NEET Physics Current Electricity MCQs Set 13

Answer: (d) \(3 ohm\)

Explanation:
1. Suppose resistance of wires are \({R}_{1}\) and \({R}_{2}\)then
2. \(\frac{6}{5}=\frac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}.\) If \({R}_{2}\) breaks then \({R}_{1}=2Ω\)
3. Hence, \(\frac{6}{5}=\frac{2×{R}_{2}}{2+{R}_{2}}⇒{R}_{2}=3Ω\)

Answer: (c) 5 :8

Explanation:
1. Resistance\(=ρ\frac{l}{A}\)
2. \(∴\frac{{R}_{1}}{{R}_{2}}=\frac{{ρ}_{1}}{{ρ}_{2}}×\frac{{l}_{1}}{{l}_{2}}×\frac{{A}_{2}}{{A}_{1}}=\frac{2}{3}×\frac{3}{4}×\frac{5}{4}=\frac{5}{8}\)

Answer: (a) \(L/2,2A\)

Explanation:
1. \(R=\frac{ρl}{A}\mathrm{or} R∝l/A.\) Thus, resistance is least in a wire of length \(L/2\) and area of cross-section \(2A\)

Answer: (d) \(\frac{G}{19}\)

Explanation:
1. \(\frac{{i}_{g}}{i}=\frac{S}{G+S }⇒\frac{5}{100}=\frac{S}{G+S}⇒S=\frac{G}{19}\)

Answer: (d) None of these

Explanation:
1. A galvanometer can be converted into an ammeter by using a low resistance wire in parallel with the galvanometer. The range of ammeter can be increased but cannot be decreased. The reason is that a series resistor cannot change the current that will produce full deflection.
2. A shunt connected in parallel is given by
3. \(S=\frac{{I}_{g}G}{I-{I}_{g}}\)
4. If \(I<{I}_{g}\), then the value of S is negative. Hence, a galvanometer cannot be converted into an ammeter of range \(I<{I}_{g}\).

Answer: (a) \(32×{10}^{4}±5\%\)

Explanation:
1. \(R=32×{10}^{4}±5\%\)

Answer: (c) 2 :1

Explanation:
1. \(H=\frac{{V}^{2}}{R}t⇒\frac{{H}_{1}}{{H}_{2}}=\frac{{R}_{2}}{{R}_{1}}=\frac{4}{2}=\frac{2}{1}\)

Answer: (d) \(0.5 ohm\)

Explanation:
1. For maximum power \(r=R\)

Answer: (c) 5.4 g

Explanation:
1. \(m=zit\)
2. \(9=z×{10}^{5}⇒z=\frac{9}{{10}^{5}}g {C}^{-1}\)
3. \(∴m=zit=9×{10}^{-5}×50×\left(20×60\right)=\)5.4 g.

Answer: (a) \(3\mathrm{min}\,20 sec\)

Explanation:
1. In parallel \(\frac{1}{{t}_{p}}=\frac{1}{{t}_{1}}+\frac{1}{{t}_{2}}⇒{t}_{p}-\frac{{t}_{1}{t}_{2}}{{t}_{1}+{t}_{2}}\)
2. \(=\frac{5×10}{5+10}=\frac{50}{15}=3.33 \mathrm{min}=3 min. 20sec\)

Answer: (a) (i) 3 W (ii) 0.4 W (iii) 2.6 W (iv) 2.6 W

Explanation:
1. (i) Rate of chemical energy consumption
2. \(=1.5×2=\)3 W
3. (ii) Rate of energy dissipation inside the cell
4. \(=2×2×0.1=0.4\) W
5. (iii) Rate of energy dissipation inside the resistor
6. \(=(3-0.4)\) W = 2.6 W
7. (iv) Power output of source \(=(3-0.4)\) W = 2.6 W
8. \(* EI\)represents rate of chemical energy consumption of the cell.
9. \(*{I}^{2}r\)represents the rate of energy dissipation inside the cell.
10. \(*(EI-{I}^{2}r)\)represents the power output of the source of emf.

Answer: (b) \(1.1×{10}^{-3}m{s}^{-1}\)

Explanation:
1. Number of free electrons per unit volume,
2. \(n=\frac{N}{M}ρ=\frac{6.023×{10}^{23}}{63.5×{10}^{-3}}×9×{10}^{3}\)
3. \(∴ {v}_{d}=\frac{i}{nAe}\)
4. \(=\frac{1.5×63.5×{10}^{-3}}{6.023×{10}^{23}×9×{10}^{3}×{10}^{-7}×1.6×{10}^{-19}}\)
5. \(=1.1×{10}^{-3}m{s}^{-1}\)

Answer: (b) Increase by about \(11\%\)

Explanation:
1. \(P∝\frac{1}{R}\mathrm{and}R∝l⇒P∝\frac{1}{l}\)
2. \(⇒\frac{{P}_{1}}{{P}_{2}}=\frac{{l}_{2}}{{l}_{1}}⇒\frac{{P}_{1}}{{P}_{2}}=\frac{(100-10)}{100}=\frac{90}{100}⇒{P}_{2}=1.11{P}_{1}\)
3. % change in power \(= \frac{{P}_{2}-{P}_{1}}{{P}_{1}}×100=11\%\)

Answer: (d) \(90 watt\)

Explanation:
1. \(\mathrm{In\ series}{P}_{S}=\frac{P}{n}⇒10=\frac{P}{3}⇒P=30 W\)
2. \(\mathrm{In\ parallel}{P}_{P}=nP=3×30=90 W\)

Answer: (a) \(0.08 Ω\)

