- Home
- Standard 11
- Physics
8.Mechanical Properties of Solids
medium
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
A
$1: \sqrt{2}$
B
$1: 2$
C
$2:1$
D
$\sqrt{2}: 1$
(JEE MAIN-2020)
Solution
$\frac{\text { Energystored }}{\text { Volume }}=\frac{1}{2} \frac{(\text { Stress })^{2}}{\mathrm{Y}}$
$\frac{\mathrm{u}_{1}}{\mathrm{u}_{2}}=\frac{1}{4} \Rightarrow 4 \mathrm{u}_{1}=\mathrm{u}_{2}$
$4 \frac{1}{2 \mathrm{Y}}\left[\frac{\mathrm{W} \cdot 4}{\pi \mathrm{d}_{1}^{2}}\right]^{2}=\frac{1}{2 \mathrm{Y}}\left[\frac{\mathrm{W} \cdot 4}{\pi \mathrm{d}_{2}^{2}}\right]^{2}$
$4=\left(\frac{\mathrm{d}_{1}}{\mathrm{d}_{2}}\right)^{4}$
$\Rightarrow \frac{\mathrm{d}_{1}}{\mathrm{d}_{2}}=\sqrt{2}: 1$
Standard 11
Physics