Physics: Post your doubts here!

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf q29

(D) Workdone = force(F) x displacement(x) hence, x = W/f
Here W = K.E of electron = 1/2 x m x v^2 and F = qE = eE where e is cahrge of electron and E is electric field strength.

 

[psychiatrist]

Q7)
The relationship needed for this question is the vector relationship F(net) = ma.

This equation tells us that the acceleration of a body is in the same direction as the vector representing the Net Force on that object. The magnitude of this
acceleration is given by F/m. Since the force on the ball is always downwards (the force of gravity, no other force acts on the ball) the acceleration is always downwards too. So, since the question tells us to use the upwards direction as the positive direction, the acceleration is always negative. This acceleration has a magnitude of (mg)/m = 9.81 ms^-2.

Therefore, the only option that has both a constant acceleration in the negative direction and a magnitude of 9.81 ms^-2 is B.

Q9)
We can either eliminate the wrong options, or immediately arrive at the right answer here:

At any point on the trajectory, the object's velocity is along the tangent to that point. So, at the highest point of it's trajectory, the tangent is perfectly horizontal and implies the object has some sort of horizontal velocity.

So, since the object is still moving at the highest point of it's trajectory (it is moving only in the horizontal direction as established above) it has both momentum (in the horizontal direction) and kinetic energy, so the last two options are eliminated. Furthermore, since we have already established that the object is moving in the horizontal direction, B cannot be right. Therefore, A has to be right.

Alternatively, if we apply Newton's Second Law here, F(net) = ma we can see that the only force on the object is in the vertical direction (gravity, exerting a downwards force). Therefore, since there is no force acting on the object that has a component in the horizontal direction, the component of acceleration in the horizontal direction will also be zero, therefore we get A.

I'll finish off the other questions in some time, when i'm free.

Hope this helped!
Good Luck for all your exams!

Thanks a lot
Can you please send me some Projectile notes, if u have!

 

[psychiatrist]

Q20) Suppose you attach two springs in series (i.e. the ends contact each other) and attach a load to the bottom spring, both springs (assuming their masses are negligible) will display the same extension - the extension of one of these connected springs will be equal to the extension of that spring as if the entire load had been applied on it.

Therefore, the total extension of two springs in series with any load is twice the extension of a single one of those springs (if their spring constants are the same) if the same load had been attached to it.

What you can say here is that since the extension is doubled, the spring-constant for the overall setup is half that of a single spring. For example, suppose you connect two springs with spring constant k in series, the combination will have a spring constant of k/2, i.e. it will be more flexible than a single one of the connected springs. The exact rule here is like connecting resistors in parallel, i.e. 1/k(combination) = 1/k + 1/k = 2/k so k(combination) = k/2.

Suppose you connect the same two springs in parallel, like in arrangement X, then the load is shared among them, so the total extension is half the extension is only one spring had supported the same load. Therefore, if two springs of constant k each are connected in parallel, the combination will have a spring constant of 2k, since they are equivalent to a single, stiff spring.

So, taking this up for the three diagrams X, Y and Z, we find out the equivalent spring constants for each combination (since the load on each is the same, the combination with the highest spring constant will display the least extension):

For X: Since two springs are connected in parallel, their equivalent spring constant is k + k = 2k.
For Y: Since four springs are connected with pairs of two being paired up, and these two parallel branches connected in series to each other, we know that each pair of two has a spring-constant of 2k. Further, 1/k(combination) = 1/(2k) + 1/(2k) = 1/k so that k(combination) = k. (this setup is equivalent to a single spring)
For Z: Three springs are connected; one spring is connected in series to two springs connected in parallel to each other. Therefore, the upper two springs together have a spring constant of 2k. Adding this and the lower single spring in series, we see that 1/k(combination) = 1/(2k) + 1/k = 3/(2k), so that k(combination) = (2k)/3.

Therefore, the smallest extension will be found for setup X (largest spring constant ==> smallest extension) and the next smallest extension will be found for Y. Therefore, the largest extension will be found for setup Z. So the answer is A.

Q22)
Both wires obey Hooke's Law - this is important, since it means they also obey the formula of Young's Modulus, which we can apply as follows:

Since both wires are made of the same material, they have the same value of Young's Modulus (Y).
Since both wires have the same extension, they have the same value of e in the formula Y = Fl/Ae (Or Fl/Ax , whichever one you prefer).

