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Physics: Post your doubts here!

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sagar65265

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DeViL gURl B) said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_1.pdf

COMPLETE EXPLANATION NEEDED FOR QUESTION 14 pleaseplease!!
Click to expand...

The concept here is pretty straightforward - at any point in an electric field, the force on a charged particle acts along the tangent to the electric field line at that point.

In other words, the force on a particle placed on a curved field line will act along the direction of the tangent of the field line at that point.

Also, the arrows on electric field lines (in diagrams only) tell you the direction of the force a positively charged particle would experience in the field - a negatively charged particle will experience a force along the opposite direction.
So the positive particle faces a force along the arrow, the negatively charged particle experiences a force in the direction opposite the arrow.

Electric Field.JPG

On the diagram above, the blue arrows represent forces on a negative charge (D,E,F) and the red arrows represent force on a positive charge(A,B,C). You'll note that both the forces at any point act along the tangent to that point, just in opposite directions (NOTE: The arrows here are not drawn to scale, so do not concern yourself with the length of the arrows and the magnitude of the field - here, just focus on the directions relative to the field lines).

The tangent represents the direction of the electric field at that point, so the electric field direction can be said to be in the direction of a force on a positive particle at that point in the field, with the force on a negative particle in the opposite direction.

So, in this question, the electric field arrows points to the right, and the lines are all straight and evenly spaced. Therefore, this is a uniform electric field, with the forces on a proton placed anywhere in this field acting to the right.

Therefore, the force on an electron (as in the question) should be in the opposite direction, i.e. to the left. The correct answer is, therefore, B.

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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meerul264 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf

Q33 please. Thanks
Click to expand...

Suppose the resistance of one of the resistors is R. Then the combined resistance of Q and R is given as follows:

1/R(Q + R) = 1/R(Q) + 1/R(R) = 1/R + 1/R = 2/R
Therefore, R(Q + R) = R/2.

This resistance is in series with R(P), and since they are in series we can add them straight away:

R(P + Q + R) = R + R/2 = 3R/2.

So the total resistance of the circuit is equal to 3R/2. Suppose the voltage is V, then we can see that the current = I = V/R:

I(circuit) = V/(3R/2) = 2V/3R Amperes

From this, since we know that the total power delivered is 12 Watts and the power given is equal to P = IV, we can write

P = 12 Watts = 2V/3R * V = 2V²/(3R)
So, we can write that

V²/R = 3/2 * 12 = 18
So V²/R = 18. Let's move on.

Since we know that the resistance of network Q + R = R(Q + R) = R/2, we can say that the potential difference across Q + R section is equal to V(Q + R) =
I * R(Q + R) =

V(Q + R) = 2V/3R * R/2 = V/3.

So the potential difference across the resistor called R is equal to V/3. Therefore, since P(resistor) = V(across resistor)²/R, we can write

P(R) = (V/3)²/R = V²/3²R = V²/(9R)

Since we know that V²/R = 18, we can write

P(R) = (18)/9 = 2 Watts = A.

Hope this helped!
Good Luck for all your exams!
 
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ShreeyaBeatz said:
Which property of a metal wire depends on its Young modulus?
A ductility
B elastic limit
C spring constant
D ultimate tensile stress
Click to expand...

The main point to understand is this - the Young's Modulus is basically a calculated value that we can find out from measurements and use later. It is akin to density - there's very few things that depend on density, but density depends on many other factors and density is also a constant for a material that we regularly use in calculations.

The ductility of a material should have nothing to do with the Young's Modulus - it reflects in the Young Modulus, but is definitely not dependent on it. The ductility of a material is difficult to quantitatively measure and is not calculated using Young's Modulus. The Young's Modulus value just tells us how well the object will stretch, not whether it will do so efficiently without wasting too much energy or whether it will spring back. A wire that springs back is not very useful as far as I can imagine (though who knows how wrong I could be!).

The elastic limit of a material cannot be calculated with the Young's modulus - think about it! The Young's Modulus simply tells us that if you have a sample of this cross section and that length, and you apply so-and-so force to it, it will have such-and-such extension - what it tells us is that with a large enough force, a large enough length, and a small enough cross sectional area, we can cause a near-infinite extension of any sample! This is obviously not true, since the Young Modulus formula implies a linear stress-strain relationship, which does not predict anything that happens after the limit of proportionality - the elastic limit, the plastic stage, the ultimate tensile strength or the fracture point. We only assume that the Young's Modulus holds in our problems, otherwise the equations to describe plastic deformation would be a very tricky set to solve, indeed!

