A sonometer wire of length 1 .5 m is made of steel. The tension in it produces an elastic strain of 1% . What is the fundamental frequency of steel if density and elasticity of steel are 7 .7 × 10 ^ 3 kgm^ - 3 and 2 .2

Correct answer: D. 178 .2 Hz f = (V)/(2 ℓ) = (1)/(2 ℓ) (T)/(μ) = (1)/(2 ℓ) (T)/(Ad) Also, Y = (T ℓ)/(A Δ ℓ) ⇒ (T)/(A) = (Y Δ ℓ)/(ℓ) ⇒ f = (1)/(2 ℓ) Y Δ ℓ ℓ d ℓ = 1.5 m, (Δ ℓ)/(ℓ) = 0.01, d = 7.7 × 1 0 ^ 3 kg / m ^ 3 Y = 2.2 × 1 0 ^ 1 1 N / m ^ 2 After solving f = (2)/(7) × 1 0 ^ 3 3 Hz f 178.2Hz

JEE Main 20137 April

Exam: JEE Main
Year: 2013
Date: 7 April
Subject: Physics
Chapter: Waves and Sound

JEE Main 2013 · 7 April · Physics · Waves and Sound

A sonometer wire of length 1 .5 m is made of steel. The tension in...

A sonometer wire of length

1 .5 m

is made of steel. The tension in it produces an elastic strain of

\text{1%}

. What is the fundamental frequency of steel if density and elasticity of steel are

7 .7 \times {10}^{3}kgm^{- 3}

and

2 .2 \times {10}^{11}Nm^{- 2}

respectively?

Options

A $200 .5 Hz$
B $770 Hz$
C $188 .5 Hz$
D $178 .2 Hz$

Correct answer

D. $178 .2 Hz$

Step-by-step solution

f = \dfrac{V}{2 ℓ} = \dfrac{1}{2 ℓ} \sqrt{\dfrac{T}{\mu}} = \dfrac{1}{2 ℓ} \sqrt{\dfrac{T}{Ad}}

Also,

Y = \dfrac{T ℓ}{A \Delta ℓ} ⇒ \dfrac{T}{A} = \dfrac{Y \Delta ℓ}{ℓ} ⇒ f = \dfrac{1}{2 ℓ} \sqrt{\dfrac{\text{Y} \Delta ℓ}{ℓ \text{d}}}

ℓ = \text{1.5 m,} \dfrac{\Delta ℓ}{ℓ} = \text{0.01, d} = \text{7.7} \times 1 {0}^{3}\text{kg} / {\text{m}}^{3}

\text{Y} = \text{2.2} \times 1 {0}^{1 1} \text{N} / {\text{m}}^{2}

After solving

f = \sqrt{\dfrac{2}{7}} \times \dfrac{1 {0}^{3}}{3} \text{H}\text{z}

f \approx \text{178.2}\text{H}\text{z}

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