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Units and Measurements question

2023 · 24 Jan · Shift 1 · Q60
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  5. /2023 · 24 Jan · Shift 1 · Q60

Units and Measurements question

2023 · 24 Jan · Shift 1 · Q60

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1

Match List I with List II

List-I
 List-II
A. Planck's constant (h) I. [M1 L2 T−2]\mathrm{[{M^1}\,{L^2}\,{T^{ - 2}}]}[M1L2T−2]
B. Stopping potential (Vs) II. [M1 L1 T−1]\mathrm{[{M^1}\,{L^1}\,{T^{ - 1}}]}[M1L1T−1]
C. Work function (ϕ\phiϕ) III. [M1 L2 T−1]\mathrm{[{M^1}\,{L^2}\,{T^{ - 1}}]}[M1L2T−1]
D. Momentum (p) IV. [M1 L2 T−3 A−1]\mathrm{[{M^1}\,{L^2}\,{T^{ - 3}}\,{A^{ - 1}}]}[M1L2T−3A−1]

Choose the correct answer from the options given below :

  1. A
    A-I, B-III, C-IV, D-II
  2. B
    A-III, B-IV, C-I, D-II
  3. C
    A-II, B-IV, C-III, D-I
  4. D
    A-III, B-I, C-II, D-IV
View written solutionFree

Correct answer: B

We match each physical quantity with its dimensional formula.

1. Planck's constant hhh

From E=hνE = h\nuE=hν So, [h]=[E][ν]=[ML2T−2][T−1]=[ML2T−1][h] = \frac{[E]}{[\nu]} = \frac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}][h]=[ν][E]​=[T−1][ML2T−2]​=[ML2T−1] Thus, A→IIIA \to IIIA→III

2. Stopping potential VsV_sVs​

Potential difference is V=workchargeV = \frac{\text{work}}{\text{charge}}V=chargework​ So, [Vs]=[ML2T−2][A T]=[ML2T−3A−1][V_s] = \frac{[M L^2 T^{-2}]}{[A\,T]} = [M L^2 T^{-3} A^{-1}][Vs​]=[AT][ML2T−2]​=[ML2T−3A−1] Thus, B→IVB \to IVB→IV

3. Work function ϕ\phiϕ

Work function is energy required to remove an electron. Hence, [ϕ]=[energy]=[ML2T−2][\phi] = [\text{energy}] = [M L^2 T^{-2}][ϕ]=[energy]=[ML2T−2] Thus, C→IC \to IC→I

4. Momentum ppp

Momentum is p=mvp = mvp=mv So, [p]=[M][LT−1]=[MLT−1][p] = [M][L T^{-1}] = [M L T^{-1}][p]=[M][LT−1]=[MLT−1] Thus, D→IID \to IID→II

Final matching

A→III,B→IV,C→I,D→IIA \to III, \quad B \to IV, \quad C \to I, \quad D \to IIA→III,B→IV,C→I,D→II

This corresponds to Option B.

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So, the answer agrees with the stored correct answer.

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