale deflection with the current Ig. Can it be converted into an ammeter of range I < Ig ?
Answer: No, a galvanometer cannot be converted into an ammeter for a range less than its full-scale deflection current (\( I < I_g \)). To convert a galvanometer, a parallel shunt resistor \( S = \frac{I_g G}{I - I_g} \) is required. If \( I < I_g \), the resistance \( S \) becomes negative, which is physically impossible.
In simple words: No, because to measure a smaller current than the meter's natural limit, you would need a physically impossible 'negative' bypass resistor.

Exam Tip: Use the shunt formula mathematically to show that the calculated resistance value becomes negative when \( I < I_g \).

Question 6. Define the term magnetic dipole moment of a current loop. Write the expression for the magnetic moment when an electron revolves at a speed ‘v’, around an orbit of radius ‘ r’ in hydrogen atom
Answer: The magnetic dipole moment of a current loop is defined as the product of the current flowing through the loop and the area enclosed by it. For \( N \) turns, \( M = N I A \). Its direction is normal to the plane of the loop.
For an electron revolving with speed \( v \) in an orbit of radius \( r \), the equivalent current is \( I = \frac{e}{T} = \frac{e v}{2\pi r} \).
The magnetic moment is:
\( M = I A = \left(\frac{e v}{2\pi r}\right) (\pi r^2) = \frac{e v r}{2} \).
In simple words: A loop's magnetic strength is the current times the loop's flat area. For a single electron circling a hydrogen atom, this simplifies to half of the charge times the speed and radius.

Exam Tip: Derive the expression step-by-step starting from basic current definitions to ensure you receive full credit.

Question 7. A wire of length L is bent round in the form of a coil having N turns of same radius. If a steady current I flows through it in a clockwise direction, find the magnitude and direction of the magnetic field produced at its centre
Answer: Since a wire of length \( L \) is bent into \( N \) turns of radius \( r \), we have:
\( L = N \times 2\pi r \implies r = \frac{L}{2\pi N} \)
The magnetic field \( B \) at the center of the circular coil of \( N \) turns is:
\( B = \frac{\mu_0 N I}{2r} \)
Substituting the value of \( r \):
\( B = \frac{\mu_0 N I}{2 (L / 2\pi N)} = \frac{\mu_0 \pi N^2 I}{L} \).
According to the right-hand rule, since the current flows in a clockwise direction, the direction of the magnetic field is perpendicular to the plane of the coil and points inwards (into the page).
In simple words: The total length of wire limits how tight the coils can be. Doing the math shows the magnetic field is proportional to the square of the turns, and the clockwise current means the field points straight down into the surface.

Exam Tip: Never forget to state both the magnitude and the direction (inwards) when asked for a magnetic field vector.

Question 8. Define current sensitivity and voltage sensitivity of a galvanometer. Increasing the current sensitivity may not necessarily increase the voltage sensitivity of a galvanometer. Justify.
Answer: Current sensitivity is the deflection of the coil per unit current flowing through it:
\( I_s = \frac{\theta}{I} = \frac{N B A}{C} \)
Voltage sensitivity is the deflection of the coil per unit voltage applied across it:
\( V_s = \frac{\theta}{V} = \frac{N B A}{C G} \)
where \( G \) is the galvanometer resistance. If we double the number of turns \( N \) to increase current sensitivity, the length of the wire doubles, which also doubles the resistance \( G \). Since both \( N \) and \( G \) double, the ratio \( \frac{N}{G} \) remains constant, leaving the voltage sensitivity unchanged.
In simple words: Current sensitivity is how much the needle moves per amp, while voltage sensitivity is the move per volt. Adding more wire loops makes the meter more sensitive to current, but the extra wire adds more resistance, keeping the voltage sensitivity exactly the same.

Exam Tip: Use the turn-doubling scenario as a clear, logical proof to justify why the two sensitivities do not always change together.

Question 9. An electron of kinetic energy 25 keV moves perpendicular to the direction of a uniform magnetic field of 0.2 milli-Tesla. Calculate the time period of rotation of the electron in the magnetic field.
Answer: The time period of rotation of a charged particle in a magnetic field is:
\( T = \frac{2\pi m}{q B} \)
Substituting the values for an electron (\( m = 9.1 \times 10^{-31}\text{ kg} \), \( q = 1.6 \times 10^{-19}\text{ C} \)) and the magnetic field (\( B = 0.2\text{ mT} = 0.2 \times 10^{-3}\text{ T} \)):
\( T = \frac{2 \times 3.14 \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 0.2 \times 10^{-3}} \approx 1.78 \times 10^{-7}\text{ s} \).
Note that this time period is completely independent of the kinetic energy of the electron.
In simple words: The time it takes for the electron to loop around in a circle depends only on its mass, charge, and the field strength. The given kinetic energy is extra information and is not needed for the calculation.

Exam Tip: Be sure to write a short note stating that the time period of the orbit is independent of the particle's speed or energy.

Question 10. Derive the relation between µ0 ,ε0 and c,symbols has their usual meanings.
Answer: We know the electrostatic and magnetic constants are given by:
\( \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2\text{/C}^2 \)
\( \frac{\mu_0}{4\pi} = 10^{-7}\text{ T m/A} \)
Multiplying these two equations together:
\( \mu_0 \varepsilon_0 = \left( \frac{\mu_0}{4\pi} \right) \cdot \left( 4\pi\varepsilon_0 \right) = 10^{-7} \times \frac{1}{9 \times 10^9} = \frac{1}{9 \times 10^{16}} \)
Since the speed of light is \( c = 3 \times 10^8\text{ m/s} \), its square is \( c^2 = 9 \times 10^{16}\text{ m}^2\text{/s}^2 \).
Therefore, we get:
\( \mu_0 \varepsilon_0 = \frac{1}{c^2} \implies c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \).
In simple words: Multiplying the electrical and magnetic constants of a vacuum gives the inverse square of the speed of light, showing that light is an electromagnetic wave.

Exam Tip: This derivation is very short but highly important. Show each substitution step clearly to ensure full credit.

