A load suspended by a massless spring produces an extention of xcm in equilibrium.When it is cut into two unequal parts ,the same load produces an extention of 7.5cm when suspended by the larger part of length 60cm .When it is suspended by the smaller part,the extention is 5cm.Then

Options

A. x=12.5cm
B. x=3.0cm
C. the length of the original spring is 90cm
D. the length of the original spring is 80cm

Explanation

To solve the problem step-by-step, we will follow the principles of elasticity and the relationships between force, extension, and length of the spring.

### Step 1: Understand the Problem
We have a spring that produces an extension \( x \) cm when a load is suspended from it. The spring is then cut into two unequal parts. The larger part (length = 60 cm) produces an extension of 7.5 cm, and the smaller part produces an extension of 5 cm.

### Step 2: Set Up the Relationships
According to Hooke's Law, the extension produced in a spring is directly proportional to the load applied and the original length of the spring. Therefore, we can write:

\[ \frac{\Delta L_1}{\Delta L_2} = \frac{L_1}{L_2} \]

Where:
- \( \Delta L_1 \) is the extension of the smaller part (5 cm).
- \( \Delta L_2 \) is the extension of the larger part (7.5 cm).
- \( L_1 \) is the length of the smaller part.
- \( L_2 \) is the length of the larger part (60 cm).

### Step 3: Calculate the Length of the Smaller Part
Let \( L_1 \) be the length of the smaller part. We know \( L_2 = 60 \) cm and \( \Delta L_2 = 7.5 \) cm. Plugging in the values:

\[ \frac{5}{7.5} = \frac{L_1}{60} \]

Cross-multiplying gives:

\[ 5 \times 60 = 7.5 \times L_1 \]

\[ 300 = 7.5 \times L_1 \]

Now, solving for \( L_1 \):

\[ L_1 = \frac{300}{7.5} = 40 \text{ cm} \]

### Step 4: Find the Total Length of the Spring
The total length of the original spring \( L_0 \) is the sum of the lengths of the two parts:

\[ L_0 = L_1 + L_2 = 40 \text{ cm} + 60 \text{ cm} = 100 \text{ cm} \]

### Step 5: Calculate the Original Extension \( x \)
Now we can find the original extension \( x \) when the entire spring of length 100 cm is used. Using the same ratio of extensions:

\[ \frac{x}{\Delta L_2} = \frac{L_0}{L_2} \]

Substituting the known values:

\[ \frac{x}{7.5} = \frac{100}{60} \]

Cross-multiplying gives:

\[ x \times 60 = 7.5 \times 100 \]

\[ 60x = 750 \]

Now, solving for \( x \):

\[ x = \frac{750}{60} = 12.5 \text{ cm} \]

### Final Answer
The original extension \( x \) of the spring is **12.5 cm**.

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