Assertion A ball tied by thread is undergoing circular motion (of radius R) in a vertical plane. (Thread always remains in vertical plane). The difference of maximum and minimum tension in thread is independent of speed (u) of ball at the lowest position `(ugtsqrt(5gR))`.
Reason For a ball of mass m tied by thread undergoing vertical circular motion (of radius R), difference in maximum and minimum magnitude of centripetal aceleraion of the ball is independent of speed (u) of ball at the lowest position `(ugrsqrt5gR))`.

Options

A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
C. If Assertion is true but Reason is false.
D. If Assertion is false but Reason is true.

Explanation

To solve the problem, we need to analyze the forces acting on a ball tied by a thread undergoing vertical circular motion. We will derive the expressions for the maximum and minimum tension in the thread and show that their difference is independent of the speed of the ball at the lowest position.

### Step 1: Understanding the Forces at the Lowest Point
At the lowest point of the circular motion, the forces acting on the ball are:
- The weight of the ball (mg) acting downwards.
- The tension in the thread (T_b) acting upwards.

Using Newton's second law, we can write the equation for the centripetal force required to keep the ball in circular motion:
\[ T_b - mg = \frac{mu^2}{R} \]
Where:
- \( T_b \) is the tension at the bottom.
- \( u \) is the speed of the ball at the lowest point.
- \( R \) is the radius of the circular path.

Rearranging gives us:
\[ T_b = mg + \frac{mu^2}{R} \]

### Step 2: Understanding the Forces at the Highest Point
At the highest point of the circular motion, the forces acting on the ball are:
- The weight of the ball (mg) acting downwards.
- The tension in the thread (T_u) also acting downwards.

For the ball to maintain circular motion at this point, the centripetal force is provided by both the tension and the weight:
\[ T_u + mg = \frac{mv^2}{R} \]
Where:
- \( T_u \) is the tension at the top.
- \( v \) is the speed of the ball at the highest point.

Rearranging gives us:
\[ T_u = \frac{mv^2}{R} - mg \]

### Step 3: Finding the Relationship Between Speeds
To find the relationship between \( u \) and \( v \), we can use the conservation of mechanical energy. The total mechanical energy at the lowest point is equal to the total mechanical energy at the highest point:
\[ \frac{1}{2} mu^2 + mg(0) = \frac{1}{2} mv^2 + mg(2R) \]
This simplifies to:
\[ \frac{1}{2} mu^2 = \frac{1}{2} mv^2 + 2mgR \]
Rearranging gives:
\[ u^2 - v^2 = 4gR \]
Thus:
\[ v^2 = u^2 - 4gR \]

### Step 4: Finding the Difference in Tension
Now we can find the difference in tension:
\[ \Delta T = T_b - T_u \]
Substituting the expressions for \( T_b \) and \( T_u \):
\[ \Delta T = \left(mg + \frac{mu^2}{R}\right) - \left(\frac{m(u^2 - 4gR)}{R} - mg\right) \]

Simplifying gives:
\[ \Delta T = mg + \frac{mu^2}{R} - \left(\frac{mu^2}{R} - 4mg\right) \]
\[ \Delta T = mg + \frac{mu^2}{R} - \frac{mu^2}{R} + 4mg \]
\[ \Delta T = 5mg \]

### Conclusion
The difference in tension \( \Delta T \) is \( 5mg \), which is independent of the speed \( u \) of the ball at the lowest position. Therefore, both the assertion and the reason given in the problem statement are true.