NEET Physics Current Electricity MCQs Set 09

Answer: (b) Thomson coefficient

Explanation:
1. \(H=σitΔθ⇒\) If\(i=1A, Δθ=1℃,t=1sec\) then\(H=σ\)

Answer: (c) \(0.12 Ω\)

Explanation:
1. \(H=\frac{{V}^{2}t}{4.2 R}\mathrm{or}\frac{H}{t}=\frac{{V}^{2}}{4.2 R}\)
2. \(⇒800=\frac{20×20}{4.2×R}⇒R=\frac{5}{42}=0.119≈0.12Ω\)

Answer: (d) \(\frac{{E}^{2}}{4r}\)

Explanation:
1. \(i=\frac{R}{r+R}\)
2. \(P={i}^{2}R\)
3. \(⟹ P=\frac{{E}^{2}R}{{\left(r+R\right)}^{2}}\)
4. Power is maximum when r=R
5. \(∴ {P}_{\mathrm{max}\,}=\frac{{E}^{2}}{4R}\)
6. \(\mathrm{or}{P}_{\mathrm{max}\,}=\frac{{E}^{2}}{4r}\)

Answer: (d) \(4 J\) and \(4 J\) respectively

Explanation:
1. Heat generated in both the cases will be same because the capacitor has the same energy initially
2. \(=\frac{1}{2}C{V}^{2}=\frac{1}{2}×200×{10}^{-6}×{\left(200\right)}^{2}=4J\)

Answer: (a) \(6.4 A/{m}^{2}\)

Explanation:
1. \(J=nqv=n\left(ze\right)v=\frac{2×{10}^{8}×2×1.6×{10}^{-19}×{10}^{5}}{{\left({10}^{-2}\right)}^{3}}=6.4A/{m}^{2}\)

Answer: (c) 93\(℃\)

Explanation:
1. \({θ}_{0}=2{θ}_{n}-{θ}_{i}=2×210-(600-273)\)
2. = \(420-327=93℃\)

Answer: (d) Remains the same

Explanation:
1. Neutral temperature is independent of temperature of cold junction.

Answer: (b) 8\(Ω\)in parallel

Explanation:
1. From the formula shunt resistance
2. \(S=\frac{{i}_{g}×g}{i-{i}_{g}}\)
3. \(⟹ S=\frac{2×12}{5-2}=8\mathrm{Ω\ (in\ parallel)}\)

Answer: (b) \(\frac{3}{2000}V/cm\)

Explanation:
1. Potential gradient \(x=\frac{V}{L}=\frac{iR}{L}\)
2. \(⇒x=\frac{2}{(15+5)}×\frac{15}{10}=\frac{3}{2000} volt/cm\)

Answer: (b) \(0.02 A\)

Explanation:
1. \(\frac{{i}_{g}}{i}=\frac{S}{G+S}⇒{i}_{g}G=\left(i-{i}_{g}\right)S\)
2. \(∴{i}_{g}G=\left(0.03-{i}_{g}\right)4r\) …(i)
3. and \({i}_{g}G=\left(0.06-{i}_{g}\right)r\) …(ii)
4. from (i) and (ii)
5. \(0.12-4{i}_{g}=0.06-{i}_{g}⇒{i}_{g}=0.02A\)

Answer: (c) 3

Explanation:
1. Resistance of galvanometer
2. \(G=50Ω\)
3. Full scale current \({i}_{g}=0.05\)
4. \(A=2.97×{10}^{-2}c{m}^{2}\)
5. \(=2.97×{10}^{-2}×{10}^{-4}{m}^{2}\)
6. \(=2.97×{10}^{-6}{m}^{2}\)
7. \(i=5A\)
8. \(ρ=5×{10}^{-7}Ωm\)
9. Required resistance to convert the galvanometer into ammeter.
10. \(R=\frac{{i}_{g}G}{i-{i}_{g}}=\frac{0.05×50}{5-0.05}=\frac{2.5}{4.95}\)
11. \(ρ\frac{l}{A}=\frac{50}{99}\)
12. \(l=\frac{50}{99}×\frac{A}{ρ}=\frac{50}{99}×\frac{2.97×{10}^{-6}}{5×{10}^{-7}}=\frac{50}{99}×\frac{29.7}{5}=10×0.3=3m\)

Answer: (d) \(111Ω\)

Explanation:
1. By using \(\frac{i}{{i}_{g}}=1+\frac{G}{S}\)
2. \(⇒\frac{i}{100×{10}^{-3}}=1+\frac{1000}{S}⇒S=\frac{1000}{9}=111Ω\)

Answer: (d) 620 W

Explanation:
1. \({R}_{200}=\frac{{R}_{800}}{1+αt}\)
2. \({R}_{200}=\frac{{R}_{800}}{1+4×{10}^{-4}×600}=\frac{{R}_{800}}{1.24}\)
3. \(\frac{{V}^{2}}{{R}_{800}}=P\) ……(i)
4. and \(\frac{{V}^{2}}{{R}_{200}}=P'\) ……(ii)
5. \(∴ \frac{P'}{P}=\frac{{R}_{800}}{{R}_{200}}=\frac{{R}_{800}}{\left(\frac{{R}_{800}}{1.24}\right)}\)
6. \({ P}^{'}=1.24P=1.24×500=620\) W

