A smooth circular tube of radius `R` is fixed in a vertical plane. A particle is projectid from its lowest point with a velocity just sufficient to carry it to th ehigest point. Show that the time taken by the particle to rach the end of the horizontal diameter is `(sqrt(R)/(g))In(1+sqrt(2))` .
Hint : `intsectheta.dtheta=In(sectheta+tantheta)`

Explanation

To solve the problem, we need to analyze the motion of a particle projected from the lowest point of a smooth circular tube of radius \( R \) and determine the time taken to reach the end of the horizontal diameter. Here’s a step-by-step solution:

### Step 1: Determine the initial velocity
The particle is projected with just enough velocity to reach the highest point of the circular path. At the highest point, the potential energy is maximum and the kinetic energy is minimum (ideally zero for just reaching the top).

Using the conservation of energy:
\[ \frac{1}{2} m v_0^2 = mg(2R) \]
where \( v_0 \) is the initial velocity, \( m \) is the mass of the particle, and \( g \) is the acceleration due to gravity.

Rearranging gives:
\[ v_0^2 = 4gR \]
Thus,
\[ v_0 = 2\sqrt{gR} \]

### Step 2: Relate the angular displacement to time
Let \( \theta \) be the angle the particle has moved from the vertical. The arc length \( s \) traveled by the particle is given by:
\[ s = R\theta \]
The time \( dt \) taken to travel this arc length at velocity \( v \) is:
\[ dt = \frac{ds}{v} = \frac{R d\theta}{v} \]

### Step 3: Find the velocity at angle \( \theta \)
Using conservation of energy again, we can find the velocity \( v \) at angle \( \theta \):
\[ \frac{1}{2} m v^2 = mg(2R - R(1 - \cos \theta)) \]
This simplifies to:
\[ \frac{1}{2} m v^2 = mg(R + R\cos \theta) \]
Thus,
\[ v^2 = 2g(R + R\cos \theta) = 2gR(1 + \cos \theta) \]
So,
\[ v = \sqrt{2gR(1 + \cos \theta)} \]

### Step 4: Substitute \( v \) into the time equation
Substituting \( v \) into the equation for \( dt \):
\[ dt = \frac{R d\theta}{\sqrt{2gR(1 + \cos \theta)}} \]
This can be rewritten as:
\[ dt = \frac{R^{1/2}}{\sqrt{2g}} \cdot \frac{d\theta}{\sqrt{1 + \cos \theta}} \]

### Step 5: Integrate to find the total time
To find the total time \( T \) taken to go from \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \):
\[ T = \int_0^{\frac{\pi}{2}} \frac{R^{1/2}}{\sqrt{2g}} \cdot \frac{d\theta}{\sqrt{1 + \cos \theta}} \]

Using the identity \( 1 + \cos \theta = 2\cos^2(\theta/2) \):
\[ \sqrt{1 + \cos \theta} = \sqrt{2}\cos(\theta/2) \]
Thus,
\[ T = \frac{R^{1/2}}{\sqrt{2g}} \int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{2}\cos(\theta/2)} = \frac{R^{1/2}}{2\sqrt{g}} \int_0^{\frac{\pi}{2}} \sec(\theta/2) d\theta \]

### Step 6: Evaluate the integral
The integral \( \int \sec(x) dx = \ln | \sec(x) + \tan(x) | + C \). Hence,
\[ \int_0^{\frac{\pi}{2}} \sec(\theta/2) d\theta = 2 \ln(\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})) = 2 \ln(\sqrt{2} + 1) \]

### Step 7: Final expression for time
Substituting back:
\[ T = \frac{R^{1/2}}{2\sqrt{g}} \cdot 2 \ln(\sqrt{2} + 1) = \frac{R^{1/2}}{\sqrt{g}} \ln(\sqrt{2} + 1) \]

Thus, the time taken by the particle to reach the end of the horizontal diameter is:
\[ T = \frac{\sqrt{R}}{g} \ln(1 + \sqrt{2}) \]