Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics sound time period, frequency and amplitude of sound amplitude, time period and frequency of a vibration nature and characteristics of sound waves

In order to reduce the loudness of a sound we have to:

  1. decrease its frequency of vibration of the sound

  2. increase its frequency of vibration of the sound

  3. decrease its amplitude of vibration of the sound

  4. increase its amplitude of vibration of the sound
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Loudness of the sound  $L = 10 \log _{10}\dfrac{I}{I _o}$
where $I \propto A^2$   ($A$ is the amplitude of vibration of sound) 
Thus the loudness of sound varies directly as the square of the amplitude of vibration of the sound source.
Hence, the in order to reduce the loudness of sound, we have to decrease the amplitude of vibration of the sound.
Multiple choice physics sound time period, frequency and amplitude of sound amplitude, time period and frequency of a vibration nature and characteristics of sound waves

A 1cm long vibrates with a fundamental frequency of 256Hz. If the length is reduced to 1/4cm keeping the tension unaltered, the new fundamental frequency will be

  1. 64

  2. 256

  3. 512

  4. 1024
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We know that $n\alpha\cfrac{1}{l}$
Thus $\cfrac{n _2}{n _1}=\cfrac{l _1}{l _1}\\n _2=\cfrac{l _1}{l _2}(n _1)=\cfrac{1\times256}{(1/4)}=256\times4=1024Hz$
Multiple choice physics sound time period, frequency and amplitude of sound amplitude, time period and frequency of a vibration nature and characteristics of sound waves

Two identical sound $A$ and $B$ each a point in the same phase. The resultant sound is $C$. The loudness of $C$ is $n$ $dB$ higher than the loudness of $A$

  1. $2$
  2. $3$
  3. $4$
  4. $6$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When two identical sound sources of the same phase are combined, the intensity doubles (I_total = 2 * I). The change in decibels is 10 * log10(I_total / I) = 10 * log10(2) = 10 * 0.3 = 3 dB. The question states the loudness is n dB higher, and 6 dB is often cited for coherent sources, but for incoherent sources, it is 3 dB. Given the options, 6 dB is likely expected if the sources are coherent.

Multiple choice physics sound time period, frequency and amplitude of sound amplitude, time period and frequency of a vibration nature and characteristics of sound waves

If the intensity of sound increases by a factor of $10^5$ , the increases in the intensity level is

  1. $5db$
  2. $10db$
  3. $25db$
  4. $50db$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The change in sound intensity level in decibels is calculated using the formula delta L = 10 log10(I2/I1). Given that the intensity increases by a factor of 10^5, substitute this into the formula to get 10 log10(10^5) = 10 * 5 = 50 dB. Therefore, the increase in the intensity level is 50 db.

Multiple choice physics sound time period, frequency and amplitude of sound amplitude, time period and frequency of a vibration nature and characteristics of sound waves

In a class of $100$ students each shouting at $100 dB$. Find noise level of class

  1. $10 dB$
  2. $100 dB$
  3. $12 dB$
  4. $120 dB$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The sound level of N identical independent sources each at level L is given by the formula L_total = L + 10 log10(N). Here, N = 100 students and each shouts at 100 dB. Calculating the addition gives 100 + 10 log10(100) = 100 + 10 * 2 = 120 dB.

Multiple choice physics sound time period, frequency and amplitude of sound amplitude, time period and frequency of a vibration nature and characteristics of sound waves

The fundamental frequency of a closed organ pipe is same as the first overtone frequency of an open pipe. The length of open pipe is $20\ cm$. The length of closed pipe is

  1. $25\ cm$
  2. $12.5\ cm$
  3. $13.33\ cm$
  4. $200\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given $:- $
Fundamental Frequency for the Open orange pipe $= ν/2l'$
where$, ν =$ Frequency and l' is the length of the open orange pipe$.$
Third Harmonic of the orange pipe $= 3ν/4l$
where$, ν =$ frequency 
and $l$ is the length of the Closed orange pipe $= 20 cm.$
Now$,$ According to the Question$,$
$3ν/4l = ν/2l'$
$⇒ 3/(4 × 20) = 1/2l'$
$⇒ 3/80 = 1/2l'$
$⇒ 2l' = 80/3$
$⇒ l' = 40/3$
$⇒ l' = 13.33 cm.$
Hence,
option $(C)$ is correct answer.