JEE Physics Current Electricity MCQs Set 25

Answer: (c) a 2% increase

Explanation:
1. Volume constant: the length rises by about 1% as the area falls by 1%.
2. \( R\propto L/A \) so \( \Delta R/R\approx1\%+1\%=2\% \) (increase).

Teacher's Note:
1. Fractional changes of L and 1/A add for small changes.
2. Constant volume links length and area.

Answer: (a) 4R

Explanation:
1. Length doubles so the area halves.
2. \( R'=\rho\dfrac{2L}{A/2}=4R \).

Teacher's Note:
1. Volume conservation links length and area.
2. Resistance rises as the square of the length factor.

Answer: (d) \( E_1r_2>E_2(R+r_1) \)

Explanation:
1. Initially \( I=\dfrac{E_1+E_2}{R+r_1+r_2} \); after shorting battery 2 (wire across it), \( I'=\dfrac{E_1}{R+r_1} \).
2. The current rises if \( E_1(R+r_1+r_2)>(E_1+E_2)(R+r_1) \), i.e. \( E_1r_2>E_2(R+r_1) \).

Teacher's Note:
1. A short across a battery removes both its emf and its internal resistance.
2. Compare the currents before and after.

Answer: (c) \( 2E\left(1-\dfrac2n\right) \)

Explanation:
1. Net emf \( (n-4)E \), total internal resistance nr, so \( I=\dfrac{(n-4)E}{nr} \).
2. A reversed cell has current flowing against its emf: \( V=E+Ir=E+\dfrac{(n-4)E}{n}=2E\left(1-\dfrac2n\right) \).

Teacher's Note:
1. For a cell being charged, the terminal voltage exceeds its emf.
2. Find the current in the whole loop first.

Answer: (b) 0.2 A

Explanation:
1. Let the junction potential be V: \( \dfrac{10-V}{10}=\dfrac{V-6}{20}+\dfrac{V-5}{30} \).
2. Multiply by 60: \( 6(10-V)=3(V-6)+2(V-5) \) gives \( V=8\ \text{V} \).
3. Current through R₁: \( \dfrac{10-8}{10}=0.2\ \text{A} \).

Teacher's Note:
1. Apply Kirchhoff's junction rule at O.
2. The current through R₁ flows into the junction.

Answer: (c) the current delivered by the battery will increase

Explanation:
1. An ideal battery keeps the potential across R fixed, so the current through R and its power are unchanged.
2. The new resistor draws extra current, and the battery supplies it: total current increases.

Teacher's Note:
1. Ideal battery means zero internal resistance, so terminal voltage is constant.
2. Parallel branches are independent when the voltage is fixed.

Answer: (b) \( P_2>P_3>P_4>P_1 \)

Explanation:
1. For equal current \( P=I^2R_{eq} \).
2. Equivalent resistances: (i) \( r/3 \), (ii) \( 3r \), (iii) \( 1.5r \), (iv) \( \dfrac{2r\cdot r}{3r}=\dfrac{2r}{3} \).
3. Order: (ii) > (iii) > (iv) > (i).

Teacher's Note:
1. At fixed current, power is proportional to the equivalent resistance.
2. Work out each equivalent resistance.

Answer: (a) 37.5 W

Explanation:
1. Resistances: \( \dfrac{200^2}{60}=667\ \Omega \) and \( \dfrac{200^2}{100}=400\ \Omega \).
2. Total 1067 Ω; \( P=\dfrac{200^2}{1067}=37.5\ \text{W} \).

Teacher's Note:
1. Each bulb's resistance comes from its rating.
2. Series total power is V squared over the total resistance.

Answer: (c) 6.7 W

Explanation:
1. Each bulb has \( R=\dfrac{120^2}{60}=240\ \Omega \).
2. In series, total 720 Ω; \( P_{total}=\dfrac{120^2}{720}=20\ \text{W} \).
3. Each bulb dissipates \( 20/3=6.7\ \text{W} \).

Teacher's Note:
1. The three bulbs share the voltage equally.
2. Each gets 40 V, one third of its rated voltage.

Answer: (c) \( \dfrac{4q^2R}{3\Delta t} \)

Explanation:
1. Linear decrease: \( i=i_0(1-t/\Delta t) \), and the charge is the area \( q=\dfrac{i_0\Delta t}{2} \), so \( i_0=\dfrac{2q}{\Delta t} \).
2. \( H=\int_0^{\Delta t}i^2R\,dt=\dfrac{i_0^2R\Delta t}{3}=\dfrac{4q^2R}{3\Delta t} \).

Teacher's Note:
1. Charge is the area under the current-time graph.
2. Heat is the integral of i squared R.

Answer: (b) 60 Ω

Explanation:
1. Voltage across the parallel set: \( 0.3\times20=6\ \text{V} \).
2. Current in 15 Ω: \( 6/15=0.4\ \text{A} \); current in R₁: \( 0.8-0.3-0.4=0.1\ \text{A} \).
3. \( R_1=6/0.1=60\ \Omega \).

