Physics: Post your doubts here!

[Wolfgangs]

Is there any copy of this year Paper 21 yet? If so, please give me a link to it

 

[ratedr0014]

5,9,15,24,30,31,33,34 help pleasee

 

[swdj24]

hey can u help me? thanks very much
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf #7,9,20,23,29,32,33,36

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_13.pdf #15,16,19,29,31,34

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf Q11,12,18,19,21,31,32,35 PLEASEEE

 

[Asad Moosvi]

Is anyone worried about paper 1?

 

[MiniSacBall]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf

Q40, Q33, Q26,

Explain plz,

I need a little bit of help in double slit experiment:
Could any one give me all the details like, what will happen if we:-

Increase the distance between the slits,
Increase the length of slits, i mean the hole
Increase the distance between the screen and slit,
Use light with higher frequency,
And vice versa. And more if you can add <3.
Suchal Riaz Thought blocker
Thanks in Advance! <3

 

[tania]

15)
Answer should be zero
Hadi Murtaza
ZaqZainab

19)
you need to find the power input... power = work done / time
work done = F x D or rather F here as Weight =mg = to find mass you use the formula density = mass/volume= 1000 x 340 = 340000... now converting it to weight x 9.8 = 3332000 ... Work done = 3332000 x distance (30 ) = 99960000
for power it has to be per second and youre given in the question per minute so you divide by 60s = 99960000/60 = 1666000 mnow thats the power in to find the power output you multiply bv the efficiency of the turbine generator = 1499400 = 1.5 MW

37)
-->After voltage passes through some resistance it will decrease
-->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

Thanku for helping guys
Q25, 23,26 14 ,33 and 40 of this paper please?

 

[Thought blocker]

I will come tomorrow, gotta sleep ._.

 

[Dr.MMM]

The pressure of a liquid at a depth below it's surface is given by the formula

Pressure = ρgh

Where ρ is the density of the liquid, g is the average gravitational field strength along that liquid column and h is the depth of the liquid column from the surface till the point where the pressure has to be measured.

So, we can write here that the pressure at a depth of 10 kilometers (which is equal to 10,000 meters) when the density of the water is 1020 kg m⁻³ and the average gravitational field strength is 9.81 ms⁻² is:

Pressure = (1020 kg m⁻³) * (9.81 ms⁻²) * (10,000 m) = 10, 006, 200 Pascals. This is equal to 1.0006 * 10⁸ Pascals = about 10⁸ Pascals = D.

Hope this helped!
Good Luck for all your exams!

Why 10km? can you please elaborate? Thanks a lot anyway

 

[Dr.MMM]

Q1: A micrometer screw gauge is used to measure the diameter of a small uniform steel sphere. The micrometer reading is 5.00 mm ± 0.01 mm.
What will be the percentage uncertainty in a calculation of the volume of the sphere, using these values?

A 0.2% B 0.4% C 0.6% D 1.2%

Ans: C

Q2: An aeroplane travels at an average speed of 600 km h–1 on an outward flight and at 400 km h–1 on the return flight over the same distance.
What is the average speed of the whole flight?

A 111 m s–1 B 167 m s–1 C 480 km h–1 D 500 km h–1

Ans: C

 

[sagar65265]

Why 10km? can you please elaborate? Thanks a lot anyway

This is the wording of the question, as posted originally:

"Q19: The Mariana Trench in the Pacific Ocean has a depth of about 10 km.
Assuming that sea water is incompressible and has a density of about 1020 kg m–3, what would
be the approximate pressure at that depth?
A 10^5 Pa B 10^6Pa C 10^7Pa D 10^8Pa
Ans D"

So since the question is asking for the pressure at "that" depth (referring to the depth of the Mariana Trench, which is 10 kilometers = 10,000 meters) I have used "that" value of depth to calculate the answer.

Good Luck for all your exams!

