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Units and Measurements question

2023 · 30 Jan · Shift 2 · Q46
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  5. /2023 · 30 Jan · Shift 2 · Q46

Units and Measurements question

2023 · 30 Jan · Shift 2 · Q46

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1

Match List I with List II:

List I List II
A. Torque I. kgm−1 s−2\mathrm{kg} \mathrm{m}^{-1} \mathrm{~s}^{-2}kgm−1 s−2
B. Energy density II. kg ms−1\mathrm{kg} \,\mathrm{ms}^{-1}kgms−1
C. Pressure gradient III. kg m−2 s−2\mathrm{kg}\, \mathrm{m}^{-2} \mathrm{~s}^{-2}kgm−2 s−2
D. Impulse IV. kg m2 s−2\mathrm{kg} \,\mathrm{m}^{2} \mathrm{~s}^{-2}kgm2 s−2

Choose the correct answer from the options given below:

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

Correct answer: A

  1. Find the dimensions/units of each quantity in List I

A. Torque

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

Since, Force=kg m s−2\text{Force} = \mathrm{kg\, m\, s^{-2}}Force=kgms−2

Therefore, Torque=kg m s−2×m=kg m2 s−2\text{Torque} = \mathrm{kg\, m\, s^{-2}} \times \mathrm{m} = \mathrm{kg\, m^2\, s^{-2}}Torque=kgms−2×m=kgm2s−2

So, A→IVA \to IVA→IV

B. Energy density

Energy density =EnergyVolume= \dfrac{\text{Energy}}{\text{Volume}}=VolumeEnergy​

Now, Energy=kg m2 s−2\text{Energy} = \mathrm{kg\, m^2\, s^{-2}}Energy=kgm2s−2 Volume=m3\text{Volume} = \mathrm{m^3}Volume=m3

Hence, Energy density=kg m2 s−2m3=kg m−1 s−2\text{Energy density} = \dfrac{\mathrm{kg\, m^2\, s^{-2}}}{\mathrm{m^3}} = \mathrm{kg\, m^{-1}\, s^{-2}}Energy density=m3kgm2s−2​=kgm−1s−2

So, B→IB \to IB→I

C. Pressure gradient

Pressure gradient =Pressuredistance= \dfrac{\text{Pressure}}{\text{distance}}=distancePressure​

Pressure is, Pressure=ForceArea=kg m s−2m2=kg m−1 s−2\text{Pressure} = \dfrac{\text{Force}}{\text{Area}} = \dfrac{\mathrm{kg\, m\, s^{-2}}}{\mathrm{m^2}} = \mathrm{kg\, m^{-1}\, s^{-2}}Pressure=AreaForce​=m2kgms−2​=kgm−1s−2

Therefore, Pressure gradient=kg m−1 s−2m=kg m−2 s−2\text{Pressure gradient} = \dfrac{\mathrm{kg\, m^{-1}\, s^{-2}}}{\mathrm{m}} = \mathrm{kg\, m^{-2}\, s^{-2}}Pressure gradient=mkgm−1s−2​=kgm−2s−2

So, C→IIIC \to IIIC→III

D. Impulse

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

Thus, Impulse=kg m s−2×s=kg m s−1\text{Impulse} = \mathrm{kg\, m\, s^{-2}} \times \mathrm{s} = \mathrm{kg\, m\, s^{-1}}Impulse=kgms−2×s=kgms−1

So, D→IID \to IID→II

  1. Final matching

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

This corresponds to Option A.

  1. Comparison with stored answer

Stored correct answer = A

My derived answer = A

So, they agree.

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