MCQ
The area of cross-section of a wire of length $1.1$ metre is $1$ $mm^2$. It is loaded with $1 \,kg.$ If Young's modulus of copper is $1.1 \times {10^{11}}\,N/{m^2}$, then the increase in length will be ......... $mm$ (If $g = 10\,m/{s^2})$
  • A
    $0.01$
  • B
    $0.075 $
  • $0.1$
  • D
    $0. 15$

Answer

Correct option: C.
$0.1$
c
(c) $l = \frac{{mgL}}{{AY}} = \frac{{1 \times 10 \times 1.1}}{{1.1 \times {{10}^{11}} \times {{10}^{ - 6}}}}m = 0.1\;mm$

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