Fluid Mechanics

JEE Main level, single correct, medium

78 of 192

Question 78 of 192

A cylindrical vessel of 4040 cm radius is completely filled with water and its capacity is 528528 dm3^3 (dm : decimeter). The vessel is placed on a solid block of exactly same height as vessel. If a small hole is made at 7070 cm below the top of water level, then horizontal range of water falling on the ground in the beginning is __________ cm.

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Answer: (B) 1402140\sqrt{2}

Volume of the cylindrical vessel, V=528V = 528 dm3=528000^3 = 528000 cm3^3

Radius of the vessel, r=40r = 40 cm

Let the height of the vessel be HH.

V=πr2HV = \pi r^2 H

528000=227×(40)2×H528000 = \dfrac{22}{7} \times (40)^2 \times H

H=528000×722×1600=105H = \dfrac{528000 \times 7}{22 \times 1600} = 105 cm

The vessel is placed on a block of height equal to the vessel, so the height of the block is 105105 cm.

Depth of the hole from the top water level, h=70h = 70 cm

Height of the hole from the ground, y=Height of block+(H−h)y = \text{Height of block} + (H - h)

y=105+(105−70)=140y = 105 + (105 - 70) = 140 cm

Velocity of efflux, v=2ghv = \sqrt{2gh}

Time of flight of water, t=2ygt = \sqrt{\dfrac{2y}{g}}

Horizontal range, R=v×t=2gh×2yg=2hyR = v \times t = \sqrt{2gh} \times \sqrt{\dfrac{2y}{g}} = 2\sqrt{hy}

R=270×140=270×70×2=1402R = 2\sqrt{70 \times 140} = 2\sqrt{70 \times 70 \times 2} = 140\sqrt{2} cm

Answer: 1402140\sqrt{2}

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