We know, Viscosity=Pascal‐second
Let [η]=PxAyTz=[M1L−1T−1]
⇒[M1L+1T−1]x[L2]y[T1]z=M1L−1T−1
⇒MxL+x+2yT−x+z=M1L−1T−1
From above, we have
x=1x+2y=−1−x+z=−1
Solving above three equations, we get
y=−1 and z=0
So, Viscosity =[P1A−1T0]
JEE Main 2022 — Physics Mechanics
If momentum [P], area [A] and time [T] are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is
Held on 25 Jul 2022 · Verified 6 Jul 2026.
[PA−1T0]
[PAT−1]
[PA−1T]
[PA−1T−1]
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.