JEE Physics Current Electricity MCQs Set 17

Answer: (d) \( (i_1=i_3)>i_2 \)

Explanation:
1. Resistance is proportional to length divided by area, so \( R_1\propto \frac{L}{A} \), \( R_2\propto \frac{1.5L}{A/2}=\frac{3L}{A} \) and \( R_3\propto \frac{L/2}{A/2}=\frac{L}{A} \).
2. With the same voltage, current varies inversely with resistance, so conductors I and III carry equal currents and conductor II carries less.

Teacher's Note:
1. Always compare l/A for conductors of the same material.
2. Halving both length and area leaves the resistance unchanged.

Answer: (c) the faces of area 6L² (separated by L)

Explanation:
1. Resistance between opposite faces is \( R=\frac{\rho d}{A} \), where d is the separation and A the face area.
2. For the three pairs, \( \frac{d}{A} \) equals \( \frac{3L}{2L^2}=\frac{1.5}{L} \), \( \frac{2L}{3L^2}=\frac{0.67}{L} \) and \( \frac{L}{6L^2}=\frac{0.17}{L} \).
3. The smallest resistance, and hence the largest current, is for the faces of area 6L².

Teacher's Note:
1. Largest area and smallest separation give the least resistance.
2. This is a quick way to compare currents in blocks of different orientation.

Answer: (b) \( \frac{(\xi-V)R}{V} \)

Explanation:
1. The current is \( I=\frac{V}{R} \) and the lost voltage is \( \xi-V=Ir \).
2. \( r=\frac{\xi-V}{I}=\frac{(\xi-V)R}{V} \).

Teacher's Note:
1. Terminal voltage V = ξ − Ir.
2. Options (c) and (d) would give a negative resistance.

Answer: (d) 0.4 Ω

Explanation:
1. Emf = 2.2 V. Current \( I=\frac{2}{4}=0.5\ \text{A} \).
2. Lost volt \( =2.2-2=0.2\ \text{V}=Ir \), so \( r=\frac{0.2}{0.5}=0.4\ \Omega \).

Teacher's Note:
1. Open-circuit voltage equals the emf.
2. Use the terminal voltage across the load to find the current.

Answer: (a) 1.5 A

Explanation:
1. One row: emf \( 5000\times0.15=750\ \text{V} \), internal resistance \( 5000\times0.25=1250\ \Omega \).
2. 100 identical rows in parallel: emf 750 V, internal resistance \( \frac{1250}{100}=12.5\ \Omega \).
3. \( I=\frac{750}{500+12.5}\approx1.5\ \text{A} \).

Teacher's Note:
1. For identical rows in parallel the emf stays the same and the internal resistance divides by the number of rows.
2. Series cells add both emf and internal resistance.

Answer: (a) 0.005 V/cm

Explanation:
1. Current \( I=\frac{2}{15+5}=0.1\ \text{A} \).
2. Potential drop across the wire \( =0.1\times5=0.5\ \text{V} \).
3. Gradient \( =\frac{0.5}{100}=0.005\ \text{V/cm} \).

Teacher's Note:
1. Potential gradient is the voltage drop per unit length of the wire.
2. Find the current in the whole series circuit first.

Answer: (b) 24 C

Explanation:
1. \( q=\int_2^3(2t+3t^2)\,dt=\left[t^2+t^3\right]_2^3 \).
2. \( =(9+27)-(4+8)=24\ \text{C} \).

Teacher's Note:
1. Charge is the time integral of current.
2. Do not forget to subtract the value at the lower limit.

Answer: (b) (n − 1)G

Explanation:
1. Full-scale current \( I_g=\frac{V}{G} \). For range nV with series resistance R: \( nV=I_g(G+R)=\frac{V}{G}(G+R) \).
2. \( n G=G+R\Rightarrow R=(n-1)G \).

Teacher's Note:
1. A voltmeter is a galvanometer with a high series resistance.
2. The total resistance must be n times the original.

Answer: (a) 220 Ω

Explanation:
1. \( \frac{55}{R}=\frac{20}{100-20}=\frac14 \).
2. \( R=4\times55=220\ \Omega \).

Teacher's Note:
1. Left gap resistance over right gap resistance equals left length over right length.
2. A short left length means the left resistance is the smaller one.

Answer: (d) \( \frac{\beta}{\alpha} \)

Explanation:
1. Change in series resistance for a small temperature rise Δt: \( R_1\alpha\,\Delta t-R_2\beta\,\Delta t \).
2. Setting it to zero gives \( R_1\alpha=R_2\beta\Rightarrow\frac{R_1}{R_2}=\frac{\beta}{\alpha} \).

