If the banking angle of curved road is given by `tan^(-1)((3)/(5))` and the radius of curvature of the road is 6 m, then the safe driving speed should not exceed (take, `g=10ms^(-2)`)

Options

A. `86.4"km h"^(-1)`
B. `43.2"km h"^(-1)`
C. `21.6"km h"^(-1)`
D. `30.4"km h"^(-1)`

Explanation

To find the safe driving speed on a banked curve, we can use the relationship between the banking angle, radius of curvature, and gravitational acceleration. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Banking angle, θ = tan^(-1)(3/5)
- Radius of curvature, r = 6 m
- Acceleration due to gravity, g = 10 m/s²

### Step 2: Calculate the value of tan(θ)
From the given banking angle:
\[ \tan(\theta) = \frac{3}{5} \]

### Step 3: Write down the formula for safe speed on a banked road
The formula relating the safe speed (v) on a banked road is given by:
\[ \tan(\theta) = \frac{v^2}{rg} \]
Where:
- v = safe speed
- r = radius of curvature
- g = acceleration due to gravity

### Step 4: Substitute the known values into the formula
Substituting the values we have:
\[ \frac{3}{5} = \frac{v^2}{6 \times 10} \]

### Step 5: Simplify the equation
This simplifies to:
\[ \frac{3}{5} = \frac{v^2}{60} \]

### Step 6: Cross-multiply to solve for v²
Cross-multiplying gives:
\[ 3 \times 60 = 5v^2 \]
\[ 180 = 5v^2 \]

### Step 7: Solve for v²
Dividing both sides by 5:
\[ v^2 = \frac{180}{5} = 36 \]

### Step 8: Take the square root to find v
Taking the square root of both sides:
\[ v = \sqrt{36} = 6 \text{ m/s} \]

### Step 9: Convert speed from m/s to km/h
To convert from meters per second to kilometers per hour, we use the conversion factor \( \frac{18}{5} \):
\[ v = 6 \times \frac{18}{5} = \frac{108}{5} = 21.6 \text{ km/h} \]

### Final Answer
The safe driving speed should not exceed **21.6 km/h**.

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