JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 3)
A pressure-pump has a horizontal tube of cross sectional area $$10 \mathrm{~cm}^{2}$$ for the outflow of water at a speed of $$20 \mathrm{~m} / \mathrm{s}$$. The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is :
[given: density of water $$=1000 \mathrm{~kg} / \mathrm{m}^{3}$$]
300 N
500 N
250 N
400 N
Explanation
$${F_w} = \rho A{v^2}$$
$$ = {10^3} \times 10 \times {10^{ - 4}} \times 20 \times 20$$
$$ = 400\,N$$
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