JEE Main 2026 — Laws of Motion Question with Solution

Let the acceleration of the system be $a$. The total mass on the right side is $m_2+m_3=4+6=10$ kg, and the mass on the left side is $m_1=4$ kg. Since $10$ kg $>4$ kg, the right side accelerates downwards and the left side accelerates upwards.

The common acceleration $a$ of the system is given by:
$a=\frac{(m_2+m_3)-m_1}{m_1+m_2+m_3}g$
$a=\frac{10-4}{4+4+6}g=\frac{6}{14}g=\frac{3}{7}g$

For mass $m_1$, the equation of motion is:
$T_1-m_{1}g=m_{1}a$
$T_1=m_1(g+a)=4\left(g+\frac{3}{7}g\right)=4\left(\frac{10}{7}g\right)=\frac{40}{7}g$

For mass $m_3$, the equation of motion is:
$m_{3}g-T_2=m_{3}a$
$T_2=m_3(g-a)=6\left(g-\frac{3}{7}g\right)=6\left(\frac{4}{7}g\right)=\frac{24}{7}g$

The ratio of the tensions is:
$\frac{T_1}{T_2}=\frac{\frac{40}{7}g}{\frac{24}{7}g}=\frac{40}{24}=\frac{5}{3}$

Answer: $5/3$