A cubical vessel of height $1$ m is full of water. The minimum work done in taking water-out from vessel will be

5000 J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1000 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $5000 J$
Given,
$l = 1 m$
$g = 10 m / s$
$ρ = 1000 k g / m^{3}$
The minimum work done is equal to the change in potential energy.
$W = F i n a l p o t e n t i a l e n e r g y - i n i t i a l p o t e n t i a l e n e r g y$
$W = m g l - \frac{m g l}{2} = \frac{m g l}{2}$
$W = \frac{A l ρ g h l}{2}$ ( $m = ρ A l$ )
$W = \frac{1^{2} \times 1 \times 1000 \times 10 \times 1 \times 1}{2}$
$W = 5000 J$
The correct option is A.

Suggest Corrections