NEET Physics Current Electricity MCQs Set 08

Answer: (c) 25W

Explanation:
1. Resistance of bulb \(R=\frac{{V}^{2}}{P}=\frac{220×220}{100}=484Ω\)
2. Power when the bulb of operated on a voltage
3. \({V}^{'}=110\mathrm{Vwill\ be}{P}^{'}=\frac{{V}^{'2}}{R}=\frac{110×110}{484}=25W\)

Answer: (c) \(8.6\%\)

Explanation:
1. \({P}_{consumed}={\left(\frac{{V}_{A}}{{V}_{R}}\right)}^{2}×{P}_{R}={\left(\frac{110}{115}\right)}^{2}×500=457.46 W\)
2. So, percentage drop in power output
3. \(=\frac{(500-457.46)}{500}×100=8.6\%\)

Answer: (a) \(11.11 ohm\)

Explanation:
1. \({i}_{g}=i\frac{S}{G+S}⇒10×{10}^{-3}=\frac{S}{100+S}×100×{10}^{-3}\)
2. \(90 S=1000⇒S=\frac{1000}{90}=11.11 Ω\)

Answer: (d) 8 :1

Explanation:
1. \({R}_{1}=ρ\frac{{l}_{1}}{{A}_{1}}\) and \({R}_{2}=ρ\frac{{l}_{2}}{{A}_{2}}⇒\frac{{R}_{1}}{{R}_{2}}=\frac{{l}_{1}}{{l}_{2}}.\frac{{A}_{2}}{{A}_{1}}=\frac{{l}_{1}}{{l}_{2}}{\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{2}\)
2. Given\(\frac{{l}_{1}}{{l}_{2}}=\frac{1}{2}\) and \(\frac{{r}_{1}}{{r}_{2}}=\frac{2}{1}\)or \(\frac{{r}_{2}}{{r}_{1}}=\frac{1}{2}⇒\frac{{R}_{1}}{{R}_{2}}=\frac{1}{8}\)
3. \(∴\)Ratio of heats \(\frac{{H}_{1}}{{H}_{2}}=\frac{{V}^{2}/{R}_{1}}{{V}^{2}/{R}_{2}}=\frac{{R}_{2}}{{R}_{1}}=\frac{8}{1}\)

Answer: (c) \(1.2 Ω\)

Explanation:
1. \(r=\frac{({l}_{1}-{l}_{2})}{{l}_{2}}×{R}^{'}=\left(\frac{60-50}{50}\right)×6=1.2 Ω\)

Answer: (b) 20

Explanation:
1. Heat produced by the heater \(H=\frac{{V}^{2}}{R} ×t\)
2. For 220V heater heat produced
3. \({H}_{1}=\frac{{\left(220\right)}^{2}}{R} ×5\)
4. For 110V heater heat produced
5. \({H}_{2}=\frac{{\left(110\right)}^{2}}{R} ×L\)
6. Now, \({H}_{1}={H}_{2}\)
7. \(\frac{110×110}{R} t=\frac{220×220×5}{R}\)
8. \(t=20 \mathrm{min}\)

Answer: (c) Same in both

Explanation:
1. Because in series current is same

Answer: (c) \(\frac{{ρ}_{1}+{ρ}_{2}}{2}\)

Explanation:
1. \(R={R}_{1}+{R}_{2}=\frac{{ρ}_{1}l}{A}+\frac{{ρ}_{2}l}{A}=\left({ρ}_{1}+{ρ}_{2}\right)\frac{l}{A}\)…(i)
2. \(R=\frac{ρ(2l)}{A}\)…(ii)
3. From Eqs.(i) and (ii), \(2ρ={ρ}_{1}+{ρ}_{2}\mathrm{or\ ρ}=\frac{{ρ}_{1}+{ρ}_{2}}{2}\)

Answer: (b) \({10}^{-2}cm/sec\)

Explanation:
1. Order of drift velocity \(={10}^{-4}m/sec={10}^{-2}cm/sec\)

Answer: (c) \(2×{10}^{-4}m/s\)

Explanation:
1. By using \({v}_{d}=\frac{i}{neA}=\frac{100}{{10}^{28}×1.6×{10}^{-19}×\frac{π}{4}×{\left(0.02\right)}^{2}}\)
2. \(=2×{10}^{-4} m/sec\)

Answer: (b) 99\(Ω\)

Explanation:
1. Shunt is connected to the galvanometer
2. \({i}_{g}=\frac{iS}{S+G}\)
3. \(1=\frac{100×1}{\left(1+G\right)}\)
4. \(⟹ G=99Ω\)

Answer: (a) Is a constant

Explanation:
1. Neutral temperature is defined as temperature of a hot junction of a thermocouple at which the electromotive force of the thermocouple attains its maximum value when cold junction is maintained at a constant temperature of \(0℃.\) Hence, for a given thermocouple neutral temperature is a constant.

