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As physics p1 MCQS YEARLY ONLY.

Thought blocker

Thought blocker

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Zepudee said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_12.pdf

Q8?
Click to expand...
Short answer, because the time taken with each speed is different.

Let's do the math behind this - suppose the distance traveled with both speeds is denoted by "s".
Suppose the time taken at 600 kmph = t(1).

Then, since Speed = Distance/Time, we can write

600 = s/t(1)
So that
t(1) = s/600

Similarly, let's do the same for the other speed. The distance traveled is still "s", the speed = 400 kmph, and the time taken to traverse this distance = t(2). So:

400 = s/t(2)
So that
t(2) = s/400

The average speed is the total distance traveled dividing by the time taken (the average velocity is different - it is equal to total displacement divided by time taken).
= (Total Distance traveled)/(Time taken to cover that distance).

Since the plane travels a distance "s" in one direction and returns the same distance "s", the total distance traveled = s+s = 2s.

The time taken = t(1) + t(2) = s/400 + s/600 = 3s/1200 + 2s/1200 = 5s/1200 = s/240

Therefore, the average speed = 2s/(s/240) = 480 kmph = C.
 
Zepudee

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Thought blocker said:
Short answer, because the time taken with each speed is different.

Let's do the math behind this - suppose the distance traveled with both speeds is denoted by "s".
Suppose the time taken at 600 kmph = t(1).

Then, since Speed = Distance/Time, we can write

600 = s/t(1)
So that
t(1) = s/600

Similarly, let's do the same for the other speed. The distance traveled is still "s", the speed = 400 kmph, and the time taken to traverse this distance = t(2). So:

400 = s/t(2)
So that
t(2) = s/400

The average speed is the total distance traveled dividing by the time taken (the average velocity is different - it is equal to total displacement divided by time taken).
= (Total Distance traveled)/(Time taken to cover that distance).

Since the plane travels a distance "s" in one direction and returns the same distance "s", the total distance traveled = s+s = 2s.

The time taken = t(1) + t(2) = s/400 + s/600 = 3s/1200 + 2s/1200 = 5s/1200 = s/240

Therefore, the average speed = 2s/(s/240) = 480 kmph = C.
Click to expand...

Aw thank you so much!!
 
Mohammed Siddiq

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Thought blocker said:
Q35)

When the switch S is closed, a current is allowed to flow through the circuit. When the current flows through the circuit, a potential difference is produced across the resistor R. This means that energy is lost in the resistor.

Also, when current goes through the internal resistance of the battery, energy is lost in that internal resistance. Therefore, some of the battery's energy and potential difference is lost in the battery itself.

However, we have to remember the definition of e.m.f - it is the total work done by the battery in driving one coulomb of current around the complete circuit. This work includes the work done by the battery in pushing the current through the internal resistance also, and it is equal to the voltage across the battery when the circuit is not closed (if you put a voltmeter across the battery when the switch S is open, the voltage measured is the rating of the battery AND the e.m.f. value).

Suppose you take internal resistance and separate it from the cells in the battery, then the energy lost in the internal resistance + the energy lost in resistor R is still equal to the e.m.f.

Remember, e.m.f. is the work done by the cells in the battery to push one coulomb through the complete circuit. They will do the same work whether there is internal resistance or not, so the e.m.f. does not change in the circuit because the cells can be assumed to be separate from the internal resistance.

So, A and B are eliminated; e.m.f. does not change at all.

But because a battery = cells + internal resistance, when S is closed, current flows through the circuit and energy is lost in the internal resistance. Also, there is a loss in potential across the internal resistance, so there is a loss in potential across the battery. Therefore, the potential across the battery changes because energy is lost in the internal resistance = C.

Note: if energy is lost in any resistance, it is because work is done against that resistance (suppose you are pushing a box on the ground, friction is the resistance, and you need to do some work against that friction resistance. Your muscles are the potential pushing the current, and some of that energy is lost when you work against friction).
Click to expand...
I'm extremely grateful to you for giving your time in answering the questions, thank you.
 
