Question

Sixteen beads in a string are placed on a smooth inclined plane of inclination ${\sin }^{-1}\left(1/3\right)$ 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

The correct option is D 10 hang vertically

Let $n$ be the numbers of balls each of mass $m$ hanging vertically.
thus, $nma=nmg-T$
here $a=\frac{g}{2}$
$nm\frac{g}{2}=nmg-T\Rightarrow T=\frac{nmg}{2}...\left(1\right)$
Again, $T-(16-n)mg\sin \theta =(16-n)ma=(16-n)m\frac{g}{2}....\left(2\right)$
using (1), (2) becomes
$\frac{nmg}{2}=(16-n)mg(\frac{1}{3}+\frac{1}{2})$
or,$\frac{n}{2}=(16-n)\frac{5}{6}$
or, $6n=160-10n$
or, $16n=160\Rightarrow n=10$