JEE MAIN - Physics (2011 - No. 19)
Water is flowing continuously from a tap having an internal diameter $$8 \times {10^{ - 3}}\,\,m.$$ The water velocity as it leaves the tap is $$0.4\,\,m{s^{ - 1}}$$ . The diameter of the water stream at a distance $$2 \times {10^{ - 1}}\,\,m$$ below the tap is close to :
$$7.5 \times {10^{ - 3}}m$$
$$9.6 \times {10^{ - 3}}m$$
$$3.6 \times {10^{ - 3}}m$$
$$5.0 \times {10^{ - 3}}m$$
Explanation
From Bernoulli's theorem,
$${P_0} + {1 \over 2}\rho v_1^2\rho gh = {P_0} + {1 \over 2}\rho v_2^2 + 0$$
$${v_2} = \sqrt {v_1^2 + 2gh} $$
$$ = \sqrt {0.16 + 2 \times 10 \times 0.2} $$
$$ = 2.03\,m/s$$
From equation of continuity
$${A_2}{v_2} = {A_1}{v_1}$$
$$\pi {{D_2^2} \over 4} \times {v_2} = \pi {{D_1^2} \over 4}{v_1}$$
$$ \Rightarrow \,\,{D_1} = {D_2}\sqrt {{{{v_1}} \over {{v_2}}}} = 3.55 \times {10^{ - 3}}m$$
$${P_0} + {1 \over 2}\rho v_1^2\rho gh = {P_0} + {1 \over 2}\rho v_2^2 + 0$$
$${v_2} = \sqrt {v_1^2 + 2gh} $$
$$ = \sqrt {0.16 + 2 \times 10 \times 0.2} $$
$$ = 2.03\,m/s$$
From equation of continuity
$${A_2}{v_2} = {A_1}{v_1}$$
$$\pi {{D_2^2} \over 4} \times {v_2} = \pi {{D_1^2} \over 4}{v_1}$$
$$ \Rightarrow \,\,{D_1} = {D_2}\sqrt {{{{v_1}} \over {{v_2}}}} = 3.55 \times {10^{ - 3}}m$$
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