The frequency n of vibration of a stretched string is a function of its tension F, the length l and the mass per unit length m. From the knowledge of dimensions, prove that, n∝1/l √f/m Pls fast

n ∝ (1/l) (√F/m)

Step-by-step explanation:

Since frequency, n of vibration of a stretched string is a function of its tension F, the length l and the mass per unit length m, and has dimensions [T]⁻¹ where T = time, its dimension must be equal to that of the combination of F, L and m

Since n = f(F,L,m)

n = kFᵃLᵇmˣ

dimension of n = (dimension of F)ᵃ × (dimension of L)ᵇ × (dimension of m)ˣ

Since n = frequency, dimension of n = [T]⁻¹

F = Force, dimension of F = [M][L][T]⁻²

Also, L = length, dimension of L = [L] and

m = mass per unit length, dimension of m = [M][L]⁻¹

So, n = FᵃLᵇmˣ

[T]⁻¹  = ([M][L][T]⁻²)ᵃ( [L] )ᵇ([M][L]⁻¹)ˣ

[T]⁻¹  = ([M]ᵃ[L]ᵃ[T]⁻²ᵃ[L]ᵇ([M]ˣ[L]⁻ˣ

[M]⁰[L]⁰[T]⁻¹  = ([M]ᵃ ⁺ˣ)([L]ᵃ ⁺ ᵇ ⁻ˣ)([T]⁻²ᵃ)

equating the exponents on both sides, we have

a + x = 0  (1 )⇒ x = -a

a + b - x = 0 (2) ⇒

-2a = -1 (3) ⇒ a = 1/2

Substituting x into 2, we have

a + b -(-a) = 0

a + b + a = 0

2a + b = 0

b = -2a

b = -2 (1/2)

b = -1

x = -a = -1/2

So, substituting the variables into n, we have

n = kFᵃLᵇmˣ

n = kFᵃLᵇmˣ

n = k/l(√F/m)

n ∝ (1/l) (√F/m)

2. angles qrt and qst both measures 156 degrees.

step-by-step explanation:  

we have been given a diagram of a circle and we are told that measure of arc qvt is 156 degrees.  

we can see from our given diagram that angle qrt and angle qst subtends to arc qvt.

since we know that angles subtended by the same arc at the circumference are always equal.

by angle subtended by same arc theorem, measure of angle qrt and angle qst will be equal to the measure of arc qvt.

therefore, angles qrt and qst both measures 156 degrees and 2nd statement is the correct choice.