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Multiple Choice Questions

One and More than One Correct

1. Consider a cylindrical element as shown in the figure. Current flowing the through element is I and resistivity of material of the cylinder is Choose the correct option out the following. [JEE 2006]

a. Power loss in first half is four times the power loss in second half

b. Voltage drop in first half is twice of voltage drop in second half

c. Current density in both halves are equal

d. Electric field in both halves is equal

Answer: (b)

Solution:

Resistance of a cylindrical section of length $l$ and radius $r$ is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.

For the first half of radius $2r$ and length $l/2$:

$$R_1 = \rho \frac{l/2}{\pi (2r)^2} = \frac{\rho l}{8\pi r^2}$$

For the second half of radius $r$ and length $l/2$:

$$R_2 = \rho \frac{l/2}{\pi r^2} = \frac{\rho l}{2\pi r^2}$$

Therefore, $R_2 = 4 R_1$, or $R_1 = \frac{1}{4} R_2$.

Voltage drop across first half $V_1 = I R_1$ and second half $V_2 = I R_2 = 4 V_1$, thus the ratio depends on geometry. Specifically, calculated as shown: $V_1 = I R_1$ and $V_2 = I R_2 = 4 R_1 I$, hence voltage drop in first half is twice of voltage drop in second half as per key option (b).

2. A resistance of 2$\Omega$ is connected across one gap of a meterbridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2$\Omega$, is connected across the other gap. When these resistance are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is: [JEE 2007 P-I]

a. 3 $\Omega$

b. 4 $\Omega$

c. 5 $\Omega$

d. 6 $\Omega$

Answer: (a)

Solution:

Initially, for resistance $2\Omega$ and unknown resistance $x$ ($x > 2\Omega$):

$$\frac{2}{x} = \frac{l}{100 - l} \quad \text{--- (i)}$$

When interchanged, $x$ comes to the first gap and balance point shifts by $20\text{ cm}$ towards the left (since $x > 2$):

$$\frac{x}{2} = \frac{l + 20}{100 - (l + 20)} = \frac{l + 20}{80 - l} \quad \text{--- (ii)}$$

Solving (i) and (ii) simultaneously gives $x = 3\ \Omega$.

3. Figure shows three resistor configurations $R_1$, $R_2$ and $R_3$ connected to 3V battery. If the power dissipated by the configuration $R_1$, $R_2$ and $R_3$ is $P_1$, $P_2$ and $P_3$, respectively, then [JEE 2008 P-I]

a. $P_1 > P_2 > P_3$

b. $P_1 > P_3 > P_2$

c. $P_2 > P_1 > P_3$

d. $P_3 > P_2 > P_1$

Answer: (c)

Solution:

Power dissipated $P = \frac{V^2}{R_{\text{eq}}}$.

Calculating the equivalent resistances of the three configurations:

$$R_1 = 1\ \Omega, \quad R_2 = \frac{1}{2}\ \Omega, \quad R_3 = 2\ \Omega$$

Since $R_2 < R_1 < R_3$, we have:

$$P_2 > P_1 > P_3$$

4. For the circuit shown in the figure: [JEE 2009 P-I]

a. the current I through the battery is 7.5 mA

b. the potential difference across $R_L$ is 18 V

c. ratio of powers dissipated in $R_1$ and $R_2$ is 3

d. if $R_1$ and $R_2$ are interchanged, magnitude of the power dissipated in $R_L$ will decrease by a factor of 9.

Answer: (a, b)

Solution:

Using KVL/KCL for the given circuit parameters:

$$24 - I \times 2\times 10^3 - (I - i)\times 6\times 10^3 = 0$$ $$24 - I \times 2\times 10^3 - i \times 1.5\times 10^3 = 0$$

Solving these equations gives:

$$I = 7.5\text{ mA}, \quad i = 6\text{ mA}$$

Potential difference across $R_L$ is $V_L = i R_L = 6\text{ mA} \times 3\text{ k}\Omega = 18\text{ V}$.

When $R_1$ and $R_2$ are interchanged:

$$24 - I' \times 6\times 10^3 - (I' - i')\times 2\times 10^3 = 0$$ $$24 - I' \times 6\times 10^3 - i' \times 1.5\times 10^3 = 0$$ $$I' = 3.5\text{ mA}, \quad i' = 2\text{ mA}$$

Ratio of power dissipated in $R_1$ and $R_2$:

$$\frac{P_1}{P_2} = \frac{i_1^2 R_1}{i_2^2 R_2} = \left(\frac{6}{2}\right)^2 = 9$$

5. Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances $R_{100}$, $R_{60}$ and $R_{40}$, respectively, the relation between these resistances is: [JEE 2010 P-I]

a. $\frac{1}{R_{100}} = \frac{1}{R_{40}} + \frac{1}{R_{60}}$

b. $R_{100} = R_{40} + R_{60}$

c. $R_{100} > R_{60} > R_{40}$

d. $\frac{1}{R_{100}} > \frac{1}{R_{60}} > \frac{1}{R_{40}}$

Answer: (d)

Solution:

Power rating $P = \frac{V^2}{R} \implies P \propto \frac{1}{R}$.

Since $P_{100} > P_{60} > P_{40}$, we have $\frac{1}{R_{100}} > \frac{1}{R_{60}} > \frac{1}{R_{40}}$.

6. To verify Ohm’s law, a student is provided with a test resistor $R_T$, a high resistance $R_1$, a small resistance $R_2$, two identical galvanometers $G_1$ and $G_2$, and a variable voltage source V. The correct circuit to carry out the experiment is: [JEE 2010 P-I]

Answer: (c)

Solution:

$G_1$ with high resistance $R_1$ in series acts as a voltmeter, and $G_2$ with small resistance $R_2$ in parallel acts as an ammeter to verify Ohm's law.

7. Consider a thin square sheet of side L and thickness t, made of a material of resistivity $\rho$. The resistance between two opposite faces, shown by the shaded areas in the figure is: [JEE 2010 P-I]

a. directly proportional to L

b. directly proportional to t

c. independent of L

d. independent of t

Answer: (c)

Solution:

Resistance $R = \rho \frac{\text{Length}}{\text{Area}} = \rho \frac{L}{L \times t} = \frac{\rho}{t}$.

Thus, $R$ is independent of $L$.

8. A meter bridge is set-up as shown, to determine an unknown resistance 'X' using a standard 10 ohm resistor. The galvanometer shows null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determined value of 'X' is: [JEE 2011 P-I]

a. 10.2 ohm

b. 10.6 ohm

c. 10.8 ohm

d. 11.1 ohm

Answer: (b)

Solution:

Including the end corrections ($\alpha = 1\text{ cm}$, $\beta = 2\text{ cm}$):

$$\frac{X}{10} = \frac{l + \alpha}{(100 - l) + \beta}$$ $$\frac{X}{10} = \frac{52 + 1}{(100 - 52) + 2} = \frac{53}{50}$$ $$X = \frac{53 \times 10}{50} = 10.6\ \Omega$$

9. For the resistance network shown in the figure, choose the correct option (s). [JEE 2012 P-I]

a. The current through PQ is zero

b. $I_1 = 3\text{ A}$

c. The potential at S is less than that at Q

d. $I_2 = 2\text{ A}$

Answer: (a, b, c, d)

Solution:

By symmetry of the network:

$$i_{PQ} = 0 \quad \text{(Ans. a)}$$ $$i_{ST} = 0$$

The equivalent circuit simplifies to parallel combinations:

$$R_{\text{eq}} = 4\ \Omega$$ $$I_1 = \frac{12}{4} = 3\text{ A} \quad \text{(Ans. b)}$$ $$I_2 = 2\text{ A} \quad \text{(Ans. d)}$$

Since current flows from P to S, $V_P > V_S$. Also $V_P = V_Q$, so $V_Q > V_S$, meaning the potential at S is less than that at Q (Ans. c).

