# Question #4dd5f

Nov 6, 2016

the ans. will be $1 o h m$.

#### Explanation:

we know R=s(l/a).where s=specific resistance of the element, l=length of the element,a=circular base area of the element.
now for the first case if resistance was ${R}_{1}$ and for the second case resistance will be ${R}_{2}$.
so ${R}_{1}$=$s \left({l}_{1} / {a}_{1}\right)$ and ${R}_{2} = s \left({l}_{2} / {a}_{2}\right)$
now the diameter of the element is doubled ,so the radius of the circular base is also doubled.$\sin c e , a r e a = \pi {r}^{2}$,the area will be 4times.so,${a}_{2} = 4 {a}_{1}$.
here the volume of the element will remain constant,so the product of the area & length remains the same i.e. ${a}_{1} {l}_{1} = {a}_{2} {l}_{2}$
again ${a}_{2} = 4 {a}_{1}$ hence ${l}_{2} = {l}_{1} / 4$ .
since the element remains same s will be same for both the cases.

now${R}_{2} = s \left({l}_{2} / {a}_{2}\right)$=$s \left(\frac{{l}_{1} / 4}{4 {a}_{1}}\right)$=$s \left({l}_{1} / \left(16 {a}_{1}\right)\right)$=${R}_{1} / 16$=$\frac{16}{16}$=$1 o h m$.