Therefore, we can equate their extensions and cancel out the value of Y or equate the values of Y and cancel out the value of e. Let's do the former.

So, for P: (Tension in P) * l / AY = extension
And for Q: (Tension in Q) * (2l) / (A/2)Y = extension

So (Tension in P) * l / AY = (Tension in Q) *2l/ (A/2)Y
Cancelling out values on both sides,

(Tension in P) * 1 = 4 * (Tension in Q)
So that (Tension in P)/(Tension in Q) = 4/1 = D.

Q24) This concerns values you need to keep in memory unless they are on the data sheets, i'm not sure. Either ways, visible light falls in the range of
400 nm< λ < 700 nm. We can take any value in this range to represent out wavelength. Suppose we take 600 nm.
Then, the number of wavelengths in 1 meter = 1 / (600 * 10^-9) = 1.6667 * 10^6. So our answer is closest to B. Therefore, that's our answer.

I'll post the rest after some more time, have some work I need to finish off

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

Thanks but in q22 How to cancel values ? ( I am very very weak )
In q24 why you took 600 nm ? Y not 400nm or any other value ?

Thanks again, And please post the rest answers.

 

[Abdulaziz T]

Guyssss plz heeeelpp , how to find v and the uncertainty of v^2 in Q2 ????
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_53.pdf

 

[hela]

Guyssss plz heeeelpp , how to find v and the uncertainty of v^2 in Q2 ????
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_53.pdf

Any Idea about Paper 51 today

what was the scope

 

[unique111]

Check this site guys! Really helpful for finding uncertainties.
http://spiff.rit.edu/classes/phys273/uncert/uncert.html
Credits: sagar65265

 

[Abdulaziz T]

Any Idea about Paper 51 today

what was the scope

My paper 51 exam is after 4 hours Plz if you know how to ans my doubt , plz share

 

[sagar65265]

Thanks but in q22 How to cancel values ? ( I am very very weak )
In q24 why you took 600 nm ? Y not 400nm or any other value ?

Thanks again, And please post the rest answers.

Let's restate that equation:

(Tension in P) * l / AY = 4 * (Tension in Q) * l/ AY
The 4 comes about because we are multiplying by 2 (2l) and dividing by 1/2 (from A/2) so together they give us multiplied by 4.
We can now divide both sides by "l". This gives us

(Tension in P) / AY = 4 * (Tension in Q) / AY

We can further multiply both sides by Y, since both wires have the same Young's Modulus value:

(Tension in P) / A = 4 * (Tension in Q) / A

Lastly, we multiply by A on both sides, since we have already taken into account the coefficients ( the 1/2 for the cross sectional area of Q is part of the 4 multiplier).

(Tension in P) = 4 * (Tension in Q)
So, (Tension in P)/(Tension in Q) = 4/1

For the next one, there's no problem in using any value; suppose we use 400 nm for our wavelength, we apply the same procedure:

1 / (400 * 10^-9) = 2, 500, 000 = 2.5 * 10^ and this power of 10 is still closest to B.

Hope this helped!

Good Luck for all your exams!

 

[Thought blocker]

Check this site guys! Really helpful for finding uncertainties.
http://spiff.rit.edu/classes/phys273/uncert/uncert.html
Credits: sagar65265

♥♥♥♥

 

[sagar65265]

Guyssss plz heeeelpp , how to find v and the uncertainty of v^2 in Q2 ????
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_53.pdf

The marking scheme is really confusing; the two tables shown in the marking scheme for Part 2 (b) show the ratio m/(m+M) in the first column and v^2 in the second column, so don't check your values of v in there, you have to check your values of v^2 there.

In the question body, it is written that the speed v was calculated by

"measuring the time t for the card to pass fully through a light gate connected to a timer."

So, since the length of the card (also given in the question body) is 0.200 meters, we can say that the speed is given by

(Length of Card)/(Time taken) = 0.200/t

So, for the first row, the value of v is approximately 0.200/0.174 = 1.15 ms^-1, and squaring that gives us 1.32 m^2 * s^-2.
For the second row, the value of v is approximately 0.200/ 0.132 = 1.52 ms^-1, and squaring that gives us 2.30 m^2 * s^-2.
For the third row, the value of v is approximately 0.200/0.112 = 1.79 ms^-1, and squaring that gives us 3.19 m^2 * s^-2.

And so on.

