NEET Physics Current Electricity MCQs Set 10

Answer: (c) 4/9

Explanation:
1. \(R=\frac{{V}^{2}}{P}\mathrm{or} R∝{V}^{2} ∴\frac{{R}_{1}}{{R}_{2}}=\frac{{\left(200\right)}^{2}}{{\left(300\right)}^{2}}=\frac{4}{9}\).
2. When bulbs are connected in series, the current \(I\) is same through each. As \(P={I}^{2} R \mathrm{or} P∝R( \mathrm{as} I\) is same in series),
3. so \(\frac{{P}_{1}}{{P}_{2}}=\frac{{R}_{1}}{{R}_{2}}=\frac{4}{9}\).

Answer: (a) 4

Explanation:
1. Let resistances are \({R}_{1}\)and \({R}_{2}\),
2. Then \(S={R}_{1}+{R}_{2}\)
3. And \(P=\frac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}\)
4. \(∴\left({R}_{1}+{R}_{2}\right)=\frac{n×{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}\left(\mathrm{From} S=nP\right)\)
5. \(⟹{\left({R}_{1}+{R}_{2}\right)}^{2}={nR}_{1}{R}_{2}\)
6. \(⟹n=\left[\frac{{R}_{1}^{2}+{R}_{2}^{2}+2{R}_{1}{R}_{2}}{{R}_{1}{R}_{2}}\right]\)
7. \(=\left[\frac{{R}_{1}}{{R}_{2}}+\frac{{R}_{2}}{{R}_{1}}+2\right]\)
8. We know,
9. Arithmetic Mean \(≥\) Geometric Mean
10. \(\frac{\frac{{R}_{1}}{{R}_{2}}+\frac{{R}_{2}}{{R}_{1}}}{2}≥\sqrt{\frac{{R}_{1}}{{R}_{2}}×\frac{{R}_{2}}{{R}_{1}}}\)
11. \(⟹\frac{{R}_{1}}{{R}_{2}}+\frac{{R}_{2}}{{R}_{1}}≥2\)
12. So, n(minimum value) =2+2=4

Answer: (d) \(4 ohm\)

Explanation:
1. \(\frac{{R}_{1}}{{R}_{2}}=\frac{(1+α{t}_{1})}{(1+α{t}_{2})}⇒\frac{5}{6}=\frac{(1+α×50)}{(1+α×100)}⇒α=\frac{1}{200}\mathrm{per\ ℃}\)
2. Again by \({R}_{t}={R}_{0}(1+αt)\)
3. \(⇒5={R}_{0}\left(1+\frac{1}{200}×50\right)⇒{R}_{0}=4Ω\)

Answer: (a) \(10Ω\)

Explanation:
1. Wheatstone bridge is balanced, therefore
2. \(\frac{P}{Q}=\frac{R}{S}\) or \(1=\frac{10}{S}⇒S=10 ohm\)

Answer: (a) \(\frac{R}{{n}^{4}}\)

Explanation:
1. \(\frac{{R}_{1}}{{R}_{2}}={\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{4}⇒\frac{R}{{R}_{2}}={\left(\frac{nr}{r}\right)}^{4}⇒{R}_{2}=\frac{R}{{n}^{4}}\)

Answer: (b) 14.4\(×{10}^{5}\) J

Explanation:
1. Energy liberated \(=\frac{{V}^{2}}{R}t\)
2. \(=\frac{{\left(120\right)}^{2}}{6}×\left(10×60\right)=14.4×{10}^{5}\) J

Answer: (b) \(2×{10}^{20}\)

Explanation:
1. \(q=it=n×2e\)
2. \(n=\frac{it}{2e}=\frac{2×32}{2×1.6×{10}^{-19}}=2×{10}^{20}\)

Answer: (c) \(0.6 Ω\)

Explanation:
1. Resistance of shunted ammeter\(=\frac{GS}{G+S}\)
2. Also\(\frac{i}{{i}_{g}}=1+\frac{G}{S}⇒\frac{GS}{G+S}=\frac{{i}_{g}.G}{i}\)
3. \(⇒\frac{GS}{G+S}=\frac{0.05×120}{10}=0.6 Ω\)

Answer: (b) \(1.56×{10}^{-2}{\mathrm{ms}}^{-1}\)

Explanation:
1. Drift velocity, \({v}_{d}=\frac{I}{neA}\)
2. \(⟹{v}_{d}=\frac{5}{\left(5×{10}^{26}\right)×\left(1.6×{10}^{-19}\right)×\left(4×{10}^{-6}\right)}\)
3. \(=\frac{1}{64}=1.56×{10}^{-2}m{s}^{-1}\)

Answer: (d) \(3×3×2\) J

Explanation:
1. Total power spend across two resistors connected
2. in parallel to battery \(=\frac{{V}^{2}}{{R}_{1}}+\frac{{V}^{2}}{{R}_{2}}\)
3. \(=\frac{3×3}{2}+\frac{3×3}{2/3}=\frac{36}{2}=18\)
4. \(=3×3×2\)J

Answer: (b) 4 and 12

Explanation:
1. If we take \({R}_{1}=4Ω\), \({R}_{2}=12Ω\), then in series resistance
2. \(R={R}_{1}+{R}_{2}=4+12=16Ω\)
3. In parallel, resistance
4. \(R=\frac{4×12}{4+12}=3Ω\)
5. \(\mathrm{So,}{R}_{1}\mathrm{4Ω\ and}{R}_{2}=12Ω\)

