$A$ solid sphere has mass $M$ and radius $R$. Its moment of inertia about a parallel axis passing through a point at a distance $\frac{R}{2}$ from its centre is

  • A
    $\frac{8 MR^2}{11}$
  • B
    $\frac{11 MR^2}{18}$
  • C
    $\frac{7 MR^2}{10}$
  • D
    $\frac{13 MR^2}{20}$

Explore More

Similar Questions

The radius of gyration $K$ of a hollow sphere of mass $M$ and radius $R$ about an axis $XY$ is equal to $R$. The distance of that axis from the center of the sphere is $h$. The value of $h$ is

Consider a uniform square plate of side '$a$' and mass '$m$'. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

According to the theorem of parallel axes $I = I_g + Md^2$,the graph between $I$ and $d$ will be

The moment of inertia of a thin uniform rod of length $L$ and mass $M$ about an axis passing through a point at a distance of $\frac{L}{3}$ from one of its ends and perpendicular to the rod is

For a uniform rectangular sheet shown in the figure,the ratio of moments of inertia about the axes perpendicular to the sheet and passing through $O$ (the centre of mass) and $O'$ (corner point) is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo