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Step-by-Step Solution
Step 1: Understand the Physical Situation
A mass of 200 g (which is $0.2\,\text{kg}$) is attached to a spring of spring constant $k = 12.5\,\text{N/m}$. The other end of the spring is fixed at point O. The mass moves in a circular path on a frictionless horizontal surface with a constant angular speed $\omega = 5\,\text{rad/s}$. Let $\ell$ be the natural length of the spring and $x$ be the extension in the spring. Therefore, the total length from O to the mass during circular motion is $(\ell + x)$.
Step 2: Write the Force Balance (Centripetal Force)
For uniform circular motion, the required centripetal force $F_{\text{centripetal}}$ is provided by the spring’s restoring force $F_{\text{spring}}$:
$$
F_{\text{spring}} = k\,x, \quad F_{\text{centripetal}} = m\,\omega^2(\ell + x).
$$
Since these two forces must be equal for uniform circular motion,
$$
k\,x = m \,\omega^2 (\ell + x).
$$
Step 3: Plug in the Numerical Values
Substitute $k = 12.5\,\text{N/m}$, $m = 0.2\,\text{kg}$, and $\omega = 5\,\text{rad/s}$ into the equation:
$$
12.5 \, x = \left(0.2\right)\,\left(5^2\right)\,(\ell + x).
$$
Notice that $0.2 = \tfrac{1}{5}$. Hence,
$$
12.5 \, x = \bigl(\tfrac{1}{5}\bigr)\,(25)\,(\ell + x).
$$
Step 4: Simplify the Equation
Simplify the right side:
$$
12.5 \, x = 5(\ell + x).
$$
Distribute the 5:
$$
12.5 \, x = 5\,\ell + 5\,x.
$$
Bring the terms involving $x$ on one side:
$$
12.5 \, x - 5\,x = 5\,\ell.
$$
$$
7.5 \, x = 5\,\ell.
$$
Divide both sides by 7.5:
$$
x = \frac{5}{7.5} \,\ell \quad\Rightarrow\quad x = \frac{2}{3}\,\ell.
$$
Step 5: Conclude the Ratio
Therefore, the ratio of the extension in the spring $(x)$ to its natural length $(\ell)$ is:
$$
\frac{x}{\ell} = \frac{2}{3}.
$$
In other words, the ratio is $2 : 3$, which matches the correct answer.
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