Physics Q34 – JEE-MAIN 2026

Home Physics Mechanical Properties of FluidsQuestion #261

A cylindrical vessel of 40 cm radius is completely filled with water and its capacity is 528 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 cm below the top of water level, then horizontal range of water falling on the ground in the beginning is _______ cm.

Clue 1 of 3: Fluid Dynamics and Projectile Motion

The problem involves two main physics concepts: the efflux velocity of water from a hole (Torricelli's Law) and the subsequent projectile motion of the water stream.

🧠 Full Solution Path

Step 1: Determine the dimensions of the vessel and the height of the hole✦ Active

Given radius $R = 40$ cm. The volume of the vessel is $V = 528 {dm}^{3}$ . Convert volume to cm $^{3}$ : $1 {dm}^{3} = 1000 {cm}^{3}$ , so $V = 528 \times 1000 = 528000 {cm}^{3}$ . The volume of a cylinder is $V = π R^{2} H_{v e s s e l}$ . We can find the height of the vessel $H_{v e s s e l}$ :

H_{v e s s e l} = \frac{V}{π R^{2}} = \frac{528000}{\frac{22}{7} \times (40)^{2}} = \frac{528000 \times 7}{22 \times 1600} = 105 cm

The hole is made at a depth $h_{1} = 70$ cm below the top water level. The vessel is placed on a solid block of exactly the same height as the vessel. Therefore, the height of the hole from the ground, $h_{2}$ , is:

h_{2} = H_{v e s s e l} - h_{1} + H_{b l o c k} = 2 H_{v e s s e l} - h_{1} = 2 (105) - 70 = 210 - 70 = 140 cm

Step 2: Calculate the velocity of efflux and the time of flight○ Expand

The velocity of efflux $v$ from the hole is given by Torricelli's Law, where $h_{1}$ is the height of the water above the hole:

v = \sqrt{2 g h_{1}} = \sqrt{2 g (70)}

The time $t$ taken for the water to fall a vertical distance $h_{2}$ to the ground is given by the equation of motion under gravity:

h_{2} = \frac{1}{2} g t^{2} ⟹ t = \sqrt{\frac{2 h_{2}}{g}} = \sqrt{\frac{2 (140)}{g}}

Step 3: Calculate the horizontal range○ Expand

The horizontal range $X$ is the product of the efflux velocity and the time of flight:

X = v \times t = \sqrt{2 g (70)} \times \sqrt{\frac{2 (140)}{g}}

Simplify the expression:

X = \sqrt{2 g (70) \times \frac{2 (140)}{g}} = \sqrt{4 \times 70 \times 140}

Further simplification:

X = \sqrt{4 \times 70 \times (2 \times 70)} = \sqrt{8 \times 70^{2}} = 70 \sqrt{8} = 70 \times 2 \sqrt{2} = 140 \sqrt{2} cm

💡 Teacher's Secret Hint

Ensure all units are consistent (e.g., cm for length, cm/s² for g if not using standard 9.8 m/s²).

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Physics Question 34 – JEE-MAIN 2026

Clue 1 of 3: Fluid Dynamics and Projectile Motion

🧠 Full Solution Path