Tag: physics

Questions Related to physics

The symbol of candela is __________.

luminous intensity measurements physics

  1. Cd

  2. C

  3. Can

  4. Ca

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Correct Option: A
Explanation:
The symbol of candela is Cd.

Light from a source is analysed by an analyser. When the analyser is rotated, the intensity of the emergent light.

luminous intensity measurements physics

  1. Does not vary

  2. Remains uniformly dark

  3. Varies between maximum and zero

  4. Varies between maximum and minimum

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Correct Option: A
Explanation:

Let the intensity of the unpolarised light be $I _o$.
An analyser is used to polarized the light in one direction.
Intensity of light after passing through the analyser  $I' = \dfrac{I _o}{2}$
Intensity of light after passing through analyser remains constant $I _o/2$ even if the analyser is rotated by some angle.

 If light falls on the surface at an angle of $60^{0}$, then illuminance will be____

luminous intensity measurements physics

  1. $12$ lux

  2. $6$ lux

  3. $3$ lux

  4. $1.5$ lux

Show answer
Correct Option: D
Explanation:

Illuminance ($E$)  is  measured  in  lux.  The  lux  is  an  SI  unit 

used  when  characterizing illumination conditions of a surface:
$E = \frac {I}{R^2}  cos \alpha     lux$
where $I$ is the luminous intensity, $R$ is the distance and $\alpha$ is the surface angle.
Given $I = 300 cd$, $R = 10 m$, $\alpha = 60^o$
$\implies E = \frac {300}{10^2}    cos  60^o= 1.5   lux$

A lamp is hanging along the axis of a circular table of radius r. At what height should the lamp be placed above the table, so that the illuminance at the edge of the table is $\displaystyle \frac{1}{8}$ of that at its centre?

luminous intensity measurements physics

  1. r/2

  2. r/$\sqrt{2}$

  3. r/3

  4. r/$\sqrt{3}$

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Correct Option: D
Explanation:

$E _2 = \displaystyle \frac{1}{8} E _1$ or $\displaystyle \frac{1}{(r^2 + h^2)} \times \frac{h}{\sqrt{r^2 + h^2}} = \frac{1}{8} \frac{1}{h^2}$
(by lambert's cosine law)
or, $(r^2 + h^2)^{3/2} = (2h)^3 $ or $(r^2 + h^2)^{1/2} = 2h$
or $r^2 + h^2 = 4h^2$
$h = r / \sqrt{3}$

The speed of a supersonic wave, as compared to that of sound is

physics properties of waves sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. $less$

  2. $more$

  3. $equal$

  4. $1/10$

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Correct Option: B
Explanation:

Prefix in the word "Supersonic" is 'super', means superior to sound waves. Supersonic waves are faster than the sound wave in air.
Option "B" is correct.

The frequency of a man's voice is $300\space Hz$. If the velocity of sound waves is $336\space ms^{-1}$, the wavelength of the sound is

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. $1.12\space m$

  2. $300\times336\space m$

  3. $330/336\space m$

  4. None of these

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Correct Option: A
Explanation:

We are given, frequency $f=300 Hz$, and velocity $v=336 ms^{-1}$,

From the relation $v=f \times \lambda$,

$\lambda = v/f =336/300=1.12 m$

Option "A" is correct.

 769Hz longitudinal wave in air has a speed of 344m/s. At a particular instant, what is the phase difference (in degrees) between two points 5.0 cm apart?

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. 30

  2. 40

  3. 45

  4. 60

Show answer
Correct Option: B
Explanation:

Given that,

The frequency of wave is $\nu = 769 Hz$

The speed of wave is $v = 344 ms^{-1}$

The wavelength of wave is given by

$\lambda = \dfrac{v}{\nu}$

$\lambda = \dfrac{344}{769}$

$\lambda = 0.447 m = 44.7 cm$

Hence, the phase difference between two points 5.0 cm apart is

$\dfrac{5}{44.7} \times 360^\circ = 0.1118 \times 360^\circ = 40.24^\circ \approx40^\circ$

A rod $70\space cm$ long is clamped from middle. The velocity of sound in the material of the rod is $3500\space ms^{-1}$. The frequency of fundamental note produced by it is :

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. $3500\space Hz$

  2. $2500\space Hz$

  3. $1250\space Hz$

  4. $700\space Hz$

Show answer
Correct Option: B
Explanation:

SInce rod is clamped at middle, therefore only nodes can form at that point, and at free end only antinode can form, therefore for fundamental mode of frequency,

$\lambda/4=L/2$

$\lambda=2L=2\times 70 cm=1.4 m$

$v=3500\  ms^{-1}$ is given, 

We know, $v=f \times \lambda$ or $f = \dfrac{v}{\lambda}=3500/1.4=2500 Hz$

Option "B" is correct.

A sonometer wire, $100\ \text{cm}$ in length has a fundamental frequency of $330\ \text{Hz}$. The velocity of propagation of transverse waves along this wire is :

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. $330\ \text{ms}^{-1}$

  2. $660\ \text{ms}^{-1}$

  3. $115\ \text{ms}^{-1}$

  4. $990\ \text{ms}^{-1}$

Show answer
Correct Option: B
Explanation:

For the fundamental frequency $(f _0=330 Hz),\ \lambda=2 L=200\ \text{cm} = 2\ \text{m}$


then from the formula, $v=f _0 \times \lambda= 330 \times 2.0= 660 \text{ms}^{-1}$

If the frequency of a sound wave is increased by 25%, then the change in its wavelength will be

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. 25% decrease

  2. 20% decrease

  3. 20% increase

  4. 25% increase

Show answer
Correct Option: B
Explanation:
The frequency  of wave is given by

$\nu = \dfrac{v}{\lambda}$

$\lambda = \dfrac{v}{\nu}$

When the frequency of a sound wave is increased by 25%, then the new 
wavelength is

$\lambda' = \dfrac{v}{\nu+\dfrac{25}{100}\nu}$

$\lambda' = \dfrac{v}{\nu+\dfrac{1}{4}\nu}$

$\lambda' = \dfrac{v}{\dfrac{5}{4}\nu}$

$\lambda' = \dfrac{4v}{5\nu}$

Hence, the percent change in wavelength is

$\dfrac{\lambda - \lambda'}{\lambda} \times 100 = \dfrac{\dfrac{v}{\nu} - \dfrac{4v}{5\nu}}{\dfrac{v}{\nu}} \times 100$

$\Rightarrow -\dfrac{1}{4} \times 100 = -20$%.

Hence, wavelength decreases by 20%