NEET Physics Current Electricity MCQs Set 04

Answer: (c) \(5 V\)

Explanation:
1. Total resistant\(=2.5+0.5=3.0 kΩ=3000 Ω\)
2. \(∴\) Current, \(i=\frac{6}{3000}A\);
3. Reading of voltmeter \(=i×\left(2.5×1000\right)\)
4. \(=\frac{6}{3000}×2500=5V\)

Answer: (c) Metals to become hard

Explanation:
1. Electroplating only provides a thin deposition of a metal on the surface which in no way can give hardness to the metal

Answer: (a) \(484 Ω\)

Explanation:
1. \(\frac{{V}^{2}}{R}=P⇒R=\frac{{V}^{2}}{P}=\frac{220×220}{100}=484Ω\)

Answer: (b) \(6.25×{10}^{15}\)

Explanation:
1. \(n=\frac{1×{10}^{-3}}{1.6×{10}^{-19}}=6.25×{10}^{15}\)

Answer: (c) 0.1 \(\mathrm{mm}{s}^{-1}\)

Explanation:
1. When A is area of cross-section of wire, and n be number of free electrons per unit volume, then relation between electric current (i) and drift velocity (\({v}_{d}\)) is
2. \(i=ne A{v}_{d}\)
3. Number of atoms in 63 g of copper is equal to Avogadro’s number ie, \(6×{10}^{23}\)
4. Volume of 63 g copper
5. \(=\frac{63}{\mathrm{density}}=\frac{63}{9}=7c{m}^{3}\)
6. \(n=\frac{6.02×{10}^{23}}{7}\mathrm{per\ c}{m}^{3}\)
7. \(=\left(\frac{6.02×{10}^{29}}{7}\right)\mathrm{per}{m}^{3}\)
8. \(\mathrm{Area\ A=π}{r}^{2}\)
9. \(=π{\left(0.5×{10}^{-3}\right)}^{2}{m}^{2}\)
10. Hence, drift velocity
11. \({v}_{d}=\frac{1.1}{\left(\frac{6.02×{10}^{29}}{7}\right)×1.6×{10}^{-19}×π×{\left(0.5×{10}^{-3}\right)}^{2}}\)
12. \({v}_{d}=0.1×{10}^{-3}m{s}^{-1}\)
13. \({v}_{d}\mathrm{=0.1\ mm}{s}^{-1}\)

Answer: (a) For metallic conductors at low temperature

Explanation:
1. Because with rise in temperature resistance of conductor increases, so graph between \(V\) and \(i\) becomes non linear

Answer: (b) The junction becomes hot when current flows from antimony to bismuth

Explanation:
1. Based on Peltier effect

Answer: (d) Resistance of \(40 W\) bulb is less than the resistance of \(60 W\) bulb

Explanation:
1. The resistance of \(40 W\) bulb will be more and \(60 W\) bulb will be less

Answer: (c) \(1.25×{10}^{-3}{\mathrm{ms}}^{-1}\)

Explanation:
1. \({v}_{d}=\frac{I}{nAe}=\frac{20}{{10}^{29}×{10}^{-6}×1.6×{10}^{-19}}=1.25×{10}^{-3}m/s\)

Answer: (b) Increase resistance in main circuit

Explanation:
1. To shift the balance point on higher length, the potential gradient of the wire is to be decreased. The same can be obtained by decreasing the current of the main circuit, which is possible by increasing the resistance in series of potentiometer wire.

Answer: (a) \(4 R Ω\)

Explanation:
1. When wire is stretched to doubled its length, its resistance becomes four times

Answer: (a) 0.12A

Explanation:
1. Surface area of earth, \(A=4π{r}^{2}\)
2. Charge entering the earth per sec per unit area
3. \(J=0.15×1.6×{10}^{-19}C{m}^{-2}{s}^{-1}\)
4. \(=0.15×1.6×{10}^{-19}×{10}^{4}C{m}^{-2}{s}^{-1}\)
5. \(∴\mathrm{Current}, i=JA=J4π{r}^{2}\)
6. \(=0.15×1.6×{10}^{-19}×{10}^{4}×4×3.14×{\left(6.4×{10}^{6}\right)}^{2}\) on solving
7. \(i=0.12 A\)

Answer: (c) \(1 ohm\)

