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Question

# If the velocity (V), acceleration (A), and force (F) are taken as fundamental quantities instead of mass (M), length (L) and time (T), the dimensions of Young's modulus (Y) would be

A
FA2V4
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B
FA2V5
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C
FA2V3
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D
FA2V2
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Solution

## The correct option is A FA2V−4Step 1: Dimension of Young's Modulus Y, Force F, Area A and Velocity V We know that, Velocity V=Distancetime=LT−1Acceleration A=Change in velocitytime=LT−2 Force F=mass×acceleration=MLT−2 Youngs Modulus Y=F/AΔl/l=MLT−2L2LL=ML−1T−2Step 2: Finding Y in terms of F, A and VIf Force F, Area A and velocity V are taken as fundamental quantities, so Youngs Modulus Y can be expressed in terms of above as follows:Let,Y=FaAbVc ....(1)Putting their dimensions in above equation⇒ ML−1T−2=[MLT−2]a [LT−2]b [LT−1]c⇒ ML−1T−2=Ma L(a+b+c) T(−2a−2b−c)Comparing powers of M, L & T in LHS and RHS in above equation, we get:a=1,−1=a+b+c ⇒ −1=1+b+c ⇒ b+c=−2 ....(2)−2=−2a−2b−c ⇒ −2=−2−2b−c ⇒ c=−2bSo, Equation (2) ⇒ b−2b=−2 ⇒ b=2⇒ c=−2b=−4Therefore, putting values of a, b and c in Equation (1), we get:Y=F A2 V−4Hence Option A is correct

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