JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
If speed \(V,\) area \(A\) and force \(F\) are chosen as fundamental units, then the dimension of Young's modulus will be :
- A \(FA ^{-1} V ^{0}\)
- B \(FA ^{2} V ^{-1}\)
- C \(FA ^{2} V ^{-3}\)
- D \(FA ^{2} V ^{-2}\)
Answer & Solution
Correct Answer
(A) \(FA ^{-1} V ^{0}\)
Step-by-step Solution
Detailed explanation
\(Y = F ^{ x } A ^{ y } V ^{z}\) \(M ^{1} L ^{-1} T ^{-2}=\left[ MLT ^{-2}\right] x \left[ L ^{2}\right] y \left[ L T ^{-1}\right]^{z}\) \(M ^{1} L ^{1} T ^{-2}=[ M ]^{ x }[ L ]^{ x +2 y +z}[ T ]^{-2 x - z }\) comparing power of \(ML\) and \(T\) \(x=1 \ldots(1)\)…
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