Home
Class 12
PHYSICS
The time of oscillation of a small drop ...

The time of oscillation of a small drop of liquid under surface tension depends upon density `rho, ` radius r and surface tension, `s,` as `T alpha rho^a s^b r^c`, then the decending order of a, b and c is

A

a gt b gt c

B

a gt c gt b

C

b gt a gt c

D

c gt a gt b

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Topper's Solved these Questions

  • UNITS AND MEASUREMENT

    AAKASH SERIES|Exercise EXERCISE - II|61 Videos
  • UNITS AND MEASUREMENT

    AAKASH SERIES|Exercise PRACTICE EXERCISE|45 Videos
  • UNITS AND MEASUREMENT

    AAKASH SERIES|Exercise PRACTICE EXERCISE|45 Videos
  • SEMICONDUCTOR DEVICES

    AAKASH SERIES|Exercise EXERCISE - II|31 Videos
  • UNITS AND MEASUREMENTS

    AAKASH SERIES|Exercise EXERCISE -3|66 Videos

Similar Questions

Explore conceptually related problems

The liquid drop of density rho , radius r and surface tension o oscillates with time period T. Which of the following expression for T^2 is correct

If the time period t of the oscillation of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula t=sqrt(r^(2b)s^(c)d^(a//2)) . It is observed that the time period is directly proportional to sqrt((d)/(s)) . The value of b should therefore be :