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ANYONE good in the subject, please helppp me out! - explain the attached questionsView attachment 27610
This is the second question,
View attachment 27612
And the last one is,
Q.34. I think for this question you will have to use R=V/I. I will straight away eliminate D because at D, current is constant while V is increasing so R will increase the most at D. (*Note: You can only use R=V/I when there is a straight line(Ohm's law)) For the rest I think you will need to look at the gradient. At C the gradient is decreasing so at C the resistance would be the least.ANYONE good in the subject, please helppp me out! - explain the attached questionsView attachment 27610
R=V/I. The gradient of the graph=I/V, which is the 1/R. At D, the gradient is smallest (because it's almost flat, and a flat line's gradient is 0), so 1/the gradient at D will be even smaller.
Intensity=Power/area=Energy/time/area, so Energy=Intensity x time x area. When A doubles, the Intensity quadruples (becomes x 4). Therefore E becomes 4E. Then, when the area is halved, 4E has to be divided by 2, and thus becomes 2E.
R=V/I. The gradient of the graph=I/V, which is the 1/R. At D, the gradient is smallest (because it's almost flat, and a flat line's gradient is 0), so 1/the gradient at D will be even smaller.
Intensity=Power/area=Energy/time/area, so Energy=Intensity x time x area. When A doubles, the Intensity quadruples (becomes x 4). Therefore E becomes 4E. Then, when the area is halved, 4E has to be divided by 2, and thus becomes 2E.
Q.34. I think for this question you will have to use R=V/I. I will straight away eliminate D because at D, current is constant while V is increasing so R will increase the most at D. (*Note: You can only use R=V/I when there is a straight line(Ohm's law)) For the rest I think you will need to look at the gradient. At C the gradient is decreasing so at C the resistance would be the least.
For the second question, the angle isn't from n=0 . its the angle b/w 1st and 2nd order. How do we know the pattern for them?
you can actually do this by just using R=V/I. At a. P.d increase more than I so you apply the formula R=V/I to guess the resistance. At b.) The P.D and I are increasing almost by the same amount. (Look at the graph, It looks almost like a straight line so we can assume it's proportional.) At c. It is not that clear but the I is increasing slightly more than the P.D so resistance is slightly lower I think. D. this will be out because the P.D is increasing while the current remain constant. So resistance will increase drastically.Sorry, I cant get it :/ Can you plz explain in detail?
and for the 1st Q the ans is C.
draw a straight line from origin to each of the points ...then see whichever has the least gradient has most resistance. Hope i helpedANYONE good in the subject, please helppp me out! - explain the attached questionsView attachment 27610
Ah, my apologies. I'll try again: The question asks for the point on the I/V graph at which resistance is smallest. We know that the gradient of an I/V graph is 1/R. When R is smallest, 1/R (the gradient) is biggest. We need to check and see which point has a tangent to it with the biggest gradient:
The line at C is the steepest of all, and so will have the biggest gradient (1/R) and the smallest R.
draw a straight line from origin to each of the points ...then see whichever has the least gradient has most resistance. Hope i helped
Thats all correct but is it necessary to draw tangents from origin? I know that it is a line touching only one point on curve. Isn't it?
theres no point of drawing tangents... refer to the examiner reportThats all correct but is it necessary to draw tangents from origin? I know that it is a line touching only one point on curve. Isn't it?
It is necessary that they pass through the origin, because you want to compare their gradients. They have to pass through a common point, otherwise they'll be separate unrelated lines. And the tangents do touch the curve at only one point; you can ignore all points other than the one you're considering.
theres no point of drawing tangents... refer to the examiner report
w -o7 Q 23 , anyone?
Q29; d*sin(angle)=n*lamda
sin(angle)=(n*lamda)/d
as the angle is same for both, sin(angle) will also be same
(3*lamda)/d=(2*600)/d
solving this eq. gives lamda=400 ( B )
Q33; R is directly proportional to length and inversly proportional to area........
As the volume is same,when length is doubled, area is halved so new resistance will be 4R ( D )
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