t45=dh1
h1=d .......(i)
d+2.464dh1+h2=tan30∘=31=0.5774
h1+h2=(3.464)d×0.5774
h1+h2=2d .......(ii)
∴h2=d
JEE Main 2020 — Physics Mechanics
A balloon is moving up in air vertically above a point A on the ground. When it is a height h1, a girl standing at a distance d (point B) from A (see figure) sees it at an angle 45∘ with respect to the vertical. When the balloon climbs up a further height h2, it is seen at an angle 60∘ with respect to the vertical if the girl moves further by a distance 2.464d (point C). Then the height h2 is (given tan30∘=0.5774):

Held on 5 Sept 2020 · Verified 6 Jul 2026.
1.464d
0.732d
0.464d
d
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.