
21mu2+[2a−G×10Mm−8aGMm]=0−8aG×16Mm−2aGM×m
⇒u=4a45GM
u=23a5GM
JEE Main 2020 — Physics Mechanics
Two planets have masses M and 16M and their radii are a and 2a, respectively. The separation between the centres of the planets is 10a. A body of mass m is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is :
Held on 6 Sept 2020 · Verified 6 Jul 2026.
2aGM
4aGM
maGM2
23a5GM
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
If $\epsilon, E$ and $t$ represent the free space permittivity, electric field and time respectively, then the unit of $\frac{\epsilon E}{t}$ will be :
When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises upto certain height $h$. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by $1 \%$ each, then the height of the liquid in the tube will change by $\%$.
The velocity of a particle is given as $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$ m/s. The magnitude of acceleration at point $(1, 2, 4)$ is _______ m/s$^2$.
Suppose there is a uniform circular disc of mass $M \mathrm{~kg}$ and radius $r \mathrm{~m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^{2}$. The value of $x$ is $\_\_\_\_$. 
A soap bubble of surface tension $0.04 \mathrm{~N} / \mathrm{m}$ is blown to a diameter of 7 cm. If $(15000-x) \mu \mathrm{J}$ of work is done in blowing it further to make its diameter 14 cm, then the value of $x$ is $\_\_\_\_$. ($\pi=22 / 7$)
Work through every JEE Main Mechanics PYQ, year by year.