When does the first right derived functor of $\otimes O_Y$ is zero? Let $X$ be a variety, $V$ is a hypersurface in $X$, and $Y$ is a closed subvariety in $X$ which intersects with $V$ transverally. Suppose the corresponding closed embeddings are:$i:V \to X,\quad j: Y \to X$ .
Then we have the following exact sequence:


$0 \to L(-V) \to O_X \to i_*O_V \to 0$


One can tensor it with $j_*O_Y$ to get an exact sequence. But I don't know how does transversality play the role to make it left exact.
 A: There is a weaker condition then being transversal (which is usually defined under smoothness condition as in the comment of Matt E): namely $V$ and $Y$ cut properly if $V$ doesn't contain any irreducible component of $Y$. If $Y$ is irreducible (or pure), this just means that $Y\cap V$ has dimension $\dim Y-1$. 
Now suppose $Y$ is integral and cuts $V$ properly, then tensoring your exact sequence by $O_Y$ is exact. Indeed, the question is local on $X$. Let $x\in Y$, let $f$ be a generator of $L(-V)_x$ (local equation of $V$ at $x$). We have to show that 
$$ fO_{X,x} \otimes_{O_{X,x}} O_{Y,x} \to O_{Y,x}$$ 
is injective. Write for simplicity $A=O_{X,x}$, $O_{Y,x}=A/I$. Then the above map is
$$ fA/fI \to A/I.$$ 
Its kernel if $(fA\cap I)/fI$. Let $g=fa\in fA\cap I$. Then in $A/I$, we have $\bar{f}\bar{a}=0$. As $A/I=O_{Y,x}$ is an integral domain and $\bar{f}\ne 0$ because $V$ doesn't contain $Y$, we find $\bar{a}=0$, hence $g\in fI$ and $fA\cap I=fI$. Thus the injectivity. 
In general (without integral hypothesis on $Y$), we see by this proof that the exactness at left is equivalent to the image of $f$ in $O_{Y,x}$ is a regular element. This condition is little stronger than $V, Y$ cut properly (we could probably say something like them are secant). 
