The value of J is ___________.
Explanation:
The value of K is ___________.
Explanation:

Just after string becomes taut; there will be no velocity along the string.
$\therefore$ ${V_ \bot } = {{P\cos \theta } \over m} = {{0.2 \times 0.9} \over {1 \times 0.1}} = 1.8$ m/s
$\therefore$ $K = {1 \over 2}mv_ \bot ^2 = {1 \over 2} \times 0.1 \times {1.8^2} = 0.162$ J
A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is IO and IP respectively. Both these axes are perpendicular to the plane of the lamina. The ratio IO/IP to the nearest integer is ____________.

Explanation:
T = Total portion
R = Remaining portion and
C = Cavity and
Then, ${m_T} = \pi {(2R)^2}\sigma = 4\pi {R^2}\sigma $
${m_C} = \pi {(R)^2}\sigma = \pi {R^2}\sigma $
For IP

${I_R} = {I_T} - {I_C}$
$ = {3 \over 2}{m_T}{(2R)^2} - \left[ {{1 \over 2}{m_C}{R^2} + {m_C}{r^2}} \right]$
$ = {3 \over 2}(4\pi {R^2}\sigma )(4{R^2}) - \left[ {{1 \over 2}(\pi {R^2}\sigma ) + (\pi {R^2}\sigma )(5{R^2})} \right]$
$ = (18.5\pi {R^4}\sigma )$
For IO
${I_R} = {I_T} - {I_C}$
$ = {1 \over 2}{m_T}{(2R)^2} - {3 \over 2}{m_C}{R^2}$
$ = {1 \over 2}(4\pi {R^2}\sigma )(4{R^2}) - {3 \over 2}(\pi {R^2}\sigma )({R^2})$
$ = 6.5\pi {R^4}\sigma $
$\therefore$ ${{{I_P}} \over {{I_O}}} = {{18.5\pi {R^4}\sigma } \over {6.5\pi {R^4}\sigma }} = 2.846$
Therefore, the nearest integer is 3.