Download NEET MCQs for NEET Physics: Current Electricity
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Chapter-wise Objective Questions: Current Electricity
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Practise 25 chapter-wise NEET Physics questions on Current Electricity in this set, each with a worked solution you can open after attempting it. The questions test Ohm's law, resistivity, combinations of resistors, Kirchhoff's laws, cells, the potentiometer and the Wheatstone bridge.
(a) \({10}^{5} ohm\)
(b) \(150 k.ohm\)
(c) \(900 k.ohm\)
(d) \(9×{10}^{6}ohm\)
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Answer: (a) \({10}^{5} ohm\)
Explanation:
1. \(R=G\left(n-1\right)=50×{10}^{3}\left(3-1\right)={10}^{5}Ω\)
(a) \(0.01 R\)
(b) \(0.1 R\)
(c) \(10 R\)
(d) \(100 R\)
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Answer: (a) \(0.01 R\)
Explanation:
1. Each part will have a resistance \(r=R/10\)
2. Let equivalent resistance be \({r}_{R},\) then
3. \(\frac{1}{{r}_{R}}=\frac{1}{r}+\frac{1}{r}+\frac{1}{r}…….\) 10 times
4. \(∴\frac{1}{{r}_{R}}=\frac{10}{r}=\frac{10}{R/10}=\frac{100}{R}⇒{r}_{R}=\frac{R}{100}=0.01R\)
(a) \(2 R\)
(b) \(0.25 R\)
(c) \(4 R\)
(d) \(0.5 R\)
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Answer: (b) \(0.25 R\)
Explanation:
1. \(R∝\frac{1}{A}⇒R∝\frac{1}{{r}^{2}}∝\frac{1}{{d}^{2}}\) [\(d=\)diameter of wire]
resistance of wire is \(1.5×{10}^{-3}{K}^{-1}\) , the temperature at which the resistance becomes 0.6\(Ω\) is
(a) 720 K
(b) 345 K
(c) 993 K
(d) 690 K
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Answer: (c) 993 K
Explanation:
1. Given, \({R}_{300}=0.3Ω, {R}_{t}=0.6Ω\)
2. \(T=300K=27°C\)
3. Temperature coefficient of resistance,
4. \(α=1.5×{10}^{-3}{K}^{-1}\)
5. \(∴{R}_{300}={R}_{0}\left(1+α×27\right)\)
6. \(0.3={R}_{0}\left(1+1.5×{10}^{-3}×27\right) …\left(i\right)\)
7. Again, \({R}_{t}={R}_{0}\left(1+αt\right)\)
8. \(0.6={R}_{0}\left(1+×{10}^{-3}×t\right) …\left(\mathrm{ii}\right)\)
9. Dividing Eq. (ii) by Eq. (i), we get
10. \(\frac{0.6}{0.3}=\frac{1+1.5×{10}^{-3}t}{1+1.5×{10}^{-3}×27 }\)
11. \(⟹2\left(1+1.5×{10}^{-3}×27\right)=1+1.5×{10}^{-3}t\)
12. \(⟹2+81×{10}^{-3}=1+1.5×{10}^{-3}t\)
13. \(⟹2+0.081=1+1.5×{10}^{-3}t\)
14. \(⟹t=\frac{1.081}{1.5×{10}^{-3}}=720°C=993K\)
(a) \({T}^{2}\frac{dE}{d{T}^{2}}\)
(b) \(T\frac{dE}{dT}\)
(c) \({T}^{3}\frac{d{E}^{2}}{dT}\)
(d) \({T}^{4}\frac{{d}^{2}E}{d{T}^{2}}\)
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Answer: (b) \(T\frac{dE}{dT}\)
Explanation:
1. \(π=T dE/dT\).
(a) Ohm’s law is applicable to all conductors of electricity
(b) In an electrolyte solution, the electric current is mainly due to the movement of electrons
(c) The resistance of an incandescent lamp is lesser when the lamp is switched on
(d) Specific resistance of a wire depends upon its dimension
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Answer: (c) The resistance of an incandescent lamp is lesser when the lamp is switched on
Explanation:
1. The temperature coefficient of the carbon is negative so the resistance of carbon decreases with the increase of temperature.
