`A` and `B` are 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 are

Options

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

Explanation

To solve the problem of finding the tensions in the shorter and longer parts of the string, we can follow these steps:

### Step 1: Understand the Setup
We have two pegs, A and B, separated by a distance of 13 cm. A body weighing 169 kg (which is equivalent to 169 kgwt) is suspended by a thread of total length 17 cm. The segments of the string connecting to the pegs are perpendicular to each other.

### Step 2: Define Variables
Let:
- \( T_1 \) = Tension in the shorter segment of the string
- \( T_2 \) = Tension in the longer segment 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 Geometry
Since the segments of the string are perpendicular, we can use the Pythagorean theorem:
\[ x^2 + (17 - x)^2 = 13^2 \]
Expanding this gives:
\[ x^2 + (17^2 - 34x + x^2) = 169 \]
\[ 2x^2 - 34x + 289 - 169 = 0 \]
\[ 2x^2 - 34x + 120 = 0 \]

### 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 \( c = 120 \).
\[ x = \frac{34 \pm \sqrt{(-34)^2 - 4 \cdot 2 \cdot 120}}{2 \cdot 2} \]
\[ x = \frac{34 \pm \sqrt{1156 - 960}}{4} \]
\[ x = \frac{34 \pm \sqrt{196}}{4} \]
\[ x = \frac{34 \pm 14}{4} \]
Calculating the two possible values:
1. \( x = \frac{48}{4} = 12 \) cm
2. \( x = \frac{20}{4} = 5 \) cm

### Step 5: Assign Values
- The shorter segment \( x = 5 \) cm
- The longer segment \( 17 - x = 12 \) cm

### Step 6: Find Angles
Using the definitions of cosine:
\[ \cos \theta_1 = \frac{5}{13}, \quad \cos \theta_2 = \frac{12}{13} \]
Using the definitions of sine:
\[ \sin \theta_1 = \frac{12}{13}, \quad \sin \theta_2 = \frac{5}{13} \]

### Step 7: Set Up Equations
From the balance of horizontal components:
\[ T_1 \cos \theta_1 = T_2 \cos \theta_2 \]
Substituting the values:
\[ T_1 \cdot \frac{5}{13} = T_2 \cdot \frac{12}{13} \]
This simplifies to:
\[ T_1 = \frac{12}{5} T_2 \]

From the balance of vertical components:
\[ T_1 \sin \theta_1 + T_2 \sin \theta_2 = 169 \]
Substituting the values:
\[ T_1 \cdot \frac{12}{13} + T_2 \cdot \frac{5}{13} = 169 \]
Multiplying through by 13:
\[ 12T_1 + 5T_2 = 169 \cdot 13 \]
\[ 12T_1 + 5T_2 = 2197 \]

### Step 8: Substitute for \( T_1 \)
Substituting \( T_1 = \frac{12}{5} T_2 \) into the vertical equation:
\[ 12 \left(\frac{12}{5} T_2\right) + 5T_2 = 2197 \]
\[ \frac{144}{5} T_2 + 5T_2 = 2197 \]
Multiplying through by 5 to eliminate the fraction:
\[ 144T_2 + 25T_2 = 10985 \]
\[ 169T_2 = 10985 \]
\[ T_2 = \frac{10985}{169} = 65 \text{ kgwt} \]

### Step 9: Find \( T_1 \)
Using \( T_1 = \frac{12}{5} T_2 \):
\[ T_1 = \frac{12}{5} \cdot 65 = 156 \text{ kgwt} \]

### Final Answer
- Tension in the shorter part of the string \( T_1 = 156 \text{ kgwt} \)
- Tension in the longer part of the string \( T_2 = 65 \text{ kgwt} \)

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