The length of simple pendulum is 1 m and mass of its bob is 50 g. The bob is given sufficient velocity so that the bob describe vertical circle whose radius equal to length of pendulum. The maximum difference in the kinetic energy of bob during one revolution is.

A bob of mass $m = 0.05\ Kg$ is performing vertical circle of radius $(r) = 1m$.

In the boundary case.

a minimum velocity of $\mu = \sqrt{5gr}$ has to provided to it

so it completes a vertical circle.

So, kinetic energy $(K_1)$ at the lowest point,

$K_1 = \dfrac{1}{2}m\mu^2 = \dfrac{5}{2}mgr$

and $PE = 0$, raking it as reference at highest point

$PE : \mu_2 =mgh = 2mgr$ $(as \, h = 2r)$

From conservation of energy.

$K_1 + \mu_1= K_2+\mu_2 \Rightarrow K_2 = K_1 - \mu_2 = \dfrac{1}{2}mgr$

Thus, loss in $KE$:

$\Delta K = K_1-K_2 = 2mgr = 2\times 0.05 \times 9.8 \times 1 = 0.98\ J$ (Ans)