The relation between length L and radius R of the cylinder, if its moment of inertia about its axis is equal to that about the equatorial axis, will be

Question

The relation between length L and radius R of the cylinder, if its moment of inertia about its axis is equal to that about the equatorial axis, will be (A) L=R (B) L=2R (C) L=3R (D) L=√3R

Tardigrade Admin · Accepted Answer

(m R2/2)=M((L2/12) + (R2/4)) or (R2/2)=(L2/12)+(R2/4) or L=√3R

Q. The relation between length $L$ and radius $R$ of the cylinder, if its moment of inertia about its axis is equal to that about the equatorial axis, will be

$L = R$

$L = 2 R$

$L = 3 R$

$L = 3 R$

$\frac{m R ^{2}}{2} = M (\frac{L ^{2}}{12} + \frac{R ^{2}}{4})$
or $\frac{R ^{2}}{2} = \frac{L ^{2}}{12} + \frac{R ^{2}}{4}$
or $L = 3 R$

Q. The relation between length L and radius R of the cylinder, if its moment of inertia about its axis is equal to that about the equatorial axis, will be

L=R

L=2R

L=3R

L=3​R

2mR2​=M(12L2​+4R2​) or 2R2​=12L2​+4R2​ or L=3​R

Q. The relation between length $L$ and radius $R$ of the cylinder, if its moment of inertia about its axis is equal to that about the equatorial axis, will be

$L = R$

$L = 2 R$

$L = 3 R$

$L = 3 R$

$\frac{m R ^{2}}{2} = M (\frac{L ^{2}}{12} + \frac{R ^{2}}{4})$
or $\frac{R ^{2}}{2} = \frac{L ^{2}}{12} + \frac{R ^{2}}{4}$
or $L = 3 R$