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The electrical resistance of a mercury column in a cylindrical container is R. When the same mercury is poured into another cylindrical container twice the radius of cross-section, the resistance of mercury column now is
- a)R/2
- b)R/4
- c)R/8
- d)R/16
Correct answer is option 'D'. Can you explain this answer?
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The electrical resistance of a mercury column in a cylindrical contain...
Explanation:
Let's consider two cylindrical containers, one with radius r and the other with radius 2r. Let the height of the mercury column in both containers be h.
1. Calculation of resistance in the first container:
The resistance of a cylinder of length l, area of cross-section A, and resistivity ρ is given by the formula:
R = (ρl)/A
Here, the length of the mercury column is h and the area of cross-section is πr². So, the resistance of the mercury column in the first container is:
R1 = (ρh)/(πr²)
2. Calculation of resistance in the second container:
When the same amount of mercury is poured into the second container, the height of the mercury column remains the same (h). However, the area of cross-section becomes π(2r)² = 4πr².
So, the resistance of the mercury column in the second container is:
R2 = (ρh)/(4πr²)
3. Calculation of ratio of resistances:
We need to find the ratio of R2/R1:
(R2/R1) = [(ρh)/(4πr²)]/[(ρh)/(πr²)] = 1/4
So, R2 = (1/4)R1
4. Final answer:
We are given that the resistance of the mercury column in the first container is R. So, substituting R1 = R in the above equation, we get:
R2 = (1/4)R
Therefore, the resistance of the mercury column in the second container is R/16 or option (D).
Let's consider two cylindrical containers, one with radius r and the other with radius 2r. Let the height of the mercury column in both containers be h.
1. Calculation of resistance in the first container:
The resistance of a cylinder of length l, area of cross-section A, and resistivity ρ is given by the formula:
R = (ρl)/A
Here, the length of the mercury column is h and the area of cross-section is πr². So, the resistance of the mercury column in the first container is:
R1 = (ρh)/(πr²)
2. Calculation of resistance in the second container:
When the same amount of mercury is poured into the second container, the height of the mercury column remains the same (h). However, the area of cross-section becomes π(2r)² = 4πr².
So, the resistance of the mercury column in the second container is:
R2 = (ρh)/(4πr²)
3. Calculation of ratio of resistances:
We need to find the ratio of R2/R1:
(R2/R1) = [(ρh)/(4πr²)]/[(ρh)/(πr²)] = 1/4
So, R2 = (1/4)R1
4. Final answer:
We are given that the resistance of the mercury column in the first container is R. So, substituting R1 = R in the above equation, we get:
R2 = (1/4)R
Therefore, the resistance of the mercury column in the second container is R/16 or option (D).
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The electrical resistance of a mercury column in a cylindrical container is R. When the same mercury is poured into another cylindrical container twice the radius of cross-section, the resistance of mercury column now isa)R/2b)R/4c)R/8d)R/16Correct answer is option 'D'. Can you explain this answer?
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The electrical resistance of a mercury column in a cylindrical container is R. When the same mercury is poured into another cylindrical container twice the radius of cross-section, the resistance of mercury column now isa)R/2b)R/4c)R/8d)R/16Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The electrical resistance of a mercury column in a cylindrical container is R. When the same mercury is poured into another cylindrical container twice the radius of cross-section, the resistance of mercury column now isa)R/2b)R/4c)R/8d)R/16Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The electrical resistance of a mercury column in a cylindrical container is R. When the same mercury is poured into another cylindrical container twice the radius of cross-section, the resistance of mercury column now isa)R/2b)R/4c)R/8d)R/16Correct answer is option 'D'. Can you explain this answer?.
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