InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1151. |
Why can’t we hear the sound of squeaks produced by bats? |
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Answer» Bat uses echolocation to sense movement and their surroundings. We cannot hear the high pitched noise they emit because it is of a frequency higher than what the human can hear. There are some sounds of bat that we can hear but we can’t hear squeaks because their frequency comes in the range of ultrasound (more than 20 kHz) and we can hear between 20 Hz to 20 kHz only. |
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| 1152. |
If a simple pendulum oscillates 10 times in 10 seconds, what would be its frequency? |
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Answer» n = 10 t = -10 s \(\text{f}=\frac{n}{t}=\frac{10}{10s}=1\text{H}Z\) |
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| 1153. |
Do we hear the sounds produced by bats? |
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Answer» We cannot hear sounds produced by bats because the sounds produced by bats have frequency more than 20,000 cycles / seconds that is more than audible range. |
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| 1154. |
Say True or False with explanation:Time taken by an object to complete one oscillation is called time period. |
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Answer» True. Time taken by an object to complete one oscillation is called time period. |
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| 1155. |
Say True or False with explanation:Unwanted or unpleasant sound is termed as music. |
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Answer» False. Unwanted, or unpleasant sounds are termed noise. Sounds which are melodious, and pleasing to ears are called music. |
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| 1156. |
Say True or False with explanation:Noise pollution may cause partial hearing impairment. |
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Answer» True. Unwanted, unpleasant sounds are called noise. If one is subjected to noise continuously for a long time, one may suffer from partial hearing impairment. |
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| 1157. |
Two identical sounds A and B reach a point in the same phase. The resultant sound is C. The loudness of C is n dB higher than the loudness of A. The value of n is (a) 2 (b) 3 (c) 4 (d) 6 |
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Answer» Correct Answer is: (d) 6 Let a = amplitude due to A and B, individually. Loudness due to A = IA = ka2 (k = constant) Loudness due to A + B = IC = k(2a)2 = 4IA. n = 10 log10 (IC/IA) = 10 log10 4 = 10 x 0.6 = 6. |
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