Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen: Overview, Questions, Preparation

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

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

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

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

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 = $\sqrt[]{\frac{3RT}{M}\times \frac{M}{\gamma RT}}$

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

(a) The equation y = 100 cos (100πt + 0.5x ) is representing a travelling wave along x-direction

(b) The equation y = 5 cos (4x ) sin (20t) represents a stationary wave, because it contains sin cos terms ., combinations of two progressive waves.

(c) The equation y = 10 cos [ (252 – 250) πt ] cos [ (252+250)πt ] involving sum and difference of two near by frequencies 252 and 250 have this equation represents beats formation.

(d) As the equation y = y = 4 sin (5x – t/2) + 3 cos (5x – t/2) involves negative sign with x, have if represents a travelling wave along x – direction.

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 $\pi$ ) $\times 1$ =550 $\pi$  m/s

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 $?\varnothing =\frac{2\pi }{\lambda }\times$ path difference

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

$=6.4\pi rad$

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

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

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

(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 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

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

Dividing above equations

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

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

n₂=2n₁ so it resonates 2^nd harmonic.

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

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

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 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

we know the speed of air  v $\propto \sqrt[]{T}$

$\frac{v_t}{v_o}=\sqrt[]{\frac{T_T}{T_o}}=\sqrt[]{\frac{T_T}{273}}$

$\frac{V_T}{V_o}=\frac{3}{1}$

3/1 = $\sqrt[]{\frac{T_T}{T_o}}=\frac{T_T}{273}=9$

T_T=273 $\times 9=2457K$

=2457-273=2184 ^oC

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

Let n₁>n₂ beat frequency $v_b=n_1-n_2$

As we know time per beats = 1/frequency

Time period beats = T_b= 1/ $v_b=1/n_1-n_2$

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

Given that length of wire l=12m

Mass of the wire m=2.10kg

T= 2.06 $\times {10}^{4}N$

Speed of transverse wave v= $\sqrt[]{\frac{T}{\mu }}$ = $\sqrt[]{\frac{2.06\times {10}^4}{2.1/12}}=\sqrt[]{\frac{2.06\times 12\times {10}^4}{2.1}}=343m/s$

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

length of pipe

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

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

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

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

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

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

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

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

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

Given frequency of the wave v=256Hz

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

(a) time taken to pass through mean position is

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

consider the frequency of tuning fork= 512Hz

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

$\lambda =4L$

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

= 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

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

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

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(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

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(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

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(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

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(b) Propagation of longitudinal waves through a medium leads to transmission of energy through the medium without matter being transmitted. So no movement of energy and matter hence momentum

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(c) When mechanical transverse wave propagates through a medium, the constituents of the medium oscillate perpendicular to wave motion causing change in shape. That is each element of the medium is subjected to shearing stress. Solids and strings have shear, that is why tey bear stress. But in fluids there is no fixed shape so transverse waves does not formed in fluid.

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(d) Due to compression and rarefactions density of the medium changes.at compressed regions density is maximum and at rarefactions density is minimum

As density is changing so boyle, s law not obeyed

Bulk modulus remains same

The time of compression and rarefaction is too small so no transfer of heat.

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(b) Amplitude of reflected wave

A_f= $\frac{2}{3}\times A_i$ =2 (0.6)/3=0.4 units

Equation of incident wave

Y_i=0.6sin2 $\pi \left(t-\frac{x}{2}\right)$

For reflected wave 

y_r=A_rsin2 $\pi$ (t+x/2+ $\pi$ )

Y_r=-0.4sin2 $\pi \left(t+\frac{x}{2}\right)$

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(b) m=2.5kg

$\mu =$ mass per unit length

=m/l =2.5/20=125/10=0.125kg/m

V= $\sqrt[]{\frac{T}{\mu }}=\sqrt[]{\frac{200}{0.125}}$

L=v $\times t$

20= $\sqrt[]{\frac{125}{2\times {10}^5}}=20\times \sqrt[]{\frac{25\times 5}{2\times {10}^5}}$

= 20 $\times \sqrt[]{\frac{25}{0.4\times {10}^5}}=\frac{1}{2}$

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(c) Let frequency of the source is n_o

Let the speed of sound wave in the medium is v

As observer is stationary

Apparent frequency n_a= $\frac{v}{v-v_s}{n}_o$

$\frac{v}{v-v_s}{n}_o=n_a{n}_o$

When the train is going away from the observer

Apparent frequency n_a= $\frac{v}{v-v_s}{n}_o$ = $n_a<n_o$

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(a, b, c) y (x, t)=30sin (36t+0.018x+ $\frac{\pi }{4}$ )

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

(a) as the equation involves positive sign with x . hence the wave is travelling from right to left. Hence option a is correct.

