Physics: Post your doubts here!

[A*****]

For 36, you have to apply kirchhoff's 2nd law. There are two points where the voltmeter is connected, I will name the point on the right, A and the one on the left, B. So at A, the pd is E because it just came out of the cell and hasn't been used up(I am treating the internal resistances as resistor so that it's easier). This means that the pd at B must also be E since the potential difference is zero between these points. This means that the E at A was used up by the r1 since the more pd(also E) came out of the new cell. So now we can form 2 equations. First one: E=Ir1(just using V=IR), second: 2E=E+I(r2+R) which simplifies to become: E=I(r2+R). Now I can solve them simultaneously. I replace the E in E=I(r2+R) with Ir1(which is my other equation, E=Ir1).
This becomes Ir1=I(r2+R) ----> r1=r2+R ----> r1-r2=R making B the answer.

Could it also be r2-r1 if it was in the options?? Bqz the same equation can be formed with r2

 

[hellodjfos;s'ff]

Both of them please

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For 37, again you apply kirchhoff's 2nd law. Two cells are supporting each other while one is opposing. 2E-E=I(R+R+R) ---> E=I(3R) ----> R=E/3I . Now, you can find the pd used up by resistor in the left part of the circuit using V=IR. V=I * E/3I ----> V=E/3 . So that means that a remaining 2E/3 pd goes to P. Now since the cells in the upper and right part of the circuit are opposing each other, their emfs are cancelled so the pd at Q becomes 0. So the potential difference between P and Q is 2E/3 - 0 making the answer 2E/3(C).

 

[A*****]

For 37, again you apply kirchhoff's 2nd law. Two cells are supporting each other while one is opposing. 2E-E=I(R+R+R) ---> E=I(3R) ----> R=E/3I . Now, you can find the pd used up by resistor in the left part of the circuit using V=IR. V=I * E/3I ----> V=E/3 . So that means that a remaining 2E/3 pd goes to P. Now since the cells in the upper and right part of the circuit are opposing each other, their emfs are cancelled so the pd at Q becomes 0. So the potential difference between P and Q is 2E/3 - 0 making the answer 2E/3(C).

Thank u sooooo muchhh!!

 

[hellodjfos;s'ff]

Could it also be r2-r1 if it was in the options?? Bqz the same equation can be formed with r2

If you work the other way around, I think you should get the same equation, regardless of that, it wouldn't matter whether it's r2-r1 or r1-r2.

 

[hellodjfos;s'ff]

Thank u sooooo muchhh!!

Your welcome

 

[A*****]

This may sound silly but hope u understand my question
This is the diffraction pattern right?? What if we had a double slit and there was an interference pattern...would it also be a horizontal pattern? Like when we draw diagrams we make the pattern as if it was vertical

 

[Kanekii]

Need help with these

 

[A*****]

Shouldn't D be correct here as the air resistance is negligible and using the equation s=ut + 0.5at² isn't it possible?

 

[A*****]

Need help with theseView attachment 64063 View attachment 64064 View attachment 64065 View attachment 64066

For 11, 2(1) +3(-4) = 5x
x= -2 and I used -ve for left side so it is 2m/s to the left

 

[A*****]

Need help with theseView attachment 64063 View attachment 64064 View attachment 64065 View attachment 64066

 

[aisya farooq]

how to solve this???

 

[hellodjfos;s'ff]

This may sound silly but hope u understand my question
This is the diffraction pattern right?? What if we had a double slit and there was an interference pattern...would it also be a horizontal pattern? Like when we draw diagrams we make the pattern as if it was vertical

Yes, there would be diffraction in this question. Yes, if you had a double slit, you would also have a horizontal pattern. In diagrams, the pattern appears to be vertical but in reality it's horizontal. If you do a practical for this you would know(CIE encourages this).

 

[hellodjfos;s'ff]

Yes, there would be diffraction in this question. Yes, if you had a double slit, you would also have a horizontal pattern. In diagrams, the pattern appears to be vertical but in reality it's horizontal. If you do a practical for this you would know(CIE encourages this).

Also, search for images of actual experiment results for double slit and diffraction grating to understand this. Both are horizontal. The diffraction grating spreads the light into the colour spectrum, horizontally. While, double slit produced dark and bright fringes horizontally as well.

 

[hellodjfos;s'ff]

Need help with theseView attachment 64063 View attachment 64064 View attachment 64065 View attachment 64066

For 9, since the force of gravity acting on her is constant, her acceleration and deceleration would be constant too. Consequently, the gradient of the graphs must be constant too, this rules out B and C. To decide between A and D, find the diagram with a completely vertical line which would be D. There would be a completely vertical line because when the girl is on the trampoline, the spring of the trampoline suddenly causes her velocity to change direction(go opposite) but remain of the same magnitude. This means that her velocity changes instantly(no time taken). So the answer is D.

