NEET Physics Current Electricity MCQs Set 19

Answer: (c) Peltier effect

Explanation:
1. Production of heat at junctions due to current is known as Peltier effect

Answer: (c) \(90 AH\)

Explanation:
1. Strength\(=5×18=90AH\)

Answer: (c) \({10}^{-2} V/m\)

Explanation:
1. Potential gradient \(\left(x\right)=\frac{iρ}{A}=\frac{0.1×{10}^{-7}}{{10}^{-6}}={10}^{-2}V/m\)

Answer: (a) From \(Fe\) to \(Cu\) through the hot junction

Explanation:
1. According to Seebeck effect

Answer: (c) \(1000 Wh\)

Explanation:
1. \(\mathrm{Energy}=\frac{{V}^{2}}{R}t=\frac{200×200×2}{80}=1000 Wh\)

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: (c) \(R/4\)

Explanation:
1. In India, power, \({P}_{1}=\frac{{V}^{2}}{R}=\frac{{220}^{2}}{R}\)
2. In USA, power, \({ P}_{2}=\frac{{V}^{'2}}{{R}^{'}}=\frac{{\left(110\right)}^{2}}{{R}^{'}}\)
3. As \({P}_{1}={P}_{2}\mathrm{,\ so}\frac{{220}^{2}}{R}=\frac{{\left(110\right)}^{2}}{{R}^{'}}\)
4. or \({R}^{'}=\frac{{\left(110\right)}^{2}}{{\left(220\right)}^{2}}R=\frac{R}{4}\)

Answer: (b) 17.616 g

Explanation:
1. The amount of chlorine
2. \(m=zlt=z\left(\frac{P}{V}\right) t \left[∵P=VI\right]\)
3. \(=0.367×{10}^{-6}×\left(\frac{100×1000}{125}\right)×60\)
4. \(=0.017616 \mathrm{kg=17.616\ g}\)

Answer: (a) \(\frac{1}{2}K{T}_{0}\)

Explanation:
1. Thermo electric power \(E=k\left(T{T}_{0}-{T}_{0}{T}_{r}-\frac{1}{2}{T}_{0}^{2}+\frac{1}{2}{T}_{r}^{2}\right)\)
2. \(S=\frac{dE}{dT}=k\left[T-0-\frac{1}{2}×2T+0\right]\)
3. \(=k[{T}_{0}-T]\)
4. At \(T={T}_{0}/2\)
5. \(S=\frac{1}{2}k{T}_{0}\)

Answer: (c) \(\sqrt{5r}\)

Explanation:
1. Resistance, R\(=\frac{ρl}{A}\)
2. For wire P, \(20=\frac{ρl}{π{r}^{2}} …\left(i\right)\)
3. Similarly for wire Q,
4. \(8=\frac{ρ\left(2l\right)}{π{\left({r}^{'}\right)}^{2}} ….\left(\mathrm{ii}\right)\)
5. Dividing Eq.(i) by Eq. (ii), we have
6. \(\frac{20}{8}=\frac{ρl}{π{r}^{2}}×\frac{π{\left({r}^{'}\right)}^{2}}{ρ\left(2l\right)}\)
7. \(⟹5={\left(\frac{{r}^{'}}{r}\right)}^{2}\)
8. \(⟹{r}^{'}=\sqrt{5}r\)

Answer: (c) \(17.6 mg\)

Explanation:
1. \(P=100 \mathrm{watt,}V=125 V\)
2. Since \(P=VI, ∴I=\frac{P}{V}=\frac{100}{125}ampere\)
3. Mass of chlorine liberated \(=zIt\)
4. \(=0.367×{10}^{-6}×\frac{100}{125}×60=0.0176×{10}^{-3}kg=17.6mg\)

Answer: (a) Length =50 cm, diameter=0.5 mm

Explanation:
1. The resistance of a conductor
2. \(∴R=ρ\frac{1}{π{r}^{2}}\)
3. \(⟹ R∝\frac{1}{{r}^{2}}\)
4. \(\left(a\right){R}_{1}=\frac{50}{{\left(0.25×{10}^{-1}\right)}^{2}}=800×{10}^{2}Ω\)
5. \(\left(b\right){R}_{2}=\frac{100}{{\left(0.5×{10}^{-1}\right)}^{2}}=400×{10}^{2}Ω\)
6. \(\left(c\right){R}_{3}=\frac{200}{{\left(1×{10}^{-1}\right)}^{2}}=200×{10}^{2}Ω\)
7. \(\left(d\right){R}_{4}=\frac{300}{{\left(1.5×{10}^{-1}\right)}^{2}}=133.3×{10}^{2}Ω\)
8. Hence, electrical resistance of first wire is maximum.

Answer: (a) \(B\) draws more current than \(A\)

Explanation:
1. The power drawn by the bulb is
2. \(P=\frac{{V}^{2}}{R}\)
3. \(⇒R=\frac{{V}^{2}}{P}\)
4. or \(R∝\frac{1}{P}\)
5. (as \(V\) is same in parallel)
6. It means that greater power will have less resistance and therefore, draws more current. Hence, current flowing in bulb \(B\) will be more.

