An open tank `10 m` long and `2m` deep is filled upto height `1.5 m` of oil of specific gravity `0.80`. The tank is accelerated uniformly from rest to a speed of `10m//sec`. The shortest time (in seconds) in which this speed may be attained without spilling anyoil (in sec). `[g = 10m//s^(2)]`

Explanation

To solve the problem step by step, we will analyze the situation and apply the principles of fluid mechanics and kinematics.

### Step 1: Understand the tank and fluid properties
- The tank is open and has dimensions: length = 10 m, depth = 2 m.
- The height of the oil in the tank is 1.5 m.
- The specific gravity of the oil is 0.80, which means its density (ρ) can be calculated as:
\[ \text{Density of oil} = \text{Specific gravity} \times \text{Density of water} = 0.80 \times 1000 \, \text{kg/m}^3 = 800 \, \text{kg/m}^3 \]

### Step 2: Determine the spilling condition
- When the tank is accelerated, the oil will tend to spill if the angle of inclination of the oil surface exceeds the horizontal.
- The condition for spilling can be derived from the balance of forces acting on the oil. The tangent of the angle (θ) formed by the oil surface with the horizontal can be expressed as:
\[ \tan(\theta) = \frac{a}{g} \]
where \( a \) is the acceleration of the tank and \( g \) is the acceleration due to gravity (10 m/s²).

### Step 3: Calculate the angle of inclination
- The height of the oil is 1.5 m, and the remaining depth of the tank is 0.5 m (2 m - 1.5 m).
- The horizontal distance (length of the tank) is 10 m.
- The height difference that causes the oil to spill is 0.5 m, so we can set up the equation:
\[ \tan(\theta) = \frac{0.5 \, \text{m}}{10 \, \text{m}} = \frac{1}{20} \]

### Step 4: Relate acceleration to spilling condition
- From the tangent equation, we have:
\[ \tan(\theta) = \frac{a}{g} \implies a = g \cdot \tan(\theta) = 10 \cdot \frac{1}{20} = 0.5 \, \text{m/s}^2 \]

### Step 5: Apply kinematic equations to find time
- We need to find the shortest time (t) to reach a speed of 10 m/s from rest with an acceleration of 0.5 m/s².
- Using the first equation of motion:
\[ V = U + aT \]
where \( V = 10 \, \text{m/s} \), \( U = 0 \, \text{m/s} \), and \( a = 0.5 \, \text{m/s}^2 \):
\[ 10 = 0 + 0.5T \implies T = \frac{10}{0.5} = 20 \, \text{seconds} \]

### Conclusion
The shortest time in which the tank may be accelerated to a speed of 10 m/s without spilling any oil is **20 seconds**.