JEE Main 2023 — Units and Dimensions Question with Solution

(A) Surface tension is related to force and length as  $S=\frac{F}{l}=\frac{ML{T}^{-2}}{ L}=M T^{-2}$

$=\mathrm{kg} {\mathrm{s}}^{-2}$ (IV)

(B) Pressure is related to force and area as $P=\frac{F}{A}$

 $=\frac{ML{T}^{-2}}{ L^2}=\mathrm{kg} {\mathrm{m}}^{-1} {\mathrm{s}}^{-2}$  (III)

(C) Viscosity in terms of force, area and velocity gradient is given by  $\eta =\frac{F}{A\left({\displaystyle \frac{dv}{dx}}\right)}=\frac{ML{T}^{-2}}{ L^2\left({\displaystyle \frac{L{T}^{-1}}{L}}\right)}$

$=M{L}^{-1} T^{-1}=\mathrm{kg} {\mathrm{m}}^{-1} {\mathrm{s}}^{-1}$ (I)

(D) Impulse is related to force and time of application as $J=\int F \mathrm{dt}=ML{T}^{-2}\times T$

$=ML{T}^{-1}=\mathrm{kg} \mathrm{m} {\mathrm{s}}^{-1}$ (II).

So A-(IV), B-(III), C-(I), D-(II)