An open tank `10 m` long and `2 m` deep is filled up to `1.5 m` height of oil of specific gravity `0.82`. The tank is uniformly accelerated along its length from rest to a speed of `20 m//s` horizontally. The shortest time in which the speed may be attained without spilling any oil is

Options

A. `20s`
B. `18s`
C. `10s`
D. `5s`

Explanation

To solve the problem, we need to determine the shortest time in which the tank can accelerate to a speed of 20 m/s without spilling any oil. Here’s a step-by-step solution:

### Step 1: Understand the problem
We have an open tank that is 10 m long and 2 m deep, filled with oil to a height of 1.5 m. The specific gravity of the oil is 0.82. The tank is accelerated horizontally, and we need to find the time it takes to reach a speed of 20 m/s without spilling the oil.

### Step 2: Calculate the density of the oil
The specific gravity (SG) of the oil is given as 0.82. The density of water is approximately 1000 kg/m³. Therefore, the density of the oil (ρ_oil) can be calculated as:
\[ \rho_{\text{oil}} = \text{SG} \times \rho_{\text{water}} = 0.82 \times 1000 \, \text{kg/m}^3 = 820 \, \text{kg/m}^3 \]

### Step 3: Determine the angle of the oil surface due to acceleration
When the tank accelerates, the oil surface will tilt. The angle (θ) can be determined using the relationship between the acceleration (a) of the tank and the gravitational acceleration (g = 9.81 m/s²):
\[ \tan(\theta) = \frac{a}{g} \]

### Step 4: Relate the tilt of the oil surface to the height difference
The height difference (h) across the length of the tank (10 m) can be expressed as:
\[ h = L \cdot \tan(\theta) = 10 \cdot \tan(\theta) \]
Substituting for tan(θ):
\[ h = 10 \cdot \frac{a}{g} \]

### Step 5: Calculate the maximum height difference before spilling
To avoid spilling, the height of the oil must not exceed the tank's height of 1.5 m. Therefore:
\[ h \leq 1.5 \, \text{m} \]
Substituting for h:
\[ 10 \cdot \frac{a}{g} \leq 1.5 \]
\[ \frac{a}{g} \leq 0.15 \]
\[ a \leq 0.15 \cdot g = 0.15 \cdot 9.81 \approx 1.4715 \, \text{m/s}^2 \]

### Step 6: Calculate the required acceleration to reach 20 m/s
Using the first equation of motion:
\[ v = u + at \]
Where:
- v = final velocity = 20 m/s
- u = initial velocity = 0 m/s
- a = acceleration
- t = time

Rearranging gives:
\[ t = \frac{v - u}{a} = \frac{20}{a} \]

### Step 7: Substitute the maximum acceleration
Substituting the maximum acceleration we found:
\[ t = \frac{20}{1.4715} \approx 13.59 \, \text{s} \]

### Conclusion
The shortest time in which the speed may be attained without spilling any oil is approximately **13.59 seconds**.

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