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Question: A small steel ball of radius r is allowed to fall under gravity through a column of a viscous liquid...

A small steel ball of radius r is allowed to fall under gravity through a column of a viscous liquid of coefficient of viscosity. After some time the velocity of the ball attains a constant value known as terminal velocity vT.v_{T}. The terminal velocity depends on (i) the mass of the ball. (ii) η\eta (iii) r and (iv) acceleration due to gravity g. which of the following relations is dimensionally correct

A

vT∝mgηrv_{T} \propto \frac{mg}{\eta r}

B

vT∝ηrmgv_{T} \propto \frac{\eta r}{mg}

C

vT∝ηrmgv_{T} \propto \eta rmg

D

vT∝mgrηv_{T} \propto \frac{mgr}{\eta}

Answer

vT∝mgηrv_{T} \propto \frac{mg}{\eta r}

Explanation

Solution

Given vTv_{T}= terminal velocity = [LT−1]\lbrack LT^{- 1}\rbrack, m = Mass = [M], g

Acceleration due to gravity = [LT−2]\lbrack LT^{- 2}\rbrack

r = Radius = [L], η\eta = Coefficient of viscosity = [η]\lbrack\eta\rbrack

By substituting the dimension of each quantity we can check the accuracy of given formula vT∝mgηrv_{T} \propto \frac{mg}{\eta r}

∴\therefore [LT−1]=[M][LT−2][ML−1T−1][L]\lbrack LT^{- 1}\rbrack = \frac{\lbrack M\rbrack\lbrack LT^{- 2}\rbrack}{\lbrack ML^{- 1}T^{- 1}\rbrack\lbrack L\rbrack}= [LT−1]\lbrack LT^{- 1}\rbrack

L.H.S. = R.H.S. i.e., the above formula is Correct**.**