Electromagnetic Induction

Question Q1. Define the term self-inductance of a coil. Give its SI unit.
Answer: Self-inductance of a coil is defined as the induced electromotive force (emf) set up in the coil when the rate of change of current through the same coil is unity (1 A/s). Alternatively, it is the ratio of magnetic flux linkage to the current (\( L = \frac{\Phi}{I} \)). The SI unit of self-inductance is the Henry (H).
In simple words: Self-inductance is a coil's natural resistance to changes in its own electric current. It is measured in units called henries.

Exam Tip: Always state both the definition and the correct SI unit (Henry) to secure all allocated marks.

Question Q2. If the rate of change of current is 2 ampere/second in a solenoid induces an emf of 40mV in the solenoid, what is the self-inductance of this solenoid?
Answer: The magnitude of induced emf in a solenoid is given by:
\( e = L \frac{dI}{dt} \)
Given \( e = 40\text{ mV} = 40 \times 10^{-3}\text{ V} \) and \( \frac{dI}{dt} = 2\text{ A/s} \):
\( 40 \times 10^{-3} = L \cdot 2 \)
\( L = \frac{40 \times 10^{-3}}{2} = 20 \times 10^{-3}\text{ H} = 20\text{ mH} \).
The self-inductance of the solenoid is \( 20\text{ mH} \).
In simple words: By dividing the induced voltage by the speed at which the current is changing, we find that the solenoid's self-inductance is 20 millihenries.

Exam Tip: Remember to convert the voltage from millivolts (mV) to standard volts (V) before performing the division.

Question Q3. Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily.
Answer: According to Lenz's law, the induced current in the rings must flow in a direction that opposes the increase in magnetic flux:

- In the ring above the wire (Ring 1), the magnetic field is directed out of the page and is increasing. To oppose this, the induced field must point into the page, creating a **clockwise** current.
- In the ring below the wire (Ring 2), the magnetic field is directed into the page and is increasing. To oppose this, the induced field must point out of the page, creating an **anticlockwise** current.
In simple words: The growing current in the wire creates a stronger magnetic field. The top ring fights this by flowing clockwise, and the bottom ring fights it by flowing counter-clockwise.

Exam Tip: Use the right-hand grip rule to find the direction of the magnetic field from the straight wire first, then apply Lenz's law to the rings.

Question Q4. State Lenz’s law in electromagnetic induction.
Answer: Lenz's law states that the direction of the induced current or electromotive force (emf) in a closed circuit is always such that it opposes the change in magnetic flux or the very cause that produces it.
In simple words: Lenz's law says that any electrical current created by a changing magnetic field will flow in a direction that fights against that magnetic change.

Exam Tip: Mention that Lenz's law is a direct consequence of the law of conservation of energy to write a comprehensive answer.

Question Q5. Write the expression of electromagnetic energy stored in an inductor of inductance L when steady current is passed through it.
Answer: The electromagnetic energy \( U \) stored in the magnetic field of an inductor is given by:
\( U = \frac{1}{2} L I^2 \)
where \( L \) is the self-inductance and \( I \) is the steady current.
In simple words: The energy stored in a coil's magnetic field is equal to half of its inductance multiplied by the square of the current flowing through it.

Exam Tip: Be sure to define what both symbols (\( L \) and \( I \)) stand for in your final expression.

Question Q6. Two circular loops are placed with their centres at fixed distance apart. How would you orient the loops to have (i) maximum (ii) minimum Mutual inductance?
Answer: (i) For maximum mutual inductance, place the two loops parallel to each other on a common axis, so that almost all magnetic flux from the first loop passes through the second.
(ii) For minimum mutual inductance, orient the loops perpendicular to each other, so that the magnetic flux lines of one loop lie parallel to the plane of the other, resulting in zero net flux linkage.
In simple words: (i) Place them flat and facing each other to share the most magnetic field. (ii) Turn one loop sideways so they are at a right angle, which prevents them from sharing any magnetic field.

Exam Tip: Explain that mutual inductance depends directly on the alignment of the magnetic flux lines between the two coils.

Question Q7. Define eddy current. Give one application of eddy current
Answer: Eddy currents are circulating loops of electrical current induced within the body of a solid metal conductor when it experiences a changing magnetic field.
A key practical application is in electromagnetic damping (such as in magnetic brakes used in trains).
In simple words: When a solid chunk of metal is exposed to a changing magnetic field, small whirlpools of electricity form inside the metal itself. These are used to create smooth magnetic brakes.

Exam Tip: Other valid applications include induction furnaces and electric power meters. Mentioning these can provide an alternative strong answer.

Question Q8. What is magnetic flux? Write its expression and SI unit.
Answer: Magnetic flux is defined as the total number of magnetic field lines passing normally through a given surface area. Its mathematical expression is:
\( \Phi_B = \vec{B} \cdot \vec{A} = B A \cos\theta \)
The SI unit of magnetic flux is the Weber (Wb).
In simple words: Magnetic flux is a measure of how much magnetic field passes through an area. It is measured in a unit called webers.

Exam Tip: Clearly state that \( \theta \) is the angle between the magnetic field vector and the normal to the surface area vector.

Dual Nature of Matter and Radiation

Question 1. The wavelength of electromagnetic radiation is doubled. What will happen to the energy of photon?
Answer: The energy of a photon is inversely proportional to its wavelength, as shown by the formula \( E = \frac{h c}{\lambda} \). Therefore, if the wavelength of the radiation is doubled, the energy of the photon will be halved.
In simple words: A photon's energy is linked to its wavelength in an opposite way. If you double the wave's length, you cut its energy in half.

Exam Tip: Always write down the mathematical relationship \( E \propto \frac{1}{\lambda} \) to justify your conclusion.

Question 2. Ultraviolet light is incident on two photosensitive materials having work function ϕ1 &ϕ2 (ϕ1 > ϕ2). In which of the case will K.E. of emitted electrons be grater? Why?
Answer: According to Einstein's photoelectric equation, the maximum kinetic energy is \( K_{\text{max}} = h\nu - \phi \). Since both materials are exposed to the same frequency \( \nu \) of ultraviolet light, the material with the smaller work function will produce electrons with higher kinetic energy. Given \( \phi_1 > \phi_2 \), the kinetic energy of emitted electrons will be greater for the material with work function \( \phi_2 \).
In simple words: The work function is the 'energy cost' to free an electron. Since the second material has a lower cost, more leftover light energy is turned into kinetic energy, making the electrons fly off faster.

