Physics · Fluid Mechanics

JEE Main 2026 — 6 April, Evening Shift — Question 7

A cylindrical vessel of $40\mathrm{cm}$ radius is completely filled with water and its capacity is $528\mathrm{dm}^3$ (dm: decimeter). The vessel is placed on a solid block of exactly same height as vessel. If a small hole is made at $70\mathrm{cm}$ below the top of water level, then horizontal range of water falling on the ground in the beginning is ______ cm.

Option A:
$120\sqrt{2}$
Option B:
$140\sqrt{2}$
Correct
Option C:
$140\sqrt{3}$
Option D:
$120\sqrt{3}$

Answer: B

Step-by-step solution

First find height of vessel: volume = $\pi r^2 h = 528\times10^{-3}$ m³, r=0.4 m, so h = 0.528/(π×0.16) = 1.05 m = 105 cm. Hole depth from top = 70 cm, so height of hole above ground = (105-70)+105 = 140 cm. Velocity of efflux = $\sqrt{2g\times0.7}=\sqrt{14}$ m/s. Time to fall = $\sqrt{2\times1.4/10}=\sqrt{0.28}=\sqrt{28}/10$ s. Range = $\sqrt{14}\times\sqrt{28}/10 = \sqrt{392}/10 = 19.8/10=1.98$ m = 198 cm = $140\sqrt{2}$ cm.

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Physics
Chapter: Fluid Mechanics
Topic: Variation of Static Pressure Inside a Liquid