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A uniform meter stick of mass 200 g is suspended from the ceiling through 2 vertical strings of equal length fixed at ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm ftom left end. Find tensions in two strings.?
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A uniform meter stick of mass 200 g is suspended from the ceiling thro...
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Tensions in the two strings can be calculated by considering the forces acting on the meter stick and the small object. Let's break down the problem step by step:

1. Identify the forces acting on the meter stick:
- Weight of the meter stick: This acts vertically downwards from the center of mass.
- Tension in the left string: This acts upwards and is denoted by T1.
- Tension in the right string: This also acts upwards and is denoted by T2.

2. Analyze the forces acting on the small object:
- Weight of the small object: This acts vertically downwards from the object's center of mass.

3. Establish equilibrium conditions:
For the meter stick to be in equilibrium, the sum of the torques about any point must be zero. We can choose the left end of the meter stick as the pivot point.

4. Calculate the torques:
- Torque due to the weight of the meter stick: Since the weight acts at the center of mass of the meter stick, its torque is zero.
- Torque due to the tension in the left string (T1): This creates a clockwise torque since it acts to the left of the pivot point.
- Torque due to the tension in the right string (T2): This creates a counterclockwise torque since it acts to the right of the pivot point.
- Torque due to the weight of the small object: This creates a clockwise torque since it acts to the left of the pivot point.

5. Set up the torque equation:
Since the meter stick is in equilibrium, the sum of the torques must be zero. Using the equation Στ = 0, we can write:
- T1 * L1 = T2 * L2 + M * g * d

In this equation, L1 and L2 represent the distances of the left and right strings from the pivot point, M is the mass of the meter stick, g is the acceleration due to gravity, and d is the distance of the small object from the pivot point.

6. Substitute the given values and solve for T1 and T2:
- L1 and L2 are equal in length since the strings are of equal length.
- M = 200 g
- g = 9.8 m/s^2
- d = 70 cm = 0.7 m

By substituting these values into the equation, we can solve for T1 and T2.

7. The final answer:
Once you calculate the values of T1 and T2, you will have the tensions in the two strings. Make sure to include the units in your final answer.

Remember to present your solution in a clear and organized manner, and label all the variables and steps you used for calculation.
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A uniform meter stick of mass 200 g is suspended from the ceiling through 2 vertical strings of equal length fixed at ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm ftom left end. Find tensions in two strings.?
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A uniform meter stick of mass 200 g is suspended from the ceiling through 2 vertical strings of equal length fixed at ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm ftom left end. Find tensions in two strings.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A uniform meter stick of mass 200 g is suspended from the ceiling through 2 vertical strings of equal length fixed at ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm ftom left end. Find tensions in two strings.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform meter stick of mass 200 g is suspended from the ceiling through 2 vertical strings of equal length fixed at ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm ftom left end. Find tensions in two strings.?.
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