Exam Tip: State Einstein's photoelectric equation at the start of your explanation to establish a robust physics-based argument.

Question 3. An increase in the intensity of incident light does not change the maximum velocity of the emitted photo electrons. Why?
Answer: According to Einstein's photoelectric theory, the maximum velocity of emitted photoelectrons depends solely on the frequency of the incident light and the work function of the metal. Increasing the intensity of the light only increases the number of photons (and thus the number of emitted electrons), but does not change the energy of any individual photon. Therefore, the maximum velocity remains unchanged.
In simple words: Making light brighter just adds more packets of light, but it doesn't make any single packet stronger. Since electrons are freed by individual packets, their top speed remains the exact same.

Exam Tip: Emphasize the difference between light intensity (number of photons) and light frequency (energy of each photon) in your answer.

Question 4. The stopping potential in an experiment on photoelectric effect is 1.6 volt.what is the maximum kinetic energy of the photoelectrons emitted?.
Answer: The maximum kinetic energy \( K_{\text{max}} \) is related to the stopping potential \( V_0 \) by:
\( K_{\text{max}} = e V_0 \)
Given \( V_0 = 1.6\text{ V} \):
\( K_{\text{max}} = 1.6\text{ eV} \)
In Joules, this is:
\( K_{\text{max}} = 1.6 \times 1.6 \times 10^{-19}\text{ J} = 2.56 \times 10^{-19}\text{ J} \).
In simple words: A stopping voltage of 1.6 volts means the most energetic electrons have exactly 1.6 electron-volts of kinetic energy.

Exam Tip: Expressing the final kinetic energy in both electron-volts (eV) and Joules (J) shows a complete and thorough understanding.

Question 5. Name the experiment which verified the wave nature of particles.
Answer: The wave nature of particles (specifically electrons) was experimentally verified by the Davisson and Germer experiment.
In simple words: The Davisson and Germer experiment proved that tiny solid particles like electrons can behave like waves.

Exam Tip: Be sure to spell 'Davisson and Germer' correctly to avoid losing easy marks on this direct recall question.

Question 6. What is the stopping potential applied to a photocell if the maximum kinetic energy of a photoelectron is 5eV?
Answer: Since the maximum kinetic energy \( K_{\text{max}} = e V_0 \), where \( V_0 \) is the stopping potential:
Given \( K_{\text{max}} = 5\text{ eV} \):
\( 5\text{ eV} = e V_0 \implies V_0 = 5\text{ V} \).
The magnitude of the stopping potential is \( 5\text{ V} \) (applied as a negative potential of \( -5\text{ V} \)).
In simple words: To stop electrons carrying 5 electron-volts of energy, you need to apply an opposing voltage of exactly 5 volts.

Exam Tip: Mentioning that the stopping potential is applied as a negative voltage on the collector plate shows excellent conceptual depth.

Question 7. Define threshold wavelength.
Answer: Threshold wavelength is the maximum wavelength of incident radiation that can cause photoelectric emission from a metal surface. Any radiation with a wavelength longer than this threshold will fail to eject photoelectrons, regardless of its intensity.
In simple words: The threshold wavelength is the longest wavelength of light that still has enough energy to knock an electron out of a metal.

Exam Tip: Make sure to use the word 'maximum' when defining threshold wavelength, as it corresponds to the minimum frequency required.

Question 8. Show graphically, the variation of de-Broglie wavelength (λ)with the momentum of an electron.
Answer: The de-Broglie wavelength is inversely proportional to momentum:
\( \lambda = \frac{h}{p} \implies \lambda \propto \frac{1}{p} \)
A plot of \( \lambda \) versus \( p \) forms a rectangular hyperbola:

In simple words: Since wavelength decreases as momentum increases, the graph curves downward, starting high on the left and flattening out to the right.

Exam Tip: Label both axes clearly, putting the de-Broglie wavelength \( \lambda \) on the vertical axis and momentum \( p \) on the horizontal axis.

Question 9. An α-particle and a proton are accelerated from rest by the same potential. Find the ratio of their de-Broglie wavelengths.
Answer: The de-Broglie wavelength of a particle accelerated through a potential \( V \) is:
\( \lambda = \frac{h}{\sqrt{2 m q V}} \)
Since the potential \( V \) is the same for both:
\( \lambda \propto \frac{1}{\sqrt{m q}} \)
Therefore, the ratio of their wavelengths is:
\( \frac{\lambda_{\alpha}}{\lambda_p} = \sqrt{\frac{m_p q_p}{m_{\alpha} q_{\alpha}}} \)
We know that the mass of an alpha particle is four times that of a proton (\( m_{\alpha} = 4 m_p \)) and its charge is twice that of a proton (\( q_{\alpha} = 2 q_p \)):
\( \frac{\lambda_{\alpha}}{\lambda_p} = \sqrt{\frac{m_p \cdot q_p}{(4 m_p)(2 q_p)}} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}} \).
The ratio of their de-Broglie wavelengths is \( 1 : 2\sqrt{2} \) (or \( 1 : \sqrt{8} \)).
In simple words: An alpha particle is much heavier and carries more charge than a proton. When accelerated by the same voltage, its larger mass and charge give it a shorter de-Broglie wavelength, resulting in a ratio of 1 to square root of 8.

Exam Tip: Clearly write down the mass and charge ratios of the two particles before plugging them into the formula to make your derivation easy to follow.

Question 10. The two lines A and B shown in the graph plot the de-Broglie wavelength λ as function of 1/ √V (V is the accelerating potential) for two particles having the same charge. Which of the two represents the particle of heavier mass?
Answer: The de-Broglie wavelength is related to the accelerating potential \( V \) by:
\( \lambda = \frac{h}{\sqrt{2 m q V}} = \left( \frac{h}{\sqrt{2 m q}} \right) \frac{1}{\sqrt{V}} \)
In a plot of \( \lambda \) versus \( \frac{1}{\sqrt{V}} \), the slope is given by \( \text{Slope} = \frac{h}{\sqrt{2 m q}} \).
Since both particles have the same charge \( q \), the slope is inversely proportional to the square root of the mass (\( \text{Slope} \propto \frac{1}{\sqrt{m}} \)). This means a flatter slope represents a heavier mass. Since line B has a smaller slope than line A, line B represents the heavier particle.
In simple words: The slope of the line on this graph is inversely related to the mass of the particle. Because line B is flatter (smaller slope), it represents the particle with the heavier mass.

