Question

The velocity of the particle at time $t$ is given by $v = at + \dfrac{b}{{t + c}}$ , where a, b and c are the constant, the dimension of a, b and c are respectively.
A.$L{T^{ - 2}},L$ $and$ $T$
B.${L^2},T$ $and$ $L{T^2}$
C.$L{T^2},LT$ $and$ $L$
D.$L,LT$ $and$ ${T^2}$

Answer 1

We use the principle of homogeneity means dimension of velocity ‘$v$’ must be equal to the dimension $at + \dfrac{b}{{t + c}}$ , where at is equal to dimension of velocity and $\dfrac{b}{{t + c}}$ is equal to dimension of velocity.
Formula Used:
$Velocity = \dfrac{{Displacement}}{{Time}}$

Complete step by step solution:
For finding the dimension of velocity. We use the formula mention below,
$Velocity = \dfrac{{Displacement}}{{Time}}$
Unit of displacement and time are the unit of meter and second.
$Velocity = \dfrac{{metre}}{{\sec ond}}$
$\therefore$Dimension of velocity ‘$v$’$= \left[ {{M^0}L{T^{ - 1}}} \right]$
Now dimension of velocity = Dimension of $at$
$\therefore \left[ {L{T^{ - 1}}} \right] = at$
Dimension of ‘$t$’ $= \left[ T \right]$

$\therefore a = \dfrac{{L{T^{ - 1}}}}{T}$
Dimension of $a = L{T^{ - 2}}$
Now, Dimension of ‘$v$’=Dimension of $\dfrac{b}{{t + c}}$
$L{T^{ - 1}} = \dfrac{b}{{t + c}}$ equation(1)
Before knowing the dimension of $b$ , we need to know the dimension of $c$, as we can see in equation(1). $c$ is added to ‘$t$’ means the dimension of ‘$e$’ will be same as like ‘$t$’ therefore ‘$c$’ dimension is $\left[ T \right]$
So, $\left[ {L{T^1}} \right] = \dfrac{b}{{\left[ T \right]}}$
$b = L{T^{ - 1}}.T$
$\therefore$Dimension of $b = \left[ L \right]$
$\therefore$The values of a, b and c are $\left[ {L{T^{ - 2}}} \right],\left[ L \right]and\left[ T \right]$

Note: Principle of homogeneity is a physical equation will be dimensionally correct if the dimension of all the terms occurring on both sides of the equation are the same. This principle is based on the fact that only the physical quantity of the same kind can be added, subtracted or compared. Thus velocity can be added to velocity not to force.