Three blocks of masses $2 \mathrm{~m}, 4 \mathrm{~m}$ and 6 m are placed as shown in figure. If $\sin 37^{\circ}=\frac{3}{5}, \sin 53^{\circ}=\frac{4}{5}$ the acceleration of the system is

• Correct answer
  A

  $a=\frac{17}{30} g$
• B

  $a=\frac{13}{30} g$
• C

  $a=\frac{13}{15} g$
• D

  $a=\frac{15}{35} g$

Solution

$\begin{aligned} & F_{n e t}=(2 m+4 m+6 m) a \\ \Rightarrow \quad & 10 \mathrm{mg} \sin 53^{\circ}-2 \mathrm{mg} \sin 37^{\circ}=12 \mathrm{ma}\end{aligned}$
$\begin{aligned} & \Rightarrow \mathrm{a}=\frac{\left(10 \times \frac{4}{5}-2 \times \frac{3}{5}\right) g}{12} \\ & \therefore \mathrm{a}=\frac{17}{30} \mathrm{~g}\end{aligned}$

Hence, the correct answer is option: 'A'.