Exam Tip: Use the slope formula to prove your conclusion mathematically, as examiners look for this specific justification.

Question 11. Deduce de Broglie wavelength of electron accelerated by potential of V volt.Expression for de- Broglie wavelength associated with Accelerated Electron:
Answer: The de-Broglie wavelength associated with an electron of momentum \( p \) is:
\( \lambda = \frac{h}{p} \) --- (1)
The relationship between momentum and kinetic energy \( E_k \) is:
\( p = \sqrt{2 m E_k} \) --- (2)
When an electron is accelerated from rest through a potential difference \( V \), its kinetic energy is:
\( E_k = e V \) --- (3)
Substituting (3) into (2), we get:
\( p = \sqrt{2 m e V} \)
Substituting this momentum into (1) yields:
\( \lambda = \frac{h}{\sqrt{2 m e V}} \)
Substituting standard values (\( h = 6.63 \times 10^{-34}\text{ J s} \), \( m = 9.1 \times 10^{-31}\text{ kg} \), \( e = 1.6 \times 10^{-19}\text{ C} \)):
\( \lambda = \frac{12.27}{\sqrt{V}}\text{ \AA} \) (or \( \frac{1.227}{\sqrt{V}}\text{ nm} \)).
In simple words: An electron gets kinetic energy from a voltage source, which determines its momentum. Since wavelength is Planck's constant divided by momentum, we combine these rules to find the wavelength in terms of the applied voltage.

Exam Tip: Memorize the simplified final expression \( \lambda = \frac{12.27}{\sqrt{V}}\text{ \AA} \), as it is highly useful for solving numerical problems quickly.

Semiconductor Devices and Communication System

Question 1. On what factors does the conductivity of metal depend?
Answer: The electrical conductivity (\( \sigma \)) of a metal depends on:
(i) The number density of free electrons (\( n \)) in the metal.
(ii) The relaxation time (\( \tau \)) of the free electrons (which is related to their drift velocity).
In simple words: A metal conducts electricity better if it has a high concentration of free electrons and if those electrons can move with fewer collisions.

Exam Tip: State the formula \( \sigma = \frac{n e^2 \tau}{m} \) to show clearly how conductivity depends on electron density and relaxation time.

Question 2. What is energy band?
Answer: An energy band is a continuous range of closely spaced electronic energy levels formed in a solid crystal due to the close proximity and interaction of its constituent atoms.
In simple words: In a solid crystal, the separate energy levels of individual atoms merge together to form broad, continuous bands of allowed energies for the electrons.

Exam Tip: Mention the interaction of atoms in a crystal lattice as the primary cause for the splitting of energy levels into bands.

Question 3. Where are donor & acceptor levels located in a semiconductor?
Answer: - In an n-type semiconductor, the donor energy level lies just below the bottom of the conduction band.
- In a p-type semiconductor, the acceptor energy level lies just above the top of the valence band.
In simple words: Donor energy levels are located extremely close to the conduction band, while acceptor levels sit just slightly above the valence band.

Exam Tip: Drawing a simple energy band diagram showing these levels relative to the conduction and valence bands can make your explanation much clearer.

Question 4. What is the value of potential barrier for a (i) Ge diode & (ii) Si diode
Answer: The values of the potential barrier are:
(i) For a Germanium (Ge) diode: \( 0.3\text{ V} \)
(ii) For a Silicon (Si) diode: \( 0.7\text{ V} \)
In simple words: To make current flow through a diode, you must overcome a small internal voltage barrier of 0.3 volts for germanium and 0.7 volts for silicon.

Exam Tip: Memorize these standard values as they are frequently used in circuit analysis questions.

Question 5. What is Zener breakdown?
Answer: Zener breakdown occurs in a highly doped p-n junction diode under high reverse-bias voltage. The strong electric field in the narrow depletion region is sufficient to break covalent bonds, liberating a large number of electron-hole pairs and causing a sharp increase in reverse current.
In simple words: When a strong reverse voltage is applied to a heavily doped diode, the powerful electric field pulls electrons out of their bonds, creating a sudden flood of electric current.

Exam Tip: Explain that Zener breakdown is characterized by a sharp rise in current while the voltage across the diode remains constant.

Question 6. What is the relation between α & β?
Answer: The relationship between the common-base current gain (\( \alpha \)) and the common-emitter current gain (\( \beta \)) is:
\( \alpha = \frac{\beta}{1 + \beta} \)
Alternatively, it can be written as:
\( \beta = \frac{\alpha}{1 - \alpha} \).
In simple words: These two factors represent how much a transistor amplifies current in different circuit setups. They are mathematically linked by a simple fraction.

Exam Tip: Be prepared to derive this relation using the basic transistor current equation: \( I_E = I_B + I_C \).

Question 7. What is phase relationship between input & output voltages of a C.B. amplifier & C.E. amplifier?
Answer: (i) In a Common-Base (C.B.) amplifier, the input and output voltages are in phase (phase difference of \( 0^\circ \)).
(ii) In a Common-Emitter (C.E.) amplifier, the input and output voltages differ in phase by \( 180^\circ \) (they are out of phase).
In simple words: For a common-base amplifier, the output signal goes up and down at the exact same time as the input. For a common-emitter amplifier, the output is flipped upside down compared to the input.

Exam Tip: Clearly highlight the \( 180^\circ \) phase reversal in C.E. amplifiers, as this is a key characteristic often tested in examinations.

Question 8. Why is germanium preferred over silicon for making semiconductor devices?
Answer: Germanium has a lower bandgap energy (\( E_g = 0.7\text{ eV} \)) compared to Silicon (\( E_g = 1.1\text{ eV} \)). This smaller bandgap allows charge carriers to be excited more easily, making Germanium useful for low-voltage switching and high-speed conduction applications.
In simple words: Germanium requires less energy to free its electrons than silicon, which can make it faster and more conductive in low-energy electronic circuits.

