NEET Physics Current Electricity MCQs Set 16

Answer: (d) \(22Ω\)

Explanation:
1. \({R}_{2}={n}^{2}R={\left(2\right)}^{2}×5.5=22Ω\)

Answer: (a) Less, grater, equal

Explanation:
1. If E is the emf of the battery, r the internal resistance, i the current drawn and V the PD across the plates of battery, then
2. \(V=E-ir\)
3. Case I When the direction of current in the battery is from –ve to +ve or outside +ve to –ve, then
4. \(i=+\mathrm{ve}\)(during discharging)
5. V<E
6. Case II When the direction of current in the battery is form +ve to –ve or outside –ve to +ve, then
7. \(i=-\mathrm{ve}\) (during discharging)
8. \(V=E+ir\)
9. \(⟹\)V>E
10. Case III When zero current, then
11. \(i=o\)
12. \(V=E\)

Answer: (c) \(0.1 mm/sec\)

Explanation:
1. \({v}_{d}=\frac{i}{nAe}=\frac{1.344}{{10}^{-2}×1.6×{10}^{-19}×8.4×{10}^{22}}\)
2. \(=\frac{1.344}{10×1.6×8.4}=0.01cm/s=0.1mm/s\)

Answer: (b) \(\frac{1}{R}\)

Explanation:
1. At constant p.d., heat produced\(=\frac{{V}^{2}}{R}i.e. H∝\frac{1}{R}\)

Answer: (b) \(10 ohm\) in parallel

Explanation:
1. \(∵{i}_{g}=10\%\) of \(i=\frac{i}{10}⇒S=\frac{G}{(n-1)}=\frac{90}{(10-1)}=10Ω\)

Answer: (c) 25 W

Explanation:
1. Resistance of electric bulb \(R=\frac{{V}^{2}}{P}\) where subscripts denote for rated parameters.
2. \(R=\frac{{\left(220\right)}^{2}}{100}\)
3. Power consumed at 110 V, \({P}_{\mathrm{consumed}}=\frac{{V}^{2}}{R}\)
4. \(∴ {P}_{\mathrm{consumed}}=\frac{{\left(110\right)}^{2}}{{\left(220\right)}^{2}/100}=25 W\)

Answer: (d) \(20 m\)

Explanation:
1. \(R∝\frac{1}{{r}^{2}}⇒\frac{{R}_{1}}{{R}_{2}}=\frac{{l}_{1}}{{l}_{2}}×\frac{{r}_{2}^{2}}{{r}_{1}^{2}}⇒\frac{1}{1}=\frac{5}{{l}_{2}}×{\left(\frac{2}{1}\right)}^{2}⇒{l}_{2}=20m\)

Answer: (d) 1.20

Explanation:
1. \(V=E-ir=1.5-2×0.15=1.20 Volt\)

Answer: (c) \(807 Ω\)

Explanation:
1. \(R=\frac{{V}^{2}}{P}=\frac{{\left(220\right)}^{2}}{60}=807 Ω\)

Answer: (c) 0.24

Explanation:
1. Resistance of a bulb \(= \frac{{\left(\mathrm{Rated\ voltage}\right)}^{2}}{\mathrm{(Rated\ power)}}\)
2. \({R}_{{B}_{1}}=\frac{{\left(220\right)}^{2}}{40}\mathrm{Ωand}{R}_{{B}_{2}}=\frac{{\left(220\right)}^{2}}{60}Ω\)
3. When the bulbs are connected in series,
4. \({R}_{S}={R}_{{B}_{1}}+{R}_{{B}_{2}}\)
5. \(=\frac{{\left(220\right)}^{2}}{40}+\frac{{\left(220\right)}^{2}}{60}={\left(220\right)}^{2}\left[\frac{1}{40}+\frac{1}{60}\right]\)
6. \(={\left(220\right)}^{2}\left[\frac{60+40}{60×40}\right]={\left(220\right)}^{2}\left(\frac{100}{2400}\right)=\frac{{\left(220\right)}^{2}}{24}\)
7. \(∴{P}_{1}=\frac{{V}_{S}^{2}}{{R}_{s}}={\left(220\right)}^{2}×\frac{24}{{\left(220\right)}^{2}}=24W\)
8. When the bulbs are connected in parallel
9. \(\frac{1}{{R}_{p}}=\frac{1}{{R}_{{B}_{1}}}+\frac{1}{{R}_{{B}_{2}}}⇒\frac{1}{{R}_{P}}=\frac{40}{{\left(220\right)}^{2}}+\frac{60}{{\left(220\right)}^{2}}\)
10. \(\frac{1}{{R}_{p}}=\frac{100}{{\left(200\right)}^{2}}\mathrm{or}{R}_{p}=\frac{{\left(220\right)}^{2}}{100}\)
11. \(∴{P}_{2}=\frac{{V}_{S}^{2}}{{R}_{p}}={\left(220\right)}^{2}×\frac{100}{{\left(200\right)}^{2}}=100 W\)
12. \(∴\frac{{P}_{1}}{{P}_{2}}=\frac{24 W}{100 W}=0.24\)

