Q. If $F = 6\pi\eta^{a}r^{b}v^{c}$, Where $F =$viscous force $\eta = \$coefficient of viscosity $r = \$radius of spherical body $v = \$terminal velocity of the body. Find the values of $a,b$ and $c$.
A) $a = 1,b = 2,c = 1$
B) $a = 1,b = 1,c = 1$
C) $a = 2,b = 1,c = 1$
D) $a = 2,b = 1,c = 2$
Explanation
According to homogeneity principle LHS=RHS Or $F = 6\pi\eta^{a}r^{b}v^{c}$ Or $\left\lbrack MLT^{- 2} \right\rbrack = \left\lbrack ML^{- 1}T^{- 1} \right\rbrack^{a}\lbrack L\rbrack^{b}\left\lbrack LT^{- 1} \right\rbrack^{c}$ Or $\left\lbrack M^{1}L^{1}T^{- 2} \right\rbrack = \lbrack M^{0}L^{- a + b + c}T^{- a - c}\ \rbrack$ $\therefore\ \ \ a = 1, - a + b + c = 1$ and $- a - c = - 2$ After solving, we get $a = 1,b = 1,c = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.