1.

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

Answer»

`12 Omega `
`3Omega`
`1.5 Omega`
`48Omega`

Solution :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 volumesare 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`


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