(b) W=36

2 $\pi v=36$

$v=\frac{36}{2\pi }$ = $\frac{18}{\pi }$

$k=0.018=\frac{2\pi }{\lambda }$

$\frac{2\pi v}{v\lambda }=0.018$

w/v=0.018

v=2000cm/s=20m/s

(c) 2 $\pi v=36$

$v=\frac{18}{\pi }$ = 5.7Hz

(d) $\frac{2\pi }{\lambda }=0.018$ , $\lambda =2\pi$ /0.0018cm=3.48cm

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(b, c) y (x, t) = 0.06 sin (2πx/3) cos (120πt)

(a) y (x, t)=asinkxcoswt

(b) w=120 $\pi$ , f=60hz

(c) k=2 $\frac{\pi }{3}=\frac{2\pi }{\lambda },\lambda =3m,f=60hz$ , v =60 (3)=180m/s

(d) since in stationary wave all particles of the medium executes SHM with varying amplitude nodes.

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(c, d) Speed of sound waves in a fluid is given by

V= $\sqrt[]{\frac{B}{\rho }}$

V= $\frac{1}{\sqrt[]{\rho }},$ V $\propto \sqrt[]{B}$ so when we increase velocity bulk modulus also increases.

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(b, c, d)

during propagation of a plane progressive mechanical wave

(i) clearly the particles O, A and B are having different phase.

(ii) Particles of the wave are having up and SHM.

(iii) For a progressive wave propagating in a fluid

V= $\sqrt[]{\frac{B}{\rho }}$ , V= $\sqrt[]{\frac{1}{\rho }}$

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(a, b, d) As we know by y (x, t) = 0.06 sin (2πx/3) cos (120πt)

By comparing the equation with general equation N denotes nodes and A denotes antinodes.

(a) Clearly frequency is common for all the points

(b) Consider all the particles between two nodes they are having same phase at given time

(c) But are having different amplitude of 0.06sin (2 $\frac{\pi }{3}x$ ) and because of different amplitudes they are having different energies.

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(a, b) v_o=400Hz, v=340m/s

V_m=10m/s

(a) as both source and observer are stationary, hence frequency observed will be same as natural frequency v_o=400Hz

(b) the speed of sound v=v+v_w= 340+10=350m/s

(c) there will be on effect on frequency because there is no relative motion between source and observer hence c, d are incorrect.

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(a, b, d, e) a) clearly every particle at x will have amplitude =asinkx=fixed

b) for mean position =0

coswt=0

wt= (2n-1) $\frac{\pi }{2}$

hence for a fixed of n all particles are having same value of time t= (2n-1) $\frac{\pi }{2w}$

c) amplitude of all the particles are asinkx which is different for different particles at different values of x

d) the energy is a stationary wave is confined between two nodes.

e) particles at different nodes are always at rest.

r? = 10αt²î + 5β (t-5)?
v? = dr? /dt = 20αtî + 5β?
As L? = m (r? × v? )
So, at t=0, L=0
given L is same at t=t as at t=0
⇒ r? × v? = 0
⇒ (10αt²î + 5β (t-5)? ) × (20αtî + 5β? ) = 0
⇒ 50αβt² (î×? ) + 100αβt (t-5) (? ×î) = 0
⇒ 50αβt² k? - 100αβt (t-5) k? = 0
⇒ 50t² - 100t (t-5) = 0
⇒ 50t² - 100t² + 500t = 0
⇒ -50t² + 500t = 0
⇒ 50t (10 - t) = 0
⇒ t = 10 second

ω = √k_eq/μ [μ = (m? )/ (m? +m? ) (Reduced mass)]
k_eq = (k × 4k)/ (k+4k) = 4k/5
ω = √ (4k/5) / (m? / (m? +m? )
= √ (4×20/5) / (0.2×0.8)/ (0.2+0.8)
= √ (16/0.16)
= 10 rad/s

F_net = -kq²/ (l+x)² + kq²/ (l-x)²
= kq² [ (l+x)² - (l-x)² / (l²-x²)² ]
= kq² [ 4lx / (l²-x²)² ]
For x << l,
ma ≈ -4kq²x / l³
a = - (4kq²/ml³)x
ω² = 4kq²/ml³
ω = √ (4kq²/ml³)
= √ (4 × 9×10? × 10 / (1×10? × 1)
= 6 × 10? rad/s
= 6000 × 10? rad/s

