JEE MAIN - Physics (2021 - 27th August Evening Shift - No. 26)
A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be __________ cm. (Take speed of sound in air as 340 ms$$-$$1)
Answer
34
Explanation
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$${\lambda \over 4}$$ = l $$\Rightarrow$$ $$\lambda$$ = 4l
f = $${V \over \lambda } = {V \over {4l}}$$
$$\Rightarrow$$ 250 = $${{340} \over {4l}}$$
$$\Rightarrow$$ l = $${{34} \over {4 \times 25}}$$ = 0.34 m
l = 34 cm
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