To calculate the uncertainty in v, we can note that there are only two terms in the formula we have used to obtain the velocity, which are 0.200 m and t seconds.
There is no uncertainty apparent in 0.200 meters; the uncertainty we can see is in t seconds, as given in the table.
Therefore, since we are dividing the values concerned, we take the (percentage uncertainty in 0.200 meters) + (percentage uncertainty in t seconds) as the final percentage uncertainty in the value of v.

For example, the percentage uncertainty in t for the first row is (0.002/0.174 * 100) = 1.149 %
Therefore, the percentage uncertainty for v in the first row is also 1.149 %.
The value of this uncertainty is (1.149/100) * 1.15 = 0.0132.

As a second example, the percentage uncertainty in t for the second row is (0.002/0.132 * 100) = 1.515 %
Therefore, the percentage uncertainty for v in the second row is also 1.515 %.
The value of this uncertainty is (1.515/100) * 1.52 = 0.0229.

So, now that that part is over, let's move on to v^2.

When we are multiplying two numbers, we add their percentage uncertainties; that sum is equal to the percentage uncertainty in the multiplied product.
So, when we are multiplying v into v to give us v^2, we add together the (percentage uncertainty in v) and the (percentage uncertainty in v) to give us
2 * (percentage uncertainty in v).

As an example, suppose we take the second row; v here is 1.52 and v^2 = 2.30.
So, the percentage uncertainty in v = 1.515 % and accordingly, the percentage uncertainty in v^2 = 2 * 1.515 = 3.030 %.
So, to get the final absolute uncertainty in v^2, we ind out what 3.030% of v^2 is:

Absolute Uncertainty in v^2 = (3.030/100) * v^2 = 0.03030 * 2.3 = 0.0695

That is the final absolute uncertainty in v^2.

Hope this helped!
Good Luck for all your exams!

 

[Abdulaziz T]

The marking scheme is really confusing; the two tables shown in the marking scheme for Part 2 (b) show the ratio m/(m+M) in the first column and v^2 in the second column, so don't check your values of v in there, you have to check your values of v^2 there.

In the question body, it is written that the speed v was calculated by

"measuring the time t for the card to pass fully through a light gate connected to a timer."

So, since the length of the card (also given in the question body) is 0.200 meters, we can say that the speed is given by

(Length of Card)/(Time taken) = 0.200/t

So, for the first row, the value of v is approximately 0.200/0.174 = 1.15 ms^-1, and squaring that gives us 1.32 m^2 * s^-2.
For the second row, the value of v is approximately 0.200/ 0.132 = 1.52 ms^-1, and squaring that gives us 2.30 m^2 * s^-2.
For the third row, the value of v is approximately 0.200/0.112 = 1.79 ms^-1, and squaring that gives us 3.19 m^2 * s^-2.

And so on.

To calculate the uncertainty in v, we can note that there are only two terms in the formula we have used to obtain the velocity, which are 0.200 m and t seconds.
There is no uncertainty apparent in 0.200 meters; the uncertainty we can see is in t seconds, as given in the table.
Therefore, since we are dividing the values concerned, we take the (percentage uncertainty in 0.200 meters) + (percentage uncertainty in t seconds) as the final percentage uncertainty in the value of v.

For example, the percentage uncertainty in t for the first row is (0.002/0.174 * 100) = 1.149 %
Therefore, the percentage uncertainty for v in the first row is also 1.149 %.
The value of this uncertainty is (1.149/100) * 1.15 = 0.0132.

As a second example, the percentage uncertainty in t for the second row is (0.002/0.132 * 100) = 1.515 %
Therefore, the percentage uncertainty for v in the second row is also 1.515 %.
The value of this uncertainty is (1.515/100) * 1.52 = 0.0229.

So, now that that part is over, let's move on to v^2.

When we are multiplying two numbers, we add their percentage uncertainties; that sum is equal to the percentage uncertainty in the multiplied product.
So, when we are multiplying v into v to give us v^2, we add together the (percentage uncertainty in v) and the (percentage uncertainty in v) to give us
2 * (percentage uncertainty in v).

As an example, suppose we take the second row; v here is 1.52 and v^2 = 2.30.
So, the percentage uncertainty in v = 1.515 % and accordingly, the percentage uncertainty in v^2 = 2 * 1.515 = 3.030 %.
So, to get the final absolute uncertainty in v^2, we ind out what 3.030% of v^2 is:

Absolute Uncertainty in v^2 = (3.030/100) * v^2 = 0.03030 * 2.3 = 0.0695

That is the final absolute uncertainty in v^2.

