The velocity v of a particle at time t is given by v=a t+(b)/(t+c) , where a, b and c are constants. The dimensions of a, b and c are:

Correct answer: B. LT^ -2 , ~L and T Using principle of homogeniety. Dimension of v= Dimension of a t= Dimenstion of (b)/(t+c) array l a t =[LT^ -1 ] |a| = [LT^ -1 ] [T] =[LT^ -2 ] And ([b])/([t]) =[LT^ -1 ] [b] =[LT^ -1 ][T]=[L] And c =T c =[T] array

NEET 2006

Exam: NEET
Year: 2006
Subject: Physics
Chapter: Units and Dimensions

NEET 2006 · Physics · Units and Dimensions

The velocity v of a particle at time t is given by v=a t+(b)/(t+c)...

The velocity

v

of a particle at time

t

is given by

v=a t+\frac{b}{t+c}

, where

a, b

and

c

are constants. The dimensions of

a, b

and

c

are:

Options

AL, LT and $\mathrm{LT}^2$
B $\mathrm{LT}^{-2}, \mathrm{~L}$ and $\mathrm{T}$
C $\mathrm{L}^2, \mathrm{~T}$ and $\mathrm{LT}^2$
D $\mathrm{LT}^2$ , LT and L.

Correct answer

B. $\mathrm{LT}^{-2}, \mathrm{~L}$ and $\mathrm{T}$

Step-by-step solution

Using principle of homogeniety. Dimension of

v=

Dimension of

a t=

Dimenstion of

\frac{b}{t+c}

\begin{array}{l} \therefore a t =\left[\mathrm{LT}^{-1}\right] \\ \Rightarrow |a| =\frac{\left[\mathrm{LT}^{-1}\right]}{[\mathrm{T}]}=\left[\mathrm{LT}^{-2}\right] \\ \text{And } \frac{[b]}{[t]} =\left[\mathrm{LT}^{-1}\right] \\ {[b]} =\left[\mathrm{LT}^{-1}\right][\mathrm{T}]=[\mathrm{L}] \\ \text{And } c =\mathrm{T} \\ \Rightarrow c =[\mathrm{T}] \end{array}

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