
dm=ρ×4πx2dx=ρ0(1−r2x2)×4πx2dx
m=4πρ∫0r(x2−R2x4)dxs
m=4πρ0∣3r3−5R2r5∣
E=r2Gm=r2G×4πρ0(3r3−5R2r5)
E=4πGρ0(3r−5R2r3)
E is maximum when dtdE=0⇒drdE=4πGρ0(31−5R23r2)=0
⇒r=35R
Emax=4πGρ0×35R[31−51×95]
Emax=2785πGρ0R
The mass density of a planet of radius R varies with the distance r from its centre as ρ(r)=ρ0(1−R2r2) Then the gravitational field is maximum at:
Held on 3 Sept 2020 · Verified 6 Jul 2026.
r=43R
r=R
r=31R
r=95R
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.