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. The number of equal parts into which a conductor having resistance R_{0} = 100 omega should be cut to obtain the resistance R = 1 Omega if the parts are connected in parallel is (C) 20 (D) 2 a (B) 10 (A) 5?
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. The number of equal parts into which a conductor having resistance R...
Understanding the problem:
To find the number of equal parts into which a conductor should be cut in order to obtain a resistance of 1 Ohm when connected in parallel.
Given data:
- Initial resistance of the conductor (R_{0}) = 100 Ohm
- Desired resistance (R) = 1 Ohm
Formula for resistors in parallel:
1/R_{total} = 1/R_{1} + 1/R_{2} + 1/R_{3} + ... + 1/R_{n}
Solution:
Let's assume the conductor is cut into 'n' equal parts.
So, resistance of each part = R_{0}/n
Now, when these 'n' parts are connected in parallel, the total resistance will be:
1/R = n/(R_{0}/n)
1/R = n^2/R_{0}
n^2 = R_{0}/R
n^2 = 100/1
n^2 = 100
n = 10
Therefore, the conductor should be cut into 10 equal parts to obtain a resistance of 1 Ohm when connected in parallel.
Hence, the correct answer is (B) 10.
To find the number of equal parts into which a conductor should be cut in order to obtain a resistance of 1 Ohm when connected in parallel.
Given data:
- Initial resistance of the conductor (R_{0}) = 100 Ohm
- Desired resistance (R) = 1 Ohm
Formula for resistors in parallel:
1/R_{total} = 1/R_{1} + 1/R_{2} + 1/R_{3} + ... + 1/R_{n}
Solution:
Let's assume the conductor is cut into 'n' equal parts.
So, resistance of each part = R_{0}/n
Now, when these 'n' parts are connected in parallel, the total resistance will be:
1/R = n/(R_{0}/n)
1/R = n^2/R_{0}
n^2 = R_{0}/R
n^2 = 100/1
n^2 = 100
n = 10
Therefore, the conductor should be cut into 10 equal parts to obtain a resistance of 1 Ohm when connected in parallel.
Hence, the correct answer is (B) 10.
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. The number of equal parts into which a conductor having resistance R_{0} = 100 omega should be cut to obtain the resistance R = 1 Omega if the parts are connected in parallel is (C) 20 (D) 2 a (B) 10 (A) 5?
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. The number of equal parts into which a conductor having resistance R_{0} = 100 omega should be cut to obtain the resistance R = 1 Omega if the parts are connected in parallel is (C) 20 (D) 2 a (B) 10 (A) 5? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about . The number of equal parts into which a conductor having resistance R_{0} = 100 omega should be cut to obtain the resistance R = 1 Omega if the parts are connected in parallel is (C) 20 (D) 2 a (B) 10 (A) 5? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for . The number of equal parts into which a conductor having resistance R_{0} = 100 omega should be cut to obtain the resistance R = 1 Omega if the parts are connected in parallel is (C) 20 (D) 2 a (B) 10 (A) 5?.
. The number of equal parts into which a conductor having resistance R_{0} = 100 omega should be cut to obtain the resistance R = 1 Omega if the parts are connected in parallel is (C) 20 (D) 2 a (B) 10 (A) 5? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about . The number of equal parts into which a conductor having resistance R_{0} = 100 omega should be cut to obtain the resistance R = 1 Omega if the parts are connected in parallel is (C) 20 (D) 2 a (B) 10 (A) 5? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for . The number of equal parts into which a conductor having resistance R_{0} = 100 omega should be cut to obtain the resistance R = 1 Omega if the parts are connected in parallel is (C) 20 (D) 2 a (B) 10 (A) 5?.
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