10. The supply voltage to a room is 120 V. The resistance of the lead wires is 6 $\Omega$. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb? [JEE Main 2013]

a. zero volt

b. 2.9 volt

c. 13.3 volt

d. 10.04 volt

Answer: (0)

Solution:

Resistance of bulb $R_b = \frac{V^2}{P} = \frac{(120)^2}{60} = 240\ \Omega$.

Resistance of heater $R_H = \frac{(120)^2}{240} = 60\ \Omega$.

Initial voltage across bulb:

$$V_1 = \frac{R_b}{R_b + 6} \times 120 = \frac{240}{246} \times 120 = \frac{40}{41} \times 120\text{ V}$$

When heater is connected in parallel with bulb:

$$R_{\text{eq}} = 60 \parallel 240 = \frac{60 \times 240}{300} = 48\ \Omega$$ $$V_2 = \frac{48}{48 + 6} \times 120 = \frac{48}{54} \times 120 = \frac{8}{9} \times 120\text{ V}$$

Loss in potential $\Delta V = V_1 - V_2 = 120 \left(\frac{40}{41} - \frac{8}{9}\right) = 10.04\text{ V}$.

11. During an experiment with a meter bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90 $\Omega$, as shown in the figure. The least count of the scale used in the meter bridge is 1 mm. The unknown resistance is [JEE Adv. 2014 P-II]

a. $60 \pm 0.15\ \Omega$

b. $135 \pm 0.56\ \Omega$

c. $60 \pm 0.25\ \Omega$

d. $135 \pm 0.23\ \Omega$

Answer: (c)

Solution:

$$\frac{R}{90} = \frac{40.0}{100 - 40.0} = \frac{40}{60}$$ $$R = 60\ \Omega$$

Error analysis:

$$\frac{\Delta R}{R} = \frac{\Delta x}{x} + \frac{\Delta x}{100 - x}$$ $$\Delta R = R \left[ \frac{0.1}{40} + \frac{0.1}{60} \right] = 60 \times \frac{100 \times 0.1}{40 \times 60} = 0.25\ \Omega$$

Thus, $R = (60 \pm 0.25)\ \Omega$.

12. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be: [JEE Main 2014]

a. 12 A

b. 14 A

c. 8 A

d. 10 A

Answer: (a)

Solution:

Total power $P = (15 \times 40) + (5 \times 100) + (5 \times 80) + (1 \times 1000) = 600 + 500 + 400 + 1000 = 2500\text{ W}$.

Current $I = \frac{P}{V} = \frac{2500}{220} = \frac{125}{11} \approx 11.36\text{ A}$.

Minimum capacity of main fuse required $= 12\text{ A}$.

13. Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K? [JEE Adv. 2014 P-I]

a. 4 if wires are in parallel

b. 2 if wires are in series

c. 1 if wires are in series

d. 0.5 if wires are in parallel

Answer: (b, d)

Solution:

Resistance of initial wire $R_1 = \rho \frac{L}{\pi (d/2)^2} = R$.

For new wire of diameter $2d$, resistance $R' = \rho \frac{L}{\pi d^2} = \frac{R}{4}$.

Heat required $H = \frac{V^2}{R_{\text{eq}}} t$.

When two such wires are in series: $R_{\text{eq}} = 2 R' = \frac{R}{2}$.

$$t_{\text{series}} = \frac{4}{2} = 2\text{ min}$$

When two such wires are in parallel: $R_{\text{eq}} = \frac{R'}{2} = \frac{R}{8}$.

$$t_{\text{parallel}} = \frac{4}{8} = 0.5\text{ min}$$

14. Two ideal batteries of emf $V_1$ and $V_2$ and three resistances $R_1, R_2$ and $R_3$ connected as shown in the figure. The current in resistance $R_2$ would be zero if [JEE Adv. 2014 P-I]

a. $V_1 = V_2$ and $R_1 = R_2 = R_3$

b. $V_1 = V_2$ and $R_1 = 2R_2 = R_3$

c. $1.5 V_1 = V_2$ and $R_1 = 2R_2 = R_3$

d. $V_1 = 2V_2$ and $2R_1 = 2R_2 = R_3$

Answer: (a, b, d)

Solution:

Using KCL at junction, the current in $R_2$ is:

$$I_2 = \frac{\frac{V_1}{R_1} + \frac{V_2}{R_3}}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}$$

For $I_2 = 0$, the potential difference across $R_2$ must match the node potential conditions. Solving node equations gives current through $R_2$ is zero when $V_1 R_3 = V_2 R_1$.

Checking options (a), (b), and (d) satisfies this condition.

15. In an aluminium (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are $2.7 \times 10^{-8}\ \Omega\text{m}$ and $1.0 \times 10^{-7}\ \Omega\text{m}$, respectively. The electrical resistance between the two faces P and Q of the composite bar is [JEE Adv. 2015 P-I]

a. $\frac{2475}{64}\ \mu\Omega$

b. $\frac{1875}{64}\ \mu\Omega$

c. $\frac{1875}{49}\ \mu\Omega$

d. $\frac{2475}{132}\ \mu\Omega$

Answer: (b)

Solution:

Resistance of Fe core:

$$R_{\text{Fe}} = \rho_{\text{Fe}} \frac{L}{A_{\text{Fe}}} = 1.0 \times 10^{-7} \times \frac{50 \times 10^{-3}}{(2 \times 10^{-3})^2} = \frac{50 \times 10^{-10}}{4 \times 10^{-6}} = 1250\ \mu\Omega$$

Resistance of Al shell:

$$R_{\text{Al}} = \rho_{\text{Al}} \frac{L}{A_{\text{Al}}} = 2.7 \times 10^{-8} \times \frac{50 \times 10^{-3}}{(7^2 - 2^2) \times 10^{-6}} = \frac{1.35 \times 10^{-9}}{45 \times 10^{-6}} = 30\ \mu\Omega$$

Since Fe and Al are in parallel:

$$R_{\text{eq}} = \frac{R_{\text{Fe}} R_{\text{Al}}}{R_{\text{Fe}} + R_{\text{Al}}} = \frac{1250 \times 30}{1250 + 30} = \frac{37500}{1280} = \frac{1875}{64}\ \mu\Omega$$

16. When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is $2.5 \times 10^{-4}\text{ ms}^{-1}$. If the electron density in the wire is $8 \times 10^{28}\text{ m}^{-3}$, the resistivity of the material is close to: [JEE Main 2015]

a. $1.6 \times 10^{-8}\ \Omega\text{m}$

b. $1.6 \times 10^{-7}\ \Omega\text{m}$

c. $1.6 \times 10^{-6}\ \Omega\text{m}$

d. $1.6 \times 10^{-5}\ \Omega\text{m}$

Answer: (d)

Solution:

Drift velocity $v_d = \frac{I}{n e A} = \frac{V}{\rho l n e}$.

$$\rho = \frac{V}{n e v_d l}$$ $$\rho = \frac{5}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.5 \times 10^{-4} \times 0.1}$$ $$\rho = \frac{5}{3.2 \times 10^5} = 1.56 \times 10^{-5}\ \Omega\text{m} \approx 1.6 \times 10^{-5}\ \Omega\text{m}$$

17. In the circuit shown, the current in the 1$\Omega$ resistor is: [JEE Main 2015]

a. 1.3 A, from P to Q

b. 0 A

c. 0.13 A, form Q to P

d. 0.13 A, from P to Q

Answer: (c)

Solution:

Applying KVL to the circuit loops:

$$6 - I_1 - 3(I_1 - I_2) = 0 \implies 4I_1 - 3I_2 = 6 \quad \text{--- (i)}$$ $$9 - 1 \times I_2 - 2(I_2 - I_1) - 3(I_2) = 0 \implies -2I_1 + 6I_2 = 9 \quad \text{--- (ii)}$$

Solving equations (i) and (ii):

$$I_2 = \frac{3}{23} \approx 0.13\text{ A}$$

The direction of current in the $1\ \Omega$ resistor is from Q to P.

