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As physics p1 MCQS YEARLY ONLY.

ahmed abdulla

ahmed abdulla

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PhyZac said:
View attachment 28709
Click to expand...

Q1 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?
answer is 1.4ms-2
can u solve it via newton second law plz?

Q2--

An electric power cable consists of six copper wires c surrounding a steel core s
1.0 km of one of the copper wires has a resistance of 10 Ω and 1.0 km of the steel core has a
resistance of 100 Ω.
What is the approximate resistance of a 1.0 km length of the power cable?
B 1.6 Ω
 
P

PhyZac

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ahmed abdulla said:
Q1 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?
answer is 1.4ms-2
can u solve it via newton second law plz?

Q2--

An electric power cable consists of six copper wires c surrounding a steel core s
1.0 km of one of the copper wires has a resistance of 10 Ω and 1.0 km of the steel core has a
resistance of 100 Ω.
What is the approximate resistance of a 1.0 km length of the power cable?
B 1.6 Ω
Click to expand...

If I would solve it i would have been, this way.
Resolve downwards
mg - T = ma
19.62 - T = 2a (1)

Resolving (>)
T - 6 = ma
T - 6 = 8a (2)

add (1) and (2)

13.62 = 10a
a = 1.362
= 1.4

But you can check this, http://answers.yahoo.com/question/index?qid=20090603001954AAbZhQb

For Q2, consider them as parallel wires. And use formula

1/R = 1/R1 + 1/R2 ...

1/R = [(1/10) x 6] + 1/100

1/R = 0.61

R = 1/0.61
= 1.63...
= 1.6
 
h4rriet

h4rriet

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magnesium said:
http://www.freeexampapers.com/#A Level/Physics/CIE/2002 JunQ 11, 21, 35 PLZ!
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11. M1v1=M2v2, so v1/v2=M2/M1.
21. A mass of a liquid of density ρ is thoroughly mixed with an equal mass of another liquid of density 2ρ. No change of the total volume occurs.
So the volume of the first liquid = m/ρ. The volume of the second liquid = m/2ρ. Add the volumes to get 3m/2ρ. So ρ of the mixture=m/v=2m/3m/2ρ=4/3ρ.
35. All you gotta do is use the formula: resistance of R1/(resistance of R1+R2+R3) x the total voltage = the PD across resistor R1.
 
bigboy

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h4rriet said:
11. M1v1=M2v2, so v1/v2=M2/M1.
21. A mass of a liquid of density ρ is thoroughly mixed with an equal mass of another liquid of density 2ρ. No change of the total volume occurs.
So the volume of the first liquid = m/ρ. The volume of the second liquid = m/2ρ. Add the volumes to get 3m/2ρ. So ρ of the mixture=m/v=2m/3m/2ρ=4/3ρ.
35. All you gotta do is use the formula: resistance of R1/(resistance of R1+R2+R3) x the total voltage = the PD across resistor R1.
Click to expand...

whats the total voltage?
 
Raweeha

Raweeha

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Nab900 said:
Assalam o Alaikum,
Can someone please explain question 36 of May/June 2012 paper 13
why is the answer C :confused:
y isnt it D

and question 6 of October/November 2012 paper 11
Please explainn these two questions as deep as you can
Thanks in advance :D
Click to expand...

Wa3alykumussalam wa ra7matullahi wa barakatuhu!
Okay for question 6 from w12 p11 (since I've already done it) :
Since the values are in degrees C, the answer cannot be A or B as they both involve percentages.
From the shape of the graph, we can tell that theta increases as X increases, therefore the actual uncertainty also increases and becomes further and further away from zero. (I think) So the answer is C.
Guys, if I'm mistaken please don't hesitate to correct me! :)
 
Raweeha

Raweeha

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Omar99 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf
please some one question no. 11,12,15 and 20.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_12.pdf
question no. 20 and 22
Click to expand...

Okay, for s12 p11:
11. The answer is B as the masses are both equal and so are their speeds. To calculate the kinetic energy, you do 1/2mv2, in which the velocity is squared so the direction of motion does not matter, and 1/2mv2 doubled is equal to mv2.
12. You have to make to simultaneous equations involving T, then solve for acceleration using s=9 (if you're stuck, tell me, and I'll re-solve the whole thing)
15. You can eliminate A and B - why would there be a horizontal force when the crane itself is not moving? Which leaves C or D. Remember that because L is moved further away, The distance RL increases and the moments on the right increases. To balance this, the moments on the left must also increase, and this is done in D where the distance WR is increased.
20. The diagrams represents molecules at a lower temp, therefore lower KE and lower speed. So the arrows must be short. That leaves C and D. Do a quick count of the molecules in both boxes - you will find a greater no. in box D. Remember than density is mass/volume, and the volume of both is equal, so the answer is D.
Hope that helps, and, if I'm mistaken, don't hesitate to correct me :)
 
Nab900

Nab900

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Raweeha said:
Wa3alykumussalam wa ra7matullahi wa barakatuhu!
Click to expand...
Raweeha said:
Okay for question 6 from w12 p11 (since I've already done it) :
Since the values are in degrees C, the answer cannot be A or B as they both involve percentages.
From the shape of the graph, we can tell that theta increases as X increases, therefore the actual uncertainty also increases and becomes further and further away from zero. (I think) So the answer is C.
Guys, if I'm mistaken please don't hesitate to correct me! :)
Click to expand...


Thanks for your time (y), appreciate that
but what I am actually concerned about is why arent we taking percentage uncertainties into consideration :unsure:
and is "X" percentage uncertainty or just uncertainty :confused:.
 
Raweeha

Raweeha

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We aren't taking % into consideration because all values are in degrees C for A, B, C and D. (either 0 deg C or 100 deg C)
X is calculated as a %, which will then be translated to actual, eg. on the graph for 1oo deg C, the % uncertainty is +/- 1%, so the actual value will range from 99-101 deg C, and obviously this will increase as X increases.
I hope you're following? :p
 
Nab900

Nab900

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Raweeha said:
We aren't taking % into consideration because all values are in degrees C for A, B, C and D. (either 0 deg C or 100 deg C)
X is calculated as a %, which will then be translated to actual, eg. on the graph for 1oo deg C, the % uncertainty is +/- 1%, so the actual value will range from 99-101 deg C, and obviously this will increase as X increases.
I hope you're following? :p
Click to expand...

Thanks a lot, (y)
that makes it much more clear now that you mention it :D
 
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