Explanation

(d)
10 volume =1 volume of $H_{2}O_{2}$ gives 10 volume of $O_{2}$ at NTP.
$\ \ \ \ \ \ 2H_{2}O_{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \rightarrow 2H_{2}O\ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ O_{2}$
$2(2 + 32) = 68\ g\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 22400\ mL\ at\ NTP$
At NTP
$\because\ \ \ \ \ \ \ \ 22400\ mL\ of\ O_{2}\ is\ obtained\ from\$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 68\ g\ H_{2}O_{2}$
$\therefore\ \ \ \ \ \ \ 10\ mL\ of\ O_{2}\ is\ obtained\ from$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{68 \times 10}{22400} = 0.03035\ g\ H_{2}O_{2}$
$1\ mL\ of\ H_{2}O_{2}\$solution contains
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.03035\ g\ H_{2}O_{2}$
$100\ mL\ of\ H_{2}O_{2}\$solution contains
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.03035 \times 100$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3.035\ g\ H_{2}O_{2}$
$\therefore$ Strength of 10 volume $H_{2}O_{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3.035 \times 10$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 30.35\ \ g/L$