NEET Physics Current Electricity MCQs Set 06

Practice MCQs for NEET Physics Current Electricity

Explore reliable objective questions for Current Electricity tailored for NEET learners. Utilizing these Physics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

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These 25 NEET Physics multiple choice questions on Current Electricity come with answers and explanations to help you check your method. Topics include Ohm's law, resistivity, combinations of resistors, Kirchhoff's laws, cells, the potentiometer and the Wheatstone bridge.

Question 1: A\({4}_{μ}\) F conductor is charged to 400 V and then its plates are joined through a resistance of 1 k\(Ω\). The heat produced in the resistance is
(a) 0.18 J
(b) 0.21 J
(c) 0.25 J
(d) 0.32 J
Show Answer & Explanation

Answer: (d) 0.32 J

Explanation:
1. The energy stored in the capacitor \(=\frac{1}{2}C{V}^{2};\)
2. This energy will be converted into heat in the resistor.
3. \(H=\frac{1}{2}×4×{10}^{-6}×400×400\)
4. \(=32×{10}^{-2}\)
5. \(=0.32\) J

Question 2: What determines the emf between the two metals placed in an electrolyte?
(a) Relative position of metals in the electro chemical series
(b) Distance between them
(c) Strength of electrolyte
(d) Nature of electrolyte
Show Answer & Explanation

Answer: (a) Relative position of metals in the electro chemical series

Explanation:
1. The relative position of metals in the electro chemical series determines the emf between the two metals placed in an electrolyte.

Question 3: A tap supplies water at 22\(℃\). A man takes 1 L of water per min at 37\(℃\) from the geyser. The power of the geyser is
(a) 525 W
(b) 1050 W
(c) 1575 W
(d) 2100 W
Show Answer & Explanation

Answer: (b) 1050 W

Explanation:
1. \(Pt=mSθ\)
2. \(P=\frac{1×4200×15}{60}\)W\(=1050\) W

Question 4: The reciprocal of resistance is
(a) Conductance
(b) Resistivity
(c) Voltage
(d) None of the above
Show Answer & Explanation

Answer: (a) Conductance

Explanation:
1. The reciprocal of resistance is called conductance

Question 5: It is easier to start a car engine on a hot day than on a cold day. This is because the internal resistance of the car battery
(a) Decreases with rise in temperature
(b) Increases with rise in temperature
(c) Decreases with a fall in temperature
(d) Does not change with a change in temperature
Show Answer & Explanation

Answer: (a) Decreases with rise in temperature

Explanation:
1. Internal resistance \(∝\frac{1}{\mathrm{Temperature}}\)

Question 6: Silver and copper voltameters are connected in parallel with a battery of emf 12 V. In 30 min 1 g of silver and 1.8 g of copper are liberated. The energy supplied by the battery is
(a) 720 J
(b) 2.41 J
(c) 24.12 J
(d) \(4.34×{10}^{4}\) J
Show Answer & Explanation

Answer: (d) \(4.34×{10}^{4}\) J

Explanation:
1. Current in silver voltmeter.
2. \({i}_{1}=\frac{{m}_{1}}{{z}_{1}{t}_{1}}=\frac{1}{\left(11.2×{10}^{-4}\right)×(30×60)}=0.5\) A
3. Current in copper voltmeter \({i}_{2}=\frac{{m}_{2}}{{z}_{2}{t}_{2}}\)
4. \({i}_{2}=\frac{1.8}{\left(6.6×{10}^{-4}\right)×(30×60)}=1.51\) A
5. So, total current given by battery \({i}_{1}+{i}_{2}=2.01\) A
6. Energy supplied by battery\(=Eit\)
7. \(W=\left(12\right)×\left(2.01\right)×\left(30×60\right)J=4.34×{10}^{4}\) A

Question 7: A potentiometer having the potential gradient of \(2 mV/cm\) is used to measure the difference of potential across a resistance of \(10 ohm.\) If a length of \(50 cm\) of the potentiometer wire is required to get the null point, the current passing through the \(10 ohm\) resistor is (in \(mA\))
(a) 1
(b) 2
(c) 5
(d) 10
Show Answer & Explanation

Answer: (d) 10

Explanation:
1. \(V=xl⇒iR=xl\)
2. \(⇒i×10=\left(\frac{2×{10}^{-3}}{{10}^{-2}}\right)×50×{10}^{-2}=0.1\)
3. \(⇒i=10×{10}^{-3}A=10 mA\)

Question 8: A wire \(100cm\) long and \(2.0 mm\) diameter has a resistance of \(0.7 ohm,\) the electrical resistivity of the material is
(a) \(4.4×{10}^{-6} ohm×m\)
(b) \(2.2×{10}^{-6}ohm×m\)
(c) \(1.1×{10}^{-6} ohm×m\)
(d) \(0.22×{10}^{-6} ohm×m\)
Show Answer & Explanation

Answer: (b) \(2.2×{10}^{-6}ohm×m\)

Explanation:
1. \(R=\frac{ρL}{A}⇒0.7=\frac{ρ×1}{\frac{22}{7}{\left(1×{10}^{-3}\right)}^{2}}\)
2. \(ρ=2.2×{10}^{-6}ohm-m\)

