The work done when system raises a column of water of radius 5.0mm through 10 cm is

Dear Student,

Please find below the solution to the asked query. 

You know that work done is given as:

w = -Mgh
​Where w = work done
M = mass
g = acceleration due to gravity = 9.8 m/s² = 980 cm/s²
h = height 

And M = v x d
where v = volume
d = density of water = 1 g/cm³

So, M = $\pi r^{2}l \times d$
here r = radius = 5 mm = 0.5 cm
l = length

So, M = 3.14 x (0.5)² x 10 x 1

Thus, w = -Mgh
             = $- 3.14 x (0.5{)}^2 \times 10 \times 1\times 980\times \frac{10}{2}$  (the centre of mass of column lies half way along its length 'l')
            = - 3.85 x 10³ Joule

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Akanksha Jain