Two wires A and B with lengths in the ratio of 3: 1, diameters in the ratio of 1:2 and resistivity in the ratio of 1:20 are joined in parallel with a source of e.m.f. 2V. Ratio of the rates of heat production of wire A with respect to wire B is

5 : 2

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2 : 5

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5 : 3

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3 : 5

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Solution

The correct option is C 5 : 3
Ratio of length, diameters and resistivity is given by,
$\frac{L_{1}}{L_{2}} = \frac{3}{1} \frac{d_{1}}{d_{2}} = \frac{1}{2} \frac{P_{1}}{P_{2}} = \frac{1}{20}$

$Q_{1} = \frac{V^{2} t}{R_{1}} Q_{2} = \frac{V^{2} t}{R_{2}}$

$\frac{Q_{1}}{Q_{2}} = \frac{R_{2}}{R_{1}} = (\frac{L_{2}}{L_{1}}) (\frac{ρ_{2}}{ρ_{1}}) (\frac{A_{1}}{A_{2}})$

$= (\frac{L_{2}}{L_{1}}) (\frac{ρ_{2}}{ρ_{1}}) (\frac{d_{1}^{2}}{d_{2}^{2}})$

$= \frac{1}{3} \times 20 \times \frac{1}{4}$

$= \frac{5}{3}$

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Two wires A and B with lengths in the ratio of 3: 1, diameters in the ratio of 1:2 and resistivity in the ratio of 1:20 are joined in parallel with a source of e.m.f. 2V. Ratio of the rates of heat production of wire A with respect to wire B is

The correct option is C 5 : 3Ratio of length, diameters and resistivity is given by,L1L2=31 d1d2=12 P1P2=120Q1=V2tR1Q2=V2tR2Q1Q2=R2R1=(L2L1)(ρ2ρ1)(A1A2) =(L2L1)(ρ2ρ1)(d21d22)=13×20×14 =53