Q.
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) $a_{1} = a_{2},a_{1} = a_{3}$
Explanation
(b) For $A$, $T = f = 2mg$
$2mg - mg = ma_{1}$
$\therefore\ a_{1} = g$
[image:5_image166.emf]
For $B$,
From force diagram shown in figure,
[image:5_image166.emf]
$2mg - mg = 3ma_{2}$
$a_{2} = \frac{g}{3}$
For $C$,
$\therefore 2mg - mg = 2ma_{3}$
$\therefore\ a_{3} = \frac{g}{2}$
[image:5_image167.jpeg]
So, $a_{1} > a_{3} > a_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.