Question

An electric motor of efficiency 75% empties a cubical
tank of length 1 = 3m completely filled with water, in 5
minutes, then (g = 10 m/s2).
(A) the power input = 1.8 x 103W
(B) the power input = 600 H
(C) the power input = 2 x 104W
(D) None​

Answer 1

answer : option (A) the power input = 1.8 × 10³ W

volume of water = volume of cubical

tank = (l)³

= (3)³ = 27 m³

we know, density of water is 1000 kg/m³

so, mass of water in the tank, m = 1000 kg/m³ × 27m³

= 27000 kg

let after time t, mass of water through out the container , dm = (m/H)dh

where H = 3m

so, potential energy of water, E = dmgh

=(m/H)g ∫h.dh

= mg/H [ h²/2]^H_0

= mgH/2

= 27000kg × 10m/s² × 3m/2

= 40.5 × 10⁴ J

power required to empty the tank, P = E/t

= 40.5 × 10⁴/(5 × 60)

= 13.5 × 10² watt

now efficiency of motor is 75%

so, the power input = 13.5 × 10²/(75/100)

= 13.5 × 10² × 4/3

= 18 × 10² watt

= 1.8 × 10³

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Answer 2

$\huge\bold\purple{Answer:-}$

(A) the power input = 1.8 x 103W