Q65 / 75 · JEE Main 2020 (09 Jan Shift 2) · Physics

Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is $1 : 4,$ the ratio of their diameters is:

Correct answer
A
$\sqrt{2} : 1$
B
$1 : 2$
C
$2 : 1$
D
$1 : \sqrt{2}$

Solution

\frac{d u}{d v} = \frac{1}{2}

stress

\times \frac{s t r e s s}{y}

= \frac{1}{2} \frac{F^{2}}{A^{2} y}

\frac{d u}{d v} α \frac{1}{d^{4}}

\frac{{(\frac{d u}{d v})}_{1}}{{(\frac{d u}{d v})}_{2}} = \frac{d_{2}^{4}}{d_{1}^{4}} = \frac{1}{4}

\frac{d_{1}}{d_{2}} = {(4)}^{1 / 4}

\frac{d_{1}}{d_{2}} = \sqrt{2} : 1

Hence, the correct answer is option: 'A'.

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