Physics MCQs thread.

[cocospaniel]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w09_qp_11.pdf

Ques.....13 how tention iz 60 ?

You need to use the radius and not the diameter. So divide 100 mm by 2 and you get 50mm.
Torque= Fd
F(tension)= Torque/d = 3/(50*10^-3) and you get 60 N

 

[arlery]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w09_qp_11.pdf

Ques.....13 how tention iz 60 ?

Moment = Force x perpendicular distance

d= radius = 0.05m
Tension x perpendicular = Torque (from center of wheel)
Tx 0.05 (radius) = 3
T = 3/0.05 = 60 N

 

[RafaySid]

this one?

Alright this one was quite difficult and took some time. Just concentrate, Q and R are in parallel. If two equal resistors are in parallel, their equivalent resistance is half of their resistance (a fact), like 2Ω and 2Ω would give a total resistance of 1Ω (try yourself out). So suppose the resistance of each is xΩ, so the total resistance would be x + x/2 Ω = 1.5xΩ. So take the ratio 0.5x/1.5x = 1/3, that means in the parallel setup 1/3 of the power is dissipated: 1/3 of 12 = 4W. As current is halved b/w the two (Q and R), 4/2 = 2W..!!!

 

[cocospaniel]

this one?

Alright this one was quite difficult and took some time. Just concentrate, Q and R are in parallel. If two equal resistors are in parallel, their equivalent resistance is half of their resistance (a fact), like 2Ω and 2Ω would give a total resistance of 1Ω (try yourself out). So suppose the resistance of each is xΩ, so the total resistance would be x + x/2 Ω = 1.5xΩ. So take the ratio 0.5x/1.5x = 1/3, that means in the parallel setup 1/3 of the power is dissipated: 1/3 of 12 = 4W. As current is halved b/w the two (Q and R), 4/2 = 2W..!!!

:shock: Nothing but respect for you

 

[ninjas4life]

help..?

 

[RafaySid]

this one?

Alright this one was quite difficult and took some time. Just concentrate, Q and R are in parallel. If two equal resistors are in parallel, their equivalent resistance is half of their resistance (a fact), like 2Ω and 2Ω would give a total resistance of 1Ω (try yourself out). So suppose the resistance of each is xΩ, so the total resistance would be x + x/2 Ω = 1.5xΩ. So take the ratio 0.5x/1.5x = 1/3, that means in the parallel setup 1/3 of the power is dissipated: 1/3 of 12 = 4W. As current is halved b/w the two (Q and R), 4/2 = 2W..!!!

:shock: Nothing but respect for you

Haha thank you thank you...

 

[arlery]

help..?

d= ut + 0.5at^2

since u = 0
d = 0.5at^2

1.25 = 0.5( 9.8 )t^2

t = 0.5s

d = vt for horizontal motion
10 = 0.5v
v = 20m/s

 

[diwash]

Can any one help me with Oct/Nov 10 P12 Q no..8?

 

[yubakkk]

@diwash plz post link or post screen picture

 

[histephenson007]

help..?

Is the answer for 18 .... C?

 

[cocospaniel]

Can any one help me with Oct/Nov 10 P12 Q no..8?

Distance travelled would be 8m. but they ask for displacement, which is a vector quantity. So 5-3= 2m. And that is the displacemet. SO the answer is A.

 

[arlery]

Use the equation of motion d = ut + 0.5at^2

equation 1 : 40 = 12u + 72a

equation 2 : 80 = 18u + 162a

[I added the distances, so that initial speed of both equations will be same]

now solve these simultaneously, and you will get 0.37 m/s^2 as the answer.

 

[diwash]

Thanks..

 

[cocospaniel]

Can anyone help me with Q1, November 2008?

 

[Mueez]

http://www.xtremepapers.me/CIE/Internat ... _qp_11.pdf
Q.19 nd Q.30 plzzz

 

[kirashinagami]

Can anyone help me with Q1, November 2008?

its quite simple, use the equation v = f x lambda

taking v as speed of light and equate for frequency

hope this helps

 

[kirashinagami]

and the answer is c

 

[mrpaudel]

speed of light = 3*10^8m/s....
and 600nm=1 wavelength
or, 3*10^17nm= 3*10^17/600 = 5*10^14 and therefore C

 

[DeathDealer]

Hey wot about mine yaaWr ?
Me jxt Asked 2 mo ques bt !

http://www.xtremepapers.me/CIE/Internat ... _qp_11.pdf
Ques 17 n 27

 

[nerd007]

Hi, can anyone please do questions number 5, 9, 13, 18, 22, 34 and 36? Thanks a lot =)

http://www.xtremepapers.me/CIE/Internat ... _qp_11.pdf