Physics MCQs thread.

[Mueez]

http://www.xtremepapers.me/CIE/Internat ... _qp_11.pdf
sum 1 help m3 with Q.27
http://www.xtremepapers.me/CIE/Internat ... _qp_12.pdf
Q.14
plzzz

 

[yubakkk]

for q.27 in up u should mugup the wavelength of electomagnetic waves
for 14in down
total work=gain in pe +work by friction
so u will get D

 

[Saturation]

Can someone pls show me the working? I tried the principle of moments, but it doesn't work!

 

[Mobeen]

@saturation
300 * 2 = W *1.25
W = 480N

 

[Mobeen]

is the answer correct ?

 

[yubakkk]

for uniform object its geometric centre is in its c.g so weight acts from 1.25m away from pivot
so taking moment about the pivot
w*1.25=300*2
so w=480
its B

 

[Saturation]

@saturation
300 * 2 = W *1.25
W = 480N

yeah it is, but why is it W*1.25 ?

 

[ShootingStar]

pls explain this question to me :unknown:

 

[Saturation]

isn't the center 1.5m ?

 

[arlery]

because centre of gravity is in the middle of the beam.

 

[Mobeen]

isn't the center 1.5m ?

the total lenght is 2.5 .. and the weight acts at the centre . centre is at 2.5/2 = 1.25

 

[arlery]

how do I solve this?

 

[Mobeen]

pls explain this question to me :unknown:

Total pressure = 17.5MPa
total pressure = pressure of oil + pressure of water
17.5 * 10^6 = 830*g*x + 1000*g*(2000-x)
solve it
x = 1270 (D)

 

[yubakkk]

any one??

 

[yubakkk]

how do I solve this?

use v=f*lamda to find f
so then find max v from given relation
and use max ke =0.5*m*(maxv)^2

 

[yubakkk]

can u give ans of my q.??

 

[Saturation]

isn't the center 1.5m ?

the total lenght is 2.5 .. and the weight acts at the centre . centre is at 2.5/2 = 1.25

omg! I didn't notice that! I thought that the total length was 2.5 + 0.5 ! I should look at the markings more often! :%)

 

[Mobeen]

hehe .. it happens .

 

[Hateexams93]

A submarine descends vertically at constant velocity. The three forces acting on the submarine
are viscous drag, upthrust and weight.
Which relationship between their magnitudes is correct?
A weight < drag
B weight = drag
C weight < upthrust
D weight > upthrust

 

[yubakkk]

for constant velocity a=0
so f=w-d-u(d is drag force u is upthrust w is weight f is net force)
0=w-d-u
w=d+u
so w>u option D
have u understand if u know ans of my q. of two tubes plz , help me..