NEET Physics Current Electricity MCQs Set 22

Answer: (b) G/19

Explanation:
1. Shunt of an ammeter,
2. \(S=\frac{{I}_{g}×G}{I-{I}_{g}}\)
3. \(=\frac{5×G}{100-5}=\frac{G}{19}\)

Answer: (a) Doubled

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

Answer: (a) 0.75A

Explanation:
1. In series, \(i=\frac{3×2}{3r+2}\)…(i)
2. In parallel, \(i=\frac{2}{r/3+2}=\frac{2×3}{r+2×3}\) …(ii)
3. From Eqs. (i) and (ii), \(3r+2=r+6 \mathrm{or} r=2Ω\)
4. From eqs. (i), \(i=\frac{3×2}{3×2+2}=0.75A\)

Answer: (d) Relaxation time decreases

Explanation:
1. Resistance of a conductor, \(R=\frac{m}{n{e}^{2}τ}\frac{l}{A}\)
2. Where the symbols have their usual meaning
3. As the temperature increases, the relaxation time \(τ\) decreases

Answer: (c) \(20 W\)

Explanation:
1. In series\({P}^{'}=\frac{P}{n}=\frac{60}{3}=20 watts\)

Answer: (b) Increases as the temperature rises

Explanation:
1. The resistance of a metallic wire at temperature \(t°C\) is given by
2. \({R}_{t}={R}_{0}\left(1+αt\right)\)
3. Where \(α\) is coefficient of expansion.
4. Hence, resistance of wire increases on increasing the temperature. Also, from Ohm’s law, ratio of \(\frac{V}{i}\) is equal to R \(ie\),
5. \(\frac{V}{i}=R\)
6. Hence, on increasing the temperature the ratio\(\frac{V}{i}\) increases.

Answer: (a) \(1.08×{10}^{4}J\)

Explanation:
1. Chemical energy reduced
2. \(=VIt\)
3. \(=6×5×6×60=10800\)
4. \(=1.08×{10}^{4}J\)

Answer: (d) When the temperature of cold junction increases, the temperature of inversion also increases

Explanation:
1. Temperature of cold junction \({T}_{c},\) temperature of inversion \({T}_{i}\) and neutral temperature \({T}_{n}\) are related as follows
2. \({T}_{n}=\frac{{T}_{i}+{T}_{c}}{2}⇒{T}_{i}=2{T}_{n}-{T}_{c}\)
3. As \({T}_{n}\) is constant for a given thermocouple \({T}_{i}\) decreases with increase in \({T}_{c}\)

Answer: (d) 30

Explanation:
1. Resistance of the lamp
2. \({R}_{0}=\frac{{V}^{2}}{P}=\frac{{\left(30\right)}^{2}}{90}=10 Ω\)
3. Current in the lamp
4. \(i=\frac{30}{10}=3 Ω\)
5. When lamp is operated on a 120 V, then resistance
6. \({R}^{'}=\frac{{V}^{'}}{i}=\frac{120}{30}=40 Ω\)
7. Thus, for proper glow, the resistance required to the put in series will be
8. \(R={R}^{'}-{R}_{0}=40-10=30 Ω\)

Answer: (d) \(27:6:1\)

Explanation:
1. Mass, \(M=\mathrm{volume×density}=Al×d\)
2. or \(A=M/ld\)
3. Resistance \(R=ρl/A=ρl/(M/ld)\)
4. \(=\frac{ρ{l}^{2}d}{M}\)
5. So \(R∝{l}^{2}/M\)
6. Thus, \({R}_{1}:{R}_{2}:{R}_{3}=\frac{{l}_{1}^{2}}{{M}_{1}}:\frac{{l}_{2}^{2}}{{M}_{2}}:\frac{{l}_{3}^{2}}{{M}_{3}}\)
7. \(=\frac{{3}^{2}}{1}:\frac{{2}^{2}}{2}:\frac{{1}^{2}}{3}=27:6:1\)

Answer: (b) \(0.8 amp\)

Explanation:
1. \({i}_{g}S=\left(i-{i}_{g}\right)G⇒{i}_{g}\left(S+G\right)=iG\)
2. \(⇒\frac{{i}_{g}}{i}=\frac{G}{S+G}=\frac{8}{2+8}=0.8\)

Answer: (d) \(0.4 ohm\)

Explanation:
1. \(E=2.2 V, R=4 Ω, V=2V\)
2. \(r=\left(\frac{E}{V}-1\right)R=\left(\frac{2.2}{2}-1\right)×4=0.1×4=0.4Ω\)

