JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 19)
In a resonance tube experiment when the tube
is filled with water up to a height of 17.0 cm
from bottom, it resonates with a given tuning
fork. When the water level is raised the next
resonance with the same tuning fork occurs at
a height of 24.5 cm. If the velocity of sound in
air is 330 m/s, the tuning fork frequency is :
2200 Hz
3300 Hz
1100 Hz
550 Hz
Explanation
$${l_1} = l - 17$$
$${l_2} = l - 24.5$$
We know, $$v = 2f({l_1} - {l_2})$$
$$ \Rightarrow $$ $$330 = 2 \times f \times [(f \times [(l - 17) - (l - 24.5)] \times 10^{ - 2}$$
$$ \Rightarrow $$ $$165 = f \times 7.5 \times {10^{ - 2}}$$
$$ \Rightarrow $$ $$f = {{165 \times 1000} \over {7.5}}$$
$$ \Rightarrow $$ $$f = 2200\,Hz$$
$${l_2} = l - 24.5$$
We know, $$v = 2f({l_1} - {l_2})$$
$$ \Rightarrow $$ $$330 = 2 \times f \times [(f \times [(l - 17) - (l - 24.5)] \times 10^{ - 2}$$
$$ \Rightarrow $$ $$165 = f \times 7.5 \times {10^{ - 2}}$$
$$ \Rightarrow $$ $$f = {{165 \times 1000} \over {7.5}}$$
$$ \Rightarrow $$ $$f = 2200\,Hz$$
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