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Physics: Post your doubts here!

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The examiner report says that the gradient is NOT the reciprocal of resistance. The ans is A. Can someone explain the reasoning behind this? And if the gradient is not the reciprocal of resistance then what does it represent? :)
Focus on the graph i.e its shape. From W to X the line keeps straight thus constant. From that point onwards The curve rises, with an apparent inc in I more than V. Thus R decreases as I is inv prop to R. Than the curve begins to dip so an inc in R cause I is dec.
The qtn, i think, should be done on basis of shape of graph rather than gradient.
 
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The examiner report says that the gradient is NOT the reciprocal of resistance. The ans is A. Can someone explain the reasoning behind this? And if the gradient is not the reciprocal of resistance then what does it represent? :)
Resistance is not the gradient or inverse of the gradient for such graphs. For V-I graphs, the resistance is calculated by taking a single point on the graph say (I, V). Then the resistance will be R = V/I.
 
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Can anyone solve my following paper 4 doubts?

M/J 2003
5(d) I know that alpha will deflect up and beta down. I dont know what is written in ms. Can anyone please draw and show it to me with explanation?

M/J 2005
6(b) No idea how to attempt. Can anyone show the graph please with explanation?

M/J 2006
4(c) Can anyone mark A and S an explain that positions on piston?
6(a) How to do this? Explanation needed.
6(c) How to write for this?


M/J 2007
3(b) I am not able to figure out number of squares to calculate area under graph. I wish I could have equation and could integrate xD :p Any trick to solve this type of questions?
7(a)(i) I am not able to write answers for this type of questions. Any help?
7(b) How to do this? I need a graph with explanation.
7(c) Again this graph question, where to start, what is natural frequency where to end? I need to see the graph.

M/J 2008
6(a) How parallel?
6(b)(i) why we not using 4.5cm?
6(b)(ii) why zero?

M/J 2009
6(c) how to calculate this? I used the same formula as b(ii) just changed the value of current and then subtract answer from 2.3g and then multiplied by 2 as now its ac. Why is this method incorrect?

Please I have my mock examinations on head, seeking for help. A lot more flood of doubts to come. Thank you. :)
??
 
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View attachment 59592 View attachment 59592
The examiner report says that the gradient is NOT the reciprocal of resistance. The ans is A. Can someone explain the reasoning behind this? And if the gradient is not the reciprocal of resistance then what does it represent? :)
Resistance is the inverse of the gradient ONLY if the graph is a straight line through the origin, i.e, when the resistance is constant. Ohmic conductors (ex: metal wires) show this relationship.
 
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for s05 6b, it follows from the current graph that as current varies in magnitude and direction the magnetic flux changes so the shape of the graph will be the same as that of the current and since induced emf is directly proportional to rate of of change of magnetic flux completing as many cycles as current in the first and shape is also same.
 
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for s05 6b, it follows from the current graph that as current varies in magnitude and direction the magnetic flux changes so the shape of the graph will be the same as that of the current and since induced emf is directly proportional to rate of of change of magnetic flux completing as many cycles as current in the first and shape is also same.
For s07 3b, I dont think there is any other way to do it. In fact divide the area under the graph into a geometrical shape and find the value for it and then for the curved part find the value of 1 box and count the boxes as accurately as you can.
For question 6a I think, you just need to know the general things about simple harmonic and then mould your answer according to the questions. Say the the motion is vibratory or to or fro and the magnet accelerates towards the mean position and since the magnet comes to rest after a few oscillations you can say that damping occurs. For the second part of the same question, I think that as resistance increases the current flowing in coil decreases and magnetic field around wire weakens and hence interacts lesser with the magnet's field so damping occurs slowly. ( not sure about this tho)
 
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For your first question A will be on the upper side of the region of field where the 1cm thing is written and B will be on the he lower line of the box next to the region of field and both come from Flemings left hand rule.
I got that. I was asking which will deflect more. :)
 
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for s05 6b, it follows from the current graph that as current varies in magnitude and direction the magnetic flux changes so the shape of the graph will be the same as that of the current and since induced emf is directly proportional to rate of of change of magnetic flux completing as many cycles as current in the first and shape is also same.
For s07 3b, I dont think there is any other way to do it. In fact divide the area under the graph into a geometrical shape and find the value for it and then for the curved part find the value of 1 box and count the boxes as accurately as you can.
For question 6a I think, you just need to know the general things about simple harmonic and then mould your answer according to the questions. Say the the motion is vibratory or to or fro and the magnet accelerates towards the mean position and since the magnet comes to rest after a few oscillations you can say that damping occurs. For the second part of the same question, I think that as resistance increases the current flowing in coil decreases and magnetic field around wire weakens and hence interacts lesser with the magnet's field so damping occurs slowly. ( not sure about this tho)
Thanks.
 
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View attachment 59607
I want the solution for this
The answer is D
First find the time taken for the stone to reach the water surface. This can be found by using the equation of motion: s = ut + 0.5at^2
s is 78 m, u is zero since it is dropped from rest and a is 9.81 m/s^2

Substitute these values into the equation and solve for 't'. You'll get t = 3.99s.

Then find the time taken for the sound waves to reach the person. This can be found by : t = s/v
t = 78/330 = 0.236s

Add these two times calculated: 0.236 + 3.99 = 4.226 ~ 4.22s

So the answer is D.
 
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First find the time taken for the stone to reach the water surface. This can be found by using the equation of motion: s = ut + 0.5at^2
s is 78 m, u is zero since it is dropped from rest and a is 9.81 m/s^2

Substitute these values into the equation and solve for 't'. You'll get t = 3.99s.

Then find the time taken for the sound waves to reach the person. This can be found by : t = s/v
t = 78/330 = 0.236s

Add these two times calculated: 0.236 + 3.99 = 4.226 ~ 4.22s

So the answer is D.
Thnx mate
 
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