JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2024
- Hard
- 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 _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\left[\mathrm{ML}^3 \mathrm{~T}^{-3}\right]\)
Step-by-step Solution
Detailed explanation
\(\mathrm{F}=\mathrm{ax}^2+\mathrm{bt}^{1 / 2}\)
\({[\mathrm{a}]=\frac{[\mathrm{F}]}{\left[\mathrm{x}^2\right]}=\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]}\)
\({[\mathrm{b}]=\frac{[\mathrm{F}]}{\left[\mathrm{t}^{1 / 2}\right]}=\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-5 / 2}\right]}\)
\({\left[\frac{\mathrm{b}^2}{\mathrm{a}}\right]=\frac{\left[\mathrm{M}^2 \mathrm{~L}^2 \mathrm{~T}^{-5}\right]}{\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]}=\left[\mathrm{M}^1 \mathrm{~L}^3 \mathrm{~T}^{-3}\right]}\)
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