MHT CET 2025 22nd April Morning Shift | Rotational Motion Question 5 | Physics

[ $\mathrm{I}_{\mathrm{h}}=$ moment of inertia of hollow sphere about an axis coinciding with its diameter, $\mathrm{I}_5=$ moment of inertia of solid sphere about an axis coinciding with its diameter]

$\mathrm{I}_{\mathrm{s}}>\mathrm{I}_{\mathrm{h}}$

$\quad I_h \geqslant I_s$

$\mathrm{I}_{\mathrm{h}}>\mathrm{I}_{\mathrm{s}}$

$I_h=I_s$

MHT CET 2025 22nd April Morning Shift

MCQ (Single Correct Answer)

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If force $\vec{F}=-3 \hat{i}+\hat{j}+5 \hat{k}$ acts along $\vec{r}=7 \hat{i}+3 \hat{j}+\hat{k}$ then the torque acting at that point is

$(14 \hat{\mathrm{i}}-38 \hat{\mathrm{j}}+16 \hat{\mathrm{k}})$

$(-14 \hat{i}+34 \hat{j}-16 \hat{k})$

$(21 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$

$(4 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

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The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is I. It is rotating with angular velocity $\omega$. Another identical ring is gently placed on it so that their centres coincide. If both the rings are rotating about the same axis then loss in kinetic energy is

$\frac{\mathrm{I} \omega^2}{16}$

$\frac{\mathrm{I} \omega^2}{8}$

$\frac{\mathrm{I} \omega^2}{4}$

$\frac{\mathrm{I} \omega^2}{2}$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

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A body is rotating about its own axis. Its rotational kinetic energy is ' x ' and its angular momentum is ' $y$ '. Hence its moment of inertia about its own axis is

$\frac{x^2}{2 y}$

$\frac{y^2}{2 x}$

$\frac{x}{2 y}$

$\frac{y}{2 x}$

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