Teacher's Note:
1. The increase in one resistor must cancel the decrease in the other.
2. Here R1 and R2 are the resistances at the starting temperature.

Answer: (d) 1.5 A

Explanation:
1. Total emf \( \sum E=1.5\sum r \) and total resistance \( \sum r \).
2. \( I=\frac{\sum E}{\sum r}=1.5\ \text{A} \).

Teacher's Note:
1. The ratio of emf to internal resistance is the same for every cell.
2. Series cells in a loop: total emf over total resistance.

Answer: (d) 1127 K

Explanation:
1. 300 K is 27 °C. With \( R_t=R_0(1+\alpha t) \): \( R_0=\frac{1}{1+0.00125\times27}=0.9674\ \Omega \).
2. \( 2=0.9674(1+0.00125\,t)\Rightarrow t=854\ ^\circ\text{C} \).
3. In kelvin, T = 854 + 273 = 1127 K.

Teacher's Note:
1. The coefficient is defined with respect to 0 °C, so find R0 first.
2. Do not use 1 Ω as the resistance at the reference temperature.

Answer: (d) \( R=R_2-R_1 \)

Explanation:
1. Current \( I=\frac{2E}{R+R_1+R_2} \).
2. Terminal voltage of the second source is zero: \( E-IR_2=0\Rightarrow I=\frac{E}{R_2} \).
3. Equating, \( R+R_1+R_2=2R_2\Rightarrow R=R_2-R_1 \).

Teacher's Note:
1. Zero terminal voltage means the whole emf is lost across the internal resistance.
2. Set the two expressions for I equal.

Answer: (d) V′ < 2V

Explanation:
1. Let the galvanometer resistance be G and full-scale current Ig: \( V=I_g(G+R) \) and \( V'=I_g(G+2R) \).
2. \( V'-2V=I_g(G+2R)-2I_g(G+R)=-I_gG<0 \), so V′ < 2V.

Teacher's Note:
1. The galvanometer's own resistance is always part of the total.
2. Doubling only the added resistance does not double the total.

Answer: (b) 10 V range with 200 kΩ resistance in series

Explanation:
1. 10 V range: \( R=\frac{10}{50\times10^{-6}}-100\approx200\ \text{k}\Omega \) in series, so option (b) is correct.
2. 50 V range would need about 1 MΩ, not 10 kΩ.
3. 10 mA range: \( S=\frac{50\times10^{-6}\times100}{10\times10^{-3}-50\times10^{-6}}\approx0.5\ \Omega \), which matches neither (c) nor (d).

Teacher's Note:
1. Voltmeter conversion uses a series resistance, ammeter conversion a parallel shunt.
2. Check each option with R = V/Ig − G or S = IgG/(I − Ig).

Answer: (b) 6

Explanation:
1. Heat produced \( =\frac{V^2}{R}t \) and temperature rise \( \Delta T=\frac{V^2t}{R\,ms} \).
2. Both the resistance R and the mass m are proportional to the length, so \( \Delta T\propto\frac{V^2}{L^2} \).
3. \( \frac{9E^2}{L^2}=\frac{N^2E^2}{4L^2}\Rightarrow N=6 \).

Teacher's Note:
1. A longer wire has more resistance and also more mass to heat.
2. Equal temperature rise in equal time needs V proportional to length.

Answer: (c) \( \frac{\rho_1V}{\rho_1d_1+\rho_2d_2} \)

Explanation:
1. In steady state the same current density j flows through both layers, so \( E_1=\rho_1j \) and \( E_2=\rho_2j \).
2. \( V=E_1d_1+E_2d_2=j(\rho_1d_1+\rho_2d_2) \).
3. \( E_1=\frac{\rho_1V}{\rho_1d_1+\rho_2d_2} \).

Teacher's Note:
1. In the steady state the layers behave as resistors in series.
2. The field in each layer is proportional to its resistivity.

Answer: (d) \( \frac{\rho_2V}{\rho_1d_1+\rho_2d_2} \)

Explanation:
1. With the common current density j, \( E_2=\rho_2j \) and \( j=\frac{V}{\rho_1d_1+\rho_2d_2} \).
2. \( E_2=\frac{\rho_2V}{\rho_1d_1+\rho_2d_2} \).

