JEE Physics Current Electricity MCQs Set 10

Answer: (b) 13.1 V

Explanation:
1. An ideal voltmeter draws no current, so the cells simply form a parallel combination.
2. Equivalent emf: \( E_{eq}=\frac{E_1/r_1+E_2/r_2}{1/r_1+1/r_2}=\frac{10/1+15/0.6}{1/1+1/0.6}=\frac{35}{2.667}=13.1\ \text{V} \)
3. The voltmeter reads this open-circuit value, 13.1 V.

Teacher's Note:
1. Cells in parallel: equivalent emf is the conductance-weighted average of the individual emfs.
2. A cell with smaller internal resistance pulls the terminal voltage closer to its own emf, so the answer must lie between 10 V and 15 V but nearer 15 V.

Answer: (a) \( \frac{ml}{ne^2\tau A} \)

Explanation:
1. Conductivity \( \sigma=\frac{ne^2\tau}{m} \), so resistivity \( \rho=\frac{m}{ne^2\tau} \).
2. \( R=\frac{\rho l}{A}=\frac{ml}{ne^2\tau A} \).

Teacher's Note:
1. Resistivity is the inverse of conductivity, and conductivity is ne²τ/m in the free-electron model.
2. R must increase with l and decrease with A, which immediately rules out options (b) and (c).

Answer: (d) 0.4 mm/s

Explanation:
1. \( I=neAv_d \Rightarrow v_d=\frac{I}{neA} \)
2. \( v_d=\frac{5.4}{(8.4\times10^{28})(1.6\times10^{-19})(10^{-6})}=\frac{5.4}{1.344\times10^{4}}\approx 4\times10^{-4}\ \text{m/s} \)
3. That is 0.4 mm/s.

Teacher's Note:
1. Drift speeds in metals are tiny, of the order of a fraction of a millimetre per second.
2. Always substitute n in m⁻³ and A in m² to keep the result in m/s.

Answer: (d) \( a^2:c^2:b^2 \)

Explanation:
1. Along x: length a, cross-section b × c, so \( R_x=\frac{\rho a}{bc} \).
2. Along y: length c, cross-section a × b, so \( R_y=\frac{\rho c}{ab} \).
3. Along z: length b, cross-section a × c, so \( R_z=\frac{\rho b}{ac} \).
4. Multiplying each by abc/ρ gives \( R_x:R_y:R_z=a^2:c^2:b^2 \).

Teacher's Note:
1. For a rectangular block, the resistance in a given direction is ρ × (length in that direction) / (area perpendicular to it).
2. Writing every resistance over the common factor ρ/(abc) makes the ratio easy to read.

Answer: (a) ±5%

Explanation:
1. Red = 2, blue = 6, orange = multiplier 10³, so the value is 26 kΩ.
2. The fourth band (gold) gives the tolerance: gold = ±5%, silver = ±10%, no band = ±20%.

Teacher's Note:
1. The first two bands are digits, the third is the multiplier and the fourth is the tolerance.
2. Remember: gold ±5%, silver ±10%, no fourth band ±20%.

Answer: (b) \( \frac{1}{13} \)th of initial

Explanation:
1. Fraction of current through the galvanometer with shunt S: \( \frac{I_g}{I}=\frac{S}{G+S} \).
2. Given \( \frac{4}{G+4}=\frac15\Rightarrow G=16\ \Omega \).
3. Two shunts in parallel: \( S'=\frac{4\times2}{4+2}=\frac43\ \Omega \).
4. \( \frac{I_g}{I}=\frac{4/3}{16+4/3}=\frac{4}{52}=\frac1{13} \).

Teacher's Note:
1. Deflection is proportional to the current through the galvanometer coil.
2. Adding a second shunt in parallel lowers the shunt resistance, so even less current goes through the coil.

Answer: (a) 20 A

Explanation:
1. Total power = 15 × 45 + 15 × 100 + 15 × 10 + 2 × 1000 = 675 + 1500 + 150 + 2000 = 4325 W.
2. \( I=\frac{P}{V}=\frac{4325}{220}\approx 19.7\ \text{A} \)
3. The fuse must carry at least 19.7 A, so the smallest suitable rating among the options is 20 A.

