Question

A pump is required to lift 1000 kg of water per minute from a well 20 m deep and eject it at a rate of 20 m/s.

(x) How much work is done in ejecting water?

(y) How much work is done in lifting water?

(z) What HP (horse power) engine is required for the purpose of lifting water?

(1) 2 $\times$${10}^5$ $\frac{j}{min}$ (2) 4 $\times$${10}^5$ $\frac{j}{min}$ (3) 3 $\times$${10}^5$ $\frac{j}{min}$

(4) 6.6 $\times$${10}^3$ $\frac{j}{s}$ (5) 8 $\times$${10}^4$ watts (6)${10}^5$ watts

Take g = $\frac{10m}{s^2}$

• A. x - 1, y - 2, z - 5
• B. x - 2, y - 1, z - 6
• C. x - 1, y - 1, z - 4
• D. x - 1, y - 1, z - 6

Answer 1

The correct option is C

x - 1, y - 1, z - 4

Work done in lifting water = gain in PE (Potential Energy)

Work = 1000 $\times$ g $\times$ 20 = 1.96 $\times$ ${10}^5$ J per minute

Work done (per minute) in giving it KE = $\frac{1}{2}$m$v^2$ = $\frac{1}{2}$(1000)${20}^2$ = 2 *${10}^5$ j per minute

Power of the engine = work per second = $\frac{1}{6}$(1.96 + 2) * ${10}^5$J = 6.6 * ${10}^3$ (watts)

Since 1HP = 746 W, Hence, Power required = 8.85 HP