Physics: Post your doubts here!

[Kanekii]

how did you know you have to make a stationary wave like this with 4 nodes?

Since the tube is open at both ends antinode will be formed at each end. For this you will have to figure out different combinations of wavelength from the fundamental frequency

 

[Canttouchthis]

When the current is zero, the voltage corresponds to the emf of the cell
Since it is decreased, so the intercept will be lower, ruling out A and C
V/I is internal resistance of battery which has increased so the gradient of new graph will increase, making the answer B

sick, thanks

 

[Kanekii]

that one is b
I am so dead for tomrrw
helppp

Sem here, i just hate waves its drawing and etc i just haige if

 

[Canttouchthis]

This is the stationary wave formed....before this those formed were not given in the options so this is corect
View attachment 64129

don't get this one
why is length=2wavelenth??

 

[Canttouchthis]

For destructive interference the path difference is 0.5λ, 1.5λ, 2.5λ, etc. For the first dark fringe i.e. n=1, it is 0.5λ so if u put the value of n in the C option so you will find that the p.d is 0.5λ, hence C is correct

genius, thanks again

 

[hellodjfos;s'ff]

View attachment 64130

View attachment 64131

Hi
need help, pls can someone explain me how to solve these questions
the answer are
27 C
28 D

27: The formula for path difference is (n-1/2)lamda
n=2, and you can find lamda using x=lamda(D)/a (the double slit formula)
Your lamda will turn out to be 4x10^-9 m
Now just insert the n and lamda into the path difference equation: (2-0.5)(4x10^-7)
Ans will be option C

28: For open tube/pipe, formula for lowest frequency is f=v/2L
v=fx(2L) ---- your open tube equation
For closed tube, formula for lowest frequency is: f=v/4L
v=fy(8L) ----- your closed tube equation
Solve the two equations algebraically now
fx(2L)=fy(8L)
fx=4Fy(D)

 

[Kanekii]

27: The formula for path difference is (n-1/2)lamda
n=2, and you can find lamda using x=lamda(D)/a (the double slit formula)
Your lamda will turn out to be 4x10^-9 m
Now just insert the n and lamda into the path difference equation: (2-0.5)(4x10^-7)
Ans will be option C

28: For open tube/pipe, formula for lowest frequency is f=v/2L
v=fx(2L) ---- your open tube equation
For closed tube, formula for lowest frequency is: f=v/4L
v=fy(8L) ----- your closed tube equation
Solve the two equations algebraically now
fx(2L)=fy(8L)
fx=4Fy(D)

How'd you figure out Q28 without drawing a wave?

 

[Canttouchthis]

27: The formula for path difference is (n-1/2)lamda
n=2, and you can find lamda using x=lamda(D)/a (the double slit formula)
Your lamda will turn out to be 4x10^-9 m
Now just insert the n and lamda into the path difference equation: (2-0.5)(4x10^-7)
Ans will be option C

28: For open tube/pipe, formula for lowest frequency is f=v/2L
v=fx(2L) ---- your open tube equation
For closed tube, formula for lowest frequency is: f=v/4L
v=fy(8L) ----- your closed tube equation
Solve the two equations algebraically now
fx(2L)=fy(8L)
fx=4Fy(D)

could you answer my question why the 2wavelength=length on previous page

 

[A*****]

how did you know you have to make a stationary wave like this with 4 nodes?

U first try with 1 node then with 2 etc but the wavelengths for those are not in the options
They are 200cm, 100cm 66.67cm and 50cm respectively

 

[Canttouchthis]

U first try with 1 node then with 2 etc but the wavelengths for those are not in the options
They are 200cm, 100cm 66.67cm and 50cm respectively

how??? why did you assume 2wavelength equal length???

 

[A*****]

how??? why did you assume 2wavelength equal length???

1 closed loop is half of wavelength we have read this during our course
In this diagram, there are 3 closed loops and 2 halves making a total of 4 complete loops...thus that is 2λ

 

[Kanekii]

27: The formula for path difference is (n-1/2)lamda
n=2, and you can find lamda using x=lamda(D)/a (the double slit formula)
Your lamda will turn out to be 4x10^-9 m
Now just insert the n and lamda into the path difference equation: (2-0.5)(4x10^-7)
Ans will be option C

28: For open tube/pipe, formula for lowest frequency is f=v/2L
v=fx(2L) ---- your open tube equation
For closed tube, formula for lowest frequency is: f=v/4L
v=fy(8L) ----- your closed tube equation
Solve the two equations algebraically now
fx(2L)=fy(8L)
fx=4Fy(D)

I think it should be fx=V/L since there are two antinodes in X
For Fy, 4Fy=V/L
Wavelength/4 as only one antinode
I think im confused.

 

[A*****]

I think it should be fx=V/L since there are two antinodes in X
For Fy, 4Fy=V/L
Wavelength/4 as only one antinode
I think im confused.

I think this will help u

 

[A*****]

I think it should be fx=V/L since there are two antinodes in X
For Fy, 4Fy=V/L
Wavelength/4 as only one antinode
I think im confused.

For fx although there are 2 antinodes but u can see in the diagram that it is 1 cmplt loop which means it is equal to half a wavelength

 

[Kanekii]

I think this will help u

View attachment 64132

Thank you soo muccch i finally understand it now

 

[A*****]

Thank you soo muccch i finally understand it now

No problem

 

[Ayesha_m01]

900x0.2=1.2xF
F=150

Bro thanks ALOT!.. :'(

 

[mariam001]

can someone explain q 36 in paper 11 may june 2017 ...

 

[Hisham Khan]

This ?

 

[A*****]

Is this statement regarding 1% correct?? But the extra 0.1 is 10%, not 1%