Answer: (d) 9975

Explanation:
1. \(R=\frac{V}{{i}_{g}}-G=\frac{100}{10×{10}^{-3}}-25=9975Ω\)

Answer: (a) Charge

Explanation:
1. Kirchhoff’s first law is based on the law of conservation of charge

Answer: (b) \(1.611 amp\)

Explanation:
1. Here two cells are in series
2. Therefore total emf \(=2E\)
3. Total resistance \(=R+2r\)
4. \(∴i=\frac{2E}{R+2r}=\frac{2×1.45}{1.5+2×0.15}=\frac{2.9}{1.8}=\frac{29}{18}=1.611 amp\)

Answer: (c) \(\frac{ev}{2πr}\)

Explanation:
1. If a charged particle of charge q resolves in a circular orbit of radius r with frequency v, then the orbital current is given by
2. \(I=qv\)
3. \(or I=q\frac{ω}{2π}\left(∵=2πv⟹v=\frac{ω}{2π}\right)\)
4. \(I=\frac{\mathrm{ev}}{2πr}\left(∵v=rω\right)\)

Answer: (b) 1500 C

Explanation:
1. Let \(q\) be the charge flowing through copper voltmeter.
2. The charge flowing through silver voltmeter \(=(2000-q)\)
3. Now, \(m={z}_{\mathrm{Cu}}q={z}_{\mathrm{Ag}}×(2000-q)\);
4. \(∴q=\frac{{Z}_{\mathrm{Ag}}}{{Z}_{\mathrm{Cu}}}×(2000-q)\)
5. \(=\frac{1.008×{10}^{-6}}{3.36×{10}^{-7}}×(2000-q)\)
6. \(q=1500\) C

Answer: (c) \(0.02 Ω\)

Explanation:
1. \(\frac{i}{{i}_{g}}=1+\frac{G}{S}⇒\frac{1}{{10}^{-3}}=1+\frac{20}{S}⇒S=\frac{20}{900}≈0.02Ω\)

Answer: (d) Resistivity

Explanation:
1. The temperature coefficient of resistivity for thermister is negative. Therefore by increase in temperature the resistivity of the thermister decreases.

Answer: (b) 2000 W

Explanation:
1. Let the resistance of the two heaters be denoted by \({R}_{1}\mathrm{and}{R}_{2}.\)
2. Then, \({R}_{1}=\frac{{V}^{2}}{{P}_{1}}\)
3. \({R}_{2}=\frac{{V}^{2}}{{P}_{2}}\)
4. In parallel combination of resistances,
5. \(\frac{1}{{R}_{P}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}\)
6. \(\frac{{P}_{P}}{{V}^{2}}=\frac{{P}_{1}}{{V}^{2}}+\frac{{P}_{2}}{{V}^{2}}\)
7. \({P}_{p}={P}_{1}+{P}_{2}\)
8. Given, \({P}_{1}=1000 W\)
9. \({P}_{2}=1000 W\)
10. \(∴ {P}_{p}=1000+1000\)
11. Hence, \({P}_{p}=2000 W\)

Answer: (a) \(10 Ω\)

Explanation:
1. \(\frac{i}{{i}_{g}}=1+\frac{G}{S}⇒\frac{i.G}{{V}_{g}}=1+\frac{G}{S}⇒\frac{100×{10}^{-3}×40}{800×{10}^{-3}}=1+\frac{40}{S}\)
2. \(⇒S=10Ω\)

Answer: (a) 2W

Explanation:
1. When two batteries are in series to the external resistance
2. \(R,\) total resistance of the circuit
3. \(= R+2r=R+2×1=R+2Ω\)
4. Total emf of batteries = 2+2 = 4 V
5. \(∴\) Joulean power across \(R\)
6. \(={\left(\frac{4}{R+2}\right)}^{2}×R=\frac{{4}^{2}R}{{\left(R-2\right)}^{2}+8 R}\).
7. Joulean power will be maximum if \(R-2=0\)
8. or \(R=2Ω\)
9. \(∴\) Maximum joulean power\(=\frac{{4}^{2}×2}{{\left(2 2\right)}^{2}+8×2}=\) 2 watt

Answer: (c) \(25 Ω\)

Explanation:
1. \(H=\frac{{V}^{2}}{R}\)
2. \(⇒R=\frac{{V}^{2}}{H}\)
3. \(=\frac{{\left(25\right)}^{2}}{25}=25 Ω\)

Answer: (d) \(\frac{1}{ρ}\)

Explanation:
1. \(P=\frac{{V}^{2}}{R}\)but\(R=\frac{ρl}{A}⇒P=\frac{{V}^{2}}{ρl / A}=\frac{A{V}^{2}}{ρl}\). Since \(\frac{A{V}^{2}}{l}\)is constant as per given condition so\(P∝\frac{1}{ρ}\)

Answer: (a) Absolute temperature of junction \(T\)

Explanation:
1. Peltier coefficient is directly proportional to absolute temperature \(T\).

Answer: (a) 2 rows of 12 cells connected in parallel

Explanation:
1. Suppose \(m\) rows are connected in parallel and each row contains \(n\) identical cells (each cell having \(E=15 V\) and \(r=2Ω\))
2. For maximum current in the external resistance \(R,\) the necessary condition is \(R=\frac{nr}{m}\)
3. \(⇒12=\frac{n×2}{m}⇒n=6m\) …(i)
4. Total cells\(=24=n×m\) …(ii)
5. On solving equations (i) and (ii) \(n=12\) and \(m=2 i.e.\) 2 rows of 12 cells are connected in parallel