Physics MCQs thread.

[ciraphla]

http://www.xtremepapers.me/CIE/Internat ... _qp_11.pdf
oct/nov 10 v1 q25

 

[MysteRyGiRl]

@MHHAIDER q.31 is A cz electrons r always presnt only wn currnt is applied do they start moving

 

[MHHaider707]

@mysterygirl : wat abt that Q. 34??

 

[diwash]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_11.pdf
oct/nov 10 v1 q25

I1 = K* a^2 (k is proportionality constant)
I2 = K * (aCos 60)^2

now equating ..
I1/ a^2 = I2 / a^2 *0.25
this implies I2 = 0.25 I1 ( I2...the intensity after polirasition and I1 before polarisation)

 

[diwash]

@deathdealer...
the answer is A..unlike other options...in A when small mass(less heavy) is added the spring with less k will extend properly( the box will not disturb its expansion) so its sensitive to less heavy mass..but then heavy mass is added...the spring with small k will extend much..and its extension is disturbed by the cage..so is less sensative...
hope u got it

Ummmmmm......yeah got it buh lila bih yaaaaWr :/
Less k meanx more Extention
Less weight small extention n Heavy Weight LARGE extention !
And da Cage Supportx da LOW k so dat it don Break
BUT
Fo dah HIGH k Less mass wud hav less Extention Where az Heavy Mass REsultx MORE EXTENTION
HOW MORE HEAVY IZ LESS SENSATIVE ?

I think u didnot got what i said..
look...the problem is with the spring with small K..because it extends very easilt (is less stiff)....
with small mass the extension is small and it can detect it easily because it is less stiff...
and now with large mass the extension will be large and the cage prevents the spring to extend properly ..so is less sensetive...

if u didnot understood my explanation leave it..i think u have already got it...

 

[ciraphla]

@ thanks diwash

 

[diwash]

u r always welcomed...

 

[mickysharif]

Can someone please explain the easiest way to solve this sortov question. I have no idea how to get to the answer..

​

 

[arlery]

It's D. Since current is divided in parallel, more current will need to be supplied.

 

[mrpaudel]

its easy man..!! here's how u do..u have resistance of ammeter which is 2 ohm...!! so get a total resistance...and suppose emf a value...u can suppose that anything..n jus look which one gets higher current..!! thats it..!!

 

[MahirLatif]

Can someone please explain the easiest way to solve this sortov question. I have no idea how to get to the answer..

​

since p.d. is constant,
by I = V/R.
I is inversely proportional to R.
Means less the resistance, more the current.
Ammeter has 2 ohm resistance.. Caluclate combined R in each case.. The Lowest R shows highest current in Ammeter (answer is D)

 

[libra94]

Can someone please explain the easiest way to solve this sortov question. I have no idea how to get to the answer..

​

take voltage as lets say 12V
then find the total resistance(i hope u know hw to do tht)
and find the current by I=V/R

 

[arlery]

http://www.xtremepapers.me/CIE/index.ph ... _qp_12.pdf

Q. 14

 

[MahirLatif]

http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s10_qp_12.pdf

Q. 14

weight * sin angle = 10-friction
20 sin angle = 6
sine angle = 6/20

acceleration = g * sine angle

 

[MahirLatif]

Acceleration is 3 m/s^2

 

[mickysharif]

ok thanks guys, what about this next one.. I hate electricity questions..

 

[arlery]

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

q. 16 why is the answer B 5.5 kW?

 

[MahirLatif]

ok thanks guys, what about this next one.. I hate electricity questions..

suppose P, Q nd R have equal resistances = 2 each
total effective Resistance = 3

Power = (Voltage of battery)^2 / total Resistance
12 = v^2/3
V= 6
now Voltage in P = (2/3)*6
voltage in parallel combination Q and R = (1/3)*6 =2 V
in parallel , voltage is same.. So Q and R both have 2 V accross them
power in R = V^2/Resistance = 2^2/2 = 2 W

 

[basoom16]

question 9 and 10
winter 2009 variant 1

 

[cocospaniel]

ok thanks guys, what about this next one.. I hate electricity questions..

Someone wrote this before so I just copied it and pasted it.

'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..!!!'