
dm=σ⋅dA=σ2πrdr
=2πkr3dr
M=∫dm=2πk∫0Rr3dr
=2πk(4R4)=2πkR4
dI=dmr2=2πkr5dr
I=2πk∫0Rr5dr=2πk6R6=3×22πkR6
since, M=2πkR4
so,
I=32MR2
A thin disc of mass M and radius R has mass per unit area σ(r)=kr2 where r is the distance from its centre. Its moment inertia about an axis going through its centre of mass and perpendicular to its plane is:
Held on 10 Apr 2019 · Verified 6 Jul 2026.
3MR2
2MR2
6MR2
32MR2
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Net gravitational force at the center of a square is found to be $F_{1}$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_{2}$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_{1}}{F_{2}}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$. 
A particle of mass 2 kg is projected vertically upward with a speed of 30 m/s. The maximum height reached by the particle is (g = 10 m/s²):
Two projectiles are projected with the same initial velocities at the $15^\circ$ and $30^\circ$ with respect to the horizontal. The ratio of their ranges is $1:x$. The value of $x$ is:
In an experiment the values of two spring constants were measured as $k_{1}=(10 \pm 0.2) \mathrm{N} / \mathrm{m}$ and $k_{2}=(20 \pm 0.3) \mathrm{N} / \mathrm{m}$. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :
The surface tension of a soap solution is $3.5 \times 10^{-2}$ N/m. The work required to increase the radius of a soap bubble from $1$ cm to $2$ cm is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is _____. ($\pi = 22/7$)
Work through every JEE Main Mechanics PYQ, year by year.