JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
If speed \((V)\), acceleration \((A)\) and force \((F)\) are considered as fundamental units, the dimension of Young’s modulus will be
- A \({V^{ - 2}}{A^2}{F^{ - 2}}\)
- B \({V^{ - 2}}{A^2}{F^2}\)
- C \({V^{ - 4}}{A^{ - 2}}F\)
- D \({V^{ - 4}}{A^{2}}F\)
Answer & Solution
Correct Answer
(D) \({V^{ - 4}}{A^{2}}F\)
Step-by-step Solution
Detailed explanation
\begin{array}{l} \frac{F}{A} = y \cdot \frac{{\Delta \ell }}{\ell }\,\,;\,\,\left[ Y \right] = \frac{F}{A}\\ Now\,from\,\dim ension\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,F = \frac{{ML}}{{{T^2}}}\,\,;\,\,L = \frac{F}{M} \cdot {T^2}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{L^2} =…
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