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A tuning fork of frequency 500 H(Z) is s...

A tuning fork of frequency `500 H_(Z)` is sounded on a resonance tube . The first and second resonances are obtined at `17 cm` and `52 cm` . The velocity of sound is

A

`170 m//s`

B

`350 m//s`

C

`520 m//s`

D

`850 m//s`

Text Solution

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The correct Answer is:
To find the velocity of sound using the given data, we will follow these steps: ### Step 1: Understand the Resonance Condition In a resonance tube, the first resonance occurs when there is a pressure antinode at the open end and a pressure node at the closed end. The distance from the closed end to the first resonance point is \( \frac{\lambda}{4} \) (where \( \lambda \) is the wavelength). ### Step 2: Identify the Resonance Lengths We are given: - First resonance length, \( L_1 = 17 \, \text{cm} \) - Second resonance length, \( L_2 = 52 \, \text{cm} \) ### Step 3: Calculate the Difference in Lengths The difference between the first and second resonance lengths gives us the distance between two consecutive nodes and antinodes: \[ \Delta L = L_2 - L_1 = 52 \, \text{cm} - 17 \, \text{cm} = 35 \, \text{cm} \] ### Step 4: Relate the Difference to Wavelength The difference in lengths corresponds to \( \frac{\lambda}{2} \): \[ \Delta L = \frac{\lambda}{2} \implies 35 \, \text{cm} = \frac{\lambda}{2} \] Thus, we can find the wavelength \( \lambda \): \[ \lambda = 2 \times 35 \, \text{cm} = 70 \, \text{cm} \] ### Step 5: Convert Wavelength to Meters Convert the wavelength from centimeters to meters: \[ \lambda = 70 \, \text{cm} = 0.70 \, \text{m} \] ### Step 6: Use the Wave Equation The velocity of sound \( v \) can be calculated using the formula: \[ v = f \lambda \] where \( f \) is the frequency of the tuning fork. Given \( f = 500 \, \text{Hz} \): \[ v = 500 \, \text{Hz} \times 0.70 \, \text{m} = 350 \, \text{m/s} \] ### Final Answer The velocity of sound is \( 350 \, \text{m/s} \). ---

To find the velocity of sound using the given data, we will follow these steps: ### Step 1: Understand the Resonance Condition In a resonance tube, the first resonance occurs when there is a pressure antinode at the open end and a pressure node at the closed end. The distance from the closed end to the first resonance point is \( \frac{\lambda}{4} \) (where \( \lambda \) is the wavelength). ### Step 2: Identify the Resonance Lengths We are given: - First resonance length, \( L_1 = 17 \, \text{cm} \) ...
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DC PANDEY-SOUND WAVES-Level 1 Objective
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  2. When interference is produced by two progressive waves of equal freque...

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  3. A tuning fork of frequency 500 H(Z) is sounded on a resonance tube . T...

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  4. A vehicle , with a horn of frequency n is moving with a velocity of 30...

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  5. How many frequencies below 1 kH(Z) of natural oscillations of air colu...

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  6. a sound source emits frequency of 180 h(Z) when moving towards a rigid...

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  7. Two sound waves of wavelengths lambda(1) and lambda(2) (lambda (2) gt ...

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  8. A, Band C are three tuning forks. Frequency of A is 350 H(Z) . Beats p...

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  9. The first resonance length of a resonance tube is 40 cm and the second...

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  10. Two identical wires are streched by the same tension of 100 N and each...

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  11. A tuning fork of frequency 340 Hz is excited and held above a cylindri...

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  12. In a closed end pipe of length 105 cm , standing waves are set up corr...

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  13. Oxygen is 16 times heavier than hydrogen. At NTP equal volumn of hydro...

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  16. One train is approaching an observer at rest and another train is rece...

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  17. Speed of sound in air is 320 m//s . A pipe closed at one end has a len...

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  18. Four sources of sound each of sound level 10 dB are sounded together i...

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  19. A longitudinal sound wave given by p = 2.5 sin.(pi)/(2) (x - 600 t) (p...

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  20. Sound waves of frequency 600 H(Z) fall normally on perfectly reflectin...

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