(a) \(nR\)
(b) \(\frac{R}{n}\)
(c) \(\frac{n}{R}\)
(d) \(\frac{R}{{n}^{2}}\)
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Answer: (d) \(\frac{R}{{n}^{2}}\)
Explanation:
1. Resistance of each part will be \(\frac{R}{n};\) such \(n\) parts are joined in parallel so \({R}_{eq}=\frac{R}{{n}^{2}}\)
(a) Becomes two times
(b) Becomes half
(c) Does not change
(d) Becomes one-fourth
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Answer: (c) Does not change
Explanation:
1. \({v}_{d}=\frac{I}{nAl}=\frac{1}{nAe}×\frac{V}{R}\)
2. \(=\frac{1}{nAe}×\frac{V}{(ρl/A)}=\frac{V}{neρl}\)
3. As \({v}_{d}\) is independent of area of cross-section hence drift velocity will not change, when diameter is doubled
(a) Is liberated at anode
(b) Is liberated at cathode
(c) Is not liberated
(d) Remains in the solution
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Answer: (b) Is liberated at cathode
Explanation:
1. Because \(H\) has positive charge
(a) \(260℃\)
(b) \(270℃\)
(c) \(265℃\)
(d) \(520℃\)
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Answer: (b) \(270℃\)
Explanation:
1. \({T}_{n}=\frac{{T}_{i}+{T}_{c}}{2}=\frac{10+530}{2}=270℃\)
(a) 388 s
(b) 776 s
(c) 400 s
(d) 800 s
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Answer: (b) 776 s
Explanation:
1. Mass of copper deposited,
2. \(m=\mathrm{volume×density}\)
3. \(=(\mathrm{area×thickness)×density}\)
4. \(=[2×(12×3)×0.002]×8.9\) g
5. \(t=\frac{m}{zI}=\frac{[2(12×3)×0.002×8.91]}{33×{10}^{-5}×5}=776\)s.
(a) \(\mathrm{sin}\,40°\)
(b) \(\mathrm{cos}\,40°\)
(c) \(\mathrm{tan}\,40°\)
(d) \(\mathrm{cot}\,40°\)
Show Answer & Explanation
Answer: (d) \(\mathrm{cot}\,40°\)
Explanation:
1. Resistance, \(R=\frac{V}{i}=\mathrm{cot}\,40°\)
(a) \(ρ\)’>\(ρ\)
(b) \(ρ\)’<\(ρ\)
(c) \({ρ}^{'}= ρ\)
(d) There is no simple relation between\(ρ\) and \({ρ}^{'}\)
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Answer: (a) \(ρ\)’>\(ρ\)
Explanation:
1. The resistivity of metal increases when it is converted into an alloy
2. \(∴ {ρ}^{'}>ρ\)
\({10}^{9}Ω\)-cm and area of cross-section \({10}^{-2}c{m}^{2}\). The potential gradient is
(a) \({10}^{9}V{m}^{-1}\)
(b) \({10}^{11}V{m}^{-1}\)
(c) \({10}^{10}V{m}^{-1}\)
(d) \({10}^{8}V{m}^{-1}\)
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Answer: (d) \({10}^{8}V{m}^{-1}\)
Explanation:
1. Potential gradient is given by
2. \(k=\frac{V}{l}=\frac{IR}{l}\left(∵V=I R\right)\)
3. \(=\frac{I×ρl/A}{l}\left(∵R=\frac{ρl}{A}\right)\)
4. \(=\frac{Iρ}{A}\)
5. \(∴ k=\frac{0.01×{10}^{-3}×{10}^{9}×{10}^{-2}}{{10}^{-2}×{10}^{-4}}={10}^{8}V{m}^{-1}\)
(a) \(60 minutes\)
(b) \(30 minutes\)
(c) \(15 minutes\)
(d) \(7.5 minutes\)
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Answer: (a) \(60 minutes\)
Explanation:
1. \({H}_{1}={H}_{2}⇒\frac{{v}^{2}}{R}{t}_{1}=\frac{{v}^{2}}{2R}{t}_{2}⇒{t}_{2}=2{t}_{1}⇒{t}_{1}=30 \mathrm{min}\)
2. \(∴{t}_{2}=60 \mathrm{min}\)
(a) \(4 Ω\)
(b) \(6 Ω\)
(c) \(8 Ω\)
(d) \(3 Ω\)
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Answer: (b) \(6 Ω\)
Explanation:
1. The bridge will be balanced when the shunted resistance of value \(2Ω ie, 2=\frac{3×S}{3+S}\). On solving \(S=6 Ω\)
(a) \(750 J\)
(b) \(650 J\)
(c) \(350 J\)
(d) \(250 J\)
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Answer: (a) \(750 J\)
