One end of a metal wire is fixed to a ceiling and a load of 2 ~kg hangs from the other end. A similar wire is attached to the bottom of the load and another load of 1 ~kg hangs from this lower wire. Then the ratio of longitudinal strain of upper wire to that of the lower wire will be \underline\hspace2cm . [Area of cross section of wire =0.005 ~cm^2, Y=2 \times 10^11 Nm^-2 and g=10 ~ms^-2 ]

Properties of Matter PYQ — JEE Main 2024 · 1 Feb · Shift 2 · Q86 with Solution

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Correct answer: 3

For a wire,

\text{strain} = \frac{\text{stress}}{Y} = \frac{F/A}{Y} = \frac{F}{AY}

Since both wires are similar, they have the same cross-sectional area $A$ and Young's modulus $Y$ .

The lower wire supports only the $1\,\text{kg}$ mass. So,

T_{\text{lower}} = 1 \times 10 = 10\,\text{N}

Hence strain in lower wire is

\epsilon_{\text{lower}} = \frac{10}{AY}

The upper wire supports both the $2\,\text{kg}$ load and the $1\,\text{kg}$ load hanging below it. So total load on upper wire is

(2+1)\,\text{kg} = 3\,\text{kg}

Therefore,

T_{\text{upper}} = 3 \times 10 = 30\,\text{N}

Hence strain in upper wire is

\epsilon_{\text{upper}} = \frac{30}{AY}

\frac{\epsilon_{\text{upper}}}{\epsilon_{\text{lower}}} = \frac{30/(AY)}{10/(AY)} = 3

The ratio of longitudinal strain of upper wire to lower wire is

\boxed{3}