Question 9: A charge of \(2×{10}^{-1}C\) move at 30 revolutions per second in a circle of diameter 80 cm. The current linked with the circuit is
(a) 0.02A
(b) 20A
(c) 0.60A
(d) 60A
Show Answer & Explanation

Answer: (c) 0.60A

Explanation:
1. \(i=q/T=qv\)
2. \(=2×{10}^{-2}×30=0.6 A\)

Question 10: Two electric bulbs whose resistances are in the ratio of \(1 :2\) are connected in parallel to a constant voltage source. The powers dissipated in them have the ratio
(a) 1 :2
(b) 1 :1
(c) 2 :1
(d) 1 :4
Show Answer & Explanation

Answer: (c) 2 :1

Explanation:
1. We know that \(\frac{{P}_{1}}{{P}_{2}}=\frac{{R}_{2}}{{R}_{1}}=\frac{2}{1}\)

Question 11: Drift velocity \({v}_{d}\) varies with the intensity of electric field as per the relation
(a) \({v}_{d}∝E\)
(b) \({v}_{d}∝\frac{1}{E}\)
(c) \({v}_{d}=\)constant
(d) \({v}_{d}∝{E}^{2}\)
Show Answer & Explanation

Answer: (a) \({v}_{d}∝E\)

Explanation:
1. \({v}_{d}=\frac{e}{m}×\frac{V}{l}τ\) or \({v}_{d}=\frac{e}{m}.\frac{El}{l}τ [∴V=El]\)
2. \(∴{v}_{d}∝E\)

Question 12: An electric kettle takes \(4 A\) current at \(220 V.\) How much time will it take to boil \(1 kg\) of water from room temperature \(20℃?\) The temperature of boiling water is \(100℃\)
(a) \(0.63 minutes\)
(b) \(6.3 minutes\)
(c) \(12.6 minutes\)
(d) \(12.8 minutes\)
Show Answer & Explanation

Answer: (b) \(6.3 minutes\)

Explanation:
1. Electric power consumed by kettle \(P=220×4W\)
2. Heat required
3. \(H=1000×1\left(100-20\right)=1000×80 cal=4200×80 J\)
4. \(P=\frac{H}{t}⇒H=P×t\)
5. \(∴220×4×t=4200×80⇒t=6.3 minutes\)

Question 13: Two identical heaters rated \(220 volt, 1000 \mathrm{watt}\) are placed in series with each other across \(220 volt\) lines. If resistance does not change with temperature, then the combined power is
(a) \(1000 watt\)
(b) \(2000 watt\)
(c) \(500 watt\)
(d) \(4000 watt\)
Show Answer & Explanation

Answer: (c) \(500 watt\)

Explanation:
1. Power of the combination \({P}_{s}=\frac{P}{n}=\frac{1000}{2}=500W\)

Question 14: A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if
(a) Both the length and radius of wire are halved
(b) Both the length and radius of wire are doubled
(c) The radius of wire is doubled
(d) The length of wire is doubled
Show Answer & Explanation

Answer: (b) Both the length and radius of wire are doubled

Explanation:
1. When length and radius both are doubled, in accordance with relation \(R=\frac{ρL}{A}\) the resistance of wire is reduced to 1/2 of its initial value. As at constant voltage the heat produced \(H∝1/R\), hence heat produced is doubled

Question 15: The range of a voltmeter of resistance 500\(Ω\) is 10V. the resistance to be connected to convert it into an ammeter of range 10A is
(a) 1\(Ω\) in parallel
(b) 1\(Ω\) in series
(c) 0.1\(Ω\) in parallel
(d) 0.1\(Ω\) in series
Show Answer & Explanation

Answer: (c) 0.1\(Ω\) in parallel

Explanation:
1. Here, \(V =10V, G=500Ω\)
2. \({i}_{g}=\frac{V}{G}=\frac{10}{500}=2×{10}^{-2}A\)
3. Now,
4. \(S=\frac{{i}_{g}G}{i-{i}_{g}}=\frac{2×{10}^{-2}×500}{10-2×{10}^{-2}}≃1\mathrm{Ω\ in\ parallel.}\)

Question 16: The resistivity of iron is \(1×{10}^{-7}ohm-m.\) The resistance of a iron wire of particular length and thickness is \(1 ohm.\) If the length and the diameter of wire both are doubled, then the resistivity in \(ohm-m\) will be
(a) \(1×{10}^{-7}\)
(b) \(2×{10}^{-7}\)
(c) \(4×{10}^{-7}\)
(d) \(8×{10}^{-7}\)
Show Answer & Explanation

Answer: (a) \(1×{10}^{-7}\)

Explanation:
1. Resistivity of a material is its intrinsic property and is constant at particular temperature. Resistivity does not depend upon shape

Question 17: A piece of fuse wire melts when a current of \(15 ampere\) flows through it. With this current, if it dissipates \(22.5 W,\)the resistance of fuse wire will be
(a) Zero
(b) \(10 Ω\)
(c) \(1 Ω\)
(d) \(0.10 Ω\)
Show Answer & Explanation

