Physics multiple choice help

[tanmaydube]

http://xtremepapers.net/CIE/index.php?d ... 2_qp_1.pdf
Q3

3 Which formula could be correct for the speed v of ocean waves in terms of the density ρ of sea- water, the acceleration of free fall g, the depth h of the ocean and the wavelength λ?

Please hep it is urgent



THANK YOU!

 

[destined007]

Here:

 

[tanmaydube]

http://xtremepapers.net/CIE/Internation ... _qp_11.pdf
Q6

6 A student finds the density of a liquid by measuring its mass and its volume. The following is a
summary of his measurements.
mass of empty beaker = (20 ± 1) g
mass of beaker + liquid = (70 ± 1) g volume of liquid = (10.0 ± 0.6) cm3
He correctly calculates the density of the liquid as 5.0 g cm–3.
What is the uncertainty in this value?
A 0.3gcm–3 B 0.5gcm–3 C 0.6gcm–3 D 2.6gcm–3


THANKS A LOT!

 

[destined007]

here:

 

[whatsmyname]

damn...cant beleive how fast ive forgotten how to do these since m/j lol

 

[destined007]

happens.

 

[rz123]

The change ∆V in the volume V of some water when the pressure on the water increases
by ∆p is given by the expression
∆p = 2.2 × 10
9
∆V
V
,
where ∆p is measured in pascal.
In many applications, water is assumed to be incompressible.
By reference to the expression, justify this assumption.

plz explain the answer.

 

[rz123]

http://xtremepapers.net/CIE/Internation ... 8_qp_2.pdf
Q.6 b part. plz some1 make me understand asnwers...

 

[hassam]

dude post question properly....formula is not clear.
i rememba' though from my memory that we did an operation using the formula which gave a very small value like 0.00000647...(e.g) so since dv(change in volume) was so small...we assumed it to be incompressible....
it would be better if u cud post the formula in a more legible way......

 

[destined007]

http://xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s08_qp_2.pdf
Q.6 b part. plz some1 make me understand asnwers...

 

[rz123]

dude post question properly....formula is not clear.
i rememba' though from my memory that we did an operation using the formula which gave a very small value like 0.00000647...(e.g) so since dv(change in volume) was so small...we assumed it to be incompressible....
it would be better if u cud post the formula in a more legible way......

sorry bro, it's Q.4 b part3

http://xtremepapers.net/CIE/Internation ... 8_qp_2.pdf

 

[destined007]

by referring to the expression (question tells you to refer to the expression):
∆p = 2.2 × 10^9∆V/V
∆V=∆pV/(2.2 x 10^9)

because 2.2 x 10^9 is in denominator, ∆V will become very small.

 

[rz123]

http://xtremepapers.net/CIE/Internation ... 8_qp_1.pdf

Q.8 how is the answer c , why not A as the acceleration of free fall is constant and the gradient shows velocity which will be increasing as football falls...kindly clear me

 

[destined007]

See this is a trick question.
It is mentioned in the question that ball falls through "air". The air word is very important. This means that ball faces air resistance. When ball falls through air resistance its acceleration decreseses with time until it reaches zero. Therefore, C is correct.
Always read the question cearfully.

 

[rz123]

oh i got it , yeah u r right bout reading question carefully . thanks mate. bro i m posting some more and will appreciate ur help. no need to answer all at once, take ur time.
All from M/J 2008 P1 :Q. 14, Q.16 (how does force decreases by Fs), q.21 ,
Q.23 (which formula will be used here 1/2 kx ?),
q.31 (how to calculate this force ? ) ,
q. 34 ( i estimated the answer to be 16 by guessing the 4l wire to have 1/4 cross sectional area as compared to the other.)

 

[destined007]

Q14) answer is A
assume the above point where there's only force F acting (on the wall) as pivot. now apply sum of clockwise moment = sum of anti clockwise moment.
clock wise moment is F x h (h is the perpendicular distance from pivot to the bottom force F) + W x a (a is the perpendicular distance from pivot)
anticlockwise moment is W x 2a (2a is the perpendicular distance from pivot to the bottom force W)

Q16) answer is A
horizontal movement of X to bottom of Y and from bottom of Y vertically to Y is the same as moving from X to Y.
moving from X to bottom of Y, the work done= Fs
moving from bottom of Y to Y, the work done= zero (because direction of force and direction of motion is perpendicular)
Therefore, total work done is Fs.
becasue the charge is +ve and it is being move in the direction of electric field, there will be a decrease in work done as feild is doing work. If it were to move against the direction of electric field, the charge would have done work and work done must have increased.

Q21) answer is A
pressure=weight (force)/area
P=mg/A
using mass= density x volume
P=pVg/A
=pAhg/A (V=Ah)
=phg

Q23) answer is A
there is compression involved instead of expansion.
So potnetial energy= 1/2Fr
=1/2(6)(100-70)x10^-3
=0.09 J

Q31) answer is C
F=Eq where E=V/d
F=Vq/d
F=5000x(5x1.6x10^-19)/(0.8x10^-2)
F=5x10^-13 N

electric field in downward, so force will be upward because the charge is -ve.

Q34) answer is C
Volume is constant
V=Al (for 1st wire)

because the length has increased by 4 for the 2nd wire, the area must decrease by 4 to keep the volume same.
V=A/4 x 4l= Al

resistance of X/resistance of Y= (pA/l)/((pA/4)/4l)
= 1/(1/16)
=16

 

[rz123]

^Thanks a lot bro .
A diffraction grating has N lines per unit length and is placed at 90° to monochromatic light of
wavelength λ.
What is the expression for θ, the angle to the normal to the grating at which the third order
diffraction peak is observed?
http://xtremepapers.net/CIE/Internation ... 8_qp_1.pdf

Q.28, 36, 37 and 38. plz answer these too with explanation like u did above..i am facing problem in comprehending diffraction gratting and potentiometer so if u have some notes or link then kindly share.

 

[destined007]

Q28) Answer is B
formula is dsinx=nλ where d is grating constant and is equals to 1/N (N is line per unit length)
1/N sinx=3λ
sinx=3Nλ

Q36) answer is D
pd= work done in moving unit charge
when the temperature of thermistor is reduced, its resistance increases.
Therefore, more work has to be done to overcome the resistance so pd increases.

Q37) answer is A
combined resistance of parallel circuit: 1/R=1/100 + 1/100
R= 50 kOhm
Total resistance of circuit: 100 + 50 = 150 kOhm
current in circuit: V/R = 6/(150 x 10^3)
= 4 x 10^-5 = 40 x 10^-6 = 40 µA
current equally divides into 2 (because the 2 resistances are the same) when it passes the parallel circuit, so current in the two 100 KOhm will be 20 µA each.

Q38) answer is B. This is the standard diagram. remember it.

 

[rz123]

dude if u have the latest Cambridge international physics book by David Sang etc... then kindly answer one question.
Ch 2 Test urself. q. 15 E part which says check ur answer to part d by calculating the car's displacement using s=ut+1/2 at^2 ,
question is why did he used U value to be 8 in order to calculate s , rather then 20 m/s as this is the initial velocity given in the graph not 8 :|

 

[destined007]

i don't have that book. If you post that question here ill look into it.