Physics — Class 12 practice questions

Worked examples for Physics — Class 12

1. Question 1

   The amplitude of a damped oscillator falls to half its initial value in $t_{h} = 20\ \text{s}$.

   The time in which its total energy falls to half its initial value is

   1. A. $20\ \text{s}$

      You gave the energy the same decay constant as the amplitude. Energy goes as the SQUARE of the amplitude, so it falls away faster and cannot halve in the same time.

   2. B. $10\ \text{s}$Correct answer

      $A = A_{0}e^{-bt/2m}$ and $E \propto A^{2}$, so $E = E_{0}e^{-bt/m}$: the energy's exponent is twice the amplitude's, so it reaches half in half the time, $10\ \text{s}$.

   3. C. $40\ \text{s}$

      You multiplied the $20\ \text{s}$ by $2$ where it must be divided by $2$. Squaring a decaying amplitude doubles the exponent, which shortens the half-time instead of lengthening it.

   4. D. $5\ \text{s}$

      You took the energy's exponent as four times the amplitude's. Squaring $A_{0}e^{-bt/2m}$ gives $e^{-bt/m}$, which is twice the exponent, not four times.

   Explanation

   In a damped oscillation the amplitude decays as $A = A_{0}e^{-bt/2m}$, and since the total energy of an oscillator goes as the square of its amplitude, $E = E_{0}e^{-bt/m}$. The energy therefore decays with exactly twice the exponent of the amplitude, so whatever fraction the amplitude needs a time $t$ to reach, the energy reaches in $\frac{t}{2}$. The damping constant $\frac{b}{m}$ never has to be evaluated — it is fixed by the given half-time and cancels.

2. Question 2

   A series LCR circuit is connected to an AC supply.

   $V_{rms} = 150\ \text{V}$

   $R = 30\ \Omega$

   $X_{L} = 20\ \Omega$

   $X_{C} = 20\ \Omega$

   The rms current in the circuit is

   1. A. $5\ \text{A}$Correct answer

      $X_{L} = X_{C}$, so the two reactances cancel and $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}} = R = 30\ \Omega$. Hence $I_{rms} = \frac{150}{30} = 5\ \text{A}$.

   2. B. $3\ \text{A}$

      You added the two reactances instead of subtracting them: $\sqrt{30^{2} + 40^{2}} = 50\ \Omega$ gives $3\ \text{A}$. The inductive and capacitive voltages are opposite in phase, so $X_{L}$ and $X_{C}$ subtract.

   3. C. $7.5\ \text{A}$

      You divided the supply voltage by a reactance, $\frac{150}{20}$, instead of by the impedance. A reactance alone is the impedance only when the resistance is absent.

   4. D. $750\ \text{W}$

      You worked out $\frac{V_{rms}^{2}}{R}$, which is the power the circuit dissipates, and offered it as the current. A rate of energy supplied is not a flow of charge.

   Explanation

   In a series LCR circuit the inductive and capacitive voltages are in antiphase, so the impedance is $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$. When the two reactances are equal the circuit is at resonance: their effects cancel exactly, the impedance falls to the resistance alone, and the current is the largest the supply can drive. The whole question is the recognition that $X_{L} = X_{C}$ leaves $Z = R$.

3. Question 3

   A moving coil galvanometer is made to work as an ammeter by joining a shunt across it.

   $G = 40\ \Omega$ (resistance of the galvanometer)

   $I_{g} = 5\ \text{mA}$ (current for full scale deflection)

   $S = 10\ \Omega$ (resistance of the shunt)

   The range of the ammeter so formed is

   1. A. $20\ \text{mA}$

      You have found the current in the shunt alone, $\frac{I_{g}G}{S}$, and stopped there. The ammeter reads the current entering the instrument, so the $5\ \text{mA}$ flowing through the galvanometer itself must be added back.

   2. B. $6.25\ \text{mA}$

      You have used $I_{g}\left(1 + \frac{S}{G}\right)$, with $G$ and $S$ interchanged. The branch of SMALLER resistance carries the larger current, so dividing by $G$ makes the shunt carry less than the galvanometer, which is the opposite of what a shunt is for.

   3. C. $25\ \text{mV}$

      The arithmetic is right but the quantity is not: the range of an ammeter is a current, in amperes, not a potential difference. A millivolt answer here belongs to a voltmeter conversion, where a large resistance goes in series instead.

   4. D. $25\ \text{mA}$Correct answer

      At full scale the shunt carries $I_{s} = \frac{I_{g}G}{S} = 20\ \text{mA}$, and the range is the total current, $I = I_{g}\left(1 + \frac{G}{S}\right) = 5 \times 5 = 25\ \text{mA}$.

