A hemispherical tank is full of water. It is connected to a pipe that empties it at a rate of 7 litres per second. How much approximate time (in minutes) will it take to empty half the tank if the internal radius of the tank is 2.1 meters? (Use π = 22⁄7 and 1 m³=1000 liters)

Correct — D. At a constant 7 litres per second, the time depends on the volume to be drained and nothing else.

Cube the radius: r³ = 2.1³ = 9.261 m³
Hemisphere: (2⁄3) × (22⁄7) × 9.261 = (44⁄21) × 9.261 = 44 × 0.441 = 19.404 m³
Halve and convert: 9.702 m³ = 9702 litres

Divide by the rate: 9702 ÷ 7 = 1386 seconds
Convert to minutes: 1386 ÷ 60 = 23.1 ≈ 23 minutes → option (d)

Why the others are wrong

• (a)
  16 minutes — 16 minutes drains 16 × 60 × 7 = 6720 litres, well short of the 9702 litres in half the tank.
• (b)
  11 minutes — 11 minutes is about a quarter of the tank: a quarter holds 4851 litres, which takes 693 s = 11.55 minutes. The stem asks for half.
• (c)
  46 minutes — 46 minutes empties the whole tank: 19,404 litres ÷ 7 = 2772 s = 46.2 minutes. The stem stops at half, which is 23.

When a pipe drains at a fixed rate, time = volume ÷ rate. The shape of the tank matters for the volume and nowhere else.

A hemisphere of radius r holds (2⁄3)πr³, half of a sphere's (4⁄3)πr³.

Keep the units in step: 1 m³ = 1000 litres, and a rate in litres per second gives an answer in seconds, which must be divided by 60 for minutes.

Key facts

• Volume of a hemisphere = (2⁄3)πr³.
• 1 m³ = 1000 litres.
• With r = 2.1 m the full tank holds 19.404 m³, or 19,404 litres.
• At a constant rate, time = volume ÷ rate.

Common traps

• Draining the full tank and answering 46 minutes when the stem asks for half.
• Using the sphere's (4⁄3)πr³ for a hemisphere.
• Reading half the tank as half the depth. A hemisphere's volume is not proportional to its depth.

No directly related past PYQ was found.