NEET Physics Current Electricity MCQs Set 14

Answer: (c) 33

Explanation:
1. Resistance of each bulb \(=\frac{{\left(220\right)}^{2}}{60}\)
2. If the number of bulbs is \(n\) then effective resistance
3. \(=\frac{{\left(220\right)}^{2}}{60n}\)
4. \(∴i=9=\frac{V}{{R}_{\mathrm{eff}}}=\frac{220×60n}{220×220} ⇒ n=33\)

Answer: (b) 6.6 \(×{10}^{-6}\) m

Explanation:
1. Thickness\(d=\frac{V}{A}=\frac{m}{ρA}=\frac{zlt}{ρA}\)
2. \(=\frac{\left(3.3×{10}^{7}\right)×\left(1.5\right)×(20×60)}{9000×(50×10×2)×{10}^{-4}}=6.6×{10}^{-6}\) m.

Answer: (c) 15.6\(Ω\)

Explanation:
1. Resistance,\(R=\frac{ρλ}{A}\)
2. For given problem,
3. \(R∝{λ}^{2}\)
4. \(∴ \frac{{R}_{1}}{{R}_{2}}=\frac{{λ}_{1}^{2}}{{λ}_{2}^{2}}\)
5. \({R}_{2}={\left(\frac{{λ}_{2}}{{λ}_{1}}\right)}^{2}×{R}_{1}=15.6 Ω\)

Answer: (c) Metal oxides with high temperature coefficient of resistivity

Explanation:
1. They are the resistors made up of semiconductors whose resistance decreases with the increase in temperature. This implies that they have negative and high temperature coefficient of resistivity.
2. They are usually made of metal oxides with high temperature coefficient of resistivity.

Answer: (d) None of the above statement is correct

Explanation:
1. Resistivity is the property of the material. It does not depend upon size and shape

Answer: (c) Current

Explanation:
1. Ammeter is used to measure the current through the circuit

Answer: (b) \(2.70×{10}^{-7}Ω\)

Explanation:
1. Resistivity of the material of the rod
2. \(ρ=\frac{RA}{l}\)
3. \(=\frac{3×{10}^{-3}×π{\left(0.3×{10}^{-2}\right)}^{2}}{1}\)
4. \(=27×{10}^{-9}π Ωm\)
5. Resistance of disc,
6. \(R=\frac{\mathrm{Resistivity\ of\ rod\ ×Thickness}}{\mathrm{Area\ of\ cross-section}}\)
7. \(=27×{10}^{-9}π×\frac{{10}^{-3}}{π×{\left(1×{10}^{-2}\right)}^{2}}\)
8. \(=2.7×{10}^{-7}Ω\)

Answer: (c) 63.3\(Ω\)

Explanation:
1. Current drawn from the cell in resistance \({R}_{1}\) will be \(I=E\)/\(({R}_{1}+r)\)
2. Therefore, heat produced in \({R}_{1} ie,\)
3. \({H}_{1}=\frac{{E}^{2}{R}_{1}t}{{\left({R}_{1}+r\right)}^{2}}\)
4. Heat produced in \({R}_{2} ie, {H}_{2}=\frac{{E}^{2}{R}_{2}t}{{\left({R}_{2}+r\right)}^{2}}\)
5. As per question \({H}_{1}={H}_{2}\)
6. or \(\frac{{E}^{2}{R}_{1}t}{{\left({R}_{1}+r\right)}^{2}}=\frac{{E}^{2}{R}_{2}t}{{\left({R}_{2}+r\right)}^{2}}\)
7. On solving we get;
8. \(r=\sqrt{{R}_{1}{R}_{2}}\)
9. \(=\sqrt{100×40}\)=63.25\(Ω\)

Answer: (a) \(\frac{1}{4}amp\)

Explanation:
1. \(P=\frac{{V}^{2}}{R}⇒100=\frac{{\left(200\right)}^{2}}{R}⇒R=\frac{4×{10}^{4}}{{10}^{2}}=400 Ω\)
2. Now,\(i=\frac{V}{R}=\frac{100}{400}=\frac{1}{4}amp\)

Answer: (c) Half of the atomic weight of Cu to the atomic weight of Ag

Explanation:
1. The ratio of the weights deposited on cathodes will be in the ratio of their chemical equivalents. The chemical equivalent of copper = At. Wt/2 and that of silver = At. Wt/1.