Teacher's Note:
1. Same method as for layer 1 with ρ2 in place of ρ1.
2. Check: E1d1 + E2d2 must equal V.

Answer: (a) \( \epsilon_1\rho_1=\epsilon_2\rho_2 \)

Explanation:
1. The free surface charge at the boundary is \( \sigma=\epsilon_2E_2-\epsilon_1E_1 \) (difference of the displacement fields).
2. With \( E_1=\rho_1j \) and \( E_2=\rho_2j \): \( \sigma=j(\epsilon_2\rho_2-\epsilon_1\rho_1) \).
3. It is zero when \( \epsilon_1\rho_1=\epsilon_2\rho_2 \).

Teacher's Note:
1. Gauss's law at the boundary relates the jump in D to the free charge.
2. Equal products of permittivity and resistivity make the leakage and charging time constants equal.

Answer: (a) 0.5 V

Explanation:
1. With K1 open, current in the wire \( I=\frac{2}{10+15}=0.08\ \text{A} \), so the drop across the wire is 0.8 V over 50 cm.
2. Potential gradient \( =\frac{0.8}{50}=0.016\ \text{V/cm} \).
3. \( \epsilon_2=0.016\times31.25=0.5\ \text{V} \).

Teacher's Note:
1. With K2 open the cell is on open circuit, so the balance point gives its emf directly.
2. Find the gradient from the current and wire resistance.

Answer: (a) 7.5 Ω

Explanation:
1. With K1 closed, R1 is shorted: current \( I=\frac{2}{10}=0.2\ \text{A} \) and gradient \( =\frac{2}{50}=0.04\ \text{V/cm} \).
2. Terminal voltage of ε2 with R2 connected \( =0.04\times5=0.2\ \text{V} \).
3. \( 0.2=\frac{0.5\times5}{5+r}\Rightarrow r=7.5\ \Omega \).

Teacher's Note:
1. Closing K2 puts R2 across the cell, so the null point gives the terminal voltage.
2. Use the potential divider relation V = εR2/(R2 + r).

Answer: (a) 12.5 cm

Explanation:
1. With K1 closed the gradient is 0.04 V/cm (as above).
2. With K2 open the balance gives the emf: \( AJ=\frac{0.5}{0.04}=12.5\ \text{cm} \).

Teacher's Note:
1. Shorting R1 raises the current in the wire and hence the gradient, so the balance point moves closer to A.
2. Balance length equals emf divided by gradient.

Answer: (c) \( \frac{\rho I}{2\pi r^2} \)

Explanation:
1. The current spreads over a hemisphere of area \( 2\pi r^2 \), so \( j=\frac{I}{2\pi r^2} \).
2. \( E=\rho j=\frac{\rho I}{2\pi r^2} \).

Teacher's Note:
1. The block is a half-space, so the area is that of a hemisphere, not a full sphere.
2. Ohm's law in vector form is E = ρj.

Answer: (a) \( \frac{\rho I}{\pi a}-\frac{\rho I}{\pi(a+b)} \)

Explanation:
1. From A: \( V(r)=\frac{\rho I}{2\pi r} \). B is at distance a and C at a + b, so \( V_B-V_C=\frac{\rho I}{2\pi}\left(\frac1a-\frac1{a+b}\right) \).
2. The current leaving at D gives \( V(r)=-\frac{\rho I}{2\pi r} \) with r measured from D. C is at distance a and B at a + b from D, so this part of \( V_B-V_C \) equals \( \frac{\rho I}{2\pi}\left(\frac1a-\frac1{a+b}\right) \), the same as the contribution from A.
3. Adding both: \( \Delta V=\frac{\rho I}{\pi}\left(\frac1a-\frac1{a+b}\right) \).

Teacher's Note:
1. Superposition: the sink at D contributes the mirror image of the source at A.
2. Potential from a source falls as 1/r, so the nearer point is at higher potential.

Answer: (b) B = 5 tesla, 75 K < TC(B) < 100 K

Explanation:
1. The critical temperature decreases steadily as the field increases, so for B between 0 and 7.5 T the value of TC(B) lies between 100 K and 75 K.
2. At 5 T (less than 7.5 T) this gives 75 K < TC(B) < 100 K, which is certain.
3. At 10 T (more than 7.5 T) TC(B) must be below 75 K, ruling out (c), and its exact value cannot be fixed, ruling out (a) and (d).

Teacher's Note:
1. The TC(B) curve falls as B rises.
2. Only an interval can be guaranteed, not an exact value.