MHT CET 2024 10th May Morning Shift | Rotational Motion Question 14 | Physics

A ring and a disc roll on horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is 6 J then total kinetic energy of the disc is

$\frac{3}{2} \mathrm{~J}$

$\frac{5}{2} \mathrm{~J}$

$\frac{7}{2} \mathrm{~J}$

$\frac{9}{2} \mathrm{~J}$

MHT CET 2024 10th May Morning Shift

MCQ (Single Correct Answer)

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A solid metallic sphere of radius ' $R$ ' having moment of inertia '$I$' about diameter is melted and recast into a solid disc of radius ' $r$ ' of a uniform thickness. The moment of inertia of a disc about an axis passing through its edge and perpendicular to its plane is also equal to '$I$'. The ratio $\frac{r}{R}$ is

$\frac{1}{\sqrt{2}}$

$\frac{2}{\sqrt{5}}$

$\frac{2}{\sqrt{10}}$

$\frac{2}{\sqrt{15}}$

MHT CET 2024 9th May Evening Shift

MCQ (Single Correct Answer)

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A particle of mass ' m ' is rotating in a circular path of radius ' $r$ '. Its angular momentum is ' $L$ ' The centripetal force acting on it is ' $F$ '. The relation between ' $F$ ', ' $L$ ', ' $r$ ' and ' $m$ ' is

$\mathrm{F}=\frac{\mathrm{L}}{\mathrm{mr}^2}$

$\mathrm{L}=\mathrm{m}^2 \mathrm{Fr}^2$

$\frac{\mathrm{L}^2}{\mathrm{~m}}=\mathrm{Fr}^3$

$\frac{\mathrm{F}}{\mathrm{L}^3}=\mathrm{mr}^2$

MHT CET 2024 9th May Evening Shift

MCQ (Single Correct Answer)

-0

Three thin rods, each mass ' 2 M ' and length ' L ' are placed along $\mathrm{x}, \mathrm{y}$ and z axis which are mutually perpendicular. One end of each rod is at origin. Moment of inertia of the system about x - axis is

$\frac{4 \mathrm{ML}^2}{3}$

$\frac{\mathrm{ML}^2}{12}$

$\frac{\mathrm{ML}^2}{6}$

$\frac{2 \mathrm{ML}^2}{3}$

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