A body of mass 'm' is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is 1 cm. If the angular velocity is doubled, the elongation in the spring is 5 cm. The original length of the spring is :

A body of mass $$ m $$ is tied to a spring and whirled in a horizontal plane with constant angular velocity. The spring elongates due to the centrifugal force acting on the body. Let the original length of the spring be $$ L $$ cm. In the first scenario, with angular velocity $$ \omega $$, the elongation is 1 cm. Therefore, the total length of the spring becomes $$ L + 1 $$ cm, which is the radius of the circular path.

The centrifugal force is given by $$ m \omega^2 r $$, where $$ r $$ is the radius. This force is balanced by the spring force, which is $$ k \times \text{elongation} $$, where $$ k $$ is the spring constant. For the first case:

$$ k \times 1 = m \omega^2 (L + 1) $$

Simplifying, we get:

$$ k = m \omega^2 (L + 1) \quad \text{(Equation 1)} $$

When the angular velocity is doubled to $$ 2\omega $$, the elongation becomes 5 cm. The total length is now $$ L + 5 $$ cm. The force balance equation becomes:

$$ k \times 5 = m (2\omega)^2 (L + 5) $$

Simplifying the right side:

$$ 5k = m \cdot 4\omega^2 (L + 5) $$

$$ 5k = 4 m \omega^2 (L + 5) \quad \text{(Equation 2)} $$

Substitute the expression for $$ k $$ from Equation 1 into Equation 2:

$$ 5 \left[ m \omega^2 (L + 1) \right] = 4 m \omega^2 (L + 5) $$

Divide both sides by $$ m \omega^2 $$ (assuming $$ m \neq 0 $$ and $$ \omega \neq 0 $$):

$$ 5 (L + 1) = 4 (L + 5) $$

Expand both sides:

$$ 5L + 5 = 4L + 20 $$

Subtract $$ 4L $$ from both sides:

$$ 5L - 4L + 5 = 20 $$

$$ L + 5 = 20 $$

Subtract 5 from both sides:

$$ L = 15 $$

Thus, the original length of the spring is 15 cm.

Hence, the correct answer is Option A.