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प्रश्न
The fundamental frequency of a closed pipe is 293 Hz when the air in it is a temperature of 20°C. What will be its fundamental frequency when the temperature changes to 22°C?
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उत्तर
Given:
The fundamental frequency of a closed pipe is 293 Hz. Let this be represented by f1.
Temperature of the air in closed pipe T1 = 20°C = 20 + 273 = 293 K
Let f2 be the frequency in the closed pipe when the temperature of the air is T2 .
∴ T2 = 22°C = 22 + 273 = 295 K
Relation between f and T : \[f \propto \sqrt{T}\]
∴\[\frac{f_1}{f_2} = \frac{\sqrt{T_1}}{\sqrt{T_2}}\]
\[ \Rightarrow \frac{293}{f_2} = \frac{\sqrt{293}}{\sqrt{295}}\]
\[ \Rightarrow f_2 = \frac{293 \times \sqrt{295}}{\sqrt{293}} = 294 \text { Hz }\]
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