A sonometer wire under a tension of N10N, produces a frequency of 250Hz when plucked. Keeping the length of the wire constant, the tension is adjusted to pr...

Answer Details

The relationship between tension (T) and frequency (f) for a string fixed at both ends, such as a sonometer wire, is given by the equation: f = (1/2L)√(T/μ) where L is the length of the string, μ is the linear mass density of the string. Since the length of the wire is kept constant, we can equate the two expressions for frequency to find the relationship between the two tensions: (1/2L)√(T1/μ) = f1 = 250 Hz (1/2L)√(T2/μ) = f2 = 350 Hz Dividing the two equations gives: (f2/f1) = √(T2/T1) (350/250) = √(T2/N10N) Squaring both sides of the equation gives: (350/250)^2 = T2/N10N Solving for T2, we get: T2 = (350/250)^2 x N10N T2 = 19.6 N (to 2 significant figures) Therefore, the new tension required to produce a frequency of 350 Hz is 19.6 N. Answer option (B) is correct.