Hypersurfaces without variable cohomology Let $X$ be a smooth projective variety over $\mathbb C$ of dimension $n+1$. If $Y$ is a smooth very ample hypersurface, we know that except $H^n(X;\mathbb Q)\rightarrow H^n(Y;\mathbb Q)$, the restriction of cohomology (rational coefficients) at other degrees are surjective. My question is the following.
What are the restrictions we can put on $X$, so that there exists a smooth very ample hypersurface $Y$ such that the restriction $H^*(X;\mathbb Q)\rightarrow H^*(Y;\mathbb Q)$ is surjective on all degrees. For example, I would love to know whether it is true if $X$ is a smooth projective toric variety (or with more restrictions). We can replace very ample by ample if it's better. Boundary divisors(torus invariant divisors) satisfies the cohomology requirements, but they might not even be ample. Other possible conditions for $X$ are welcomed as well.
(Should this question be made into community wiki?)

p.s.
My motivation is that in my project, I want to find an ample hypersurface all of whose cohomology classes can be lifted to the ambient space. But uniqueness is not needed. I realized such a hypersurface may not be found for just any variety (e.g., hypersurface in $\mathbb P^n$ if the degree is sufficiently large). So I want to put a reasonable condition on $X$. I have tried to decompose $H^n(Y;\mathbb Q)$ into $H^n(Y;\mathbb Q)=H^n_{fix}(Y;\mathbb Q)\oplus H^n_{var}(Y;\mathbb Q)$ where $H^n_{fix}(Y;\mathbb Q)$ is the image of the restriction from $X$, and $H^n_{var}(Y;\mathbb Q)$ is generated by the vanishing cycles in a Lefschetz pencil. I kind of feel that requiring the monodromy to be trivial does not simplify the problem. But I also realized that if the monodromy $H^n_{var}(Y;\mathbb Q)$ is "very big", I will be equally happy in my project. But what I want for "very big" is quite a messy condition. Any discussion along this line is also welcomed in the comment because there are definitely nice results about monodromy that I'm not aware about. 
 A: Suppose $X$ is a regular surface, i.e. $H^1(X)=0$, e.g. a toric surface. Then $Y$ is a smooth rational curve. So $X$ has a very ample smooth rational curve. This immediately requires $X$ to be rational by taking a pencil of smooth rational curves. Does every rational surface have such a curve? It's ok for projective space (lines) and Hirzebruch surfaces (high degree sections).

Not for $\mathbb P^2$ blown up at $k$ points for $k \geq 2$. Let $D = n H - \sum_{i=1}^k m_i E_i$. Then $D \cdot D = n^2 - \sum_{i=1}^k m_i^2$ and $K = -3 H + \sum_{i=1}^k E_i$ so $D \cdot K = -3n + \sum_{i=1}^k m_i$
By adjunction we have
$$-2= D \cdot D + D \cdot K = n^2 - \sum_{i=1}^k m_i^2  -3n + \sum_{i=1}^k m_i$$
$$ = \left(n- \sum_{i=1}^k m_i\right) \left(n + \sum_{i=1}^k m_i\right) +\sum_{1 \leq i< j \leq k} m_i m_j -3n + \sum_{i=1}^k m_i$$
$$ = \left(n- \sum_{i=1}^k m_i-1\right) \left(n + \sum_{i=1}^k m_i-2\right) +\sum_{1 \leq i< j \leq k} m_i m_j -2$$
Thus
$$\left(n- \sum_{i=1}^k m_i-1\right) \left(n + \sum_{i=1}^k m_i-2\right) +\sum_{1 \leq i< j \leq k} m_i m_j=0$$
But in fact it is positive:
$\left(n- \sum_{i=1}^k m_i-1\right)>0$ by ampleness because it is the intersection number with the line through the points.
$m_i\geq 1$ by ampleness because it's the intersection number with $E_i$, so  $\left(n + \sum_{i=1}^k m_i-2\right) >0$  and $m_im_j>0$
This is a contradiction so this is impossible. When $k=2$ this variety is toric, showing that being toric is not a sufficient condition.
