Question

Three blocks of masses $2\text{kg}$, $4\text{kg}$ and $6\text{kg}$ are connected by strings and placed on a fricitionless incline of ${53}^{\circ }$ as shown. A force of $120\text{N}$ is applied upward along the incline on the $6\text{kg}$ block so that system accelerates. If the strings are ideal, the ratio tension in the strings $\frac{T_1}{T_2}$ will be $(g=10{\text{ms}}^{-2})$

• A. $1:1$
• B. $1:2$
• C. $1:3$
• D. $1:4$

Answer 1

The correct option is C $1:3$

It is given that the masses are,

$m_1=2\text{kg},m_2=4\text{kg},m_3=6\text{kg}$

hence we know that the acceleration of the system will be

$a_{sys}=\frac{\text{applied force}-\text{opposing force}}{\text{total mass}}$

Opposing force will be component of weight along the incline. So, opposing force$=12g\times \sin {53}^{\text{o}}$ down the incline.

Hence acceleration will be,
$a_{sys}=\frac{120-12\times 10\times \frac{4}{5}}{12}=\frac{24}{12}=2{\text{ms}}^{-2}$

By the free body diagram of the blocks, we get



Now writing force equation for the block of mass $m_1=2\text{kg}$

$T_1-m_{1}g\sin {53}^{\circ }=m_1{a}_{sys},T_1=2\times 2+2\times 10\times \frac{4}{5}=20N$

Now writing force equation for the block $m_3=6\text{kg}$

$F-T_2-m_{3}g\sin {53}^{\text{o}}=6a_{sys}\Rightarrow 120-T_2-6\times 10\times \frac{4}{5}=6\times 2$

We get $T_2=60\text{N}$ and $T_1=20\text{N}$

$\frac{T_1}{T_2}=\frac{1}{3}$

Hence option C is the correct answer.