Answer: (a) Zero charge

Explanation:
1. As steady current is flowing through the conductor, hence the number of electrons entering from one end and outgoing from the other end of any segment is equal. Hence charge will be zero

Answer: (b) About 2.4

Explanation:
1. \(\frac{{r}_{\mathrm{iron}}}{{r}_{\mathrm{Copper}}}=\sqrt{\frac{{ρ}_{\mathrm{iron}}}{{ρ}_{\mathrm{copper}}}}=\sqrt{\frac{1×{10}^{-7}}{1.7×{10}^{-8}}}=2.4\)

Answer: (b) 1.5 V

Explanation:
1. Here , \(m=1 g={p}^{-3}\mathrm{kg};\)
2. \(z=1.044×{10}^{-8}\mathrm{kg}{C}^{-1}\)
3. \(H=34 \mathrm{k\ cal.=34×1000×4.2\ J}\)
4. \(q=\frac{m}{z}=\frac{{10}^{-3}}{1.044×{10}^{-8}}=\frac{{10}^{5}}{1.044}\) C
5. and\(V=\frac{H}{q}=\frac{34×1000×4.2}{({10}^{5}/1.044)}=1.5\) V.

Answer: (b) 60 W bulb

Explanation:
1. Resistance of 40 W bulb \(=\frac{240 ×240}{40}\)
2. \(=1440 Ω\)
3. It’s safe current \(=\frac{240}{1440}=0.167 A\)
4. Resistance of 60 W bulb \(=\frac{240 ×240}{60}\)
5. \(=960 Ω\)
6. It’s safe current \(=\frac{240}{960}=0.25 A\)
7. When connected in series to 420 V supply, then the current
8. \(I=\frac{420}{1440+960}=\frac{420}{2400}\)
9. \(=0.175 A\)
10. Thus, current is greater for 40 W bulb, so it will fuse.

Answer: (d) Flow from Antimony to Bismuth at the cold junction

Explanation:
1. When a circuit is made up on any two metals in thermoelectric series, the current flows across the cold junction from the later occurring metal in the series to the one occurring earlier. In thermoelectric series Bismuth comes earlier than Antimony. So, at cold junction current flows from Antimony to Bismuth and at hot junction it flows from bismuth to Antimony.

Answer: (d) 0.1835 g

Explanation:
1. \(I=\frac{P}{V}=\frac{{10}^{5}}{200}=500\) A
2. and \(W=zlt=0.367×{10}^{-3}×500×1=0.1835\) g

Answer: (d) \(2450 Ω\) in series

Explanation:
1. Full deflection current \({i}_{g}=25×4×{10}^{-4}=100×{10}^{-4}A\)
2. Using \(R=\frac{V}{{I}_{g}}-G=\frac{25}{100×{10}^{-4}}-50=2450Ω\) in series

Answer: (c) \(560Ω±10\%\)

Explanation:
1. \(R=56×10±10\%=560+10\%\)

Answer: (a) Same

Explanation:
1. In the first case, \(Zit=m\)
2. In the second case, \(Z×\frac{i}{4}×4t=m\)

Answer: (d) \(50 cm\)

Explanation:
1. \(S=\left(\frac{100-l}{l}\right).R\)
2. Initially, \(30=\left(\frac{100-l}{l}\right)×10⇒l=25cm\)
3. Finally, \(10=\left(\frac{100-l}{l}\right)×30⇒l=75cm\)
4. So, shift \(=50cm\)

Answer: (d) 2 : 1

Explanation:
1. The resistance are connected in parallel hence, voltage will remain constant
2. \(E=\frac{{V}^{2}t}{R}\)
3. \(⟹E∝\frac{1}{R}\)
4. \(\frac{{E}_{1}}{{E}_{2}}=\frac{{R}_{2}}{{R}_{1}}=\frac{2R}{R}=\frac{2}{1}\)

Answer: (d) Potentiometer does not take current from the circuit

Explanation:
1. Potentiometer works on null deflection method. In balance condition no current flows in secondary circuit.

Answer: (c) The electrons move in random direction with a speed of the order close to that of velocity of light

Explanation:
1. At room temperature, the free electrons in a conductor move randomly with speed of the order of \({10}^{5}m{s}^{-1}\). Since, the motion of the electrons is random there is no net charge flow in any direction.