Step 1: Calculate Unknown Resistance X:
Using Meter Bridge principle: $\frac{X}{R} = \frac{l}{100-l}$.
$R = 15 \Omega$, $l = 40$ cm.
\[ \frac{X}{15} = \frac{40}{60} = \frac{2}{3} \implies X = 10 \Omega \]
Step 2: Calculate Total Circuit Resistance:
The battery supplies current to the entire bridge circuit. Usually, the wire and the resistor gaps are in parallel across the battery in standard diagrams where the battery is connected to the ends of the wire A and C.
Current $I = 280$ mA = 0.28 A. Voltage $V = 2$ V.
Total External Resistance $R_{ext} = \frac{V}{I} = \frac{2}{0.28} = \frac{200}{28} \approx 7.14 \Omega$.
Step 3: Determine Wire Resistance ($R_w$):
The circuit consists of the wire resistance $R_w$ in parallel with the series combination of gap resistors $(X + R)$.
$X + R = 10 + 15 = 25 \Omega$.
\[ \frac{1}{R_{ext}} = \frac{1}{R_w} + \frac{1}{X+R} \]
\[ \frac{1}{7.14} = \frac{1}{R_w} + \frac{1}{25} \]
Or simpler: $\frac{1}{R_w} = \frac{I}{V} - \frac{1}{25} = \frac{0.28}{2} - 0.04 = 0.14 - 0.04 = 0.1$.
\[ R_w = \frac{1}{0.1} = 10 \Omega \]
Step 4: Calculate Resistance per unit length ($\lambda$):
Length of wire = 100 cm.
\[ \lambda = \frac{R_w}{100} = \frac{10}{100} = 0.1 \, \Omega cm^{-1} \]
Step 5: Final Answer:
The resistance per unit length is $0.1 \Omega cm^{-1}$.