Physics MCQs thread.

[gexceln]

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

The answer is D. weight > upthrust.

Since the submarine is descending at constant velocity this means that upward forces = downward forces => upthrust + drag = weight. Therefore weight is greater than both upthrust and drag individually and D is the only answer that corresponds to this situation.

 

[yubakkk]

hey gexceln can u xplain my q. of about tubes in page94

 

[ShootingStar]

another one..

 

[ShootingStar]

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)

ah i see! thnx

 

[yubakkk]

hey cant u solve??

 

[gexceln]

hey gexceln can u xplain my q. of about tubes in page94

Sure. What you have is a case of stationary wave formation.

In a closed tube, the fundamental resonance occurs when a stationary wave with one node and one antinode (i.e. a quarter wavelength of the fundamental wave is setup in the tube). For the case of a tube that is open on both ends, fundamental resonance occurs when you have a wave with two antinodes (one at each end of the tube) and one node (in the middle) is setup, hence one half of a wavelength is setup in the tube. Since both waves are sound waves traveling with equal speeds in air this means that the frequency of the wave in the open ended tube has to be twice the frequency of the wave in the open tube to compensate for the second wave's shorter wavelength.

Hence you need to use a tuning fork with twice the frequency of the first one (2f), hence the answer is D.

Take a look at this

http://www.revisesmart.co.uk/physics/waves/standing-waves.html

 

[grumpy]

hint:
the length of the tubes and speed of the sound in this question is considered same for both tubes.

 

[gexceln]

Hey, yubakkk , I have a question I want to ask and I was wondering how one attaches the images like what you did for the stationary waves question. Thanks!

 

[arlery]

Why is the answer B?

 

[arlery]

Hey, yubakkk , I have a question I want to ask and I was wondering how one attaches the images like what you did for the stationary waves question. Thanks!

convert your image to .png format. Then click on Post reply and below the preview and submit option, there is option for upload attachment. Click there and upload your file.

 

[yubakkk]

Hey, yubakkk , I have a question I want to ask and I was wondering how one attaches the images like what you did for the stationary waves question. Thanks!

post ur q. here

 

[yubakkk]

Why is the answer B?

dsin45=3lamda eqn 1
d sin 90=n*lamda eqn 2
solve it 2 get n
90 iz used becauz max order is seen on 90degree

 

[Zizu1992]

Please explain http://www.xtremepapers.me/CIE/Internat ... _qp_11.pdf
Q 15

 

[arlery]

Please explain http://www.xtremepapers.me/CIE/Internat ... _qp_11.pdf
Q 15

m1v1 = m2v2
2(2) = 1(v2)

v2 = 4 m/s

KE = (.5x2x2^2) + (.5x1x4^2)

= 4 + 8 = 12J

Hope that helps.

 

[daasmanher]

help question 15
http://www.xtremepapers.me/CIE/Internat ... _qp_12.pdf

 

[yubakkk]

Please explain http://www.xtremepapers.me/CIE/Internat ... _qp_11.pdf
Q 15

total momentun before=after
m1v1 = m2v2
2(2) = 1(v2)
v2 = 4 m/s
KE = (.5x2x2^2) + (.5x1x4^2)
= 4 + 8 = 12J

 

[yubakkk]

Please explain http://www.xtremepapers.me/CIE/Internat ... _qp_11.pdf
Q 15

total momentun before=after
m1v1 = m2v2
2(2) = 1(v2)
v2 = 4 m/s
KE = (.5x2x2^2) + (.5x1x4^2)
= 4 + 8 = 12J

 

[gexceln]

Why is the answer B?

For the 3rd order diffraction dsin45 = 3λ => (sqrt2/6)*d = λ

sinθ is less than or equal to 1

For n=5 for example : dsinθ = 5λ => dsinθ = (5*sqrt2/6)*d
=> sinθ = (5*sqrt2/6)

(5*sqrt2/6) > 1 and is thus out of range. The last permissible value for n is 4 since (4*sqrt2/6) < 1.

 

[arlery]

help question 15
http://www.xtremepapers.me/CIE/Internat ... _qp_12.pdf

25000 V - 100% Voltage

80% voltage = 20000 V

P = VI
4x10^6 = 20000 x I
I = 200 A [C]

 

[daasmanher]

help question 15
http://www.xtremepapers.me/CIE/Internat ... _qp_12.pdf

25000 V - 100% Voltage

80% voltage = 20000 V

P = VI
4x10^6 = 20000 x I
I = 200 A [C]

wont we take 80% of the power?