Answer: (b) \(1.0 ohm\)

Explanation:
1. \(3=1.5\left(1+r\right)⇒r=1Ω\)

Answer: (d) In the ratio \({\left(50\right)}^{2}:{\left(49\right)}^{2}\) in the first and second case respectively

Explanation:
1. If resistance of each bulb be \(R\), then bias voltage \(V\) is same in both cases, hence \(P=\frac{{v}^{2}}{{R}_{\mathrm{total}}}\)
2. In the case \({R}_{\mathrm{total}}=50R \mathrm{and\ in\ second\ case} 49 R\)
3. \(∴\frac{{P}_{1}}{{P}_{2}}={\left(\frac{49}{50}\right)}^{2}⇒{P}_{2}={\left(\frac{50}{49}\right)}^{2}.{P}_{1}\)
4. or \({P}_{2}>{P}_{1}\)

Answer: (b) Non-\(ohmic\)

Explanation:
1. This is because of secondary ionisation which is possible in the gas filled in it

Answer: (c) Zero

Explanation:
1. \(I=\frac{E}{R+r}\)
2. \(I=\frac{E}{R}=\mathrm{constant}\)
3. where, R=external resistance
4. r= internal resistance =0

Answer: (c) \(-200℃\)

Explanation:
1. \(E=40t+\frac{1}{10}{t}^{2} at\)
2. Inversion, \(E\) will be minimum
3. \(\frac{dE}{dt}=0\)
4. \(\frac{d}{dt}\left(40t+\frac{1}{10}{t}^{2}\right)=0\)
5. \(40+\frac{1}{5} t=0\)
6. \(t=-40×5\)
7. \(t=-200℃\)

Answer: (a) \(100 W\)

Explanation:
1. \({P}_{Rated}=\frac{{V}_{Rated}^{2}}{R}⇒R∝\frac{1}{{P}_{Rated}}\mathrm{[V-constant]}\)
2. So bulb of high power will have less resistance

Answer: (c) \(2 ampere\)

Explanation:
1. \(i=\frac{q}{t}=\frac{4}{2}=2 ampere\)

Answer: (c) 10 A

Explanation:
1. \(i=\frac{ne}{t}=\frac{62.5×{10}^{18}×1.6×{10}^{-19}}{1}=10 ampere\)

Answer: (c) 4 : 1

Explanation:
1. \({R}_{s}=R+R=2R \mathrm{and}{ R}_{P}=\frac{R×R}{R+R}=R/2\)
2. \(\frac{{H}_{1}}{{H}_{2}}=\frac{{I}^{2}{R}_{S}}{{I}^{2}{R}_{P}}=\frac{{R}_{S}}{{R}_{P}}=\frac{2R}{R/2}=\frac{4}{1}\).

Answer: (d) \(16:1\)

Explanation:
1. Let \(d\) be the density of the material of copper wire. Let\({l}_{1},{l}_{2}\) be the lengths of copper wires of diameter 1 mm and 2 mm respectively. As
2. Mass=volume\(×\mathrm{density}=\)(\(π{D}^{2}/4)l.d\)
3. So, \(1=\frac{π{\left({10}^{-3}\right)}^{2}}{4}{l}_{1}×d=π×\frac{{\left(2×{10}^{-3}\right)}^{2}}{4}{l}_{2}×d\)
4. or \({l}_{1}=4{l}_{2}\)
5. Now, \(R=\frac{ρl}{π{D}^{2}/4}ie,R∝\frac{l}{{D}^{2}}\)
6. \(∴ \frac{{R}_{1}}{{R}_{2}}=\frac{{l}_{1}}{{l}_{2}}×\frac{{D}_{2}^{2}}{{D}_{1}^{2}}=4×{2}^{2}=16\)

Answer: (a) 160 MW

Explanation:
1. Work done in delivering \(q\) coulomb of charge from clouds to ground.
2. \(W=Vq\)
3. \(=4×{10}^{6}×4=16×{10}^{6} J\)
4. The power of lighting strike is
5. \(P=\frac{W}{t}=\frac{{16×10}^{6}}{0.1}=160×{10}^{6}W\)
6. \(=160 \mathrm{MW}\)

Answer: (d) \(0.1 m/s\)

Explanation:
1. Current density of drinking electrons \(j=nev\)
2. \(n=5×{10}^{7} c{m}^{-3}=5×{10}^{7}×{10}^{6}{m}^{-3}\)
3. \(v=0.4 m{s}^{-1}, e=1.6×{10}^{-19}C⇒j=3.2×{10}^{-6} A{m}^{-2}\)
4. Current density of ions \(=\left(4-3.2\right)×{10}^{-6}=0.8×{10}^{-6}\frac{A}{{m}^{2}}\)
5. This gives \(v\) for ions \(=0.1 m{s}^{-1}\)

Answer: (b) Current

Explanation:
1. The resistance of ammeter is very low, so when it is used in parallel through a circuit then excess current will flow through it thus, damaging it.

Answer: (d) \(1.1 m\)

Explanation:
1. \(l=\frac{Rπ{r}^{2}}{ρ}=\frac{4.2×3.14×{\left(0.2×{10}^{-3}\right)}^{2}}{48×{10}^{-8}}=1.1m\)

Answer: (c) 5\(Ω\)

Explanation:
1. \(\frac{{V}^{2}}{R}+\frac{{V}^{5}}{5}=4×\left(\frac{{V}^{2}}{R+5}\right)\mathrm{or}\frac{R+5}{5R}=\frac{4}{R+5}\)
2. On solving, we get \(R=5Ω.\)