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Physics: Post your doubts here!

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sagar65265

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Thought blocker said:
meko v ni aya ._.
Click to expand...

Suppose light goes from the source L, it has an amplitude A (suppose). Then, some waves pass through the upper slit and some waves pass through the lower slit (on the diagram. In real life, this is a top-down view, so the slits are right next to each other horizontally).

The waves that pass through the upper slit (on the diagram) have amplitude A (diffraction doesn't change amplitude), and the waves that pass through the lower slit (on the diagram) also have amplitude A.

Therefore, the waves that pass through either slit have an amplitude of A. At the center of the screen S, one set of waves from each slit (one set from upper slit, one set from lower slit) overlap to form a central maxima. Since each of those sets has wavelength A, the superimposing constructive interference means that the waves at the center have amplitude A + A = 2A (since the waves at the center are in phase and so interfere constructively, their amplitudes add up).

So, the Intensity here is I. Since I = kA², we can write for this situation

I = k * (2A)²
I = k * (4A²)
I = 4kA²

So that k = I/(4A²).

This constant k remains the same for any particular wave, and since the wave does not change when one slit is blocked, this can be applied in the final condition as well.

In the final condition, only waves coming out of one slit reach the center. This set of waves has an amplitude of A, and since they do not interfere with themselves (and since there are no waves coming from the other slit to interfere with them), the final amplitude in this final situation is equal to A.

We again write, using I = kA² for this situation,

I(final) = k * (A²)

Since k = I/(4A²), we put this value into the equation above to get

I(final) = I/(4A²) * (A²)
I(final) = I/4 = D.

Hope this helped!
Good Luck for all your exams!
 
Thought blocker

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sagar65265 said:
Suppose light goes from the source L, it has an amplitude A (suppose). Then, some waves pass through the upper slit and some waves pass through the lower slit (on the diagram. In real life, this is a top-down view, so the slits are right next to each other horizontally).

The waves that pass through the upper slit (on the diagram) have amplitude A (diffraction doesn't change amplitude), and the waves that pass through the lower slit (on the diagram) also have amplitude A.

Therefore, the waves that pass through either slit have an amplitude of A. At the center of the screen S, one set of waves from each slit (one set from upper slit, one set from lower slit) overlap to form a central maxima. Since each of those sets has wavelength A, the superimposing constructive interference means that the waves at the center have amplitude A + A = 2A (since the waves at the center are in phase and so interfere constructively, their amplitudes add up).

So, the Intensity here is I. Since I = kA², we can write for this situation

I = k * (2A)²
I = k * (4A²)
I = 4kA²

So that k = I/(4A²).

This constant k remains the same for any particular wave, and since the wave does not change when one slit is blocked, this can be applied in the final condition as well.

In the final condition, only waves coming out of one slit reach the center. This set of waves has an amplitude of A, and since they do not interfere with themselves (and since there are no waves coming from the other slit to interfere with them), the final amplitude in this final situation is equal to A.

We again write, using I = kA² for this situation,

I(final) = k * (A²)

Since k = I/(4A²), we put this value into the equation above to get

I(final) = I/(4A²) * (A²)
I(final) = I/4 = D.

Hope this helped!
Good Luck for all your exams!
Click to expand...
:**********
 
K

Kamihus

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sweetym said:
I also Have doubt in this question 11,12,13.. Sorry..but Thank you fr ur help..:)
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Thought blocker said:
12)
The angle that T1 makes with R is greater than the angle that T2 makes with R.
R has to be balanced by the horizontal componts of T1 and T2
Let the angle be θ
R= T1cosθ + T1cosθ and R= T2cosθ + T2cosθ
We can see that for cos,, the greater the angle.. less the value of cosθ
so in order to cancel out R the value of T1 must be greater as cos θ is less in diagram 1. Therefore T1>T2

13)
distance of upper string from centre opf Q = 100/2 = 50mm=0.05m
distance of upper string from centre of p = 150/2=75mm=0.075m
tension in upper string = torque/distance = 3 Nm / 0.05 m = 60 N
torque on P = force*distance = 60 * 0.075 = 4.5

14)
The horizontal velocity will remain the same.
The vertical velocity will be zero at that time.
So the kinetic energy will be due to horizontal velocity alone which is v*cos*45 = v √2/2
Kinetic energy at max point = 1/2 * m * (v √2/2)² = [1/2 m v² ]* 1/2 (the thing in square brackets is equal to kinetic energy at the beginning) so k(max)=0.5K(initial)
Got it :) ?