Hope this helped!
Good Luck for all your exams!

Thaaaaaaaaaannkkkkk youuuu very much <3 it helped a lot .. Thanks and good luck to you too

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_qp_1.pdf

Question 7 I don't understand it!!
Please explain!
Thank you

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf

Question 7 I don't understand it!!
Please explain!
Thank you

The relationship needed for this question is the vector relationship F(net) = ma.

This equation tells us that the acceleration of a body is in the same direction as the vector representing the Net Force on that object. The magnitude of this
acceleration is given by F/m. Since the force on the ball is always downwards (the force of gravity, no other force acts on the ball) the acceleration is always downwards too. So, since the question tells us to use the upwards direction as the positive direction, the acceleration is always negative. This acceleration has a magnitude of (mg)/m = 9.81 ms^-2.

Therefore, the only option that has both a constant acceleration in the negative direction and a magnitude of 9.81 ms^-2 is B.

 

[sagar65265]

Thanks but in q22 How to cancel values ? ( I am very very weak )
In q24 why you took 600 nm ? Y not 400nm or any other value ?

Thanks again, And please post the rest answers.

25) There is an equation that states that "the intensity of a wave is proportional to the square of its amplitude". Therefore, doubling the amplitude increases the intensity by a factor of 4; tripling the amplitude increases the intensity by a factor of 9.

So, for a particular point, we can write

I = k * A²

using the variables in this situation.
To find the answer we are being asked for, we rewrite the equation, assuming that the new intensity is 2I, the constant k is still the same (it always is for a particular wave) and the amplitude A has increased by a factor of C (i.e. the amplitude at the relevant point is CA):

2I = k * (CA)² and dividing both sides by 2, we get
I = [k * (CA)²]/2

Since the magnitude of I is the same in both equations, we can equate the first equation and this most recent equation to get

[k * (CA)²]/2 = k * A²

Dividing both sides by k, we get

[(CA)²]/2 = A²

Multiplying both sides by 2 and opening out the terms in the brackets. we get

C² * A² = 2 * A²

Dividing both sides by A², we get

C² = 2

Taking the square root on both sides, we get C = √2.

Therefore, the amplitude in the new, relevant position is CA = √2 * A = √2A = B.

Q27)

You need to know how a standing wave looks; this is a very important topic not only in the MCQ paper but in the theory paper as well, so revise that and this problem will become easy. I'm going to assume you know the concept; if you have any doubts after reading this, be sure to revise and ask in the forums until you've grasped the concept. From what I recall, suchalriaz has done a really good post on this but there's a little drought in my head on where it is.

Either ways, on a standing wave, there is a distance of half a wavelength between any two maxima; from one maxima/anti-node to another/ from one minima/node to another is a distance equal to half a wavelength.

Therefore, here half a wavelength is 15 mm = 0.015 meters.
In a last step, we can calculate that if λ/2 = 0.015, λ = 0.030 meters.

Since the waves concerned are electromagnetic waves, they travel at the speed of light (approximated as 3.0 * 10^8 ms^-1) and follow the equation v = fλ. So, we can write the following equation:

3 * 10^8 = 0.030λ
Therefore, λ = 1 * 10^10 meters = C.

Q33)
The formula that relates the resistance of a sample of material to it's physical properties is what we need to know here:

R = ρL/A

Where R is the resistance of the sample, ρ is a property of the material that sample is made of, L is the length of cross-section through which the current flows, and A is the cross-sectional area through which current flows in that sample.

We can rearrange this to give

ρ = RA/L

Therefore, the left side is constant for any material, so we can equate the piece of soft metal before rolling and after rolling out to give us the answer.

Note, that part about the volume remaining constant is very, very important. Supposing that the initial length is L and the initial cross sectional area is A.
The initial volume, therefore, is AL.
Taking a look at the final case, we see that the final length is 2L, and the final cross sectional area is some variable say A(2).
In this case, the volume is 2LA(2).

