StemCET Logo

Physics Question 29 – JEE-MAIN 2026

Three masses m1=4 kg, m2=4 kg and m3=6 kg are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of T1/T2 is _______ (take g=10 m/s2)

Recognize the two main blocks of mass connected by the main string over the pulley.

Step 1: Determine System Acceleration✦ Active

Identify the effective masses on each side of the pulley. On the left, ML=m1=4 kg. On the right, MR=m2+m3=4+6=10 kg. Since MR>ML, the system accelerates such that MR moves down and ML moves up. The acceleration a of the system is given by:

a=(MRML)(MR+ML)g

Substitute the given values (g=10 m/s2):

a=(104)(10+4)×10=614×10=307 m/s2
Step 2: Calculate Tension T1○ Expand

Tension T1 is in the main string. Consider the forces on mass m1. It is accelerating upwards. Applying Newton's second law:

T1m1g=m1a

Solve for T1:

T1=m1(g+a)=4(10+307)=4(70+307)=4(1007)=4007 N
💡 Teacher's Secret Hint

Remember to consider the direction of acceleration when setting up the force equations.

Step 3: Calculate Tension T2 and the Ratio T1/T2○ Expand

Tension T2 is in the string connecting m2 and m3. Consider the forces on mass m3. It is accelerating downwards with acceleration a. Applying Newton's second law:

m3gT2=m3a

Solve for T2:

T2=m3(ga)=6(10307)=6(70307)=6(407)=2407 N

Finally, calculate the ratio T1/T2:

T1T2=40072407=400240=4024=53
💡 Teacher's Secret Hint

Ensure you apply Newton's second law correctly for each mass, considering its direction of motion.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.