A sonometer wire of length 100cm under a tension of 10N, has a frequency of 250Hz. Keeping the length of the wire constant, the tension is adjusted to produce a new frequency of 350Hz. The new tension is
- 5.1N
- 7.1N
- 14.0N
-
19.6N
The correct answer is: D
Explanation
\(\frac{F_1}{F_2}\) = \(\sqrt\frac{T_1}{T_2}\)\(\frac{250}{350}\) = \(\sqrt\frac{10}{T_2}\)
T2 = 19.6N