Answer: (b) 2.0 W

Explanation:
1. For maximum power \({R}_{\mathrm{ext}}={R}_{\mathrm{int}}\)
2. When batteries are connected in parallel.
3. \({R}_{\mathrm{int}}=\frac{1}{2}=0.5Ω\)
4. Current \(I=\frac{E}{{R}_{\mathrm{ext}}+{R}_{\mathrm{int}}}=\frac{2}{0.5+0.5}=2Ω\)
5. Maximum power is given by
6. \({P}_{\mathrm{max}}={i}^{2}{R}_{\mathrm{ext}}\)
7. \({ ⇒P}_{\mathrm{max}}={\left(2\right)}^{2}×0.5=\)2 W
8. If we assume batteries to be connected in series, then
9. \({ R}_{\mathrm{int}}=1+1=2Ω\)
10. Current \(I=\frac{2E}{{R}_{\mathrm{ext}}+{R}_{\mathrm{int}}}=\frac{2×2}{2+2}=1\) A
11. So, maximum power is now given by
12. \({P}_{\mathrm{max}}={i}^{2}{R}_{\mathrm{ext}}=1×2=\) 2 W
13. In either case \({P}_{\mathrm{max}}=\) 2 W

Answer: (a) 5 ampere

Explanation:
1. Current, \(I=\frac{q}{t}=\frac{3000C}{10×60s}=5A\)

Answer: (b) 300C

Explanation:
1. \(i=\frac{dQ}{dt}=10+4t \mathrm{or} dQ=\left(10+4\right)dt\)
2. Integrating it, we get \(Q=10t+\frac{4{t}^{2}}{2}=10t+2{t}^{2}\)
3. When \(t=10s, Q=10×10+2×{10}^{2}=300 C\)

Answer: (b) Seebeck effect

Explanation:
1. Production of \(e.\)m.f. by temperature difference is known as Seebeck effect

Answer: (d) 5.16

Explanation:
1. In series \(i=\frac{2E}{2 + R}\)
2. \(∴ {J}_{1}={i}^{2}R={\left(\frac{2E}{2+R}\right)}^{2} .R\)
3. In parallel
4. \(i=\frac{E}{0.5+R}\)
5. \({J}_{2}={i}^{2}R={\left(\frac{E}{0.5+R}\right)}^{2} .R\)
6. \(\frac{{J}_{1}}{{J}_{2}}=2.25=\frac{4{\left(0.5 +R\right)}^{2}}{{\left(2 +R\right)}^{2}}\)
7. Solving we get, \(R=4 Ω\)

Answer: (b) \(400°C\)

Explanation:
1. Let resistance for bulb filament at \(o°C\) be \({R}_{0}\) and at a temperature \(θ°C\) its value be \(200Ω\). Then, \(100={R}_{0}\left(1+α×100\right)={R}_{0}(1+0.005×100)\)
2. \(={R}_{0}(1.5)\) …(i)
3. and \(200={R}_{0}\left(1+α×θ\right)={R}_{0}(1+0.005×θ)\)
4. \(={R}_{0}(1.005θ)\) …(ii)
5. Dividing Eq. (ii) by Eq.(i), we get \(2=\frac{1+0.005θ}{1.5}\)
6. \(3=1+0.005θ\)
7. \(⇒ θ=\frac{2}{0.005}=400°C\)

Answer: (c) 7.5 cal

Explanation:
1. From Joule’s law, for a current carrying conductor at a definite temperature the rate of production of heat is given by
2. \(H=Vit\)
3. Where \(i\) is current, \(V\) the potential difference and \(t\) the time.
4. Given, \(V=1.5 \mathrm{volt,}i=2.1 A, t=1s\)
5. \(∴H=15×2.1×1=31.5 J\)
6. Also, \(1 \mathrm{cal=4.2\ J}\)
7. \(∴H=\frac{31.5}{4.2}=7.5 \mathrm{cal}\)

Answer: (d) Current only

Explanation:
1. Only current through the conductor of non-uniform area of cross-section is constant. Drift velocity or drift speed vary inversely with the area of cross-section of the conductor

Answer: (d) \(1800 coulomb\)

Explanation:
1. Change \(q=it=0.5 A×3600 \mathrm{sec=1800}coulomb\)

Answer: (b) Energy

Explanation:
1. Kirchhoff’s second law is based on the law of conservation of energy

Answer: (b) 111\(Ω\)

Explanation:
1. Shunt
2. \(S=\frac{{I}_{g}G}{I-{I}_{g}}\)
3. Here, \(\frac{{I}_{g}}{I}=\frac{1}{34}\)
4. \(∴\frac{1}{34}=\frac{S}{S+G}\)
5. \(∴ S=\frac{G}{33}=\frac{3663}{33}=111Ω\)

Answer: (d) 30mA

Explanation:
1. Current for 50 divisions,
2. \({I}_{g}=\frac{1×50}{10}=2\mathrm{mA}\)
3. \(∴ \frac{{I}_{g}}{I}=\frac{S}{S+G }\)
4. \(⟹I=\left(\frac{S+G}{S}\right){I}_{g}\)
5. Or
6. \(I=\left(\frac{4+20}{4}\right)\mathrm{5mA=30mA}\)

Answer: (a) \(±\)5%

Explanation:
1. The tolerance level of resistance is mostly 1%, 2%, 5% and 10%. In old days 20% was also common, but these are now rare. Now a days 5% tolerance in treands.