Answer: (d) \(0.10 Ω\)

Explanation:
1. \(P={i}^{2}R⇒22.5={\left(15\right)}^{2}×R⇒R=0.10 Ω\)

Question 18: The electric current passing through a metallic wire produces heat because of
(a) Collisions of conduction electrons with each other
(b) Collisions of the atoms of the metal with each other
(c) The energy released in the ionization of the atoms of the metal
(d) Collisions of the conduction electrons with the atoms of the metallic wires
Show Answer & Explanation

Answer: (d) Collisions of the conduction electrons with the atoms of the metallic wires

Explanation:
1. Colliding electrons lose their kinetic energy as heat

Question 19: When a resistance of \(2ohm\) is connected across the terminals of a cell, the current is \(0.5 ampere.\) When the resistance is increased to \(5 ohm,\) the current is \(0.25 ampere.\) The internal resistance of the cell is
(a) \(0.5 ohm\)
(b) \(1.0 ohm\)
(c) \(1.5 ohm\)
(d) \(2.0 ohm\)
Show Answer & Explanation

Answer: (b) \(1.0 ohm\)

Explanation:
1. Let the e.m.f. of cell be \(E\) and internal resistance be \(r\)
2. Then \(0.5=\frac{E}{(r+2)}\) and \(0.25=\frac{E}{(r+5)}\)
3. On dividing, \(2=\frac{5+r}{2+r}⇒r=1Ω\)

Question 20: The relation between Faraday’s constant \(F,\) electron charge \(e\) and avogadro number \(N\) is
(a) \(F=N/e\)
(b) \(F=Ne\)
(c) \(N={F}^{2}\)
(d) \(F={N}^{2}e\)
Show Answer & Explanation

Answer: (b) \(F=Ne\)

Explanation:
1. Faraday constant \(=1\) mole electron charge \(=Ne\)
2. \(=6.02×{10}^{23}×1.6×{10}^{-19}=96500\)

Question 21: A cell of internal resistance \(r\) is connected to an external resistance \(R.\) The current will be maximum in \(R,\) if
(a) \(R=r\)
(b) \(R(c) \(R>r\)
(d) \(R=r/2\)
Show Answer & Explanation

Answer: (a) \(R=r\)

Explanation:
1. Current through \(R\) is maximum when total internal resistance of the circuit is equal to external resistance

Question 22: The current density(number of free electrons per \({m}^{3}\))in metallic conductor is of the order of
(a) \({10}^{22}\)
(b) \({10}^{24}\)
(c) \({10}^{26}\)
(d) \({10}^{28}\)
Show Answer & Explanation

Answer: (d) \({10}^{28}\)

Explanation:
1. The current density of electrons in a metallic conductor \(={10}^{22}c{m}^{-3}={10}^{28}{m}^{-3}\)

Question 23: A fuse wire with a radius of 1 mm blows at 1.5 A. If the fuse wire of the same material should blow at 3.0 A, the radius of the fuse wire must be
(a) \({4}^{1/3}\) mm
(b) \(\sqrt{2}\)mm
(c) 0.5 mm
(d) 8.0 mm
Show Answer & Explanation

Answer: (a) \({4}^{1/3}\) mm

Explanation:
1. The temperature of the wire increases to such a value at which, the heat produced per second equals heat lost per second due to radiation \(ie\),
2. \({I}^{2}\left(\frac{ρl}{π{r}^{2}}\right)=H×2πrl\), where \(H\) is heat lost per second per unit area due to radiation.
3. Hence, \({I}^{2}∝{r}^{3}\)
4. So \(\frac{{I}_{1}^{2}}{{I}_{2}^{2}}=\frac{{r}_{1}^{3}}{{r}_{2}^{3}}\mathrm{or}{r}_{2}={r}_{1}{\left({I}_{2}/{I}_{1}\right)}^{2/3}\)
5. \(=1×{\left(3.0/1.5\right)}^{2/3}={4}^{1/3}\) mm.

Question 24: A steady current of 1.5 A flows through a copper voltameter for 10 min. If the electrochemical equivalent of copper is \(30×{10}^{-5}g {C}^{-1},\) the mass of copper deposited on the electrode will be
(a) 0.40 g
(b) 0.50 g
(c) 0.67 g
(d) 0.27 g
Show Answer & Explanation

Answer: (d) 0.27 g

Explanation:
1. According to Faraday’s law of electrolysis
2. \(m=zit\)
3. Here, \(i=1.5 A, t=10 \mathrm{min=10×60\ s}\)
4. \(z=30×{10}^{-5}{\mathrm{gC}}^{-1}\)
5. Hence, mass of copper deposited on the electrode
6. \(m=30×{10}^{-5}×1.5×10×60=27×{10}^{-2}=0.27 g\)

Question 25: The maximum current that flows through a fuse wire before it blows out varies with its radius as
(a) \({r}^{3/2}\)
(b) \(r\)
(c) \({r}^{2/3}\)
(d) \({r}^{1/2}\)
Show Answer & Explanation

Answer: (a) \({r}^{3/2}\)

Explanation:
1. It is called safe current and is proportional to \({r}^{3/2}\)

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