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_1.pdf
Q3 and Q5 please

Q3) The volt is defined in a very specific manner; it is not defined by the equation "V = IR", but by a different equation. The definition itself is

"The Volt (as it pertains to an electric current) is the electric potential energy per unit charge, i.e. per coulomb, at any point in the circuit."

In other words, the Volt is the Energy per unit Charge, which suggests the following formula:

(Volts or Voltage at any one point in a circuit) = (Energy of certain amount of charge at that point)/(Magnitude of that amount of charge).

In symbols, V = W/Q. Since Q is charge and has the unit of As (Q = I * t, and since I = current has units of "A" - Ampere - and t = time has units of "s" - Second - the units of Q are As) and W = energy and has units Nm = Newton * Meter = kg ms⁻² * m = kg m²s⁻², we can write

Units of V = (Units of W)/(Units of Q) = (kg m²s⁻²)/(As) = kg m²s⁻³ A⁻¹ = D.

Q5)

When you are multiplying or dividing a set of numbers in any order, you can find the percentage uncertainty in the final result by adding up the percentage uncertainties in the individual numbers themselves. In this example, we need to find the mass and the volume of the block in order to calculate the density.

Since the volume is the product of 3 numbers (Length, Breadth and Height) we can write

Volume = (5.00 ± 0.01) cm * (2.00 ± 0.01) cm * (1.00 ± 0.01) cm

The % uncertainty in the length = (0.01 * 100)/5.00 = 0.2 %
The % uncertainty in the breadth = (0.01 * 100)/2.00 = 0.5 %
The % uncertainty in the height = (0.01 * 100)/1.00 = 1.0 %

Adding these up, we get 0.2 % + 0.5 % + 1.0 % = 1.7 %
Therefore, the % uncertainty in the volume is 1.7%.
So, the volume = (10.0 ± 1.7%) cm³.

Density = Mass/Volume, and so to get the % uncertainty in Density we will have to add the % uncertainties in Mass and Volume. As we know, the percentage uncertainty in volume is equal to 1.7 %. The % uncertainty in mass is equal to (0.1 * 100)/(25.0) = 0.4 %.
So, adding these up, we know that the % uncertainty in the Density is equal to 1.7% + 0.4% = 2.1%.

So, since we know that the Density is calculated to have a value of 2.50, we can calculate the final uncertainty as

(2.1/100) * 2.5 = 5.25/100 = 0.05 (to 1 significant figure, as is recommended by examiners) = ± 0.05 g cm⁻³ = C.

Hope this helped!
Good Luck for all your exams!

 

[sagar65265]

Q1: A micrometer screw gauge is used to measure the diameter of a small uniform steel sphere. The micrometer reading is 5.00 mm ± 0.01 mm.
What will be the percentage uncertainty in a calculation of the volume of the sphere, using these values?

A 0.2% B 0.4% C 0.6% D 1.2%

Ans: C

Q2: An aeroplane travels at an average speed of 600 km h–1 on an outward flight and at 400 km h–1 on the return flight over the same distance.
What is the average speed of the whole flight?

A 111 m s–1 B 167 m s–1 C 480 km h–1 D 500 km h–1

Ans: C

View attachment 45220

View attachment 45222View attachment 45222

First Question)

Okay, so what we need to find out is the percentage uncertainty in the value of the volume of the sphere with a radius measurement of r = 5.00 mm ± 0.01 mm.

The formula for volume of a sphere as a function of radius is 4πr³/3. Therefore, we are multiplying r into r into r to get r³, and this is the only value in the equation that has any uncertainty - there is no uncertainty in 4/3 and there is no uncertainty in π, so r is the only quantity we need to concern ourselves with.

The rule for uncertainties when any number of quantities are multiplied goes as follows:
"When one multiplies or divides several measurements together, one can often determine the fractional (or percentage) uncertainty in the final result simply by adding the uncertainties in the several quantities."

