Question

Two metallic wires whose masses are 12 g and 18 g have lengths in the ratio of 3 : 4. Their resistances are in the ratio of

• A. 4 : 3
• B. 9 : 18
• C. 18 : 27
• D. 27 : 32

Answer 1

Answer: (D)

27 : 32

Given : $m_1=12$ g $m_2=18$ g $L_1:L_2=3:4$

Resistance of wire $R=\frac{\rho L^{2}d}{m}$ where d (density ) and $\rho$ (resistivity) are constants

$⟹$ $\frac{R_1}{R_2}=\frac{m_2}{m_1}(\frac{L_1}{L_2}{)}^2$

$\therefore$ $\frac{R_1}{R_2}=\frac{18}{12}(\frac{3}{4}{)}^2=\frac{27}{32}$