JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 14)
Match List I with List II
Choose the correct answer from the options given below:
| LIST I | LIST II | ||
|---|---|---|---|
| A. | Angular momentum | I. | $\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]$ |
| B. | Torque | II. | $\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right]$ |
| C. | Stress | III | $\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]$ |
| D. | Pressure gradient | IV. | $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$ |
Choose the correct answer from the options given below:
A - I, B - IV, C - III, D - II
A - III, B - I, C - IV, D - II
A - IV, B - II, C - I, D - III
A - II, B - III, C - IV, D - I
Explanation
$\vec{L}=\vec{r} \times \vec{p} \Rightarrow[\mathrm{L}]=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\right]\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\right]$
$$ =\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right] $$
$$ \begin{aligned} \vec{\tau}=\vec{r} \times \vec{F} \Rightarrow[\tau] & =\left[\mathrm{L}^{1}\right]\left[\mathrm{MLT}^{-2}\right] \\\\ & =\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right] \end{aligned} $$
Stress $\equiv$ Pressure $=\frac{F}{A} \Rightarrow[$ Stress $]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$
Pressure Gradient $=\frac{d P}{d x} \Rightarrow[$ Pressure Gradient $]$ $$ =\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right] $$
$$ =\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right] $$
$$ \begin{aligned} \vec{\tau}=\vec{r} \times \vec{F} \Rightarrow[\tau] & =\left[\mathrm{L}^{1}\right]\left[\mathrm{MLT}^{-2}\right] \\\\ & =\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right] \end{aligned} $$
Stress $\equiv$ Pressure $=\frac{F}{A} \Rightarrow[$ Stress $]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$
Pressure Gradient $=\frac{d P}{d x} \Rightarrow[$ Pressure Gradient $]$ $$ =\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right] $$
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