The expression given below shows the variation of velocity ( $v$ ) with time ( t ), $v=\mathrm{At}^{2}+\frac{\mathrm{Bt}}{\mathrm{C}+\mathrm{t}}$. The dimension of ABC is :

(A) $\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-3}\right]$

(B) $\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-3}\right]$

(C) $\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right]$

(D) $\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]$

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Units & Measurement

The expression given below shows the variation of velocity (

v

) with time ( t ),

v = At^{2} + \frac{Bt}{C + t}

.
The dimension of ABC is :

Answer: (A)

Given

v = A t^{2} + \frac{B t}{C + t}

, and

v

has dimensions

[L T^{- 1}]

.

Since

C

is added to

t

[C] = [T]

.

From the first term,

[A] [T]^{2} = [L T^{- 1}] ⟹ [A] = [L T^{- 3}]

From the second term, since

[C + t] = [T]

\frac{[ B ] [ T ]}{[ T ]} = [L T^{- 1}] ⟹ [B] = [L T^{- 1}]

Multiplying,

[A B C] = [L T^{- 3}] [L T^{- 1}] [T] = [L^{2} T^{- 3}] = [M^{0} L^{2} T^{- 3}]

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option