Units And Dimensions Question 6
Question 6 - 2024 (31 Jan Shift 1)
A force is represented by $\mathrm{F}=a \mathrm{ax}^{2}+b \mathrm{t}^{1 / 2}$
Where $\mathrm{x}=$ distance and $\mathrm{t}=$ time. The dimensions of $b^{2} / a$ are :
(1) $\left[\mathrm{ML}^{3} \mathrm{~T}^{-3}\right]$
(2) $\left[\mathrm{MLT}^{-2}\right]$
(3) $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]$
(4) $\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]$
Show Answer
Answer: (1)
Solution:
$\mathrm{F}=\mathrm{ax}^{2}+\mathrm{bt}^{1 / 2}$
$[\mathrm{a}]=\frac{[\mathrm{F}]}{\left[\mathrm{x}^{2}\right]}=\left[\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]$
$[\mathrm{b}]=\frac{[\mathrm{F}]}{\left[\mathrm{t}^{1 / 2}\right]}=\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-5 / 2}\right]$
$\left[\frac{b^{2}}{a}\right]=\frac{\left[\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-5}\right]}{\left[\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]}=\left[\mathrm{M}^{1} \mathrm{~L}^{3} \mathrm{~T}^{-3}\right]$
