NEET Physics Current Electricity MCQs Set 17

Answer: (d) \(\frac{8}{3} ohm\)

Explanation:
1. Effective resistance between the points \(A\) and \(B\) is
2. \(R=\frac{32}{12}=\frac{8}{3}Ω\)

Answer: (c) 53A

Explanation:
1. \(i=\frac{dQ}{dt}=\frac{d}{dt}\left(5{t}^{2}+3t+1\right)=10t+3\)
2. When \(t=5s, i=10×5+3=53 A\)

Answer: (b) \(\frac{3}{2}K {T}_{0}\)

Explanation:
1. \(E=K\left(T-{T}_{r}\right){T}_{0}+\frac{1}{2}K({T}^{2}-{T}_{r}^{2})\)
2. \(\frac{dE}{dT}=K{T}_{0}+\frac{1}{2}K×2T=K{T}_{0}+KT\)
3. At temperature \(T={T}_{0}\)/2,
4. Thermo-electric power is \(\frac{dE}{dT}=K{T}_{0}+K\frac{{T}_{0}}{2}=\frac{3}{2}K{ T}_{0}\).

Answer: (c) 1.0%

Explanation:
1. On stretching, volume (\(V\)) remains constant. So
2. \(V=Al \mathrm{or} l=V/A.\)
3. Now, \(R=\frac{ρl}{A}=\frac{ρV}{{A}^{2}}=\frac{ρV}{{π}^{2}{D}^{4}/16}=\frac{16ρV}{{π}^{2}{D}^{4}}\)
4. Taking logarithm of both the side and differentiating it we get
5. \(\frac{∆R}{R}=-4\frac{∆D}{D}\mathrm{or}\frac{∆R}{R}=-4×0.25=1.0\%\)

Answer: (a) \(\frac{GR}{ρ}\)

Explanation:
1. Conductivity \(σ=\frac{1}{ρ}\) …(i)
2. and conductance \(G=\frac{1}{R}\)
3. \(⇒GR=1\) ….(ii)
4. From equation (i) and (ii) \(σ=\frac{GR}{ρ}\)

Answer: (d) 1.5\(Ω\)

Explanation:
1. Resistance of original wire is
2. \(R=ρ\frac{l}{A}\)
3. \(ρ\),being the specific resistance of wire. When the wire is cut in two equal halves then resistance becomes
4. \({R}^{'}=\frac{ρl/2}{A}=\frac{R}{2}\)
5. Thus, the net resistance of parallel combination of two halves is given by
6. \({R}_{\mathrm{net}}=\frac{{R}^{'}×{R}^{'}}{{R}^{'}+{R}^{'}}\)
7. \(=\frac{{R}^{'}}{2}=\frac{R}{2×2}=\frac{6}{4}=1.5Ω\)

Answer: (c) Ammeter is placed in parallel across the conductor in a circuit

Explanation:
1. Ammeter is always connected in series with circuit

Answer: (d) \(\frac{1}{10}\)

Explanation:
1. \(\frac{{i}_{g}}{i}=\frac{S}{G+S}=\frac{4}{36+4}=\frac{4}{40}=\frac{1}{10}\)

Answer: (a) Increase

Explanation:
1. From Joule’s law, the heat produced is given by
2. \(P=\frac{{V}^{2}}{R}\)
3. Where \(V\)is potential and \(R\) the resistance.
4. When one bulb is fused, the total resistance of the circuit decreases. Hence, \((P)\) illumination increases.

