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JEE Main 202331 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
31 January
Shift
Evening Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2023 · 31 January · Evening Shift · Physics · Units and Dimensions

Match List I with List II List I List II A Angular momentum I [ ML...

Match List I with List II

List I List II
A Angular momentum I [ML2T−2]\left[{ML}^{2}{T}^{- 2}\right]
B Torque II \left[{ML}^{- 2}{T}^{- 2}\right]</td></tr><tr><td>C</td><td>Stress</td><td>III</td><td>[ML2T−1]</td> </tr> <tr> <td>C</td> <td>Stress</td> <td>III</td> <td>\left[{ML}^{2}{T}^{- 1}\right]
D Pressure gradient IV [ML−1T−2]\left[{ML}^{- 1}{T}^{- 2}\right]

Choose the correct answer from the options given below :

Options

  1. AA-I, B-IV, C-III, D-II
  2. BA-III, B-I, C-IV, D-II
  3. CA-II, B-III, C-IV, D-I
  4. DA-IV, B-II, C-I, D-III

Correct answer

B. A-III, B-I, C-IV, D-II

Step-by-step solution

As we know, angular momentum is given by, L→=r→×p→⇒[L]=[M0L1T0][M1L1T−1]\overset{\to}{L} = \overset{\to}{r} \times \overset{\to}{p} ⇒ \left[L\right] = \left[{M}^{0}{L}^{1}{T}^{0}\right]\left[{M}^{1}{L}^{1}{T}^{- 1}\right]

=[M1L2T−1]= \left[{M}^{1}{L}^{2}{T}^{- 1}\right]

Torque is given by,

τ→=r→×F→⇒[τ]=[L1][MLT−2]\overset{\to}{τ} = \overset{\to}{r} \times \overset{\to}{F} ⇒ \left[τ\right] = \left[{L}^{1}\right]\left[{MLT}^{- 2}\right]

=[ML2T−2]= \left[{ML}^{2}{T}^{- 2}\right]

Also, Stress ≡≡ Pressure =FA= \dfrac{F}{A}

Hence, [Stress] =[ML−1T−2]= \left[{ML}^{- 1}{T}^{- 2}\right]

Pressure Gradient =dPdx= \dfrac{dP}{dx}

Therefore, [Pressure Gradient]=[ML−2T−2]= \left[{ML}^{- 2}{T}^{- 2}\right]

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