Teacher's Note:
1. Parallel branches share the same voltage.
2. Use the junction rule for the main current.

Answer: (a) (1), (3) and (4) only

Explanation:
1. The balance condition \( P/Q=R/S \) does not involve the emf, so (1) holds.
2. Multiplying all resistances by 5 keeps the ratios: (3) holds; adding 10 Ω to all changes the ratios: (2) fails.
3. Interchanging the battery and galvanometer leaves the bridge balance condition unchanged: (4) holds.

Teacher's Note:
1. Balance depends only on resistance ratios.
2. The conjugate arms property makes the interchange valid.

Answer: (d) a voltmeter of range 10 V

Explanation:
1. A voltmeter of range V has resistance about \( V/I_g \): 10 kΩ for 10 V and 5 kΩ for 5 V.
2. An ammeter has a very small resistance (the shunt). So the largest is the 10 V voltmeter.

Teacher's Note:
1. Voltmeter resistance grows with its range, ammeter resistance falls as range increases.
2. Use V = I_g R_total.

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. For charging, the charger voltage must exceed the emf: \( V_{AB}=E+Ir>E \), so (i) and (ii) hold.
2. Current enters the battery's positive terminal and leaves through the negative: (iii) holds, so (iv) fails.

Teacher's Note:
1. Charging current runs opposite to the direction in which the battery normally drives current.
2. The charger's positive terminal is connected to the battery's positive terminal.

Answer: (a) (i) and (iii) only

Explanation:
1. Power in the 2 Ω resistor is \( I^2\times2 \), greatest when the total resistance is least.
2. The parallel part is least (zero) when R = 0 (shorting the 8 Ω), giving \( I=\dfrac{12}{2}=6\ \text{A} \).
3. \( P=6^2\times2=72\ \text{W} \).

Teacher's Note:
1. Maximize the current through the fixed resistor, not the power in R.
2. A short across the parallel part gives the largest current.

Answer: (a) (i) and (iii) only

Explanation:
1. \( R_1=\dfrac{10}{0.01}=1\ \text{k}\Omega \) and \( R_2=\dfrac{220}{0.05}=4.4\ \text{k}\Omega \): (i) true.
2. The filament is hotter at 220 V, so its resistance is higher (a non-ohmic device): (iii) true, (ii) false.

Teacher's Note:
1. Ohm's law holds only at constant temperature.
2. A tungsten filament's resistance rises with temperature.

Answer: (a) (i) and (ii) only

Explanation:
1. The potential drop along the wire must be able to balance E₂, so E₁ > E₂: (i).
2. The opposing emfs must act in the same sense along the wire, so like terminals go to the same end (either positive or negative, not only positive): (ii) true, (iii) false.
3. There is no requirement on the galvanometer's resistance: (iv) false.

Teacher's Note:
1. The measured cell must oppose the driver cell in the wire.
2. The galvanometer only detects the null point.

Answer: (c) \( 1.56\times10^{19} \)

Explanation:
1. Charge \( q=It=0.5\times10=5\ \text{C} \).
2. Each Cu²⁺ ion carries \( 2e=3.2\times10^{-19}\ \text{C} \).
3. Number \( =\dfrac{5}{3.2\times10^{-19}}=1.56\times10^{19} \).

Teacher's Note:
1. Each doubly charged ion carries two elementary charges.
2. Charge is current times time.

Answer: (a) 25 V/m

Explanation:
1. \( V=IR=10\times5=50\ \text{V} \).
2. \( E=V/L=50/2=25\ \text{V/m} \).

Teacher's Note:
1. The field in a uniform wire is potential difference over length.
2. Do not forget to compute V first.

Answer: (b) \( R/9 \)

Explanation:
1. Each piece has resistance \( R/3 \).
2. Three pieces side by side (in parallel): \( \dfrac{R/3}{3}=\dfrac R9 \).

Teacher's Note:
1. Side by side means the cross-section is three times as large and the length one third.
2. Resistance falls by 3 × 3 = 9.

Answer: (d) 27 : 6 : 1

Explanation:
1. \( R=\dfrac{\rho L}{A}=\dfrac{\rho L^2}{V}\propto\dfrac{L^2}{m} \).
2. Ratios \( \dfrac{9}{1}:\dfrac42:\dfrac13=9:2:\dfrac13=27:6:1 \).

Teacher's Note:
1. Volume is proportional to mass for the same metal.
2. Express resistance in terms of length and volume.

Answer: (d) 7.5 V

Explanation:
1. \( V=E-Ir=12-90\times0.05=7.5\ \text{V} \).

Teacher's Note:
1. Terminal voltage drops by Ir when current is drawn.
2. A large starter current causes a big drop.

Answer: (b) 24 mA

Explanation:
1. Maximum current is on short-circuit: \( I=\dfrac{E}{r}=\dfrac{12}{500}=0.024\ \text{A}=24\ \text{mA} \).

Teacher's Note:
1. The internal resistance limits the short-circuit current.
2. This is why an old battery cannot start a car.