Tension in a uniform string of length L varies with length uniformly. At one end A tension is `T_(0)` and at other end B it is `16T_(0)`. If mass per unit length of string is `mu`. Choose the correct options (s)

Options

A. Time taken by a pulse to move from end A and B is `(2)/(5)Lsqrt((mu)/(T_(0)))`
B. Time taken by a pulse to move from end A to end B is `(2)/(15)sqrt((muL)/(T_(0)))(sqrt(15)-1)`
C. Tension in the string as function of distance x from end A is `T=T_(0)+(15T_(0))/(L)x`
D. Tension in the string as function of distance x from end A is `T=T_(0)+(15T_(0))/(L)x`

Explanation

To solve the problem, we need to analyze the tension in the uniform string and derive the relationship for tension as a function of distance along the string. We will also determine the time taken for a wave pulse to travel along the string.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The tension at one end (A) of the string is \( T_0 \) and at the other end (B) is \( 16T_0 \).
- The length of the string is \( L \) and the mass per unit length is \( \mu \).

2. **Finding the Tension as a Function of Distance**:
- Since the tension varies uniformly from \( T_0 \) to \( 16T_0 \), we can express the tension \( T(x) \) at a distance \( x \) from end A as:
\[ T(x) = T_0 + \left( \frac{16T_0 - T_0}{L} \right) x \]
- Simplifying this gives:
\[ T(x) = T_0 + \frac{15T_0}{L} x \]
- Therefore, the tension at any point \( x \) along the string is:
\[ T(x) = T_0 + \frac{15T_0}{L} x \]

3. **Finding the Wave Velocity**:
- The velocity \( v \) of a wave in a string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
- Substituting \( T(x) \) into this equation gives:
\[ v(x) = \sqrt{\frac{T_0 + \frac{15T_0}{L} x}{\mu}} \]

4. **Time Taken for a Pulse to Travel**:
- The time taken \( dt \) for a small distance \( dx \) can be expressed as:
\[ dt = \frac{dx}{v(x)} = \frac{dx}{\sqrt{\frac{T_0 + \frac{15T_0}{L} x}{\mu}}} \]
- Rearranging gives:
\[ dt = \frac{dx \sqrt{\mu}}{\sqrt{T_0 + \frac{15T_0}{L} x}} \]
- To find the total time \( T \) taken for the pulse to travel from \( x = 0 \) to \( x = L \), we integrate:
\[ T = \int_0^L \frac{\sqrt{\mu}}{\sqrt{T_0 + \frac{15T_0}{L} x}} \, dx \]

5. **Evaluating the Integral**:
- Let \( u = T_0 + \frac{15T_0}{L} x \), then \( du = \frac{15T_0}{L} dx \) or \( dx = \frac{L}{15T_0} du \).
- Changing the limits accordingly, when \( x = 0 \), \( u = T_0 \) and when \( x = L \), \( u = 16T_0 \).
- The integral becomes:
\[ T = \int_{T_0}^{16T_0} \frac{\sqrt{\mu}}{\sqrt{u}} \cdot \frac{L}{15T_0} \, du \]
- Evaluating this integral gives:
\[ T = \frac{L \sqrt{\mu}}{15T_0} \left[ 2\sqrt{u} \right]_{T_0}^{16T_0} = \frac{L \sqrt{\mu}}{15T_0} \left( 2\sqrt{16T_0} - 2\sqrt{T_0} \right) \]
- Simplifying this results in:
\[ T = \frac{L \sqrt{\mu}}{15T_0} \cdot 2(4 - 1)\sqrt{T_0} = \frac{6L \sqrt{\mu T_0}}{15T_0} = \frac{2L \sqrt{\mu}}{5\sqrt{T_0}} \]

### Final Answers:
- The tension in the string as a function of distance \( x \) is:
\[ T(x) = T_0 + \frac{15T_0}{L} x \]
- The time taken for a pulse to travel from A to B is:
\[ T = \frac{2L \sqrt{\mu}}{5\sqrt{T_0}} \]