Explanation:
1. \(\frac{1}{R}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=\frac{3}{1}⇒R=\frac{1}{3} ohm\)
2. Now such three resistance are joined in series, hence total
3. \(R=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1ohm\)

Answer: (c) \(I/2\)

Explanation:
1. From \({v}_{d}=\frac{i}{neA}⇒i∝{v}_{d}A⇒i∝{v}_{d}{r}^{2}\)

Answer: (a) \(2.5×{10}^{-4}ohm\)

Explanation:
1. \(S=\frac{G}{\frac{i}{{i}_{g}}-1}=\frac{25}{\frac{5}{50×{10}^{-6}}-1}=\frac{25}{{10}^{5}-1}=\frac{25}{{10}^{5}}=2.5×{10}^{-4}Ω\)

Answer: (b) \(50/99 Ω\) in parallel

Explanation:
1. \(S=\frac{{i}_{g}×G}{i-{i}_{g}}=\frac{10×{10}^{-3}×50}{1-{10}^{-3}×10}=\frac{50}{99}Ω\) in parallel

Answer: (a) \(\frac{{σ}_{i}}{G}={σ}_{v}\)

Explanation:
1. \({σ}_{i}=\frac{θ}{i}=\frac{θ}{iG}.G={σ}_{V}G⟹\frac{{σ}_{i}}{G}={σ}_{v}\)

Answer: (c) \({v}_{P}=\frac{1}{4}{v}_{Q}\)

Explanation:
1. Drift velocity \({v}_{d}=\frac{i}{neA}⇒{v}_{d}∝\frac{1}{A}\) or \({v}_{d}∝\frac{1}{{d}^{2}}\)
2. \(⇒\frac{{v}_{P}}{{v}_{Q}}={\left(\frac{{d}_{Q}}{{d}_{P}}\right)}^{2}={\left(\frac{d/2}{d}\right)}^{2}=\frac{1}{4}⇒{v}_{P}=\frac{1}{4}{v}_{Q}\)

Answer: (c) 4.1

Explanation:
1. \(It=\frac{m}{z}=\frac{5×{10}^{-3}}{3.387×{10}^{-7}}\)
2. \(=\frac{5×{10}^{4}}{3.387×60×60}\) Ah = 4.1 Ah

Answer: (c) \(15.6 Ω\)

Explanation:
1. Given, \({l}_{1}=l+\frac{25}{100}l=\frac{5l}{4}\). Since volume or wire remains unchanged on increasing length, hence
2. \(Al-{A}_{1}×\frac{5l}{4 \mathrm{or}{A}_{1}}=4A/5\)
3. Given, \(R=10=ρl/A, \mathrm{and}\)
4. \({R}_{1}=\frac{ρ{l}_{1}}{{A}_{1}}=\frac{ρ5l/4}{4A/5}=\frac{25ρl}{16A}\)
5. \(\mathrm{or}{R}_{1}=\frac{25}{16}×10=\frac{250}{16}=15.6Ω\)

Answer: (b) \(R/4\)

Explanation:
1. \(R=\frac{ρl}{a}\) for first wire and \({R}^{'}=\frac{ρl}{4a}=\frac{R}{4}\) for second wire

Answer: (a) 240 V

Explanation:
1. Resistance of \({V}_{1}=80×200\)
2. =16000\(Ω\)
3. Resistance of \({V}_{2}\)=32000\(Ω\)
4. The current flowing in the circuit is given by
5. \(i=\frac{80}{16000}=\frac{1}{200}A\)
6. Total resistance of the circuit
7. =16000+32000=48000\(Ω\)
8. Line voltage\(=iR\)
9. \(=\frac{48000}{200}=240V\)

Answer: (a) 7 :13

Explanation:
1. Using Wheatstone principle\(\frac{P}{Q}=\frac{R}{S}=\frac{R}{100-l}\)
2. \(=\frac{35}{100-35}=\frac{35}{65}=\frac{7}{13}\)

Answer: (d) \(0.25 Ω\)

Explanation:
1. \(\frac{X}{1}=\frac{20}{80}⇒X=\frac{1}{4}Ω=0.25Ω\)

Answer: (c) \(80 watt\)

Explanation:
1. \({P}_{p}=nP=2×40=80 W\)