Answer: (a) \(21×{10}^{3}±10\%\)

Explanation:
1. Red, brown, orange, silver red and brown represents the first two significant figures
2. \(∴R=21×{10}^{3}±10\%\)

Answer: (b) \(12℃\)

Explanation:
1. \({i}^{2}Rt=Cθ=3C;C=\) Thermal capacity
2. when\({i}_{1}=2i⇒C{θ}_{1}=4{i}^{2}Rt=4×3C⇒{θ}_{1}=12℃\)

Answer: (c) 1 :1

Explanation:
1. The bulbs are in series, hence they will have the same current through them

Answer: (c) 280 m

Explanation:
1. Length of the wire is
2. \(l=\frac{RA}{ρ}\)
3. \(∴l=\frac{4×π×{\left(0.7×{10}^{-3}\right)}^{2}}{2.2×{10}^{-8}}=280m\)

Answer: (b) 8

Explanation:
1. Let the original resistance is R \(Ω\).
2. ∴ \(V=IR\)
3. \(V=5×R=5R …\)(i)
4. When 2 \(Ω\) resistance is inserted, then total resistance
5. =(R+2)\(Ω\)
6. ∴ \(V = I’ (R+2)=4(R+2)\) …(ii)
7. From Eqs. (i) and (ii), we get
8. \(5R = 4(R+2)\)
9. ∴ R=8 \(Ω\)

Answer: (c) \(9 V\)

Explanation:
1. \(V=E-ir=12-60×5×{10}^{-2}=9V\)

Answer: (d) Copper strip decreases and that of germanium increases

Explanation:
1. For conductors, resistance \(∝\) Temperature and for semiconductor, resistance \(∝\frac{1}{\mathrm{Temperature}}\)

Answer: (b) 1/3

Explanation:
1. Given that,
2. \(\frac{{l}_{1}}{{l}_{2}}=\frac{4}{3}\mathrm{and}\frac{{r}_{1}}{{r}_{2}}=\frac{2}{3}\)
3. Here, \({l}_{1}\)and \({l}_{2}\)are the length of the wires while \({r}_{1}\)and \({r}_{2}\) are the radii of the wires.
4. Now, we know that
5. \(V=IR⟹IR\mathrm{=constant}\)
6. \(⟹{I}_{1}{R}_{1}={I}_{2}{R}_{2}\)
7. \(\mathrm{or}\frac{{I}_{1}}{{I}_{2}}=\frac{{R}_{2}}{{R}_{1}} ….\left(i\right)\)
8. But we know that the resistance of the wire is
9. \(R=\frac{ρl}{A}\)
10. Hence, from Eq (i)
11. \(\frac{{I}_{1}}{{I}_{2}}=\frac{ρ{l}_{2}/{A}_{2}}{ρ{l}_{1}/{A}_{1}}\)
12. Here,
13. \({R}_{1}=\frac{ρ{l}_{1}}{{A}_{1}},{R}_{2}=\frac{ρ{l}_{2}}{{A}_{2}}and {ρ}_{1}={ρ}_{2}=ρ\)
14. Because both wires are of same material.
15. \(∴ \frac{{I}_{1}}{{I}_{2}}=\frac{{l}_{2}{A}_{1}}{{l}_{1}{A}_{2}}\)
16. \(⟹\frac{{I}_{1}}{{I}_{2}}=\frac{{l}_{2}π{r}_{1}^{2}}{{l}_{1}π{r}_{2}^{2}}\)
17. \(⟹\frac{{I}_{1}}{{I}_{2}}=\frac{{l}_{2}{r}_{1}^{2}}{{l}_{2}{r}_{2}^{2}}\)
18. Here, \(\frac{{l}_{2}}{{l}_{1}}=\frac{3}{4} and\frac{{r}_{1}}{{r}_{2}}=\frac{2}{3}\)
19. \(⟹ \frac{{I}_{1}}{{I}_{2}}=\frac{3}{4}×{\left(\frac{2}{3}\right)}^{2}\)
20. Or \(\frac{{I}_{1}}{{I}_{2}}=\frac{1}{3}\)

Answer: (d) \(6.4×{10}^{3}s\)

