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Physics MCQs for NEET — Practice Questions with Answers

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A parallel combination of three resistors takes a current of 7.5 A form a 30 V supply, It the two resistors are $ 10 \Omega $ and $ 12 \Omega $. find which is the third one?

A. $ 4 \Omega $ B. $ 15 \Omega $ C. $ 12 \Omega $ D. $ 22 \Omega $

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Explanation

$ I = 7.5 A , V = 30 V, R_1 = 10 \Omega , R_2 = 12 \Omega , R_3 = ? $ $ R_{net} = { V \over I}$ $ \Rightarrow { 1 \over Rnet }= {I \over V} = { 1 \over R_1 } + { 1 \over R_2 } + { 1 \over R_3 } $

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A potentiometer wire has length 10 mand resistance $ 20 \Omega $. A 2.5V battery of negligible internal resistance is connected across the wire with an $ 80 \Omega $ series resistance. The potential gradient on the

A. $ 2.5 \times 10^{-4} V/cm $ B. $ 0.62 \times 10^{-4} V/mm $ C. $ 1 \times 10^{-5} V/mm $ D. $ 5 \times 10^{-5} V/mm $

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Explanation

$ V = ( {E \over R_S + R_i}) Rl$ $ =({ 2.5 \over 80+20}) \times 20= 0.5 volt $ $ Potential gradient = { V/ l} = { 0.5 volt / 10 metre } = 5 \times 10 ^ {-2} Vm^{-1} = 5 \times 10^{-5} V /mm$

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The drift velocity of free electrons through a conducting wire of radius r, carrying current I, is if the same current is passed through a conductor of radius 2r what will be the drift velocity?

A. $ {V_d \over 4 } $ B. $ {V_d } $ C. $ {2V_d} $ D. $ {24V_d} $

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Explanation

$ I = nAV_dq$ $V_d \alpha 1/ r^2$ $ \therefore V_d^1 = Vd /4 $

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A carbon resistor has a set of coaxial coloured rings in the order brown, violet brown and silver. The value of resistance (in ohms) is.

$ (27 \times 10 ) \pm 5 \%$

$ (27 \times 10 ) \pm 10 \%$

$ (17 \times 10 ) \pm 5 \%$

$ (17 \times 10 ) \pm 10 \%$

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Explanation

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A cross a wire of length l and thickness d, a p.d of V is applied. If the p.d is doubled the dirft velocity becomes....

A. becomes double B. becomes half C. Close not change D. becomes Zero

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Explanation

$ I =V/R= nAV_dq$ $ \therefore V_d= V $

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The masses of three wires of copper are in the ratio of 1:3:5 and their lengths are in the ratio of 5:3:1. The ratio of their electrical resistance is:

A. 1:01:01 B. 1:03:05 C. 5:03:01 D. 125:15:01

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Explanation

$R \alpha {l^2 \over m} $ $\therefore R_1 : R_2:R_3= {25 \over 1} : { 9 \over 3} : { 1 \over 5 } =125 : 15:1$

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Resistivity of material of a conducting wire is $ 4 \times 10^{â€“8} \Omega m$ volume of the wire is $4m^3$ and its resistance is$ 4 \Omega $ Therefore its length will be.

A. 500m B. 5000m C. 20,000m D. $4 \times 10^{-5}m$

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Explanation

$ l^2 ={ RV \over \rho}$

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How would you arrange 48 cells each of e.m.f 2V and inteanal resistance $1.5 \Omega $ so as to pass maximum current through the external resistance of $2 \Omega$ ?

2 cells in 24 groups

4 cells in 12 groups

8 cells in 6 groups

3 cells in 16 groups

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Explanation

[{No of rows}/{no of cells in each row}] = (m/n) = (r/R) where r = internal resistance of each cell

R = external resistance

∴ (m/n) = {1.5/2} = (3/4)

also total number of cells = 48 m × n = 48

∴ (3/4) n × n = 48 n2 = {(48 × 4)/3} n = 8

∴ m = (48/8) = 6

∴ 6 rows with 8 cell in each of row

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How many dry cells, each of emf 1.5V and internal resistance $0.5 \Omega $, much be joined in series with a resistor of $ 20 \Omega $ to give a current of 0.6A in the circuit ?

A. 2 B. 8 C. 10 D. 12

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Explanation

$ l = { n \varepsilon \over R +nr } \Leftrightarrow 0.6 = { n \times 1.5 \over 20 \times 0.5 n } \Leftrightarrow 12 + 0.3 n = 1.5 n \therefore 12 = 1.2 n $ $ \therefore n = 10 $cells to be connected in series

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Two electric bulbs whose resistances are in the ratio of 1:2 are connected in parallel to a constant voltage source the power dissipated in them have the ratio.

A. 1:02 B. 1:01 C. 2:01 D. 1:04

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Explanation

$ {P_1/P_2 } = {V^2 / R_1 \over V^2 / R_2 } = {R_2 \over R_1} = 2 /1 $

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