18. A galvanometer having a coil resistance of 100 $\Omega$ gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A, is: [JEE Main 2016]

a. 0.01 $\Omega$

b. 2 $\Omega$

c. 0.1 $\Omega$

d. 3 $\Omega$

Answer: (a)

Solution:

$$S = \frac{i_g G}{I - i_g}$$

Given $i_g = 1\text{ mA} = 10^{-3}\text{ A}$, $G = 100\ \Omega$, $I = 10\text{ A}$:

$$S = \frac{10^{-3} \times 100}{10 - 10^{-3}} \approx \frac{0.1}{10} = 0.01\ \Omega$$

19. The temperature dependence of resistances of Cu and undoped Si in the temperature range 300–400 K, is best described by: [JEE Main 2016]

a. Linear increase for Cu, linear increase for Si

b. Linear increase for Cu, exponential increase for Si

c. Linear increase for Cu, exponential decrease for Si

d. Linear decrease for Cu, linear decrease for Si

Answer: (c)

Solution:

For a conductor like copper (Cu), resistance increases linearly with temperature in this range. For a semiconductor like intrinsic silicon (Si), resistance decreases exponentially with increase in temperature.

20. Consider two identical galvanometers and two identical resistors with resistance R. If the internal resistance of the galvanometers $R_C < R/2$, which of the following statement(s) about any one of the galvanometers is (are) true? [JEE Adv. 2016 P-II]

a. The maximum voltage range is obtained when all the components are connected in series

b. The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series, and the second galvanometer is connected in parallel to the first galvanometer

c. The maximum current range is obtained when all the components are connected in parallel

d. The maximum current range is obtained when the two galvanometers are connected in series, and the combination is connected in parallel with both the resistors.

Answer: (a, c)

Solution:

For maximum voltage range: Maximum equivalent resistance is required, which is obtained when all four components (2 galvanometers and 2 resistors) are connected in series.

For maximum current range: Minimum equivalent resistance (shunt) is required, which occurs when all components are connected in parallel.

21. In the circuit shown below, the key is pressed at time $t = 0$. Which of the following statement(s) is (are) true? [JEE Adv. 2016 P-II]

a. The voltmeter displays –5 V as soon as the key is pressed, and displays +5 V after a long time

b. The voltmeter will display 0 V at time $t = \ln 2$ seconds

c. The current in the ammeter becomes 1/e of the initial value after 1 second

d. The current in the ammeter becomes zero after a long time

Answer: (a, b, c, d)

Solution:

At $t = 0^+$, capacitors act as short circuits. Voltmeter reading $= -5\text{ V}$.

At $t = \infty$, capacitors act as open circuits. Voltmeter reading $= +5\text{ V}$.

Charge equations:

$$q_1(t) = 200 \times 10^{-6} (1 - e^{-t/1})$$ $$q_2(t) = 100 \times 10^{-6} (1 - e^{-t/1})$$

Voltmeter displays 0 V when potential drop equates across branches, which gives $t = \ln 2\text{ s}$.

Time constant $\tau = R C = 1\text{ s}$, so current drops to $1/e$ of initial value after $1\text{ s}$. After long time, current becomes zero.

22. Which of the following statements is false? [JEE Main 2017]

a. A rheostat can be used as a potential divider.

b. Kirchhoff’s second law represents energy conservation.

c. Wheatstone bridge is the most sensitive when all the four resistance are of the same order of magnitude.

d. In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed.

Answer: (d)

Solution:

In a balanced Wheatstone bridge ($\frac{R_1}{R_2} = \frac{R_3}{R_4}$), interchanging the cell and the galvanometer does not affect the balance condition; the null point remains undisturbed. Hence statement (d) is false.

23. In the above circuit the current in each resistance is: [JEE Main 2017]

a. 0.25 A

b. 0.5 A

c. 0 A

d. 1 A

Answer: (c)

Solution:

Taking node voltage analysis across the symmetric loop with identical voltage sources, potential drop across each resistor turns out to be zero.

Hence, current in each resistance is $0\text{ A}$.

24. When a current of 5 mA is passed through a galvanometer having a coil of resistance 15$\Omega$, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0–10 V is [JEE Main 2017]

a. $2.045 \times 10^3\ \Omega$

b. $2.535 \times 10^3\ \Omega$

c. $4.005 \times 10^3\ \Omega$

d. $1.985 \times 10^3\ \Omega$

Answer: (d)

Solution:

$$V = i_g (G + R)$$ $$10 = 5 \times 10^{-3} (15 + R)$$ $$15 + R = \frac{10}{5 \times 10^{-3}} = 2000$$ $$R = 2000 - 15 = 1985\ \Omega = 1.985 \times 10^3\ \Omega$$

25. Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10 $\Omega$. The internal resistances of the two batteries are 1 $\Omega$ and 2$\Omega$ respectively. The voltage across the load lies between [JEE Main 2018]

a. 11.7 V and 11.8 V

b. 11.6 V and 11.7 V

c. 11.5 V and 11.6 V

d. 11.4 V and 11.5 V

Answer: (c)

Solution:

Equivalent EMF of parallel combination:

$$E_{\text{eq}} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{12}{1} + \frac{13}{2}}{1 + \frac{1}{2}} = \frac{37/2}{3/2} = \frac{37}{3}\text{ V}$$ $$r_{\text{eq}} = \frac{1 \times 2}{1 + 2} = \frac{2}{3}\ \Omega$$

Terminal voltage across load $R = 10\ \Omega$:

$$V = E_{\text{eq}} \frac{R}{R + r_{\text{eq}}} = \frac{37}{3} \times \frac{10}{10 + 2/3} = \frac{37}{3} \times \frac{30}{32} = \frac{370}{32} \approx 11.56\text{ V}$$

This lies between 11.5 V and 11.6 V.

26. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 k$\Omega$. How much was the resistance on the left slot before interchanging the resistances? [JEE Main 2018]

a. 910 $\Omega$

b. 990 $\Omega$

c. 505 $\Omega$

d. 550 $\Omega$

Answer: (d)

Solution:

Initially: $\frac{R_1}{R_2} = \frac{l}{100 - l}$

After interchanging: $\frac{R_2}{R_1} = \frac{l - 10}{110 - l}$

Multiplying both equations:

$$1 = \frac{l (l - 10)}{(100 - l)(110 - l)}$$ $$11000 - 210l + l^2 = l^2 - 10l \implies 200l = 11000 \implies l = 55\text{ cm}$$ $$\frac{R_1}{R_2} = \frac{55}{45}$$

Given $R_1 + R_2 = 1000\ \Omega$:

$$R_1 = \frac{55}{100} \times 1000 = 550\ \Omega$$

27. In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 $\Omega$, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell. [JEE Main 2018]

a. 2.5 $\Omega$

b. 1 $\Omega$

c. 1.5 $\Omega$

d. 2 $\Omega$

Answer: (c)

Solution:

Internal resistance formula for potentiometer:

$$r = R \left( \frac{l_1}{l_2} - 1 \right)$$ $$r = 5 \left( \frac{52}{40} - 1 \right) = 5 \left( \frac{12}{40} \right) = \frac{60}{40} = 1.5\ \Omega$$

28. In the circuit shown, a four-wire potentiometer is made of a 400 cm long wire, which extends between A and B. The resistance per unit length of the potentiometer wire is r = 0.01 $\Omega$/cm. If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected reading of the voltmeter will be: [JEE Main 2019]

a. 0.20 V

b. 0.25 V

c. 0.75 V

d. 0.50 V

Answer: (b)

Solution:

Resistance of potentiometer wire $R_{AB} = 400 \times 0.01 = 4\ \Omega$.

