There are two wires P and Q made of the same material and both have the same mass . The radius of wire P is half the radius of the wire Q. the resistance of P is `24 Omega`. Find the resistance of Q.

Explanation

Let m be the mass and d the density of material. Let r be the radius and l be the length.
`m= (pi r^(2) l) d rArr l= (m)/(pi r^(2) d)` …(i)
We know that resistance is given by the following relation.
`R= rho (l)/(A) rArr R= rho (l)/(pi r^(2))` ...(ii)
Substituting l from equation (i) in equation (ii) we get
`R= rho (m)/(pi^(2) r^(4) d)` ...(iii)
In equation (iii) it is only the radius which is different for P and Q and rest of the parameters are same. Hence their resistances are inversely proportional to fourth power of the radius of wire
`(R_(P))/(R_(Q)) = ((r_(Q))/(r_(P)))^(4) rArr (R_(P))/(R_(Q)) = ((2r_(P))/(r_(P)))^(4) = 16`
`rArr (R_(P))/(R_(Q)) = 16 rArr R_(Q) = (R_(P))/(16) rArr R_(Q) = (24)/(16) = 1.5 Omega`