A force is represented by $\mathrm{F}=\mathrm{ax}^{2}+\mathrm{bt}^{1 / 2}$ Where $\mathrm{x}=$ distance and $\mathrm{t}=$ time. The dimensions of $b^{2} / a$ are :

(A) $\left[\mathrm{ML}^{3} \mathrm{~T}^{-3}\right]$

(B) $\left[\mathrm{MLT}^{-2}\right]$

(C) $\left[M L^{-1} T^{-1}\right]$

(D) $\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]$

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Units & Measurement

A force is represented by

F = ax^{2} + bt^{1/2}

Where

x =

distance and

t =

time. The dimensions of

b^{2} / a

are :

Answer: (A)

Given

F = a x^{2} + b t^{1/2}

, where

x

is distance and

t

is time.

From

a x^{2} = [F]

[a] = \frac{[ F ]}{[ x ^{2} ]} = \frac{M L T ^{- 2}}{L ^{2}} = M L^{- 1} T^{- 2}

From

b t^{1/2} = [F]

[b] = \frac{[ F ]}{[ t ^{1/2} ]} = \frac{M L T ^{- 2}}{T ^{1/2}} = M L T^{- 5/2}

\frac{b ^{2}}{a} = \frac{( M L T ^{- 5/2} ) ^{2}}{M L ^{- 1} T ^{- 2}} = \frac{M ^{2} L ^{2} T ^{- 5}}{M L ^{- 1} T ^{- 2}} = M L^{3} T^{- 3}

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option