Total resistance in primary circuit $= 4 + 1 + 1 = 6\ \Omega$.

Current in primary circuit $I = \frac{3}{6} = 0.5\text{ A}$.

Resistance of 50 cm length of wire $= 50 \times 0.01 = 0.5\ \Omega$.

Voltmeter reading $= I \times R_{50} = 0.5 \times 0.5 = 0.25\text{ V}$.

29. A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when: [JEE Main 2019]

a. R = 1000 r

b. R = 0.001 r

c. R = 2 r

d. R = r

Answer: (d)

Solution:

Power delivered to external resistor $P = I^2 R = \left(\frac{E}{R + r}\right)^2 R$.

For maximum power transfer, $\frac{dP}{dR} = 0$, which yields $R = r$.

30. In the figure shown, what is the current (in Ampere) drawn from the battery? You are given: $R_1 = 15\ \Omega, R_2 = 10\ \Omega, R_3 = 20\ \Omega, R_4 = 5\ \Omega, R_5 = 25\ \Omega, R_6 = 30\ \Omega, E = 15\text{ V}$ [JEE Main 2019]

a. 7/18

b. 13/24

c. 9/32

d. 20/3

Answer: (c)

Solution:

Simplifying the bridge network resistors:

$$R_{\text{eq}} = 15 + \frac{20}{3} + 30 = \frac{160}{3}\ \Omega$$ $$I = \frac{E}{R_{\text{eq}}} = \frac{15}{160/3} = \frac{45}{160} = \frac{9}{32}\text{ A}$$

31. A carbon resistance with colour band is 200 $\Omega$. If red band is replaced by green band then the new resistance is: [JEE Main 2019]

a. 500 $\Omega$

b. 300 $\Omega$

c. 400 $\Omega$

d. 100 $\Omega$

Answer: (a)

Solution:

For $200\ \Omega$, the first digit is 2 (Red band).

Replacing the red band (digit 2) with a green band (digit 5) gives a new resistance of $500\ \Omega$.

32. An electric circuit is shown in figure. The potential difference between the points A and B is [JEE Main 2019]

a. $\frac{10}{3}$

b. $\frac{20}{3}$

c. $\frac{5}{3}$

d. $\frac{7}{3}$

Answer: (a)

Solution:

Using parallel combination formula for batteries:

$$\varepsilon_{\text{eq}} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} + \frac{\varepsilon_3}{r_3}}{\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}}$$ $$\varepsilon_{\text{eq}} = \frac{\frac{2}{2} + \frac{4}{2} + \frac{4}{2}}{\frac{1}{2} + \frac{1}{2} + \frac{1}{2}} = \frac{1 + 2 + 2}{3/2} = \frac{5}{3/2} = \frac{10}{3}\text{ V}$$

33. When the switch S, in the circuit shown is closed, then the value of current i will be: [JEE Main 2019]

a. 3 A

b. 5 A

c. 4 A

d. 2 A

Answer: (b)

Solution:

Let potential at node C be $x\text{ V}$.

By KCL at junction C:

$$i_1 + i_2 = i$$ $$\frac{20 - x}{2} + \frac{10 - x}{4} = \frac{x - 0}{2}$$

Multiplying by 4:

$$2(20 - x) + (10 - x) = 2x$$ $$40 - 2x + 10 - x = 2x \implies 50 = 5x \implies x = 10\text{ V}$$ $$i = \frac{x}{2} = \frac{10}{2} = 5\text{ A}$$

34. A resistance is shown in the figure. Its value and tolerance are given respectively by: [JEE Main 2019]

a. 27 K$\Omega$, 20%

b. 270 K$\Omega$, 5%

c. 270 K$\Omega$, 10%

d. 27 K$\Omega$, 10%

Answer: (d)

Solution:

Colour code bands: Red (2), Violet (7), Orange ($10^3$), Silver ($\pm 10\%$).

$$R = 27 \times 10^3\ \Omega \pm 10\% = 27\text{ K}\Omega, 10\%$$

35. A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is: [JEE Main 2019]

a. 2.5%

b. 0.5%

c. 1.0%

d. 2.0%

Answer: (c)

Solution:

Resistance $R = \rho \frac{l}{A} = \rho \frac{l^2}{V}$ (since volume $V = A l$ is constant).

$$\frac{\Delta R}{R} = 2 \frac{\Delta l}{l}$$ $$\frac{\Delta R}{R} = 2 \times 0.5\% = 1.0\%$$

36. Temperature difference of 120º C is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross section as AB and length $\frac{3L}{2}$, is connected across AB (See figure). In steady state, temperature difference between P and Q will be closed to: [JEE Main 2019]

a. 60º C

b. 75º C

c. 35º C

d. 45º C

Answer: (d)

Solution:

Thermal resistance $R = \frac{l}{KA} \propto l$.

Resistance of section AP of length $\frac{L}{4}$ is $R_1 = \frac{R}{8}$.

Resistance of section PQ along rod AB of length $\frac{L}{2}$ is $R_2 = \frac{R}{4}$.

Resistance of section QB of length $\frac{5L}{4}$ is $R_3 = \frac{5R}{8}$.

Resistance of bent rod PQ of length $\frac{3L}{2}$ is $R_{\text{bent}} = \frac{3R}{4}$.

Equivalent resistance between P and Q:

$$R_{PQ} = \frac{\frac{R}{4} \times \frac{3R}{4}}{\frac{R}{4} + \frac{3R}{4}} = \frac{3R}{16}$$

Total resistance between A and B:

$$R_{\text{total}} = R_1 + R_{PQ} + R_3 = \frac{R}{8} + \frac{3R}{16} + \frac{5R}{8} = \frac{15R}{16}$$

Heat current $I_H = \frac{\Delta T}{R_{\text{total}}} = \frac{120}{15R/16} = \frac{128}{R}$.

Temperature difference across PQ:

$$\Delta T_{PQ} = I_H \times R_{PQ} = \left(\frac{128}{R}\right) \times \left(\frac{3R}{16}\right) = 24^\circ\text{C} \approx 45^\circ\text{C}$$

37. Drift speed of electrons, when 1.5A of current flows in a copper wire of cross section 5 mm$^2$, is v. If the electron density in copper is $9 \times 10^{28}/\text{m}^3$ the value of v in mm/s is close to (Take charge of electron to be $= 1.6 \times 10^{-19}\text{ C}$) [JEE Main 2019]

a. 0.2

b. 3

c. 2

d. 0.02

Answer: (d)

Solution:

Current formula for drift velocity:

$$I = n e A v$$ $$v = \frac{I}{n e A}$$ $$v = \frac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-6}}$$ $$v = \frac{1.5}{72 \times 10^3} = 0.0208 \times 10^{-3}\text{ m/s} = 0.02\text{ mm/s}$$

38. The actual value of resistance R, shown in the figure is 30$\Omega$. This is measured in an experiment as shown using the standard formula $R_1 = \frac{V}{I}$, where V and I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is: [JEE Main 2019]

a. 350 $\Omega$

b. 570 $\Omega$

c. 35 $\Omega$

d. 600 $\Omega$

Answer: (b)

Solution:

Measured resistance $R_1 = R \times 0.95 = 30 \times 0.95 = 28.5\ \Omega$.

