Question

A uniform length of wire $1.2$ m has a resistance of $16\Omega$. It is stretched such that its resistance is $36\Omega$. The new length is

• A. $1.8m$
• B. $2.4m$
• C. $2.7m$
• D. $3m$

Answer 1

Answer: (A)

$1.8m$

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. $R\propto l^2$.
So $\frac{l_2}{l_1}=\sqrt{\frac{R_2}{R_1}}\Rightarrow l_2=1.2\sqrt{\frac{36}{16}}=1.8$m.