A given wire of resistance 1 Ω is stretched to double its length. What will be its new resistance?

Question

Numerical

Solution

Let ‘I’ be the length and ‘a’ be the area of cross – section
of the resistor with resistance, R = 1Ω

`"R" = ∝ l/r^2`

when the wire is stretched to double its length,
the new length I’ = 2I and the new area of cross section,

`a^1 = a/2 "and" l^1 = 2l`

`"R"^1 ∝ l^1/a^1`

∴ `"R"^1 ∝ (2l)/(a/2)`

`"R"^1 ∝ (4l)/a`

R¹ ∝ 4R

Resistance increases to 4 times original resistance = R¹ = 4Ω

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Q 9.Q 8. (ii)Q 10.

Chapter 8: Current Electricity - EXERCISE - 8(A) [Page 187]

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Selina Concise Physics [English] Class 10 ICSE

Chapter 8 Current Electricity
EXERCISE - 8(A) | Q 9. | Page 187

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