Let's do this by example. We are multiplying r by itself three times (to get r³) and so to get the percentage uncertainty in the final result, we resort to "adding the uncertainties in he several quantities". In other words, we add the (percentage uncertainty in r) to the (percentage uncertainty in r) to the
(percentage uncertainty in r).

Basically we multiply the percentage uncertainty in r by 3.
The percentage uncertainty in r = (0.01 mm/5.00 mm) * 100 = 0.2 %.
Multiplying this by 3, we get the uncertainty in the volume of the sphere with this radius to be 0.2 * 3 = 0.6% = C.

(more information here: http://spiff.rit.edu/classes/phys273/uncert/uncert.html)

Second Question)

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.

Third Question - Q22)

https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-554#post-826497

Fourth Question - Q23)

From the earlier question I think you might see a pattern - the equivalent spring constant of springs in series (end-to-end attachment) is found in a similar manner to the way the equivalent resistance of resistors connected in parallel is found:

1/k(equivalent) = 1/k(1) + 1/k(2) + 1/k(3)..........+ 1/k (when n springs of spring constant k(1), k(2), k(3) until k are connected in series)

Similarly, the equivalent spring constant of springs connected in parallel (connected side to side at the same level) is found in a similar manner to the way the equivalent resistance of resistors connected is series is found:

k(equivalent) = k(1) + k(2) + k(3).............+ k (when n springs of spring constant k(1), k(2), k(3) until k are connected in parallel)

So, what we can do here is find the equivalent spring constant for each option, and see which one turns out to have the largest extension.

Let's take A first. There are two coils, which implies that there are two springs in that combination, connected in series. Also, both those springs have the same spring constant, let's say "k". Therefore, we can add their spring constants like so:

1/k(equivalent) = 1/k + 1/k = 2/k
Therefore, since 1/k(equivalent) = 2/k, k(equivalent) = k/2.
So the spring constant for the setup shown in option A is equal to k/2.

Onto option B. There are three coils, so we can assume that there are three springs connected in series, and each one of them has the same spring constant, "k" - the same spring constant as the one used for the previous option. So, their equivalent spring constant is given by

1/k(equivalent) = 1/k + 1/k + 1/k = 3/k
Since 1/k(equivalent) = 3/k, k(equivalent) = k/3.
So the spring constant for the setup shown in option B is equal to k/3.

Option C: We have two springs in parallel. Finally, an easier calculation!
Either ways, each one those springs has the spring constant "k", again the same as before. Since they are attached in parallel, we can find their equivalent spring constant as follows:

k(equivalent) = k + k = 2k.

Option D: Happy days, another simple calculation!
Okay, so there are three springs in parallel, and all of them have the spring constant k. So, their equivalent spring constant is given by

k(equivalent) = k + k + k = 3k.

Alright. Now for the extensions.
Each of those springs follows the formula |F| = kx, or (Magnitude of Stretching Force) = (Spring Constant) * (Extension). So, we can rewrite the equation as
x = |F|/k. Let's calculate these ratios for each of the options.

Option A: x = (2 Newtons)/(k/2) = (4/k) meters.
Option B: x = (1 Newton)/(k/3) = (3/k) meters.
Option C: x = (6 Newtons)/(2k) = (6/2k) = (3/k) meters.
Option D: x = (8 Newtons)/(3k) = (8/3k) = (2.667/k) meters.

Clearly, out of all these, the largest extension comes about for Option A, since the rest all have the same denominator but a smaller numerator. So, A.

Hope this helped!
Good Luck for all your exams!

 

[Snackbox86]

Thank you

9)

If only one person is using the elevator, the mass of the person+elevator = 80 kg + 520 kg = 600 kg.
Suppose we take both the person and the elevator as our system. Let's take a look of at the forces acting on them.

i) The gravitational force of mg = (600 kg) * (9.81 ms⁻²) = 5886 kg ms⁻², acting downwards.
ii) The Tension force of magnitude T, acting upwards.

