NEET Physics Current Electricity MCQs Set 12

Answer: (d) \(1:4\)

Explanation:
1. \({v}_{d}=\frac{i}{nAe}=\frac{i×4}{nπ{D}^{2}e}ie, {v}_{d}∝\frac{1}{{D}^{2}}\)
2. \(∴ \frac{v{d}_{1}}{v{d}_{2}}=\frac{{D}_{2}^{2}}{{D}_{1}^{2}}={\left(\frac{1}{2}\right)}^{2}=\frac{1}{4}\)

Answer: (a) 1 : 4

Explanation:
1. Let the resistance of each heater wire is \(R.\) When two wires are connected in series, the heat developed is
2. \({H}_{1}=\frac{{V}^{2}t}{2R} …(i)\)
3. When two heater wires are connected in parallel, the heat developed is
4. \({H}_{2}=\frac{{V}^{2}t}{R/2}=\frac{2{V}^{2}t}{R} …\mathrm{(ii)}\)
5. Dividing Eq. (i) by Eq. (ii), we get
6. \(\frac{{H}_{1}}{{H}_{2}}=\frac{1}{4}\mathrm{or}{H}_{1} :{H}_{2}=1 :4\)

Answer: (b) The collisions of the conducting electrons with the lattice consisting of the ions of the metal increases

Explanation:
1. The resistance of a metal increases with increasing temperature this is because, with increase in temperature the ions of the conductor vibrate with greater amplitude and the collision between ions and electrons becomes more frequent.

Answer: (c) \(0.5 ohm\)

Explanation:
1. Short circuit current \({i}_{SC}=\frac{E}{r}⇒3=\frac{1.5}{r}⇒r=0.5Ω\)

Answer: (c) Manganin

Explanation:
1. Manganin or constantan are used for making the potentiometer wire

Answer: (a) \(4.5 V\)

Explanation:
1. \(i=\frac{E}{R+r}=\frac{5}{4.5+0.5}=1A\)
2. \(V=E-ir=5-1×0.5=4.5 \mathrm{Volt}\)

Answer: (c) 2 : 9

Explanation:
1. In series combination, the net resistance
2. \({R}_{s}=R+2R=3R\)
3. Heat produced in \({R}_{s},\)
4. \({H}_{s}=\frac{{V}^{2}}{{R}_{s}}=\frac{{V}^{2}}{3R} …(i)\)
5. I parallel combination, the net resistance
6. \({R}_{P}=\frac{R×2R}{R+2R}=\frac{2{R}^{2}}{3R}=\frac{2}{3} R\)
7. Heat produced in \({R}_{P},\)
8. \({H}_{P}=\frac{{V}^{2}}{{R}_{P}}=\frac{{V}^{2}}{2R/3}=\frac{3{V}^{2}}{2R} …(\mathrm{ii)}\)
9. Dividing Eq. (i) by Eq. (ii), we obtain
10. \(\frac{{H}_{s}}{{H}_{p}}=\frac{{V}^{2}/3R}{3{V}^{2}/2R}=\frac{2}{9}\)

Answer: (d) \(55 Ω\)

Explanation:
1. \(\frac{H}{t}=\frac{{V}^{2}}{R}⇒\frac{H}{t}∝\frac{1}{R}\)

Answer: (c) \(8l\)

Explanation:
1. In a meter bridge the ratio of two resistances is
2. \(\frac{R}{{R}^{'}}=\frac{l}{{l}^{'}}\)
3. Where \(l\) and \(l'\) are balancing lengths.
4. Resistance \(R=\frac{ρl}{A}=\frac{ρl}{π{r}^{2}}\)
5. In material remains same \(ρ={ρ}^{'}\)
6. Given, \({l}^{'}=2l\)
7. \({r}^{'}=\frac{r}{2}\)
8. \(∴ {R}^{'}=\frac{ρ{l}^{'}}{{A}^{'}}=\frac{ρ2l}{π{\left(\frac{r}{2}\right)}^{2}}=\frac{8ρl}{π{r}^{2}}\)
9. \({R}^{'}=8R\)
10. Therefore, the new balancing point is expected to be 8\(l\).

Answer: (a) \(8\mathrm{min}\,20 s\)

Explanation:
1. Heat gained by water = Heat supplied by container heat lost
2. \(⇒mS∆θ=1000t-160t\)
3. \(⇒t=\frac{2×4.2×1000×50}{840}=8\mathrm{min}\,20 sec\)

Answer: (b) \(2 a/3b\)

Explanation:
1. \(\frac{dT}{dt}=\frac{d}{dt}\left(a{t}^{2}-b{t}^{3}\right)=2at-3b{t}^{2}\)
2. When \(t={t}_{n}(ie,\)neutral temperature), \(\frac{dE}{dt}=0\)
3. \(∴0=2a{t}_{n}-3b{t}_{n}^{2}\mathrm{or}{t}_{n}=\frac{2a}{3b}\).

