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Chapterwise Solutions of HC Verma

HC Verma Physics Part 1 Solutions for Chapter 16 Sound Waves

5/21/2012

46 Comments

 
Download HC Verma Solutions for HC Verma Physics Part 1 Solutions Chapter 16 Sound Waves solved by our expert teachers. We have curated solutions for all questions of chapter 16. You can download the Solution of HC Verma - Sound Waves and prepare for your upcoming competitive examinations.

Solution of HC Verma Concept of Physics -1- Chapter 16 Sound Waves

Download HC Verma Solutions for Chapter 16 for Free in PDF

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46 Comments
admin
11/3/2012 03:44:54 pm

This is attempt to provide free education to children who are preparing for engineering entrance examination.
If you are having any doubt regarding solving questions of HC Verma or understanding any topic related to this, you may leave your comment.
we will give your answer within 24 hours.
Thank You

Reply
mohan
8/14/2014 12:44:26 am

Part 1 of H C Verma. Sound waves,In Q83-b, frequency due to Doppler effect is given as 2000 Hz,where as in Q86-b and 86-d different frequencies are given even though in both cases source is moving perpendicular.Is there transverse Doppler effect ?

Reply
Suyash
2/4/2013 08:25:35 pm

A cylindrical tube, open at both ends, has a fundamental frequency F in air. The tube is dipped vertically into water such that half of its length is inside water. The fundamental frequency of air column is now
3F/4
F
F/2
2F

Reply
Admin
2/5/2013 06:54:22 pm

Dear Suyush,
In this question first we will calculate the fundamental frequency of the cylindrical tube.
Let there is a node at both the ends of the cylindrical tube. Then
distance between two nodes = λ/2 = l
or λ = 2l
where l = length of the cylindrical tube
the natural frequency v = speed of sound / λ = speed of sound / (2l)
Now there is water at half the distance of the tube. This means that the anti node is forming at the water surface.
so, distance between node and anti node = λ/4
length of tube covered = l/2
so, λ/4 = l/2
λ = 2l

frequency = speed of sound / λ = speed of sound = 2l
again we are getting the same frequency. This means that the frequency remains same.

Reply
shivam
4/26/2014 05:28:24 pm

F

Reply
kashaf
10/22/2014 07:46:42 pm

F ... frequency is a rigid factor

Reply
ayush
2/18/2013 07:18:34 pm

plz explain ques.31(hcv)

Reply
sam
4/18/2013 05:11:32 pm

how is that when loudness is 120 dB , the intensity is 1 W/(m^2) ?
(ques. 21)

Reply
theyatin
4/18/2013 10:48:38 pm

dear,
dB= 10 log10 (I/Io)
according to this intensity I is approx 1 W/m^2

Reply
kashaf
10/22/2014 07:51:40 pm

how ??? i0 is taken as 10^-12 for reference

kashaf
10/22/2014 07:50:32 pm

suno yaar ... dB=10log10(i/i0) hai ?
120=10log10(i/i0)
120=120log10(i) (dB=120 given
i0=10^-12 2/m^2 reference ..... refer to HCV theory)
10=i
ab i ko 10 rakh kar solve kar answer aa jaye ga .... galat solve kare hai ye log yaar

Reply
aman sangal
5/6/2013 07:34:44 pm

you haven't solved question 33. question 34 is marked as 33 in your answer.

Reply
theyatin
5/6/2013 08:53:37 pm

dear,
thank you for pointing out mistake. . .

Reply
Arindam Nandi
1/25/2014 11:40:41 pm

Sir, in Q.24 why is wave length/4=2.5?
Shouldn't it be =5.0 cm. since in Quincke's apparatus, when the movable part is pulled out, the total increase in path length is 5.0cm.

Reply
kashaf
10/22/2014 07:53:39 pm

distance b/w consecutive node and antinode lamda/4 hi hoti hai

Reply
shivani sharma
2/6/2014 12:56:40 am

sir,in Q.-6,why it is written that sound waves spread in all directions if wavelength of sound is much larger than the diameter of the speaker?

Reply
theyatin
2/10/2014 02:17:59 am

dear,
it is due to diffraction.
just like light coming from small aperture diffracts to spread.

Reply
shivani sharma
2/6/2014 06:00:24 am

sir can we say that in resonance there are no beats produced?

Reply
theyatin
2/10/2014 02:15:38 am

dear,
resonance is continuous sound
beat is high amplitude and low frequency sound clip.

