Question

5. Nichrome wire of length l and radius ‘r’ has resistance of 10Ω. How would the resitance of the wire change when (i) only the diameter is doubled? (ii) only length of the wire is doubled?

Answer 1

Answer:

(i) Resistance = 2.5 Ohm

(ii) Resistance = 20 Ohm

Explanation:

Under normal conditions :

R = δ l /A = 10Ω

CASE 1

(i) A = π $r^{2}$

       =  π $[\frac{d}{2} ]^{2}$   = π $d^{2}$ / 4

When the diameter is doubled:

A' = π $d^{2}$               A' = 4 A

R' = δ l /A'

   = δ l /4A

   = $\frac{1}{4}$ x δ l /A

   = $\frac{1}{4}$ x $10$

R' = 2.5 Ω

(ii) When length of the wire is doubled :

l' = 2l

R' = δ l' /A

   = 2 x δ l /A

   = 2 x 10

R' = 20 Ω

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