NEET Physics Current Electricity MCQs Set 21

Answer: (d) \(180.0 ohm\)

Explanation:
1. In case of stretching of wire \(R∝{l}^{2}\)
2. \(⇒\) If length becomes 3 times so Resistance becomes 9 times \(i.e. {R}^{'}=9×20=180Ω\)

Answer: (b) 727K

Explanation:
1. \({R}_{2}={R}_{o}\left(1+α{t}_{2}\right)\mathrm{and}{R}_{{t}_{1}}={R}_{o}(1+α{t}_{1})\)
2. \(∴\frac{{R}_{{t}_{2}}}{{R}_{{t}_{1}}}=\frac{1+α{t}_{2}}{1+α{t}_{1}}\)
3. or \(\frac{1.5}{1}=\frac{1+0.00125×{t}_{2}}{1+0.00125×27}\)
4. On solving we get;\({t}_{2}=454°C=454+273=727K\)

Answer: (a) 1\(Ω\)

Explanation:
1. Internal resistance of the cell
2. \(r=\left[\frac{E}{V}-1\right]R\)
3. \(r=\left[\frac{1.5}{1.4}-1\right]14=1Ω\)

Answer: (a) \(7.5 Ω\)

Explanation:
1. For one wire cable,
2. Resistance, \({R}^{'}=ρl/π{\left(9×{10}^{-3}\right)}^{2}=5Ω\)
3. For other wire of cable,
4. Resistance, \({R}^{'}=ρl/π{\left(3×{10}^{-3}\right)}^{2}\)
5. \(={9}^{2}×5/{3}^{2}=45Ω\)
6. When six wires each of resistance \(R’\) are connected in parallel, their effective resistance will be
7. \({R}_{p}=\frac{R'}{6}=\frac{45}{6}=7.5Ω\)

Answer: (c) \(57 Ω\)

Explanation:
1. Potential gradient \(x=\frac{e}{(R+{R}_{h}+r)}.\frac{R}{L}\)
2. \(⇒\frac{{10}^{-3}}{{10}^{-2}}=\frac{2}{(3+{R}_{h}+0)}×\frac{3}{1}⇒{R}_{h}=57Ω\)

Answer: (b) \(19989 Ω\)

Explanation:
1. \(V=i.R.=\frac{e}{(R+{R}_{h}+r)}.R⇒{10}^{-3}=\frac{2}{(10+R+r)}×10\)
2. \(⇒R=19,989Ω\)

Answer: (d) 386000 C

Explanation:
1. \({V}_{2}=\frac{{P}_{1}{V}_{1}}{{P}_{2}}=\frac{22.4×2}{1}=44.8\) L
2. 11.2 L of \({H}_{2}\) is liberated by charge \(=96500\) C
3. 44.8 L of\({H}_{2}\) is liberated by charge
4. \(=\frac{96500}{11.2}×44.8=386000\)C

Answer: (a) \(36×{10}^{5} joules\)

Explanation:
1. \(1 kWh=1000 W×3600 sec=36×{10}^{5}W\)-\(\mathrm{sec}\,(\mathrm{or}J)\)

Answer: (d) \(96500 coulomb\)

Explanation:
1. Equivalent weight of aluminium \(=\frac{27}{3}=9\)
2. So \(1 faraday=96500 C\) are required to liberate \(9 gm\) of \(Al\)

Answer: (c) \(15 amp\)

Explanation:
1. Current capacity of a fuse wire should be slightly greater than the total rated load current

Answer: (b) \({R}_{A}<{R}_{B}\)

Explanation:
1. In parallel\({P}_{consumed}∝\) Brightness \(∝\frac{1}{R}\)
2. \({P}_{A}>{P}_{B}\left[\mathrm{Given}\right] ∴{R}_{A}<{R}_{B}\)

Answer: (a) 25 W bulb

Explanation:
1. Resistance of 25 W bulb \(=\frac{220 ×220}{25}=1936 Ω\)
2. It’s safe current \(=\frac{220}{1936}=0.11 A\)
3. Resistance of 100 W bulb \(=\frac{220 × 220}{100}=484 Ω\)
4. It’s safe current \(=\frac{220}{484}=0.48 A\)
5. when connected in series to 440 V supply, then the current
6. \(i=\frac{440}{\left(1936 +484\right)}=0.18 A\)
7. Thus, current is greater for 25 W bulb, so it will fuse.