I = I? e? /?
= (20/10000) e^- (1×10? / 10×10? ³)
= 2 × 10? ³ e? ¹
The provided solution calculates as:
= 2 × 10? ³ e? ¹
= 2e? ¹ mA
= 2 × 0.37mA
= 0.74 mA = 74/100 mA
(Note: There seems to be a calculation discrepancy in the source image, the steps shown lead to I = 2e? ¹ mA)

(1/2)mu² = (1/2)mv² + mgl (1-cosθ)
⇒ v² = u² - 2gl (1-cosθ)
⇒ v² = 3² - 2×10×0.5× (1 - 1/2)
⇒ v² = 9 - 5 = 4
v = 2m/s

From conservation of momentum:
2×4 = 2v? + mv?
Given v? = 1 m/s (interpreted from intermediate steps)
8 = 2 (1) + mv?
mv? = 6 . (i)
From coefficient of restitution (e=1 for elastic collision):
e = (v? - v? )/ (u? - u? )
1 = (v? - v? )/ (4 - 0)
-1 = (v? - v? )/ (0 - 4)  (as written in the image)
⇒ 4 = v? - 1
⇒ v? = 5 . (ii)
Put (2) in (1), m (5) = 6
m = 1.2kg

Given: Power in R_B = 200 W, R_B = 50 Ω, V_source = 200 V
I² R_B = 200
I² (50) = 200
I² = 4 ⇒ I = 2A
Also, I = V_source / (R + R_B)
2 = 200 / (R + 50)
2 (R + 50) = 200
R + 50 = 100
R = 50Ω

Difference in Reading = Positive Zero Error - Negative Zero Error
= (+5) - [- (100-92)]
= 5 - [-8]
= 13

We know, μ? = 1 + χ?
B? ∝ μ?
B ∝ μ
(ΔB/B? ) = (B-B? )/B? = (μ-μ? )/μ? = μ/μ? - 1 = μ? - 1 = χ?
% (ΔB/B) = χ? × 100 = 2.2 × 10? × 100 = 22/10?
(Note: The value for χ? in the source image appears to have a typo as 2.2 × 10? , it has been corrected to 2.2 × 10? to match the final answer.)

ε = |dΦ/dt| = A|dB/dt|
Given B (t), dB/dt = (4/π) × 10? ³ × (-1/100)
ε = (π × 1²) × | (4/π) × 10? ³ × (-1/100)|
= 4 × 10? V
To find when B=0:
B = 0 ⇒ 1 - t/100 = 0
⇒ t = 100 second
Energy Dissipated, E = P × t = (ε²/R) × t
E = (4×10? )² / (2×10? ) × 100 = 80mJ

$\frac{1}{20}=(1.55-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{f}=(1.50-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Divide (1) by (2), $\frac{f}{20}=\frac{0.55}{0.50}$

$f=22\mathrm{c}\mathrm{m}$

$\mathrm{\Delta }{E}_1=\frac{12400\mathrm{e}\mathrm{V}\text{?}}{1085\text{?}}=11.43\mathrm{e}\mathrm{V}$

$\mathrm{\Delta }{E}_2=\frac{12400}{304}\mathrm{e}\mathrm{V}=40.80\mathrm{e}\mathrm{V}$

$\mathrm{\Delta }{E}_1+\mathrm{\Delta }{E}_2=54.4\mathrm{e}\mathrm{V}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\Rightarrow 52.23=54.4\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

If $n_1=1\mathrm{}{n}_2=5$

Acceleration of the combined system along the incline is given by

$a=g\mathrm{s}\mathrm{i}\mathrm{n}?{37}^{?}a=\frac{3g}{5}$

With respect to block M :

$\begin{array}{ll}\therefore \mathrm{}& N=mg-ma\mathrm{s}\mathrm{i}\mathrm{n}?{37}^{?}=mg-\frac{3mg}{5}\cdot \frac{3}{5}\\ & N=\frac{16}{25}mg\\ & \mu N=ma\mathrm{c}\mathrm{o}\mathrm{s}?{37}^{?}\\ \Rightarrow \mathrm{}& \mu \frac{16}{25}mg=\frac{3mg}{5}\cdot \frac{4}{5}\Rightarrow \mu =\frac{3}{4}=\frac{750}{1000}\Rightarrow I=750\end{array}$

