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A jeweler is holding a gold chain of uniform mass per unit length hanging vertically just above a weighing scale. He offers to charge the customer for half the max reading of the scale, after he releases the chain. What % more than the actual price does the customer pay if her agrees to the offer? 50 5 25 20 Answer is 50% pls explain how?
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A jeweler is holding a gold chain of uniform mass per unit length hang...
Solution:

Given, the jeweler is holding a gold chain of uniform mass per unit length hanging vertically just above a weighing scale. He offers to charge the customer for half the max reading of the scale, after he releases the chain.

Let the actual weight of the gold chain be W and let the length of the gold chain be L.

The weight of the gold chain = W

The weight of the gold chain per unit length = W/L

When the jeweler releases the chain, it will fall freely under the influence of gravity and the weight of the chain will be transferred to the weighing scale. The maximum weight that will be recorded on the weighing scale will be when the chain is just about to touch the scale.

The weight of the gold chain when it is just about to touch the scale = W/L * L = W

So, the jeweler offers to charge the customer for half the max reading of the scale, which is W/2.

Let the actual price of the gold chain be P.

So, the customer pays P/2 for the gold chain.

The % more than the actual price the customer pays = [(P/2 - P)/P] * 100%

= -50%

Therefore, the customer pays 50% less than the actual price of the gold chain.

Answer: 50% less.
Community Answer
A jeweler is holding a gold chain of uniform mass per unit length hang...
Let the mass of the gold chain be m, its length be L, and let λ = m/L. 
Consider the chain to consist of sections of length dx. 
After the chain has fallen a distance x, the sections that have not yet reached the table have velocity v=(2gx)^1/2
The scale has to support a section of the string of length x and has to reduce the momentum of a section of length dx from λdxv to zero in a time interval dt

The scale has to exert a force F = λxg + λv^2(in upward direction)= λxg + λ2gx = 3λxg.
The maximum force is exerted when the last link is stopped by the scale.  F max = 3λLg = 3mg
1/2 F max  = 1.5 mg The customer pays 50% more than the actual charge
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A jeweler is holding a gold chain of uniform mass per unit length hanging vertically just above a weighing scale. He offers to charge the customer for half the max reading of the scale, after he releases the chain. What % more than the actual price does the customer pay if her agrees to the offer? 50 5 25 20 Answer is 50% pls explain how?
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A jeweler is holding a gold chain of uniform mass per unit length hanging vertically just above a weighing scale. He offers to charge the customer for half the max reading of the scale, after he releases the chain. What % more than the actual price does the customer pay if her agrees to the offer? 50 5 25 20 Answer is 50% pls explain how? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A jeweler is holding a gold chain of uniform mass per unit length hanging vertically just above a weighing scale. He offers to charge the customer for half the max reading of the scale, after he releases the chain. What % more than the actual price does the customer pay if her agrees to the offer? 50 5 25 20 Answer is 50% pls explain how? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A jeweler is holding a gold chain of uniform mass per unit length hanging vertically just above a weighing scale. He offers to charge the customer for half the max reading of the scale, after he releases the chain. What % more than the actual price does the customer pay if her agrees to the offer? 50 5 25 20 Answer is 50% pls explain how?.
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