Physics MCQs thread.

[nerd007]

http://www.xtremepapers.me/CIE/Internat ... _qp_12.pdf

question 14?

step by step please. someone posted the answer before but it wasn't very clear.

 

[mrpaudel]

can u xplain me of q.18 of this year??
i m doing 2 give ur ans.....
plz if u know 18 then...

u mean O/N10 12 Q18?
well, its quite simple cause the molecules will always have a range of energies
evaporation occurs at all times, meaning molecules will leave the surface of water.
and besides gas which have negligible attractive force, liquid and solid have significant attractive forces.

i m taking about may 2010 ppr 11 q.18

here's the thing..!! Ice has a cage structure in which atoms are arranged..so when it melts...the atoms comes closer..dont think its coming from solid to liquid..eventhough ice is solid, it has a hollow structure inside which makes atoms distance a bit more than the water..!! and u know..density is max at 4C...so answer is B!!

 

[yubakkk]

total work= gain in pe+work by friction
this is like a formula

 

[cocospaniel]

Q32, Nov 2008, variant 1, anyone?

 

[histephenson007]

why is the answer B? =S

Oh, this is a tough one...
You have to make Simultaneous equations...

d(x/L) = l
after rearranging...
L = dx/l

Then,

with the first set of info ...
you can get
L = d(x/l) = 1700d
2nd set of info...
L+2 = d(x/l) = 5000d

1700d + 2 = 5000d
Therefore,
d = 6.0 *10^-4

Answer is B

 

[mrpaudel]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_12.pdf

question 14?

step by step please. someone posted the answer before but it wasn't very clear.

the Direction of force must be in the direction in which person pushing..!! so...downward is 200 and for horizontal force is zero...downward is 200 means upward is also 200...so calculate the force in the required direction by this process...
200=xSin30 which gives u x=400N...this is the force in the direction in which person's pushing..!! now...work done jus to lift that load=400*1.5N =600N...its jus the work done if no friction was present..so due to friction..it needs other 150N to overcome ..thus 600N+150N=750N..i dont know how far i m able to describe!!

 

[arlery]

Q32, Nov 2008, variant 1, anyone?

Since all the wires are in parallel. So (6 x 1/10) +1/100
= 1.6 ohms

 

[cocospaniel]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_12.pdf

question 14?

step by step please. someone posted the answer before but it wasn't very clear.

Gain in Pe: Mgh = 200 * 1.5 = 300 J

You need to find the work done by friction, it is Fd. You need to d by using trigonomatry. (d is parallel to the ramp btw). so d=1.5/sin30 = 3
Therefore 150*3= 450.

Total work done = gain in pe + WD against friction = 750J

Answer is D

 

[Xthegreat]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_12.pdf

question 14?

step by step please. someone posted the answer before but it wasn't very clear.

the distance of the slope
sin 30 = 1.5 / d
d = 3m

following the diagram below, find F
F = 200 / sin 30
F = 400 N



400 - friction = 250 N
work done = force x distance
W = 250 x 3 = 750

 

[aliya_zad]

IM gettin it as 4 I0, pls help

 

[MHHaider707]

IM gettin it as 4 I0, pls help

Its B....
coz frequency is halved so intensity also decrease by factor of 4
amplitude doubles and intensity should increase by factor of 4.....this decrease and increase cancels out and I remains same!!

 

[wonderland]

guys, explain how to do may june 2005 question no 22....... nd the answer is D..

 

[cocospaniel]

IM gettin it as 4 I0, pls help

I0 is proportional to X0^2 and t0^2

FOr the wave Q
is proportional to 4X0^2 and 1/4t0^2 because frequence is halved, not doubled. So everything cancels out and you get I0.

So the answer is B.

 

[wonderland]

guys, explain how to do may june 2005 question no 22....... nd the answer is D..

 

[cocospaniel]

guys, explain how to do may june 2005 question no 22....... nd the answer is D..

Use the young Modulus formula. E= Fl/Ae
Equation 1 - Ee= F(tension)*l/A
Equation 2 - Ee= F(tension)*2l/0.5A = 4Fl/A

Since it's the same wires, young modulus is the same, (e) is the same. So equate both equations.
Fl/A=4Fl/A Do some algebraic manipulation and you'll get 4/1

 

[Mueez]

http://www.xtremepapers.me/CIE/Internat ... 5_qp_1.pdf
Q.7
http://www.xtremepapers.me/CIE/Internat ... 8_qp_1.pdf
Q.37

 

[wonderland]

URGENT HOW TO DO QUESTION 3 MAY JUNE 2005 AND THE ANSWER IS B

 

[cocospaniel]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w05_qp_1.pdf
Q.7
http://www.xtremepapers.me/CIE/Internat ... 8_qp_1.pdf
Q.37

Q7, you use s=ut+0.5at^2, u=0 so s=0.5at^2

s=h and t=t2-t1

Then you make a the subject of the formula so have 2h/(t2^2-t1^2)

The answer is D

 

[nerd007]

Does anyone know how to solve questions like these? I always get them wrong.

Two wires P and Q are made from the same material.
Wire P is initially twice the diameter and twice the length of wire Q. The same force, applied to
each wire, causes the wires to extend elastically.
What is the ratio of the extension in P to that in Q?
A.1/2 B.1 C.2 D.4

Thanks a lot. You guys have been great =D

 

[cocospaniel]

Does anyone know how to solve questions like these? I always get them wrong.

Two wires P and Q are made from the same material.
Wire P is initially twice the diameter and twice the length of wire Q. The same force, applied to
each wire, causes the wires to extend elastically.
What is the ratio of the extension in P to that in Q?
A.1/2 B.1 C.2 D.4

Thanks a lot. You guys have been great =D

I have wrote this many times
Use the young Modulus formula. E= Fl/Ae
Equation 1 - Ee= F(tension)*l/A
Equation 2 - Ee= F(tension)*2l/0.5A = 4Fl/A

Since it's the same wires, young modulus is the same, (e) is the same. So equate both equations.
Fl/A=4Fl/A Do some algebraic manipulation and you'll get 4/1