Physics MCQs thread.

[girlscampisra]

My answer is 15 Nm .. whats the answer?

 

[Mobeen]

i have the same doubts as arlery's
and plus this one !

damn ! i cant figure what would be the perpendicular distance of each force from P . e.g for force 10N would it be 2m or the 4m ?

 

[xHazeMx]

Why is the answer A?

Apparently, the thicker spring has higher k, and the thinner spring has lower k. the spring with the lower k can be affected ( extended ) by small amount of force such as wind which gives an error in determining the mass. in that case, it should be kept in a rigid box to insulate it from any other external forces ( wind ). whereas in the thicker spring, the external forces are negligible because it needs much more force to be extended ( since it has a high k value). So A is the answer.

Ok, my reason is completely different! The one with the lower constant is kept in the box,cause : Picture this, as you add small amounts of weight, the spring with the lower k will stretch, instead of the higher k. So as we keep adding small masses, which keep increasing in weight, the spring with the lower K keeps stretching, until a point where it comes in contact with the box, preventing it from stretching further. So when larger masses ( the masses which are added after the spring with the lower K has fully stretched) are added, the spring with the HIGHER k , will start to stretch, which will result in lower sensitivity for the "large" masses.
The opposite would have happened with arrangement C, where the spring with the lower K will keep on stretching and eventually break!

Hope you got it! :friends:

thats what i thought as well in the beginning but i guess this concept is wrong because at higher weights, the spring in the box DOESN'T STOP extending !! .. it moves through the hole of the box. don't forget that the lower part of the spring is just like a wire with a circle to connect springs which CAN move out the box also !

 

[aamir12345]

@hassam plz tell me the answers of above questions and plz explain that grating question

was it A for the first one and B for the second
plz confirm

 

[arlery]

C is the answer for TRAIN question!

How??

 

[unique840]

x=(u^2)/2a
2a is constant so x is directly proportional to u^2
20% inc in speed is 6/5 u.
new x= (6/5)^2

 

[MysteRyGiRl]

i have the same doubts as arlery's
and plus this one !

ok lsn...frst of all 10N and 5N wil act anticlockwise and 20N acts clockwise.and its force*PERPENDICULAR dist so form an eqn.
20N*3=(10N*2)+(5N*2) (in 2*10, the 2 comes from da perpndicular 2 which is on da pivot side)
its 60=30
subtract 30 frm 60 n da answer is A hope its clear

 

[libra94]

can someone plz help me with q31 nov 03
here's the link: http://www.xtremepapers.me/CIE/Internat ... 3_qp_1.pdf
im getting 2 A
i first found the total current which is 3A then found the ratio of 6ohm resister and 3ohm resister which is 2:1
then found the current going in 6ohm resister will be 2/3 * 3 which is 2A, can someone please explain me this!!

 

[MysteRyGiRl]

@arlery cd u plz post da year??of da train q i mn

 

[arlery]

November 2010 V1 Q.9

 

[arlery]

can someone plz help me with q31 nov 03
here's the link: http://www.xtremepapers.me/CIE/Internat ... 3_qp_1.pdf
im getting 2 A
i first found the total current which is 3A then found the ratio of 6ohm resister and 3ohm resister which is 2:1
then found the current going in 6ohm resister will be 2/3 * 3 which is 2A, can someone please explain me this!!

Maybe the markscheme is wrong. I'm getting 2A as well.

 

[MysteRyGiRl]

^yea gttng 2A too

 

[unique840]

can someone plz help me with q31 nov 03
here's the link: http://www.xtremepapers.me/CIE/Internat ... 3_qp_1.pdf
im getting 2 A
i first found the total current which is 3A then found the ratio of 6ohm resister and 3ohm resister which is 2:1
then found the current going in 6ohm resister will be 2/3 * 3 which is 2A, can someone please explain me this!!

i didnt understnd how u did it. bt mene phle 6 n 3 k parallel combination ka sum nikala that is 2 ohms. so the voltage will be 6v in d 2 ohm resistor. n 6v across d 6 n 3 ohm resistors respectively. we have the voltage n resistance. so v=ir, 6=6i. so current is 1A

 

[arlery]

^ wow you're so smart MashAllah. I feel so stupid right now. -__-

 

[arlery]

Okay I get it now.

So basically the parallel comibation is (6*3)/(6+3) = 2ohms
V1 = 12 * 2/4 = 6V
I = V/R = 6/6 = 1A

 

[girlscampisra]

C is the answer for TRAIN question!

How??

v2 is V square ok?

2as/v2-u2 = 2as/v2-u2
equate both ..
cancel out "a"
and keep v2 10^2 in first equation
and keep v2 12^ in second ..

Ive assumed 10 and 12 as increase of 20% ..
12^2/10^2 .. you get the answer in terms of "x"

p.s : i know i havent explained well, but i can't !

 

[girlscampisra]

oh i mean i can't explain it here ..

 

[unique840]

C is the answer for TRAIN question!

How??

x=(u^2)/2a
2a is constant so x is directly proportional to u^2
20% inc in speed is 6/5 u.
new x= (6/5)^2

 

[Talha]

Plz explain da correct method to get da answer! Thnx

 

[Hateexams93]

xHazemx i'm confused more now :s can u draw a diagram explaining what u said

recheck my first answer, i edited it. tell me if u don't understand it

at the highest point, (theta) = 90 why ? i just didn't get this point