Solved physics Paper 5??

SararaIH

SararaIH

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radhika bhandari said:
velocity is measured by the time taken for the card to pass along a light gate..means distace =0.200 mwhich is the lengthof card.got it??:)
Click to expand...

Could you show me your working for at least one of the values on the table please kindly? Sorry to ask so many times :(
 
littlecloud11

littlecloud11

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Soldier313 said:
Thanx a ton :)
PS: If you do get time, I'd greatly appreciate it if you upload the procedure too. I find it sorta difficult to understand the ms for this qn :/
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No problem. :)

Procedure:
Set up the apparatus as shown in the diagram. The monochromatic filters are used to ensure only a particular wavelength reaches the observer.
Here, the wavelength used in the experiment is the independent variable and the value of theta at which the two sources appear as one is the dependent variable.
A meter scale is used to measure the distance of the observer from the source and the separation of the sources. When the sources are too close together a vernier caliper should be used for more accurate measurement.
The two sources are brought closer together until they appear as one. The value of theta at this distance is calculated.

Measurements :
Measure the distance between the two sources and the separation of the source (the distances to be measured are shown by the yellow lines in the diagram)
The wavelength of the light source also has to be measured. This can be done using diffraction grating using the same filter in the diagram.
The value of theta is calculated using the formula tanθ/2 = (1/2 the separation)/ perpendicular distance from observer to source [you can see this from the diagram]

Control:
Ensure that both light sources have the same power or intensity.
Carry out the experiment in a dark room to ensure there are no other sources of light.
the sources should be view with the same eye for each experiment.
Repeat the experiment several times for each wavelength and take the average value of theta to eliminate errors.

Analysis:
Plot a graph of theta against wavelength to determine the relationship.
if the graph is a straight line through origin then the relationship is valid and the two variable are proportional.

Safety:
The light sources will be hot so care should be taken while handling them. Heat resistant gloves should be worn.
 
radhika bhandari

radhika bhandari

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SararaIH said:
Could you show me your working for at least one of the values on the table please kindly? Sorry to ask so many times :(
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its fine.look for t=174exponential-3 v=0.2 upon 174exponential-3 that gives 1.15 and its square gives 1.32.i think now u got it??ryt??:LOL:
 
Soldier313

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littlecloud11
JazakAllah khair sis!
May Allah bless you immensely! :)

PS: just one clarification, the wavelength of the light sources after placing the filter, can just be read off the filter right? I mean, we don't need to do the whole diffraction grating explanation yeah?
 
littlecloud11

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Soldier313 said:
littlecloud11
JazakAllah khair sis!
May Allah bless you immensely! :)

PS: just one clarification, the wavelength of the light sources after placing the filter, can just be read off the filter right? I mean, we don't need to do the whole diffraction grating explanation yeah?
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Ameen.
Nope. That's just an alternative. The filter already has the wavelength quoted on it.
 
SararaIH

SararaIH

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radhika bhandari said:
its fine.look for t=174exponential-3 v=0.2 upon 174exponential-3 that gives 1.15 and its square gives 1.32.i think now u got it??ryt??:LOL:
Click to expand...

Yes I got it! I just wasn't understanding what you meant by 'upon'. It means o.2 divided by 174 right?
Thank you so much!
And I'm so sorry for the hassle :(
 
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abdullahaftab

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Guys Help Needed !!
Paper 5 May/June/10 Variant 53
Q2
e (ii)
The question where we have to find Uncertainty I got my answer 4.85% But another Person who solved this Question above Got 2%.
Plzz can anyone explain how to do this Part. These types of Questions are almost in every paper :'(
Thanks
 
meobunny

meobunny

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Someone please send me a snap shot or a picture of mayjune 05 2008 Q1 experiment? Diagram please?
 
blueberryyums

blueberryyums

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meobunny said:
Someone please send me a snap shot or a picture of mayjune 05 2008 Q1 experiment? Diagram please?
Click to expand...


Hey, I have posted a picture of this. It should be the 126th post. I didn't save the diagram on my laptop so it will be time exhausting to upload it from my phone again.. BUT it is here :D
 
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abdullahaftab

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daniyal007 said:
It basically depends on the magnitude of the best fit and worst fit... We always need a difference for uncertainity, therefore, the greater magnitude need to be placed first so that the difference is positive. if magnitude of wors fit is greater than , uncertainity= worst fit- best fit......if magnitude of best fit is greater than ,,uncertainity=best fit - worst fit....Hope u got my point :)
Click to expand...

Guys Help Needed !!
Paper 5 May/June/10 Variant 53
Q2
e (ii)
The question where we have to find Uncertainty I got my answer 4.85% But another Person who solved this Question above Got 2%.
Plzz can anyone explain how to do this Part. These types of Questions are almost in every paper :'(
Thanks.
Maybe U can help me As u solved this Paper..................
 
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