8.Mechanical Properties of Solids
medium

Young's modulus of rubber is ${10^4}\,N/{m^2}$ and area of cross-section is $2\,c{m^2}$. If force of $2 \times {10^5}$ dynes is applied along its length, then its initial length $l$ becomes

A

$3L$

B

$4L$

C

$2L$

D

None of the above

Solution

(c) $Y = {10^4}N/{m^2},A = 2 \times {10^{ – 4}}{m^2},F = 2 \times {10^5}dyne = 2N$

$l = \frac{{FL}}{{AY}} = \frac{{2 \times L}}{{2 \times {{10}^{ – 4}} \times {{10}^4}}} = L$

 Final length $=$ initial length $+$ increment $= 2L$

Standard 11
Physics

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