Question

When the string of a conical pendulum makes an angle of $45^{\circ}$ with the vertical, its time period is $T_{1} .$ When the string makes an angle of $60^{\circ}$ with the vertical, its time period is $T_{2}$. Then, $\frac{T_{1}^{2}}{T_{2}^{2}}$ is

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\sqrt{5}$
• D. $\sqrt{7}$

Answer 1

Answer: (A)

$T_{\text {conical }}=2 \pi \sqrt{\frac{l \cos \theta}{g}} $
$\Rightarrow \frac{T_{1}}{T_{2}}=\sqrt{\frac{\cos 45^{\circ}}{\cos 60^{\circ}}}$
$\frac{T_{1}^{2}}{T_{2}^{2}}=\left(\frac{1}{\sqrt{2}}\right) \div\left(\frac{1}{2}\right)=\sqrt{2}$