We have two wires A and B of same mass and same material. The diameter of wire A is half of that of B. If the resistance of the wire A is 24 `Omega` , then the resistance of the wire B will be

Options

A. `12 Omega `
B. `3Omega`
C. `1.5 Omega`
D. `48Omega`

Answer

C

Explanation

Here ` D_1 = (D_2)/(2) " or " r_1 = (r_2)/(2) " or " A_1 = (A_2)/(4)` . As mass and material of two wires are same , their volumes are same i.e., `A_1l_1 = A_2l_2 " So " l_1 = (A_2l_2)/(A_1) = 4l_2`
` therefore A_1 = (rhol_1)/(A_1) = 24Omega " and " R = (rho l_2)/(A_2) = (rho ( (l_1)/(4)) )/(4 A_1) = 1/16 . (rho l_1)/(A_1) = (R_1)/(16) = 24/16 = 1.5 Omega`