Question

Calculate the work done in raising a stone of mass 5kg and specific gravity 3, lying at the bed of a lake through a hight of 4m.

Ans=130.7 J

Answer 1

Loss of weight in water = weight in air / specific gravity = 5/3

Weight of stone in water = 5-5/3 = 10/3

Force = 10/3 x 9.8 = 98/3

Work done = 5 x 98/3 = 163.3

Answer 2

$\maltese \: \red{\mid{\underline{\overline{\textbf{Answer}}}\mid}}$

Archimedes principle says that an object immersed in liquid faces a loose in its weight which is equal to weight of liquid displaced by it.

Weight of water displaced = loss of weight of object in water.

By this principle we can say :

Relative density = Weight of object in air / loss of weight of object in water.

Loss of weight of object in water = Weight of object in air / Specific gravity.

After inputting the values for this case we, get :

$\maltese \: loss \: of \: weigth = \frac{5}{3}$

Now it's weight in water = Weight in air - loss of weight.

$\maltese \: weight \: in \: water \: = 5 - \frac{5}{3} = \frac{10}{3}$

Now the force on this object will be :

$\maltese \: f = \frac{10}{3} \times 9.8 = \: \frac{98}{3}$

It is given object is displaced by 4m.Now the work done :

$\maltese \: w = fs$

$\maltese \: w = \frac{98}{3} \times 4$

$\maltese \: w = 130.7joule$

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BY SCIVIBHANSHU

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