Q. An object is being weighed on a spring balance moving around a curve of radius 100 m at a speed $7\ ms^{- 1}$. The object has a weight of 60 kg-wt. The reading registered on the spring balance would be

A) 60.075 kg-wt
B) 60.125 kg-wt
C) 60.175 kg-wt
D) 60.225 kg-wt
Explanation

(a) Here, $r = 100m,v = 7ms^{- 1},m = 60kg$
Reading registered= resultant force =?
Two force are acting, weight $mg$ and centripetal force $\frac{mv^{2}}{r}$ at $90{^\circ}$ to each other
$\therefore$ Resultant force $= \sqrt{(mg)^{2} + \left( \frac{mv^{2}}{r} \right)^{2}} = mg\left\lbrack 1 + \left( \frac{v^{2}}{rg} \right)^{2} \right\rbrack^{1/2}$
$= 60 \times 9.8\left\lbrack 1 + \left( \frac{7 \times 7}{100 \times 9.8} \right)^{2} \right\rbrack^{1/2}$
$= 60.075 \times 9.8$ N = 60.075 kg-wt

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.