Question

The expression for centripetal force $(F)$ depends upon mass of body $(m)$, speed $(v)$ of the body and the radius ( $r$ ) of circular path will be expression for centripetal force

• A. $F=\frac{m v^{2}}{2 r^{3}}$
• B. $F=\frac{m v^{2}}{r}$
• C. $F=\frac{m v^{2}}{r^{2}}$
• D. $F=\frac{m^{2} v^{2}}{2 r}$

Answer 1

Answer: (B)

According to question,
$F \propto m^{ a } v^{b} r^{c}$
$F=k m^{a} v^{b} r^{c}$
$k$, being a dimensionless constant.
From homogeneity of dimensions,
$LHS = RHS$
$\left[ MLT ^{-2}\right] =[ M ]^{a}\left[ LT ^{-1}\right]^{b}[ L ]^{c}$
Or $\left[ MLT ^{-2}\right] =\left[ M ^{a} L ^{b+ c} T ^{-b}\right]$
Comparing the powers, we obtain
$a=1$
$b+c=1$
$-b=-2 \Rightarrow b=2$
$\therefore 2+c=1$
$\Rightarrow c=-1$
Therefore, $F=k m v^{2} r^{-1}=\frac{k m v^{2}}{r}$.
The experimental value of $k$ is found to be here
$\therefore F=\frac{m v^{2}}{r}$