Equating these two, since the volume is constant in both situations, we see that

AL = 2LA(2)

Dividing both sides by 2L, we get

A/2 = A(2)

Therefore, not only does the length change, but the cross sectional area also changes. So, for the final situation, we get (where R(2) is the final resistance, the variable we want to obtain):

ρ = R(2) * (A/2) / (2L)
ρ = R(2)A/(4L)

Since ρ remains constant in both situations, we equate the first equation and this most recent one to get

RA/L = R(2)A/(4L)

Dividing both sides by A, we get

R/L = R(2)/(4L)

Multiplying both sides by 4L, we get

4R = R(2)

So the final resistance is 4R = D.

Hope this helped!
Good Luck for all your exams!

 

[Thought blocker]

25) There is an equation that states that "the intensity of a wave is proportional to the square of its amplitude". Therefore, doubling the amplitude increases the intensity by a factor of 4; tripling the amplitude increases the intensity by a factor of 9.

So, for a particular point, we can write

I = k * A²

using the variables in this situation.
To find the answer we are being asked for, we rewrite the equation, assuming that the new intensity is 2I, the constant k is still the same (it always is for a particular wave) and the amplitude A has increased by a factor of C (i.e. the amplitude at the relevant point is CA):

2I = k * (CA)² and dividing both sides by 2, we get
I = [k * (CA)²]/2

Since the magnitude of I is the same in both equations, we can equate the first equation and this most recent equation to get

[k * (CA)²]/2 = k * A²

Dividing both sides by k, we get

[(CA)²]/2 = A²

Multiplying both sides by 2 and opening out the terms in the brackets. we get

C² * A² = 2 * A²

Dividing both sides by A², we get

C² = 2

Taking the square root on both sides, we get C = √2.

Therefore, the amplitude in the new, relevant position is CA = √2 * A = √2A = B.

Q27)

You need to know how a standing wave looks; this is a very important topic not only in the MCQ paper but in the theory paper as well, so revise that and this problem will become easy. I'm going to assume you know the concept; if you have any doubts after reading this, be sure to revise and ask in the forums until you've grasped the concept. From what I recall, suchalriaz has done a really good post on this but there's a little drought in my head on where it is.

Either ways, on a standing wave, there is a distance of half a wavelength between any two maxima; from one maxima/anti-node to another/ from one minima/node to another is a distance equal to half a wavelength.

Therefore, here half a wavelength is 15 mm = 0.015 meters.
In a last step, we can calculate that if λ/2 = 0.015, λ = 0.030 meters.

Since the waves concerned are electromagnetic waves, they travel at the speed of light (approximated as 3.0 * 10^8 ms^-1) and follow the equation v = fλ. So, we can write the following equation:

3 * 10^8 = 0.030λ
Therefore, λ = 1 * 10^10 meters = C.

Q33)
The formula that relates the resistance of a sample of material to it's physical properties is what we need to know here:

R = ρL/A

Where R is the resistance of the sample, ρ is a property of the material that sample is made of, L is the length of cross-section through which the current flows, and A is the cross-sectional area through which current flows in that sample.

We can rearrange this to give

ρ = RA/L

Therefore, the left side is constant for any material, so we can equate the piece of soft metal before rolling and after rolling out to give us the answer.

Note, that part about the volume remaining constant is very, very important. Supposing that the initial length is L and the initial cross sectional area is A.
The initial volume, therefore, is AL.
Taking a look at the final case, we see that the final length is 2L, and the final cross sectional area is some variable say A(2).
In this case, the volume is 2LA(2).

Equating these two, since the volume is constant in both situations, we see that

AL = 2LA(2)

Dividing both sides by 2L, we get

A/2 = A(2)

Therefore, not only does the length change, but the cross sectional area also changes. So, for the final situation, we get (where R(2) is the final resistance, the variable we want to obtain):

ρ = R(2) * (A/2) / (2L)
ρ = R(2)A/(4L)

Since ρ remains constant in both situations, we equate the first equation and this most recent one to get

RA/L = R(2)A/(4L)

Dividing both sides by A, we get

R/L = R(2)/(4L)

Multiplying both sides by 4L, we get

4R = R(2)

So the final resistance is 4R = D.

Hope this helped!
Good Luck for all your exams!

How do you get symbols like, lambda, rho.. etc etc ?

 

[DeViL gURl B)]

Quest

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf

Question 7 I don't understand it!!
Please explain!
Thank you

Question 4 as well please..