Exam Tip: State the specific bandgap values (\( 0.7\text{ eV} \) for Ge and \( 1.1\text{ eV} \) for Si) to make your comparison precise.

Question 9. Why is germanium preferred over silicon for making semiconductor devices?
Answer: Germanium is chosen in specific low-power applications due to its smaller bandgap energy (\( 0.7\text{ eV} \) versus \( 1.1\text{ eV} \) for Silicon). This lower barrier results in a lower threshold voltage, allowing devices to operate more efficiently at lower operating potentials.
In simple words: Because germanium has a smaller energy gap, it conducts electricity more easily at lower voltages compared to silicon.

Exam Tip: State that the lower bandgap in Germanium is the primary reason it is preferred for low-voltage switching devices.

Question 10. Why is p-n junction also called a junction diode?
Answer: A p-n junction is called a junction diode because it has a unidirectional conduction property, allowing current to flow easily in only one direction (forward bias) while blocking it in the other (reverse bias). This behavior is highly similar to the operation of a traditional vacuum diode.
In simple words: It is called a diode because, like an old vacuum tube diode, it acts as a one-way street for electrical current.

Exam Tip: Draw a comparison with the vacuum diode's unidirectional current property to justify the naming convention.

Question 11. What are intrinsic and extrinsic semiconductors. How many types of extrinsic semiconductor are there.
Answer: An intrinsic semiconductor is a pure semiconductor material without any added impurities. An extrinsic semiconductor is a pure semiconductor that has been deliberately doped with specific impurity atoms to alter its electrical properties.
There are two types of extrinsic semiconductors:
- n-type semiconductors
- p-type semiconductors
In simple words: Intrinsic semiconductors are completely pure, while extrinsic ones have had tiny amounts of other elements mixed in to boost their conductivity. Extrinsic types are divided into n-type and p-type.

Exam Tip: Define both terms clearly and list 'p-type' and 'n-type' as the two categories of extrinsic semiconductors.

Question 12. Which of the transistors p-n-p & n-p-n is more useful?
Answer: The n-p-n transistor is more useful than the p-n-p transistor because its majority charge carriers are electrons, which have significantly higher mobility than the holes that act as majority carriers in a p-n-p transistor. This results in faster switching speeds and better performance.
In simple words: NPN transistors are better because they run on moving electrons, which travel much faster through the material than the 'holes' used in PNP transistors.

Exam Tip: Always highlight 'higher electron mobility' as the core physics reason why NPN transistors are superior to PNP transistors.

Question 13. In a transistor connected on C.E. mode Rc = 4kΩ, Ri =1kΩ, Ic = 1mA &Ib =20μA. Find the voltage gain?
Answer: The current gain \( \beta \) in common-emitter mode is:
\( \beta = \frac{I_C}{I_B} = \frac{1\text{ mA}}{20\text{ }\mu\text{A}} = \frac{10^{-3}}{20 \times 10^{-6}} = 50 \)
The voltage gain \( A_v \) is given by:
\( A_v = \beta \frac{R_c}{R_i} \)
Substituting the values:
\( A_v = 50 \times \frac{4\text{ k}\Omega}{1\text{ k}\Omega} = 50 \times 4 = 200 \).
The voltage gain of the transistor is \( 200 \).
In simple words: First, we calculate the current amplification factor (beta), which is 50. Then we multiply beta by the ratio of output resistance to input resistance to find that the voltage is boosted by 200 times.

Exam Tip: Remember to convert all current and resistance values to their standard SI units (amperes and ohms) before performing calculations.

Question 14. In a n-p-n transistor the collector current is 10mA. If 90% of electrons emitted reach the collector find base and emitter current?
Answer: Since 90% of the emitted electrons reach the collector, the collector current \( I_C \) is 90% of the emitter current \( I_E \):
\( I_C = 0.90 \times I_E \)
Given \( I_C = 10\text{ mA} \):
\( I_E = \frac{10\text{ mA}}{0.90} \approx 11.11\text{ mA} \).
Using the transistor current relation:
\( I_E = I_B + I_C \implies I_B = I_E - I_C \)
\( I_B = 11.11\text{ mA} - 10\text{ mA} = 1.11\text{ mA} \).
Thus, the emitter current is \( 11.11\text{ mA} \) and the base current is \( 1.11\text{ mA} \).
In simple words: The collector collects 90 percent of the total current starting from the emitter. This means the emitter current must be about 11.11 mA, leaving the remaining 1.11 mA to escape through the base.

Exam Tip: Use the fundamental relation \( I_E = I_B + I_C \) to calculate the base current after finding the emitter current.

Question 15. the output of an OR gate is connected to both the inputs of NAND gate write its tooth table.
Answer: Let the inputs of the OR gate be \( A \) and \( B \). The output of the OR gate is \( Y' = A + B \). Since both inputs of the NAND gate are connected to \( Y' \), the final output \( Y \) is:
\( Y = \overline{Y' \cdot Y'} = \overline{Y'} = \overline{A + B} \).
This combination acts as a NOR gate. The truth table is:

In simple words: Connecting an OR gate to a NAND gate with tied inputs creates a NOR gate. The output is only 1 when both inputs are 0; otherwise, the output is always 0.

Exam Tip: Clearly explain the logic expression of the combined gate (\( Y = \overline{A + B} \)) to show that the setup acts as a NOR gate.

Communication System

Question 1. Which basic modes of communication are used for telephonic communication?
Answer: Telephonic communication utilizes the point-to-point mode of communication, where a dedicated link is established between a single transmitter and a single receiver.
In simple words: Telephone calls use point-to-point communication, which means the signal goes directly from one specific caller to one specific listener.

Exam Tip: Clearly distinguish point-to-point communication from broadcast mode (like radio or TV) where one transmitter sends to many receivers.

Question 2. what is change in amplitude modulation?
Answer: In amplitude modulation, the amplitude of the high-frequency carrier wave is varied in accordance with the instantaneous amplitude of the modulating (audio) signal, while its frequency and phase remain constant.
In simple words: In AM radio, the height (amplitude) of the carrier wave is changed to match the pattern of the sound wave we want to send.