   Explanation

   A shunt across a galvanometer is a current divider: the same potential difference sits across both branches, so they carry currents in inverse proportion to their resistances, and at full scale the shunt carries $I_{g}G/S$. The range of the instrument is the total current drawn at that moment, $I = I_{g} + I_{g}G/S = I_{g}(1 + G/S)$, so the galvanometer's own share is never dropped. A small shunt beside a large galvanometer resistance therefore widens the range in the ratio $(G + S)/S$, which is exactly why the shunt is made small.

4. Question 4

   A pipe closed at one end is sounded in its first overtone together with a tuning fork, and beats are heard. The note of the pipe is the lower of the two.

   $f_{s} = 404\ \text{Hz}$ (frequency of the fork)

   $b = 4\ \text{Hz}$ (beat frequency)

   $v = 320\ \text{m/s}$ (speed of sound in air)

   The length of the pipe is

   1. A. $0.20\ \text{m}$

      You put the $400\ \text{Hz}$ into $L = \frac{v}{4f}$, which is the FUNDAMENTAL of a closed pipe. The pipe is sounding one mode higher, at three times its fundamental.

   2. B. $400\ \text{Hz}$

      You stopped at the frequency of the pipe, which is only the first move. A length was asked for, so the closed-pipe relation $L = \frac{3v}{4f}$ still has to be used.

   3. C. $0.40\ \text{m}$

      You read the first overtone as the second harmonic, $\frac{2v}{4L}$. A pipe closed at one end has a node at the closed end and an antinode at the open end, so it supports only odd harmonics and its first overtone is $3f_{0}$.

   4. D. $0.60\ \text{m}$Correct answer

      The pipe is the lower note, so it sounds at $404 - 4 = 400\ \text{Hz}$. The first overtone of a closed pipe is its third harmonic, $f = \frac{3v}{4L}$, giving $L = \frac{3 \times 320}{4 \times 400} = 0.60\ \text{m}$.

   Explanation

   Two notes sounded together beat at the difference of their frequencies, so a pipe heard as the lower note against a $404\ \text{Hz}$ fork with $4$ beats per second is sounding at $400\ \text{Hz}$. That frequency is then the input to a second relation: a pipe closed at one end carries a node at the closed end and an antinode at the open end, admits odd harmonics only, and so has its first overtone at $\frac{3v}{4L}$. The stem states which note is lower precisely so that the beat relation has a single reading and the whole work lies in chaining the two laws.

5. Question 5

   The magnetic flux linked with a closed conducting loop of fixed resistance is changed by the same amount twice over: once in a short time, and once in a long time.

   The quantity that has the same value in both cases is

   1. A. the emf induced in the loop

      Faraday's law is about a rate: $\varepsilon = \frac{\Delta\Phi}{\Delta t}$. You used the flux change itself in place of the rate, so the quick change gives the larger emf, not an equal one.

   2. B. the heat produced in the loop

      The heat is $\frac{\varepsilon^{2}}{R}\Delta t = \frac{(\Delta\Phi)^{2}}{R,\Delta t}$. One factor of $\Delta t$ survives in the denominator, so the quicker change dissipates more heat — a faster change costs more work.

   3. C. the average current in the loop

      You applied the charge's time-independence to the current. $I = \frac{\Delta\Phi}{R,\Delta t}$ still carries $\Delta t$, so the current is larger for the quicker change even though the charge is the same.

   4. D. the charge that circulates in the loopCorrect answer

      $q = I,\Delta t = \frac{\Delta\Phi}{R,\Delta t}\times\Delta t = \frac{\Delta\Phi}{R}$. The time cancels, so the same flux change drives the same charge round the same loop however fast it is made.

   Explanation

   Faraday's law fixes the induced emf as the rate of change of flux, so emf, current and the heat they produce all depend on how long the change takes. The charge is the current integrated over that time, and the two time factors cancel, leaving $q = \frac{\Delta\Phi}{R}$ — a quantity set by the total flux change and the resistance alone. This is why a ballistic galvanometer, which responds to charge, measures a flux change without any timing being needed.

6. Question 6

   A steady current flows through a coil. The current is then raised, and the energy stored in the coil's magnetic field rises.

   $U_{1} = 9\ \text{J}$ (stored energy before the rise)

   $\Delta I = 2\ \text{A}$ (rise in the current)

   $U_{2} = 25\ \text{J}$ (stored energy after the rise)

   The self-inductance of the coil is

   1. A. $1\ \text{H}$

      You found $I_{1} = 3\ \text{A}$ and then used $U = LI^{2}$. The stored energy is $\frac{1}{2}LI^{2}$, so the inductance is twice what you wrote.

   2. B. $2\ \text{H}$Correct answer

      $U \propto I^{2}$, so $\frac{I_{2}}{I_{1}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$. With $I_{2} = I_{1} + 2\ \text{A}$ this gives $3(I_{1} + 2) = 5I_{1}$, so $I_{1} = 3\ \text{A}$; then $L = \frac{2U_{1}}{I_{1}^{2}} = \frac{18}{9} = 2\ \text{H}$.

