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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

  1. A \({V^{ - 2}}{A^2}{F^{ - 2}}\)
  2. B \({V^{ - 2}}{A^2}{F^2}\)
  3. C \({V^{ - 4}}{A^{ - 2}}F\)
  4. D \({V^{ - 4}}{A^{2}}F\)
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Correct Answer

(D) \({V^{ - 4}}{A^{2}}F\)

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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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