Register

Using acebook

Using +

Or, use the form below

Enter your name Enter your e-mail address Enter password Enter Phone number

Already a registered user? Click here to Login

Login

Using acebook

Using +

Or, use the form below

Enter your e-mail address Enter password

Forgot Password?

Not a registered user? Click here to Register

Forgot Password

Enter your e-mail address

Not a registered user? Click here to Register

Already a registered user? Click here to Login

Report an Error

Image for question is missing or is incorrect

Answers choices for this question are incorrect

Question is incorrect or has incomplete data

Explanation does not seem okay

Please add some additional information:

Time Spent - 00:00

Laws of Motion

A pendulum is released from $\theta_0=60 ^{\circ}$. The rate of change of speed of bob at $\theta=30 ^{\circ}$ is $(g=10 m/s^2)$

Like / Save

Ask a doubt

Report Error

Discuss this question

$10\; m/s^2 $

$2.5\; m/s^2$

$5\; m/s^2$

$5 \sqrt 3\; m/s^2$

You are currently solving

Laws of Motion practice questions for JEE

Submit

Next

Skip

End Practice