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QJEEM16Ph5b55: JEE Main 2016 Physics - Class 11 JEEM16Ph5b55 Solution: Consider a water jar of radius R that has water filled up to…

JEE Main 2016 (Online) 9th April Morning Slot Physics - Properties of Matter Question 228 English
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 :

Multiple Choice Options

  • A x=r(HH+h)x = r\left( {{H \over {H + h}}} \right)
  • B x=r(HH+h)12x = r{\left( {{H \over {H + h}}} \right)^{{1 \over 2}}}
  • C x=r(HH+h)14x = r{\left( {{H \over {H + h}}} \right)^{{1 \over 4}}}
  • D x=r(HH+h)2x = r{\left( {{H \over {H + h}}} \right)^{{2}}}

Step-by-Step Verified Solution

v1 = velocity of water when it leak from hole

v2 = velocity of water when it reach the ground.

From Bernoulli's principle,

12ρv12+ρgh{1 \over 2}\rho {v_1}^2 + \rho gh = 12ρv22{1 \over 2}\rho {v_2}^2

⇒\Rightarrow     v12{v_1}^2 + 2gh = v22{v_2}^2

From Torricelli's theorem,

v1 = 2gH\sqrt {2gH}

∴\therefore     v22{v_2}^2 = 2gh + 22gH

From continuity equation,

A1v1{A_1}{v_1} = A2v2{A_2}{v_2}

⇒\Rightarrow     πr2×2gH\pi {r^2} \times \sqrt {2gH} = πx22g(h+H)\pi {x^2}\sqrt {2g\left( {h + H} \right)}

⇒\Rightarrow    x2  =  r2 HH+g\sqrt {{H \over {H + g}}}

⇒\Rightarrow    x = r (HH+g)14{\left( {{H \over {H + g}}} \right)^{{1 \over 4}}}

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