Answer: (a) 4.9\(Ω\)

Explanation:
1. When a constant current is passed through a wire of uniform cross-section, the potential difference across any portion of the wire is directly proportional to the length of that portion.
2. Potential gradient=\(\mathrm{0.2\ mVc}{m}^{-1}\)
3. ∴ Potential difference across potentiometer wire
4. =0.02 × 1= 0.02V
5. Total resistance \(=r+R=490+R\)
6. \(∴ \frac{{V}_{1}}{{V}_{2}}=\frac{{R}_{1}}{{R}_{2}}\)
7. \(⟹\frac{0.02}{2}=\frac{R}{490+R}\)
8. \(⟹R=4.9Ω\)

Answer: (c) \(48 Ω\)

Explanation:
1. \({i}_{g}=\frac{iS}{S+G}⇒10=\frac{50×12}{12+G}⇒12+G=60⇒G=48Ω\)

Answer: (a) 1 :2

Explanation:
1. Drift velocity, \({v}_{d}=\frac{eτV}{mL}\left[∵E=\frac{V}{L}\right]\)
2. Where the symbols have their usual meaning
3. If the temperature are not same, \(τ\) cannot be same. Then noneof the given options is correct
4. If temperatures are same, then \(\frac{{v}_{{d}_{1}}}{{v}_{{d}_{2}}}=\frac{{V}_{1}}{{V}_{2}}=\frac{1}{2}\)

Answer: (d) \(\frac{4{π}^{2}m{e}^{5}}{{h}^{3}}\)

Explanation:
1. \(i=\frac{e}{t}=\frac{e}{2πr/v}=\frac{ev}{2πr}\)
2. Here, \(v=\frac{{e}^{2}}{h}\mathrm{and} r={h}^{2}/m{e}^{2}\)
3. \(∴i=\frac{e(\frac{{e}^{2}}{h})}{2π({h}^{2}/m{e}^{2})}=\frac{{e}^{3}×m{e}^{2}}{2π{h}^{3}}=\frac{m{e}^{5}}{2π{h}^{3}}\)
4. \(i=\frac{4{π}^{2}m{e}^{5}}{{h}^{3}}\)

Answer: (d) \({10}^{29}\)

Explanation:
1. Density of copper,
2. \(ρ=9×{10}^{3}\mathrm{kg/}{m}^{3} =9×{10}^{6}g/{m}^{3}\)
3. Avogadro number, \({N}_{A}=6.023×{10}^{23}\)
4. Mass of 1 mole of copper atom, M=63.5g
5. Thus, number of free electrons per volume is
6. \(n=\frac{{N}_{A}}{M}ρ=\frac{6.023×{10}^{23}}{63.5}×9×{10}^{6}=8.5×{10}^{28}{m}^{-3}\)

Answer: (b) \(\frac{10}{9}Ω\)

Explanation:
1. Given, \(E=2.2V, R=5Ω\)
2. \(V=1.8V, r=?\)
3. Now current in the circuit
4. \(i=\frac{E}{R+r}\)
5. \(\mathrm{or}\frac{V}{R}=\frac{E}{R+r}\)
6. \(\frac{1.8}{5}=\frac{2.2}{5+r}\)
7. \(\mathrm{or} r=\frac{10}{9}Ω\)

Answer: (b) \(150 sec\)

Explanation:
1. \(W=JH⇒P×t=J×ms ∆θ\)
2. \(⇒t=\frac{J×m×s∆θ}{P} [\mathrm{For\ water} 1 litre=1 kg]\)
3. \(⇒t=\frac{4.2×1×1000×(40-10)}{836}=150 sec\)
4. Short Trick : use formula \(t=\frac{4200×m×∆θ}{P}\)

Answer: (d) 50 W, 200 W

Explanation:
1. Tow bulbs of 100 W are connected in series, the total power
2. \({P}^{'}=\frac{100×100}{100+100}=\frac{10000}{200}\)
3. \({P}^{'}=50 W\)
4. When two bulbs are connected in parallel the total power
5. \({P}^{''}=100+100=200 W\)

Answer: (d) \(2.4×{10}^{-8}Ωm\)

Explanation:
1. The resistance R of a particular conductor is related to the resistivity \(ρ\) of its material by
2. \(R=\frac{ρl}{A}\)
3. Or
4. \(ρ=\mathrm{resistivity} =\frac{RA}{l}\)
5. Given, R=0.072\(Ω\)
6. \(A=2mm×2mm=4×{10}^{-6}{m}^{2}, l\)=12
7. \(∴ ρ=\frac{0.072×4×{10}^{-6}}{12}\)
8. \(=2.4×{10}^{-8}Ωm\)

Answer: (b) Electrochemical equivalent of the substance

Explanation:
1. The amount of decomposition (\(ie\), mass of the substance liberated during electrolysis) is proportional to ECE of the substance.

Answer: (c) 15 kJ

Explanation:
1. \(R=\frac{230×230}{100}=529 Ω\)
2. \(∴H=\frac{{V}^{2}}{R}×t=\frac{115×115}{529}×10×60=15 \mathrm{kJ}\)

Answer: (b) \(1.42×{10}^{-3}\)V

Explanation:
1. \({{E}_{\mathrm{Ni}}}_{\mathrm{-Cu}}=at+b{t}^{2}\)
2. \(=\left(16.3×{10}^{-6}\right)\left(100\right)+\left(-0.021×{10}^{-6}\right)×{\left(100\right)}^{2}\)
3. \(=1.42×{10}^{-3}\)V

Answer: (b) 760\(Ω\)

Explanation:
1. Potential gradient along wire
2. \(=\frac{\mathrm{potential\ difference\ along\ wire}}{\mathrm{length\ of\ wire}}\)
3. \(∴0.1×{10}^{-3}=\frac{I×40}{1000}\mathrm{Vc}{m}^{-1}\)
4. Current in wire.
5. \(I=\frac{1}{400}A \mathrm{or} I=\frac{E}{R+{R}^{'}}\)
6. \(∴ \frac{2}{40+R}=\frac{I}{400}\)
7. Or \(R=800 - 40 =760Ω\)