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Two blocks m1 = 5 gm and m2 = 10 gm are hung vertically over a light frictionless pulley as shown here. What is the acceleration of the masses when they are left free? [2000] (where g is acceleration due to gravity)
- a)g/3
- b)g/2
- c)g
- d)g/5
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Two blocks m1 = 5 gm and m2 = 10 gm are hung vertically over a light f...
Let T be the tension in the string.
Adding (i) and (ii),
Most Upvoted Answer
Two blocks m1 = 5 gm and m2 = 10 gm are hung vertically over a light f...
Acceleration Calculation:
Given:
m1 = 5 gm = 0.005 kg
m2 = 10 gm = 0.01 kg
To find:
Acceleration of the masses when they are left free.
Assumptions:
1. The pulley is light and frictionless.
2. The string is light and inextensible.
The system consists of two masses connected by a string passing over a pulley. When the masses are left free, they experience a net force in the direction of the larger mass due to the force of gravity.
Net Force Calculation:
The net force on each mass can be calculated using Newton's second law of motion:
F = m * a
For m1:
F1 = m1 * g
= 0.005 * 9.8
= 0.049 N (upwards)
For m2:
F2 = m2 * g
= 0.01 * 9.8
= 0.098 N (downwards)
Since the masses are connected by a string passing over a pulley, the tension in the string will be the same on both sides.
Tension Calculation:
Let T be the tension in the string.
For m1:
T - F1 = m1 * a1
T - 0.049 = 0.005 * a1
For m2:
F2 - T = m2 * a2
0.098 - T = 0.01 * a2
Since the masses are connected, their accelerations are equal, so a1 = a2 = a.
Solving the equations:
T - 0.049 = 0.005 * a
0.098 - T = 0.01 * a
Adding the equations:
T - 0.049 + 0.098 - T = 0.005 * a + 0.01 * a
0.049 = 0.015 * a
a = 0.049 / 0.015
a = 3.27 m/s²
Therefore, the acceleration of the masses when they are left free is approximately 3.27 m/s².
Answer:
Option a) g/3
Given:
m1 = 5 gm = 0.005 kg
m2 = 10 gm = 0.01 kg
To find:
Acceleration of the masses when they are left free.
Assumptions:
1. The pulley is light and frictionless.
2. The string is light and inextensible.
The system consists of two masses connected by a string passing over a pulley. When the masses are left free, they experience a net force in the direction of the larger mass due to the force of gravity.
Net Force Calculation:
The net force on each mass can be calculated using Newton's second law of motion:
F = m * a
For m1:
F1 = m1 * g
= 0.005 * 9.8
= 0.049 N (upwards)
For m2:
F2 = m2 * g
= 0.01 * 9.8
= 0.098 N (downwards)
Since the masses are connected by a string passing over a pulley, the tension in the string will be the same on both sides.
Tension Calculation:
Let T be the tension in the string.
For m1:
T - F1 = m1 * a1
T - 0.049 = 0.005 * a1
For m2:
F2 - T = m2 * a2
0.098 - T = 0.01 * a2
Since the masses are connected, their accelerations are equal, so a1 = a2 = a.
Solving the equations:
T - 0.049 = 0.005 * a
0.098 - T = 0.01 * a
Adding the equations:
T - 0.049 + 0.098 - T = 0.005 * a + 0.01 * a
0.049 = 0.015 * a
a = 0.049 / 0.015
a = 3.27 m/s²
Therefore, the acceleration of the masses when they are left free is approximately 3.27 m/s².
Answer:
Option a) g/3
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Two blocks m1 = 5 gm and m2 = 10 gm are hung vertically over a light frictionless pulley as shown here. What is the acceleration of the masses when they are left free? [2000] (where g is acceleration due to gravity)a)g/3b)g/2c)gd)g/5Correct answer is option 'A'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two blocks m1 = 5 gm and m2 = 10 gm are hung vertically over a light frictionless pulley as shown here. What is the acceleration of the masses when they are left free? [2000] (where g is acceleration due to gravity)a)g/3b)g/2c)gd)g/5Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two blocks m1 = 5 gm and m2 = 10 gm are hung vertically over a light frictionless pulley as shown here. What is the acceleration of the masses when they are left free? [2000] (where g is acceleration due to gravity)a)g/3b)g/2c)gd)g/5Correct answer is option 'A'. Can you explain this answer?.
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