Question

In figure, the blocks $A, B$ and $C$ each of mass $m$ have acceleration $a_{1}, a_{2}$ and $a_{3}$ respectively. $F_{1}$ and $F_{2}$ are external forces of magnitude $2\, mg$ and $mg$ respectively. Then

• A. $a_{1}=a_{2}=a_{3}$
• B. $a_{1} > a_{3} > a_{2}$
• C. $a_{1}=a_{2}, a_{2}=a_{3}$
• D. $a_{1}=a_{2}, a_{1}=a_{3}$

Answer 1

Answer: (B)

For $A, T=f=2 m g $
$ 2 m g-m g =m a_{1} $
$\therefore a_{1}=g$
For $B$,
From force diagram shown in figure,

$2\,mg - mg = 3\, ma_2$
$a_2 = \frac{g}{3}$
For discussion of motion of the point mass, we assume that for $C$

$\therefore 2 m g-m g=2 m a_{3}$
$\therefore a_{3}=\frac{g}{2}$
So, $a_{1} > a_{3} > a_{2}$