$I_{AB}=m(0{)}^2+2m{\left(\frac{a}{2}\right)}^2+3m{\left(\frac{a}{2}\right)}^2$

$\begin{array}{ll}{I}_{AB}=\frac{m{a}^2}{2}+\frac{3m{a}^2}{4}=\frac{5m{a}^2}{4}& \\ I_{AB}=\frac{25m{a}^2}{20}=\frac{N}{20}m{a}^2& \therefore \mathrm{}\mathrm{N}=25\end{array}$

Kindly consider the following Image 

Kindly consider the following Image 

 Newton's law of cooling, $\begin{array}{rr}& -\frac{dT}{dt}=k\left(T-T_0\right)\Rightarrow -\int \frac{dT}{T-T_0}=k\int dt\\ & \Rightarrow \mathrm{}{\left[-\mathrm{l}\mathrm{n}?\left|T-T_0\right|\right]}_T^{\left(T+T_0\right)/2}=kt\mathrm{}\left(?\text{Heat}\propto \left(T-T_0\right)\right)\\ & \Rightarrow \mathrm{}-\mathrm{l}\mathrm{n}?\left|\frac{\left(T-T_0\right)/2}{T-T_0}\right|=kt\mathrm{}\Rightarrow \mathrm{}-\mathrm{l}\mathrm{n}?\left|\frac{1}{2}\right|=kt\mathrm{}\Rightarrow \mathrm{}t=\frac{\mathrm{l}\mathrm{n}?2}{k}=\frac{\mathrm{l}\mathrm{n}?4}{2k}\mathrm{}\therefore \mathrm{}n=4\end{array}$

$M={\chi }_{m}H$ and ${}_{}\frac{}{}$ ${\chi }_m=\frac{C}{T}($ Curie law $$ $)$ ,

$$ $M=\frac{CH}{T}$  or $\frac{M_1}{M_2}=\frac{C{H}_1/T_1}{C{H}_2/T_2}=\left(\frac{H_1}{H_2}\right)\left(\frac{T_2}{T_1}\right)$ $\frac{{}_{}}{{}_{}}\frac{{}_{}{}_{}}{{}_{}{}_{}}\left(\frac{{}_{}}{{}_{}}\right)\left(\frac{{}_{}}{{}_{}}\right)$

$M_2=M_1\left(\frac{H_2}{H_1}\right)\left(\frac{T_1}{T_2}\right)=12(\mathrm{A}/\mathrm{m})\left[\frac{0.2T}{0.6T}\right]\left[\frac{4\mathrm{K}}{16\mathrm{K}}\right]$

$=1\mathrm{A}/\mathrm{m}$

${\eta }_1=1-\frac{T_2}{T_1},{\eta }_2=1-\frac{T_3}{T_2}$

as ${\eta }_1={\eta }_2\Rightarrow \frac{T_2}{T_1}=\frac{T_3}{T_2}\Rightarrow T_2={\left(T_1{T}_3\right)}^{\frac{1}{2}}=\sqrt[]{400\times 900}=600\mathrm{K}$

$v=\frac{fu}{u-f}=\frac{(-10)(-15)}{-15+10}\mathrm{c}\mathrm{m}=-30\mathrm{c}\mathrm{m}$

$v_{\text{image}}=\frac{dv}{dt}=-{\left(\frac{v}{u}\right)}^2\frac{du}{dt}=-{\left(\frac{-30}{-15}\right)}^2\left(10\right)\mathrm{c}\mathrm{m}{\mathrm{s}}^{-1}=-40\mathrm{c}\mathrm{m}{\mathrm{s}}^{-1}$

$V_{\text{object}}=\frac{du}{dt}=+10\mathrm{c}\mathrm{m}{\mathrm{s}}^{-1}$

Relative velocity $=v_{\text{object}}-v_{\text{image}}$

$=10-(-40)=50\mathrm{c}\mathrm{m}{\mathrm{s}}^{-1}$

Total oscillation = $\frac{1}{2}+\frac{1}{8}=\frac{5}{8}$ $\frac{}{}\frac{}{}\frac{}{}$

$\Rightarrow timetaken=\frac{T}{2}+\frac{T}{12}=\frac{7T}{12}$

= 7

$V=k{T}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$T{V}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=K$

$\Rightarrow \gamma -1=-3/2$

$\Rightarrow \gamma =-1/2$

Work done = $\frac{nR\Delta T}{-y+1}=\frac{1\times R\times 90}{3/2}=6R$