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)

$T_{\text{conical }}=2\pi \sqrt{\frac{l\cos \theta }{g}}$
$\Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{\cos 45^{\circ }}{\cos 60^{\circ }}}$
$\frac{T_1^2}{T_2^2}=\left(\frac{1}{\sqrt{2}}\right)÷\left(\frac{1}{2}\right)=\sqrt{2}$