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Physics: Post your doubts here!

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PLEASE HELP! HOW DO YOU CALCULATE THIS ?? I COULDNT REACH THE ANSWER :(
A student wishes to determine the density ρ of lead. She measures the mass and diameter of a
small sphere of lead:
mass = (0.506 ± 0.005) g
diameter = (2.20 ± 0.02) mm.
What is the best estimate of the percentage uncertainty in her value of ρ ?
A 1.9% B 2.0% C 2.8% D 3.7%
 
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can any1 explain hall probe thing

Hall probe is used to to measure Magnetic flux density and for this it uses hall effect.

F(magnetic) = eE

F(magnetic)=F(electric)
qvB=eE

E=V(Hall) / d

so qvB=eE becomes
qvB=e.V(Hall) / d

Hall voltage =vBd.
...........................................

hall probe is made of semi conductor material and two wires are connected to both ends of the semiconductor which is supplying a constant currect I. there are two more wires connected to the opposite edges of semiconductor and a volt meter is also connected to the terminals to measure Hall p.d.

as Hall voltage =vBd
V(hall) directly proportional to B

Then hall probe is placed in magnetic felid whose flux density (B) is unknown and then corresponding HAll p.d V is noted

B= (Voltage/unknown voltage) .B not
 
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http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_53.pdf
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_51.pdf
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_53.pdf
thnx a bunch..... if possible plz post the daigrams or the above pprs the first questions.... in the first link the question says to use a hall probe along with the the daigram can u give a brief explanation and for the other links too it wud be v.helpful for me thnx in advance...May Allah bless u...
Hall probe is used to to measure Magnetic flux density and for this it uses hall effect.

F(magnetic) = eE

F(magnetic)=F(electric)
qvB=eE

E=V(Hall) / d

so qvB=eE becomes
qvB=e.V(Hall) / d

Hall voltage =vBd.
...........................................

hall probe is made of semi conductor material and two wires are connected to both ends of the semiconductor which is supplying a constant currect I. there are two more wires connected to the opposite edges of semiconductor and a volt meter is also connected to the terminals to measure Hall p.d.

as Hall voltage =vBd
V(hall) directly proportional to B

Then hall probe is placed in magnetic felid whose flux density (B) is unknown and then corresponding HAll p.d V is noted

B= (Voltage/unknown voltage) .B not
 
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when it is rotated the angle b/w B and probe is varying and cz of this the Hall p.d is changing from max +ve --> max -ve ....just see that when magnetic felid is at 90 degree to the probe the V(hall) gets max +Ve ...
Did u get or you are still confused with it ?
n when is it max -ve? somewht im starting to get the concept
actually this topic im doing self study no help so its tough fo me to get sorry
 
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_53.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_51.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_53.pdf
thnx a bunch..... if possible plz post the daigrams or the above pprs the first questions.... in the first link the question says to use a hall probe along with the the daigram can u give a brief explanation and for the other links too it wud be v.helpful for me thnx in advance...May Allah bless u...


For 's10_qp_53.pdf'
u have to draw a hall probe .....first make a 3.D rectangle and from one side u have to attach the voltmeter to measure hall voltage ...from one side the wires are attached having the current I ...and the third side is used to oass the magnetic feild lines .....*u have to explain about its calibration too*
make sure that u draw in such a way that B is having a 90 degree angle to probe.....

P.s for attempting these questions u have to study Hall effect + hall probe and its calibration in detail otherwise its too difficult to solve these questions.
 
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For 's10_qp_53.pdf'
u have to draw a hall probe .....first make a 3.D rectangle and from one side u have to attach the voltmeter to measure hall voltage ...from one side the wires are attached having the current I ...and the third side is used to oass the magnetic feild lines .....*u have to explain about its calibration too*
make sure that u draw in such a way that B is having a 90 degree angle to probe.....

P.s for attempting these questions u have to study Hall effect + hall probe and its calibration in detail otherwise its too difficult to solve these questions.
what abt the calibeation?oh god this is hard
 
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thats what i meant with alignment ....when the magnetic substances are gets their proper alignment having proper direction of north and south pole then they are magnetized and thats the reason why they are attracted to the magnets..
View attachment 35048
Why aren't other substances aligned? What's so special about ferromagnetic materials that they are the only ones aligned due to a magnet? Also what exactly in iron is aligned?? The atoms or electrons???
 
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I have a doubt about wires in a single cable.
If you go through the examiner report for ON2007 P2 Q6, it states that the two copper wires were to be taken in series not parallel. But then for Q32 in ON 2008 P1 the examiner report for that exam says that the wires were to be taken in parallel not series.
Both the cables look the same to me so why the different solutions?
 
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A circuit has 3 branches. Two of the branches have resistors but one is empty. Will the current in the empty branch be the full current due to the battery OR will it be the divided current ? And if it's divided, how will it be? Keep in mind, the branch has no resistor or device.
 
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Solution:
Assuming the table to be a smooth surface (frictionless).
Loss in Gravitational Potential Energy = Gain in Kinetic Energy
Loss in G.P.E. = K.E. of mass + K.E. of trolley
0.5*9.8*1 = 0.5(0.5)(v)^2 + 0.5(1.5)(v)^2
0.25v^2 + 0.75v^2 = 4.9
v^2 = 4.9

Considering trolley,
K.E. = 0.5(1.5)(4.9) = 3.675 ~ 3.7 J
 
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