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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