Three blocks are connected by massless strings on a frictionless inclined plane of $30^{\circ}$ as shown in the figure. A force of $104 \mathrm{~N}$ is applied upward along the incline to mass $m_3$ causing an upward motion of the blocks. What is the acceleration of the blocks? (Assume, acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

• A

  $6.0 \mathrm{~m} / \mathrm{s}^2$
• B

  $4.5 \mathrm{~m} / \mathrm{s}^2$
• C

  $3.0 \mathrm{~m} / \mathrm{s}^2$
• Correct answer
  D

  $1.5 \mathrm{~m} / \mathrm{s}^2$

Solution

The given situation is shown in the figure below,
If the system moves upward with acceleration $a \mathrm{~m} / \mathrm{s}^2$ on the application of $104 \mathrm{~N}$, then the equation of motion for $8 \mathrm{~kg}$ block is $$ \begin{aligned} & 104-T_1-8 g \sin 30^{\circ}=8 a \\ & \Rightarrow \quad 104-T_1-8 \times 10 \times \frac{1}{2}=8 a \quad\left[\because g=10 \mathrm{~m} / \mathrm{s}^2\right] \\ & \quad 64-T_1=8 a \end{aligned} $$ Equation of motion for $5 \mathrm{~kg}$ block is $$ \begin{array}{r} T_1-T_2-5 g \sin 30^{\circ}=5 a \\ T_1-T_2-25=5 a \end{array} $$ Equation of motion for $3 \mathrm{~kg}$ block is $$ \begin{aligned} T_2-3 g \sin 30^{\circ} & =3 a \\ T_2-15 & =3 a \end{aligned} $$ Adding Eqs. (i), (ii) and (iii), we get $$ 24=16 a \Rightarrow a=1.5 \mathrm{~m} / \mathrm{s}^2 $$ Hence, acceleration $a$ of block is $1.5 \mathrm{~m} / \mathrm{s}^2$.

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