Physics Paper 1

[Lemon]

Hey Guys,
I searched all through this forum thinking maybe someone must have already asked these questions that i am stuck on , well,,. .. but to no avail. No one has asked these , meaning you all must have already solved it. So could you please kindly help me with it :| Not much, just 10 of them. won't take long...Please help
Here it is:

2008 may/jun - Q 14
2008 oct/nov - Q11, 32
2007 oct/nov - Q9

2006 may/jun - Q15,22, 24(in 24 the answer should have been A since twice the intensity so twice the amplitude, ain't it?)

2006 oct/nov- Q 15, 16, 21

 

[Babuzar]

Hey Guys,
I searched all through this forum thinking maybe someone must have already asked these questions that i am stuck on , well,,. .. but to no avail. No one has asked these , meaning you all must have already solved it. So could you please kindly help me with it :| Not much, just 10 of them. won't take long...Please help
Here it is:

2008 may/jun - Q 14
2008 oct/nov - Q11, 32
2007 oct/nov - Q9

2006 may/jun - Q15,22, 24(in 24 the answer should have been A since twice the intensity so twice the amplitude, ain't it?)

2006 oct/nov- Q 15, 16, 21

Im in a hurry so cant answer the questions now. But i also had the same problem with 2008 Oct/Nov Q32, it was strange, you feel as if its C, and feel pretty sure, dodnt really know how its B. And btw its I proportional to the square of A.

 

[intel1993]

in questn 32 all of the copper wireas and steel core r in parralell....

6(1/10)+ 1/100 = 1/r

solve for R.........
u will get the ans 2 be B(1.6)

 

[Babuzar]

OH! parallel. Got it thanks!

 

[Anomaly]

Hey Guys,
I searched all through this forum thinking maybe someone must have already asked these questions that i am stuck on , well,,. .. but to no avail. No one has asked these , meaning you all must have already solved it. So could you please kindly help me with it :| Not much, just 10 of them. won't take long...Please help
Here it is:

2008 may/jun - Q 14
2008 oct/nov - Q11, 32
2007 oct/nov - Q9

2006 may/jun - Q15,22, 24(in 24 the answer should have been A since twice the intensity so twice the amplitude, ain't it?)

2006 oct/nov- Q 15, 16, 21

about may june 06 Q 24, the amplitude doesnt double as the intensity doubles. Say for example the initial waveform had and Intensity I with A =5 thus I=k(5)^2 (since Intensity depends on the square of the amplitude- hence the constant k comes out to be I/25. Now in the second wave the 2I=kA^2. Put in the value of k found previously and you'll get A=7.7 not 10

 

[ds29]

M/J 06 Q9

This is a typical graph of Simple Harmonic Motion.
The Answer is D

This is because consider an object at the equilibrium position. It has displace zero and MAXIMUM velocity. Let's say it goes upwards then once it reaches the highest point it's velocity will be 0 and it's acceleration will be down. So it's velocity has changed from going upwards to going downwards later on.

Take the same situation at the bottom it has passed the Equilibrium Position with maximum downward Velocity (Point C) and then comes to the lowest position with 0 velocity. Point D.
Therefore Point D is the answer.

 

[ds29]

M/J 06 Q15

Consider moments anticlockwise at Pivot = 200 * 0.8 = 160 Nm
Consider moments clockwise at Pivot = 300 * (1.2 - 0.8) = 120 Nm
Therefore we need 40Nm in clockwise direction.

Therefore answer is A

 

[ds29]

MJ 06 Q22

The Work Done is the area under the graph. We are treating the curved part as a straight line as said by the Question,
So area under graph Uptil X = 0.5 * 10e-3 * 500 = 2.5 J
Area under graph from X - Y = 1.05 J
Therefore total work Done = 3.55J

Answer is B

 

[ds29]

MJ 06 Q24

Intensity directly proportional to Amplitude^2
Therefore when Intensity doubles Amplitude increases by a multiple of square root 2.
Either B or D.
Half the frequency means it's B.

Answer is B

 

[ds29]

ON 06 Q15

Resolve Vertically upwards = 10Cos(30) + 10Cos(30) = 17.3 N
Resolving Horizontally = 0 (10Sin(30) - 10Sin(30))
Therefore answer is C

 

[ds29]

ON 06 Q16
Apply MGH
80 * 10 * (2.5 * 6) because each flight is 2.5m and there are 6 flights
= 12,000 J

Answer is D

 

[ds29]

ON 06 Q21

Ok P = h$g where P = Pressure h = height $ = density and g = gravitational

We know total Pressure is 17.5 * 10^6 Pa

We first consider Pressure by Oil
P = x * 830 * 9.81
P = 8142.3x

Next we consider Pressure by water
Here we have to consider height as (2000-x) since we are given total length is 2000m

P = (2000-x) * 1000 * 9.81
= 1.962 * 10^7 - 9810x

Now both of these pressures = Total Pressure
Therefore
8142.3X - 9810X = 17.5*10^6 - 1.962 * 10^7
-1667.7X = -2.12* 10^6
X = 1271m
Approximate equal to 1270 and therefore answer is D

 

[ds29]

ON 07 Q9
They are asking for Vertical displacement after 5 seconds.

Consider the first 3 seconds. Displacement is Area under Graph = 0.5 * 3 * 30 = 45 m
Consider the next 2 seconds. It's NEGATIVE displacement. Area under graph = 0.5 * 2 * 20 = 20m

Therefore final displacement = 45 m - 20 m = 25 m
Answer is B

 

[ds29]

ON 08 Q11
We need to form 2 simultaneous equations.
First we consider the 2k object.
Apply F = ma
T = Tension
20 - T = 2a ---- (1)

Then we consider the 8kg object.
T - 6 = 8a ----- (2)

Solving the simlutaneous equations gives us a as 1.4
Answer is A

 

[ds29]

ON 08 Q 32

Everything is connect in parallel.
There are 6 copper wires each with 10 ohm resistance.
If you consider that in parallel you get 10 / 6 = 1.67 Ohms
Steel Core = 100 Ohms
Both Parallel as well. Therefore ( 100 inverse + 1.67 inverse ) whole thing inverse
You get 1.63 Ohms which is B
Answer is B

 

[1992]

ON 08 Q 32

Everything is connect in parallel.
There are 6 copper wires each with 10 ohm resistance.
If you consider that in parallel you get 10 / 6 = 1.67 Ohms
Steel Core = 100 Ohms
Both Parallel as well. Therefore ( 100 inverse + 1.67 inverse ) whole thing inverse
You get 1.63 Ohms which is B
Answer is B

how do u know its in parelles, i mean, i calculate it using as if they were series so i got another asn, and its wrong
but in the quuestion in doesnt says so :S

 

[ds29]

The diagram shows it as parallel.

If it was series then the resistance should be end on end. Like cables connected end of end. Therefore instead 6 1k cables we'll be having 1 6km cable.