DIP202 : APPLIED PHYSICS
UNIT- IV
1. A wave eqn.is given by ( ) Where time is in second. Then calculate amplitude, frequency, wave length, velocity, displacement.
Ans: ( )
Comparing it with standard eqn. ( )
( ) ( )
( ) ( )
m
Velocity = = =78.5 m/s Displacement = 0.05 m
2. A person observes the smoke from a gun 1. 4 sec before he hears the bang. If the gun is 476 m away from the person, find the speed of sound?
Ans: the speed of sound v= d/t = 476 /1.4 = 340m/s
3. A sound wave of frequency 5000 Hz travel in air with speed of 350m/sec. Calculate the wave length.
Ans: v = n = 𝑣 / = 350/5000 = 0.07m
4. If 2000 ripples produced in 5 sec in a pond find time period and frequency of ripples formation.
Ans: No. of ripples in 5 s = 2000
No. of ripples in 1 s = 2000 / 5 = 400 Frequency = 400 Hz
Time period = 1 𝑓 = 1 400 s = 0.0025 sec
5. A bat emits ultrasonic wave of frequency 30 Khz..If its speed is 350m/s and bats hear its echo after 0.6 sec after emitting the , find how far is bat from obstacle and wave length of wave?
Ans: f = 30 Khz = 30000hz
𝑣
𝑣 𝑓
6. Calculate the wavelength of a sound whose frequency is 220 Hz and speed is 440m/s in a given medium.
Ans: v = 440 m/s ; f = 220 Hz
𝑣 𝑓
7. An echo is heard in 3 sec after the emission of sound. If speed of sound in air is 342 m/s, what is the distance of the reflecting surface from the source?
Ans: 𝑣
8. A person standing near the cliff fires the gun and heard the echo after 1.5 sec. If the speed of sound in air is 340m/sec, how far is person from the cliff?
Ans: 𝑣
9. Meera is standing between two hills. She shouted loudly and hears first echo after 0.5 sec and second echo after 1 sec. what is distance between two hills?
Ans: Let the distance b/w
nearest clif and Meera = x m distant clif and Meera = y m Distance b/w two clif = (x + y) m
Total distance covered by sound to produce first echo = 2x m and time = 0.5 sec
Total distance covered by sound to produce 2nd echo = 2y m and time = 1 sec
So, Distance b/w two cliff = (85 + 170) = 255 m
10. A man standing in a valley between two parallel mountains fires a gun and hears echo at an interval of 2 s and 3.5 s. What is (a) The distance between two mountains (b) the location of the man with respect to the mountain?
Ans: speed of sound = 340 m/sec
Let the distance b/w nearest clif and man = x m and the distance b/w distant clif and man = y m
Distance b/w two clif = (x + y) m
Total distance covered by sound to produce first echo = 2x m and time = 2 sec
( )
( )
So, Distance b/w two cliff = (340 + 595) = 935 m
11. A stone is dropped from a 500 m tall building into a pond. When is sound splash heard? Given g =10 m/s2, speed of sound = 340 m/s.
Ans: Time taken by stone to reach pond = t :
t = 10 sec
Time taken by sound to travel 500 m t = D/v = 500/340 = 1.47 sec
Total time to hear splash = 10 sec + 1.47sec = 11.47 sec
12.Find reverberation time for a hall of dimensions 40’ x 30’ x 20’ having average absorption coefficient of 0.15.
Ans: Given: L = 40'; W = 30'; H= 20'; a = 0.15
V o l u m e o f t h e h a l l , V = 40 × 30 × 20 = 24000 ft3 S u r f a c e a r e a o f h a l l , S = 2[(40 × 30) + (30 × 20) + (20 × 40)]
= 2(1200+600+800)
= 5200 ft2
A = aS = 0.15 × 5200 = 780 OWU
13. If a University lecture hall is heavily damped with absorption coefficient 0.3. Calculate its reverberation time if approximation is applied.
Ans: Given: L = 15m; W = 8 m; H = 3m
V o l u m e o f t h e h a l l , V = 15 × 8 × 3 = 369 m3
S u r f a c e a r e a o f h a l l , S = 2[(15 × 8) + (8 × 3) + (3 × 15)]
= 2(120+24+45) = 2 (189)
= 378 m2
A = aS = 0.3 × 378 = 756 OWU
Ans: E= =
=
=
E =
E= 1.58 ev
15. The human eye’s optic nerve needs 2 × 10-17
joules of energy to trigger a series of impulses to signal the brain there is something to see. How many photons of 475 nm blue light is needed to trigger this response?
Ans: λ = 475 nm = 475 × 10-9 m
E = n hc / λ = (6.626 × 10-34 ) × (3 × 108 m/s) / 4.75 x 10-7 m = 4.18 x 10-19 J
( ) ( )
( ) ( )
16. If the energy of a photon is 350×10−10J, determine the wavelength of that photon. Ans: E = 350×10−10J; c = 3×108m/s; h = 6.626×10−34Js
Photon energy formula is given by, E = hc / λ
λ = hc /E = 6.626×10−34×3×108 / 350×10−10 = 19.87 x 10-28/350 x 10−10 = 0.056 x 10-16 m
17. What is the wavelength of an electron moving at 5.31 x 106 m/sec? (mass of electron = 9.11 x 10-31 kg; h = 6.626 x 10-34 J·s)
Ans: de Broglie's equation is λ = h/mv = 6.626 x 10-34 J·s/ 9.11 x 10-31 kg x 5.31 x 106 m/sec λ = 6.626 x 10-34
/4.84 x 10-24 = 1.37 x 10-10 m = 1.37 Å
18. What is the momentum of a photon with a de Broglie wavelength of 4.6 m? ; h = 6.626 x 10-34 J·s.
Ans: p = h/ λ = (6.63 × 10-34) /4.6 = 1.44 × 10-34 kg∙m/s
19.A ray of light is incident on the surface of the plate of glass of refractive index 1.5 at the polarising angle. Then find the angle of refraction.
Ans: By Brewster’s law 1.5 = 56.3°
Angle of refraction, – –
20.When light is incident on a plane surface of a material at a glancing angle 30°, the reflected light is found to be completely planed polarized. Then find the angle of refraction and refractive index of the medium.
Ans: Angle of refraction, – – By Brewster’s law = √
21.Two polaroid, A and B are kept with their transmission axes at an angle θ with respect to one another. If the transmitted intensity of light It=0.75I0 where I0 is the intensity of light incident on the system then find θ.
Ans:
22.If the angle between the pass axis of polarizer and analyzer is , write the ratio of the intensities of original light and the transmitted light after passing through the analyser Ans: The intensity of light after passing through polarizer is reduced to half.
The light intensity passing from analyzer
