Using equation of continuity
40VA=20VB
⇒2VA=VB
Using Bernoulli’s equation
PA+21ρVA2=PB+21ρVB2
⇒PA−PB=21(VB2−VA2)
⇒ΔP=211000(VB2−4VB2)
⇒ΔP=500×43VB2
⇒VB=1500(ΔP)×4=1500(700)×4sm
Volume flow rate =20×100×VB=2732cm3/s
JEE Main 2020 — Physics Mechanics
Water flows m a horizontal tube (see figure). The pressure of water changes by 700Nm−2 between A and B where the area of cross section are 40cm2 and 20cm2, respectively. Find the rate of flow of water through the tube. (density of water =1000kgm−3 )

Held on 9 Jan 2020 · Verified 6 Jul 2026.
3020cm3/s
2720cm3/s
2420cm3/s
1810cm3/s
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.