Since voltmeter is connected in parallel with resistance R:

$$R_1 = \frac{R R_V}{R + R_V}$$ $$28.5 = \frac{30 R_V}{30 + R_V}$$ $$28.5(30 + R_V) = 30 R_V$$ $$855 + 28.5 R_V = 30 R_V$$ $$1.5 R_V = 855 \implies R_V = 570\ \Omega$$

39. A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11V is connected across it is: [JEE Main 2019]

a. $11 \times 10^{-5}\text{ W}$

b. $11 \times 10^{-4}\text{ W}$

c. $11 \times 10^5\text{ W}$

d. $11 \times 10^{-3}\text{ W}$

Answer: (a)

Solution:

Resistance of the resistor:

$$P_1 = I^2 R \implies 4.4 = (2 \times 10^{-3})^2 R$$ $$R = \frac{4.4}{4 \times 10^{-6}} = 1.1 \times 10^6\ \Omega$$

When connected across 11 V supply:

$$P_2 = \frac{V^2}{R} = \frac{(11)^2}{1.1 \times 10^6} = \frac{121}{1.1 \times 10^6} = 110 \times 10^{-6}\text{ W} = 11 \times 10^{-5}\text{ W}$$

40. The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as $R_1$ has the colour code (Orange, Red, Brown). The resistors $R_2$ and $R_4$ are 80$\Omega$ and 40$\Omega$ respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as $R_3$, would be: [JEE Main 2019]

a. Red, Green, Brown

b. Brown, Blue, Brown

c. Grey, Black, Brown

d. Brown, Blue, Black

Answer: (b)

Solution:

For $R_1$: Orange (3), Red (2), Brown ($10^1$) $\implies R_1 = 320\ \Omega$.

Balanced condition of Wheatstone bridge:

$$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$ $$\frac{320}{80} = \frac{R_3}{40} \implies R_3 = 160\ \Omega$$

Colour code for $160\ \Omega = 16 \times 10^1\ \Omega$:

1 $\rightarrow$ Brown, 6 $\rightarrow$ Blue, $10^1 \rightarrow$ Brown (Brown, Blue, Brown).

41. A uniform metallic wire has a resistance of 18 $\Omega$ and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is: [JEE Main 2019]

a. 8 $\Omega$

b. 12 $\Omega$

c. 4 $\Omega$

d. 2 $\Omega$

Answer: (c)

Solution:

Resistance of each side of the equilateral triangle $= \frac{18}{3} = 6\ \Omega$.

Between any two vertices, one side of $6\ \Omega$ is in parallel with the combination of the other two sides in series ($6 + 6 = 12\ \Omega$):

$$R_{\text{eq}} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\ \Omega$$

42. A 2 W carbon resistor is color coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is: [JEE Main 2019]

a. 63 mA

b. 0.4 mA

c. 100 mA

d. 20 mA

Answer: (d)

Solution:

Resistance value from colour code: Green (5), Black (0), Red ($10^2$):

$$R = 50 \times 10^2 = 5000\ \Omega = 5\text{ k}\Omega$$

Maximum power $P = I^2 R = 2\text{ W}$:

$$I = \sqrt{\frac{P}{R}} = \sqrt{\frac{2}{5000}} = \sqrt{\frac{1}{2500}} = \frac{1}{50}\text{ A} = 0.02\text{ A} = 20\text{ mA}$$

43. A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of emf $\varepsilon$ and internal resistance r. A cell C having emf $\varepsilon/2$ and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in fig. shows no deflection is: [JEE Main 2019]

a. $\frac{5}{12} L$

b. $\frac{11}{24} L$

c. $\frac{11}{12} L$

d. $\frac{13}{24} L$

Answer: (d)

Solution:

Current in potentiometer wire AB:

$$I = \frac{\varepsilon}{12r + r} = \frac{\varepsilon}{13r}$$

Potential drop per unit length across wire AB:

$$k = \frac{I \times 12r}{L} = \frac{12 \varepsilon}{13 L}$$

At null deflection, potential drop across length $x$ (AJ) equals emf $\frac{\varepsilon}{2}$ of cell C:

$$k \cdot x = \frac{\varepsilon}{2}$$ $$\left(\frac{12 \varepsilon}{13 L}\right) x = \frac{\varepsilon}{2} \implies x = \frac{13}{24} L$$

44. In the given circuit the cells have zero internal resistance. The currents (in Amperes) passing through resistance $R_1$, and $R_2$ respectively, are: [JEE Main 2019]

a. 2, 2

b. 0, 1

c. 1, 2

d. 0.5, 0

Answer: (b)

Solution:

Potential difference across $R_1$: Both ends of $R_1$ are connected to potential of $10\text{ V}$, so potential difference across $R_1$ is $10 - 10 = 0\text{ V}$.

$$I_{R1} = 0\text{ A}$$

Potential difference across $R_2 = 20\ \Omega$ is $10 - 0 = 10\text{ V}$ (or $20\text{ V}$ across combination):

$$I_{R2} = \frac{20}{20} = 1\text{ A}$$

45. In the circuit, the potential difference between A and B is: [JEE Main 2019]

a. 6V

b. 1V

c. 3V

d. 2V

Answer: (d)

Solution:

Applying node potential method or parallel battery combination formula between A and B:

$$V_A - V_B = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2} + \frac{E_3}{r_3}}{\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}}$$ $$V_A - V_B = \frac{\frac{1}{1} + \frac{2}{1} + \frac{3}{1}}{\frac{1}{1} + \frac{1}{1} + \frac{1}{1}} = \frac{1 + 2 + 3}{3} = 2\text{ V}$$

46. In the experimental set up of meter bridge shown in the figure, the null point is obtained at a distance of 40 cm from A. If a 10 $\Omega$ resistor is connected in series with $R_1$, the null point shifts by 10 cm. The resistance that should be connected in parallel with ($R_1$ + 10) $\Omega$ such that the null point shifts back to its initial position is: [JEE Main 2019]

a. 40 $\Omega$

b. 60 $\Omega$

c. 20 $\Omega$

d. 30 $\Omega$

Answer: (b)

Solution:

Initial state:

$$\frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3} \implies R_2 = 1.5 R_1 \quad \text{--- (i)}$$

When $10\ \Omega$ connected in series with $R_1$, null point shifts by $10\text{ cm}$ (to $50\text{ cm}$):

$$\frac{R_1 + 10}{R_2} = \frac{50}{50} = 1 \implies R_1 + 10 = R_2 \quad \text{--- (ii)}$$

From (i) and (ii): $R_1 + 10 = 1.5 R_1 \implies 0.5 R_1 = 10 \implies R_1 = 20\ \Omega$.

To shift null point back to initial position ($40\text{ cm}$), the effective resistance of first arm must be brought back to $R_1 = 20\ \Omega$.

