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As physics p1 MCQS YEARLY ONLY.

haha101

haha101

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can someone tell mei which variants from 11 12 13 are similar and which are different ?
 
h4rriet

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papersxtreme said:
june 2012/11 mcq 12

june 2012/12 mcq 13
Help. please
Click to expand...
12. Assume the man and the barrel meet halfway. Calculate the potential energy lost by the barrel and the potential energy gained by the man. Minus this from the initial energy of the whole system (which is the potential energy of the barrel). What you'll get is the k.e. Use 1/2mv^2, and m=mass of man+mass of barrel.
13. Draw a free-body diagram for the box. The driving force will be the weight of the 2 kg mass, and the frictional force will be 6. Find the resultant force, and find a using a=F/m.
 
A*(a*)

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h4rriet said:
Yeah actually you're right, it'll be 3/4.
Click to expand...

but these values do NOT support any graph, could u tell me if any graph is supported?
My sir said, he thinks there seems to be a lil problem with the question
 
h4rriet

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A*(a*) said:
but these values do NOT support any graph, could u tell me if any graph is supported?
My sir said, he thinks there seems to be a lil problem with the question
Click to expand...

Well if 0.5:0.75:1 can't be simplified into whole numbers, then there must be something wrong with the question.
 
R

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leosco1995 said:
June 2005
=========

Q1. C.

Fact, basic stuff.

Q2. C

Rearrange the equation to make k the subject, so k = F/rv.

= (kgms^-1)/(m * ms^-1)
= kgm^-1s^-1

Q3.B

A reasonable estimate for an athelete running a 100m race is approximately 10 seconds.

K.E = 1/2 * 80 * 10^2
= 4000 J

Q4. C

Principle speed value = 16
Uncertainty = (0.1/40) + (0.05/2.50) * 16 = 0.36

We always round off the uncertainty value to one s.f., so it becomes 0.4

So the answer is 16 +/ 0.4

Q5. A

The length of the pulse is how long you see the change in the Y-axis. It's for 2cm, meaning 2 μs.

Q6. D

Easy stuff. To find acceleration you take the gradient of a velocity-time graph.

Q7. B

The acceleration during the motion of a falling ball will always be constant, i.e. 9.81 ms^-2. Since they told us to take upwards as positive and the gravitational force acts downwards, this is actually -9.81 ms^-2.

Q8. D

Distance is area of the graph. Class 4 stuff. :p

Q9. A

Acceleration doesn't act in the horizontal direction so it's value is therefore 0.

B is wrong because the object has velocity throughout the motion.
C is wrong because of B, the resultant velocity will be non-zero because of the horizontal velocity being non-zero.
D is just nonsense.

Q10. A

B, C and D are Newton's 2nd, 3rd and 1st law respectively.

Q11. A

Momentum is always conserved so we use that formula. Also when the objects stick on impact, the total mass will be the sum of the individual masses. Let 'm' be the mass of one of the objects.

60m + (40 * -m) = 2mx (where x is the speed of the masses after impact).
20m = 2mx
x = 10 because m cancel out.

Q12. C

Fact. D is wrong because it gravity is the point through which gravity APPEARS to act.

Q13. A

The forces are shown in this picture: http://www.xtremepapers.com/community/attachments/moe-png.12069/ (Thanks a ton to Unicorn for this).

(5*2) + (2*10) - (3*20) = 30 Nm anti-clockwise.

Q14. D

Resolve the horizontal 4N and vertical 3N component to get a 5N component parallel to the diagonal 4N component. Since the 5N force would be greater, the resultant force would be 1N in its direction.

Q15.B

K.E will be constant because the velocity and mass are constant (velocity beacause it says in the question).

P.E will start from a high value and decrease uniformly because the height is decreasing uniformly.

Q16. C

The gradient of an energy/time graph is power since P = E/t. So we are looking for the point where the gradient is the steepest. This is from 2s-3s, so the gradient there is (40-10)/1 = 30W.

Q17. B

P.E = mgh.

They have given the density and volume from which we can calculate the mass. g is 9.81 and h is 3.0m.

Q18. A

Fact.

Q19. C

Brownian motion, the molecules of liquid collide with the molecules of the pollen grains.

Q20. A

Let a regular extension be 'x'. In parallel, the extension is divided by the # of springs and the opposite for a 'series' extension.

Extension in X is e/2.
Extension in Y is e/2 + e/2 = e.
Extension in Z is e/2 + e = 1.5e.

The order is X -> Y -> Z.

Q21. D

You need to know these graphs. Brittle (glass) is just a straight steep line, rubber is like that of graph X (note that they don't obey Hooke's law) and Y is that of steel, a ductile material.

Q22. D

Let their Young Modulus be equal to 'E'. They have to have the same YM (same material), so..

E = FL/Ax (where x is extension).
F = EAx/L (E and x don't matter here because they're constant).

For P, F = A/l
For Q, F = 0.5A/2l

Ratio is 4:1.

Q23. A

Fact, all transverse waves travel at the same speed in a vaccuum.

