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Q.

One end of a horizontal oil pipe is closed using a valve and the reading of the barometer attached to the pipe is $2 \times 10^{5} Pa$ . When the valve in the pipe is opened the reading of the barometer falls to $10^{5} Pa$ . The velocity of oil flowing through the pipe will be (density of oil = 800 kg m^-3) in ms^–1

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a

5

b

10

c

2.5

d

$5 \sqrt{10}$

answer is D.

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Detailed Solution

$\begin{array}{l} P_{1} = 2 \times 10^{5} Pa; V_{1} = 0; V_{2} = ? \\ P_{1} - P_{2} = \frac{ρ}{2} (V_{2}^{2} - V_{1}^{2}); 2 \times 10^{5} - 10^{5} = \frac{800}{2} (V_{2}^{2}) \\ 10^{5} = 400 (V_{2}^{2}) \Rightarrow \frac{10^{3}}{4} = V_{2}^{2} \Rightarrow \frac{1000}{4} = V_{2}^{2} \\ V_{2} = 5 \sqrt{10} m / s \end{array}$

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