Question

A spherical ball of mass m and radius r is allowed to fall in a medium of viscosity. The time in which the velocity of the body increases from 0 to 0.63 times the terminal velocity is called the time constant. Dimensionally, it can be represented by
A. $\dfrac{{m{r^2}}}{6}$
B. $\dfrac{{mr}}{6}$
C. $\dfrac{m}{{6r}}$
D. none of these

Answer 1

Hint: As, the dimension of time constant is the dimension of time so find the dimensions of the given options by using the dimensions for example dimensionality for mass is =[M], dimensionality of radius is = [L], and dimensionality of gravity is = [LT$^{-2}$] and then see which has the same dimension of time.

Complete step-by-step answer:
As, it is given that a spherical ball of mass m and radius is r and is allowed to fall in a medium of viscosity. So when an object moves through a viscous medium such as air or water, the object will experience a velocity dependent retarding force and it may be proportional to the first or the second power of velocity.
Mass of body is = m
Radius of spherical body = r
Velocity increases = 0 to 0.63
As we know the dimensionality of time constant = [T]
Dimensionality of mass =[M]
Dimensionality of radius r = [L]
Dimensionality of gravity g = [LT$^{-2}$]
Now, we calculate the dimensionality of all the options.
For option A $\dfrac{{m{r^2}}}{6} = \left[ {M{L^2}} \right]$
For option B $\dfrac{{mr}}{6} = \left[ {ML} \right]$
For option C $\dfrac{m}{{6r}} = \dfrac{{\left[ M \right]}}{{\left[ L \right]}} = \left[ {M{L^{ - 1}}} \right]$
As we observe, no option has the same dimension as that of time constant. It means none of the options is correct.

 Hence, we can say that the D option is correct.

Note: Care must be taken for writing the dimensionality of all the entities. And these dimensional formulas always write in the square brackets. After putting the dimensionalities of all the entities then solve their powers carefully otherwise, we will not get the correct results.