Answer: (d) Depends upon the relative values of internal and external resistances

Explanation:
1. To draw maximum current from a combination of cells, external resistance \(R=\frac{nr}{m}.\)Therefore, grouping of cells depends upon the relative values of internal and external resistance.

Answer: (d) Shape of cross-section

Explanation:
1. Resistance of a conductor
2. \(R=\frac{ρl}{A} ………………..\left(i\right)\)
3. As \(ρ\) depends on the material,
4. soR depends on the material.
5. According to the given formula in Eq.(i), it depends on length. Moreover resistance \(∝\)temperature.
6. If \({R}_{0}\)= resistance of conductor at \(0° C,\)
7. \({R}_{t}\)= resistance of conductor at \(t°C\),
8. and \(a,β\) =temperature coefficient of resistance, then
9. \({R}_{t}={R}_{0}\left(1+αt+β{t}^{2}\right)\)
10. The resistance of a straight conductor does not depend on shape of cross-section.

Answer: (d) \(2.5 Ω\)

Explanation:
1. Case (i)\(E+E=(r+r+5)×1.0\)
2. or \(2E=2r+5\) …(i)
3. Case (ii) \(E=\left(\frac{r×r}{r+r}+5\right)×0.8\)
4. or \(E=\left(\frac{r}{2}+5\right)0.8 \mathrm{or} E=0.4r+4.0\) …(ii)
5. Multiplying Eq.(ii) by 2 and equating with Eq.(i), we get
6. \(2r+5=00.8r+8\)
7. or \(1.2=3 \mathrm{or} r=\frac{3}{1.2}=2.5Ω\)

Answer: (b) \(\frac{n{e}^{2}τ}{{m}_{e}}\)

Explanation:
1. Resistivity, \(ρ=\frac{{m}_{e}}{n{e}^{2}t}\)
2. Conductivity, \(σ=\frac{1}{p}=\frac{n{e}^{2}τ}{{m}_{e}}\)

Answer: (d) \(400℃\)

Explanation:
1. Comparing the given equation with standard equation
2. \(E=αt+\frac{1}{2}β{t}^{2}\)
3. \(α=40 \mathrm{and}\frac{1}{2}β=-\frac{1}{20}⇒β=-\frac{1}{10}\)
4. Hence neutral temperature \({t}_{n}=-\frac{α}{β}=\frac{-40}{-1/10}\)
5. \(⇒{t}_{n}=400℃\)
6. Alternate solution : We know at neutral temperature
7. \(\frac{dE}{dθ}=0\)
8. \(E=40θ-\frac{{θ}^{2}}{20}⇒40-\frac{2θ}{20}=0⇒θ=400℃\)

Answer: (a) \({P}_{1}+{P}_{2}watt\)

Explanation:
1. If resistances of bulbs are \({R}_{1}\) and \({R}_{2}\) respectively then in parallel
2. \(\frac{1}{{R}_{P}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}⇒\frac{1}{\left(\frac{{V}^{2}}{{P}_{p}}\right)}=\frac{1}{\left(\frac{{V}^{2}}{{P}_{1}}\right)}+\frac{1}{\left(\frac{{V}^{2}}{{P}_{2}}\right)}\)
3. \(⇒{P}_{P}={P}_{1}+{P}_{2}\)

Answer: (c) \(1/100 Ω\)

Explanation:
1. Lowest resistance will be in the case when all the resistors are connected in parallel
2. \(\frac{1}{R}=\frac{1}{0.1}+\frac{1}{0.1}….\)10 times
3. \(\frac{1}{R}=10+10……\)10 times
4. \(\frac{1}{R}=100 i.e. R=\frac{1}{100}Ω\)