22)
use the formula for E
E = F L / Ax
rearrange to get the ratio x / L on one side (change in length / original length)
you'll get is x / L = F / E A : Where - (A = pi r ^2 )
= 20 / 2 * 10^(11) * pi x (2.5 * 10^(-4))^2
= 5.1 * 10^(-4)
multiply this by a 100 to get the percentage
5.1 x 10^-4 x 100 = 5.1 x 10^-2 %
So answer is B.
Click to expand...
sagar65265 said:
Q12)

Suppose we take the threads and the picture as the system. Then, the only forces on our "system" are the weight of the system (the force of gravity on it) and the tension force from the uppermost string. In other words, the forces in both cases are

i) Weight of the system, acting downwards; in both case, the magnitude of this is equal to the weight of the picture (threads don't weigh a significant amount)
ii) Tension in the uppermost string, acting upwards; in the first case, the magnitude is R₁ and in the second case it is R₂.

Therefore, in the first case, since the picture is balanced, the forces are also balanced and (Weight of the system) = (Tension in uppermost thread). So, W = R₁.
In the second case, the picture is again balanced, so the forces are also balanced and (Weight of the system) = (Tension in the uppermost thread). So, W = R₂.

Since the weight is the same in both cases, R₁ = R₂, narrowing down our options to A or B.

Let's continue this analysis. Suppose we take the picture alone as our system, the sum of the vertical components of the tension forces on it should be equal to it's weight. In other words, since the two threads exert the same tension force, 2 * the vertical component of one tension force = weight of the picture (otherwise the picture is not balanced).

So, the vertical component of the tension T₁ should be equal to the vertical component of the tension T₂.
The vertical component of the tension should be equal to T₁ * cos(θ₁) OR T₂ * cos(θ₂) where θ is the angle the tension force makes with the vertical.
You can see that the angle θ₂ for T₂ is smaller than the angle θ₁ for T₁, and since cos(θ) decreases as θ increases, cos(θ₂) > cos(θ₁) (check that line again, and confirm to yourself that the inequality holds).

But if the vertical components of the tension have to be the same, T₁ has to exert a larger force to compensate for the smaller value of cos(θ₁) than T₂, therefore we can say that T₂ < T₁, which means that B is the correct answer.

I'll try posting the others later, have some stuff to deal with.

Good Luck for all your exams!
Click to expand...
Thought blocker , why didn't you take 0.075 distance and took 0.5m in Q13?
 
Asad Moosvi

Asad Moosvi

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Thought blocker said:
How 2 amplitudes and then 1 ? o_O ?
Is it given in book ._- ?
Click to expand...

They superpose. Let's say the waves coming from the slits have an amplitude of 1. They constructively interfere at the center, so the resultant amplitude is 2. However, in the second case, waves are only coming from one slit with an amplitude of 1.
 
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sagar65265

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papajohn said:
Please its an EMERGENCY!!
Q19 & 13 . Thought blocker and anyone can help
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_11.pdf
Click to expand...

Can't reply now, but i've answered Q19 earlier, so i'm pasting that here:

19)

Take the load on the right-hand side string. Since that load is in equilibrium, the tension force on it has to be equal to it's weight, which we are given is 100 Newtons.

Now take the scale on the right. The scale will tell us how hard the string is pulling on it, which is the definition of the tension in the string. So the tension in the left-hand string is given to us by the reading on the scale. Since this is 20 Newtons, the tension in the left-hand side string is also 20 Newtons.

Whenever the disc rotates a little, it loses energy because the tension in the string does a little work on the disc - take the right-hand string. The disc rotates in an anti-clockwise direction, such that the point of contact with the right-hand string moves upwards (Imagine it - the disc rotates, so the point on the right moves a little bit up). Since the force of tension acts downwards, the work done by the force is negative (because force and displacement are in the opposite direction, F.s becomes negative).

The work done by the force in 1 second is the (Magnitude of Force) * (distance traveled). The distance traveled by the point where the string contacts the disc is equal to 50 revolutions (every second, it goes around 50 times). This distance = 50 * 2πr.
We are told that circumference = 0.30 meters. Since circumference = 2πr, we can say that 2πr = 0.3 meters.
So distance = 50 * 0.3 = 15 meters.

Therefore, the work done by this 100 Newton force = - 100 * 15 = - 1500 Joules per Second. = -1500 Watts.

On the other side, the tension in the left-hand side string does positive work (because it acts downwards, and the point in contact with the string also moves downwards with the rotation). The distance traveled is the same, 15 meters, but the work done is positive because both the distance and force are in the same direction (downwards).

This work = + 20 Newtons * 15 meters = 300 Joules per Second = 300 Watts.

Therefore, the net power by external force = -1500 + 300 = -1200 Watts.
Since the disc keeps spinning at a constant rate and is not slowing down, the motor has to provide this much power per second to counter the effects of the external force. Therefore, the power of the motor is 1200 Watts = 1.2 kW = B.

Good Luck for all your exams!
 
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