Explanation:
1. Internal resistance,
2. \(r=R\left(\frac{{l}_{1}-{l}_{2}}{{l}_{2}}\right)=1\left(\frac{540-500}{500}\right)=\frac{40}{500}=0.08 Ω\)

Answer: (a) \(0.008/℃\)

Explanation:
1. Using \({R}_{{T}_{2}}={R}_{{T}_{1}}[1+α({T}_{2}-{T}_{1})]\)
2. \(⇒{R}_{100}={R}_{50}[1+α(100-50)]\)
3. \(⇒7=5\left[1+\left(α×50\right)\right]⇒α=\frac{(7-5)}{250}=0.008/℃\)

Answer: (a) 1050 W

Explanation:
1. Power is the amount of work done or energy \(\left(Q\right)\) transferred per unit of time \(\left(t\right).\)
2. \(∴ P=\frac{H}{t}\)
3. \(H=mc ∆θ\)
4. Where \(m\) is mass, \(c\) the specific heat and \(∆θ\) the temperature difference.
5. \(∴P=\frac{mc ∆θ}{t}\)
6. \(∆θ=\left(35-20\right)℃=15℃\)
7. \(C=1000, 4.2 \mathrm{J=1\ cal}\)
8. \(t=1\mathrm{min}\,=60s\)
9. \(∴P=\frac{1×1000×15×4.2}{60}{\mathrm{Js}}^{-1}=1050 W\)

Answer: (b) 30 : 3 : 10 :1

Explanation:
1. Let \(I\) be the total current passing through balanced wheat
2. stone bridge. Current through arms of resistances \(P\) and
3. \(Q\) in series is
4. \({I}_{1}=\frac{I×300}{330+110}=\frac{3}{4}I\) and current through arms of
5. Resistances \(R\) and \(S\) in series is \({I}_{2}=\frac{I×110}{330+110}=\frac{1}{4}I\)
6. \(∴\) Ratio of heat developed per sec
7. \({H}_{P} :{H}_{Q} :{H}_{R} :{H}_{S}\)
8. \(={\left(\frac{3}{4}I\right)}^{2}×100 :{\left(\frac{3}{4}I\right)}^{2}×10 :{\left(\frac{1}{4}I\right)}^{2}×300 :{\left(\frac{1}{4}I\right)}^{2}×30\)
9. \(=30 :3 :10 :1\).

Answer: (c) 1400 K

Explanation:
1. The resistance of a metallic wire at temperature \(t°C\)is given by
2. \({R}_{t}={R}_{0}\left(1+α t\right) …\left(i\right)\)
3. Given, \(α=0.00125{K}^{-1}\)
4. \({R}_{300}=1Ω\)
5. From Eq. (i), we have
6. \(1={R}_{o}\left(1+0.00125×300\right) ..\left(\mathrm{ii}\right)\)
7. and \(2={R}_{0}\left(1+0.00125×T\right) ..\left(\mathrm{iii}\right)\)
8. \(∴ \frac{2}{1}=\frac{1+0.00125×T}{1+0.00125×300}\)
9. or 2.75=1+0.00125\(×\)T
10. or \(T=\frac{1.75}{0.00125}=1400K\)

Answer: (a) \(\frac{zIt}{m}=\mathrm{constant}\)

Explanation:
1. \(m=Zit⇒\frac{m}{Zit}=1[\mathrm{constant}]\)

Answer: (c) 900\(Ω\)

Explanation:
1. When a resistance of 100\(Ω\) is connected in series current,
2. \(i=\frac{V}{100+R} ……….\left(i\right)\)
3. When a resistance of 1000\(Ω\) is connected in series, the its range double
4. Current, \(i=\frac{2V}{1100+R}\) ……(ii)
5. From Eqs. (i) and (ii)
6. \(\frac{V}{100+R}=\frac{2V}{100+R}\)
7. \(R=900Ω\)

Answer: (d) \(12 Ω\)

Explanation:
1. \(R=\frac{ρl}{A}=10\)
2. Now length \({l}_{1}=l+l/10=11 l/10\)
3. \(∴\) New area \({A}_{1}=Al/{l}_{1}\)
4. \(∴\) New resistance,
5. \({R}_{1}=\frac{ρ{l}_{1}}{{A}_{1}}=ρ(11/10)/(10/11)A\)
6. \(=\frac{121}{100}\frac{ρl}{A}=\frac{121}{100}×10=12.1 Ω\)

Answer: (d) \(A\) will increase and \(V\) will decrease

Explanation:
1. After connecting a resistance \(R\) in parallel with voltmeter its effective resistance decreases. Hence less voltage appears across it \(i.e. V\) will decreases. Since overall resistance decreases so more current will flow \(i.e. A\) will increase

Answer: (d) \(10.8 gm\)

Explanation:
1. Equivalent weight of copper \(=\frac{64}{2}=32\)
2. \(\frac{\mathrm{Equivalent\ weight\ of}Cu}{\mathrm{Equivalent\ weight\ of}Ag}=\frac{\mathrm{Weight\ of}Cu \mathrm{deposited}}{\mathrm{Weight\ of}Ag \mathrm{deposited}}\)
3. Weight of copper deposited \(=\frac{10.8×32}{108}=3.2gm\)

Answer: (b) \(83℃\)

Explanation:
1. \({R}_{t}={R}_{0}(1+αt)\)
2. Initially, \({R}_{0}\left(1+30α\right)=10Ω\)
3. Finally, \({R}_{0}\left(1+αt\right)=11Ω\)
4. \(∴\frac{11}{10}=\frac{1+αt}{1+30α}\)
5. \(⇒10+\left(10×0.002×t\right)=11+330×0.002\)
6. \(⇒0.02t=1+0.66=1.66⇒t=\frac{1.66}{0.02}=83℃\)