JEE PYQ: Units & Measurements - Question ID fa0049b818b2 (JEE Main 2021)

ID: fa0049b818b2JEE Main 2021Single Correct MCQ
Match List - I with List - II :

List I List II
(a) h (Planck's constant) (i) [MLT1][ML{T^{ - 1}}]
(b) E (kinetic energy) (ii) [ML2T1][M{L^2}{T^{ - 1}}]
(c) V (electric potential) (iii) [ML2T2][M{L^2}{T^{ - 2}}]
(d) P (linear momentum) (iv) [ML2I1T3][M{L^2}{I^{ - 1}}{T^{ - 3}}]


Choose the correct answer from the options given below :

Select Option

Step-by-step Explanation

Core Formula & Concept:

In the chapter Units & Measurements, every physical quantity is expressed in terms of the fundamental dimensions: Mass (MM), Length (LL), Time (TT), and Electric Current (II). The key formulas we use are:

  • Kinetic energy: E=12mv2[E]=[M][L2T2]=[ML2T2]E = \tfrac{1}{2}mv^2 \quad \Rightarrow \quad [E] = [M][L^2T^{-2}] = [ML^2T^{-2}].
  • Linear momentum: P=mv[P]=[M][LT1]=[MLT1]P = mv \quad \Rightarrow \quad [P] = [M][LT^{-1}] = [MLT^{-1}].
  • Planck’s constant: E=hνh=Eν[h]=[ML2T2][T1]=[ML2T1]E = h\nu \quad \Rightarrow \quad h = \tfrac{E}{\nu} \quad \Rightarrow \quad [h] = \tfrac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}].
  • Electric potential: V=Wq=Force×distancecurrent×time[V]=[MLT2][L][I][T]=[ML2I1T3]V = \tfrac{W}{q} = \tfrac{Force \times distance}{current \times time} \quad \Rightarrow \quad [V] = \tfrac{[MLT^{-2}][L]}{[I][T]} = [ML^2I^{-1}T^{-3}].
Step-by-Step Derivation:

Step 1: Dimension of Planck’s constant hh

From E=hνE = h\nu, we get h=Eνh = \tfrac{E}{\nu}. Since [E]=[ML2T2][E] = [ML^2T^{-2}] and [ν]=[T1][\nu] = [T^{-1}], we have [h]=[ML2T2][T1]=[ML2T1].[h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]. This matches List II entry (ii).

Step 2: Dimension of kinetic energy EE

E=12mv2E = \tfrac{1}{2}mv^2 gives [E]=[M][L2T2]=[ML2T2].[E] = [M][L^2T^{-2}] = [ML^2T^{-2}]. This matches List II entry (iii).

Step 3: Dimension of electric potential VV

V=Wq=Force×distancecurrent×timeV = \tfrac{W}{q} = \tfrac{Force \times distance}{current \times time}. [V]=[MLT2][L][I][T]=[ML2I1T3].[V] = \frac{[MLT^{-2}][L]}{[I][T]} = [ML^2I^{-1}T^{-3}]. This matches List II entry (iv).

Step 4: Dimension of linear momentum PP

P=mvP = mv gives [P]=[M][LT1]=[MLT1].[P] = [M][LT^{-1}] = [MLT^{-1}]. This matches List II entry (i).

Step 5: Matching the lists

We obtain:

  • (a) hh \to (ii)
  • (b) EE \to (iii)
  • (c) VV \to (iv)
  • (d) PP \to (i)

This corresponds exactly to option A.

Common Traps & Exam Tip:

Students often confuse the dimensions of energy ([ML2T2][ML^2T^{-2}]) with those of Planck’s constant ([ML2T1][ML^2T^{-1}]), or mistake electric potential ([ML2I1T3][ML^2I^{-1}T^{-3}]) for momentum ([MLT1][MLT^{-1}]). A quick check is to remember:

  • Energy has T2T^{-2} while Planck’s constant has T1T^{-1}.
  • Electric potential involves current II in its dimension, momentum does not.

Always derive each dimension from its defining formula rather than memorizing the list.

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