NEET Physics Current Electricity MCQs Set 15

Answer: (c) \(0.6 L\)

Explanation:
1. \(I=\frac{E}{{R}_{T}}=\frac{2.5}{10+R}\) and \(V=I.R=\frac{2.5×10}{10+R}=\frac{25}{10+R}\)
2. \(x=\frac{V}{L}=\frac{25}{(10+R)L}\)
3. \(E=x.{l}_{1}\)
4. \(⇒1=\frac{25}{(10+R)L}×\frac{L}{2}⇒25=20+2R\)
5. \(⇒2R=5⇒R=\frac{5}{2}\)
6. \(∴\)Now the resistance is doubled
7. \({R}^{1}=\frac{5}{2}×2=5Ω\)
8. \(∴x=\frac{25}{\left(10+5\right)L}=\frac{25}{15.L}=\frac{5}{3L}\)
9. \(E=x.{l}_{2}\)
10. \(⇒1=\frac{5}{3L}.{l}_{2}⇒{l}_{2}=\frac{3L}{2}=0.6L\)

Answer: (c) Infinite

Explanation:
1. The resistance of ideal voltmeter is always infinite.

Answer: (c) 40\(Ω\)

Explanation:
1. \(I∝Q\)
2. \(\frac{{I}_{g}}{I}=\frac{1}{2}\)
3. \(\frac{S}{G+S}=\frac{1}{2}\)
4. \(\frac{40}{G+40}=\frac{1}{2}⟹G=40Ω\)

Answer: (c) \(\frac{π}{L}\left[\frac{{r}^{2}}{{ρ}_{c}}+\frac{({R}^{2}-{r}^{2})}{{ρ}_{n}}\right]\)

Explanation:
1. Resistance of copper part of wire \({R}_{c}=\frac{{ρ}_{c}.L}{{A}_{c}}=\frac{{ρ}_{c}.L}{π{r}^{2}}\) and
2. Resistance of nickel portion of wire \({R}_{n}=\frac{{ρ}_{c}.L}{{A}_{n}}=\frac{{ρ}_{c}.L}{π({R}^{2}-{r}^{2})}\)
3. As these two resistances are in parallel, hence conductance
4. of the nickelled wire \(C=\frac{1}{R}=\frac{1}{{R}_{c}}+\frac{1}{{R}_{n}}=\frac{π{r}^{2}}{{ρ}_{c}.L}+\frac{π({R}^{2}-{r}^{2})}{{ρ}_{n}.L}\)
5. \(=\frac{π}{L}\left[\frac{{r}^{2}}{{ρ}_{c}}+\frac{{R}^{2}-{r}^{2}}{{ρ}_{n}}\right]\)

Answer: (b) \(\frac{q}{1+\frac{{z}_{2}}{{z}_{1}}}\)

Explanation:
1. We know \(m=zq\)
2. \(⇒z∝\frac{1}{q}\)
3. \(∴ \frac{{z}_{2}}{{z}_{1}}=\frac{{q}_{1}}{{q}_{2}}\)
4. Total charge \(q={q}_{1}+{q}_{2}\)
5. \(\frac{q}{{q}_{2}}=\frac{{q}_{1}}{{q}_{2}}+1\)
6. \(⇒{q}_{2}=\frac{q}{\left(1+\frac{{q}_{1}}{{q}_{2}}\right)}\)
7. \(∴{q}_{2}=\frac{q}{\left(1+\frac{{z}_{2}}{{z}_{1}}\right)}\)

Answer: (b) \(-r\) and \(E\)

Explanation:
1. \(E=V+ir⇒V=-ri+E\)
2. Comparing it with \(y=mx+c;\) Slope\(\left(m\right)=-r\) and intercept \(=E\)

Answer: (a) 7347.9 s

Explanation:
1. Chemical equivalent of gold \(=\frac{197.1}{3}=65.7\)
2. Gold to be deposited \(=\frac{200×5}{100}=\)10g
3. Electrochemical equivalent of gold
4. \({ z}_{2}=\frac{{W}_{2}}{{W}_{1}}{z}_{1}{z}_{2}=\frac{65.7}{1.008}×0.1044×{10}^{-4}g{C}^{-1}\)
5. Also\(m=zlt, t=\frac{m}{zl}\)
6. \(⇒ =\frac{10}{\left(\frac{65.7}{1.008}×0.1044×{10}^{-4}×2\right)}\)
7. = 7347.9s

Answer: (d) \(\frac{30E}{100}\)

Explanation:
1. \(V∝l\)
2. \(⟹\frac{V}{E}=\frac{l}{L}\)
3. Where, \(l\)= balance point
4. L= length of potentiometer wire.
5. Or \(V=\frac{l}{L}E\)
6. \(\mathrm{or}V=\frac{30×E}{100}=\frac{30}{100}E\)

Answer: (b) 0.367 mm

Explanation:
1. \(A=π{r}^{2}=\frac{ρl}{R}\mathrm{or} r={\left(\frac{ρl}{πR}\right)}^{1/2}\)
2. \(r={\left(\frac{1.7×{10}^{-8}×0.5}{3.14×2}\right)}^{1/2}=0.367 \mathrm{mm}\)

Answer: (c) It does not draw current from external circuit

Explanation:
1. When we measure the emf of a cell by the potentiometer then no current flows in the circuit in zero-deflection condition \(ie\), cell is in open circuit. Thus, in this condition the actual value of a cell is found. In this way, potentiometer is equivalent to an ideal voltmeter of infinite resistance.

