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Physics: Post your doubts here!

ZaqZainab

ZaqZainab

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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
 
ZaqZainab

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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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Ahmed H. Al-Neel said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_1.pdf
question 25 please!
Click to expand...
25)
Use dsinθ = nλ
So given, θ = 45 and n = 3
so d x sin45 = 3λ ---> equation 1
We need to find highest order yes so check at which values of θ the sinθ has its maximum value, that is sin90 = 1
so we get other equation :
d x sin90 = nλ ---> equation 2
Now divide Equation 1 by 2 so that d and λ gets cut and you get n
so sin45 = 0.7 and sin90 = 1 ; (d x 0.7 = 3λ) / (d = nλ) this will give :
0.7 = 3/n
so n = 3 / 0.7 ≈ 4
so you get your answer as 4th :)
 
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