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Question

Physics Question on Current electricity

A wire of length ll and resistance RR is stretched to get the radius of cross-section r2\frac{r}{2} . Then the new value of RR is

A

16 R

B

4 R

C

8 R

D

5 R

Answer

16 R

Explanation

Solution

R=ρlπr2R=\frac{\rho l}{\pi {{r}^{2}}} Since, V1=V2{{V}_{1}}={{V}_{2}} \Rightarrow l×π(r2)2=l×πr2l' \times \pi {{\left( \frac{r}{2} \right)}^{2}}=l\times \pi {{r}^{2}} \Rightarrow l4=l\frac{l'}{4}=l or l=4ll'=4l \therefore R=ρ4lπr2×4=16RR'=\frac{\rho 4l}{\pi {{r}^{2}}}\times 4=16R