   3. C. $8\ \text{H}$

      You set the energy increase equal to $\frac{1}{2}L(\Delta I)^{2}$. Energy is not linear in the current, so the extra energy is $\frac{1}{2}L\left(I_{2}^{2} - I_{1}^{2}\right)$, which is not $\frac{1}{2}L(\Delta I)^{2}$ — the two differ by $LI_{1}\Delta I$.

   4. D. $3\ \text{A}$

      This is the original current, which you had to find on the way, offered as the answer. The question asks for the inductance, which is measured in henry; one more substitution into $U_{1} = \frac{1}{2}LI_{1}^{2}$ finishes the work.

   Explanation

   The energy stored in an inductor goes as the square of the current, so the two energy readings fix the ratio of the two currents even though neither current is stated. Combining that ratio with the stated increase in current recovers the original current, and one substitution into $U = \frac{1}{2}LI^{2}$ then gives the inductance. The point of the question is that the withheld current can be eliminated rather than guessed.

7. Question 7

   A conducting rod slides without friction at a constant speed along two long parallel rails of negligible resistance, closed at one end by a resistor. A uniform magnetic field acts perpendicular to the plane of the rails.

   $l = 0.4\ \text{m}$ (length of the rod)

   $v = 5\ \text{m/s}$ (speed of the rod)

   $R = 2\ \Omega$ (resistance of the resistor)

   $B = 0.5\ \text{T}$ (magnetic field)

   The external force that must be applied to the rod to keep its speed constant is

   1. A. $0.5\ \text{W}$

      This is the power dissipated in the resistor, $\frac{\varepsilon^{2}}{R}$, not the force. It is the rate at which your force does work; dividing it by the speed $v$ returns the force.

   2. B. $0.2\ \text{N}$

      You found the current $I = 0.5\ \text{A}$ correctly, then wrote the force as $Il$ and left the field out of it. The force on a current-carrying rod is $BIl$, so the field enters the answer twice — once through the emf and once through the force.

   3. C. $0.25\ \text{N}$

      You squared the field but not the length: $\frac{B^{2}lv}{R}$. The length appears once in $\varepsilon = Blv$ and again in $F = BIl$, so it is squared exactly as the field is.

   4. D. $0.1\ \text{N}$Correct answer

      The rod develops $\varepsilon = Blv = 1\ \text{V}$, which drives $I = \frac{\varepsilon}{R} = 0.5\ \text{A}$, and the field then opposes the motion with $F = BIl = \frac{B^{2}l^{2}v}{R} = 0.1\ \text{N}$. The applied force must balance it.

   Explanation

   A rod moving across a magnetic field is a source of emf $Blv$; the closed circuit turns that emf into a current $\frac{Blv}{R}$, and the field exerts a force $BIl$ on that current which, by Lenz's law, opposes the motion. Since the speed is constant the applied force exactly balances this opposing force, giving $\frac{B^{2}l^{2}v}{R}$. The field and the length each enter the chain twice, which is why both appear squared.

8. Question 8

   A cell is balanced against a potentiometer wire with nothing connected across it. A resistance box is then joined across the same cell, and the balancing length falls.

   $l_{1} = 75\ \text{cm}$ (balancing length with nothing across the cell)

   $R = 4\ \Omega$ (resistance set on the box)

   $l_{2} = 50\ \text{cm}$ (balancing length with the box joined across the cell)

   The internal resistance of the cell is

   1. A. $6\ \Omega$

      From $\frac{E}{V} = \frac{l_{1}}{l_{2}}$ the internal resistance is $r = R\left(\frac{l_{1}}{l_{2}} - 1\right)$; you reported $R\frac{l_{1}}{l_{2}} = 6\ \Omega$ and never subtracted the $R$ that stands for the load itself.

   2. B. $1.33\ \Omega$

      You divided the difference by the OPEN-circuit length, $4 \times \frac{25}{75}$. The shortfall $l_{1} - l_{2}$ belongs to the drop inside the cell and must be compared with $l_{2}$, the length that measures the terminal voltage.

   3. C. $2\ \Omega$Correct answer

      The same gradient $k$ serves both readings, so $\frac{E}{V} = \frac{l_{1}}{l_{2}} = \frac{75}{50}$. With $V = \frac{ER}{R + r}$ this gives $r = R\frac{l_{1} - l_{2}}{l_{2}} = 4 \times \frac{25}{50} = 2\ \Omega$.

   4. D. $8\ \Omega$

      You turned the fraction over, $4 \times \frac{50}{25}$. That makes the internal resistance twice the load it is feeding, which no cell balanced at two-thirds of its open-circuit length can have.

   Explanation

   A potentiometer reads a potential difference as a length, through a gradient the stem never states. Taking both readings on the same wire with the same driving current makes that gradient common to them, so dividing the open-circuit reading by the loaded one eliminates it and leaves a pure ratio of lengths. Setting that ratio equal to $\frac{E}{V} = \frac{R + r}{R}$ then gives the internal resistance from the known load alone.