 

[psychiatrist]

Let's restate that equation:

(Tension in P) * l / AY = 4 * (Tension in Q) * l/ AY
The 4 comes about because we are multiplying by 2 (2l) and dividing by 1/2 (from A/2) so together they give us multiplied by 4.
We can now divide both sides by "l". This gives us

(Tension in P) / AY = 4 * (Tension in Q) / AY

We can further multiply both sides by Y, since both wires have the same Young's Modulus value:

(Tension in P) / A = 4 * (Tension in Q) / A

Lastly, we multiply by A on both sides, since we have already taken into account the coefficients ( the 1/2 for the cross sectional area of Q is part of the 4 multiplier).

(Tension in P) = 4 * (Tension in Q)
So, (Tension in P)/(Tension in Q) = 4/1

For the next one, there's no problem in using any value; suppose we use 400 nm for our wavelength, we apply the same procedure:

1 / (400 * 10^-9) = 2, 500, 000 = 2.5 * 10^ and this power of 10 is still closest to B.

Hope this helped!

Good Luck for all your exams!

ty

 

[psychiatrist]

25) There is an equation that states that "the intensity of a wave is proportional to the square of its amplitude". Therefore, doubling the amplitude increases the intensity by a factor of 4; tripling the amplitude increases the intensity by a factor of 9.

So, for a particular point, we can write

I = k * A²

using the variables in this situation.
To find the answer we are being asked for, we rewrite the equation, assuming that the new intensity is 2I, the constant k is still the same (it always is for a particular wave) and the amplitude A has increased by a factor of C (i.e. the amplitude at the relevant point is CA):

2I = k * (CA)² and dividing both sides by 2, we get
I = [k * (CA)²]/2

Since the magnitude of I is the same in both equations, we can equate the first equation and this most recent equation to get

[k * (CA)²]/2 = k * A²

Dividing both sides by k, we get

[(CA)²]/2 = A²

Multiplying both sides by 2 and opening out the terms in the brackets. we get

C² * A² = 2 * A²

Dividing both sides by A², we get

C² = 2

Taking the square root on both sides, we get C = √2.

Therefore, the amplitude in the new, relevant position is CA = √2 * A = √2A = B.

Q27)

You need to know how a standing wave looks; this is a very important topic not only in the MCQ paper but in the theory paper as well, so revise that and this problem will become easy. I'm going to assume you know the concept; if you have any doubts after reading this, be sure to revise and ask in the forums until you've grasped the concept. From what I recall, suchalriaz has done a really good post on this but there's a little drought in my head on where it is.

Either ways, on a standing wave, there is a distance of half a wavelength between any two maxima; from one maxima/anti-node to another/ from one minima/node to another is a distance equal to half a wavelength.

Therefore, here half a wavelength is 15 mm = 0.015 meters.
In a last step, we can calculate that if λ/2 = 0.015, λ = 0.030 meters.

Since the waves concerned are electromagnetic waves, they travel at the speed of light (approximated as 3.0 * 10^8 ms^-1) and follow the equation v = fλ. So, we can write the following equation:

3 * 10^8 = 0.030λ
Therefore, λ = 1 * 10^10 meters = C.

Q33)
The formula that relates the resistance of a sample of material to it's physical properties is what we need to know here:

R = ρL/A

Where R is the resistance of the sample, ρ is a property of the material that sample is made of, L is the length of cross-section through which the current flows, and A is the cross-sectional area through which current flows in that sample.

We can rearrange this to give

ρ = RA/L

Therefore, the left side is constant for any material, so we can equate the piece of soft metal before rolling and after rolling out to give us the answer.

Note, that part about the volume remaining constant is very, very important. Supposing that the initial length is L and the initial cross sectional area is A.
The initial volume, therefore, is AL.
Taking a look at the final case, we see that the final length is 2L, and the final cross sectional area is some variable say A(2).
In this case, the volume is 2LA(2).

Equating these two, since the volume is constant in both situations, we see that

AL = 2LA(2)

Dividing both sides by 2L, we get

A/2 = A(2)

Therefore, not only does the length change, but the cross sectional area also changes. So, for the final situation, we get (where R(2) is the final resistance, the variable we want to obtain):

ρ = R(2) * (A/2) / (2L)
ρ = R(2)A/(4L)

Since ρ remains constant in both situations, we equate the first equation and this most recent one to get

RA/L = R(2)A/(4L)

Dividing both sides by A, we get

R/L = R(2)/(4L)

Multiplying both sides by 4L, we get

4R = R(2)

So the final resistance is 4R = D.

