If speed (V), acceleration

Physics Units, Dimensions and Measurement NEET Previous Year Question MCQ (Single Correct)

and force (F) are considered as fundamental units, the dimension of Young's modulus will be : V –4 A 2 F

V –2 A 2 F –2

V –4 A –2 F

V –2 A 2 F 2

Text Solution

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The correct answer is:

Let Y = f(V, F, A)

∴ Y = K V x F y A z , K → Unit less

∴ [Y] = [V] x [F] y [A] z

∴ [ML –1 T –2 ] = [LT –1 ] x [MLT –2 ] y [LT –2 ] z

∴ [M] [L –1 ] [T –2 ] = [M] y [L x+y+z ] [T –x–2y–2z ]

∴ y = 1, x + y + z = –1 ; –x – 2y – 2z = 2

∴ x + z = –2 ∴ x + 2y + 2z = 2

∴ z = 2, x = –4 ∴ x + 2z = 0 ∴ [Y] = [V –4 F A 2 ]

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