A bucket full of water weighs 5 kg , it is pulled from a well 20 m deep. There is a small hole in the bucket through which water leaks at a constant rate. If it is observed that for every metre the bucket loses 0.2 kg mass of water, then the total work done in pulling the bucket up from the well is [ g=10 m s- 2 ]

Question

A bucket full of water weighs 5 kg , it is pulled from a well 20 m deep. There is a small hole in the bucket through which water leaks at a constant rate. If it is observed that for every metre the bucket loses 0.2 kg mass of water, then the total work done in pulling the bucket up from the well is [ g=10 m s- 2 ] (A) 600 J (B) 400 J (C) 100 J (D) 500 J

Tardigrade Admin · Accepted Answer

The weight of bucket when it has been pulled up a distance

x i s (5 - 0.2 x)

Hence, the required work is

W = \int_{x = 20}^{x = 0} - (5 - 0.2 x) \times 10 \times d x

= [50 x]_{x = 0}^{x = 20} - [2 \frac{x ^{2}}{2}]_{x = 0}^{x = 20}

W = 50 \times 20 - (20)^{2} = 600 J

$600 J$

$400 J$

$100 J$

$500 J$

The weight of bucket when it has been pulled up a distance $x i s (5 - 0.2 x)$
Hence, the required work is
$W = \int_{x = 20}^{x = 0} - (5 - 0.2 x) \times 10 \times d x$
$= [50 x]_{x = 0}^{x = 20} - [2 \frac{x ^{2}}{2}]_{x = 0}^{x = 20}$
$W = 50 \times 20 - (20)^{2} = 600 J$

600J

400J

100J

500J

The weight of bucket when it has been pulled up a distance xis(5−0.2x) Hence, the required work is W=∫x=20x=0​−(5−0.2x)×10×dx =[50x]x=0x=20​−[22x2​]x=0x=20​ W=50×20−(20)2=600J

$600 J$

$400 J$

$100 J$

$500 J$

The weight of bucket when it has been pulled up a distance $x i s (5 - 0.2 x)$
Hence, the required work is
$W = \int_{x = 20}^{x = 0} - (5 - 0.2 x) \times 10 \times d x$
$= [50 x]_{x = 0}^{x = 20} - [2 \frac{x ^{2}}{2}]_{x = 0}^{x = 20}$
$W = 50 \times 20 - (20)^{2} = 600 J$