NEET Physics Current Electricity MCQs Set 20

Answer: (d) 0.1V/m

Explanation:
1. Potential gradient of a potentiometer
2. \(K=\frac{Iρ}{A}=\frac{0.2×4×{10}^{-7}}{8×{10}^{-7}}\)
3. \(=0.1V/m\)

Answer: (b) 2\(Ω\)

Explanation:
1. Here, \(r=\frac{{l}_{1}-{l}_{2}}{{l}_{2}}×2Ω\)
2. Where \({l}_{1}\mathrm{=240\ cm,}{l}_{2}\mathrm{=120\ cm}\)
3. \(∴ r=\frac{240-120}{120}×2\)
4. \(=\frac{120}{120}×2×2Ω\)

Answer: (c) \((n-1)G\)

Explanation:
1. Voltmeter is an instrument which measure the potential difference between two points. A high resistance is connected in series with coil of the galvanometer to convert it into voltmeter. This resistance is either 2000\(Ω\) or more than that.
2. Resistance connected in series is given by
3. \(R=\frac{V}{{I}_{g}}-G ..(i)\)
4. Eq. (i) is the value of the resistance required to convert the galvanometer to voltmeter of range 0 to V.
5. From the relation
6. \(R=\frac{V}{{I}_{g}}-G\)
7. \(R=\frac{nV}{\left(\frac{V}{G}\right)}-G\)
8. \(\mathrm{or} R=nG-G=G\left(n-1\right)\)
9. Hence, \(R=(n-1) G\)

Answer: (b) 15 V

Explanation:
1. Force \(=\)Electric intensity \(×\)charge
2. \(=\frac{\mathrm{Potential\ diffsrence}}{\mathrm{distance}}\mathrm{×charge}\)
3. \(∴ 4.8×{10}^{-19}=\frac{V}{5}×{1.6×10}^{-19}\)
4. or \(V=15\) volt

Answer: (b) 2%

Explanation:
1. \(P={i}^{2}R⇒\frac{∆P}{P}=\frac{2∆i}{i} [R→\mathrm{Constant}]\)
2. \(⇒\)% change in power \(=2×\%\)change in current
3. \(=2×1=2\%\)

Answer: (c) \(8 l\)

Explanation:
1. If \(l\) is the balancing length for \({R}_{1},\) and \(l'\) for \({R}_{2},\)
2. \(\frac{{V}_{0}}{{L}_{0}}.l=\frac{E{R}_{1}}{(r+{R}_{1}+{R}_{2})}⇒\frac{{V}_{0}}{{L}_{0}}.l=\frac{E{R}_{1}}{\left({r}_{1}+9{R}_{1}\right)}\) as \({R}_{2}=8{R}_{1}\)
3. \(\frac{{V}_{0}}{{L}_{0}}{l}^{'}=\frac{E.{R}_{2}}{({r}_{1}+9{R}_{1})}=\frac{E.8{R}_{1}}{({r}_{1}+9{R}_{1})}\)
4. \(∴\frac{{l}^{'}}{l}=\frac{{R}_{2}}{{R}_{1}}=\frac{8{R}_{1}}{{R}_{1}}⇒{l}^{'}=8 l\)

Answer: (b) 2,4

Explanation:
1. \(\frac{P}{Q}=\frac{1/3}{1-1/3}=\frac{1/3}{2/3}\)
2. \(⟹P:Q=1:2\)
3. \(P=k, Q=2k\)
4. \(\frac{P+6}{Q}=\frac{\frac{2}{3}}{1-\frac{2}{3}}\)
5. \(⟹\frac{P+6}{Q}=\frac{2}{1}\)
6. \(\frac{k+6}{2k}=\frac{2}{1}\)
7. \(⟹k+6=4k\)
8. \(⟹k=2\)
9. \(∴ P=2Ω, Q=4Ω\)

Answer: (a) \(19.8×{10}^{-7}\mathrm{kg}\)

Explanation:
1. Mass of substance liberated at cathode \(m=zit\)
2. Where, \(z=\) electro chemical equivalent
3. \(= 3.3×{10}^{-7}\mathrm{kg\ .}{C}^{-1}\)
4. \(i=\) current flowing = 3 A,
5. \(t=2\)s
6. \(m=3.3×{10}^{-7}×3×2\)
7. \(=19.8×{10}^{7}\) kg

Answer: (c) \(14.5 volt\)

Explanation:
1. \(V=E-IR=15-10×0.05=14.5V\)

Answer: (c) \(2 Watt\)

Explanation:
1. \({P}_{\mathrm{max}}=\frac{{E}^{2}}{4r}=\frac{{\left(2\right)}^{2}}{4×0.5}=2W\)

Answer: (b) 6

Explanation:
1. \(R=\frac{ρL}{A}; {R}_{1}=\frac{ρ2L}{A}=2R\)
2. \(m=A L d;{m}_{1}=A 2 L d=2m\)
3. Now, \(\frac{{\left(3V\right)}^{2}t}{R}=mc∆T\) …(i)
4. and \(\frac{{\left(NV\right)}^{2}t}{2R}=2mc∆T\) …(ii)
5. Solving (i) and (ii), we get \(N=6\).

