Question

Two blocks of masses 8 kg and 12 kg are connected at the two ends of a light inextensible string. The string passes over a frictionless pulley. The acceleration of the system is:

• A. $\frac{g}{4}$
• B. $\frac{g}{5}$
• C. $\frac{g}{8}$
• D. $\frac{g}{6}$

Answer 1

Answer: (B)

$\frac{g}{5}$

Let a be the common acceleration of the system and T be tension of the string.
The equations of motion of two blocks are
$T-8g=8a$ ........(i)

And $12g-T=12a$ ......(ii)

Adding (i) and (ii), we get
$4g=20a$

or $a=\frac{g}{5}$