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Units and Measurements question

2021 · 20 Jul · Shift 2 · Q57
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  5. /2021 · 20 Jul · Shift 2 · Q57

Units and Measurements question

2021 · 20 Jul · Shift 2 · Q57

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
If time (t), velocity (v), and angular momentum (l) are taken as the fundamental units. Then the dimension of mass (m) in terms of t, v and l is :
  1. A
    [t−1v1l−2][{t^{ - 1}}{v^1}{l^{ - 2}}][t−1v1l−2]
  2. B
    [t1v2l−1][{t^1}{v^2}{l^{ - 1}}][t1v2l−1]
  3. C
    [t−2v−1l1][{t^{ - 2}}{v^{ - 1}}{l^1}][t−2v−1l1]
  4. D
    [t−1v−2l1][{t^{ - 1}}{v^{ - 2}}{l^1}][t−1v−2l1]
View written solutionFree

Correct answer: D

  1. Write dimensions of the given fundamental quantities in terms of M,L,TM, L, TM,L,T.
  • Time: [t]=T[t] = T[t]=T
  • Velocity: [v]=LT−1[v] = LT^{-1}[v]=LT−1
  • Angular momentum: [l]=ML2T−1[l] = ML^2T^{-1}[l]=ML2T−1

We need to express mass [m]=M[m] = M[m]=M in terms of t,v,lt, v, lt,v,l.

  1. Assume
[m]=[t]a[v]b[l]c[m] = [t]^a [v]^b [l]^c[m]=[t]a[v]b[l]c

So,

M=Ta(LT−1)b(ML2T−1)cM = T^a (LT^{-1})^b (ML^2T^{-1})^cM=Ta(LT−1)b(ML2T−1)c
  1. Expand the RHS dimensions
Ta LbT−b McL2cT−cT^a \, L^b T^{-b} \, M^c L^{2c} T^{-c}TaLbT−bMcL2cT−c =McLb+2cTa−b−c= M^c L^{b+2c} T^{a-b-c}=McLb+2cTa−b−c
  1. Compare powers with M1L0T0M^1L^0T^0M1L0T0.

Thus,

  • For MMM: c=1c = 1c=1
  • For LLL: b+2c=0⇒b+2=0⇒b=−2b + 2c = 0 \Rightarrow b + 2 = 0 \Rightarrow b = -2b+2c=0⇒b+2=0⇒b=−2
  • For TTT: a−b−c=0⇒a−(−2)−1=0⇒a+1=0⇒a=−1a - b - c = 0 \Rightarrow a - (-2) - 1 = 0 \Rightarrow a + 1 = 0 \Rightarrow a = -1a−b−c=0⇒a−(−2)−1=0⇒a+1=0⇒a=−1
  1. Substitute values
[m]=[t]−1[v]−2[l]1[m] = [t]^{-1}[v]^{-2}[l]^1[m]=[t]−1[v]−2[l]1

So the dimension of mass is:

[t−1v−2l][t^{-1}v^{-2}l][t−1v−2l]
  1. Match with options

This corresponds to Option D: [t−1v−2l1][{t^{-1}}{v^{-2}}{l^1}][t−1v−2l1].

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

Hence, they agree.

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