Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

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

Frequency of vibrations produced by a stretched wire

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

Mass per unit length = mass/length = $\frac{\pi r^{2}l\rho }{l}=\pi r^2\rho$

$v=\frac{n}{2L}\sqrt[]{\frac{T}{\pi r^2\rho }}=v=\sqrt[]{\frac{1}{r^2}}$

$v\propto \frac{1}{r}$ so when radius is tripled v will be 1/3 rd of previous value.

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

Displacement of an elastic wave y =3sinwt+4coswt

3= acos $\varnothing$

4=asin $\varnothing$

On dividing above equation

tan $\varnothing$ =4/3

$\varnothing ={tan}^{-}\left(\frac{4}{3}\right)$

a²cos^{2 $\varnothing$} +a²sin^{2 $\varnothing$} = 3²+4²

a² (cos^{2 $\varnothing +{sin}^2\varnothing$} )=25

a².1=25

a=5

Y= 5cos $\varnothing sinwt$ +5sin $\varnothing coswt$

= 5 [cos $\varnothing sinwt+sin\varnothing coswt$ ]=5sin (wt+ $\varnothing$ )

$\varnothing ={tan}^{-1}\left(\frac{4}{3}\right)$

Hence amplitude =5cm

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

Frequency of tuning fork A

$v_A=512$ Hz

Probable frequency of tuning fork B

${\mathrm{v}}_{\mathrm{B}}={\mathrm{v}}_{\mathrm{A}}+5=512\pm 5=517$ Hz or 507Hz

When B is loaded its frequency reduces .

If it is 517Hz it might reduced to 507Hz given again a beat of 5Hz

If it 507Hz reduction will always increase the beat frequency, hence $v_B=517$ Hz

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

As the organ pipe is open at both ends, hence for first harmonic

l= $\frac{\lambda }{2}$

$\lambda =2l$

V=c/2l

For pipe closed at one end

V'=c/4L'

Hence v=v'

c/2L =c/4L'

L'/L=2/4=1/2

L'=L/2

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

Wire of twice the length vibrates in its second harmonic . thus, if the tuning fork resonates at L, it will resonate at 2L

So the sonometer frequency is

$v=\frac{n}{2l}\sqrt[]{\frac{T}{m}}$

Now if it vibrates with length L we assume $v=v_1$

n=n₁

$v=\frac{n}{2L}\sqrt[]{\frac{T}{m}}$

When length is doubled then

$v_2=\frac{n_1}{2*2L}\sqrt[]{\frac{T}{m}}$

Dividing above equations

$\frac{v_1}{v_2}=\frac{n_1}{n_2}*2$

To Keep the resonance $\frac{v_1}{v_2}=1=\frac{n_1}{n_2}*2$

n₂=2n₁ so it resonates 2^nd harmonic.

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

Wave functions y= 2cos2 $\pi$ (10t-0.0080x + 3.5)

= 2cos(20 $\pi t-0.016\pi x+7\pi$ )

Now standard equation of travelling wave can be written as

Y= a cos (wt-kx+ $\varnothing$ )

So by comparing with above equation

a= 2cm

w=20 $\pi rad/s$

k=0.016 $\pi$

path difference =4cm

(a) phase difference $?\varnothing =\frac{2\pi }{\lambda }*$ path difference

$?\varnothing =0.016\pi *4*100$

$=6.4\pi rad$

$(b)\; ?\varnothing =\frac{2\pi }{\lambda }*\left(0.5*100\right)=0.8\pi rad$

$(c)\; ?\varnothing =\frac{2\pi }{\lambda }*\frac{\lambda }{2}=\pi rad$

$(d)\; ?\varnothing =\frac{2\pi }{\lambda }*\frac{3\lambda }{4}=\frac{3\pi }{2}rad$

(e) T= $2\pi$ /w=2 $\pi$ /20 $\pi =1/10s$

At x=100cm

t=T

${\varnothing }_1$ =20 $\pi T-0.016\pi \left(100\right)+7\pi$

= 20 $\pi \left(\frac{1}{10}\right)-1.6\pi +7\pi =2\pi -1.6\pi +7\pi$

Again at x=100cm t=5s

${\varnothing }_2$ =20 $\pi T-0.016\pi \left(100\right)+7\pi$

=100 $\pi -1.6\pi +7\pi$

From above two equation phase difference ${\varnothing }_2-{\varnothing }_1$

=(100 $\pi -1.6\pi +7\pi$ )- $2\pi -1.6\pi +7\pi$

= 100 

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

standard equation of a progressive wave is given by

Y=asin(wt-kx+ $\varnothing$ )

 This is travelling along positive x- direction

Given equation is y=5sin(100 $\pi t-0.4\pi x$ )

Comparing with standard equation

(a) amplitude =5m

(b) k=2 $\frac{\pi }{\gamma }$ =0.4x

wavelength =2 $\frac{\pi }{k}$ = $\frac{2\pi }{0.4\pi }=\frac{20}{4}=5m$

(c) w=10 $\pi$

w=2 $\pi v=100\pi$

frequency v= 100 $\pi /2\pi$ =50Hz

(d) wave velocity $v=\frac{w}{k}\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{}\mathrm{k}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{w}\mathrm{a}\mathrm{v}\mathrm{e}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{}\mathrm{k}=2\mathrm{\pi }/\mathrm{\lambda }$

=100 $\pi /0.4\pi$ =1000/4

250m/s

(e) y= 5sin(100 $\pi t-0.4\pi x$ )

dy/dt = particle velocity

dy/dt = 5(100 $\pi )$ cos[100 $\pi t-0.4\pi x$ ]

for particle velocity amplitude (dy/dt)_max which will be for cos[100 $\pi t-0.4\pi x$ ]_max=1

so particle velocity amplitude =(dy/dt)_max =5(100 $\pi$ ) $*1$ =550 $\pi$  m/s

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

we know that rms speed of molecules of a gas

C= $\sqrt[]{\frac{3p}{\rho }}=\sqrt[]{\frac{3RT}{M}}$

 Where M is the molar mass of the gas

Speed of sound wave in gas v= $\sqrt[]{\frac{\gamma p}{\rho }}=\sqrt[]{\frac{\gamma RT}{M}}$

On dividing above equation we get

c/v = $\sqrt[]{\frac{3RT}{M}*\frac{M}{\gamma RT}}$

c/v = $\sqrt[]{\frac{3}{\gamma }}$

where $\gamma$ = adiabatic constant for diatomic gas

$\gamma =\frac{7}{5}$

c/v=constant

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

Speed of wave in solid =8km/s

Speed of wave in liquid = 5km/s

Required time = [ $\frac{1000-0}{8}+\frac{3500-1000}{5}+\frac{6400-3500}{8}$ ] $*2$

= [ $\frac{1000}{8}+\frac{2500}{5}+\frac{2900}{8}$ ] $*2$

= [125+500+362.5] $*2$ =1975       (diameter = radius $*2$ )

As we are considering at diametrically opposite point, hence there is a multiplication of 2