A) \[\left[ M{{L}^{-1}}{{T}^{-2}} \right]\]
B) \[\left[ {{M}^{0}}L{{T}^{-2}} \right]\]
C) \[\left[ ML{{T}^{-2}} \right]\]
D) \[\left[ M{{L}^{2}}{{T}^{-2}} \right]\]
Correct Answer: A
Solution :
When strain is small, the ratio of the longitudinal tress to the corresponding longitudinal strain is called Young?s modulus of the material of the body. \[Y=\frac{stress}{strain}=\frac{\frac{F}{A}}{\frac{l}{L}}\] Where \[F\] is force, \[A\] is area, \[l\] and \[L\]are change in length and original length. Dimensions of \[Y=\frac{\left[ M{{L}^{-1}}\,{{T}^{-2}} \right]}{\left[ {{M}^{\circ }}{{L}^{\circ }}\,{{T}^{\circ }} \right]}=\left[ M{{L}^{-1}}{{T}^{-2}} \right]\] Alternative: The SI unit of \[Y\] is \[N{{m}^{-2}}\] or Pascal. Therefore, Dimensions of \[Y=\frac{newton}{metr{{e}^{2}}}=\frac{\left[ ML{{T}^{-2}} \right]}{\left[ {{L}^{2}} \right]}\]
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