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Solved physics Paper 5??

nishan2052

nishan2052

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when we divide quantity by something.what will be the uncertainty of new value? mj/10/52 question number two.plz someone help
 
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s.asg259

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sagar65265

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Soldier313 said:
Aoa wr wb
Can someone please provide a detailed diagram and explanation for qn 1 of this paper please?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_51.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_51.pdf

littlecloud11
geek101
sagar65265
biba

And everyone else :)
Thank you.
Click to expand...

I'll try to upload an image later, but hopefully, for now, the explanation is okay - sorry!

The independent variable in the investigation is the field strength B at the center of the coil and the dependent variable is, then, the radius of the coil. Since the strength of the magnetic field at the center of the coil depends upon the current in the coil and the number of turns in the coil, these two variables need to be kept constant; no matter what the radius is, the same number of turns should be used (the more the coils, the greater the field strength, so it is a control variable) per coil. The current can be kept constant by placing a rheostat in series with the coil and using an ammeter to measure the current through the coil. If the current decreases, the resistance of the rheostat can be decreased to bring up the current. If the current increases (sudden surge, any other reason) the resistance of the Rheostat can be increased to bring the current down.

The experiment can be carried out by first taking a length of wire and wrapping it firmly around a cylindrical object; this will give it a coil-like shape with an approximately constant radius. To increase the radius later, a wider cylinder can be used. A ruler can be used to measure the diameter of the coil, and the result divided by 2 to get the radius. This measurement can be repeated around the coil along several diameters and the values of r averaged.

The coil can be connected to a d.c. power source, an ammeter and a rheostat, all in series (diagram) and the coil can be hung (after being flattened) from a clamp/ retort stand holder. The center of the coil can be found using a ruler, and a stack of books with a track/ruler placed on top of them can be aligned with the coil such that the ruler is perpendicular to the plane of the coil and is going into the plane of the coil. This track can be used to move the hall probe towards the center of the coil. The Hall probe should be connected to a calibrated galvanometer/ voltmeter, and the maximum reading shown can be noted down as the value of B. Several readings of B can be taken for each radius r and averaged to plot into the graph.

About 10 readings can be plotted into a graph of B on the y-axis versus (1/r) on the x-axis. If the relationship is indeed true, the graph should be a straight line through the origin (the equation is the form of B = k/r where k is a constant, so the gradient is B * r = k, so gradient is constant).

The coil may heat up due to the current passing through it, so it is advisable to let the coil cool down between experiment before replacing it/ using heat resistant gloves while doing so.

A large current can be used to ensure a large value of B; this reduces percentage uncertainties (any that might creep into the readings) and also gives a large value of B. Similarly, a large number of coils can also be used, but kept constant throughout the experiment. Any external currents and magnetic fields will have to be eliminated, so the experiment should be performed in an isolated location. Again, the current can be kept constant using a rheostat (as mentioned above) and if the same Hall Probe setup is being used for all the experiments, the probe and voltmeter can be calibrated in a magnetic field of known strength.

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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nishan2052 said:
when we divide quantity by something.what will be the uncertainty of new value? mj/10/52 question number two.plz someone help
Click to expand...

If you are referring to dividing t by 10 to give T, the uncertainty is also divided by 10.

The reason is that when you multiply/divide two quantities together, you add their percentage uncertainties to find the percentage uncertainty in the resulting value.
In this case, if you want to find the percentage uncertainty in the value of T, it will be the (% uncertainty in t) + (% uncertainty in 10) = (% uncertainty in t), since 10 is a constant, absolute value which is completely accurate and has no percentage error.

As an example, if we take the first value of t, 18.9, the percentage uncertainty is:

0.1/18.9 = 0.00529 (0.5 %).
Therefore, the uncertainty in the final answer is 0.5% = 0.00529:
The absolute uncertainty is 1.89 * 0.00529 = 0.01.

You can see that the uncertainty in 1.89 is 10 times less than the uncertainty in 18.9, so the uncertainty has been divided by 10, the same value that divided t to give T.

Hope this helped!

Good Luck for all your exams!
 
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moinul

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ANYONE CAN U PLZ HELP ME WITH THIS QUESTION.
A value is given under (L/10^-2) is 54.5 +-0.5
now (1/L) needs to be calculated. For 54.5 its (1/54.5*10^-2) = 1.83
BUT FOR THIS VALUE HOW TO CALCULATE THE ABSOLUTE UNCERTAINTY(+-0.5 )
 
Hirdayesh.B.Shrestha

Hirdayesh.B.Shrestha

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If the error bars to be plotted are for the values on the x-axis, then, I suppose the error bars must also be drawn horizontally, right?
 
