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A uniform metre rod weighing 100 g is loaded with 10,30, 50, 70, 90 g at 10, 30, 50, 70, 90 cm marks respectively Find at what mark on it will it balance, if placed wedge ?
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A uniform metre rod weighing 100 g is loaded with 10,30, 50, 70, 90 g ...
At 70cm bcoz when object on both sides are 90g it's weight balanced can be analyzed by drawing diagram
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A uniform metre rod weighing 100 g is loaded with 10,30, 50, 70, 90 g ...
Introduction:
In this problem, we have a uniform meter rod with different weights placed at specific marks on the rod. We need to determine at what mark the rod will balance when placed on a wedge.
Given information:
- Weight of the meter rod = 100 g
- Weights placed at marks: 10 g at 10 cm, 30 g at 30 cm, 50 g at 50 cm, 70 g at 70 cm, and 90 g at 90 cm.
Approach:
To solve this problem, we need to consider the moments of the weights on the rod. The moment of a weight is calculated by multiplying its weight with its distance from the pivot point.
Calculating moments:
1. The moment of the 10 g weight at 10 cm mark = 10 g * 10 cm = 100 g.cm
2. The moment of the 30 g weight at 30 cm mark = 30 g * 30 cm = 900 g.cm
3. The moment of the 50 g weight at 50 cm mark = 50 g * 50 cm = 2500 g.cm
4. The moment of the 70 g weight at 70 cm mark = 70 g * 70 cm = 4900 g.cm
5. The moment of the 90 g weight at 90 cm mark = 90 g * 90 cm = 8100 g.cm
Determining the balance point:
To find the mark where the rod balances, we need to ensure that the sum of the moments on one side of the pivot is equal to the sum of the moments on the other side.
Let's assume the balance point is at distance 'x' cm from the pivot.
1. Moments on the left side of the pivot = 100 g * x cm
2. Moments on the right side of the pivot = 100 g * (100 - x) cm + 100 g.cm + 900 g.cm + 2500 g.cm + 4900 g.cm + 8100 g.cm
Setting up the equation:
100 g * x cm = 100 g * (100 - x) cm + 100 g.cm + 900 g.cm + 2500 g.cm + 4900 g.cm + 8100 g.cm
Simplifying the equation:
100x = 10000 - 100x + 100 + 900 + 2500 + 4900 + 8100
200x = 19000
x = 95 cm
Conclusion:
The rod will balance at the 95 cm mark when placed on the wedge. At this point, the sum of the moments on the left side of the pivot will be equal to the sum of the moments on the right side of the pivot.
In this problem, we have a uniform meter rod with different weights placed at specific marks on the rod. We need to determine at what mark the rod will balance when placed on a wedge.
Given information:
- Weight of the meter rod = 100 g
- Weights placed at marks: 10 g at 10 cm, 30 g at 30 cm, 50 g at 50 cm, 70 g at 70 cm, and 90 g at 90 cm.
Approach:
To solve this problem, we need to consider the moments of the weights on the rod. The moment of a weight is calculated by multiplying its weight with its distance from the pivot point.
Calculating moments:
1. The moment of the 10 g weight at 10 cm mark = 10 g * 10 cm = 100 g.cm
2. The moment of the 30 g weight at 30 cm mark = 30 g * 30 cm = 900 g.cm
3. The moment of the 50 g weight at 50 cm mark = 50 g * 50 cm = 2500 g.cm
4. The moment of the 70 g weight at 70 cm mark = 70 g * 70 cm = 4900 g.cm
5. The moment of the 90 g weight at 90 cm mark = 90 g * 90 cm = 8100 g.cm
Determining the balance point:
To find the mark where the rod balances, we need to ensure that the sum of the moments on one side of the pivot is equal to the sum of the moments on the other side.
Let's assume the balance point is at distance 'x' cm from the pivot.
1. Moments on the left side of the pivot = 100 g * x cm
2. Moments on the right side of the pivot = 100 g * (100 - x) cm + 100 g.cm + 900 g.cm + 2500 g.cm + 4900 g.cm + 8100 g.cm
Setting up the equation:
100 g * x cm = 100 g * (100 - x) cm + 100 g.cm + 900 g.cm + 2500 g.cm + 4900 g.cm + 8100 g.cm
Simplifying the equation:
100x = 10000 - 100x + 100 + 900 + 2500 + 4900 + 8100
200x = 19000
x = 95 cm
Conclusion:
The rod will balance at the 95 cm mark when placed on the wedge. At this point, the sum of the moments on the left side of the pivot will be equal to the sum of the moments on the right side of the pivot.
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A uniform metre rod weighing 100 g is loaded with 10,30, 50, 70, 90 g at 10, 30, 50, 70, 90 cm marks respectively Find at what mark on it will it balance, if placed wedge ?
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A uniform metre rod weighing 100 g is loaded with 10,30, 50, 70, 90 g at 10, 30, 50, 70, 90 cm marks respectively Find at what mark on it will it balance, if placed wedge ? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A uniform metre rod weighing 100 g is loaded with 10,30, 50, 70, 90 g at 10, 30, 50, 70, 90 cm marks respectively Find at what mark on it will it balance, if placed wedge ? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform metre rod weighing 100 g is loaded with 10,30, 50, 70, 90 g at 10, 30, 50, 70, 90 cm marks respectively Find at what mark on it will it balance, if placed wedge ?.
A uniform metre rod weighing 100 g is loaded with 10,30, 50, 70, 90 g at 10, 30, 50, 70, 90 cm marks respectively Find at what mark on it will it balance, if placed wedge ? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A uniform metre rod weighing 100 g is loaded with 10,30, 50, 70, 90 g at 10, 30, 50, 70, 90 cm marks respectively Find at what mark on it will it balance, if placed wedge ? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform metre rod weighing 100 g is loaded with 10,30, 50, 70, 90 g at 10, 30, 50, 70, 90 cm marks respectively Find at what mark on it will it balance, if placed wedge ?.
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