Teacher's Note:
1. Add the powers of all appliances to get the maximum load, then divide by the supply voltage.
2. Choose the nearest rating that is not below the calculated current.

Answer: (c) \( \frac{G}{99} \)

Explanation:
1. Let the total current be I. The galvanometer takes I/100 and the shunt takes 99I/100.
2. Same potential drop: \( \frac{I}{100}G=\frac{99I}{100}S \).
3. \( S=\frac{G}{99} \)

Teacher's Note:
1. The shunt and the galvanometer are in parallel, so they have the same potential difference.
2. If a fraction 1/n of the current goes through the galvanometer, then S = G/(n − 1).

Answer: (b) \( 12.5\times10^{-3}\ \Omega \)

Explanation:
1. Full-scale current: 150 / 10 = 15 mA. Full-scale voltage: 150 / 2 = 75 mV.
2. \( G=\frac{75\ \text{mV}}{15\ \text{mA}}=5\ \Omega \)
3. \( S=\frac{I_gG}{I-I_g}=\frac{0.015\times5}{6-0.015}=1.253\times10^{-2}\ \Omega\approx12.5\times10^{-3}\ \Omega \)

Teacher's Note:
1. Current sensitivity gives the full-scale current and voltage sensitivity gives the full-scale voltage; their ratio is the coil resistance.
2. Then apply the shunt formula S = IgG/(I − Ig).

Answer: (d) 30 Ω

Explanation:
1. Rated current of the lamp: \( I=\frac{P}{V}=\frac{90}{30}=3\ \text{A} \).
2. The resistor must drop 120 − 30 = 90 V at 3 A.
3. \( R=\frac{90}{3}=30\ \Omega \)

Teacher's Note:
1. The lamp must still carry its rated current, so the extra voltage has to be dropped across the series resistor.
2. Find the current first, then apply Ohm's law to the resistor.

Answer: (b) 2.62 g

Explanation:
1. Heat produced in 1 s: \( H=\frac{V^2}{R}t=\frac{210^2}{50}=882\ \text{J} \).
2. In calories: 882 / 4.2 = 210 cal.
3. \( m=\frac{210}{80}=2.62\ \text{g} \)

Teacher's Note:
1. All the electrical energy is converted to heat, which melts ice at 80 cal per gram.
2. Convert joules to calories before dividing by the latent heat.

Answer: (a) 4 W

Explanation:
1. The emfs oppose, so the net emf is 10 − 4 = 6 V and \( I=\frac{6}{6}=1\ \text{A} \).
2. The 10 V cell drives current through the 4 V cell in the direction opposite to its own emf, so the 4 V cell is being charged and consumes power.
3. \( P=EI=4\times1=4\ \text{W} \)

Teacher's Note:
1. A cell traversed against its emf absorbs power E × I (it is charging); a cell traversed along its emf supplies power.
2. Internal resistances are neglected here.

Answer: (c) 3 V

Explanation:
1. Emf is proportional to the balance length: \( \frac{E_2}{E_1}=\frac{l_2}{l_1} \).
2. \( E_2=2\times\frac{60}{40}=3\ \text{V} \)

Teacher's Note:
1. In a potentiometer, E = (potential gradient) × (balance length), and the gradient stays the same for both cells.
2. So the emfs are in the ratio of the balance lengths.

Answer: (d) 60 cm

Explanation:
1. Initially \( \frac{P}{Q}=\frac{20}{80}=\frac14 \).
2. After interchange the ratio of the left gap to the right gap is \( \frac{Q}{P}=4 \), so \( \frac{l}{100-l}=4\Rightarrow l=80\ \text{cm} \).
3. Shift = 80 − 20 = 60 cm.

Teacher's Note:
1. In a meter bridge the balance condition is (left resistance)/(right resistance) = l/(100 − l).
2. Interchanging the resistances replaces l by (100 − l).