Reply
Rohan
2/13/2014 02:19:04 pm

What do we mean by resonance condition in all these cases ??

Reply
theyatin link
2/13/2014 11:38:26 pm

dear,
i cant understand what you are asking?

i am providing a link to study more about resonance.

Reply
rahul
2/21/2014 12:52:19 am

sound wave question 87

Reply
theyatin
2/25/2014 02:37:06 pm

dear,
87. Let the distance between the source and the observer is ‘x’ (initially)
So, time taken for the first pulse to reach the observer is t1 = x/v
and the second pulse starts after T (where, T = 1/v)
and it should travel a distance [x-1/2at^2]
(S=ut+1/2at^2 => ut=S-1/2at^2)
[t=(S-1/2at^2)/ u]
now t2=T+t

Reply
zhenith gour
2/25/2014 03:20:13 am

if the pipe open at both ends resonates to a frequency n1 and a pipe closed at one end resonates to a frequency n2, if they joint to make a pipe closed at one end, then the fundamental frequency will be
n1n2/ 2 n2 + n1
2 n2 n1/ 2 n2 + n1
2 n2 n1 / n1 + n2
n2 + 2n1/ n1 n2

Reply
plz sir, reply
2/26/2014 03:00:07 am

Reply
when will i hav my ans sir?
3/8/2014 03:25:31 am

Alok Kumar Sharma
3/11/2014 11:33:47 pm

I Think the ans is the first option n1n2/2 n2 + n1, let me explain:-
let "L1" be the length of the pipe open at both ends and "L2" be the length of the pipe closed at one end. and let "v" be the speed of wave in the medium
for the pipe open at both ends
fundamental frequency = n1 = v/2 L1
=> L1=v/2 n1
for the pipe closed at one end
fundamental frequency =n2 = v/4 L2
=> L2=v/4 n2

now, total length of the pipe = L1+L2
=> (v/2 n1) +(v/4 n2) let this be equal to L
now, as the combined pipe is also a pipe closed at one end, therefore the fundamental frequency of this pipe is = v/4L
putting the value of L in the above expression, we get the answer as, (n1*n2) / (2 n2 + n1)

Rahul gaydhane
3/4/2014 09:49:59 pm

R.7.

Reply
Manshi
3/15/2014 06:21:36 pm

not getting Q.no 21 of chapter 16 i.e of sound waves. can u help me out?

Reply
theyatin
3/17/2014 01:59:00 pm

dear,
intensity=power/ area
power is 2W and area is of sphere

Reply
zhenith gour
4/24/2014 08:06:49 pm

Thanx sir, u r great

Reply
Vismay
6/11/2014 10:27:17 pm

Sir in que 32 why it is mentioned that S1S2=S2S3

Reply
Vismay
6/11/2014 10:42:40 pm

And sir why there is no information given about frequency or wavelength ...do there is any derivation from the statement given about distances b/w sources for frequency to be equal...

Reply
sai
7/2/2014 12:25:40 pm

67 th question calculation is wrong

Reply
mahesh
8/28/2014 12:21:57 pm

33 question isn't explained and skipped to 34
I request a solution to the 33Q

Reply
me
12/11/2014 02:07:47 am

will you pllz upload ans of the latest edition...i guess the ans given r for a older edition n doesnt match with my book

Reply
suyash
3/6/2015 10:11:00 pm

ques 28

Reply
shashi
4/25/2015 03:47:52 pm

In the. 58th question why did it change to Tb / Ta

Reply
gaziya
10/18/2015 01:39:03 am

what is the unit of force

Reply
mahesh
1/16/2016 08:17:35 am

In que 43 the rod acts as open organ pipe but it is clamped in middle so even harmonic should be skipped

Reply
mahesh
1/16/2016 08:21:44 am

Sir please reply at earliest

Reply
Ryan link
3/17/2016 09:00:55 pm

Sir/Madam,
Could you explain the calculus part of Question 14 of Sound Waves?

Thanks

Reply
Sneha
11/26/2016 08:14:03 pm

Please can u explain ques 83 to me

Reply
sakshi
1/11/2017 08:14:57 pm

Ques 7th part c

Reply
Prashant Patil
2/6/2017 07:50:36 am

Are plz explain H.C.V ch no.16 question no 3

Reply
mrityunjoy gope
4/20/2017 12:42:24 am

please explain Q15 NO

Reply



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