NEET Physics Current Electricity MCQs Set 07

Answer: (b) \(-α\)

Explanation:
1. \(E=αT+\frac{1}{2}β{T}^{2}\)
2. For inversion temperature
3. \(E=0, T=\frac{-2α}{β}\)
4. Thermo electric power
5. \(P=\frac{dE}{dT}=α+\frac{1}{2}β.\left(2T\right)⇒α+β.T\)
6. \(α+β\left(\frac{-2α}{β}\right)=α-2α=-α\)

Answer: (c) The magnetic field should be increased

Explanation:
1. For a galvanometer
2. \(NIAB=C θ\)
3. \(⟹ \frac{θ}{I}=\frac{NAB}{C}\)
4. Here, \(\frac{θ}{I}\) is called the sensitivity of galvanometer so to increase the sensitivity of galvanometer, C should be decreased and N, A and B should be increased

Answer: (b) \(3/\sqrt{ρ}\)

Explanation:
1. \(R=\frac{ρl}{A}=\frac{ρl}{V/l}=\frac{ρ{l}^{2}}{V}\)
2. or \(l={\left(\frac{RV}{ρ}\right)}^{1/2}={\left(\frac{3×3}{ρ}\right)}^{1/2}=\frac{3}{\sqrt{ρ}}\)

Answer: (a) 0.5

Explanation:
1. \(E=V+ir\)
2. After short-circuiting, \(V=0⇒r=\frac{E}{i}=\frac{2}{4}=0.5Ω\)

Answer: (d) 95%

Explanation:
1. Let \(x\) watt be the power loss in transmission line in the form of heat
2. \(∴x={\left(\frac{P}{V}\right)}^{2}R={\left(\frac{10×1000}{200}\right)}^{2}×0.2\)
3. \(=500\mathrm{W=0.5\ kW}\)
4. Efficiency of transmission
5. \(=\frac{\mathrm{power\ delivered\ by\ line}}{\mathrm{power\ supplied\ to\ line}}=\frac{\mathrm{10\ kW}}{\mathrm{10\ kW+}x \mathrm{kW}}\)
6. \(=\frac{10}{10+0.5}=\frac{10}{10.5}=0.95=95\%\)

Answer: (d) \(\frac{R}{16}\)

Explanation:
1. \(R=\frac{ρl}{A}\)
2. \(\mathrm{or} R∝l\)
3. \(∴ \frac{{R}_{1}}{{R}_{2}}=\frac{{l}_{1}}{{l}_{2}}=\frac{L}{L/4}=4\)
4. Or \({R}_{2}=\frac{R}{4}\left(∵{R}_{1}=R\right)\)
5. In parallel combination of such four resistances.
6. \(\frac{1}{{R}^{'}}=\frac{1}{{R}_{2}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{2}}\)
7. \(\mathrm{or}\frac{1}{{R}^{'}}=\frac{1}{R/4}+\frac{1}{R/4}+\frac{1}{R/4}+\frac{1}{R/4}\)
8. \(\mathrm{or}\frac{1}{{R}^{'}}=\frac{4}{R}+\frac{4}{R}+\frac{4}{R}+\frac{4}{R}\)
9. \(\mathrm{or}\frac{1}{{R}^{'}}=\frac{16}{R}\)
10. \(\mathrm{or}{R}^{'}=\frac{R}{16}\)

Answer: (a) \(12×{10}^{3}J\)

Explanation:
1. Heat produced by heater is given by
2. \(H=Pt\)
3. Given, \(P=100 W, t=2 \mathrm{min=2×60s=12s}\)
4. \(∴H=100×120=12×{10}^{3}J\)

Answer: (c) High resistance in series in series with its coil

Explanation:
1. Voltmeter has high resistance and is always connected in parallel with the circuit. So, to convert a galvanometer into voltmeter, a high resistance must be connected in series with it so that it draws negligible current from the circuit.

Answer: (b) Greater than \(1.85 A\)

Explanation:
1. In general, ammeter always reads less than the actual value because of its resistance

Answer: (d) \(1 mJ\)

Explanation:
1. By using \(H=σQ∆θ\)
2. \(⇒H=\left(10×{10}^{-6}\right)×10×\left(60-50\right)={10}^{-3}J=1 mJ\)

Answer: (c) \(2.75 m\)

Explanation:
1. \(L∝l\)
2. \(\frac{{L}_{1}}{{L}_{2}}=\frac{{l}_{1}}{{l}_{2}}\)
3. \(\frac{10}{11}=\frac{2.5}{{l}_{2}}\)
4. \(10{l}_{2}=2.5×11\)
5. \({l}_{2}=\frac{2.5×11}{10}=2.75M\)

Answer: (c) Both bulbs will fuse

Explanation:
1. Because given voltage is very high

Answer: (a) \(4Ω\)

Explanation:
1. \(\mathrm{Resistance=}\frac{\mathrm{Potential\ difference}}{\mathrm{Current}}\)

Answer: (b) \((8/3) Ω\)

Explanation:
1. The equivalent resistance between two corners of equilateral triangle having resistance \(R\) in each arm \(=2R/3=2×4/3 =8/3Ω\)

Answer: (b) More with 39 bulbs than with 40

Explanation:
1. When 1 bulb fuses, the total resistance of the circuit decreases hence the current increases. Since \(P={i}^{2}R,\) therefore illumination increases

Answer: (d) The electron number density of the conductor is enormous

Explanation:
1. In a conductor, the electron number density \(ie\), number of electrons per unit volume of a conductor is very large \((≈{10}^{28}{m}^{-3})\), so large current in a conductor is obtained irrespective of their small drift speed.

