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Physics MCQs for NEET — Practice Questions with Answers

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The sensitivity of a potentiometer can be increased by

increasing the potential gradient

decreasing the length of potentiometer wire

decreasing the current in the potentiometer wire

all of these

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Explanation

$(3) Sensitivity \propto \frac{1}{Potential gradient}$

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A current of 1 mA is flowing through a copper wire. How many electrons will pass a given point in one second $[e = 1.6 \times 10^{- 19} Coulomb]$

$6.25 \times 10^{19}$

$6.25 \times 10^{15}$

$6.25 \times 10^{31}$

$6.25 \times 10^{8}$

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Explanation

$(2) I = ne n = \frac{1}{e} = \frac{10^{- 3}}{1.6 \times 10^{- 19}} = 6.25 \times 10^{18} \times 10^{- 3} = 6.25 \times 10^{15}$

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The drift velocity of free electrons in a conductor is 'v' when a current 'i' is flowing in it. If both the radius and current are doubled, then drift velocity will be :

$\frac{v}{2}$

$\frac{v}{4}$

$\frac{v}{8}$

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Explanation

$(2) I = {AV}_{d} ne - - - - - - - (1) 2 I = 4 A V_{d_{1}} ne - - - - - - - (2) \frac{1}{2} = \frac{1}{4} \frac{V_{d}}{V_{d_{1}}} \Rightarrow V_{d_{1}} = \frac{V_{d}}{2}$

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A battery has e.m.f. 4V and internal resistance r. When this battery is connected to an external resistance of 2 ohms, a current of 1 amp. flows in the circuit. How much current will flow if the terminals of the battery are connected directly

1 Amp

2 Amp

4 Amp

Infinite

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Explanation

$(2) I = \frac{E}{r + 2} I = \frac{4}{r + 2} r = 2 Ω I = \frac{E}{r} = \frac{4}{2} = 2 Amp$

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Three copper wires have their lengths in the ratio 5:3:1 and their masses are in the ratio 1:3:5. Their electrical resistance will be in the ratio of :

5:3:1

1:3:5

125:15:1

1:15:125

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Explanation

$M a s s m = ρ \times A \times l A r e a A = \frac{m}{ρ l}$

$R e s i s \tan c e R = \frac{ρ l}{A} ∴ R \propto \frac{l^{2}}{m} R_{1} : R_{2} : R_{3} = \frac{25}{1} : \frac{9}{3} : \frac{1}{5} = 125 : 15 : 1$

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The current in a wire varies with time according to the relation i= (3+2t)A. The amount of charge passing a cross section of the wire in the time interval t=0 to t=4.0 sec would be (where t is time in seconds)

28 C

30.5 C

8 C

82 C

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Explanation

$i = (3 + 2 t) \frac{dq}{dt} = 3 + 2 t \int_{o}^{q} dq = \int_{o}^{4} (3 + 2 t) dt q = {(3 t + t^{2})}_{o}^{4} q = 12 + 16 = 28 C$

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The resistivity of iron is 1 × 10^–7 ohm – m. The resistance of 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 :

1 × 10^–7

2 × 10^–7

4 × 10^–7

8 × 10^–7

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Explanation

Resistivity of some material is its intrinsic property and is constant at particular temperature. Resistivity does not depend upon shape.

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The temperature coefficient of resistance for a wire is 0.00125/°C. At 300K its resistance is 1 ohm. The temperature at which the resistance becomes 2 ohm is

1154 K

1100 K

1400 K

1127 K

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Explanation

$\frac{ρ_{1}}{ρ_{2}} = \frac{(1 + α t_{1})}{(1 + α t_{2})} \Rightarrow \frac{1}{2} = \frac{(1 + 0 .00125 \times 27)}{(1 + 0 .00125 \times t)}$

$\Rightarrow t = 854 ° C \Rightarrow T = 1127 K$

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When the length and area of cross-section both are doubled, then its resistance :

Will become half

Will be doubled

Will remain the same

Will become four times

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Explanation

$R_{1} \propto \frac{l}{A} \Rightarrow R_{2} \propto \frac{2 l}{2 A} i . e . R_{2} \propto \frac{l}{A}$

$∴ R_{1} = R_{2}$

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The resistivity of a wire :

Increases with the length of the wire

Decreases with the area of cross-section

Decreases with the length and increases with the cross-section of the wire

None of the above statement is correct

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Explanation

Resistivity is the property of the material. It does not depend upon size and shape.

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