A wire of length 'l' is gradually stretched so that its length increases to $3l$. If its original resistance is $R$, then its new resistance will be :

From the NCERT chapter, resistance of a conductor of length $l$ and area of cross-section $A$ is $$R=\rho\dfrac{l}{A}$$ where $\rho$ is the resistivity of the material — a fixed property of the material that does not change on stretching (only $l$ and $A$ change).

When the wire is stretched, its volume $V=Al$ stays constant (no material is added or removed). If the length increases from $l$ to $3l$, the new area of cross-section $A'$ must satisfy $$A'(3l)=Al \implies A'=\dfrac{A}{3}$$

The new resistance is $$R'=\rho\dfrac{3l}{A/3}=\rho\dfrac{9l}{A}=9\left(\rho\dfrac{l}{A}\right)=9R$$

Correct option: C) $9R$. Physically, $R\propto l$ and $R\propto 1/A$; volume conservation gives $A\propto 1/l$ on stretching, so combined, $R\propto l^2$ — tripling the length increases resistance ninefold.

Why the distractors are wrong:

• (A) $3R$ — results if a student wrongly assumes the area stays unchanged while only length triples ($R'=\rho\cdot3l/A=3R$), ignoring that stretching must conserve volume.
• (B) $6R$ — has no valid physical derivation from the resistance formula; it likely comes from an arithmetic slip while combining the length and area scale factors.
• (D) $27R$ — arises if a student wrongly assumes the area shrinks to $A/9$ (treating length and area as scaling by the cube of the stretch factor) instead of correctly applying volume conservation, which gives $A'=A/3$, not $A/9$.