8.Mechanical Properties of Solids
easy

In $CGS$ system, the Young's modulus of a steel wire is $2 \times {10^{12}}$. To double the length of a wire of unit cross-section area, the force required is

A

$4 \times {10^6}$ dynes

B

$2 \times {10^{12}}$ dynes

C

$2 \times {10^{12}}$ newtons

D

$2 \times {10^8}$ dynes

Solution

(b)To double the length of wire,

Stress = Young's modulus

 $\frac{F}{A} = 2 \times {10^{12}}\frac{{dyne}}{{c{m^2}}}.$

If $A = 1$ then $F = 2 × 10^{12}$ dyne

Standard 11
Physics

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