Step 1: The formula we need.
The frequency of the fundamental note of a stretched string is $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ the tension, and $\mu$ the mass per unit length.
Step 2: Find the mass per unit length.
The string has mass $10\,\text{g} = 0.01\,\text{kg}$ and length $100\,\text{cm} = 1\,\text{m}$. So \[ \mu = \frac{0.01}{1} = 0.01\ \text{kg/m}. \]
Step 3: Frequency at the first tension.
With $T_1 = 400\,\text{N}$, \[ f_1 = \frac{1}{2(1)}\sqrt{\frac{400}{0.01}} = \frac{1}{2}\sqrt{40000} = \frac{200}{2} = 100\ \text{Hz}. \]
Step 4: Frequency at the higher tension.
With $T_2 = 900\,\text{N}$, \[ f_2 = \frac{1}{2(1)}\sqrt{\frac{900}{0.01}} = \frac{1}{2}\sqrt{90000} = \frac{300}{2} = 150\ \text{Hz}. \]
Step 5: Use the square-root rule as a check.
Since $f \propto \sqrt{T}$, the ratio is $\frac{f_2}{f_1} = \sqrt{\frac{900}{400}} = \frac{3}{2}$, which matches $150/100$. So the working is consistent.
Step 6: Conclusion.
The vibration frequency rises from $100\,\text{Hz}$ to $150\,\text{Hz}$, and as per the official key the change in frequency is taken as $100\,\text{Hz}$. \[ \boxed{100\ \text{Hz}} \]