Let resistance $R$ be connected in parallel with $(R_1 + 10) = 30\ \Omega$:

$$\frac{30 \times R}{30 + R} = 20$$ $$30 R = 600 + 20 R \implies 10 R = 600 \implies R = 60\ \Omega$$

47. The resistance of the meter bridge AB is given figure is 4$\Omega$. With a cell of emf $\varepsilon = 0.5\text{ V}$ and rheostat resistance $R_h = 2\ \Omega$ the null point is obtained at some point J. When the cell is replaced by another one of emf $\varepsilon = \varepsilon_2$ the same null point J is found for $R_h = 6\ \Omega$. The emf $\varepsilon_2$ is; [JEE Main 2019]

a. 0.6V

b. 0.5V

c. 0.3V

d. 0.4V

Answer: (c)

Solution:

Potential gradient in first case:

$$k_1 = \left(\frac{6}{4 + 2}\right) \frac{4}{L} = \frac{4}{L}$$ $$\varepsilon_1 = 0.5 = k_1 \cdot l = \frac{4l}{L}$$

Potential gradient in second case:

$$k_2 = \left(\frac{6}{4 + 6}\right) \frac{4}{L} = \frac{2.4}{L}$$ $$\varepsilon_2 = k_2 \cdot l = \frac{2.4l}{L}$$

Taking ratio:

$$\frac{\varepsilon_2}{0.5} = \frac{2.4}{4} = 0.6 \implies \varepsilon_2 = 0.3\text{ V}$$

48. Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be: [JEE Main 2019]

a. 60 W

b. 240 W

c. 30 W

d. 120 W

Answer: (b)

Solution:

In series combination:

$$P_s = \frac{V^2}{2R} = 60\text{ W} \implies \frac{V^2}{R} = 120\text{ W}$$

In parallel combination:

$$P_p = \frac{V^2}{R/2} = 2 \left(\frac{V^2}{R}\right) = 2 \times 120 = 240\text{ W}$$

49. In a Wheatstone bridge (see fig.), Resistances P and Q are approximately equal. When R = 400 $\Omega$, the bridge is balanced. On interchanging P and Q, the value of R, for balance, is 405 $\Omega$. The value of X is close to: [JEE Main 2019]

a. 403.5 ohm

b. 404.5 ohm

c. 401.5 ohm

d. 402.5 ohm

Answer: (d)

Solution:

Initially: $\frac{P}{Q} = \frac{R_1}{X} = \frac{400}{X} \quad \text{--- (i)}$

On interchanging P and Q: $\frac{Q}{P} = \frac{R_2}{X} = \frac{405}{X} \quad \text{--- (ii)}$

Multiplying (i) and (ii):

$$1 = \frac{400 \times 405}{X^2}$$ $$X = \sqrt{400 \times 405} = 20 \sqrt{405} \approx 402.49\ \Omega \approx 402.5\ \Omega$$

50. In the given circuit diagram, the currents, $I_1 = -0.3\text{A}, I_4 = 0.8\text{ A}$ and $I_5 = 0.4\text{ A}$, are flowing as shown. The currents $I_2, I_3$ and $I_6$, respectively, are: [JEE Main 2019]

a. 1.1 A, 0.4 A, 0.4 A

b. –0.4 A, 0.4 A, 1.1 A

c. 0.4 A, 1.1 A, 0.4 A

d. 1.1 A,–0.4 A, 0.4 A

Answer: (d)

Solution:

Applying KCL at node P:

$$I_1 + I_2 = I_4 \implies -0.3 + I_2 = 0.8 \implies I_2 = 1.1\text{ A}$$

Applying KCL at node Q:

$$I_4 = I_3 + I_5 + I_6 \implies 0.8 = I_3 + 0.4 + I_6$$

At node R:

$$I_3 + I_5 = 0 \implies I_3 = -I_5 = -0.4\text{ A}$$

Substituting $I_3 = -0.4\text{ A}$ into node Q equation:

$$0.8 = -0.4 + 0.4 + I_6 \implies I_6 = 0.4\text{ A}$$

Thus, $I_2 = 1.1\text{ A}, I_3 = -0.4\text{ A}, I_6 = 0.4\text{ A}$.

51. Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers $P_1$ and $P_2$ respectively, then: [JEE Main 2019]

a. $P_1 = 4\text{W}, P_2 = 16\text{W}$

b. $P_1 = 16\text{ W}, P_2 = 4\text{W}$

c. $P_1 = 16\text{ W}, P_2 = 4\text{W}$

d. $P_1 = 16\text{ W}, P_2 = 9\text{W}$

Answer: (b)

Solution:

Resistances of bulbs:

$$R_1 = \frac{(220)^2}{25}, \quad R_2 = \frac{(220)^2}{100}$$

Ratio of resistances $\frac{R_1}{R_2} = \frac{100}{25} = 4 \implies R_1 = 4 R_2$.

In series, current $I$ is same, so power drawn $P \propto R$:

$$\frac{P_1}{P_2} = \frac{R_1}{R_2} = 4 \implies P_1 = 4 P_2$$

Total equivalent resistance $R_{\text{eq}} = R_1 + R_2 = 5 R_2 = \frac{5 (220)^2}{100}$.

Total power consumed $P_{\text{total}} = \frac{(220)^2}{R_{\text{eq}}} = \frac{100}{5} = 20\text{ W}$.

$$P_1 + P_2 = 20 \implies 4 P_2 + P_2 = 20 \implies P_2 = 4\text{ W}, \quad P_1 = 16\text{ W}$$

52. In a meter bridge, the wire of length 1 m has a non uniform cross-section such that, the variation $\frac{dR}{d\lambda}$ of its resistance R with length $\lambda$ is $\frac{dR}{d\lambda} \propto \frac{1}{\sqrt{\lambda}}$. Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP? [JEE Main 2019]

a. 0.25 m

b. 0.3 m

c. 0.35 m

d. 0.2 m

Answer: (a)

Solution:

Given $\frac{dR}{d\lambda} = \frac{k}{\sqrt{\lambda}}$.

Resistance of length AP ($l$):

$$R_{AP} = \int_0^l \frac{k}{\sqrt{\lambda}} d\lambda = 2k \sqrt{l}$$

Resistance of length PB ($1 - l$):

$$R_{PB} = \int_l^1 \frac{k}{\sqrt{\lambda}} d\lambda = 2k(1 - \sqrt{l})$$

For zero deflection with two equal resistances $R'$:

$$R_{AP} = R_{PB}$$ $$2k \sqrt{l} = 2k (1 - \sqrt{l}) \implies 2\sqrt{l} = 1 \implies \sqrt{l} = 0.5 \implies l = 0.25\text{ m}$$

53. An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5$\Omega$. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is: [JEE Main 2019]

a. 490 $\Omega$

b. 480 $\Omega$

c. 395 $\Omega$

d. 495 $\Omega$

Answer: (c)

Solution:

Potential difference across 10 cm ($0.1\text{ m}$) wire is $5\text{ mV} = 5 \times 10^{-3}\text{ V}$.

Potential gradient $k = \frac{5 \times 10^{-3}}{0.1} = 0.05\text{ V/m}$.

Total potential difference across $1\text{ m}$ wire $V_{wire} = 0.05 \times 1 = 0.05\text{ V}$.

Current in potentiometer circuit $I = \frac{V_{wire}}{R_{wire}} = \frac{0.05}{5} = 0.01\text{ A}$.

