AS Physics P1 MCQs Preparation Thread.

[A*(a*)]

This is the second question,

 

[A*(a*)]

And the last one is,

 

[h4rriet]

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.

This is the second question,
View attachment 27612

And the last one is,

View attachment 27614

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.

 

[Frankuzi]

ANYONE good in the subject, please helppp me out! - explain the attached questionsView attachment 27610

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.

 

[A*(a*)]

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.

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?

 

[A*(a*)]

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.

and for the 1st Q the ans is C.

 

[A*(a*)]

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.

Sorry, I cant get it :/ Can you plz explain in detail?

 

[TaffsAsLevel]

please help somebody 71. how is it B?

 

[h4rriet]

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?

To get the angle between the 1st and 2nd orders, you first find the 1st order angle, then the 2nd, then you do minus the 1st from the 2nd. But you're not required to do that in the question. All you need to know is that if the number of visible orders increases, then the angles between all the orders will decrease, to make space for all the extra orders that are now visible.

 

[Frankuzi]

Sorry, I cant get it :/ Can you plz explain in detail?

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.
Hope you can get this~

 

[h4rriet]

and for the 1st Q the ans is C.

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.

 

[bogus]

ANYONE good in the subject, please helppp me out! - explain the attached questionsView attachment 27610

draw a straight line from origin to each of the points ...then see whichever has the least gradient has most resistance. Hope i helped

 

[A*(a*)]

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.

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?

 

[A*(a*)]

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?​

 

[h4rriet]

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?

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.

 

[bogus]

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 report

 

[A*(a*)]

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

Thanks I got it

 

[lunafawkes]

w -o7 Q 23 , anyone?

 

[h4rriet]

w -o7 Q 23 , anyone?

v=f x lambda. f=8/50=0.16 Hz. v=2 x pi x 2 x 0.16. K.E.=1/2xmx(v^2).

 

[prettypearlshy]

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 )

thank you so much!i was confused with Q29 as well you made it so simple!