When a resistor of 2 ohm stretched to double the original length new r

1.	When a resistor of 2 ohm stretched to double the original length new resistance
Answer» Answer: Lets read the statement. When a resistor of 2Ω is stretched... So we aren't GIVEN any new wire... We stretched the wire. Lets assume the original length as x. After stretching the wire, let the length be 2×x. If length is 2x, then Cross-sectional area=1/2 (Simple equation: If length INCREASES, area decreases. Or, in other words, l∝1/A) Now, resistance= 2Ω. Resistance of unstretched wire of the unstretched wire (Original resistance): $R=\frac{\rhoL}{A}$ Now, we derived that l=2x and a=1/2 for the new wire. SUBSTITUTE these values in the above equation. Resistance of new wire: $R_{new} =\frac{\rho2l}{\frac{A}{2} }$ The 2 below A will GO to the numerator. Further simplifying: $R_{new} =\frac{\rho22l}{A}$ $R_{new} =\frac{\rho4l}{A }$ $R_{new} =4(\frac{\rhol}{A })$ We know that $\frac{\rho*L}{A}=R$ . Substitute this in the above equation. $R_{new} =4(R)$ Thus, from the above derivation, we conclude that New resistance= 4×(Original resistance) Now, original resistance is 2Ω. New resistance=4×2 New resistance=8Ω. TIP: Dont memorise, understand. HOPE THIS HELPS :D