Explanation:
1. \(m=Zi t ⇒i=\frac{m}{Zt}=\frac{0.99}{0.00033×1200}=2.5A\)
2. Hence heat generated in the coil is
3. \(H={i}^{2}Rt={\left(2.5\right)}^{2}×0.1×1200=750 J\)
(a) \({H}_{2}O\)
(b) \({H}_{2}S{O}_{4}\)
(c) \({H}_{2}O \mathrm{and}{H}_{2}S{O}_{4}\mathrm{both}\)
(d) \({H}_{2}\)and \({O}_{2}\)
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Answer: (a) \({H}_{2}O\)
Explanation:
1. Because \({H}_{2}O\) is used as electrolyte
(a) \(15 minutes\)
(b) \(12 minutes\)
(c) \(10 minutes\)
(d) \(8 minutes\)
Show Answer & Explanation
Answer: (c) \(10 minutes\)
Explanation:
1. \(H=\frac{{V}^{2}}{R}t\)
2. Since supply voltage is same and equal amount of heat is produced, therefore
3. \(\frac{{R}_{1}}{{t}_{1}}=\frac{{R}_{2}}{{t}_{2}}\mathrm{or}\frac{{R}_{1}}{{R}_{2}}=\frac{{t}_{1}}{{t}_{2}} …(i)\)
4. But\(R∝l⇒\frac{{R}_{1}}{{R}_{2}}=\frac{{l}_{1}}{{l}_{2}}\)…(ii)
5. By (i) and (ii)\(, \frac{{l}_{1}}{{l}_{2}}=\frac{{t}_{1}}{{t}_{2}}\)…(iii)
6. Now\({l}_{2}=\frac{2}{3}{l}_{1}⇒\frac{{l}_{1}}{{l}_{2}}=\frac{3}{2}\)
7. \(∴\) By equation\(\left(\mathrm{iii}\right),\frac{3}{2}=\frac{15}{{t}_{2}}⇒{t}_{2}=10 minutes\)
(a) 8
(b) 4
(c) 2
(d) None of these
Show Answer & Explanation
Answer: (b) 4
Explanation:
1. Let the length of various edges in increase order be\(l,x,2l\) respectively
2. \({R}_{\mathrm{max}}=\frac{ρ2l}{xl}=\frac{2ρ}{x};{R}_{\mathrm{min}}=\frac{ρl}{2lx}=\frac{ρ}{2x}\)
3. \(∴ \frac{{R}_{\mathrm{max}}}{{R}_{\mathrm{min}}}=4\)
(a) 0.42 J \({s}^{-1}\)
(b) 0.84 J \({s}^{-1}\)
(c) 1 J \({s}^{-1}\)
(d) 5 J \({s}^{-1}\)
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Answer: (c) 1 J \({s}^{-1}\)
Explanation:
1. Chemical energy consumed per sec = heat energy produced per sec.
2. \({I}^{2}\left(R+r\right)={\left(0.2\right)}^{2}\left(21+4\right)=1 J {s}^{-1}\)
(a) \(R=r/2\)
(b) \(R=r\)
(c) \(R=2r\)
(d) \(R=0\)
Show Answer & Explanation
Answer: (a) \(R=r/2\)
Explanation:
1. For maximum energy, we have
2. External resistance of the circuit
3. =Equivalent internal resistance of the circuit \(i.e. R=\frac{r}{2}\)
(a) Unchanged
(b) Halved
(c) Quadrupled
(d) 1/4 of the original value
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Answer: (b) Halved
Explanation:
1. Conductance \(C=\frac{1}{R}=\frac{A}{ρl}⇒C∝\frac{1}{l}\)
(a) \(4 Ω\)
(b) \(3 Ω\)
(c) \(\frac{12}{7} Ω\)
(d) \(\frac{7}{12} Ω\)
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Answer: (b) \(3 Ω\)
Explanation:
1. \(\frac{7}{12}=\frac{1}{4}+\frac{1}{R}⇒R=3Ω\)
(a) \(0.1 W\)
(b) \(1 W\)
(c) \(10 W\)
(d) \(100 W\)
Show Answer & Explanation
Answer: (b) \(1 W\)
Explanation:
1. \(P=Vi⇒i=\frac{2.2×{10}^{3}}{22000}=\frac{1}{10}A\)
2. \(\mathrm{Now\ loss\ of\ power=}{i}^{2}R={\left(\frac{1}{10}\right)}^{2}×100=1 W\)
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FAQs
You can get most exhaustive NEET Physics Current Electricity MCQs Set 05 for free on StudiesToday.com. These MCQs for NEET Physics are updated for the 2026-27 academic session as per NEET examination standards.
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Yes, Physics MCQs for NEET have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused NEET exams.
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