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Units and Measurements question

2021 · 31 Aug · Shift 1 · Q57
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Units and Measurements question

2021 · 31 Aug · Shift 1 · Q57

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
Match List - I with List - II.

List - I List - II
(a) Torque (i) MLT −1^{ - 1}−1
(b) Impulse (ii) MT −2^{ - 2}−2
(c) Tension (iii) ML 2^{ 2}2 T −2^{ - 2}−2
(d) Surface Tension (iv) MLT −2^{ - 2}−2


Choose the most appropriate answer from the option given below :
  1. A
    (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
  2. B
    (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
  3. C
    (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
  4. D
    (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
View written solutionFree

Correct answer: A

  1. Write dimensions of each physical quantity

    We match each item in List-I with its dimensional formula.

    (a) Torque

    Torque =Force×distance= \text{Force} \times \text{distance}=Force×distance

    Since dimensions of force are: [F]=MLT−2[F] = MLT^{-2}[F]=MLT−2

    Therefore, [τ]=[F][L]=MLT−2⋅L=ML2T−2[\tau] = [F][L] = MLT^{-2} \cdot L = ML^2T^{-2}[τ]=[F][L]=MLT−2⋅L=ML2T−2

    So, Torque →\to→ (iii).


    (b) Impulse

    Impulse =Force×time= \text{Force} \times \text{time}=Force×time

    [J]=[F][T]=MLT−2⋅T=MLT−1[J] = [F][T] = MLT^{-2} \cdot T = MLT^{-1}[J]=[F][T]=MLT−2⋅T=MLT−1

    So, Impulse →\to→ (i).


    (c) Tension

    Tension is a force.

    Therefore, [Tension]=MLT−2[\text{Tension}] = MLT^{-2}[Tension]=MLT−2

    So, Tension →\to→ (iv).


    (d) Surface Tension

    Surface tension =Forcelength= \dfrac{\text{Force}}{\text{length}}=lengthForce​

    [S]=MLT−2L=MT−2[S] = \frac{MLT^{-2}}{L} = MT^{-2}[S]=LMLT−2​=MT−2

    So, Surface Tension →\to→ (ii).

  2. Final matching

    (a)→(iii),(b)→(i),(c)→(iv),(d)→(ii)(a)\to(iii), \quad (b)\to(i), \quad (c)\to(iv), \quad (d)\to(ii)(a)→(iii),(b)→(i),(c)→(iv),(d)→(ii)
  3. Compare with options

    This corresponds to Option A.

  4. Compare with stored correct answer

    Stored correct answer is A, which matches our derived answer.

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