Explanation:
1. Current in wire \(i=Ane{v}_{d}\)
2. Here, \(i=1A, n=8×{10}^{28}\mathrm{electron/}{m}^{-3}\)
3. \(A=5×{10}^{-7}{m}^{2}\)
4. \(∴1=8×{10}^{28}×1.6×{10}^{-19}×5×{10}^{-7}×{v}_{d}\)
5. \(\mathrm{or}{v}_{d}=\frac{1}{8×{10}^{28}×1.6×{10}^{-19}×5×{10}^{-7}}\)
6. Now,
7. \(t=\frac{1}{{v}_{d}}=8×{10}^{28}×1.6×{10}^{-19}×5×{10}^{-7}\)
8. \(=64×{10}^{2}=6.4×{10}^{3}s\)

Answer: (c) 35 min

Explanation:
1. From Joule’s law,
2. \(H=\frac{{V}^{2}t}{RJ}\mathrm{cal}\)
3. Where \(V\) is potential difference, \(t\) the time and \(R\) the resistance.
4. \(R={R}_{1}+{R}_{2}\)
5. Let \(\frac{t}{R}=\frac{HJ}{{V}^{2}}= \mathrm{constant=}k\)
6. \(⇒{R}_{1}=k{t}_{1}\mathrm{and}{R}_{2}=k{t}_{2}\)
7. \(∴ kt=k{t}_{1}+k{t}_{2}\)
8. Or \(t=15+20\)
9. \(t=35 \mathrm{min}\)

Answer: (c) \(30 C\)

Explanation:
1. \(i=\frac{dQ}{dt}⇒dQ=idt⇒Q=\int_{{t}_{1}}^{{t}_{2}} idt=\int_{0}^{5} \left(1.2t+3\right)dt\)
2. \(={\left[\frac{1.2{t}^{2}}{2}+3t\right]}_{0}^{5}=30C\)

Answer: (b) 66.7 W

Explanation:
1. When two similar bulbs of different powers are connected in series, then
2. \(\frac{1}{P}=\frac{1}{{P}_{1}}+\frac{1}{{P}_{2}}\)
3. Given, \({P}_{1}=200 W, {P}_{2}=100 W\)
4. \(⇒\frac{1}{P}=\frac{1+2}{200}\)
5. \(⇒P=\frac{200}{3}=66.7 W\)

Answer: (a) \(1.38 V\)

Explanation:
1. \(P=\frac{W}{t}=Vi⇒V=\frac{W}{it}=\frac{1000}{2×6×60}=1.38 V\)

Answer: (c) \(R=r\)

Explanation:
1. In series combination of cells current, \(i=\frac{nE}{nr+R}\)
2. In parallel combination of cell, \({i}^{'}=\frac{E}{(\frac{r}{n)+R}}\)
3. If \(i={i}^{'}\mathrm{then}\frac{nE}{nr+R}=\frac{E}{(\frac{r}{n)+R}}=\frac{nE}{r+nR}\)
4. It will be so if \(r=R\)

Answer: (a) \(\frac{{ρ}_{1}{l}_{1}+{ρ}_{2}{l}_{2}}{{l}_{1}+{l}_{2}}\)

Explanation:
1. \({R}_{1}=\frac{{ρ}_{1}{l}_{1}}{A}\)and \({R}_{2}=\frac{{ρ}_{2}{l}_{2}}{A}\). In series \({R}_{eq}={R}_{1}+{R}_{2}\)
2. \(\frac{{ρ}_{eq.}({l}_{1}+{l}_{2})}{A}=\frac{{ρ}_{1}{l}_{1}}{A}+\frac{{ρ}_{2}{l}_{2}}{A}⇒{ρ}_{eq}=\frac{{ρ}_{1}{l}_{1}+{ρ}_{2}{l}_{2}}{{l}_{1}+{l}_{2}}\)

Answer: (a) \(999 Ω\)

Explanation:
1. \(R=\frac{V}{{i}_{g}}-G=\frac{10}{10×{10}^{-3}}-1=999Ω\)