Answer: (b) Avoid the production of eddy currents

Explanation:
1. The moving coil galvanometer have their coil wound on a metallic (copper or aluminium) frame, so as to make the motion dead beat due to the production of eddy currents. In the ballistic galvanometer, on the other hand, the damping is to be reduced to the minimum and hence the frame is of a non-conducting material \(eg,\) paper or bamboo.

Answer: (b) The resistance will be halved and the specific resistance will remain unchanged

Explanation:
1. \(R∝\frac{l}{{r}^{2}}⇒\frac{{R}_{2}}{{R}_{1}}=\frac{{l}_{2}}{{l}_{1}}×\frac{{r}_{1}^{2}}{{r}_{2}^{2}}=\left(\frac{2}{1}\right)×{\left(\frac{1}{2}\right)}^{2}=\frac{1}{2}\)
2. \(⇒{R}_{2}=\frac{{R}_{1}}{2},\) specific resistance doesn’t depend upon length, and radius

Answer: (b) 193000 coulomb

Explanation:
1. 11.2 L of \({H}_{2}\) is liberated by 96500 C 22.4 L of \({H}_{2}\) is liberated by
2. 96500\(×2=193000\) C.

Answer: (d) 1.21 times, same

Explanation:
1. In Stretching of wire \(R∝{l}^{2}⇒\frac{{R}_{1}}{{R}_{2}}={\left(\frac{{l}_{1}}{{l}_{2}}\right)}^{2}\)
2. If \({l}_{1}=100\), then \({l}_{2}=110 ⇒\frac{{R}_{1}}{{R}_{2}}={\left(\frac{100}{110}\right)}^{2}\)
3. \(⇒{R}_{2}=1.21 {R}_{1}\)
4. Resistivity doesn’t change with stretching

Answer: (d) \(\mathrm{cot}\,θ\)

Explanation:
1. At an instant approach the student will choose \(\mathrm{tan}\,θ\) will be the right answer. But it is to be seen here the curve makes the angle \(θ\) with the \(V\)-axis. So it makes an angle \((90-θ)\) with the \(i\)-axis
2. So resistance=slope\(=\mathrm{tan}\,\left(90-θ\right)=\mathrm{cot}\,θ\)

Answer: (b) \(2×{10}^{19}\)

Explanation:
1. Current \(I=\frac{ne}{l}\)
2. \(3.2=\frac{n×1.6×{10}^{-19}}{1}\)
3. \(n=\frac{3.2}{1.6×{10}^{-19}}=2×{10}^{19}\)

Answer: (a) \(1×{10}^{-6}\)

Explanation:
1. \(R=\frac{V}{i}=ρ\frac{l}{A}⇒\frac{2}{4}=ρ\frac{50×{10}^{-2}}{{\left(1×{10}^{-3}\right)}^{2}}⇒ρ=1×{10}^{-6}Ωm\)

Answer: (d) \(3×{10}^{8} m{s}^{-1}\)

Explanation:
1. The speed at which current travels through the conductors means the speed of electric effect travelling through conductor which is at a speed of light

Answer: (b) 14.4 min

Explanation:
1. We know, \(\frac{{V}^{2}t}{4.2 R}=Q=ms dθ\)
2. Let,\(N=\) initial number of turns
3. \(R=\) resistance of the coil
4. \(⇒R=ρ\frac{L}{A}=\frac{ρ×N×2πr}{A}\)
5. \(\frac{{V}^{2}t}{4.2×ρ×N×2πr}=Q=ms dθ\)
6. \(\frac{t}{ N}=\) Constant
7. \(∴\frac{{t}_{1}}{{N}_{1}}=\frac{{t}_{2}}{{N}_{2}}\)
8. \(⇒ =\frac{{N}_{2}}{{N}_{1}}×{t}_{1}=\frac{9}{10}×16=\)14.4 min.

Answer: (b) 200

Explanation:
1. \({t}_{n}=\frac{α}{β}=\left(\frac{500}{5}\right)=100℃\)
2. Also\({t}_{n}=\frac{{t}_{i}+{t}_{c}}{2}⇒100=\frac{{t}_{i}+0}{2}⇒{t}_{i}=200℃\)

Answer: (d) -7%

Explanation:
1. \(m=zq=\frac{E}{F}It=\frac{M}{pF}It \mathrm{or} I=\frac{mpF}{Mt}\)
2. \(∴{ I}^{'}=\frac{2.60×1×96500}{108×(40×60)}=0.968\) A
3. % error in reading of ammeter
4. \(=\left(\frac{I-{I}^{'}}{{I}^{'}}\right)×100=\left(\frac{0.90-0.986}{0.968}\right)×100\)
5. \(=-7\%\)

Answer: (a) Be doubled

Explanation:
1. Drift velocity is defined as the average velocity with which free electrons get defined towards the positive end of the conductor under the influence of an external electric field.
2. Drift velocity is given by
3. \({v}_{d}=\frac{eEτ}{m}\)
4. \(But E=\frac{V}{l}\)
5. (if \(l\) is length of the conductor and V is constant potential difference applied across the ends of the conductor)
6. \(∴{v}_{d}=\frac{eVτ}{ml}⟹{v}_{d}∝V\)
7. So, when the potential difference is doubled the drift velocity will be doubled.