Exam Tip: Specify that the frequency and phase of the carrier wave must remain unchanged during amplitude modulation.

Question 3. What is meaning of the term ‘attenuation’ used in communication system?
Answer: Attenuation refers to the gradual loss of strength or intensity of a signal as it propagates through a transmission medium.
In simple words: Attenuation is the weakening of a signal as it travels a long distance through a wire or through the air.

Exam Tip: Note that attenuation is usually measured in decibels (dB) per unit length.

Question 4. What is the full form of ‘GPS’?
Answer: The full form of GPS is Global Positioning System.
In simple words: GPS stands for Global Positioning System, which helps find locations using satellites.

Exam Tip: Ensure proper capitalization of 'Global Positioning System' to write a clean, complete answer.

Question 5. Arrange the following network in increasing order of the number of computers that may be present in the network:Internet; LAN; WAN
Answer: The correct increasing order of networks based on the number of connected computers is:
LAN (Local Area Network) < WAN (Wide Area Network) < Internet
In simple words: A local network (LAN) has the fewest computers, a wide network (WAN) has more, and the Internet connects almost every computer in the world.

Exam Tip: Use the less-than symbol (<) to clearly show the increasing order of network sizes.

Question 6. A TV tower has a height of 71 m. What is the maximum distance up to which TV transmission can be received? Given that the radius of the earth = 6.4 × 106 m.
Answer: The maximum distance \( d \) for receiving transmissions from a tower of height \( h \) is given by the formula:
\( d = \sqrt{2 R h} \)
Substituting the given values:
\( R = 6.4 \times 10^6\text{ m} \)
\( h = 71\text{ m} \)
\( d = \sqrt{2 \times 6.4 \times 10^6 \times 71} = \sqrt{9.088 \times 10^8} \approx 30146\text{ m} \) (or \( 30.15\text{ km} \)).
Thus, the maximum reception distance is \( 30146\text{ m} \).
In simple words: Using the curve of the Earth and the height of the tower, we calculate that the television signal can travel a maximum distance of about 30.1 kilometers before the Earth's curve blocks it.

Exam Tip: State the formula \( d = \sqrt{2 R h} \) clearly and show the steps of calculation to ensure you get full credit.

Question 7. Explain why high frequency carrier waves are needed for effective transmission of signal. A message signal of 12 KHz and peak voltage 20 V is used to modulate a carrier wave of frequency 12 MHz and peak voltage 30 V. Calculate the (i) modulation index (ii) side-band frequencies.
Answer: High-frequency carrier waves are required for effective signal transmission because:
1. They require practical antenna lengths (since antenna size is proportional to wavelength \( \lambda = \frac{c}{\nu} \)).
2. They carry higher energy, allowing them to propagate long distances with less attenuation.
3. They avoid mixing of signals from different transmitters by permitting multiplexing.

Numerical Calculation:
Given:
Modulating signal frequency, \( f_m = 12\text{ kHz} \)
Peak voltage of modulating signal, \( A_m = 20\text{ V} \)
Carrier wave frequency, \( f_c = 12\text{ MHz} = 12000\text{ kHz} \)
Peak voltage of carrier wave, \( A_c = 30\text{ V} \)

(i) The modulation index \( \mu \) is:
\( \mu = \frac{A_m}{A_c} = \frac{20}{30} \approx 0.67 \)

(ii) The side-band frequencies are:
Upper Side Band (USB) \( = f_c + f_m = 12000\text{ kHz} + 12\text{ kHz} = 12012\text{ kHz} \) (or \( 12.012\text{ MHz} \))
Lower Side Band (LSB) \( = f_c - f_m = 12000\text{ kHz} - 12\text{ kHz} = 11988\text{ kHz} \) (or \( 11.988\text{ MHz} \))
In simple words: High-frequency waves keep antennas small and help signals travel farther without fading. For this specific signal, the modulation index is about 0.67, and the sideband frequencies are 12012 kHz and 11988 kHz.

Exam Tip: When explaining why high-frequency carriers are needed, always list antenna size and signal power as the two primary reasons.

Question 8. What is the range of frequencies used in satellite communication? What is common between these waves and light waves?
Answer: The frequencies used in satellite communication are in the gigahertz (GHz) range, typically:
(i) Downlink frequency range: \( 3.7 \text{ to } 4.2\text{ GHz} \)
(ii) Uplink frequency range: \( 5.9 \text{ to } 6.4\text{ GHz} \)
Both these satellite communication waves and light waves are electromagnetic waves. Consequently, they both travel at the speed of light in vacuum (\( 3 \times 10^8\text{ m/s} \)) and propagate in a straight line.
In simple words: Satellite signals use super-high frequency radio waves around 4 to 6 gigahertz. Just like light, these signals are electromagnetic waves that travel in straight lines at the speed of light.

Exam Tip: State both the uplink and downlink frequency ranges and clearly specify their shared electromagnetic nature to secure full marks.

Question 9. What is ground wave communication? On what factors does the maximum range of propagation in this mode depend?
Answer: Ground wave (or surface wave) communication is a method of radio propagation where waves travel along the surface of the Earth from the transmitter to the receiver.
The maximum range of ground wave propagation depends on:
(i) The frequency of the transmitted signal (lower frequencies suffer less absorption).
(ii) The electrical properties (conductivity and permittivity) of the ground or soil.
(iii) The power of the transmitter.
In simple words: Ground wave communication involves sending radio signals that follow the curve of the Earth's surface. The range depends on how much power the transmitter has, the quality of the ground, and using low frequencies so the ground doesn't absorb the signal.

Exam Tip: Highlight that ground wave propagation is only effective for frequencies below a few MHz because higher frequencies are rapidly absorbed by the ground.

Question. Write the functions of the following in communication systems:
(i) Transducer
(ii) Repeater
Answer: (i) **Transducer:** A device that converts one form of energy into another. In communication, it converts non-electrical signals (such as sound or voice) into electrical signals so they can be processed and transmitted.
(ii) **Repeater:** A system that combines a receiver, amplifier, and transmitter. It receives weak signals from a transmitter, boosts their power, and retransmits them to extend the total range of the communication system.
In simple words: (i) A transducer acts like a translator, turning sound into electrical signals. (ii) A repeater acts like a relay runner, catching a weak signal, making it strong again, and throwing it further along.

