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How was Physics P42

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I didn't get that either.
My Kinetic energy change was positive and I said the velocity increases as V^2 is proportional to Kinetic energy.
 
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I don't remember exactly. Kinetic energy was positive and potential was negative is all that I remember.
 
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what was the attenuation per length? , what was the graph like for comparator? , how did u derive the expression for K.e in 1st or second question. was it simply equating mv^2/ 4 = gmm/r^2 .......and then later obtaining the same expression? , which area of the rectangle in the magnetic field question you shaded for the electron upper face or lower face ? , , , what were the reasons for particle effect etc for that last question of quantum?

kindly share ur answers apart from predicting stupid gt :|
 
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shit, should have scored full marks for such a paper!! :"( something went wrong for two parts. Btw, can anyone inbox me a msg of how the graph for the opamp comparator should be?
Did any1 inbox it to u? Pls inbox it to me a well!
 
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what was the attenuation per length? , what was the graph like for comparator? , how did u derive the expression for K.e in 1st or second question. was it simply equating mv^2/ 4 = gmm/r^2 .......and then later obtaining the same expression? , which area of the rectangle in the magnetic field question you shaded for the electron upper face or lower face ? , , , what were the reasons for particle effect etc for that last question of quantum?

kindly share ur answers apart from predicting stupid gt :|
Derive using gravational force = Cent. Force then x1/2

I did get change in kinetic energy as far I could remember was 1.0 x 10^8 J But then after retrying i got another value.
 
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was the k.e 9 into 10^7 j something?
Yes the K.E was 9.something x 10^7 J
The graph for the comparator was a square curve with maximum and minimum amplitude at the input supply.
The attenuation per length was 17dB/Km
The face was not upper or lower it was the side of the rectangle which was into the page.

The reasons were:

1- Increasing the intensity doesnt increase number of electron emitted from the surface
2- One photon can interact with and hence exchange energy with only one electron. There is a one-to-one interaction.
3- If the frequency is below the threshold frequency there will be no emission of electron.
 
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Nobody's answers are matching :(
I got change in Ek 1.0 x 10 ^8
I did it on a scientific calculator just put the values in the equation cant get it wrong.
And attenuation was 52dB/Km
 
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Yes the K.E was 9.something x 10^7 J
The graph for the comparator was a square curve with maximum and minimum amplitude at the input supply.
The attenuation per length was 17dB/Km
The face was not upper or lower it was the side of the rectangle which was into the page.

The reasons were:

1- Increasing the intensity doesnt increase number of electron emitted from the surface
2- One photon can interact with and hence exchange energy with only one electron. There is a one-to-one interaction.
3- If the frequency is below the threshold frequency there will be no emission of electron.

Besides the comparator, I got the same answers! For the reasons, there is also the fact that the photo-current or rate of emission of electrons increases linearly with increasing intensities of incident radiation.
 
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