Hope this helped!
Good Luck for all your exams!

25)the amplitude A has increased by a factor of C <--- Where does this came from Is it written in question and I can't see or its a mistake ?
27) Isn't Lambda / 2 for 1st harmonic, I know what you mean by "Either ways, on a standing wave, there is a distance of half a wavelength between any two maxima; from one maxima/anti-node to another/ from one minima/node to another is a distance equal to half a wavelength." But, How did you know we dont use 2nd or 3rd harmonic i.e :
Lambda or Lambda / 4 ?
33) I din't understood "Therefore, the left side is constant for any material, so we can equate the piece of soft metal before rolling and after rolling out to give us the answer.

Note, that part about the volume remaining constant is very, very important. Supposing that the initial length is L and the initial cross sectional area is A.
The initial volume, therefore, is AL.
Taking a look at the final case, we see that the final length is 2L, and the final cross sectional area is some variable say A(2).
In this case, the volume is 2LA(2).

Equating these two, since the volume is constant in both situations, we see that

AL = 2LA(2)

Dividing both sides by 2L, we get

A/2 = A(2)

Therefore, not only does the length change, but the cross sectional area also changes. So, for the final situation, we get (where R(2) is the final resistance, the variable we want to obtain):

ρ = R(2) * (A/2) / (2L)
ρ = R(2)A/(4L)

Since ρ remains constant in both situations, we equate the first equation and this most recent one to get

RA/L = R(2)A/(4L)

Dividing both sides by A, we get

R/L = R(2)/(4L)

Multiplying both sides by 4L, we get

4R = R(2)

So the final resistance is 4R = D."
:'( I am so weak in Electronics part.

 

[DeViL gURl B)]

The relationship needed for this question is the vector relationship F(net) = ma.

This equation tells us that the acceleration of a body is in the same direction as the vector representing the Net Force on that object. The magnitude of this
acceleration is given by F/m. Since the force on the ball is always downwards (the force of gravity, no other force acts on the ball) the acceleration is always downwards too. So, since the question tells us to use the upwards direction as the positive direction, the acceleration is always negative. This acceleration has a magnitude of (mg)/m = 9.81 ms^-2.

Therefore, the only option that has both a constant acceleration in the negative direction and a magnitude of 9.81 ms^-2 is B.

Thank you

 

[sagar65265]

25)the amplitude A has increased by a factor of C <--- Where does this came from Is it written in question and I can't see or its a mistake ?
27) Isn't Lambda / 2 for 1st harmonic, I know what you mean by "Either ways, on a standing wave, there is a distance of half a wavelength between any two maxima; from one maxima/anti-node to another/ from one minima/node to another is a distance equal to half a wavelength." But, How did you know we dont use 2nd or 3rd harmonic i.e :
Lambda or Lambda / 4 ?
33) I din't understood "Therefore, the left side is constant for any material, so we can equate the piece of soft metal before rolling and after rolling out to give us the answer.

Note, that part about the volume remaining constant is very, very important. Supposing that the initial length is L and the initial cross sectional area is A.
The initial volume, therefore, is AL.
Taking a look at the final case, we see that the final length is 2L, and the final cross sectional area is some variable say A(2).
In this case, the volume is 2LA(2).

Equating these two, since the volume is constant in both situations, we see that

AL = 2LA(2)

Dividing both sides by 2L, we get

A/2 = A(2)

Therefore, not only does the length change, but the cross sectional area also changes. So, for the final situation, we get (where R(2) is the final resistance, the variable we want to obtain):

ρ = R(2) * (A/2) / (2L)
ρ = R(2)A/(4L)

Since ρ remains constant in both situations, we equate the first equation and this most recent one to get

RA/L = R(2)A/(4L)

Dividing both sides by A, we get

R/L = R(2)/(4L)

Multiplying both sides by 4L, we get

4R = R(2)

So the final resistance is 4R = D."
:'( I am so weak in Electronics part.

26) C is just a random value I included; we know that the value of A changes by a factor; from A it becomes (something) * A, not (something) + A. So, I just decided to make that (something) C - it isn't part of the question, it is just something I put it. You will get the same answer by replacing (CA) with A(final). It doesn't matter, you can put whatever you want there.

27) The harmonics do not matter here;

I'll try posting the rest, I have to go right now.