1. Let the hydrocarbon be $C_xH_y$.

2. Use the combustion equation $C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$

3. Use volume ratios At the same temperature and pressure, gaseous volumes are proportional to moles.

   Given:

   • $10\ \text{mL}$ hydrocarbon requires $55\ \text{mL}$ $O_2$
   • $40\ \text{mL}$ $CO_2$ is formed

4. Find $x$ from $CO_2$ volume From the equation, 1 volume of $C_xH_y$ gives $x$ volumes of $CO_2$.

   So, $x = \frac{40}{10} = 4$

   Hence the hydrocarbon is of the form $C_4H_y$.

5. Find $y$ from $O_2$ volume From the equation, 1 volume of $C_xH_y$ requires $\left(x + \frac{y}{4}\right)$ volumes of $O_2$.

   Therefore, $4 + \frac{y}{4} = \frac{55}{10} = 5.5$

   $\frac{y}{4} = 1.5$

   $y = 6$

6. Thus the hydrocarbon formula is $C_4H_6$

7. Check options

   • A: $C_4H_7Cl$ → not a hydrocarbon
   • B: $C_4H_6$ → correct
   • C: $C_4H_8$ → would need $4 + 8/4 = 6$ volumes of $O_2$, not $5.5$
   • D: $C_4H_{10}$ → would need $4 + 10/4 = 6.5$ volumes of $O_2$, not $5.5$

Hence, the correct answer is B: $C_4H_6$.