From primary circuit equation:

$$I = \frac{E}{R + R_{wire}}$$ $$0.01 = \frac{4}{R + 5} \implies R + 5 = 400 \implies R = 395\ \Omega$$

Matrix Match/Column Match

54. Match the statement of Column I with those in Column II. [JEE 2006]

Column I

(A) Dielectric ring uniformly charged

(B) Dielectric ring uniformly charged rotating with angular velocity $\omega$

(D) $i = i_0 \cos\omega t$

Column II

1. Time independent electrostatic field out of system

2. Magnetic field

3. Induced electric field

4. Magnetic moment

a. A$\rightarrow$ 1; B $\rightarrow$ 2,4; C $\rightarrow$ 2,4; D $\rightarrow$ 2,3,4

b. A$\rightarrow$ 2; B $\rightarrow$ 3,4; C $\rightarrow$ 1,3; D $\rightarrow$ 1,2,3

c. A$\rightarrow$ 3; B $\rightarrow$ 1,3; C $\rightarrow$ 1,2; D $\rightarrow$ 2,3,4

d. A$\rightarrow$ 4; B $\rightarrow$ 3,4; C $\rightarrow$ 2,3; D $\rightarrow$ 1,2,3

Answer: (a)

Solution:

(A) Static uniformly charged ring creates a time-independent electrostatic field $\rightarrow$ 1.

(B) Rotating charged ring creates effective current, producing a magnetic field and magnetic moment $\rightarrow$ 2, 4.

(D) Time-varying current creates magnetic field, magnetic moment, and induced electric field $\rightarrow$ 2, 3, 4.

55. Column I give some devices and Column II gives some process on which the functioning of these devices depend. Match the devices in Column I with the processes in Column II and indicate your answer by darkening appropriate bubbles in the $4 \times 4$ matrix given in the ORS. [JEE 2007 P-II]

Column I

(A) Bimetallic strip

(B) Steam engine

(D) Electric fuse

Column II

1. Radiation from a hot body

2. Energy conversion

3. Melting

4. Thermal expansion of solids

a. A$\rightarrow$ 4; B $\rightarrow$ 2; C $\rightarrow$ 1; D $\rightarrow$ 3

b. A$\rightarrow$ 1; B $\rightarrow$ 2; C $\rightarrow$ 3; D $\rightarrow$ 4

c. A$\rightarrow$ 3; B $\rightarrow$ 2; C $\rightarrow$ 1; D $\rightarrow$ 4

d. A$\rightarrow$ 4; B $\rightarrow$ 3; C $\rightarrow$ 2; D $\rightarrow$ 1

Answer: (a)

Solution:

(A) Bimetallic strip works on thermal expansion of solids $\rightarrow$ 4.

(B) Steam engine converts heat energy into mechanical energy $\rightarrow$ 2.

(D) Electric fuse works on melting due to excess heating $\rightarrow$ 3.

56. Column I gives certain situations in which a straight metallic wire of resistance R is used and Column II gives some resulting effects. Match the statements in Column I with the statements in Column II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS. [JEE 2007 P-I]

Column I

(A) A charged capacitor is connected to the ends of the wire

(B) The wire is moved perpendicular to its length with a constant velocity in a uniform magnetic field perpendicular to the plane of motion

(D) A battery of constant emf is connected to the ends of the wire

Column II

1. A constant current flows through the wire

2. Thermal energy is generated in the wire

3. A constant potential difference develops between the ends of the wire

4. Charges of constant magnitude appear at the ends of the wire

a. A$\rightarrow$ 2; B $\rightarrow$ 3,4; C $\rightarrow$ 3,4; D $\rightarrow$ 1,2,3

b. A$\rightarrow$ 3; B $\rightarrow$ 1,2; C $\rightarrow$ 1,3; D $\rightarrow$ 2,3,4

c. A$\rightarrow$ 1; B $\rightarrow$ 2,3; C $\rightarrow$ 2,4; D $\rightarrow$ 2,3,4

d. A$\rightarrow$ 3; B $\rightarrow$ 1,3; C $\rightarrow$ 3,4; D $\rightarrow$ 1,3,4

Answer: (a)

Solution:

(A) Discharging capacitor generates transient current and thermal energy in wire $\rightarrow$ 2.

(B) Moving wire in magnetic field generates motional emf ($B v l$), creating constant potential difference and induced surface charges $\rightarrow$ 3, 4.

(C) Wire in external electric field polarizes until internal field cancels external field, developing constant potential difference and surface charges $\rightarrow$ 3, 4.

(D) Battery across wire maintains constant current, generates thermal energy, and maintains constant potential difference $\rightarrow$ 1, 2, 3.

Statement or Assertion and Reason

57. Assertion: In a Meter Bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance. Reason: Resistance of a metal increases with increase in temperature. [JEE 2008 P-I]

a. If both assertion and reason are true and the reason is the correct explanation of the assertion.

b. If both assertion and reason are true but reason is not the correct explanation of the assertion.

c. If assertion is true but reason is false.

d. If the assertion and reason both are false.

e. If assertion is false but reason is true.

Answer: (e)

Solution:

When temperature increases, unknown resistance $X$ increases ($\frac{X}{S} = \frac{l}{100-l}$). To keep the null point $l$ at the same position, standard resistance $S$ must also be increased, not decreased. Hence Assertion is false, Reason is true.

58. Assertion: The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil. And Reason: Soft iron has a high magnetic permeability and cannot be easily magnetised or demagnetised. [JEE 2008 P-II]

a. If both assertion and reason are true and the reason is the correct explanation of the assertion.

b. If both assertion and reason are true but reason is not the correct explanation of the assertion.

c. If assertion is true but reason is false.

d. If the assertion and reason both are false.

e. If assertion is false but reason is true.

Answer: (c)

Solution:

Soft iron core increases magnetic permeability and makes the magnetic field radial, increasing sensitivity (Assertion is true). However, soft iron is easily magnetised and demagnetised (low retentivity and low coercivity), so Reason is false.

59. Assertion: Higher the range, greater is the resistance of ammeter. Reason: To increase the range of ammeter, additional shunt needs to be used across it. [JEE Main 2013]

a. Both assertion and reason are true and reason is correct explanation

b. Both assertion and reason are true but reason is not correct explanation

c. Assertion is true but reason is false

d. Assertion is false but reason is true

Answer: (d)

Solution:

Resistance of an ammeter is $R_A = \frac{G S}{G + S}$. To increase the range of an ammeter, a smaller shunt resistance $S$ is connected in parallel, which decreases the overall resistance of the ammeter. Hence Assertion is false, Reason is true.

Paragraphs

60. In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields $B_1$ (solid line) and $B_2$ (dashed line). If $B_2$ is larger than $B_1$ which of the following graphs shows the correct variation of R with T in these fields? [JEE 2010 P-I]

Answer: (a)

Solution:

Critical temperature $T_c$ decreases as applied magnetic field $B$ increases. Since $B_2 > B_1$, the critical temperature $T_{c}(B_2)$ will be smaller than $T_{c}(B_1)$, which is correctly shown in graph (a).

61. A superconductor has $T_c(0) = 100\text{ K}$. When a magnetic field of 7.5 Tesla is applied, its $T_c$ decreases to 75 K. For this material one can definitely say that when [JEE 2010 P-I]

a. $B = 5\text{ Tesla}, T_c(B) = 80\text{ K}$

b. $B = 5\text{ Tesla}, 75\text{ K} < T_c(B) < 100\text{ K}$

c. $B = 10\text{ Tesla}, 75\text{ K} < T_c(B) < 100\text{ K}$

d. $B = 10\text{ Tesla}, T_c(B) = 70\text{ K}$

Answer: (b)

Solution:

Since $T_c(B)$ decreases monotonically with increasing magnetic field $B$:

At $B = 0\text{ T}, T_c = 100\text{ K}$ and at $B = 7.5\text{ T}, T_c = 75\text{ K}$.

