MCQ
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$2:1$
- ✓$\sqrt{2}: 1$
$\frac{\mathrm{u}_{1}}{\mathrm{u}_{2}}=\frac{1}{4} \Rightarrow 4 \mathrm{u}_{1}=\mathrm{u}_{2}$
$4 \frac{1}{2 \mathrm{Y}}\left[\frac{\mathrm{W} \cdot 4}{\pi \mathrm{d}_{1}^{2}}\right]^{2}=\frac{1}{2 \mathrm{Y}}\left[\frac{\mathrm{W} \cdot 4}{\pi \mathrm{d}_{2}^{2}}\right]^{2}$
$4=\left(\frac{\mathrm{d}_{1}}{\mathrm{d}_{2}}\right)^{4}$
$\Rightarrow \frac{\mathrm{d}_{1}}{\mathrm{d}_{2}}=\sqrt{2}: 1$
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$(A)$ $\omega=\frac{3 v x}{ L ^2+3 x^2}$
$(B)$ $\omega=\frac{12 v x}{L^2+12 x^2}$
$(C)$ $x_M=\frac{L}{\sqrt{3}}$
$(D)$ $\omega_M=\frac{v}{2 L} \sqrt{3}$