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Two wires of the metal have the same length but their cross-sections are in the ratio 3:1 They are joined in series: The resistance of the thicker wire is $ 10 \Omega $ . The total resistance of the combination will be

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Explanation

$ R_1 = { \delta l \over 3 } = 10 \Omega , R _2 ={ \delta l \over 1} = 30 \Omega $ $ Rs = R_1 + R_2 = 10 \Omega + 30 \Omega = 40 \Omega $

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The ratio of angular speeds of minute hand and hour hand of a watch is

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Explanation

ω=2π/T

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A wire 50cm long and $1 mm^2$ in cross-section carries a curent of 4A when connected to a 2V battery. The resistivity of the wire is:

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Explanation

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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?

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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

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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?

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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.

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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....

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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:

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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^{–8} \Omega m$ volume of the wire is $4m^3$ and its resistance is$ 4 \Omega $ Therefore its length will be.

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Explanation

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

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