A sphere of radius R is cut from a larger sol | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the density of the sphere be $\rho$. The mass of the larger sphere of radius $2R$ is $M = \rho \cdot \frac{4}{3} \pi (2R)^3 = 8 \rho \cdot \fr

Vedclass · Accepted Answer

$\frac{7}{57}$

Vedclass · Answer

(C) Let the density of the sphere be $\rho$. The mass of the larger sphere of radius $2R$ is $M = \rho \cdot \frac{4}{3} \pi (2R)^3 = 8 \rho \cdot \frac{4}{3} \pi R^3$. Let $m_0 = \rho \cdot \frac{4}{3} \pi R^3$ be the mass of the smaller sphere. Then $M = 8m_0$,so $m_0 = M/8$.The moment of inertia of the larger sphere about the $Y$-axis (passing through its center) is $I_1 = \frac{2}{5} M (2R)^2 = \frac{8}{5} MR^2$.The smaller sphere has radius $R$ and its center is at $x = R$. Its moment of inertia about the $Y$-axis is calculated using the parallel axis theorem: $I_2 = I_{cm} + m_0 d^2 = \frac{2}{5} m_0 R^2 + m_0 R^2 = \frac{7}{5} m_0 R^2$.Substituting $m_0 = M/8$,we get $I_2 = \frac{7}{5} (M/8) R^2 = \frac{7}{40} MR^2$.The moment of inertia of the remaining part is $I_{rem} = I_1 - I_2 = \frac{8}{5} MR^2 - \frac{7}{40} MR^2 = \frac{64 - 7}{40} MR^2 = \frac{57}{40} MR^2$.The ratio of the moment of inertia of the smaller sphere to that of the rest part is $\frac{I_2}{I_{rem}} = \frac{7/40 MR^2}{57/40 MR^2} = \frac{7}{57}$.

$A$ sphere of radius $R$ is cut from a larger solid sphere of radius $2R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the $Y$-axis is

Similar Questions

The moment of inertia of a circular ring about an axis perpendicular to its plane and passing through its center is $200 \, gm \cdot cm^2$. The moment of inertia about its diameter is ....... $gm \cdot cm^2$.

$A$ solid sphere of mass $5 \text{ kg}$ and a disc of mass $4 \text{ kg}$ have the same radius. The ratio of the moment of inertia of the sphere about its tangent to the moment of inertia of the disc about a tangent in its plane will be $x : y$. The value of $x$ and $y$ respectively is

If the moment of inertia of a uniform solid cylinder about the axis of the cylinder is $\frac{1}{n}$ times its moment of inertia about an axis passing through its midpoint and perpendicular to its length, then the ratio of the length and radius of the cylinder is

For the given square frame $ABCD$ with center $O$,which of the following relations is correct?

The moment of inertia of a uniform semicircular wire of mass $m$ and radius $r$,about an axis passing through its centre of mass and perpendicular to its plane is ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module