JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 20)
The disc of mass M with uniform surface mass density $$\sigma$$ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position $${x \over 3}{a \over \pi },{x \over 3}{a \over \pi }$$ where x is _____________. (Round off to the Nearest Integer).
[a is an area as shown in the figure]
_17th_March_Evening_Shift_en_20_1.png)
[a is an area as shown in the figure]
Answer
4
Explanation
_17th_March_Evening_Shift_en_20_2.png)
$$dm = \sigma {1 \over 2}R \times Rd\theta $$
$$dm = {{\sigma {R^2}d\theta } \over 2}$$
$${x_{cm}} = {{\int {x\,dm} } \over {\int {dm} }} = {{\int\limits_0^{\pi /2} {{{\sigma {R^2}} \over 2}d\theta \left( {{{2R} \over 3}\cos \theta } \right)} } \over {\int\limits_0^{\pi /2} {{{\sigma {R^2}} \over 2}d\theta } }}$$
$$ = {{2R} \over 3}{{\int\limits_0^{\pi /2} {\cos \theta d\theta } } \over {\int\limits_0^{\pi /2} {d\theta } }}$$
$$ = {{2R} \over 3}\left( {{2 \over \pi }} \right)$$
$$ = {{4R} \over {3\pi }}$$
So the value of x = 4
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