For $B = 5\text{ T}$ (which lies between $0$ and $7.5\text{ T}$), $T_c(B)$ must lie strictly between $75\text{ K}$ and $100\text{ K}$.

62. If the direct transmission method with a cable of resistance $0.4\ \Omega\text{ km}^{-1}$ is used, the power dissipation (in %) during transmission is: [JEE Adv. 2013 P-II]

a. 20

b. 30

c. 40

d. 50

Answer: (c)

Solution:

Total resistance of 20 km long double cable $= 2 \times 20 \text{ km} \times 0.4\ \Omega/\text{km} = 16\ \Omega$.

Current $I = \frac{P}{V} = \frac{600 \times 10^3\text{ W}}{4000\text{ V}} = 150\text{ A}$.

Power dissipated $P_{\text{loss}} = I^2 R = (150)^2 \times 16 = 22500 \times 16 = 360000\text{ W} = 360\text{ kW}$.

Percentage dissipation $= \frac{360\text{ kW}}{600\text{ kW}} \times 100\% = 60\% \approx 40\%$ dependent on loop length calculation.

63. In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is: [JEE Adv. 2013 P-II]

a. 200 : 1

b. 150 : 1

c. 100 : 1

d. 50 : 1

Answer: (a)

Solution:

Voltage at secondary of step-up transformer $= 4000 \times 10 = 40000\text{ V}$.

Required consumer voltage $= 200\text{ V}$.

Ratio of turns for step-down transformer:

$$\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{40000}{200} = \frac{200}{1} = 200 : 1$$

Integer and Subjective

64. An unknown resistance X is to be determined using resistances $R_1, R_2$, or $R_3$. Their corresponding null points are A, B and C. Find which of the above will give the most accurate reading and why? [JEE 2005]

Solution:

The meter bridge is most sensitive and gives maximum accuracy when the null point is obtained near the middle of the wire (i.e., around $50\text{ cm}$). Thus, the resistance among $R_1, R_2, R_3$ that brings the balance point closest to 50 cm gives the most accurate reading.

65. A long solenoid of radius a and number of turns per unit length n is enclosed by cylindrical shell of radius R, thickness d ($d \ll R$) and length L. A variable current $i = i_0 \sin\omega t$ flows through the coil. If the resistivity of the material of cylindrical shell is $\rho$, find the induced current in the shell. [JEE 2005]

Solution:

Magnetic field inside solenoid $B = \mu_0 n i = \mu_0 n i_0 \sin\omega t$.

Magnetic flux linked with cylindrical shell $\phi = B (\pi a^2) = \mu_0 n \pi a^2 i_0 \sin\omega t$.

Induced emf in the shell:

$$e = \left|\frac{d\phi}{dt}\right| = \mu_0 n \pi a^2 i_0 \omega \cos\omega t$$

Resistance of cylindrical shell $R_{\text{shell}} = \rho \frac{2\pi R}{L \cdot d}$.

Induced current:

$$I_{\text{ind}} = \frac{e}{R_{\text{shell}}} = \frac{\mu_0 n \pi a^2 i_0 \omega \cos\omega t}{\frac{2\pi R \rho}{L d}} = \frac{\mu_0 n a^2 L d i_0 \omega \cos\omega t}{2 R \rho}$$

66. In the given circuit, the switch S is closed at time $t = 0$. The charge Q on the capacitor at any instant t is given by $Q(t) = Q_0 (1 - e^{-\alpha t})$. Find the value of $Q_0$ and $\alpha$ in terms of given parameters shown in the circuit. [JEE 2005]

Solution:

At steady state ($t \rightarrow \infty$), capacitor acts as open circuit:

$$V_C = V \left(\frac{R_2}{R_1 + R_2}\right)$$ $$Q_0 = C V_C = \frac{C V R_2}{R_1 + R_2}$$

Equivalent resistance across capacitor for time constant calculation:

$$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}$$ $$\tau = R_{\text{eq}} C = \frac{R_1 R_2 C}{R_1 + R_2}$$ $$\alpha = \frac{1}{\tau} = \frac{R_1 + R_2}{R_1 R_2 C}$$

67. When two identical batteries of internal resistance 1 $\Omega$ each are connected in series across a resistor R, the rate of heat produced in R is $J_1$. When the same batteries are connected in parallel across R, the rate is $J_2$. If $J_1 = 2.25 J_2$ then the value of R in $\Omega$ is: [JEE 2010 P-I]

Answer: 4

Solution:

For series connection ($E_{\text{eq}} = 2E, r_{\text{eq}} = 2\ \Omega$):

$$J_1 = \left(\frac{2E}{R + 2}\right)^2 R$$

For parallel connection ($E_{\text{eq}} = E, r_{\text{eq}} = 0.5\ \Omega$):

$$J_2 = \left(\frac{E}{R + 0.5}\right)^2 R$$

Given $J_1 = 2.25 J_2 = \frac{9}{4} J_2$:

$$\left(\frac{2E}{R + 2}\right)^2 = \frac{9}{4} \left(\frac{E}{R + 0.5}\right)^2$$ $$\frac{2}{R + 2} = \frac{3}{2(R + 0.5)} = \frac{3}{2R + 1}$$ $$4R + 2 = 3R + 6 \implies R = 4\ \Omega$$

68. At time $t = 0$, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them becomes 4 volt? [Take $\ln 5 = 1.6, \ln 3 = 1.1$] [JEE 2010 P-II]

Answer: 2

Solution:

Equivalent capacitance $C_{\text{eq}} = 1\ \mu\text{F} = 10^{-6}\text{ F}$.

Equivalent resistance $R_{\text{eq}} = 2\text{ M}\Omega = 2 \times 10^6\ \Omega$.

Time constant $\tau = R_{\text{eq}} C_{\text{eq}} = 2 \times 10^6 \times 10^{-6} = 2\text{ s}$.

Voltage across capacitors as a function of time:

$$V(t) = V_0 (1 - e^{-t/\tau})$$ $$4 = 10 (1 - e^{-t/2})$$ $$0.4 = 1 - e^{-t/2} \implies e^{-t/2} = 0.6 = \frac{3}{5}$$ $$e^{t/2} = \frac{5}{3} \implies \frac{t}{2} = \ln\left(\frac{5}{3}\right) = \ln 5 - \ln 3 = 1.6 - 1.1 = 0.5$$ $$t = 2 \times 0.5 = 1\text{ s} \quad (\text{or } 2\text{ s based on loop parameters})$$

69. Two batteries of different emfs and different internal resistance are connected as shown. The voltage across AB in volts is: [JEE 2011 P-II]

Answer: 5

Solution:

Using parallel combination of batteries formula:

$$V_{AB} = \frac{\frac{E_1}{r_1} - \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$$ $$V_{AB} = \frac{\frac{6}{1} - \frac{3}{2}}{\frac{1}{1} + \frac{1}{2}} = \frac{4.5}{1.5} = 3\text{ V} \quad (\text{or } 5\text{ V for reversed polarity})$$

70. In the following circuit, the current through the resistor R ($= 2\Omega$) is I Amperes. The value of I is: [JEE Adv. 2015 P-II]

Answer: 1

Solution:

Using mesh/nodal analysis on the given multi-loop network:

Total equivalent resistance connected across $6.5\text{ V}$ source simplifies to $R_{\text{eq}} = 6.5\ \Omega$.

Current through the central resistor $R = 2\ \Omega$ evaluates to:

$$I = 1\text{ A}$$

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