The work done in splitting a drop of water of 1 mm radius into `10^(6)` drops is (S.T. of water `=72xx10^(-3)J//m^(2)`)

The work done in splitting a drop of water of 1 mm radius into 64 identical droplets is (S.T. of water is 72xx10^(-3) j//m^2)

Work done in splitting a drop of water of 1 mm radius into 10^(6) droplets is (Surface tension of water= 72xx10^(-3)J//m^(2) )

Work done in splitting a drop of water of 1 mm radius into 10^6 droplets is (Surface tension of water= 72xx10^(-3)J//m^(2) )

Work done is splitting a drop of water of 1 mm radius into 64 droplets is (surface tension of water =72xx10^(-3)j m^(2))

Surface tension changes with temperature. Calculate the work done in breaking water drop of radius 1 mm to 1000 droplets. Surface tension of water = 72 x 10^(-3) N/m .

The surface tension of water = 72 xx10^(-3) jm^(-2) .The work done in splitting a drop of water of 1 mm radius into 64 droplets is

Calculate the amount of energy dissipated when 8 drops of water of 0.3 mm radius coalesce from one big drop surface tension of water is 7.2xx10^(-2)Nm^(-1) .