Question

15) A uniform length of wire 1.2 m has 5
a resistance of 16. It is stretched such
that its resistance is 36. The new
length is 1) 1.8 m 2) 2.4 m 3) 2.7 m 4) 3​

Answer 1

Answer:

we know that Under stretch if the volume of the wire remains constant then the resistance is proportional to the square of its length.

i.e.

R ∝ l^2

So

$\frac{ l_{2}}{ l_{1}} = \sqrt{ \frac{ r_{2} }{ r_{1}} } \\ \\ l_{2} = 1.2 \sqrt{ \frac{36}{16} } \\ \\ l_{2}= 1.2 \times \frac{6}{4} \\ \\ l_{2} = 0.3 \times 6 \\ \\ l_{2} = 1.8$

therefore,

correct answers is 1.8

NOTE : r = resistance

l = length

r2 = 36

r1 = 16

l1 = 1.2

l2 = ?

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