Answer:
Option B
Explanation:
Given
Rate old= k[A]x .........(i)
If concentration of reaction 'A' is increased by 10 times that means,
Ratenew = k[10A]x
rate of reaction becomes 100 times,
100 x rateold =k[10A]x ...........(ii)
from Eq.(i) and (ii)
$100\times k[A]^{x}=k[10A]^{x}$
100=10x
$10^{2}=10^{x}\Rightarrow x=2$