A highly rigid cubical block $A$ of small mass $M$ and side $L$ is fixed rigidy on the othe-cubical block of…
- $2 \pi \sqrt{M \pi L}$
- $\quad 2 \pi \sqrt{(M n / L)}$
- $2 \pi \sqrt{M L / n}$
- $\quad 2 \pi \sqrt{M / \eta L}$
Solution
Modulus of rigidity $\eta=\frac{\mathrm{F}}{\mathrm{A} \theta}$
Here, $\mathrm{A}=\mathrm{L}^{2}$
and $\theta=\frac{\mathrm{x}}{\mathrm{L}}$ for small $\theta$
$\therefore$ Restoring force $=\mathrm{F}=-\eta \mathrm{A} \theta$
or acceleration, $a=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\eta \mathrm{L}}{\mathrm{M}} \mathrm{x} \quad$ equation $(1)$
$\because \mathrm{a} \propto(-\mathrm{x})$
$\therefore$ Time period, $\mathrm{T}=2 \pi \sqrt{\left|\frac{\mathrm{x}}{\mathrm{a}}\right|}$
$\Longrightarrow \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{\eta} \mathrm{L}}}$ /
Asked in: JEE Mains - Units and Dimensions - Test 2