One end of a steel wire of radius ‘r’ is fixed to a ceiling and a load of 3 kg is attached to the free end of the wire. Another wire made of copper of radius ‘2r’ is attached to the bottom of 3 kg load and a 2 kg load is attached to the free end of the copper wire. The ratio of longitudinal strains produced in copper and steel wires is
(Young modulus of steel = 20×10¹⁰ Nm⁻²)
(Young modulus of copper = 12×10¹⁰Nm⁻²)

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3 : 2

6 : 1

1 : 6

2 : 3

answer is B.

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Detailed Solution

Given,

Young modulus of steel = 20×10¹⁰ Nm⁻²

Young modulus of copper = 12×10¹⁰Nm⁻²

From the question,

$e_{c} = \frac{∆ l_{c}}{l_{c}}$

$e_{s} = \frac{∆ l_{s}}{l_{s}}$

$∆ l = \frac{F l}{AY}$

$e = \frac{∆ l}{l} = \frac{F}{{πr}^{2} Y}$

$e_{s} = \frac{(m_{1} + m_{2}) g}{{πr}^{2} Y_{s}}_______(1)$

$e_{c} = \frac{m_{2} g}{π {(2 r)}^{2} Y_{c}}________(2)$

equation 2 divided by 2

$\frac{e_{c}}{e_{s}} = \frac{\frac{m_{2} g}{4 {πr}^{2} Y_{c}}}{\frac{(m_{1} + m_{2}) g}{{πr}^{2} Y_{s}}} = \frac{m_{2}}{m_{1} + m_{2}} \times \frac{1}{4} \times \frac{Y_{s}}{Y_{c}} = \frac{2}{5} \times \frac{1}{4} \times \frac{20 \times 10^{10}}{12 \times 10^{10}} = \frac{1}{6}$