Question

Three blocks are suspended as shown in the figure. If all the strings and pulleys are ideal and friction is absent everywhere, then the acceleration of the $500g$ block is

• A. $\frac{6g}{13}$ downwards
• B. $\frac{7g}{13}downwards$
• C. $\frac{8g}{13}downwards$
• D. $\frac{9g}{13}$ upwards

Answer 1

Answer: (C)

$500g-T=500a\dots \dots (1)$
$T-100gsin30^{\circ }-T^{\prime \prime }=100a$
Or, $T-T^{\prime \prime }-50g=100a\dots \dots (2)$

Again, $T^{\prime \prime }-50g=50a\dots \dots (3)$
From equations (2) and (3),
$T-100g=150a\dots \dots (4)$
Adding equations (1) and (4), $400g=650a$ or $a=\frac{400g}{650}=\frac{8g}{13}$
This acceleration is downwards.