1. Find moles of the hydrocarbon

Given hydrocarbon: $\mathrm{C_{10}H_{16}}$

Molar mass: $M = 10(12) + 16(1) = 120 + 16 = 136\,\text{g mol}^{-1}$

Mass given: $17\,\text{mg} = 0.017\,\text{g}$

Moles of hydrocarbon: $n_{\text{HC}} = \frac{0.017}{136} = 1.25 \times 10^{-4}\,\text{mol}$

2. Find moles of hydrogen absorbed

Hydrogen volume at STP = $8.40\,\text{mL}$

At STP, $1$ mol gas occupies $22400\,\text{mL}$.

So, $n_{\mathrm{H_2}} = \frac{8.40}{22400} = 3.75 \times 10^{-4}\,\text{mol}$

3. Relate hydrogen uptake to number of $\pi$ bonds reduced

Each double bond consumes $1$ mole of $\mathrm{H_2}$ on hydrogenation.

If number of double bonds per molecule is $x$, then $x = \frac{n_{\mathrm{H_2}}}{n_{\text{HC}}} = \frac{3.75\times10^{-4}}{1.25\times10^{-4}} = 3$

So, the hydrocarbon contains: $\boxed{3}$ double bonds.

4. Check with degree of unsaturation

For $\mathrm{C_{10}H_{16}}$, degree of unsaturation is $\text{IHD} = \frac{2C+2-H}{2} = \frac{2(10)+2-16}{2} = \frac{6}{2} = 3$

Thus total unsaturation is $3$. Since the compound takes up $3$ moles of $\mathrm{H_2}$ per mole of hydrocarbon, all three unsaturations are double bonds.

Hence, the number of double bonds is $\boxed{3}$.

5. Comparison with stored answer

Stored correct answer = $3$

Derived answer = $3$

They agree.