Question

Two masses \( M_1 \) and \( M_2 \) are accelerated uniformly on a frictionless surface as shown in figure. The ratio of the tensions \( \left( \frac{T_1}{T_2} \right) \) is

• A. \( \frac{M_2}{M_1} \)
• B. \( \frac{M_1}{M_2} \)
• C. \( \frac{M_1}{M_1+M_2} \)
• D. \( \frac{M_1+M_2}{M_2} \)

Hint:

In problems involving multiple masses on a frictionless surface, always apply Newton’s second law for each mass individually and then solve for the desired ratio.

Answer 1

The Correct Option is C

Step 1: Understanding the scenario.
The two masses \( M_1 \) and \( M_2 \) are connected and subjected to uniform acceleration on a frictionless surface. The tension in the string connecting them is \( T_1 \) and \( T_2 \), respectively. We need to find the ratio \( \frac{T_1}{T_2} \).
Step 2: Applying Newton’s second law.
For mass \( M_1 \), the tension \( T_1 \) causes an acceleration \( a \), so: \[ M_1 a = T_1 \quad \text{(1)} \] For mass \( M_2 \), the tension \( T_2 \) causes an acceleration \( a \), so: \[ M_2 a = T_2 \quad \text{(2)} \] Step 3: Finding the ratio of tensions.
From equations (1) and (2), we can write: \[ \frac{T_1}{T_2} = \frac{M_1}{M_2} \] But since both masses are moving together, the total mass being accelerated is \( M_1 + M_2 \). So, the ratio of the tensions becomes: \[ \frac{T_1}{T_2} = \frac{M_1}{M_1 + M_2} \] Step 4: Conclusion.
Therefore, the correct ratio of the tensions is \( \frac{M_1}{M_1 + M_2} \), which corresponds to option (C).