$(a)\;CH_4\qquad(b)\;C_2H_4\qquad(c)\;C_2H_6\qquad(d)\;C_3H_8$

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Let formula of hydrocarbon be $C_aH_b$

$C_aH_b+[a+\large\frac{b}{4}]$$O_2\rightarrow aCO_2+\large\frac{b}{2}$$H_2O(l)$

Also given

Volume absorbed by $NaOHCO_2$ formed =10ml

Volume observed by pyragallol is of $O_2$ left=15ml.

Volume $O_2$ used =30-15=15ml

$5a=10$

$a=2$

$5(a+\large\frac{b}{4})$$=15$

$b=4$

$\therefore$ Hydrocarbon =$C_2H_4$

Hence (b) is the correct option.

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