Home
Class 11
PHYSICS
The work done in splitting a drop of wat...

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)`)

A

`9.58 xx10^(-5) J`

B

`8.95 xx10^(-5) J`

C

`5.89 xx10^(-5) J`

D

`5.98 xx10^(-6) J`

Text Solution

Verified by Experts

The correct Answer is:
B
`4/3 piR^3 =n4/3pi(r)^3implies r=R/(n^(1//3))`
`W=4piT[n(R/(n^(1//3)))^2 -R^2]=4piR^2T[n^(1//3)-1]`
`=4pixx(1xx10^(-3))^2 xx72 xx10^(-3) (10^(6//3) -1) =8.95 xx10^(-5) J`
Promotional Banner

Similar Questions

Explore conceptually related problems

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))

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

The diameter of one drop of water is 0.2 cm. The work done in breaking one drop into 1000 equal droplets will be :- (surface tension of water =7xx10^(-2)N//m )

Calculate the work done in spraying a spherical drop of mercury of 1 mm radius into a million droplets of equal size. The surface tension of mercury =550 xx10^(-3) Nm^(-1)

The work done is breaking a water drop of radius 1 cm into 64 equal droplets (Surface tension of water is 0.07 N/m)

The diameter of one drop of water is 0.2 cm. The work done in breaking one drop into 1000 droplets will be (Given, S_("water") = 7 xx 10^(-2) Nm^(-1) )

The surface tension of water is 7xx10^(-2) N/m . The work required to break a drop of water of radius 0.5 cm into identical drops , each of radius 1 mm is

The radius of a spherical drop of water is 1 mm. If surface tension of water be 70 xx 10^(-3) Nm^(-1) , the pressure difference between inside and outside the drop will be

Work done in blowing a soap bubble of diameter 2 cm , is (S.T = 3xx10^(-2) N/m)