Question:
Three masses $m_1 = 4$ kg, $m_2 = 4$ kg and $m_3 = 6$ kg are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of $T_1/T_2$ is: (take $g = 10 \text{ m/s}^2$)}
Three masses $m_1 = 4$ kg, $m_2 = 4$ kg and $m_3 = 6$ kg are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of $T_1/T_2$ is: (take $g = 10 \text{ m/s}^2$)}
Updated On: Apr 12, 2026
- 5/3
- 2/3
- 3/5
- 3/5
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
This is a multiple-block pulley system. We need to find the acceleration of the system and then determine the tensions in the two strings.
Step 2: Key Formula or Approach:
Net acceleration $a = \frac{\text{Net Pulling Force}}{\text{Total Mass}}$.
Tension equations: $T = m(g \pm a)$.
Step 3: Detailed Explanation:
Let's consider the right side as having mass $(m_2 + m_3) = 10$ kg and the left side as $m_1 = 4$ kg.
The 10 kg mass is heavier, so the system accelerates downwards on the right.
Acceleration $a = \frac{(m_2 + m_3)g - m_1 g}{m_1 + m_2 + m_3} = \frac{10g - 4g}{4 + 4 + 6} = \frac{6g}{14} = \frac{3g}{7}$.
Now, find $T_1$ (tension in the main string passing over the pulley):
Looking at $m_1$ (accelerating upwards):
$T_1 - m_1 g = m_1 a \implies T_1 = m_1(g + a) = 4(g + \frac{3g}{7}) = 4(\frac{10g}{7}) = \frac{40g}{7}$.
Now, find $T_2$ (tension in the string between $m_2$ and $m_3$):
Looking at $m_3$ (accelerating downwards):
$m_3 g - T_2 = m_3 a \implies T_2 = m_3(g - a) = 6(g - \frac{3g}{7}) = 6(\frac{4g}{7}) = \frac{24g}{7}$.
The ratio $T_1 / T_2$:
$\frac{T_1}{T_2} = \frac{40g/7}{24g/7} = \frac{40}{24} = \frac{5}{3}$.
Step 4: Final Answer:
The ratio $T_1/T_2$ is 5/3.
This is a multiple-block pulley system. We need to find the acceleration of the system and then determine the tensions in the two strings.
Step 2: Key Formula or Approach:
Net acceleration $a = \frac{\text{Net Pulling Force}}{\text{Total Mass}}$.
Tension equations: $T = m(g \pm a)$.
Step 3: Detailed Explanation:
Let's consider the right side as having mass $(m_2 + m_3) = 10$ kg and the left side as $m_1 = 4$ kg.
The 10 kg mass is heavier, so the system accelerates downwards on the right.
Acceleration $a = \frac{(m_2 + m_3)g - m_1 g}{m_1 + m_2 + m_3} = \frac{10g - 4g}{4 + 4 + 6} = \frac{6g}{14} = \frac{3g}{7}$.
Now, find $T_1$ (tension in the main string passing over the pulley):
Looking at $m_1$ (accelerating upwards):
$T_1 - m_1 g = m_1 a \implies T_1 = m_1(g + a) = 4(g + \frac{3g}{7}) = 4(\frac{10g}{7}) = \frac{40g}{7}$.
Now, find $T_2$ (tension in the string between $m_2$ and $m_3$):
Looking at $m_3$ (accelerating downwards):
$m_3 g - T_2 = m_3 a \implies T_2 = m_3(g - a) = 6(g - \frac{3g}{7}) = 6(\frac{4g}{7}) = \frac{24g}{7}$.
The ratio $T_1 / T_2$:
$\frac{T_1}{T_2} = \frac{40g/7}{24g/7} = \frac{40}{24} = \frac{5}{3}$.
Step 4: Final Answer:
The ratio $T_1/T_2$ is 5/3.
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