Thought blocker

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Zepudee said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_12.pdf

Q10,35,37? :D

Thought blocker
Click to expand...
I want you to check out my signature.. :) Its really helpful
If you don't know many concepts than visit Khan academy :)
So here is your solutions, sorry for late as I was on date with chemistry ;)


10)
Does weight change if a force is exerted on it?
No - Basic thing :) Answer would not change :)


35)
Whats wrong, we just need to compare :¬

For original wire :
Area = Π * r ^ 2 right ? ( I assumed d/2 as r )
length = l
resisitivity = ρ
Resistance = R

For replacement wire :
Area = Π * 4r ^ 2
length = l
resistivity = 2ρ

Now compare original with replacement wire :
Original :
ρl/Π * r ^ 2 --> R

so for replacement wire :
2ρl/ Π * 4r ^ 2 = R/2


37) This was also an easy one ;)
  1. After voltage passes through some resistance it will decrease
  2. The higher the resistance the greater the decrease in voltage
-->So it will decrease less after passing through 2 ohm than when passing through 4 ohm
Hence it is D
Zepudee said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_12.pdf

Q10,35,37? :D

Thought blocker
Click to expand...
ote="Zepudee, post: 814292, member: 97182"]http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_12.pdf

Q8,36? :D ANYONEEEE[/quote]
8)
I hope that your are familiar with Equations for linear motion. ( Do not confuse with Newton's laws of motion). They are:
1) v = u + a*t
2) s = u*t + 1/2 *a*t^2
3) 2*a*s = v^2 - u^2
Here v stands for final velocity, u stands for initial velocity, s means displacement, t stands for time and a means acceleration.
We are given that:
t = 0.5 seconds
a= 9.8 meters per second square ( As the experiment is taking place on earth, so we consider acceleration due to gravity)
u = 0 ( as the feather is dropped from rest )
Now we apply second equation of motion.
s = u*t + 1/2 *a*t^2
s = 0*0.5 + 1/2*(9.8)*0.5*0.5
s = 1.225 m, which is approximately 1.3.

36) This is also a concept, we dont have fix values of currents, it varies, so we took a mean of power :) :¬

Find power at -1 and then at 2
For -1 :
P = 100W
For 2 :
P = 400W

Now as current is varying we have to take mean value of power : i.e 400 +100 = 500 / 2 = 250W

Shamajh aya :p ?
Just be calm when you solve physics, they have so many easy twists :)
 
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Thought blocker

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Kamran Nagi said:
9702_w10_qp_12.pdf
Q 8 and 9 thank you :)
Click to expand...
8)
lets see the motion in XY
initial velocity at x=v1
final velocity at y=v2
average velocity=distance/time
(V1+V2)/2=40/12 equation 1
simplifying it we get V1+V2=6.67

now consider motion in XZ
initial velocity at x=v1
final velocity=v3
V1+V2/2=(40+40)/(12+6)
simplifying it we get V1+V3=8.89 equation 2
subtract equation 1 from 2
i mean equation 2-equation 1
V1 is cancelled and we get V3-V2=2.22
since acceleration is constant so
acceleration in yz = acceleration in xz
= change in velocity/time
=2.22/6
=0.37

9)
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 will 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...
 
Kamran Nagi

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Thought blocker said:
8)
lets see the motion in XY
initial velocity at x=v1
final velocity at y=v2
average velocity=distance/time
(V1+V2)/2=40/12 equation 1
simplifying it we get V1+V2=6.67

now consider motion in XZ
initial velocity at x=v1
final velocity=v3
V1+V2/2=(40+40)/(12+6)
simplifying it we get V1+V3=8.89 equation 2
subtract equation 1 from 2
i mean equation 2-equation 1
V1 is cancelled and we get V3-V2=2.22
since acceleration is constant so
acceleration in yz = acceleration in xz
= change in velocity/time
=2.22/6
=0.37

9)
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 will 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...
Click to expand...
Thank you so much. I did get Q 9. But Q8 main I have one issue, for the second equation why didn't we use the the distance YZ instead of XZ, if we do, kiya uss se answer same aaye ga?
 
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