Physics: Post your doubts here!

[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%

 

[A*****]

This ?

 

[A*****]

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

The switch is always connected to the positive terminal of the battery so C and D are ruled out
We need a voltage between 5 and 7V so if we look at A, the output is 1/2 x 9= 4.5V which is not bw the specified range
In B the output is 2/3 x 9= 6V which is within the range so B is the correct option

 

[Hisham Khan]

The switch is always connected to the positive terminal of the battery so C and D are ruled out
We need a voltage between 5 and 7V so if we look at A, the output is 1/2 x 9= 4.5V which is not bw the specified range
In B the output is 2/3 x 9= 6V which is within the range so B is the correct option

Thanks bruh!

 

[Kanekii]

is there any other method to do this?

 

[mariam001]

The switch is always connected to the positive terminal of the battery so C and D are ruled out
We need a voltage between 5 and 7V so if we look at A, the output is 1/2 x 9= 4.5V which is not bw the specified range
In B the output is 2/3 x 9= 6V which is within the range so B is the correct option

ohhh..thank you i never knew about the switch thing
, but what did they mean when they said that a pd of 0v to its input at one switch position

 

[Hisham Khan]

View attachment 64137 is there any other method to do this?

d = 100m
u = 10m/s
v = 20m/s

v² = u² + 2ad
so
a = (v² - u²)/(2d) = (20² - 10²)/(2*100) = 1.5m/s²