JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 25)
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Consider a water jar of radius R that has water filled up to height H and is kept on astand of height h (see figure). Through a hole of radius r (r << R) at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is x. Then :
$$x = r\left( {{H \over {H + h}}} \right)$$
$$x = r{\left( {{H \over {H + h}}} \right)^{{1 \over 2}}}$$
$$x = r{\left( {{H \over {H + h}}} \right)^{{1 \over 4}}}$$
$$x = r{\left( {{H \over {H + h}}} \right)^{{2}}}$$
Explanation
v1 = velocity of water when it leak from hole
v2 = velocity of water when it reach the ground.
From Bernoulli's principle,
$${1 \over 2}\rho {v_1}^2 + \rho gh$$ = $${1 \over 2}\rho {v_2}^2$$
$$ \Rightarrow $$ $${v_1}^2$$ + 2gh = $${v_2}^2$$
From Torricelli's theorem,
v1 = $$\sqrt {2gH} $$
$$ \therefore $$ $${v_2}^2$$ = 2gh + 22gH
From continuity equation,
$${A_1}{v_1}$$ = $${A_2}{v_2}$$
$$ \Rightarrow $$ $$\pi {r^2} \times \sqrt {2gH} $$ = $$\pi {x^2}\sqrt {2g\left( {h + H} \right)} $$
$$ \Rightarrow $$ x2 = r2$$\sqrt {{H \over {H + g}}} $$
$$ \Rightarrow $$ x = r $${\left( {{H \over {H + g}}} \right)^{{1 \over 4}}}$$
v2 = velocity of water when it reach the ground.
From Bernoulli's principle,
$${1 \over 2}\rho {v_1}^2 + \rho gh$$ = $${1 \over 2}\rho {v_2}^2$$
$$ \Rightarrow $$ $${v_1}^2$$ + 2gh = $${v_2}^2$$
From Torricelli's theorem,
v1 = $$\sqrt {2gH} $$
$$ \therefore $$ $${v_2}^2$$ = 2gh + 22gH
From continuity equation,
$${A_1}{v_1}$$ = $${A_2}{v_2}$$
$$ \Rightarrow $$ $$\pi {r^2} \times \sqrt {2gH} $$ = $$\pi {x^2}\sqrt {2g\left( {h + H} \right)} $$
$$ \Rightarrow $$ x2 = r2$$\sqrt {{H \over {H + g}}} $$
$$ \Rightarrow $$ x = r $${\left( {{H \over {H + g}}} \right)^{{1 \over 4}}}$$
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