A large open top container of negligible mass and uniform cross sectional area `A` a has a small hole of cross sectional area `A//100` in its side wall near the bottom. The container is kept on a smooth horizontal floor and contains a liquid of density `rho` and mass `M_(0)`. Assuming that the liquid starts flowing out horizontally through the hole at `t=0`, calculate a the acceleration of the container and its velocity when `75%` of the liquid has drained out.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the initial conditions
- The initial mass of the liquid in the container is \( M_0 \).
- The cross-sectional area of the hole is \( A/100 \).
- The density of the liquid is \( \rho \).

### Step 2: Calculate the height of the liquid when 75% has drained out
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