Exam Tip: Make sure to describe a repeater as a combination of three components: a receiver, an amplifier, and a transmitter.

Optics

Question 1. An object is held at the principal focus of a concave lens of focal length f. Where is the image formed?
Answer: For a concave lens, when an object is placed at its focus (\( u = -f \)), we use the lens formula:
\( \frac{1}{v} - \frac{1}{u} = \frac{1}{f'} \)
where the focal length is \( f' = -f \).
\( \frac{1}{v} - \frac{1}{-f} = -\frac{1}{f} \implies \frac{1}{v} + \frac{1}{f} = -\frac{1}{f} \implies \frac{1}{v} = -\frac{2}{f} \implies v = -\frac{f}{2} \).
Therefore, a virtual and erect image is formed at a distance of \( \frac{f}{2} \) from the optical center, on the same side of the lens as the object (between the optical center and the focus).
In simple words: A concave lens always makes a smaller, upright, virtual image. If you place an object at its focus, the image will appear exactly halfway between the lens and the focus on the same side.

Exam Tip: Use the lens formula to mathematically prove the position of the image. This rigorous approach guarantees full credit.

Question 2. What is the geometrical shape of the wavefront when a plane wave passes through a convex lens?
Answer: When a plane wavefront passes through a refracting convex lens, it converges toward the principal focus. Consequently, the wavefront emerges as a spherical wavefront of decreasing radius.
In simple words: A flat wavefront passing through a magnifying lens gets bent inward. It turns into a curved, spherical wave that shrinks as it heads toward a single focus point.

Exam Tip: Draw a simple diagram showing parallel lines (representing the plane wave) passing through the lens and bending to a point to visually support your spherical wavefront explanation.

Question 3. A diverging lens of focal length ‘F’ is cut into two identical parts each forming a plano-concave lens. What is the focal length of each part?
Answer: According to the lens maker's formula, the focal length \( f \) of a lens is related to the radii of curvature. When a symmetrical concave lens of focal length \( F \) is cut vertically along its principal axis into two identical plano-concave lenses, the focal length of each individual part becomes twice that of the original lens:
\( F' = 2F \).
In simple words: Cutting a concave lens in half makes each piece flatter on one side. This cuts its bending power in half, so its focal length doubles.

Exam Tip: Use the lens maker's formula to show mathematically why the focal length doubles when one surface becomes flat (\( R_2 = \infty \)).

Question 4. How the angular separation of interference fringes in Young’s double slit experiment change when the distance between the slits and screen is doubled?
Answer: The angular separation \( \theta \) of interference fringes is given by the formula \( \theta = \frac{\lambda}{d} \), where \( \lambda \) is the wavelength of light and \( d \) is the distance between the slits. Since the distance \( D \) between the slits and the screen does not appear in this expression, the angular separation is independent of \( D \). Therefore, the angular separation remains unchanged when the screen distance is doubled.
In simple words: The angle between the light stripes doesn't care how far away the screen is. Moving the screen back makes the stripes bigger but keeps the angle between them exactly the same.

Exam Tip: Do not confuse linear fringe width (\( \beta = \frac{\lambda D}{d} \)), which depends on \( D \), with angular fringe width (\( \theta = \frac{\lambda}{d} \)), which does not.

Question 5. Two thin lenses of power +6 D and – 2 D are in contact. What is the focal length of the combination?
Answer: The total power \( P \) of the lens combination is the algebraic sum of the individual powers:
\( P = P_1 + P_2 = +6\text{ D} + (-2\text{ D}) = +4\text{ D} \)
The focal length \( f \) of the combination is:
\( f = \frac{1}{P} = \frac{1}{4}\text{ m} = 0.25\text{ m} = 25\text{ cm} \).
Thus, the focal length is \( 25\text{ cm} \).
In simple words: Adding a +6 diopter lens and a -2 diopter lens together gives a net power of +4 diopters. Dividing 100 by this power gives a combined focal length of 25 centimeters.

Exam Tip: Always ensure you write the unit of focal length in meters first before converting it to centimeters to show a systematic calculation.

Question 6. Two thin lenses of power +5 D and –2.5 D are in contact. What is the focal length of the combination?
Answer: The combined power \( P \) of the two lenses is:
\( P = P_1 + P_2 = +5\text{ D} + (-2.5\text{ D}) = +2.5\text{ D} \)
The equivalent focal length \( f \) is:
\( f = \frac{1}{P} = \frac{1}{2.5}\text{ m} = 0.4\text{ m} = 40\text{ cm} \).
Thus, the combined focal length is \( 40\text{ cm} \).
In simple words: Combining these two lenses gives a total power of +2.5 diopters, which translates to a focal length of 40 centimeters.

Exam Tip: State that the positive sign of the combined power indicates that the lens combination behaves as a converging lens.

Question 7. A converging lens is kept co-axially in contact with a diverging lens – both the lenses being of equal focal lengths. What is the focal length of the combination?
Answer: Let the focal length of the converging lens be \( +f \) and that of the diverging lens be \( -f \).
The total power of the combination is:
\( P = P_1 + P_2 = \frac{1}{f} + \left(-\frac{1}{f}\right) = 0 \)
The equivalent focal length \( F \) of the combination is:
\( F = \frac{1}{P} = \frac{1}{0} = \infty \).
Therefore, the focal length of the combination is infinite, and it behaves as a simple plane glass plate.
In simple words: Since one lens bends light inward and the other bends it outward by the exact same amount, they cancel each other out completely. Light passes straight through without bending, which is like having an infinite focal length.

Exam Tip: Explain that a combination with zero power and infinite focal length acts physically like a flat glass sheet.

Question 8. When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a decrease in the energy carried by the light wave? Justify your answer.
Answer: No, the decrease in speed does not imply a decrease in the energy carried by the light wave. According to quantum theory, the energy of a photon depends entirely on its frequency (\( E = h \nu \)). When light transitions from a rarer to a denser medium, its frequency remains constant while its speed and wavelength decrease. Consequently, the energy carried by the wave remains unchanged.
In simple words: No. The energy of light depends on its frequency (color), not its speed. Since frequency does not change when passing into glass or water, the light keeps all its energy.

