Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics beats in sound waves

The maximum time period for a sound wave that the audible to normal human being is

  1. $0.5s$
  2. $5s$
  3. $0.05s$
  4. $0.005s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The maximum time period corresponds to the minimum frequency. For a human, the minimum frequency is 20 Hz. T = 1/f = 1/20 = 0.05 seconds.

Multiple choice physics beats in sound waves

The musical interval between tube two tine of frequency $400\ Hz$ and $200\ Hz$ is:

  1. $2$
  2. $200$
  3. $1$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The musical interval is defined as the ratio of the two frequencies: 400 Hz / 200 Hz = 2.

Multiple choice physics beats in sound waves

A key of mechanical piano is first struck gently and then struck again but much harder this time. In the second case :

  1. sound will be louder but pitch will not be different

  2. sound will be louder and the pitch will also be higher

  3. sound will be louder but pitch will be lower

  4. both loudness and pitch will remain unaffected
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Loudness or softness of a sound is determined basically by its amplitude. The amplitude of the sound wave depends upon the force with which an object is made to vibrate. If we strike a table lightly, we hear a soft sound because we produce a sound wave of less energy (amplitude). Pitch depends upon frequency received by the human ear.

Multiple choice physics beats in sound waves

The characteristics of sound with the help of which we can distinguish between a SHRILL note and a GRAVE note is

  1. Loudness

  2. Pitch

  3. Quality

  4. Intensity
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Pitch is the characteristic of sound that allows us to distinguish between a shrill note (high frequency) and a grave note (low frequency).

Multiple choice physics beats in sound waves

How Jal Tarang produces sound?

  1. In Jal-tarang musical instrument the cup containing minimum water produces the sound of lowest frequency.

  2. In Jal-tarang musical instrument the cup containing minimum water produces the sound of lowest pitch.

  3. In Jal-tarang musical instrument the cup containing minimum water produces the sound of lowest frequency. As the amount of water in the cup goes on increasing, the frequency of the sound produced also goes on increasing.

  4. all
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

n Jal-tarang musical instrument the cup containing minimum water produces the sound of lowest frequency or lowest pitch. As the amount of water in the cup goes on increasing, the frequency (or the pitch) of the sound
produced also goes on increasing.

Multiple choice physics beats in sound waves

Beats are heard at the rate of $12$ every $5$ seconds when two open organ pipes of lengths $84\ cm$ and $85\ cm$ are sounded together in their fundamental modes. Find the velocity of sound in air. (An open organ pipe has an antinode at each end.

  1. $342.7\ m/s$.
  2. $300.7\ m/s$.
  3. $350.7\ m/s$.
  4. $400.7\ m/s$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an open organ pipe, fundamental frequency f = v / 2L. f1 = v / (2 * 0.84), f2 = v / (2 * 0.85). Beat frequency |f1 - f2| = (v/2) * (1/0.84 - 1/0.85) = 12/5 = 2.4. (v/2) * (0.01 / (0.84 * 0.85)) = 2.4. v = 2.4 * 2 * 0.84 * 0.85 / 0.01 = 342.72 m/s.

Multiple choice physics beats in sound waves

An under water swimmer sends a sound signal to the surface. It is produces 5 betas/sec when compared with fundamental tone of a pipe of $20$ cm length closed at one end what is wavelength of sound in water. (take V water=$1500 m/sec$ Vair=$360 m/sec$)

  1. $3.3 m$ or $3.37 m$
  2. $4.4 m$ or $4.47 m $
  3. $2.5 m$ or $2.7 m$
  4. $1 m$ or $1.7 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A pipe of length 0.2 m closed at one end has fundamental frequency f = v_air / (4L) = 360 / (4 * 0.2) = 450 Hz. The swimmer's signal frequency f_s satisfies |f_s - 450| = 5, so f_s = 455 or 445 Hz. Wavelength in water lambda = v_water / f_s = 1500 / 455 = 3.296 m or 1500 / 445 = 3.37 m.

Multiple choice physics beats in sound waves

identify the part which vibrates to produce sound in the following instruments.

  1. tabla or dholak

  2. flute

  3. tuning fork

  4. sitar
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A tuning fork produces sound through the vibration of its metal prongs. While the other instruments listed also produce sound through vibration, the tuning fork is the classic example of a simple vibrating object used in physics.

Multiple choice surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

What is the change in surface energy, when a mercury drop of radius $r$ splits up into 1000 droplets of radius $r$?

  1. $8pr^2t$
  2. $28pr^2t$
  3. $16pr^2t$
  4. $36pr^2t$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let bigger drop has radius $R$

Volume will remain same. 
 $\dfrac{4}{3}\pi R^3 = 1000\times \dfrac{4}{3} \pi r^3  \implies   r= \dfrac{R}{10}$

Change in surface energy-
 $A _2T - A _1T = 1000\times 4\pi r^2T-4\pi R^2T=4\pi[10R^2-R^2]T=36\pi R^2T$

Multiple choice surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

Two soap bubbles have radii in the ratio of $4:3$. What is the ratio of work done to blow these bubbles?

  1. 4:3

  2. 16:9

  3. 9:16

  4. 3:4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Work done to brust a soap bubble is change in surface energy which is equal surface Tension $\times $ Area.

Workdone $\alpha $ Area $\alpha $ ${ r }^{ 2 }$
hence $\dfrac { { w } _{ 1 } }{ { w } _{ 2 } } =\dfrac { { r } _{ 1 }^{ 2 } }{ { r } _{ 2 }^{ 2 } } ={ \left( \dfrac { 4 }{ 3 }  \right)  }^{ 2 }=\dfrac { 16 }{ 9 } $