50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m³, then find the rise in the water level in the tank.

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Solution

Total volume of water displaced = $(4 \times 50) m^{3} = 200 m^{3} Rise in water level = \frac{200}{40 \times 20} m = 0.25 m = 25 cm$

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50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m3, then find the rise in the water level in the tank.

Total volume of water displaced = (4 × 50) m3 = 200 m3Rise in water level = 20040 × 20 m = 0.25 m = 25 cm

50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m³, then find the rise in the water level in the tank.

Total volume of water displaced = $(4 \times 50) m^{3} = 200 m^{3} Rise in water level = \frac{200}{40 \times 20} m = 0.25 m = 25 cm$