A wire of length 1 and resistance R is stretched to get the radius of

1.	A wire of length 1 and resistance R is stretched to get the radius of cross section halved. What is the new resistance of the wire after stretching?
Answer» Given: A wire of length l and resistance R is stretched to get the radius of cross SECTION halved. To find: NEW resistance of wire. Calculation: Since the wire has been stretched , its volume will remain CONSTANT; $\therefore$ V1 = V2 $= > \pi {r}^{2} l = \pi {( \dfrac{r}{2} )}^{2} l_{2}$ $= > l = \dfrac{ l_{2} }{4}$ $= > l_{2} = 4l$ Initial resistance: $R = \rho \times ( \dfrac{l}{a} )$ $= > R = \rho \times ( \dfrac{l}{\pi {r}^{2} } )\:\:\:\:.............(1)$ FINAL resistance: $R_{2}= \rho \times ( \dfrac{4l}{a_{2}} )$ $= > R_{2}= \rho \times \bigg \{\dfrac{4l}{\pi {( \frac{r}{2}) }^{2} } \bigg \}$ $= > R_{2}= \rho \times \bigg \{\dfrac{4l}{ \frac{\pi {r}^{2} }{4} } \bigg \}$ $= > R_{2}= \rho \times \bigg \{\dfrac{16l}{ \pi {r}^{2} } \bigg \}$ $= > R_{2}= 16\times \bigg \{ \rho \times \dfrac{l}{ \pi {r}^{2} } \bigg \}$ $= > R_{2}= 16\times \bigg \{R \bigg \}$ $= > R_{2}= 16R$ So, final answer is: $\boxed{ \sf{ \red{R_{2}= 16R }}}$