Exam Tip: Emphasize that frequency is a fundamental property of the source of light and remains invariant during refraction.

Question 9. How does the angular separation between fringes in single-slit diffraction experiment change when the distance of separation between the slit and screen is doubled?
Answer: The angular separation \( \theta \) of the diffraction fringes is given by \( \theta = \frac{\lambda}{d} \), where \( \lambda \) is the wavelength of light and \( d \) is the width of the slit. Since the distance \( D \) between the slit and the screen is not present in this equation, the angular separation does not depend on \( D \). Hence, doubling the screen distance has no effect on the angular separation.
In simple words: The angle of the spread depends only on the wavelength of light and the slit size, not on how far away the wall is. Therefore, moving the wall back won't change this angle.

Exam Tip: Clearly distinguish between the angular separation \( \theta = \frac{\lambda}{d} \) and the linear width of the central maximum (\( \beta = \frac{2\lambda D}{d} \)) which does depend on \( D \).

Question 10. For the same value of angle incidence, the angles of refraction in three media A, B and C are 15°,25° and 35° respectively. In which medium would the velocity of light be minimum?
Answer: According to Snell's law, the refractive index of a medium is \( n = \frac{\sin i}{\sin r} = \frac{c}{v} \), which gives \( v \propto \sin r \) for a constant angle of incidence \( i \). Since the angle of refraction \( r \) is smallest in medium A (\( 15^\circ \)), the sine value \( \sin(15^\circ) \) is also the smallest. This implies that medium A has the highest refractive index, and consequently, the velocity of light will be minimum in medium A.
In simple words: The more light bends (meaning a smaller angle of refraction), the slower it is traveling. Since light bends the most in medium A, that is where it travels the slowest.

Exam Tip: State the relationship \( v \propto \sin r \) explicitly to construct a clear, logically sound explanation.

Question 11. In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
Answer: The linear width of the central maximum in a single-slit diffraction pattern is given by \( w = \frac{2 D \lambda}{d} \).
(i) If the slit width \( d \) is doubled, the width of the central maximum is halved, making the central band narrower and more focused.
(ii) The intensity of the diffraction pattern is proportional to the square of the slit width (\( I \propto d^2 \)). Since the slit width is doubled, the intensity of the central band increases by a factor of four.
In simple words: Doubling the width of the slit squeezes the central bright spot to half its original size, but makes it four times brighter because more light is let through.

Exam Tip: Explain both parts of the question separately: the geometric size change (halved) and the intensity change (increased by four times).

Question 12. How does the fringe width, in Young’s double-slit experiment, change when the distance ofseparation between the slits and screen is doubled?
Answer: The fringe width \( \beta \) in Young's double-slit experiment is given by the formula:
\( \beta = \frac{D \lambda}{d} \)
Here, \( D \) is the distance between the slits and the screen. Since fringe width is directly proportional to \( D \) (\( \beta \propto D online ), doubling the distance \( D \) will double the fringe width.
In simple words: The width of the light bands is directly tied to the distance to the screen. If you double that distance, the bands on the screen will become twice as wide.

Exam Tip: Write down the fringe width formula \( \beta = \frac{D \lambda}{d} \) to clearly justify the direct proportionality.

Question 13. Draw a labeled ray diagram to show the image formation in a refracting type astronomical telescope. Why should the diameter of the objective of a telescope be large?
Answer: An astronomical telescope requires a large objective diameter for two key reasons:
1. **Light Gathering Power:** A larger objective lens collects more light from faint, distant celestial bodies, producing a much brighter and clearer image.
2. **Resolving Power:** The resolving power of a telescope is directly proportional to its aperture diameter (\( \text{R.P.} = \frac{D}{1.22 \lambda} \)). A larger diameter allows the telescope to resolve and distinguish closely spaced celestial objects.

Ray Diagram:

In simple words: A larger front lens acts like a bigger bucket for light, catching more light from dim, faraway stars to make them look brighter. It also improves the telescope's ability to see fine details.

Exam Tip: Be sure to practice drawing this ray diagram in normal adjustment (where the final image is formed at infinity) as it is a highly popular exam question.

Question 14. Define resolving power of a compound microscope. How does the resolving power of a compound microscope change when (i) Refractive index of the medium between the object and objective lens increases? (ii) Wavelength of the radiation used is increased?
Answer: The resolving power of a compound microscope is defined as the reciprocal of the minimum distance between two point objects that can be seen as separate and distinct. The expression is:
\( \text{Resolving Power} = \frac{2 n \sin\theta}{1.22 \lambda} \)
(i) When the refractive index \( n \) of the medium between the object and the objective lens increases, the resolving power increases because \( \text{Resolving Power} \propto n \).
(ii) When the wavelength \( \lambda \) of the incident radiation is increased, the resolving power decreases because \( \text{Resolving Power} \propto \frac{1}{\lambda} \).
In simple words: Resolving power is how well a microscope can separate two tiny spots that are close together. Using a liquid with a higher refractive index improves this detail, but using light with a longer wavelength makes the image blurrier.

Exam Tip: State the mathematical formula for the resolving power of a microscope to serve as the foundation for your explanations.

Question 15. Define resolving power of a telescope. How does it get affected on (i) Increasing the aperture of the objective lens? (iii) Increasing the focal length of the objective lens?
Answer: The resolving power of a telescope is defined as the reciprocal of the smallest angular separation between two distant objects that can just be resolved. The formula is:
\( \text{Resolving Power} = \frac{D}{1.22 \lambda} \)
where \( D \) is the aperture of the objective lens.
(i) **Increasing the aperture (diameter \( D \)) of the objective lens:** This increases the resolving power of the telescope because \( \text{Resolving Power} \propto D \).
(iii) **Increasing the focal length of the objective lens:** The resolving power remains completely unaffected because the formula for resolving power is entirely independent of the focal length of the lens.
In simple words: A telescope's resolving power is its ability to separate two close-together stars. Making the front lens wider improves this ability, but changing the lens's focal length has no effect on it.

Exam Tip: Point out that while focal length affects the magnification of the telescope, it has no bearing on its resolving power.