NEET Physics Current Electricity MCQs Set 24

Answer: (a) \(-0.2 V\)

Explanation:
1. Current through resistance \(P\) and \(Q\),
2. \({i}_{1}=\frac{4}{90+110}=\frac{1}{50}A\)
3. \({V}_{A}-{V}_{B}=P{i}_{1}=90×\frac{1}{50}=1.8 V\)
4. Current through resistance \(R\) and \(S\),
5. \({i}_{2}=\frac{4}{40+60}=\frac{1}{25}A\)
6. \({V}_{A}-{V}_{D}=R{i}_{2}=40×\frac{1}{25}=1.6 V\)
7. \({V}_{B}-{V}_{D}=\left({V}_{A}-{V}_{D}\right)-({V}_{A}-{V}_{B})\)
8. \(=1.6-1.8=-0.2V\)

Answer: (b) \(62.5 Ω\)

Explanation:
1. \(R=\frac{{V}^{2}}{P}=\frac{{\left(250\right)}^{2}}{{10}^{3}}=62.5Ω\)

Answer: (a) \(0.5 Ω\)

Explanation:
1. \(0.9\left(2+r\right)=0.3\left(7+r\right)⇒6+3r=7+r⇒r=0.5 Ω\)

Answer: (d) 0.36\(\mathrm{mm}{s}^{-1}\)

Explanation:
1. Drift velocity is given by
2. \({v}_{d}=\frac{I}{nqA}\)
3. Where \(I\) is current, n the number of electrons, A the area, q the charge. Given
4. \(\frac{I}{A}=\frac{480A}{c{m}^{2}}, q=1.6×{10}^{-19}C\)
5. \(n=\frac{6×{10}^{23}×9}{64}\)
6. \(∴{v}_{d}=480×\frac{64}{6×{10}^{23}×9×1.6×{10}^{-19}}\)
7. \(⟹{v}_{d}=\frac{480×64}{6×9×1.6×10000}cm{s}^{-1}\)
8. \(⟹{v}_{d}=\frac{32}{900}cm{s}^{-1}\)
9. \(=\frac{32×10}{900}mm{s}^{-1}\)
10. \(=0.36mm{s}^{-1}\)

Answer: (c) 9

Explanation:
1. Let S be the large and R be the smaller resistance.
2. From formula for meter bridge
3. \(S=\left(\frac{100-l}{l}\right)R=\frac{100-20}{20}R=4R\)
4. Again, \(S=\left(\frac{100-l}{100}\right)\left(R+15\right)\)
5. \(=\frac{100-40}{40}\left(R+15\right)=\frac{3}{2}\left(R+15\right)\)
6. \(∴ 4R=\frac{3}{2}\left(R+15\right)\)
7. \(\frac{8R}{3}-R=15\)
8. \(⟹\frac{5R}{3}=15\)
9. \(R=9Ω\)

Answer: (b) \(560°C\)

Explanation:
1. Neutral temperature, \({t}_{n}=\frac{{t}_{i} +{ t}_{c}}{2}\)
2. \(⇒ 285°= \frac{{t}_{i} +10°}{2}\)
3. \(570°={t}_{i}+10°\)
4. or \({t}_{i}=560°\)

Answer: (a) Fall of potential per unit length of the wire

Explanation:
1. \(\mathrm{Potential\ gradient=}\frac{\mathrm{Potential\ difference}}{\mathrm{Length}}\)

Answer: (c) \(x g\)

Explanation:
1. According to Faraday’s first law of electrolysis, mass deposited
2. \(m=ZIt\)
3. Where \(Z=\) electrochemical equivalent of substance
4. \(∴ \frac{{m}_{1}}{{m}_{2}}=\frac{{I}_{1}}{{I}_{2}}×\frac{{t}_{1}}{{t}_{2}}\)
5. Or \(\frac{xg}{{m}_{2}}=\frac{4A}{6A}×\frac{\mathrm{60\ min}}{\mathrm{40\ min}}\mathrm{or}{m}^{2}=xg\)

Answer: (d) \(200℃;1 \mathrm{mV}\)

Explanation:
1. Given, \(V=10×{10}^{-6}t-\frac{1}{40}×{10}^{-6}{t}^{2}\)
2. At neutral temperature
3. \(\frac{dV}{dt}=0\)
4. \(∴ 10×{10}^{-6}-\frac{1}{20}×{10}^{-6}{t}_{n}=0\)
5. or \({t}_{n}=200℃\)
6. Also at neutral temperature, thermo-emf is maximum.
7. Thus,
8. \({V}_{max}=10×{10}^{-6}\left(200\right)-\frac{1}{40}×{10}^{-6}{\left(200\right)}^{2}\)
9. \(=2×{10}^{-3}-1×{10}^{-3}=1 \mathrm{mv}\)

Answer: (c) Independent of the area of cross-section

Explanation:
1. Current through a conductor is constant at even cross-section of the conductor

Answer: (b) \(E=jk\)

Explanation:
1. Specific resistance \(k=\frac{E}{j}\)

Answer: (b) \({10}^{21}\)

Explanation:
1. Power, \(P=\frac{{V}^{2}}{R}\)
2. \(R=\frac{{V}^{2}}{P}=\frac{{\left(60\right)}^{2}}{160}=22.5Ω\)
3. Now, according to Ohm’s law
4. V=IR
5. \(∴ I=\frac{60}{22.5}\)
6. \(⟹I=2.6A\)
7. Here, \(t=60s\)
8. As \(I=\frac{ne}{t}\)
9. \(⟹ n=\frac{I×t}{e}\)
10. \(=\frac{26×60}{1.6×{10}^{-19}}≈{10}^{21}\)

Answer: (d) 4960

Explanation:
1. To convert galvanometer into voltmeter, the necessary value of resistance to be connected in series with the galvanometer is
2. \(R=\frac{V}{{I}_{g}}-G\)
3. \(=\frac{50}{10×{10}^{-3}}-40\)
4. \(=5000-40=4960Ω\)

Answer: (a) Increase

Explanation:
1. Near room temperature, the electric resistance of a typical metal conductor increases linearly with temperature.
2. \(R={R}_{0}\left(1+αT\right)\)
3. Where \(α\) is the thermal resistance coefficient.

