physics ncert solutions class 11th

This is a multiple choice answer as classified in NCERT Exemplar

(c) Speed of sound wave in a medium v $\propto \sqrt[]{T}$

Clearly when temperature changes speed also changes

As v= $v\lambda$ as frequency remains constant and speed changes so wavelength changes

This is a multiple choice answer as classified in NCERT Exemplar

(c) Due to presence of moisture density of air decreases.

As we know speed of sound in air v= $\sqrt[]{\frac{\gamma P}{\rho }}$

So v $\propto \frac{1}{\sqrt[]{\rho }}$

v_2/v₁= $\sqrt[]{\frac{{\rho }_2}{{\rho }_1}}$

_{$\rho$ 2} (moist air)< $\rho$ ₁ (dry air)

So v₂>v₁

So speed of sound wave increases with increase in humidity

This is a multiple choice answer as classified in NCERT Exemplar

(c) Let frequency of the two will be  $vandv\text{'}$  and it is same in both medium

So v/ $\lambda$ = v'/ $\lambda$

$\lambda \text{'}$ = v' $\lambda$ /v

$\lambda \text{'}$ = 2v $\lambda$ /v=2 $\lambda$

Where $\lambda$ and $\lambda$ ' and v and v' are wavelength and speeds of first and second medium respectively

This is a multiple choice answer as classified in NCERT Exemplar

(b) Water waves produced by a motorboat are both longitudinal and transverse because it produce both lateral and transverse wave In the medium.

This is a short answer type question as classified in NCERT Exemplar

Let there are n number of loops in the string

length corresponding each loop is $\frac{\lambda }{2}$

So we can write L= $\frac{n\lambda }{2}$

$\lambda =\frac{2L}{n}$

v/ $v$ =2L/n

so frequency $v=\frac{n}{2L}$ v= $\frac{n}{2L}\sqrt[]{\frac{T}{\mu }}$

v $\propto n$

so $v_1:v_2:v_3:v_4=n_1:n_2:n_3:n_4$

= 1:2:3:4

This is a short answer type question as classified in NCERT Exemplar

consider the frequency of tuning fork= 512Hz

(a) L= $\frac{\lambda }{4}$

$\lambda =4L$

V= $v\lambda$ = 512 $*4*17*{10}^{-2}$

= 348.16m/s

(b) we know that v $\propto \sqrt[]{T}$

$\frac{V_{20}}{V_0}$ = $\sqrt[]{\frac{273+20}{273}}=\sqrt[]{\frac{293}{273}}$

$\frac{V_{20}}{V_0}=\sqrt[]{1.073}=1.03$

V_o= $\frac{V_{20}}{1.03}=\frac{348.16}{1.03}=338m/s$

(c) Resonance will be observed at 17cm length of air column only intensity of sound heard may be greater due to more complete reflection of the sound waves at the mercury surface because mercury is more denser than water.

This is a short answer type question as classified in NCERT Exemplar

Given frequency of the wave v=256Hz

          T= $\frac{1}{v}=\frac{1}{256}=3.9*{10}^{3}s$

(a) time taken to pass through mean position is

t=T/4=1/40 =3.9 $*\frac{{10}^{-3}}{4}s=$ 9.8 $*$ 10^-4s

(b) nodes are A, B, C, D, E (having zero displacement)

antinodes are A' and C' (having maximum displacement)

(c) it is clear from diagram A' and C' are forming antinodes have $\lambda$ separation= v/v'=360/256=1.41m

This is a short answer type question as classified in NCERT Exemplar

We have to observe the displacement and position of different points, then accordingly nature of two waves decided

Points on positions x= 10,20,30,40 never move, always at mean position with respect to time . these are forming nodes which forms a stationary wave

Distance between two successive nodes = $\frac{\lambda }{2}$

$\lambda =2($ node to node distance)

=2 (20-10)

= 2 (10)=20cm

This is a short answer type question as classified in NCERT Exemplar

As the source is moving towards the observer hence apparent frequency observed is more than the natural frequency

Frequency of whistle $v=400Hz$

Speed of train = v_t= 10m/s

Velocity of sound in air =v= 330m/s

Apparent frequency when source is moving v_app= (v/v-v_t)v

= ( $\frac{330}{330-10})400$

= $\frac{330}{320}*400=412.5Hz$

This is a short answer type question as classified in NCERT Exemplar

length of pipe

L=20cm =20 $*{10}^{-2}m$

$v_{close}=\frac{v}{4L}=\frac{330}{4*20*{10}^{-2}}$

$v_{close}=330*\frac{100}{80}=412.5Hz$

$\frac{v_{given}}{v_{closed}}=1237.5/412.5$ =3

Hence 3^rd harmonic node of the pipe is resonantly excited by the source of the given frequency.