Answer: (a) Rs 347

Explanation:
1. If a motor of \(12 HP\) works for 10 days at the rate of 8 h/day then
2. Energy consumption = \(\mathrm{Power\ ×time}\)
3. \(=12×746\frac{J}{s}×(80×60×60)s\)
4. \(=12×746×80×60×60J=2.5×{10}^{9}J\)
5. Rate of energy\(=50\frac{paise}{kWh}\)
6. \(∴3.6×{10}^{6}J\)energy cost \(\mathrm{=Rs\ 0.5}\)
7. \(∴2.5×{10}^{9}J\)energy cost \(=\frac{2.5×{10}^{9}}{2×3.6×{10}^{6}}=\mathrm{Rs} 347\)

Answer: (a) \(960℃\)

Explanation:
1. Let the temperature of molten metal is \(t℃\)
2. The thermo-\(e\)mf \(e=10×{10}^{-6}t volt\)
3. Current in the circuit \(\frac{e}{R+{R}_{G}}=\frac{{10}^{-5}t}{8+1.6}=\frac{{10}^{-5}t}{9.6}amp\)
4. But\(i=\frac{V}{{R}_{G}}=\frac{8×{10}^{-3}}{8}\)
5. \(∴\frac{{10}^{-5}t}{9.6}=\frac{8×{10}^{-3}}{8}\mathrm{or} t=\frac{9.6×{10}^{-3}}{{10}^{-5}}=960℃\)

Answer: (c) 1 : 15 : 125

Explanation:
1. Given, \({l}_{1}=1K, {l}_{2}=3K,{l}_{3}=5K\)
2. or \({m}_{1}=5m,{m}_{2}=3m,{m}_{3}=1m\)
3. We knows
4. \(R=\frac{ρl}{A}\)
5. So \({R}_{1}:{R}_{2}:{R}_{3}=\frac{{l}_{1}}{{A}_{1}} :\frac{{l}_{2}}{{A}_{2}} :\frac{{l}_{3}}{{A}_{3}}\)
6. \(=\frac{{l}_{1}^{2}}{{V}_{1}} :\frac{{l}_{2}^{2}}{{V}_{2}} :\frac{{l}_{3}^{2}}{{V}_{3}}\)
7. \(=\frac{{l}_{1}^{2}}{{m}_{1}} :\frac{{l}_{2}^{2}}{{m}_{2}} :\frac{{l}_{3}^{2}}{{m}_{3}}\)
8. \(=\frac{1}{5} :\frac{9}{3} :\frac{25}{1}=1 :15 :125\)

Answer: (b) 11\(Ω\)

Explanation:
1. Let \(R=100Ω\)
2. \(∴ {R}^{'}=100+100×\frac{10}{100}\)
3. \(=110Ω\)
4. \(∴ ∆R={R}^{'}-R\)
5. \(=110-99=11Ω\)

Answer: (c) 6.4 A towards left

Explanation:
1. An\(α-\)particle has a charge equal to 2 protons. Motion of \(α\) particle to the left, motion of proton towards left and motion of electrons towards right, all will produce conventional current towards left. The total current will be
2. \(i={10}^{19}×(2×1.6×{10}^{-19})\)
3. \(+{10}^{19}×\left(1.6×{10}^{-19}\right)+{10}^{19}×(1.6×{10}^{-19})\)
4. \(=6.4 A\)

Answer: (c) The resistance of the filament in \(40 watt\) bulb is more than the resistance in \(100 watt\) bulb

Explanation:
1. \(P=\frac{{V}^{2}}{R}⇒{R}_{1}=\frac{{V}_{1}^{2}}{{P}_{1}}=\frac{{\left(200\right)}^{2}}{40}=1000Ω\)
2. and \({R}_{2}=\frac{{V}_{2}^{2}}{{P}_{2}}=\frac{{\left(200\right)}^{2}}{100}=400Ω\)

Answer: (a) Infinite

Explanation:
1. A particular temperature, the resistance of a superconductor is zero
2. \(⇒G=\frac{1}{R}=\frac{1}{0}=∞\)

Answer: (c) \(20℃\)

Explanation:
1. Since \(H∝{i}^{2},\) so on doubling the current, the heat produced and hence the rise in temperature becomes four times

Answer: (d) \(1.41 V\)

Explanation:
1. \(E=\frac{V}{l};E\) is constant (volt. gradient)
2. \(⇒\frac{{V}_{1}}{{l}_{1}}=\frac{{V}_{2}}{{l}_{2}}⇒\frac{1.1}{140}=\frac{V}{180}⇒V=\frac{180×1.1}{140}=1.41 V\)