Answer: (d) 2000 W

Explanation:
1. When a single heater (resistance \({R}_{1}=R\)) is connected to 220 V, then it will consume a power \({P}_{1}=\)1000 W. If two such identical heaters are connected in parallel (total resistance \({R}_{2}={R}_{1}/2\) = \(R/2)\) to some source, then it will consume power \({P}_{2}\).
2. \(\frac{{P}_{2}}{{P}_{1}}=\frac{{R}_{1}}{{R}_{2}}\)
3. \(⇒ {P}_{2}=2{P}_{1}\)
4. \({ P}_{2}=\)2000 W

Answer: (a) 65 W

Explanation:
1. \({R}_{1}=\frac{{\left(220\right)}^{2}}{100}=484 Ω\)
2. And \({R}_{2}=\frac{{\left(220\right)}^{2}}{200}=242 Ω\)
3. So, \(i=\frac{220}{484+242}=0.3 A\)
4. Hence, total power consumed will be
5. \(P={i}^{2}R={\left(0.3\right)}^{2}×\left(484+242\right)=65.34≈65 W\)

Answer: (a) Quantity of charges

Explanation:
1. \(q=it=\) current \(×\) time

Answer: (a) \(4 Ω\) when the resistance of wire \(A\) is \(4.25 Ω\)

Explanation:
1. \(\frac{{R}_{A}}{{R}_{B}}={\left(\frac{{r}_{B}}{{r}_{A}}\right)}^{4}⇒\frac{{R}_{A}}{{R}_{B}}={\left(\frac{1}{2}\right)}^{4}=\frac{1}{16}⇒{R}_{B}=16{R}_{A}\)
2. When \({R}_{A}\) and \({R}_{B}\) are connected in parallel then equivalent resistance
3. \({R}_{eq}=\frac{{R}_{A}{R}_{B}}{({R}_{A}+{R}_{B})}=\frac{16}{17}{R}_{A}\)
4. If \({R}_{A}=4.25Ω\) then \({R}_{eq}=4Ω i.e.\) option (a) is correct

Answer: (b) Its length

Explanation:
1. An electric fuse of length \(l\), radius \(r\) when used in series of the circuit can withstand only if the rate of heat produced due to current in it is equal to the rate of heat lost due to radiation. If \(H\) is the rate of lost per unit area of the fuse wire, then \(H×2πrl={I}^{2}R={I}^{2}ρl/π{r}^{2}\)
2. or\(H=\frac{{I}^{2}ρ}{2{π}^{2}{r}^{3}} ie, H\) is independent of \(l\).

Answer: (b) 1.95V

Explanation:
1. Here, the resistance of 400\(Ω\)and 10000\(Ω\) are in parallel, their effective resistance \({R}_{p}\)will be
2. \({R}_{P}=\frac{400×10,000}{400+10,000}=\frac{5000}{13}Ω\)
3. Total resistance of circuit
4. \(=\frac{5000}{13}+800=\frac{15400}{13}Ω\)
5. Current in the circuit, \(i=\frac{6}{15400/13}=\frac{39}{7700}A\)
6. Potential difference across voltmeter=\(i{R}_{P}=\frac{39}{7700}×\frac{5000}{13}\)
7. \(=1.95 A\)

Answer: (a) 1A

Explanation:
1. Effective resistance \({R}_{P}\) of 4\(Ω, 6Ω\) and \(12 Ω\) in parallel will be
2. \(\frac{1}{{R}_{p}}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=\frac{6}{12}=\frac{1}{2}\mathrm{or}{R}_{p}=2Ω\)
3. Total resistance of circuit \(=2+2=4 Ω\)
4. The battery current, \(i=4/4=1A\)

Answer: (a) \(r\)

Explanation:
1. For maximum joule heat produced in resistor external resistance = Internal resistance

Answer: (a) \({T}_{n}=\frac{{T}_{i}}{2}\)

Explanation:
1. It is found that temperature of inversion \(\left({T}_{i}\right)\) is as much above the neutral temperature \(\left({T}_{n}\right)\) as neutral temperature is above the temperature of the cold junction \(\left(T\right), ie,\)
2. \({T}_{i}-{T}_{n}={T}_{n}-T \mathrm{or}{T}_{i}=2{T}_{n}-T\)
3. But, here the cold junction is kept at \(0℃,\) hence \(T=0.\)
4. Thus, \({T}_{i}=2{T}_{n}\mathrm{or}{T}_{n}=\frac{{T}_{i}}{2}\)

Answer: (c) \(20W\)

Explanation:
1. \(\frac{H}{t}=P=\frac{{V}^{2}}{R}⇒P∝\frac{1}{R}\mathrm{also}R∝\frac{1}{A}∝\frac{{l}^{2}ρ}{A.lρ}\)
2. \(⇒R∝\frac{{l}^{2}}{m}⇒R∝{l}^{2}\) [for same mass]
3. So \(\frac{{P}_{A}}{{P}_{B}}=\frac{{l}_{B}^{2}}{{l}_{A}^{2}}=\frac{4}{1}⇒{P}_{A}=20W\)

Answer: (d) 1 :1

Explanation:
1. Specific resistance doesn’t depend upon length and area