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Units and Measurements question

2019 · 11 Jan · Shift 2 · Q53
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Units and Measurements question

2019 · 11 Jan · Shift 2 · Q53

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young,s modulus will be:
  1. A
    V −-− 2A2F2
  2. B
    V −-− 4A −-− 2F
  3. C
    V −-− 4A2F
  4. D
    V −-− 2A2F −-− 2
View written solutionFree

Correct answer: C

  1. Write dimensions of the given fundamental quantities in terms of M,L,TM, L, TM,L,T.
  • Speed: [V]=LT−1[V] = LT^{-1}[V]=LT−1
  • Acceleration: [A]=LT−2[A] = LT^{-2}[A]=LT−2
  • Force: [F]=MLT−2[F] = MLT^{-2}[F]=MLT−2

We want the dimension of Young's modulus in terms of V,A,FV, A, FV,A,F.

  1. Dimension of Young's modulus

Young's modulus YYY has the same dimension as stress: [Y]=ForceArea=MLT−2L2=ML−1T−2[Y] = \frac{\text{Force}}{\text{Area}} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}[Y]=AreaForce​=L2MLT−2​=ML−1T−2

  1. Assume [Y]=[V]x[A]y[F]z[Y] = [V]^x [A]^y [F]^z[Y]=[V]x[A]y[F]z

Substitute dimensions: ML−1T−2=(LT−1)x(LT−2)y(MLT−2)zML^{-1}T^{-2} = (LT^{-1})^x (LT^{-2})^y (MLT^{-2})^zML−1T−2=(LT−1)x(LT−2)y(MLT−2)z

Expanding RHS: =MzLx+y+zT−x−2y−2z= M^z L^{x+y+z} T^{-x-2y-2z}=MzLx+y+zT−x−2y−2z

  1. Compare powers of M,L,TM, L, TM,L,T.

  • For MMM: z=1z = 1z=1

  • For LLL: x+y+z=−1x + y + z = -1x+y+z=−1 Since z=1z=1z=1, x+y+1=−1  ⟹  x+y=−2x + y + 1 = -1 \implies x+y=-2x+y+1=−1⟹x+y=−2

  • For TTT: −x−2y−2z=−2-x-2y-2z = -2−x−2y−2z=−2 With z=1z=1z=1, −x−2y−2=−2  ⟹  −x−2y=0  ⟹  x+2y=0-x-2y-2=-2 \implies -x-2y=0 \implies x+2y=0−x−2y−2=−2⟹−x−2y=0⟹x+2y=0

  1. Solve the equations

From x+y=−2x+y=-2x+y=−2 and x+2y=0x+2y=0x+2y=0 Subtracting, y=2y=2y=2 Then, x+2(2)=0  ⟹  x=−4x+2(2)=0 \implies x=-4x+2(2)=0⟹x=−4

Also, z=1z=1z=1.

So, [Y]=V−4A2F[Y] = V^{-4}A^2F[Y]=V−4A2F

  1. Match with options

This corresponds to Option C.

Hence, the dimension of Young's modulus is: V−4A2F\boxed{V^{-4}A^2F}V−4A2F​

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