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Units and Measurements question

2020 · 2 Sep · Shift 2 · Q55
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Units and Measurements question

2020 · 2 Sep · Shift 2 · Q55

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
If momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is
  1. A
    [P2AT–2]
  2. B
    [P12AT−1]\left[ {{P^{{1 \over 2}}}A{T^{ - 1}}} \right][P21​AT−1]
  3. C
    [PA12T−1]\left[ {P{A^{{1 \over 2}}}{T^{ - 1}}} \right][PA21​T−1]
  4. D
    [PA–1T–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:
  • Momentum: [P]=[MLT−1][P] = [M L T^{-1}][P]=[MLT−1]
  • Area: [A]=[L2][A] = [L^2][A]=[L2]
  • Time: [T]=[T][T] = [T][T]=[T]
  1. Assume energy can be written as
[E]=[PxAyTz][E] = [P^x A^y T^z][E]=[PxAyTz]

We know the dimensional formula of energy is

[E]=[ML2T−2][E] = [M L^2 T^{-2}][E]=[ML2T−2]
  1. Substitute dimensions of P,A,TP, A, TP,A,T:
[PxAyTz]=(MLT−1)x(L2)y(T)z[P^x A^y T^z] = (M L T^{-1})^x (L^2)^y (T)^z[PxAyTz]=(MLT−1)x(L2)y(T)z =MxLx+2yT−x+z= M^x L^{x+2y} T^{-x+z}=MxLx+2yT−x+z
  1. Compare powers with [ML2T−2][M L^2 T^{-2}][ML2T−2]:
  • For MMM: x=1x = 1x=1
  • For LLL: x+2y=2x + 2y = 2x+2y=2 1+2y=2⇒y=121 + 2y = 2 \Rightarrow y = \frac{1}{2}1+2y=2⇒y=21​
  • For TTT: −x+z=−2-x + z = -2−x+z=−2 −1+z=−2⇒z=−1-1 + z = -2 \Rightarrow z = -1−1+z=−2⇒z=−1
  1. Hence
[E]=[P1A1/2T−1][E] = [P^1 A^{1/2} T^{-1}][E]=[P1A1/2T−1]

So the dimensional formula for energy is

[PA1/2T−1]\boxed{[P A^{1/2} T^{-1}]}[PA1/2T−1]​
  1. Check options:
  • A: [P2AT−2][P^2 A T^{-2}][P2AT−2] ❌
  • B: [P1/2AT−1][P^{1/2} A T^{-1}][P1/2AT−1] ❌
  • C: [PA1/2T−1][P A^{1/2} T^{-1}][PA1/2T−1] ✅
  • D: [PA−1T−2][P A^{-1} T^{-2}][PA−1T−2] ❌

Therefore, the correct option is C.

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