NEET Physics Current Electricity MCQs Set 11

Answer: (a) Zero

Explanation:
1. An ideal cell has zero resistance

Answer: (c) \(270℃, 500℃\)

Explanation:
1. No charge in neutral temperature but temperature of inversion is
2. \({t}_{i}=2{t}_{n}-{t}_{c}⇒{t}_{i}=2×270-40=500℃\)

Answer: (b) \(4R\)

Explanation:
1. \(R∝{l}^{2}⇒\) If \(l\) doubled then \(R\) becomes 4 times

Answer: (d) 400, 800

Explanation:
1. \(\frac{dE}{dT}=4-\frac{2T}{200}4-\frac{T}{100};\)
2. At neutral point, \(T={T}_{n}.\)
3. \(\frac{dT}{dT}=0=4-{T}_{n}\)/100
4. or \({T}_{n}=400℃, {T}_{i}=2{T}_{n}-{T}_{0}\)
5. \(=2×400-0=800℃\)

Answer: (d) 1.5\(Ω\)

Explanation:
1. Let \(n\) cells be in series and \(m\) in parallel, then
2. \(\frac{nE}{R+nr}=\frac{E}{R+\frac{r}{m}}\)
3. \(⟹ n\left[R+\frac{r}{m}\right]=R+nr\)
4. \(⟹nRm+nr=Rm+mnr\)
5. \(⟹6+2r=3+4r\)
6. \(⟹2r=3\)
7. \(⟹ r=1.5Ω\)

Answer: (d) \(\frac{{i}_{2}{R}_{2}-{i}_{1}{R}_{1}}{{i}_{1}-{i}_{2}}\)

Explanation:
1. \(E={i}_{1}\left({R}_{1}+r\right)={i}_{2}({R}_{2}+r)\)
2. On solving, \(r=\frac{{i}_{2}{R}_{2}-{i}_{1}{R}_{1}}{({i}_{1}-{i}_{2})}\)

Answer: (d) \(193000 coulomb\)

Explanation:
1. \({V}_{2}=\frac{22.4×1}{1}=22.4 litre\)at NTP
2. \(∵11.2 litre\) of\({H}_{2}\)is liberated by \(96,500 C\)
3. \(∴22.4 litre\)of\({H}_{2}\)is liberated by \(96500×2=1,93,000 C\)

Answer: (a) \({10}^{-2}volt/m\)

Explanation:
1. Potential gradient\(=\frac{V}{L}=\frac{iR}{L}=\frac{iρL}{AL}=\frac{iρ}{A}\)
2. \(=\frac{0.2×40×{10}^{-8}}{8×{10}^{-6}}={10}^{-2}V/m\)

Answer: (c) 240

Explanation:
1. \(P×t=ms ∆t\)
2. \(⇒ 2.1×{10}^{3}×t\)
3. \(=1.5×4200×\left[\left(100+273\right)-\left(20+273\right)\right]\)
4. \(⇒ =\frac{1.5×4200×80}{2.1×{10}^{3}}=240s\)

Answer: (d) 16 :1

Explanation:
1. \(H={i}^{2}RT={i}^{2}\left(\frac{ρl}{A}\right)t=\frac{{i}^{2}ρVt}{{A}^{2}} [V=\mathrm{volume}=Al]\)
2. \(⇒H∝\frac{1}{{r}^{4}}⇒\frac{{H}_{1}}{{H}_{2}}={\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{4}={\left(\frac{2}{1}\right)}^{4}=\frac{16}{1}\)

Answer: (c) Is zero

Explanation:
1. Slope is zero at neutral temperature

Answer: (a) 20

Explanation:
1. The balance condition of a meter bridge experiment
2. \(\frac{R}{S}=\frac{{l}_{1}}{\left(100-{l}_{1}\right)}\)
3. Here, \(R={R}_{1}, S={R}_{2}\)
4. \(∴ \frac{{R}_{1}}{{R}_{2}}=\frac{{l}_{1}}{\left(100-{l}_{1}\right)}\)
5. Ist case
6. \(\frac{{R}_{1}+10}{{R}_{2}}=\frac{50}{50} ………\left(i\right)\)
7. \(⟹{R}_{1}+10={R}_{2}\)
8. IInd case
9. \(\frac{{R}_{1}}{{R}_{2}}=\frac{40}{60}\)
10. \(⟹{R}_{2}=\frac{60}{40}{R}_{1} …..\left(\mathrm{ii}\right)\)
11. So, Eqs. (i) and (ii) give
12. \({R}_{1}+10=\frac{60}{40}{R}_{1}\)
13. \(⟹\frac{60}{40}{R}_{1}-{R}_{1}=10\)
14. \(⟹\frac{20}{40}{R}_{1}=10\)
15. \(⟹{R}_{1}=\frac{10×40}{20}\)
16. \(∴ {R}_{1}=20Ω\)