Since the set-up accelerates upwards, let's take upwards as the positive direction. Then tension is positive and weight is negative, and we can write

T - 5886 = ma
T - 5886 = 600a
where m is the mass of the system (600 kg) and a is the acceleration of the system.

Let's repeat for the weight. It will accelerate downwards, but the magnitude of the acceleration will be the same for the weight as for the elevator+person (because the rope is assumed to be unstretchable and this means any acceleration on one side will result in the same acceleration magnitude on the other side as well).

i) The gravitational force on the weight is (640 kg) * (9.81 ms⁻²) = 6278.4 kg ms⁻² downwards.
ii) Tension T (same magnitude as above) acting upwards.

Since the weight is accelerating downwards, we take downwards as the positive direction for the weight, and write

6278.4 - T = 640a

where a is the acceleration magnitude of the weight.

We can add both these equations as so, and this eliminates the variable of T:

T - 5886 + 6278.4 - T = 640a + 600a
392.4 = 1240a

So, a = 392.4/1240 = 0.3165 = 0.32 ms⁻² = B.

10)

Let's say the mass of a particle from the nucleus (also known as a nuclide) is "m". We'll keep this variable for use throughout this question.
Let's also select the atom as our system, since the atom on a whole is isolated, which allows us to apply the law of conservation of momentum.
Then, we can also write that the mass of the system is "Am", since m is the mass of 1 nuclide and A is the number of nucleons.

Initially, the atom has no velocity. Therefore, it has no momentum in any direction.

When the decay occurs, a proton of mass "m" is emitted with a velocity v in any direction. At the same time, the new nucleus is propelled in the opposite direction with a velocity u.
So, to conserve mass, the (mass of new nucleus) + (mass of proton) = (mass of original nucleus).
In other words, (mass of new nucleus) + m = Am. Rearranging, we get (mass of new nucleus) = Am - m = m(A - 1)
In this situation, momentum is conserved. Therefore, the momentum of proton is should be the opposite direction to the momentum of the new nucleus (check!) and the magnitude of this momentum should be the same as the magnitude of the new nucleus's momentum.

The momentum of the proton is mv. The momentum of the new nucleus is (A - 1)mu.
These have to be the same in magnitude, so we can set them equal to get

mv = (A - 1)mu.
Cancelling out the m's, we get

v = (A - 1)u = B.

15)

If the spring follows Hooke's Law, then we can write |F| = kx (Where |F| is the magnitude of the spring force, k is the spring constant and x is the extension of the spring, which is equal to [length of spring with load] - [length of spring without any load]).

If a load of 16 Newtons is applied, we can say that the spring exerts and equal and opposite force on it, so that 16 Newtons = kx.
If the final length is 5 times the initial length (40mm/1000 = 0.04 meters is the initial length) the extension =

(final length) - (initial length) = 5 * (initial length) - (initial length) = 4 * (initial length)

Since the initial length is 0.04 meters, the extension = 4 * 0.04 = 0.16 meters.

Putting this into the equation we get

16 Newtons = k * 0.16 = 0.16k so that k = 100 N/m.

The energy in a spring is given by the formula kx²/2 = (1/2)kx², where the symbols mean the same as in the earlier formula. Therefore, the energy stored is equal to

E = 0.5 * k * x² = 0.5 * 100 * 0.16² = 50 * 0.0256 = 1.28 Joules = 1.3 Joules = A.

34)

The wires are connected in parallel - that's the first thing you need to notice, and the best way of noticing this is looking at the arrangement - none of the wires are connected directly to each other, but each one of them has the same starting point and the same ending point. This usually means they have the same potential difference applied across their ends, which means they are connected in parallel.

So each cable has a resistance as a function of length, and for 1.0 km, we need to calculate the overall resistance.
The resistance formula for multiple resistances connected in parallel to each other is

1/R(equivalent) = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ + 1/R₅ + 1/R₆.........

