Question

A body of mass 25gm is under water at a depth of
50cm. If the specific gravily of material of body is s
the work necessary to lift the body slowly to
the the surface is
4) 98 *10^5erg​

Answer 1

Given that,

Mass of body = 25 gm

Depth = 50 cm

We need to calculate the work necessary to lift the body slowly to  the surface

Using formula of work done

$W=mgh$

Where, m = mass of body

g = acceleration due to gravity

h = depth

Put the value into the formula

$W=25\times980.6\times50$

$W=1225750\ erg$

$W=12.25\times10^{5}\ erg$

Hence, The work necessary to lift the body slowly to  the surface is $12.25\times10^{5}\ erg$

Answer 2

Answer:

Answer->0.098J ,or, 9.8×10^5erg

Explanation:

RD = (density of an object) / (density of water)

THEREFORE,

( Relative density of object / 5 )=density of water.

Apparent weight = Actual weight - Buoyant force

= mg - (mg/5)

=0.196

so, work done= 0.196×height

= 0.196×0.5

=0.098 J

=9.8×10^5 erg