Soldier313

Soldier313

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Hirdayesh.B.Shrestha said:
If the error bars to be plotted are for the values on the x-axis, then, I suppose the error bars must also be drawn horizontally, right?
Click to expand...

yeah, they have to be horizontal in that case

chocolatelover said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_53.pdf
Can anyone tell me the value of % uncertainty in d ii and the value of e withs its absolute uncertainity.
TYSM
Click to expand...

d ii ) i got 2.9%
e ) 3001 +/- 86.9 ohms

s.asg259 said:
Hey.
In 51/M/J/11.
Part e (ii)
When we're supposed to calculate the percentage uncertainty in h, are going to use 1/12 for the fractional error in Temperature or 1/278?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_51.pdf
Click to expand...



1/278 , that's what i used
 
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leosco1995

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moinul said:
ANYONE CAN U PLZ HELP ME WITH THIS QUESTION.
A value is given under (L/10^-2) is 54.5 +-0.5
now (1/L) needs to be calculated. For 54.5 its (1/54.5*10^-2) = 1.83
BUT FOR THIS VALUE HOW TO CALCULATE THE ABSOLUTE UNCERTAINTY(+-0.5 )
Click to expand...


(0.5/54.5) * 1.83
 
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leosco1995

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hamna0006

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nishan2052 said:
when we divide quantity by something.what will be the uncertainty of new value? mj/10/52 question number two.plz someone help
Click to expand...
no matter what the question, if u subtract the origibal value from the max value, u'll get the uncertainty.
For instance, thy give you length, L, and ask you to find 10/L and uncertainty in that one.
L= 12+/-0.5
10/L= 12/10=1.2
uncertainty= {10/(12+0.5)}-{10/12)
Hope that helps :)
 
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freezingfires

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Can someone plz post the diagram of solved october november 2007 q1 p5.Its urgent.Any help will be highly appreciated!!
 
Hirdayesh.B.Shrestha

Hirdayesh.B.Shrestha

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Hey
sagar65265 said:
I'll try to upload an image later, but hopefully, for now, the explanation is okay - sorry!

The independent variable in the investigation is the field strength B at the center of the coil and the dependent variable is, then, the radius of the coil. Since the strength of the magnetic field at the center of the coil depends upon the current in the coil and the number of turns in the coil, these two variables need to be kept constant; no matter what the radius is, the same number of turns should be used (the more the coils, the greater the field strength, so it is a control variable) per coil. The current can be kept constant by placing a rheostat in series with the coil and using an ammeter to measure the current through the coil. If the current decreases, the resistance of the rheostat can be decreased to bring up the current. If the current increases (sudden surge, any other reason) the resistance of the Rheostat can be increased to bring the current down.

The experiment can be carried out by first taking a length of wire and wrapping it firmly around a cylindrical object; this will give it a coil-like shape with an approximately constant radius. To increase the radius later, a wider cylinder can be used. A ruler can be used to measure the diameter of the coil, and the result divided by 2 to get the radius. This measurement can be repeated around the coil along several diameters and the values of r averaged.

The coil can be connected to a d.c. power source, an ammeter and a rheostat, all in series (diagram) and the coil can be hung (after being flattened) from a clamp/ retort stand holder. The center of the coil can be found using a ruler, and a stack of books with a track/ruler placed on top of them can be aligned with the coil such that the ruler is perpendicular to the plane of the coil and is going into the plane of the coil. This track can be used to move the hall probe towards the center of the coil. The Hall probe should be connected to a calibrated galvanometer/ voltmeter, and the maximum reading shown can be noted down as the value of B. Several readings of B can be taken for each radius r and averaged to plot into the graph.

About 10 readings can be plotted into a graph of B on the y-axis versus (1/r) on the x-axis. If the relationship is indeed true, the graph should be a straight line through the origin (the equation is the form of B = k/r where k is a constant, so the gradient is B * r = k, so gradient is constant).

The coil may heat up due to the current passing through it, so it is advisable to let the coil cool down between experiment before replacing it/ using heat resistant gloves while doing so.

A large current can be used to ensure a large value of B; this reduces percentage uncertainties (any that might creep into the readings) and also gives a large value of B. Similarly, a large number of coils can also be used, but kept constant throughout the experiment. Any external currents and magnetic fields will have to be eliminated, so the experiment should be performed in an isolated location. Again, the current can be kept constant using a rheostat (as mentioned above) and if the same Hall Probe setup is being used for all the experiments, the probe and voltmeter can be calibrated in a magnetic field of known strength.

Hope this helped!
Good Luck for all your exams!
Click to expand...
thanks for the solution...by the way, where is the diagram?
 
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Binod Shakya

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minie23 said:
Soldier313
Here you go :) November 2002 P5 Question 2.

Please add the safety precautions and further design which is found in the marking scheme :)
Click to expand...
mass should be kept constant or we have to vary for each experiment.
I think independent variable is tension and dependent is frequency
But i am really confused.
 
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