Question

A spherical body of mass m and radius r is allowed to fall in a medium of viscosity η. The time in which the velocity of the body increases from zero to 0.63 times the terminal velocity (v) is called time constant τ. Dimensionally τ can be represented by
(a) $\frac{mr^{2}}{6\pi\eta}$(b) $\sqrt{ \bigg \lgroup \frac{6\pi mr \eta}{g^{2}}\bigg \rgroup}$(c) $\frac{m}{6\pi\eta r v}$
(d) None of these

Answer 1

as no option has dimension of time so answer is

none of these

Answer 2

Answer:

D) None of these

Explanation:

When an object moves through a viscous medium such as air or water, the object experiences a velocity dependent retarding force. The retarding force may be proportional to the first (linear) or the second power (quadratic) of the velocity.

Mass of the spherical body = m

Radius of the spherical body = r

Velocity increase = 0 to 0.63

The movement of an object moving through a viscous medium with the differential equation -  m.dV/dt = mg - kV

where m is the mass of the object,  k is the scaling factor that accounts for the area experiencing the viscous force, dimensional formula is MT.

Time constant = t  =  [T]

Viscosity η  =  {ML~-1T~-1]

For options (1), (2) and (3) dimensions are not matching with time constant. hence the answer will be none of the above.