Answer: (d) Sixteen times

Explanation:
1. In stretching of wire \(R∝\frac{1}{{d}^{4}},\) where \(d=\) Diameter of wire

Answer: (a) 2.5\(Ω\)

Explanation:
1. When r is internal resistance of the battery and \(i\)the charging current, then
2. \(V=E+ir\)
3. \(V=2+5×0.1=2.5V\)

Answer: (c) \(1.25×{10}^{19}\)

Explanation:
1. \(i=q/t=ne/t\)
2. or \(n=\frac{it}{e}=\frac{2×1}{1.6×{10}^{-19}}=1.25×{10}^{19}\)

Answer: (a) \(A\) has more loss of weight

Explanation:
1. \(R=ρ\frac{l}{A}⇒\frac{{R}_{1}}{{R}_{2}}=\frac{{A}_{2}}{{A}_{1}}\left[ρ,L \mathrm{constant}\right]⇒\frac{{A}_{1}}{{A}_{2}}=\frac{{R}_{2}}{{R}_{1}}=2\)
2. Now, when a body dipped in water, loss of weight
3. \(=V{σ}_{L}g=AL{σ}_{L}g\)
4. So, \(\frac{{\left(\mathrm{Loss\ of\ weight}\right)}_{1}}{{\left(\mathrm{Loss\ of\ weight}\right)}_{2}}=\frac{{A}_{1}}{{A}_{2}}=2\);so \(A\) has more loss of weight

Answer: (c) \(61V, 1Ω\)

Explanation:
1. Given \({V}_{1}=50 volt, {i}_{1}=11A;{V}_{2}=60 volt, {i}_{2}=1A\)
2. If \(e.m.f.\) and internal resistance of battery are \(E\) and \(r\) respectively then \(P.D.\) across terminals of battery,
3. \(V=E-ir\)
4. We have \(50=E-11r\) …(i)
5. And \(60=E-1r\) …(ii)
6. From (i) and (ii),
7. \(E=61V\) and \(r=1Ω\)

Answer: (b) The position of cathode

Explanation:
1. Positive ions get deposited on cathode

Answer: (a) \(200 W\) bulb will glow more

Explanation:
1. In parallel\({P}_{Consumed}∝{P}_{Rated}\)

Answer: (d) \(0.888 A\)

Explanation:
1. \(I=\frac{E}{R+r/4}=\frac{2}{2+1/4}=\frac{2}{2.25}=0.888A\)

Answer: (b) \({4}^{1/3}\mathrm{mm}\)

Explanation:
1. The temperature of the wire increase to such a value at which the heat produced per second equals heat lost per second due to radiation.
2. \({i}^{2}\left(\frac{ρl}{π{r}^{2}}\right)=H×2πrl\)
3. Where \(H\) is heat lost per second per unit area due to radiation.
4. Hence, \({i}^{2}∝{r}^{3}\)
5. So, \(\frac{{i}_{1}^{2}}{{i}_{2}^{2}} =\frac{{r}_{1}^{3}}{{r}_{2}^{3}}\)
6. Or \({r}_{2}={r}_{1}{\left(\frac{{i}_{2}}{{i}_{1}}\right)}^{\frac{2}{3}}\)
7. \(\left[\mathrm{Here\ :}{r}_{1}=1 \mathrm{mm,}{i}_{1}=1.5 A, {i}_{2}=3 A\right]\)
8. \({r}_{2}=1×{\left(\frac{3}{1.5}\right)}^{2/3}={4}^{1/3}\mathrm{mm}\)

Answer: (b) 20\(Ω\)

Explanation:
1. \(P=VI, I=\frac{P}{V}\)
2. or \(I=\frac{500 W}{100 V}=5A\)
3. Now, \(5 R=100\)
4. or \(R=20Ω\).

Answer: (d) Between 81 and 90 W

Explanation:
1. Let the resistance of the lamp filament be \(R.\) Then \(100={\left(220\right)}^{2}/R.\) When the voltage drops, expected power is
2. \(P={\left(220 ×0.3\right)}^{2}/R'\)
3. Here, \(R'\) will be less than \(R,\) because now the rise in temperature will be less. Therefore, \(P\) is more than \({\left(220 ×0.9\right)}^{2} R=81 W.\) But it will not be 90% of earlier value, because fall in temperature is small. Hence, option (d) is correct.

Answer: (b) The potential drop across the 60 W bulb is greater than the potential drop across the 200 W bulb

Explanation:
1. In series, the current \(I\), is same in two bulbs.
2. Resistance\(R=\frac{{V}^{2}}{P}\)
3. And potential drop \(\left(V\right)=IR\)
4. \(∴\) Potential difference across 60 W bulb is greater than the potential difference across 200 W bulb.

Answer: (c) Peltier effect

Explanation:
1. When corrent is passed through a junction of two different metals, the heat is either evolved or absorbed at the junction. This effect is known as Peltier effect.