A force is represented by F=a x^2+bt^ 1 / 2 . Where x= distance a… - Vidyadip

MCQ

A force is represented by $\mathrm{F}=a \mathrm{x}^2+\mathrm{bt}^{1 / 2}$. Where $\mathrm{x}=$ distance and $\mathrm{t}=$ time. The dimensions of $\mathrm{b}^2 / \mathrm{a}$ are :

✓
$\left[\mathrm{ML}^3 \mathrm{~T}^{-3}\right]$
B
$\left[\mathrm{MLT}^{-2}\right]$
C
$\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]$
D
$\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right]$

✓

Answer

Correct option: A.

$\left[\mathrm{ML}^3 \mathrm{~T}^{-3}\right]$

a
$\mathrm{F}=\mathrm{ax}^2+\mathrm{bt}^{1 / 2}$

${[\mathrm{a}]=\frac{[\mathrm{F}]}{\left[\mathrm{x}^2\right]}=\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]}$

${[\mathrm{b}]=\frac{[\mathrm{F}]}{\left[\mathrm{t}^{1 / 2}\right]}=\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-5 / 2}\right]}$

${\left[\frac{\mathrm{b}^2}{\mathrm{a}}\right]=\frac{\left[\mathrm{M}^2 \mathrm{~L}^2 \mathrm{~T}^{-5}\right]}{\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]}=\left[\mathrm{M}^1 \mathrm{~L}^3 \mathrm{~T}^{-3}\right]}$

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