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  • A body of mass is tied to a spring of spring constant <img alt="\mathrm{12.5 N/m,}" src="ht

A body of mass \mathrm{200 g} is tied to a spring of spring constant \mathrm{12.5 N/m,} while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed \mathrm{5rad/s}. Then the ratio of extension in the spring to its natural length will be :

Option: 1

2:5


Option: 2

1:1


Option: 3

2:3


Option: 4

1:2


Answers (1)

best_answer

\begin{aligned} & \mathrm{kx}=\mathrm{m} \omega^2\left(\ell_0+\mathrm{x}\right) \\ \end{aligned}

\begin{aligned} & \frac{\mathrm{k}}{\mathrm{m} \omega^2}=\frac{\ell_0}{\mathrm{x}}+1 \\ \end{aligned}

\begin{aligned} & \frac{12.5}{0.2 \times 25}=\frac{\ell_0}{\mathrm{x}}+1 \\ \end{aligned}

\begin{aligned} & \frac{125}{50}-1=\frac{\ell_0}{\mathrm{x}} \\ \end{aligned}

\begin{aligned} & \frac{3}{2}=\frac{\ell_0}{\mathrm{x}} \\ \end{aligned}

\begin{aligned} & \frac{\mathrm{x}}{\ell_0}=\frac{2}{3} \end{aligned}

Posted by

Suraj Bhandari

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