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Q. 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 $500 \, g$ block is

Question

NTA AbhyasNTA Abhyas 2020Laws of Motion

Solution:

$500g-T=500a\ldots \ldots (1)$
$T-100gsin 30^\circ -T^{′′}=100a$
Or, $T-T^{′′}-50g=100a\ldots \ldots (2)$
Solution
Again, $T^{′′}-50g=50a\ldots \ldots (3)$
From equations (2) and (3),
$T-100g=150a\ldots \ldots (4)$
Adding equations (1) and (4), $400g=650a$ or $a=\frac{400 g}{650}=\frac{8 g}{13}$
This acceleration is downwards.