Question

Sixteen beads in a string are placed on a smooth inclined plane of inclination sin${}^{-1}$ (1/3) such that some of them lie along the incline whereas the rest hang over the top of the plane. If acceleration at first bead is g/2, the arrangement of beads is that:

• A. 12 hang vertically
• B. 10 lie along inclined plane
• C. 8 lie along inclined plane
• D. 10 hang vertically

Answer 1

Answer: (D)

10 hang vertically

If n balls each of mass ma are hanging vertically then, $nmg-T=nma=\frac{nmg}{2}$
or $T=\frac{nmg}{2}$ .... (i)
also $T-(16-n)mgsin\theta$
$=(16-n)mg/2$
on $\frac{nmg}{2}-(16-n)mg\frac{1}{3}=(16-n)\frac{mg}{2}$
(using (i))
or $n\frac{4}{3}=\frac{80}{6}$
or $n=\frac{80}{6}\times \frac{3}{4}=10$