Answer: (b) 3 : 1

Explanation:
1. No. of ions liberated \(\left(n\right)=\frac{I×t}{e×\mathrm{Valency}(p)}\)
2. \(ie, n∝\frac{1}{P}\)
3. \(∴\frac{{n}_{\mathrm{Ag}}}{{n}_{\mathrm{Al}}}=\frac{{P}_{\mathrm{Al}}}{{P}_{\mathrm{Ag}}}=\frac{3}{1}\)

Answer: (b) \(0.06 Ω\)

Explanation:
1. Let the internal resistance of cell be \(r,\) then
2. \(i=\frac{E}{R+r}⇒15=\frac{1.5}{0.04+r}⇒r=0.06Ω\)

Answer: (b) Maximum current ratings in the ratio of 1 :4

Explanation:
1. \(R=\frac{{V}^{2}}{P}⇒{R}_{1}=\frac{200×200}{100}=400 \mathrm{Ωand}\)
2. \({R}_{2}=\frac{100×100}{200}=50 Ω.\) Maximum current rating\(i=\frac{P}{V}\)
3. So\({i}_{1}=\frac{100}{200}\) and \({i}_{2}=\frac{200}{100}⇒\frac{{i}_{1}}{{i}_{2}}=\frac{1}{4}\)

Answer: (c) May increases or de decreases

Explanation:
1. The graph between thermo \(emf\) and temperature of hot junction is parabolic in shape

Answer: (b) 2.62 g

Explanation:
1. Heat produced \(=\frac{{V}^{2}t}{4.2 R}=mL\)
2. or \(m=\frac{{V}^{2}}{4.2 R L}=\frac{{\left(210\right)}^{2}×1}{4.2×50×80}=2.62\)g

Answer: (c) \(2.4 gm\)

Explanation:
1. \(\frac{{m}_{1}}{{m}_{2}}=\frac{{Z}_{1}}{{Z}_{2}}⇒{m}_{2}=\frac{{m}_{1}{Z}_{2}}{{Z}_{1}}=\frac{14×1.2×{10}^{-6}}{7×{10}^{-6}}=2.4g\)

Answer: (c) \(2.4 μm\)

Explanation:
1. Mass deposited \(=\) density \(×\) volume of the metal
2. \(m=p×A×X\) …(i)
3. Hence from Faraday’s first law \(m=Zit\)…(ii)
4. So from equation (i) and (ii)
5. \(Zit=ρ×Ax⇒x=\frac{Zit}{ρA}\)
6. \(=\frac{3.04×{10}^{-4}×{10}^{-3}×1×3600}{9000×0.05}=2.4×{10}^{-6}m=2.4μm\)

Answer: (b) Neutral temperature

Explanation:
1. Thermo-electric power \(P=\frac{dE}{dθ};\) at \({t}_{n}, E→\) maximum
2. So \(P→\) zero

Answer: (d) 2

Explanation:
1. Let \(\left({ρ}_{A},{l}_{A},{r}_{A},{A}_{A}\right)\)and \(({ρ}_{B},{l}_{B},{r}_{B},{A}_{B})\)are specific resistances, lengths, radii and areas of wires A and B respectively.
2. Resistance of
3. \(A={R}_{A}=\frac{{ρ}_{A}{l}_{A}}{{A}_{A}}=\frac{{ρ}_{A}{l}_{A}}{π{R}_{A}^{2}}\)
4. Resistance of
5. \(B={R}_{B}=\frac{{ρ}_{B}{l}_{B}}{{A}_{B}}=\frac{{ρ}_{B}{l}_{B}}{π{r}_{B}^{2}}\)
6. For given information
7. \({ρ}_{B}=2{ρ}_{A}\)
8. \({r}_{B}=2{r}_{A}\)
9. And \({R}_{A}={R}_{B}\)
10. \(∴ \frac{{ρ}_{A}{l}_{A}}{π{r}_{A}^{2}}=\frac{{ρ}_{B}{l}_{B}}{π{r}_{B}^{2}}\)
11. \(⟹ \frac{{ρ}_{A}{l}_{A}}{π{r}_{A}^{2}}=\frac{2{ρ}_{A}×{l}_{B}}{π{\left(2{r}_{A}\right)}^{2}}\)
12. \(⟹ \frac{{l}_{B}}{{l}_{A}}=\frac{2}{1}=2:1\)

Answer: (c) Sb-Bi

Explanation:
1. Seebeck arranged the metals in a certain sequence are called the thermoelectric series. The direction of current at the hot junction is from metal occurring earlier in series to one occurring later in series. Some of the substances in series are Bi, Ni, Co, Pd, Pt, Cu, Mn, Hg, Pb, Sn, Au, Ag, Zn, Cd, Fe, Sb, Te.
2. The more separated the metals are in the series, the greater is the value of thermo-emf generated. Thus, emf developed in Sb-Bi thermocouple is higher than other thermocouples.

Answer: (c) 3 :1

Explanation:
1. \(\frac{{E}_{1}}{{E}_{2}}=\frac{{l}_{1}+{l}_{2}}{{l}_{1}-{l}_{2}}=\frac{58+29}{58-29}=\frac{3}{1}\)

Answer: (c) 500

Explanation:
1. \(E=aT+b{T}^{2}\)
2. At temperature of inversion, \(E=0,\)
3. \(∴a{T}_{i}+b{T}_{i}^{2}=0\)
4. \(⇒{T}_{i}=-\frac{a}{b}\)
5. \(⇒{T}_{i}=-\frac{10×{10}^{-6}}{(0.02×{10}^{-6})}=500℃\)

Answer: (b) Conversion of energy

Explanation:
1. Kirchhoff’s second law is \(∑V=0\)
2. It states that the algebric sum of the potential differences in any loop including those associated emf’s and those of resistive elements, must equal zero.
3. This law represents ‘conservation of energy’.

Answer: (b) \(5 A\)

Explanation:
1. \(i=\frac{12}{\left(1+1\right)+0.4}=5A\)