'A' and 'B' are the two pegs separated by 13 cm. A body of 169 Kgwt is suspended by thread of 17 cm connecting to A & B, such that the two segments of strings are perpendicular. Then tensions in shorter and longer parts of string having are

Options

A. 100 kgwt, 69 kgwt
B. 65 kgwt, 156. kgwt
C. 156 kgwt, 65 kgwt
D. 69 kgwt, 100 kgwt

Answer

To solve the problem, we need to find the tensions in the shorter and longer parts of the string connecting pegs A and B. Let's break this down step-by-step.

### Step 1: Understand the Geometry
We have two pegs A and B that are 13 cm apart. A body weighing 169 kg is suspended by a string of total length 17 cm, with the segments of the string being perpendicular to each other.

### Step 2: Define Variables
Let:
- \( T_1 \) = tension in the shorter part of the string
- \( T_2 \) = tension in the longer part of the string
- \( X \) = length of the shorter segment of the string
- \( 17 - X \) = length of the longer segment of the string

### Step 3: Apply Pythagorean Theorem
Since the segments of the string are perpendicular, we can apply the Pythagorean theorem:
\[ X^2 + (17 - X)^2 = 13^2 \]
Expanding this:
\[ X^2 + (289 - 34X + X^2) = 169 \]
Combining like terms:
\[ 2X^2 - 34X + 120 = 0 \]

### Step 4: Solve the Quadratic Equation
Using the quadratic formula \( X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 2 \), \( b = -34 \), and \( c = 120 \).
\[ X = \frac{34 \pm \sqrt{(-34)^2 - 4 \cdot 2 \cdot 120}}{2 \cdot 2} \]
Calculating the discriminant:
\[ 34^2 - 960 = 1156 - 960 = 196 \]
So,
\[ X = \frac{34 \pm 14}{4} \]
Calculating the two possible values:
\[ X_1 = \frac{48}{4} = 12 \quad \text{and} \quad X_2 = \frac{20}{4} = 5 \]

### Step 5: Determine Cosine Values
Now, we can find \( \cos \theta_1 \) and \( \cos \theta_2 \):
- For \( X = 5 \): \( \cos \theta_1 = \frac{5}{13} \) and \( \cos \theta_2 = \frac{12}{13} \)
- For \( X = 12 \): \( \cos \theta_1 = \frac{12}{13} \) and \( \cos \theta_2 = \frac{5}{13} \)

### Step 6: Set Up the Tension Equations
From the equilibrium of forces in the horizontal direction:
\[ T_1 \cos \theta_1 = T_2 \cos \theta_2 \]
Substituting the cosine values:
\[ T_1 \cdot \frac{5}{13} = T_2 \cdot \frac{12}{13} \]
Cancelling \( \frac{13}{13} \):
\[ T_1 = \frac{12}{5} T_2 \quad \text{(Equation 1)} \]

### Step 7: Set Up the Vertical Force Balance
From the equilibrium of forces in the vertical direction:
\[ T_1 \sin \theta_1 + T_2 \sin \theta_2 = 169 \]
Substituting the sine values:
\[ T_1 \cdot \frac{12}{13} + T_2 \cdot \frac{5}{13} = 169 \]
Multiplying through by 13:
\[ 12T_1 + 5T_2 = 169 \cdot 13 \quad \text{(Equation 2)} \]

### Step 8: Substitute Equation 1 into Equation 2
Substituting \( T_1 = \frac{12}{5} T_2 \) into Equation 2:
\[ 12\left(\frac{12}{5} T_2\right) + 5T_2 = 169 \cdot 13 \]
This simplifies to:
\[ \frac{144}{5} T_2 + 5T_2 = 169 \cdot 13 \]
Converting \( 5T_2 \) to a fraction:
\[ \frac{144}{5} T_2 + \frac{25}{5} T_2 = 169 \cdot 13 \]
Combining terms:
\[ \frac{169}{5} T_2 = 169 \cdot 13 \]
Cancelling \( 169 \):
\[ T_2 = 13 \cdot 5 = 65 \text{ kg} \]

### Step 9: Find \( T_1 \)
Now substituting \( T_2 \) back into Equation 1:
\[ T_1 = \frac{12}{5} \cdot 65 = 156 \text{ kg} \]

### Final Result
Thus, the tensions in the shorter and longer parts of the string are:
- \( T_1 = 156 \text{ kg} \)
- \( T_2 = 65 \text{ kg} \)