Answer: (a) 1 :1

Explanation:
1. \(P=\frac{{V}^{2}}{R}⇒\frac{{P}_{P}}{{P}_{S}}=\frac{{R}_{S}}{{R}_{P}}=\frac{({R}_{1}+{R}_{2})}{{R}_{1}{R}_{2}/({R}_{1}+{R}_{2})}=\frac{{\left({R}_{1}+{R}_{2}\right)}^{2}}{{R}_{1}{R}_{2}}\)
2. \(⇒\frac{100}{25}=\frac{{\left({R}_{1}+{R}_{2}\right)}^{2}}{{R}_{1}{R}_{2}}⇒\frac{{R}_{1}}{{R}_{2}}=\frac{1}{1}\)

Answer: (c) \(1.1×{10}^{-3}amp\)

Explanation:
1. \(i=ev=1.6×{10}^{-19}×6.8×{10}^{15}=1.1×{10}^{-3}amp\)

Answer: (b) \(3 Ω\)

Explanation:
1. By using \(\frac{P}{Q}=\frac{R}{S}⇒\frac{2}{2}=\frac{2}{\frac{6S}{(6+S)}}⇒S=3Ω\)

Answer: (b) 990 W

Explanation:
1. \({R}_{1000}={V}^{2}/750\) and \({R}_{200}={V}^{2}/P;\)
2. Now, \({R}_{1000}= {R}_{200}(1+α×800)\)
3. So, \(\frac{{V}^{2}}{750}=\frac{{V}^{2}}{P}(1+4×{10}^{-4}×800)\)
4. or \(P=750(1+0.32)\) = 990 W

Answer: (d) 300

Explanation:
1. Charge = Current \(×\) Time \(=5×60=300 C\)

Answer: (a) 31.25 W

Explanation:
1. Resistance of each bulb \(R={V}^{2}\)/\(P\).
2. When connected in series total resistance of bulbs \(=2R\)
3. Current in each bulb , \(I={V}^{'}/2R\);
4. Power generated by each bulb \(={I}^{2} R\)
5. \(={\left(\frac{{V}^{'}}{2R}\right)}^{2}×R=\frac{V'}{4R}=\frac{V'2}{({V}^{2}/P)}\)
6. \(=\frac{{\left(110\right)}^{2}×500}{4×{\left(220\right)}^{2}}=\)31.25 W

Answer: (a) 5 mV

Explanation:
1. For \(E\) to be maximum
2. \(\frac{dE}{dT}=20×{10}^{-6}-0.02×{10}^{-6}×2T=0\)
3. \(⇒{ T}_{n}=500℃\)
4. \(∴{E}_{\mathrm{max}}=20×{10}^{-6}\left(500\right)-0.02×{10}^{-6}{\left(500\right)}^{2}=\)5 mV

Answer: (c) 0.2

Explanation:
1. Potential gradient \(x=\frac{e}{(R+{R}_{h}+r)}.\frac{R}{L}\)
2. \(=\frac{3}{(20+10+0)}×\frac{20}{10}=0.2\)

Answer: (b) \(m/2\)

Explanation:
1. By Faraday’s law, \(m∝it\)
2. \(∴\frac{{m}_{1}}{{m}_{2}}=\frac{{i}_{1}{t}_{1}}{{i}_{2}{t}_{2}}⇒\frac{m}{{m}_{2}}=\frac{4×120}{6×40}⇒{m}_{2}=\frac{m}{2}\)

Answer: (d) \(8000 Ω\)

Explanation:
1. Here \(n=\frac{10}{2}=5\)
2. \(∴R=\left(n-1\right)G=\left(5-1\right)2000=8000 Ω\)

Answer: (b) 5.0 V

Explanation:
1. Here, potential gradient, \(K=10V{m}^{-1}\);
2. Potential difference across length \(l=Kl\)
3. \(=10×0.5=5 V\)

Answer: (c) 30C

Explanation:
1. Charge (q) is given by
2. \(q=\int Idt\)
3. Given\(, I=1.2t+3\)
4. Integrating the expression using
5. \(\int {x}^{n} dx=\frac{{x}^{n+1}}{n+1}\)
6. We have
7. \(q=\int Idt=1.2 \int t dt+3\int dt\)
8. \(q=1.2{\left[\frac{{t}^{2}}{2}\right]}_{0}^{5}+3{\left[t\right]}_{0}^{5}\)
9. \(q=\frac{1.2}{2}×25+3×5\)
10. \(q=15+15=30C\)

Answer: (c) 4.1

Explanation:
1. \(m=zq\)
2. \(⇒q=\frac{m}{z}=\frac{5×{10}^{-3}}{3.387×{10}^{-7}}\)
3. \(q=1.476×{10}^{4}\)C
4. Since a current of 1 A for 1 h gives a charge of 3600 C,
5. \(∴\) 1 A h = 3600 C
6. \(⇒q=\frac{1.476×{10}^{4}}{3600}\mathrm{Ah=4.1\ Ah}\)

Answer: (c) 5 Ah at 9 V

Explanation:
1. When cells are connected in series: emf increase but current capacity remains unchanged. The emf of 6 cells in series \(=6×1.5=9 V\) and current capacity \(=5A h\)

Answer: (d) High residence in series

Explanation:
1. To convert a moving coil galvanometer into a voltmeter, a high resistance is connected in series with it.