So we have 1 resistance of 100Ω (the steel core) and 6 resistances of 10Ω (the copper wires). Let's put this into the equation:

1/R(equivalent) = 1/100 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10
= 1/100 + 6/10
= 1/100 + 60/100
= 61/100
= 0.61
So since 1/R(equivalent) = 0.61, R(equivalent) = 1/0.61 = 1.63Ω = 1.6Ω = B.

Hope this helped!

Good Luck for all your exams!

Thank you so much, this is very appreciated

 

[Snackbox86]

excellent i

3)
ye ni ata ?

5)
There are total ten lines b/w 100 to 1000 soo we can assume each line to be 100 yes so, on 40 degree, look it is the fourth line of resistance, hence 400

14)
(0.35*3*9.81)+(0.1*1.4*9.81=(0.15*6*9.81)+ans
ans=2.8Nm

17)
mgsinteta*speed=40% of total power mgsinteta=force,,,,,,force*velocity=power
(total m)(9.81)sin30)==40%
100%=ans

20)
Take the leftmost column of liquid.

There is a rule in fluid-statics that "the pressure in a sample of liquid is the same at any vertical height, as long as the fluid at that height is part of the whole."

What that says is that in any single sample of liquid, the pressure at any height is the same. Here, it means that the pressure at the top of the leftmost column (connected to the bulb) is the same as the pressure in the next column on the right at the same depth/height. (Neat info here)

So, take the unbroken fluid column on the left. At the top of the leftmost tube (which exposes the fluid to the pressure of the bulb), the pressure at the surface is simply 16,000 Pascals, because that is the pressure in the bulb and is also the pressure with which the gas pushes on any section of the surface around it. The liquid forms part of this surface, so the pressure on the liquid is also 16,000 Pascals.

Now let's move to the next column on the right, where the liquid column is exposed to pressure P. Where it is exposed to pressure P, the surface faces a pressure of P Pascals. This is to be expected. However, as we descend down the liquid, the pressure increases by (ρgh) at a depth h, and continues decreasing with depth.

When we reach depth h1, we note that this level coincides with the vertical level of the fluid in the leftmost column. The pressure on the right side column is
P + ρgh1, while the pressure on the left hand column is 16,000 Pascals. Since these are equal, we can write

P + ρgh1 = 16,000.

Repeating that calculation on the right side, we get

P + ρgh2 = 8,000

Eliminating P,

16,000 - ρgh1 = 8,000 - ρgh2
13,600 * 9.8 * (h1-h2) = 8000
h1 - h2 = 0.06

So the difference is 0.06 meters = 6 centimeters. The only option that agrees is D.

Excellent in detail answers,
Thank you

 

[Snackbox86]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_12.pdf
Can i get help with question 29 and 35
Thank you

 

[swdj24]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf
Can i get help with question 29 and 35
Thank you

For number 30, you'll use the formula E=V/d ; note that 8μm=8x10^-6.. so, E=1.5/8x10^-6=187500, which will then be rounded off to 1.9x10^5

hope it helps ^^

 

[Wolfgangs]

Guys, I have a question about the paper 21 of this year. Do you remember the waves question? What was the speed and the unit of the speed of the wave? Please do reply as soon as possible

 

[Student12]

Question 3, 4 , 5 , 8 , 11 , 14 , 15 , 19 , 24 , 33 , 35 , 37 , 38 PLEASEEEE!http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_12.pdf

 

[Wolfgangs]

Guys, I have a question about the paper 21 of this year. Do you remember the waves question? What was the speed and the unit of the speed of the wave? Please do reply as soon as possible

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_1.pdf
Q3 and Q5 please

3)
the derived unit of a volt is defined with units W.A-1, and the derived unit of a watt (W) is defined in J.s-1
Now do it your self

5)
Uncertainty unique111