\sqrt{n}-\sqrt{n+1}<.01
\Rightarrow (\sqrt{n})^2<(\sqrt{n+1}+.01)^2
\Rightarrow n<n-1+.02\sqrt{n-1}+.0001
\Rightarrow 1<.02\sqrt{n-1}+.0001
\Rightarrow 50<\sqrt{n-1}+.005
Edit: sorry the ans. is 2501.
Thank you @rishabh
\hspace{-16}$Find the smallest positive Integer $\mathbf{n}$ such that \\\\ $\mathbf{\sqrt{n}-\sqrt{n-1}<0.01}$
\sqrt{n}-\sqrt{n+1}<.01
\Rightarrow (\sqrt{n})^2<(\sqrt{n+1}+.01)^2
\Rightarrow n<n-1+.02\sqrt{n-1}+.0001
\Rightarrow 1<.02\sqrt{n-1}+.0001
\Rightarrow 50<\sqrt{n-1}+.005
Edit: sorry the ans. is 2501.
Thank you @rishabh
@arnab.
i think under the root it should have been n-1 rather then n+1 in the 2nd step RHS.