Answer: (d) \( \frac{1}{75}\ {}^\circ\text{C}^{-1} \)

Explanation:
1. Take \( R_t=R_0(1+\alpha t) \): \( \frac{1+15\alpha}{1+37.5\alpha}=\frac45 \).
2. Cross-multiplying: 5 + 75α = 4 + 150α, so 75α = 1.
3. \( \alpha=\frac1{75}\ {}^\circ\text{C}^{-1} \)

Teacher's Note:
1. Write each resistance in terms of the resistance at 0 °C and divide to eliminate it.
2. Check the answer: at 37.5 °C, 1 + 37.5/75 = 1.5 and at 15 °C, 1.2; the ratio 1.2/1.5 = 4/5.

Answer: (b) 1980 Ω

Explanation:
1. Total resistance needed: \( \frac{V}{I_g}=\frac{20}{0.01}=2000\ \Omega \).
2. \( R=2000-20=1980\ \Omega \)

Teacher's Note:
1. A voltmeter is made by connecting a large resistance in series with the galvanometer.
2. The series resistance equals V/Ig − G.

Answer: (a) to the left of C (nearer A)

Explanation:
1. Balance length is proportional to emf: \( E\propto l \).
2. Since E2 < E1, the balance length for E2 is shorter than AC.
3. The null point therefore moves towards end A, to the left of C.

Teacher's Note:
1. Potentiometer emf comparison: E1/E2 = l1/l2.
2. A smaller emf balances against a shorter length of wire.

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

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

Teacher's Note:
1. Terminal voltage = emf − I × internal resistance.
2. Setting the terminal voltage to zero gives I = E/R2, which fixes R.

Answer: (b) 4 and 12

Explanation:
1. For coils a and b: single values a and b, series a + b, parallel ab/(a + b).
2. Try 4 and 12: singly 4 and 12; in series 16; in parallel \( \frac{4\times12}{16}=3\ \Omega \).
3. All four values 3, 4, 12, 16 are obtained.

Teacher's Note:
1. The largest value (16) must be the series combination and the smallest (3) the parallel one.
2. Check the pair by computing the parallel value: 4 × 12/16 = 3.

Answer: (a) \( 1.08\times10^{4}\ \text{J} \)

Explanation:
1. Energy delivered = E I t.
2. \( =6\times5\times360=1.08\times10^{4}\ \text{J} \)

Teacher's Note:
1. The chemical energy lost equals the emf times the charge passed (E × I × t).
2. Convert minutes to seconds first: 6 min = 360 s.

Answer: (d) 2 Ω

Explanation:
1. Maximum current through the meter itself is 100 A; the shunt carries 750 − 100 = 650 A.
2. Same potential drop: \( 100\times13=650\times S \).
3. \( S=2\ \Omega \)

Teacher's Note:
1. The shunt is in parallel with the ammeter, so the voltage across both is the same.
2. Current in the ammeter at full scale is 100 A.

Answer: (c) 300 Ω

Explanation:
1. Current through R3 and R4 (in series): \( I_1=\frac{5}{500}=0.01\ \text{A} \).
2. Voltage across the parallel block: 0.01 × (100 + 500) = 6 V.
3. Voltage across R1: 18 − 6 = 12 V, so the total current is \( \frac{12}{400}=0.03\ \text{A} \).
4. Current through R2: 0.03 − 0.01 = 0.02 A, hence \( R_2=\frac{6}{0.02}=300\ \Omega \).

Teacher's Note:
1. The R3-R4 branch is parallel to R2, so both have the same potential difference.
2. Find the total current through R1 from the remaining battery voltage.

Answer: (a) 7.5 Ω

Explanation:
1. Area of each new wire is (3/9)² = 1/9 of the original, so each has resistance 5 × 9 = 45 Ω.
2. Six wires in parallel: \( R=\frac{45}{6}=7.5\ \Omega \)

Teacher's Note:
1. Resistance varies inversely with the cross-sectional area, which varies as the square of the radius.
2. Identical resistances in parallel divide by their number.

Answer: (d) 6 A

Explanation:
1. Three equal cells in parallel: emf 12 V and internal resistance \( \frac{3}{3}=1\ \Omega \).
2. The load is 1 Ω.
3. \( I=\frac{12}{1+1}=6\ \text{A} \)

Teacher's Note:
1. Identical cells in parallel keep the same emf but divide the internal resistance by the number of cells.
2. Then treat the combination as one cell driving the 1 Ω resistor.