NEET UG 2019 — Physics Mechanics

When a body rolls i.e. have rotational motion, the total kinetic energy of the system will be $$\mathrm{KE}=\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right)$$KE=21​mv2(1+R2k2​) where, $\mathrm{m}=$m= mass of body, $\mathrm{v}=$v= velocity and $\mathrm{k}=$k= radius of gyration

Given, $\mathrm{m}=2 \mathrm{~kg}, \theta=30^{\circ}, \mathrm{v}=4 \mathrm{~ms}^{-1}$m=2 kg,θ=30∘,v=4 ms−1 Let $h$h be the height of the inclined plane, then from law of conservation of energy, $$\begin{aligned} \mathrm{KE} & =\mathrm{PE} \\ \frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right) & =\mathrm{mgh} \end{aligned}$$KE21​mv2(1+R2k2​)​=PE=mgh​ Substituting the given values in the above equation, we get $$\begin{aligned} & \frac{1}{2} \times 2 \times 16\left(1+\frac{1}{2}\right)=2 \times 10 \times \mathrm{h}\left[\because \text { For cylinder } \frac{\mathrm{k}^2}{\mathrm{R}^2}=\frac{1}{2}\right] \\ & \Rightarrow \quad 8 \times \frac{3}{2}=10 \mathrm{~h} \Rightarrow \mathrm{h}=1.2 \mathrm{~m} \end{aligned}$$​21​×2×16(1+21​)=2×10×h[∵ For cylinder R2k2​=21​]⇒8×23​=10 h⇒h=1.2 m​ From the above diagram $$\begin{gathered} \sin \theta=\frac{\mathrm{h}}{\mathrm{x}} \\ \Rightarrow \mathrm{x}=\frac{\mathrm{h}}{\sin \theta}=\frac{1.2}{\sin 30^{\circ}}=1.2 \times 2=2.4 \mathrm{~m}\left[\because \sin 30^{\circ}=\frac{1}{2}\right] \end{gathered}$$sinθ=xh​⇒x=sinθh​=sin30∘1.2​=1.2×2=2.4 m[∵sin30∘=21​]​