Q24. B

You need to know a reasonable estimate of the wavelength of visible light, e.g. 500 nm.

# of wavelengths in ONE metre is 1/(500 nm) 2.0 * 10^6. This is in the order of 10^6, so B is right.

Q25. B

Use ratio of intensity and amplitude.

(I1/I2) = (a1/a2)^2

1/2 = (A/x)^2

x = √2A

Q26. B

Fact, sort of.

Q27. D

Distance between 2 maxima = 0.5λ.
So 1λ = 30 mm.

F = v/λ
F = 3.0 * 10^8 / 30 mm
F = 2.0 * 10 Hz.

Q28. B

Formula is x = λr/q

According to this equation, decreasing 'q' will increase 'x'. A has nothing to do with 'x'.

Q29. B

For 2nd order, d sin θ = 600 nm * 2, which is

d sin θ = 1200 nm

For 3rd order,

d sin θ = 3λ

Since d has to be the same and the angle is also the same, we can equate the 2 equations.

3λ = 1200 nm
λ = 400 nm.

Q30. D

E = V/d.
= 900 / (4 mm)
= 2.3 * 10^3 N/C

Q31. C

Fact.

Q32. C

The area is irrelevant to this question, because Q = I * t (there is nothing to do with area in this formula).

Q = 10 * 1 = 10 C

Since one electron has a charge of 1.6 * 10^-19 C, 10 C has 6.3 * 10^19 electrons.

Q33. D

Originally, R = ρL/A

Now, the length is doubled BUT the volume is the same. This means the area has to be halved. Mathematically proving this:

Volume is length * breadth * height, so:

2lbh = v

Since Area = lb
A = 2lb
lb = 0.5A

Anyway, new resistance will be 2ρL/0.5 = 4R.

Q34. D

A is wrong because Q is a thermistor/semi-conductor/etc.
B is wrong because the resistance decreases.
C is wrong because the resistances are the same at 1.9 (same V:I) ratio.
D is right because using I^2 * R proves this is correct.

Q35. B

Fact.

Q36. D

In parallel, voltage is the same so V2 = V3.

And terminal voltage V = V1 + V3

Rearranging this gives

V - V1 = V3

Q37. At X the voltmeter is connected directly so it gets the full 4V. At Y we use the potential divider formula to find the voltage:

V = 4 * (10/20) = 2V

B is the only graph that shows this correctly.

Q38. C

Easy stuff.

Q39. B

The range of α particles is approximately 0 - 5cm. In this question, they've given us values in mm, so we can say the range is 0 - 500 mm.

B is the safest maximum range.

Q40. C

The nucleon number decreases by (4+4+0) so it becomes 209.
The proton number decreases by (2+2-1) so it becomes 82.
Click to expand...
But 27 is C in the mark scheme
 
talhanoor44

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applepie1996 said:
q9 for these type of calculation u r always suppose 2 use a straight so u make a straight line from 4m/s 0sec till 8m/s 3sec and find acceleration u get 1.33m/s^2

q 12 in da question it is given dat The angles at which the forces act can
vary.so they form a closed triangle then the resultant is zero.. and a triangle in eqilibrrium has 0 resultant force

q 28) for dis quest my teacher told me dat workdone in a circular path is 0

q31)voltage formula=p.d.=e.m.f so both use thsame formula energy/charge

q32)this is quiet simple the answer os D because 1/1+1/2+1/2=2 take the reciprocal u get 1/2ohms u find resistance of alll the circuits this way and imagine the voltage is 5V for all of dem cux the quest says when the same potential difference
is applied between points P and Q?

nov 10
q7)acceleration remains constant then when the ball bounces acceleration decreases and then wen it cums bac it is da same again

Q)8 u first find speed between points X and Y which is 3.33m/s and find speed between Y nd Z which 6.67m/s
then u find average of the time taken which is 12+6/2 which is 9 sec
then u find acceleration v-u/t so 6.67-3.33/9 = 0.37m/s^2

Q9)initial momentum of the ball is 2mv before collision... if collision was perfectly elastic one, all the momentum would have been gained by the ball... and this would have made the change in the momentum equal to 4mv.. 2mv-(-2mv)... now it is given that collision is inelastic, it wil lose some of its momentum to the wall.. so now change in momentum will be less than 4mv but cant be less than the initial momentum... so the answer is C.. i think you cannot verify the answer by the calculations

Q15) F=kv and F=mg are same
so kv=mg so v=mg/k
ke=1/2 m x v^2 substitute v and u get m^3 x g^2/2k^2

Q21)I have wrote this many times
Use the young Modulus formula. E= Fl/Ae
Equation 1 - Ee= F(tension)*l/A
Equation 2 - Ee= F(tension)*2l/0.5A = 4Fl/A

Since it's the same wires, young modulus is the same, (e) is the same. So equate both equations.
Fl/A=4Fl/A Do some algebraic manipulation and you'll get 4/1
Click to expand...






but qs 21 answer is A 1/2
 
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