AS Physics P1 MCQs Preparation Thread.

[sagar65265]

http://www.freeexampapers.com /#A Level/Physics/CIE/2002 Nov Q14?

It's basically just balancing moments; Since the entire set up can be assumed to be stationary once the 50g mass is added, the tension in the rope is

Tension = (20/1000) * 9.81 = 0.1962 N

This means that the moment caused by the suspended mass is

Moment = Tension * Distance from pivot = 0.1962 * 0.6 = 0.11772 Nm^-1

This is an Anti-Clockwise moment because the force will pull the left side upwards (and thus in the ACW direction) if it is allowed to act alone.
The weight of the ruler can be assumed to act from its center (at the 50 cm mark), and is a Clockwise moment, because the force will pull the right side of the ruler down (and thus CW) if it is allowed to act alone.
The value of this moment is :

0.1 * 9.81 * 0.1 = 0.0981 Nm^-1

So if the system has to be in equilibrium, then the net moment has to be zero. But the ACW moment due to the 20g mass is more than the CW moment due to the mass of the ruler, so we need to position the 50g mass in such a place that it will exert a CW moment to make the CW moment equal to the ACW moment. This means that the 50g mass will have to be positioned to the right of the pivot, and it will have to exert a moment of magnitude

0.11772 - 0.981 = 0.01962 Nm^-1 (this is the difference in moments in the current situation).

Therefore,

(50/1000) * 9.81 * D = 0.01962
D = 0.04 m = 4 cm to the right of the pivot.

Since the pivot is at 40 cm, the mass will be placed at the 44 cm mark.

Hope this helped!
Good Luck for all your exams!

 

[itallion stallion]

If you write down the equations of motion for the two objects (barrel coming down and man going up), you get:

Barrel:

Tension = 120g - 120 * acceleration
Tension = 1200 - 120a

Man:

Tension = 80g + 80a
Tension = 800 + 80a

Equating the two,

1200 - 120a = 800 + 80a
400 = 200a
a = 2 ms^-2.

Since the two objects both start from rest and have the same accelerations (they are connected by the string, so if one moves a distance x, so does the other, in the same time interval), the mans head will be in line with the bottom of the barrel after travelling 9 meters. Therefore,

u = 0 ms^-1 , a = 2 ms^-2 , s = 9m:

v^2 = 0 + 2 * 2 * 9
v^2 = 36
v = 6 ms^-1 = A

Hope this helped!
Good Luck for all your exams!

understood it!thanks alot and best of luck for your exams too!!

 

[TaffsAsLevel]

I think nobody can solve this question in MCQ, go to nov 2002 q.36. What the hell is that?

 

[hope4thebest]

Anyone....What is the logic in this question (Ans:B)

 

[Lyfroker]

13: B..W is in the middle of the rod..so ( W x d=F x d) so W = (300 x 2)/1.25 = 480
18: A..( 9000 x 40) - ( 20000 x 12)= 120Kj
28: D ... the distance between the slits and the screen is indirectly proportional to the wavelength..so if the wavelength is decreased so the distance must increase and so D is the answer cause it is the only one greater than 1.00m
33: C... R= pl/A = pl/(pie) (0.5d)^2..then p x 2l / (pie) x (2d/2)^2.....from 0.5d squared to 2d/2 squared is multiplication by 4 so R will be 0.5 then I=V/R so 2V/0.5R then 4I so V

thnx a lot

 

[abruzzi]

Q. 10 :

If a net force acts on an object, it will cause a change in momentum of the object (usually the change in is in velocity, either in terms of direction or magnitude). The average force between any two times is the total change in momentum that occurs between those two times divided by the time taken for the change in momentum (just like the average speed between two points is the total distance divided by the time taken to travel that distance). In general, the force at any one time is the gradient of the momentum-time graph at that point, but since the average force is being asked for, it's just the gradient of the line joining those two points, in other words, it is B, since that shows the change in momentum divided by the time taken.

Q. 20

Hmmmm. How to explain this? Okay, imagine that the beam actually bends under the stress applied by the hanging load. In that case, can you see that the top of the rod stretches and the bottom of the rod decreases in length - hopefully the diagram attached will help, so this basically means that both X and Y are going to be pulled on, towards the load, while Z will be pushed backwards, i.e. compression - sorry, but I'm not sure how else to explain this!

Hope this helped!
Good Luck for all your exams!

Thanks a lot for the detailed explanation!

 

[hope4thebest]

help required!

 

[Tkp]

welcome

thnx a lot

 

[Tkp]

b.well i memorised this

View attachment 28695
help required!

 

[Tkp]

Anyone....What is the logic in this question (Ans:B)
View attachment 28687

i did this 1 by elimination method
see by using more layers of spring increases extension.
and using smaller spring constant also increase extension.f=kx
so f/k=x
so smaller spring constant increase the extension
and y.m=stress/x/l
so e=stress*l/x
s0x=stress*l/e
so smaller young modulus increases extension
so the correct ans is B
Read the question carefully.they said which change will nt have the desired effect that means less x

 

[ZainH]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf

Q13,14,15 and 34.

 

[Lyfroker]

q#3, 10, 17, 20, 22, 30 & 40
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s03_qp_1.pdf

 

[h4rriet]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf

Q13,14,15 and 34.

13. All you gotta do is choose a pivot and work out the moments on either side of it...
14. mgh=1/2m(v^2). mgh/2 = 1/2m(x^2). Eliminate m, g, and 1/2. Cross multiply.
15. Find the angle to the horizontal of the slope. Then find the component of the weight parallel to the slope. Then use F=ma to find the acceleration. Then use a suvat. You can also do it using 1/2m(v^2)=mgh.
34. Resistance = ql/A. A=lxl, so Resistance=ql/lxl. l and l cancel out; R=q/l. l^3=V, so l = cube root of V = V^(1/3), so R=q/V^(1/3).

 

[Lyfroker]

q#31
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w03_qp_1.pdf

 

[magnesium]

http://www.freeexampapers.com/#A Level/Physics/CIE/2002 JunQ 11, 21, 35 PLZ!

 

[magnesium]

q#31
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w03_qp_1.pdf

Total current in the circuit is 3A .....V=IR where V=12 and I =12/4=3A
In the loop containing 6ohm resister less current will pass as u know R is inversely proportional to I .....

 

[ahmed abdulla]

Total current in the circuit is 3A .....V=IR where V=12 and I =12/4=3A
In the loop containing 6ohm resister less current will pass as u know R is inversely proportional to I .....

A box of mass 8.0 kg rests on a horizontal, rough surface. A string attached to the box passes

over a smooth pulley and supports a 2.0 kg mass at its other end​

When the box is released, a friction force of 6.0 N acts on it.
What is the acceleration of the box?

A 1.4 m s–2​

 

[Lyfroker]

q#3, 10, 17, 20, 22, 30 & 40
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf

 

[magnesium]

A box of mass 8.0 kg rests on a horizontal, rough surface. A string attached to the box passes
over a smooth pulley and supports a 2.0 kg mass at its other end
When the box is released, a friction force of 6.0 N acts on it.
What is the acceleration of the box?

A 1.4 m s–2

Can u plz tell the year..i need to see the diagram..is there any?

 

[magnesium]

https://www.xtremepapers.com/commun...-p1-mcqs-yearly-only.17330/page-4#post-323269
look 4 da solutions here