All this only leaves us with C - the spring constant is relevant to the stretchiness of the material, and is a constant that is very close to the Young Modulus. Both make the assumption that the limit of proportionality is not reached, and neither are the plastic stage or the fracture point. This makes sense, since wires that stretch easily will make sensitive springs (but probably not always the best of wires).

It is a rather intuitive question, but I hope this at least contributed to your thoughts on the topic.

Good Luck for all your exams!
 
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NIM

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Thought blocker said:
8)
okay, the equation we consider here is → s = 1/2 * g * t^2
When time = T, the dist = s.
However when the time = 0.5 T →time^2 = (0.5 T)^2 →0.25 T
So, the dist is proportional to the square of time.
Thus, when the time reduces to half its original value, dist reduces to 1/4th its original value.
So, the dist covered in time = 0.5T = 0.25L.
Ans = B

12)
we'll take this problem in terms of the forces acting on the barrel and the man
for the barrel: the force due to the barrel - the tension in the rope(T) = ma (because the barrel move's down, the force due to the barrel is greater than the tension)
(120*10) - T = 120a (we dont know the acceleration of the man and the barrel)
for the man: the tension (T) - the force due to the man = ma
T - (80*a) = 80a
find the value of a using simultaneous equations, which is 2 ms^-2
using v^2=u^2 + 2as
find the final velocity of the man at 9 m, taking u as 0 ms^-1

17)
F=α*V2
so
α=F÷(v2)
α=800÷20=2kg.m−1
P=W÷t=(F*d)÷t=F*v
F=(2kg.m−1)×(40m.s−1)2=3200N
and finally
P=40m.s−1*3200=128000Watts=128KW

18)
Efficiency is given by (Useful Output Work)/(Total Input Energy) =
(Useful Output Work per Second)/(Total Input Energy per Second)
The useful output work per second = useful power output = 150 * 10^3 Joules Per Second.
Therefore, Efficiency = (150 * 10^3 Joules/Second)/(Total Input Energy per second)
Every hour, 20 liters of fuel are consumed. Therefore, every second, 20/(60 * 60) liters of fuel are consumed.
The energy in these liters is equal to [20/(60 * 60)] * 40 * 10^6 Joules.
Therefore, the efficiency is equal to
(150 * 10^3)/[20/(60 * 60)] * 40 * 10^6 = (150 * 10^3 * 60 * 60)/(20 * 40 * 10^6) = B.

24)
http://www.s-cool.co.uk/a-level/phy...finitions-of-stress-strain-and-youngs-modulus
We know that E is inversely proportional to the area, and directly proportional to the length.
Now, when we increase the length to 3L, the extension also increases 3 times → 3x.
But becoz we increase the diameter to 2D, we have increased the area by 4 times. A = π(d/2)^2 → A is now 4A.
And since extension is INVERSELY proportional to area, the extension now falls → x/4

So, the new wire's extension = 3x/4
Ans = B

34)
img-20121122-00179-jpg.18658


35)
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...
thnx a lot dude!!
i just realized i got the 12th question rite but for the height i used 18m... why did u use 9m as height :D
n yeah can u explain me question 17 again ... i couldnt understand which formula did u use :p
 
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crazytaylorfanXD

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Can someone help me in a few questions from:
JUN 12 P12 Qs ( 6 , 8 , 10 , 14 , 15 , 24 , 26 , 29 , 35 , 36) I know i said few and these are a lot :p
NOV 12 P12 Qs ( 5 , 19 , 27 , 30 , 36 , 38)
i think june 12 p12 is the hardest paper ive attempted , i got 27 out of 40 :/
 
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crazytaylorfanXD said:
Can someone help me in a few questions from:
JUN 12 P12 Qs ( 6 , 8 , 10 , 14 , 15 , 24 , 26 , 29 , 35 , 36) I know i said few and these are a lot :p
NOV 12 P12 Qs ( 5 , 19 , 27 , 30 , 36 , 38)
i think june 12 p12 is the hardest paper ive attempted , i got 27 out of 40 :/
Click to expand...
s12_12
6)
1µs --> 1 cm
so, each pulses are in 2 cm
so 2cm ---> 2µs