Answer: (c) 120\(℃\)

Explanation:
1. Given, Thomson’s coefficient,\(σ=3μ\)V/\(℃\)
2. \(σ=\frac{dV}{dT}\)
3. \(\mathrm{or} d V=σdT\)
4. so \(V=σ({T}_{2}-{T}_{1})\)
5. or \(3×{10}^{-4}=3×{10}^{-6}({T}_{2}-20)\)
6. or \(100+20={T}_{2}\)
7. or \({ T}_{2}=120°\)C

Answer: (a) \({B}_{1}>{B}_{2}>{B}_{3}\)

Explanation:
1. Resistance of a bulb \(= \frac{{\left(Rated voltage\right)}^{2}}{Rated power}\)
2. For a given voltage\(, R∝\frac{1}{P}\)
3. \(∴{R}_{40}>{R}_{60}>{R}_{100}\)
4. Rate of heat produced, \(H={I}^{2}R\)
5. When the bulbs are connected in series, the current flowing through each bulb is same
6. \(∴H∝R .\) As\({R}_{40}>{R}_{60}>{R}_{100}\)
7. \(∴{H}_{40}>{H}_{60}>{H}_{100}⇒{B}_{1}>{B}_{2}>{B}_{3}\)

Answer: (d) \(5×{10}^{-3}\)

Explanation:
1. Drift velocity=mobility× intensity of electric field
2. or \({v}_{d}=μE \mathrm{or} μ=\frac{{v}_{d}}{E}\)
3. \(\mathrm{or} μ=\frac{{v}_{d}}{V/l}\left(∵E=\frac{V}{l}\right)\)
4. \(∴μ=\frac{0.5×2}{220}=4.5×{10}^{-3}\)
5. \(≈5×{10}^{-3}{m}^{2}{V}^{-1}{s}^{-1}\)

Answer: (a) \(544Ω\)

Explanation:
1. \(R=ρ\frac{l}{A}\) and mass \(m=\mathrm{volume} (V)×\mathrm{density}(d)=(A l)d\)
2. Since wires have same material so \(ρ\) and \(d\) is same for both
3. Also they have same mass \(⇒Al=\mathrm{constant}⇒l∝\frac{1}{A}\)
4. \(⇒\frac{{R}_{1}}{{R}_{2}}=\frac{{l}_{1}}{{l}_{2}}×\frac{{A}_{2}}{{A}_{1}}={\left(\frac{{A}_{2}}{{A}_{1}}\right)}^{2}={\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{4}\)
5. \(⇒\frac{34}{{R}_{2}}={\left(\frac{r}{2r}\right)}^{4}⇒{R}_{2}=544 Ω\)

Answer: (a) \(16℃\)

Explanation:
1. Thermo-emf of thermocouple \(=25\frac{μV}{℃}.\)
2. Let \(θ\) be the smallest temperature difference.
3. Therefore, after connecting the thermocouple with the galvanometer, thermo-emf
4. \(E=\left(25\frac{μV}{℃}\right)×θ(℃)\)
5. \(=25θ×{10}^{-6} V\)
6. Potential drop developed across the galvanometer
7. \(=iR={10}^{-5}×40=4×{10}^{-4}V\)
8. \(∴4×{10}^{-4}=25θ×{10}^{-6}\)
9. \(∴θ=\frac{4}{25}×{10}^{2}=16℃\)

Answer: (a) \(\frac{Iρ}{A}\)

Explanation:
1. Potential gradient \(x=\frac{V}{L}=\frac{iR}{L}=\frac{i\left(\frac{ρL}{A}\right)}{L}=\frac{iρ}{A}\)

Answer: (b) 2\(Ω\)

Explanation:
1. In series, current,
2. \({i}_{1}=\frac{2E}{2+2r}\)
3. In parallel, current,
4. \({i}_{2}=\frac{E}{2+\frac{r}{2}}=\frac{2E}{4+r}\)
5. According to the question
6. \({i}_{1}={i}_{2}\)
7. \(⟹ \frac{2E}{4+r}=\frac{2E}{2+2r}\)
8. \(⟹ r=2Ω\)

Answer: (c) High resistance in series

Explanation:
1. A voltmeter is a high resistance device and is always connected in parallel with the circuit. While an ammeter is a low resistance device and is always connected is series with the circuit. So, to use voltmeter in place of ammeter a high resistance must be connected in series with the circuit.