Tag: physics

Questions Related to physics

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

physics beats in sound waves

  1. $0.5s$

  2. $5s$

  3. $0.05s$

  4. $0.005s$

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Correct Option: C

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

physics beats in sound waves

  1. $2$

  2. $200$

  3. $1$

  4. $none\ of\ these$

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Correct Option: A

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

physics beats in sound waves

  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

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Correct Option: A
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.

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

physics beats in sound waves

  1. Loudness

  2. Pitch

  3. Quality

  4. Intensity

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Correct Option: B

How Jal Tarang produces sound?

physics beats in sound waves

  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

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Correct Option: D
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.

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.

physics beats in sound waves

  1. $342.7\ m/s$.

  2. $300.7\ m/s$.

  3. $350.7\ m/s$.

  4. $400.7\ m/s$.

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Correct Option: A

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=$360 m/sec$)

physics beats in sound waves

  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 $1.7 m$

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Correct Option: A

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

physics beats in sound waves

  1. tabla or dholak

  2. flute

  3. tuning fork

  4. sitar

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Correct Option: C

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

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

  1. $8pr^2t$

  2. $28pr^2t$

  3. $16pr^2t$

  4. $36pr^2t$

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Correct Option: D
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-
 $A _2T - A _1T = 1000\times 4\pi r^2T-4\pi R^2T=4\pi[10R^2-R^2]T=36\pi R^2T$

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

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

  1. 4:3

  2. 16:9

  3. 9:16

  4. 3:4

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Correct Option: B
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 } $