Physics - NLM Question with Solution | TestHub

500g$-T=500\mathrm{a\; -(1)}$

$T-100\mathrm{g}\sin 30°-T^{\prime \prime }=100a$

or$T-T^{\prime \prime }-50\mathrm{g}=100\mathrm{a\; -(2)}$



Again,$T^{\prime \prime }-50\mathrm{g}=50\mathrm{a\; -(3)}$

From Eqs. (ii) and (iii)

$T-100\mathrm{g}=150\mathrm{a\; -(4)}$

Adding Eqs. (i) and (iv), 400g = 650$a$or$a=\frac{400\mathrm{g}}{650}=\frac{8\mathrm{g}}{13}$

This acceleration is downwards