JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 14)
Consider two uniform discs of the same thickness and different radii R1
= R and
R2 = $$\alpha $$R made of the same material. If the ratio of their moments of inertia I1 and I2 , respectively, about their axes is I1 : I2 = 1 : 16 then the value of $$\alpha $$ is :
R2 = $$\alpha $$R made of the same material. If the ratio of their moments of inertia I1 and I2 , respectively, about their axes is I1 : I2 = 1 : 16 then the value of $$\alpha $$ is :
$$\sqrt 2 $$
2
$$2\sqrt 2 $$
4
Explanation
Moment of inertia of disc, $$I = {{M{R^2}} \over 2} = {{\left[ {p\left( {\pi {R^2}} \right)t} \right]{R^2}} \over 2}$$
$$I = K{R^4}$$
$${{{I_1}} \over {{I_2}}} = {\left( {{{{R_1}} \over {{R_2}}}} \right)^4}$$
$${1 \over {16}} = {\left( {{R \over {\alpha R}}} \right)^4} \Rightarrow \alpha = {\left( {16} \right)^{{1 \over 4}}} = 2$$
$$I = K{R^4}$$
$${{{I_1}} \over {{I_2}}} = {\left( {{{{R_1}} \over {{R_2}}}} \right)^4}$$
$${1 \over {16}} = {\left( {{R \over {\alpha R}}} \right)^4} \Rightarrow \alpha = {\left( {16} \right)^{{1 \over 4}}} = 2$$
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