JEE MAIN - Physics (2023 - 10th April Evening Shift - No. 26)
Figure below shows a liquid being pushed out of the tube by a piston having area of cross section $$2.0 \mathrm{~cm}^{2}$$. The area of cross section at the outlet is $$10 \mathrm{~mm}^{2}$$. If the piston is pushed at a speed of $$4 \mathrm{~cm} \mathrm{~s}^{-1}$$, the speed of outgoing fluid is __________ $$\mathrm{cm} \mathrm{s}^{-1}$$
_10th_April_Evening_Shift_en_26_1.png)
Answer
80
Explanation
By equation of continuity
$$ \begin{aligned} & \mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2 \\\\ & \mathrm{~V}_2=\frac{2 \times 4}{10 \times 10^{-2}}=80 \mathrm{~cm} / \mathrm{s} \end{aligned} $$
$$ \begin{aligned} & \mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2 \\\\ & \mathrm{~V}_2=\frac{2 \times 4}{10 \times 10^{-2}}=80 \mathrm{~cm} / \mathrm{s} \end{aligned} $$
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