Question

The tension in a piano wire is increased by \(25\ \%\). Its frequency becomes "\(nf\)", where f is the original frequency, then, the value of n is nearly

• A. \(\sqrt{0.8}\)
• B. \(\sqrt{1.25}\)
• C. \(\sqrt{1.12}\)
• D. \(\sqrt{1.56}\)

Hint:

Frequency of a stretched wire is proportional to the square root of the tension.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The fundamental frequency of a stretched wire is \(f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}\), so \(f\propto\sqrt{T}\) when \(l\) and \(\mu\) are fixed.

Step 2: Change in tension
A \(25\%\) increase makes the new tension \(T'=1.25\,T\).

Step 3: Calculation
\(\dfrac{f'}{f}=\sqrt{\dfrac{T'}{T}}=\sqrt{1.25}\)

Step 4: Result
So \(n=\sqrt{1.25}\), option (B).

Final Answer:
The value of n is \(\sqrt{1.25}\). \[ \boxed{\text{(B) }\sqrt{1.25}} \]