Question

$\frac{\alpha }{t^2}=Fv+\frac{\beta }{x^2}$ Find the dimension formula for $\left[\alpha \right]$ and $\left[\beta \right]$ (here t time, F = force, v = velocity, x = distance).

• A. $\left[\beta \right]=M^1{L}^2{T}^{-3};\left[\alpha \right]=M^1{L}^2{T}^{-1}$
• B. $\left[\beta \right]=M^1{L}^3{T}^{-3};\left[\alpha \right]=M^1{L}^3{T}^{-1}$
• C. $\left[\beta \right]=M^1{L}^4{T}^{-3};\left[\alpha \right]=M^1{L}^2{T}^{-1}$
• D. $\left[\beta \right]=M^1{L}^5{T}^{-3};\left[\alpha \right]=M^1{L}^2{T}^{-1}$

Answer 1

The correct option is C $\left[\beta \right]=M^1{L}^4{T}^{-3};\left[\alpha \right]=M^1{L}^2{T}^{-1}$

The given relation is:

$\frac{\alpha }{t^2}=FV+\frac{\beta }{x^2}$

Rearrange the relation and substitute the dimension of respective quantities:

$FV=\left[{MLT}^{-2}\right].[L.T^{-1}]$

$=\left[{ML}^2{T}^{-3}\right]$

$\Rightarrow$ $\alpha =\left[{ML}^2{T}^{-3}\right]\left[T^2\right]=\left[{ML}^2{T}^{-1}\right]$

$\beta =\left[{ML}^2{T}^{-3}\right]\left[L^2\right]=\left[{ML}^4{T}^{-3}\right]$