Answer: (b) \(3/11 amp\)

Explanation:
1. \(P=Vi⇒i=\frac{P}{V}=\frac{60}{220}=\frac{3}{11}amp\)

Answer: (b) The electrical signal is carried by electromagnetic waves moving with the speed of light

Explanation:
1. It is the electric field that is set up which moves with the velocity of light in that medium

Answer: (b) 112 cc

Explanation:
1. \(\frac{{{m}_{H}}_{2}}{{M}_{\mathrm{Cu}}}=\frac{{{W}_{H}}_{2}}{{W}_{\mathrm{Cu}}}=\frac{1}{\left(\frac{63.5}{2}\right)}\)
2. \(⇒{{m}_{H}}_{2}=\frac{2}{63.5}×0.3175=0.01\)g
3. 2g of\({H}_{2}\) occupies volume at NTP = 22.4 L
4. \(⇒0.\mathrm{01\ g\ of}{H}_{2}\) occupies volume at NTP given by:
5. \(V=\frac{22.4}{2}×0.01\)L
6. \(⇒\mathrm{V=11.2×0.01×1000\ cc=112\ cc}\)

Answer: (d) None of these

Explanation:
1. In series, effective resistance,
2. \({R}_{eff}={R}_{1}+{R}_{2}+{R}_{3}⇒\frac{1}{{σ}_{eff}}=\frac{1}{{σ}_{1}}+\frac{1}{{σ}_{2}}+\frac{1}{{σ}_{3}}\)
3. \(=\frac{{σ}_{2}{σ}_{3}+{σ}_{1}{σ}_{3}+{σ}_{1}{σ}_{2}}{{σ}_{1}{σ}_{2}{σ}_{3}}\)
4. \(∴{σ}_{eff}=\frac{{σ}_{1}{σ}_{2}{σ}_{3}}{{σ}_{2}{σ}_{3}+{σ}_{1}{σ}_{3}+{σ}_{1}{σ}_{2}}\)

Answer: (b) The potentiometer does not draw any current from the source of the potential

Explanation:
1. Potentiometer is based on null deflection

Answer: (a) A

Explanation:
1. Current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer when a unit current flows through it.
2. If \(θ\) is the deflection in the galvanometer when current \(I\) is passed through it, then
3. \(\mathrm{Current\ Sensitivity}{I}_{s}=\frac{θ}{I}=\frac{nBA}{k}\)
4. where\(k\) be restoring torque per unit twist, \(n\) be number of turns in the coil, B is strength of magnetic field in which coil is suspended, Abe area of coil.
5. Since, restoring per unit twist (torsional constant) is minimum for galvanometer A, hence more sensitive.

Answer: (c) \(3.35 hrs\)

Explanation:
1. \(m=Zi⇒t=\frac{m}{Zi}=\frac{m×F}{E×i}\left[∵Z=\frac{E}{F}\right]\)
2. \(t=\frac{27×96500}{108×2}=12062.5\mathrm{sec}\,=\frac{12062.5}{3600}hr=3.35 hr\)

Answer: (a) Increase by 0.2%

Explanation:
1. \(R=\frac{Sl}{A}=\frac{S{l}^{2}}{V}\)
2. \(∴ \frac{∆R}{R}=2\frac{∆l}{l}=+2.0\%\)

Answer: (d) 75%

Explanation:
1. Resistance of bulb,
2. \(R=\frac{{V}^{2}}{P}=\frac{220×220}{300}=\frac{484}{3}Ω\)
3. New power, \({P}^{'}=\frac{{\left(110\right)}^{2}}{R}=\frac{110×110}{484/3}=75\) watt reduction
4. Of power \(=\frac{300-75}{300}×100=75\%\)

Answer: (b) \(24 C\)

Explanation:
1. \(dQ=Idt⇒Q=\int_{t=2}^{t=3} Idt=\left[2\int_{2}^{3} tdt+3\int_{2}^{3} {t}^{2}dt\right]\)
2. \(={\left[{t}^{2}\right]}_{2}^{3}+{\left[{t}^{3}\right]}_{2}^{3}=\left(9-4\right)+\left(27-8\right)=5+19=24C\)

Answer: (c) \(2Ω\) resistance in series with the parallel combination of \(3Ω\) and \(6Ω\)

Explanation:
1. These questions are done by hit and trial method only. You check all the options one by one till you get the final desired result

Answer: (c) 2.87

Explanation:
1. In closed loop ABGFEHA
2. \(10-{i}_{2}×1+{i}_{1}×0.5-6=0\)
3. \(0.5{i}_{1}-{i}_{2}=-4\)
4. In closed loop BCDEB
5. \(\left({i}_{1}+{i}_{2}\right)×12+{i}_{2}×1-10=0\)
6. \(12{i}_{i}+13{i}_{2}=10\)
7. From Eqs. (i) and (ii)
8. \({i}_{2}=2.87A\)

Answer: (c) \(16R\)

Explanation:
1. Resistance of wire given by
2. \(R=ρ\frac{l}{A}\)
3. Also, volume (V)=Length\(\left(l\right)\)×Area \((A)\)
4. Where, \(A=π{r}^{2}\) (r is radius)
5. When the wire is stretched its volume (V) remains constant.
6. Hence,
7. \(R=\frac{ρV}{{π}^{2}{r}^{4}} ….\left(i\right)\)
8. When radius is halved
9. \({R}^{'}=\frac{pV}{π{r}^{2}{\left(\frac{r}{2}\right)}^{4}} ……………\left(ii\right)\)
10. \(∴\frac{{R}^{'}}{R}=\frac{16ρV}{{π}^{2}{r}^{4}}×\frac{{π}^{2}{r}^{4}}{ρV}=16\)
11. \(⟹{R}^{'}=16R\)
12. Hence, new resistance increases to sixteen times its original value.