Instead of first computing the depreciation rate as a percentage and reapplying it, it's quicker to build the two-year multiplier directly from the ratio S/P using indices.
Step 1: Write the book value formula symbolically. Under the declining balance method, after $k$ years the book value is $BV_k = P(1-r)^k$, where $r = 1-(S/P)^{1/n}$. Since $(1-r) = (S/P)^{1/n}$, this means $BV_k = P \times (S/P)^{k/n}$. This is a shortcut: we never actually need $r$ on its own, since $(S/P)^{1/n}$ raised to the $k$th power is exactly $(S/P)^{k/n}$.
Step 2: Plug in the numbers for year 2. Here $P = 20$, $S = 2$, $n = 10$ and $k = 2$, so $S/P = 2/20 = 0.1$ and $k/n = 2/10 = 0.2$.
$BV_2 = 20 \times (0.1)^{0.2}$
$(0.1)^{0.2} = 0.6310$, so $BV_2 = 20 \times 0.6310 = 12.62$ crores.
Step 3: Cross check against the year-by-year method. Working year by year gives $r = 1-(0.1)^{0.1} = 0.2057$, $BV_1 = 20(0.7943) = 15.89$ crores, and $BV_2 = 15.89(0.7943) = 12.62$ crores. Both routes land on the same number, since $(1-r)^2 = ((S/P)^{1/n})^2 = (S/P)^{2/n}$.
It's worth being clear about what "depreciated cost in its 2nd year" means here. Taken as the book value remaining at the end of year 2, the answer is 12.62 crores. Taken instead as just the depreciation charged during year 2 alone (the drop from $BV_1$ to $BV_2$), the answer would be $15.89 - 12.62 = 3.27$ crores. This solution answers the first reading, the remaining depreciated value of the asset.
Let's summarize:
So the depreciated cost of the shovel at the end of year 2 is Rs. 12.62 crores.