JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 2)
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Surface tension | I. | $$\mathrm{kg~m^{-1}~s^{-1}}$$ |
| B. | Pressure | II. | $$\mathrm{kg~ms^{-1}}$$ |
| C. | Viscosity | III. | $$\mathrm{kg~m^{-1}~s^{-2}}$$ |
| D. | Impulse | IV. | $$\mathrm{kg~s^{-2}}$$ |
Choose the correct answer from the options given below :
A-IV, B-III, C-I, D-II
A-IV, B-III, C-II, D-I
A-I, B-I, C-III, D-IV
A-III, B-IV, C-I, D-II
Explanation
$$
\begin{aligned}
\text { (A) } \text { Surface Tension }=\frac{\mathrm{F}}{\ell} & =\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}}=\mathrm{ML}^{0} \mathrm{~T}^{-2} \\\\
& =\mathrm{kg\,s}^{-2}(\mathrm{IV})
\end{aligned}
$$
$$ \begin{aligned} & \text { (B) Pressure }=\frac{F}{\mathrm{~A}}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2} \\\\ & =\mathrm{kg} \,\mathrm{m}^{-1} \mathrm{~s}^{-2}(\mathrm{III}) \end{aligned} $$
$\begin{aligned} \text { (C) Viscosity } & =\frac{\mathrm{F}}{\mathrm{A}\left(\frac{\mathrm{dV}}{\mathrm{dz}}\right)}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2\left(\frac{\mathrm{LT}^{-1}}{\mathrm{~L}}\right)} \\\\ & =\mathrm{ML}^{-1} \mathrm{~T}^{-1}=\mathrm{kg} \,\mathrm{m}^{-1} \mathrm{~s}^{-1}(\mathrm{I})\end{aligned}$
$$ \begin{aligned} \text { (D) } \text { Impulse } & =\int F d t=\mathrm{MLT}^{-2} \times \mathrm{T} \\\\ & =\mathrm{MLT}^{-1}=\mathrm{kg\,ms}^{-1} \text { (II) } \end{aligned} $$
$$ \begin{aligned} & \text { (B) Pressure }=\frac{F}{\mathrm{~A}}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2} \\\\ & =\mathrm{kg} \,\mathrm{m}^{-1} \mathrm{~s}^{-2}(\mathrm{III}) \end{aligned} $$
$\begin{aligned} \text { (C) Viscosity } & =\frac{\mathrm{F}}{\mathrm{A}\left(\frac{\mathrm{dV}}{\mathrm{dz}}\right)}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2\left(\frac{\mathrm{LT}^{-1}}{\mathrm{~L}}\right)} \\\\ & =\mathrm{ML}^{-1} \mathrm{~T}^{-1}=\mathrm{kg} \,\mathrm{m}^{-1} \mathrm{~s}^{-1}(\mathrm{I})\end{aligned}$
$$ \begin{aligned} \text { (D) } \text { Impulse } & =\int F d t=\mathrm{MLT}^{-2} \times \mathrm{T} \\\\ & =\mathrm{MLT}^{-1}=\mathrm{kg\,ms}^{-1} \text { (II) } \end{aligned} $$
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