Question

A piece of silver wire has a resistance of 1 ohms.What will be the resistance of a manganin wire of ⅓ the length and ⅓ the diameter if the specific resistance of maganin is thirty times of silver.

Answer 1

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HERE'S YOUR ANSWER


Let, the silver wire be l m long and the area of cross section be πr^2

So, 1 = p1 l/πr^2
=> p1 = πr^2/l .....(i)

Now, the Manganin wire will be l/3 m long and area if criss section will be πr^2/9.

So, Resistance = p2*(l/3)/(πr^2/9)
= p2 * 3l/πr^2

Also, p2 = 30×p1

So, Resistance = 30 * (πr^2/l) * (3l/πr^2)
= 30 × 3

Therefore, the resistance of the Manganin wire will be 90 ohms.


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Answer 2

Answer:we have given the resistance of a silver  and find the resistance of  constanton