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AS Physics P1 MCQs Preparation Thread.

Tkp

Tkp

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leosco1995 said:
Brittle materials can't sustain plastic deformation, they just break so B and C are wrong.
Of aluminium and steel, metals usually have a lower plastic deformation while ductile materials like copper and aluminum will deform much more. So A is right.
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steel is made of two materials iron and carbon.thats y al is right
 
confused123

confused123

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Knight said:
Extention is directly proportional to length and inversly proportional to area......
so for same extention for both wires P and Q, tention in P will be F for length l and tention in Q will be F/2 for length 2l, tention in P will be F for area A and tention in Q will be F/2 for area A/2.............
so the total ratio due to length l and 2l and area A and A/2 will be........
(tention in P)/(tention in Q)=4/1 ( D )
Click to expand...

i didn't got it how did u assumed tension to be F and then F/2. i am getting 1/4 for this mcq every time i try to solve..
 
Tkp

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confused123 said:
i didn't got it how did u assumed tension to be F and then F/2. i am getting 1/4 for this mcq every time i try to solve..
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m
fr this one ym and extension is same.so fl=a.do by this formula and u will get the tension
 
confused123

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A.ELWY 7 said:
Q 24: Wave P has half the amplitude but double the frequency..it have a frequency of 2t and amplitude of X0 which gives an intensity of I0....Wave Q have half the frequency and double the amplitude...so 2 times X0^2 is 4I0...and 0.5 times t^2 = 0.25I0...so 4x0.25=1 Io so answer is B
Q31: force =EQ..and E=V/d, so F =V/d x Q...substitute 200/5x10^-3 x 1.60x10^-19 = 6.4 x 10^-15 so the answer is B....the drop was an electron and as the 2nd page of the exam sas...the charge of the electron is 1.60x10^-19...and to convert 5mm to m we multiply by 10^-3
Q37: light and temperature are inversely proportional to resistance and the resistance is proportional to the volt..both have high voltage..so ...for Vt, the temperature must be high so tht the resistance of the thermistor is low and becuse it is a potential divider..if the thermistor resistance is low the other will be high and so the voltage across it...for the VL..the light have to be low to have high resistance and so high V... so the answer is C...wish i helped
Click to expand...

nice explanation bro. thanks, but can you elaborate on this '' becuse it is a potential divider..if the thermistor resistance is low the other will be high and so the voltage across it...for the VL''
 
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sagar65265

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Hassi123 said:
MAY/JUNE 2003: Q10 ans D and Q40 ans D​
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Ok, buddy, i'll answer these first:

Q10: Since the question says the surface is smooth , no friction acts anywhere. Okay, now for this, you need to know how to draw free body diagrams. Firstly, on the block X, there are 2 forces acting, the force F that is pushing it and the Newton's Third Law force - since the block X is exerting a force on the block Y, the block Y exerts an equal and opposite force on Block X. So:
If the force exerted by Block X on Block Y = X,

F - X = ma for block X. According to F = ma (better to say (net resultant force) = ma)

For block Y, mass = 3m and the only force acting on it is the force from block X which has earlier been defined as the force on Block Y from Block X and vice versa. For Block Y:

X = 3ma (since mass = 3m and the acceleration can be assumed constant it is the same variable)

Transposing for acceleration since acceleration is assumed to be the same,

a = (F - X)/m and a = X/3M
Equating the 2 equations as of now, we get:

3(F - X) = X
3F - 3X = X
3F = 4X So X = 3F/4 = D

Q40: We know that a proton has a charge of +e, so we choose the option in such a way that the multiples of the charge add up to give +1:
The first option will give us -1 since we multiply the charge on the down quark by 3 but then without a positive charge at all the answer is invalid, so the answer is wrong.

The second option gives us (2/3) *1 + (-1/3)*1 which is not equal to +1.

The third option gives us (2/3) * 2 + (-1/3)*2, which gives us zero and could possibly be the configuration of quarks in a neutron!

The fourth option gives us (2/3) * 2 + (-1/3) * 1 which gives us the +1 we wanted! = D
 
fathimaash

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resultant force means ma.. so mass here is the total mass which is 10kg.. u c both masses r attached so total mass should be considered.. i guss... if im wrong pls tell me but im sure about it (pretty similar to question in mechanics)
 
fathimaash

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q14
#s=s u=0 v=u a=10
use eqn v2=u2+2as
so u2=20s
#s=s/2 v=0 u=v a=10
use same eqn
so v2=10s
den v2/u2 i= 10s/20s ( so both s gets cancelld off)
v2/u2= 1/2
and v/u = 1/squareroot2
hope u get its difficult to write it in words sorrry abut it
 
fathimaash

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Tkp said:
mj11(3)-36,14,19
pls help me
Click to expand...
q19
find da mass of both alchol and watr seperately
watr=1.5*1000=1500kg
alcohol=0.5*800=400kg
total mass=1500+400=1900 kg
density of mixture will be=1900/2=950kgm-3
 
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sagar65265

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Hassi123 said:
MAY/JUNE 2004: Q21 ans C Q25 ans A Q26 ans B Q39 ans C​
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Q21: If the density of a substance P is greater than the density of a substance Q, it has to have more mass per unit volume, which is where the formula for density appears. Due to this:

Mass(P)/Volume(P) > Mass(Q)/Volume(Q)

In this case, the number of atoms per unit volume have units of volume^-1

Due to this, Mass X Number of atoms per unit volume is equal to density of substance.
This should make sense because the more number of atoms per unit volume you have, the lower the volume taken for a fixed number of atoms to be present, i.e. 1 mole of atoms/molecules will take less volume than 1 mole of atoms/ molecules of another substance with less atoms per unit volume. Thus the answer should be C, since P has a greater mass per unit volume.

Q25: If you take the waveform as it is and transpose it a slight bit towards the right of the screen, you will see that P has moved down a tiny bit but Q will remain in the same position since it has already reached maximum amplitude and the particle stays there for some time before going back down - the top of a crest and the bottom of a trough are sometimes slightly flat, if you may have noticed, thus it is A.

Q26: For this answer, you need 2 formulae:

Power = Intensity * Surface Area AND Intensity / Amplitude^2 = constant

In this case, Intensity = Power/Surface Area = E/S
Then, Intensity = K(Amplitude)^2, so E/S = kA^2
Next, k = E/SA^2, which remains constant all the time for this wave.

After the modifications have been made:
k = New Intensity/4A^2
E/SA^2 = New Intensity/4A^2 So cancelling out the A^2, we get:
New Intensity = 4E/S
Now we use the equation Power = Intensity * Surface Area:

Power = 4E/S * S/2 Giving us 2E = B

Q39: In all reality, a gold nucleus contains protons AND neutrons - this experiment can't tell us that there are neutrons in the nucleus - neutrons have no charge and they therefore won't interact with alpha particles. So the answer is not D.
Now for A. It is pretty clear that the gold nucleus wasn't deconstructed by this experiment so the binding energy could not be discovered, although i'm not sure if it is impossible to do so.
Honestly, it should be pretty clear the answer is C since the reflected of the alpha particles can only occur if the size of the nucleus is miniscule. I'll try to explain it better next time, but i guess this may be the best I can do - sorry!

I hope this has helped!
 
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