Home
Class 11
PHYSICS
A mercury drop of radius 1 cm is sprayed...

A mercury drop of radius 1 cm is sprayed into `10^(6)` drops of equal size. The energy expended in joule is (surface tension of mercury is `(460xx10^(-3)N//m)`

A

0.057

B

5.7

C

`5.7xx10^(-4)`

D

`5.7xx10^(-6)`

Text Solution

Verified by Experts

The correct Answer is:
A
`W=4piR^(2)T(n^(1//3)-1)`
Promotional Banner

Topper's Solved these Questions

  • MECHANICAL PROPERTIES OF FLUIDS

    NARAYNA|Exercise EXERCISE - I - (C.W) (CAPILLARITY & CAPILLARY RISE)|10 Videos

  • MECHANICAL PROPERTIES OF FLUIDS

    NARAYNA|Exercise EXERCISE - I - (C.W) (EXCESS PRESSURE INSIDE A LIQUID DROP AND SOAP BUBBLE)|3 Videos

  • MECHANICAL PROPERTIES OF FLUIDS

    NARAYNA|Exercise EXERCISE - I - (C.W) (TERMINAL VELOCITY & VISCOUS FORCE)|2 Videos

  • MATHEMATICAL REVIEW & PHYSICAL WORLD

    NARAYNA|Exercise C.U.Q|13 Videos

  • MECHANICAL PROPERTIES OF SOLIDS

    NARAYNA|Exercise LEVEL-II (H.W)|24 Videos

Similar Questions

Explore conceptually related problems

A mercury drop of radius 1 cm is sprayed into 10^(6) drops of equil size. The energy expended in joule is (surface tension of mercury is (460xx10^(-3)N//m)

A mercury drop of radius 1 cm is sprayed into 10^(6) dropletes of equal size. Calculates the energy expanded if surface tension of mercury is 35 xx 10^(-3) N//m .

A mercury drop of radius 1.0 cm is sprayed into 10^(6) droplets of equal sizes. The energy expended in this process is (Given, surface tension of mercury is 32 xx 10^(-2) Nm^(-1) )

A menisus drop of radius 1 cm sprayed into 10^(5) droplets of equal size . Calculate the energy expended if surface tension of mercury is 435 xx10^(-3) N/m

A mercury drop of radius 1.0 cm is sprayed into 10^(8) droplets of equal size. Calculate the energy expanded. (Surface tension of mercury = 32 xx 10^(-2) N//m ).

A mercury drop of radius R is sprayed into n droplets of equal size. Calculate the energy expanded if surface tension of mercury is T .

A mecury drop of radius 1 cm is sprayed into 10^(5) droplets of equal size. Calculate the increase in surface energy if surface tension of mercury is 35xx10^(-3)N//m .

A mercury drop of radius 1 cm is broken into 10^6 droplets of equal size. The work done is (T=35xx10^-2(N)/(m) )