Answer: (c) The drift velocity will decrease

Explanation:
1. \(i=nAe{v}_{d}\)
2. or \({v}_{d}=\frac{i}{nAe}ie, {v}_{d}∝\frac{1}{A}\)
3. As \(A\) increases \({v}_{d}\) decreases, because \(i\) remains constant

Answer: (d) No neutral temperature is possible for this thermocouple

Explanation:
1. \(E=aθ+b{θ}^{2}\) (given)
2. For neutral temperature \(\left({θ}_{n}\right),\frac{dE}{dθ}=0\)
3. \(⇒ a+2b {θ}_{n}=0\)
4. \(⇒ {θ}_{n}=-\frac{a}{2b}\)
5. \(∴{θ}_{n}=-\frac{700}{2}\left(∵\frac{a}{b}=700℃\right)\)
6. \(=-350℃<0℃\)
7. But neutral temperature can never be negative (less than zero), \(ie, {θ}_{n}≮0℃.\)
8. Hence, no neutral temperature is possible for this thermocouple.

Answer: (d) \(19995 Ω\)

Explanation:
1. By using \(R=\frac{V}{{i}_{g}}-G⇒R=\frac{100}{5×{10}^{-3}}-5=19,995Ω\)

Answer: (c) 8

Explanation:
1. Suppose \(n\) resistors are used for the required job. Suppose equivalent resistance of the combination is \(R'\) and according to energy conservation it’s current rating is \({i}^{'}\)
2. Energy consumed by the combination \(=n×\) (Energy consumed by each resistance)
3. \(⇒{i}^{'2}{R}^{'}=n×{i}^{2}R⇒n={\left(\frac{{i}^{'}}{i}\right)}^{2}×\left(\frac{{R}^{'}}{R}\right)={\left(\frac{4}{1}\right)}^{2}×\left(\frac{5}{10}\right)=8\)

Answer: (b) 0.8 A

Explanation:
1. Drift velocity,
2. \({v}_{d}=\frac{i}{neA}\)
3. Where I is current, n the number of electrons, e the electron charge and A the area of cross-section of wire.
4. Number of electrons per unit volume
5. \(n=\frac{2×{10}^{21}}{A×100 }\)
6. So, current in the wire
7. \(i=neA{v}_{d}\)
8. \(i=\frac{2×{10}^{21}}{A×100}×1.6×{10}^{-19}×A×0.25\)
9. =0.8A

Answer: (b) 1 : 2

Explanation:
1. \(\frac{{R}_{1}}{{R}_{2}}=\frac{{m}_{1}}{{a}_{1}^{2}}×\frac{{a}_{2}^{2}}{{m}_{2}}\)
2. \(=\left(\frac{m1}{m2}\right){\left(\frac{{a}_{2}}{{a}_{1}}\right)}^{2}=\left(\frac{1}{2}\right){\left(1\right)}^{2}=\frac{1}{2}\)
3. \(\frac{{Q}_{1}}{{Q}_{2}}=\frac{{I}^{2}{R}_{1}t}{{I}^{2}{R}^{2}t}=\frac{1}{2}\)

Answer: (c) Rs 18

Explanation:
1. Cost \(=\left(\frac{60×8×30}{1000}\right)×1.25=\)Rs 18.

Answer: (c) \(9P\)

Explanation:
1. When bulbs are connected in series, \(P=\frac{{V}^{2}}{{R}^{'}}=\frac{{V}^{2}}{3R}\)
2. When bulbs are connected in parallel
3. \({P}^{'}=\frac{{V}^{2}}{R"}=\frac{{V}^{2}×3}{R}=3×3P=9P\)

Answer: (b) \(3 Ω\)

Explanation:
1. The internal resistance of battery is given by
2. \(r=\left(\frac{E}{V}-1\right)R=\left(\frac{40}{30}-1\right)×9=\frac{9×10}{30}=3Ω\)

Answer: (a) \({10}^{-6} ohm\)

Explanation:
1. \(R=\frac{ρl}{A}=50×{10}^{-8}×\frac{50×{10}^{-2}}{{\left(50×{10}^{-2}\right)}^{2}}={10}^{-6}Ω\)

Answer: (a) \(0.1 watt\)

Explanation:
1. For power transmission power loss in line \({P}_{L}={i}^{2}R\)
2. If power of electricity is \(P\) and it is transmitted at voltage \(V,\) then \(P=Vi⇒i=\frac{P}{V}\)
3. \({P}_{L}={\left(\frac{P}{V}\right)}^{2}R=\frac{{P}^{2}R}{{V}^{2}}=\frac{2.2×{10}^{3}×2.2×{10}^{3}×10}{22000×22000}=0.1W\)

Answer: (a) 608 W

Explanation:
1. Let \({R}_{0}\) = resistance of filament at room temperature
2. \({R}_{t}\) = resistance of filament at 2500\(℃\)
3. Similarly powers, \({P}_{0}\mathrm{and}{P}_{t}.\)
4. Here, voltage remains the same.
5. \({ P}_{0}=\frac{{V}^{2}}{{R}_{0}}\)
6. or \({ R}_{0}=\frac{{V}^{2}}{{P}_{0}}, {R}^{t}=\frac{{V}^{2}}{{P}_{t}}\)
7. also \({R}_{t}={R}_{0}\left[1+α\left(2500-20\right)\right]\)
8. and \({ P}_{0}={P}_{t}[1+α\left(2500-20\right)]\)
9. \(=50[1+4.5×{10}^{-3}\left(2500-20\right)]\)
10. = 608W