8)
Use S = ut + 0.5 at^2
for distance X :
S = 0t1 + 0.5 at1^2
Here S = x
so x = 0.5 x a x t1^2 ---> equation 1

Now distance x + h
so
s = 0t2 + 0.5 at2^2
Here S = x + h
so x = 0.5 x a x t2^2 - h --> equation 2

So As x = x : equation 1 = equation 2 :
1/2 x a x t1^2 = 1/2 x a x t2^2 - h
a x t1^2 = 2( 0.5 x a x t2^2 - h)
a x t1^2 = a x t2^2 - 2h
2h = a x t2^2 - a x t1^2
2h = a ( t2^2 - t1^2 )
a = 2h / ( t2^2 - t1^2)

10)
Keep this thing in mind when you are asked for projectile thingy :
  • horizontal component of velocity = constant
  • vertical component of acceleration constant = constant
  • At top most, vertical velocity is zero
Now look it practically,
you throw a ball, when it reaches max height (top) it stops for a while
KE cannot be zero
KE = 0.5mv^2
v = root(square(vx) + square(vy))
vy = 0
but vx = constant
so v is nonzeror and hence KE canot be zero
similarly momentum canot be zero
P=mv

14)
the ladder is not stable, so the force P will not be exactly horizontal, but the other two forces will remain vertical

15)
since tourqe is equal to force*distance,
the distance moved by the spindle will be the torque,
i.e L/4 means, WL/4 and will always be clockwise in this condition
ps. this was a stupid question just remember the fact that the distance moved by
the spindle will be multiplied by the weight of the cube W.

24)
967791_804846696201706_1397227511_n.jpg

Take area of triangle and rectangle then add them :)
Simple as that :p

26)
As t = d / v = 150 * 3 / 3 * 10 ^8 = D

29)
We know c = f *λ
so f = 1/t
t = λ / c
so as we can see from figure it took 3 wavefronts to reach PX to Y
so t = 3λ / c

35)
Look at the loop in which battery is attached.
Apply Krichoff's Law
Voltage is not divided in parallel, only in series.
EMF = V
3 equations could be possible :¬
V = V1 + V2
V = V1 + V3
V2 = V3

So look! D satisfies our equation 2. :)

36)
R decreases as more light intensity fall on LDR
So obviously If LED's R would decrease the Resistance R of other resistor would increase
R = V / I so p.d across LDR decrases as R is proportional to p.d(v)
 
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crazytaylorfanXD said:
Can someone help me in a few questions from:
JUN 12 P12 Qs ( 6 , 8 , 10 , 14 , 15 , 24 , 26 , 29 , 35 , 36) I know i said few and these are a lot :p
NOV 12 P12 Qs ( 5 , 19 , 27 , 30 , 36 , 38)
i think june 12 p12 is the hardest paper ive attempted , i got 27 out of 40 :/
Click to expand...
w12_12
5)
Simple concept, use your brain :)

19)
Use Pressure = force / area and find force ( dont forget to keep basic units of area )
Then w.d = f x d = A ( distance in meter)

27)
untitled-png.44799


30)
Apply d*sin thetha = n*lamida
d = 1/n where n is no. of line/m. Metre not millimetre.
Grating's got 300lines/mm. So this means it has 300*1000 lines/m
(Coz 1m = 1000mm)
d = 1/n ie 1/300,000
U wanna see all the maximas possible, so to get max. no. of maximas on one half of the screen, u put thetha = 90
lamida is given. Convert it into m from nm.

U get n = 4
This means 4 maximas on one half of the screen.
On whole (both halves) of screen, maximas are = 4*2 = 8

BUT, u also have to consider the central maxima.
So total maximas are 8+1 = 9

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

38)
remember when two resistor are in series, more voltage drop is at resistor with more value, so their a flashing light then R = 5 mega ohm when dark, then 5M is very large than 1 K of other resistor, then almost all voltage drop at 5M, hope you get this, it is a very simple concept.
 
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