A solid sphere of radius R has moment of inertia I about its geometrical axis. It is melted into a disc of radius r and thickness t. If it's moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to

Question

A solid sphere of radius R has moment of inertia I about its geometrical axis. It is melted into a disc of radius r and thickness t. If it's moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to (A) (2/ √ 15) R  (B) (2/ √ 5) R  (C) (3/√15) R  (D) (√ 3/√15) R

Tardigrade Admin · Accepted Answer

(2/5) MR2 = ( 1/2) Mr2 + Mr2 or (2/5) MR2 (3/2)Mr2 ∴ r= (2/ √ 15) R

Q. A solid sphere of radius R has moment of inertia I about its geometrical axis. It is melted into a disc of radius r and thickness t. If it's moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to

$\frac{2}{1 5} R$

$\frac{2}{5} R$

$\frac{3}{1 5} R$

$\frac{3}{1 5} R$

$\frac{2}{5} M R^{2} = \frac{1}{2} M r^{2} + M r^{2}$
or $\frac{2}{5} M R^{2} \frac{3}{2} M r^{2}$
$∴ r = \frac{2}{15} R$

Q. A solid sphere of radius R has moment of inertia I about its geometrical axis. It is melted into a disc of radius r and thickness t. If it's moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to

1​52​R

5​2​R

1​53​R

1​53​​R

52​MR2=21​Mr2+Mr2 or 52​MR223​Mr2 ∴r=15​2​R

$\frac{2}{1 5} R$

$\frac{2}{5} R$

$\frac{3}{1 5} R$

$\frac{3}{1 5} R$

$\frac{2}{5} M R^{2} = \frac{1}{2} M r^{2} + M r^{2}$
or $\frac{2}{5} M R^{2} \frac{3}{2} M r^{2}$
$∴ r = \frac{2}{15} R$