A*** already explained 11 and 24.

For 26, you find the period of the wave using T=1/f . The period will turn out to be 4 ms. Now in a stationary wave, the wave goes up(max displacement), zero displacment, down(minimum displacement). So using this concept, for this wave, the wave will go to zero displacement in 1st ms, then up in the 2nd ms, then zero displacement in the 3rd ms, then down to the original position in the 4th ms, and then back to zero displacement in the 5th ms. That means after 5ms, the wave will have zero displacement giving the answer B.

 

[hellodjfos;s'ff]

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Shouldn't D be correct here as the air resistance is negligible and using the equation s=ut + 0.5at² isn't it possible?







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15: It is A simply because the ball's energy is being converted from kinetic to gravitational, since there is no air resistance, none of the ball's energy is being lost to the air as heat. So, the ball's energy is conserved. B is incorrect because the option says that the momentum of the ball is conserved. This is incorrect because the momentum of the ball and AIR/EARTH is being conserved. The momentum being lost by the ball is being gained by the air on earth(the ball is pushing the air causing it to flow with a velocity meaning it is getting momentum). C is incorrect because at max height, the velocity is zero so the kinetic energy must be zero. D is incorrect because the potential energy(they mean the gravitational energy) is increasing but not at a constant rate since the ball is slowing down(due to gravity). So the rate at which the ball is gaining height is decreasing until it reaches a max, so the ball's potential energy is increasing but not at a constant rate.

8: There is acceleration so there must be a resultant force. This rules out A which states that the upward and downward forces are equal. The force exerted by the man on the floor is an upward force, contact force. Since the acceleration is downwards, the resultant force must also be acting downwards. This means that the downward force is greater than the upward force(weight>contact force). The option that states this idea is D, so that's the answer.

 

[hellodjfos;s'ff]

how to solve this???

When finding the gravitational energy of regular objects, you take their height from the centre of mass. Initially, the centre of mass of the block is 0.3m above the edge of the step. So the initial gravitational energy is 500(0.8+0.3)=550J. After the block falls, its centre of mass is 0.2m above the surface. Now the gravitational energy is 500(0.2)=100J. The change in gravitational energy is 550-100=450J (C)

 

[Kanekii]

For 9, since the force of gravity acting on her is constant, her acceleration and deceleration would be constant too. Consequently, the gradient of the graphs must be constant too, this rules out B and C. To decide between A and D, find the diagram with a completely vertical line which would be D. There would be a completely vertical line because when the girl is on the trampoline, the spring of the trampoline suddenly causes her velocity to change direction(go opposite) but remain of the same magnitude. This means that her velocity changes instantly(no time taken). So the answer is D.

A*** already explained 11 and 24.

For 26, you find the period of the wave using T=1/f . The period will turn out to be 4 ms. Now in a stationary wave, the wave goes up(max displacement), zero displacment, down(minimum displacement). So using this concept, for this wave, the wave will go to zero displacement in 1st ms, then up in the 2nd ms, then zero displacement in the 3rd ms, then down to the original position in the 4th ms, and then back to zero displacement in the 5th ms. That means after 5ms, the wave will have zero displacement giving the answer B.

Thanks alot but whats the ratio of up zero and down and how to find it?

 

[hellodjfos;s'ff]

Thanks alot but whats the ratio of up zero and down and how to find it?

It's not really a ratio. You got to understand how stationary waves move and work. Check this gif here to understand: http://www.sengpielaudio.com/StandingWaves.htm . There are 5 stages to a cycle/period of a stationary wave. First is the point/state in which it starts usually when it is up, this is at 0 degrees. Then at 90 degrees, it goes down to 0 zero displacement. Then at 180 degrees, it goes down. At 270 degrees back to zero displacement. At 360 degrees its back to its original position(at the start). In the question you asked, it was asking about it in terms of the period not degrees. So 4ms for the whole cycle. At 0 ms, the wave is down. At 1 ms, it is at zero displacement. At 2 ms, it is up. At 3 ms, it is at zero displacement. At 4 ms, it is back down(original position). At 5 ms, it is back at zero displacement. So the final answer is 0(B).

 

[aisya farooq]

When finding the gravitational energy of regular objects, you take their height from the centre of mass. Initially, the centre of mass of the block is 0.3m above the edge of the step. So the initial gravitational energy is 500(0.8+0.3)=550J. After the block falls, its centre of mass is 0.2m above the surface. Now the gravitational energy is 500(0.2)=100J. The change in gravitational energy is 550-100=450J (C)

thankyouu

 

[Kanekii]

Help.