A hemispherical tank full of water is emptied by a pipe at the rate of 3 4 7 litres per second. How much time will it take to empty half the tank , if it is 3 m in diameter?

Radius of base of tank = D 2 = 3 2 m Volume of tank = 2 3 πr 3 = 2 3 × 22 7 × 3 2 3 = 99 14 m 3 = 99 14 × 1000 l i t r e s = 99000 14 l i t r e s Volume of water to be emptied = 1 2 × volume of tank = 1 2 × 99000 14 l i t r e s = 99000 28 l i t r e s Time taken to empty 3 4 7 = 25 7 l i t r e s = 1 s e c Time taken to empty 1 l i t r e s = 7 25 s e c Time taken to empty 99000 28 l i t r e s = 99000 28 × 7 25 = 69300 700 = 990 s e c = 990 60 m i n = 16 . 5 m i n Time taken to empty half tank is 16 minute and 5 second.

A cone of base radius 4 cm is divided into two parts by drawing a plane through the mid-point of its height and parallel to its base. Compare the volume of the two parts.

Given, radius of base of cone = 4 cm The cone is divided in two parts through the height of the cone. Let, the height of the cone is h cm . So, the volume of the cone = 1 3 × π × 4 2 × h cm 3 = 16 πh 3 cm 3 ∴ Height of the smaller cone is h 2 cm and radius is 4 2 cm = 2 cm ∴ Volume of smaller cone = 1 3 π × 2 2 × h 2 cm 3 = 2 πh 3 cm 3 ∴ Volume of lower frustum of the cone = Total volume of cone - Volume of shorter cone = 16 πh 3 - 2 πh 3 cm 3 = 16 πh - 2 πh 3 cm 3 = 14 πh 3 cm 3 ∴ Ratio of two parts of the cone = 2 πh 3 : 14 πh 3 = 2 : 14 = 1 : 7 The ratio of volumes of two parts of the cone is 1 : 7 .