Physics: Post your doubts here!

[umerx_xwazdorf]

View attachment 64128 help ANS is B

W acts ats at centre of radius which is 0.20m. Now to lift Weight, you would need upward force. To provide that upward force, you would need torque o a couple as there are two forces acting in opposite directions but you only require one of them to act in direction opposing the Weight (ACWM).

 

[umerx_xwazdorf]

Is the answer B? View attachment 64138

Hi can u explain me how u got the answer as B

 

[umerx_xwazdorf]

Thank youuu,i suck at physics lol

what about 17 tho

 

[iquaes]

Hi everyone, AsSalamoAlaikum Wr Wb...

To get things organized in a better way, I am making this thread. As othewise, some queries remain unanswered!

So post your PHYSICS doubts in this thread. InshaAllah other people here will help me and you all.

NOTE: If any doubts in the pastpapers, please post the link! You can find links here!

Any Physics related notes and links will be added here in this post. Feel free to provide the links to your notes around the forum, or any other websites!
Thanks!
Jazak Allah Khair!

Physics Notes:

Some links & Notes - by destined007

As physics p1 MCQS YEARLY Solved [explaination]
Physics Practical Tips - by arlery

Notes for A2 Direct Sensing (Applications) - shared by sweetiepie

Physics Summarised Notes (Click to download)

AS and A-Level Physics Definitions

A2: Physics Revision notes - by smzimran

Paper:5 Finding uncerainty in log - by XPFMember

Physics Paper 5 tips - by arlery

Physics Compiled Pastpapers: <Credits to CaptainDanger for sharing this..>

Here are the compiled A level topical Physics questions in PDF form...

Paper 1 : http://www.mediafire.com/?tocg6ha6ihkwd

Paper 2 & Paper 4 : http://www.mediafire.com/?g65j51stacmy33c

(Source : http://www.alevelforum.com/viewtopic.php?f=15&t=14)

Can anybody give me please The Answer of Cambridge International AS A Level Physics Coursebook 3rd Edition EXAM-STYLE QUESTIONS?

 

[Abdallah Divker]

Can anybody attach the answers for self assesment and exam style questions for

Cambridge International AS & A Level Physics: Coursebook, Third Edition​

Thank you

 

[redox233]

CAN ANYONE HELP ME SOLVE THIS QUESTION?

 

[PlanetMaster]

View attachment 70291

CAN ANYONE HELP ME SOLVE THIS QUESTION?

Since this is an MCQ, lets first do a quick sanity check. The cube is floating in equilibrium in mercury so the density of mercury should be heaver than the cube itself i.e the answer is either C or D.
Now, the cube floats 42% above the surface and therefore the immersed part i.e. the part that displaces mercury is 100-42% = 58% or 0.58.

Now we can have p1/v1 = p2/v2 where
p1 is the density of mercury,
p2 is the density of iron,
and therefore p1 = p2(v1/v2)
This gives us p1 = 7900/0.58 = 13,621 kg/m^3

 

[redox233]

Since this is an MCQ, lets first do a quick sanity check. The cube is floating in equilibrium in mercury so the density of mercury should be heaver than the cube itself i.e the answer is either C or D.
Now, the cube floats 42% above the surface and therefore the immersed part i.e. the part that displaces mercury is 100-42% = 58% or 0.58.

Now we can have p1/v1 = p2/v2 where
p1 is the density of mercury,
p2 is the density of iron,
and therefore p1 = p2(v1/v2)
This gives us p1 = 7900/0.58 = 13,621 kg/m^3

thankyou but can you please tell which formula is p1v1=p2v2 and what do v1 and v2 signify here?

 

[PlanetMaster]

thankyou but can you please tell which formula is p1v1=p2v2 and what do v1 and v2 signify here?

Since this is a MCQ question, I skipped a lot of steps and used just ratios and the ratio based on simply density=mass/vol. This is where p1v1=p2v2 comes from.

To elaborate the steps, lets first consider the immersed part where weight of solid equals the weight of liquid displaced.
[imath]W_{s}=W_{ld}[/imath]
And since w=mg, we get
[imath]m_{s}g=m_{ld}*g[/imath]\
And since density=mass/vol, we get
[imath]P_{s}V_{s}=P_{ld}V_{ld}[/imath] as g cancels out
Now volume of liquid displaced is equal to the volume of the immersed portion, so lets write it as
[imath]P_{s}V_{s}=P_{ld}V_{imm}[/imath]
Now we rearrange as
[imath]\frac{P_{s}}{P_{l}}=\frac{V_{imm}}{V_{s}}=0.58[/imath]
And so
[imath]P_{l}=\frac{P_{s}}{0.58}=\frac{7900}{0.58}=13621\:kg\:m\:^{-3}[/imath]

Hope that helps!

 

[redox233]

Since this is a MCQ question, I skipped a lot of steps and used just ratios and the ratio based on simply density=mass/vol. This is where p1v1=p2v2 comes from.

To elaborate the steps, lets first consider the immersed part where weight of solid equals the weight of liquid displaced.
[imath]W_{s}=W_{ld}[/imath]
And since w=mg, we get
[imath]m_{s}g=m_{ld}*g[/imath]\
And since density=mass/vol, we get
[imath]P_{s}V_{s}=P_{ld}V_{ld}[/imath] as g cancels out
Now volume of liquid displaced is equal to the volume of the immersed portion, so lets write it as
[imath]P_{s}V_{s}=P_{ld}V_{imm}[/imath]
Now we rearrange as
[imath]\frac{P_{s}}{P_{l}}=\frac{V_{imm}}{V_{s}}=0.58[/imath]
And so
[imath]P_{l}=\frac{P_{s}}{0.58}=\frac{7900}{0.58}=13621\:kg\:m\:^{-3}[/imath]

Hope that helps!

thankyou so much

 

[redox233]

CAN SOMEONE PLEASE EXPLAIN THIS QUESTION. THE ANSWER IS D

 

[PlanetMaster]

View attachment 70388

CAN SOMEONE PLEASE EXPLAIN THIS QUESTION. THE ANSWER IS D

Resistance of wire [imath]R = \frac{ρL}{A}[/imath]
And cross sectional area [imath]A = \frac{πd^{2}}{4}[/imath]

When the wire is stretched, the diameter is reduced to 94%.
This causes the area of stretched wire to be
[imath]A_{s} = \frac{π(0.94d)^{2}}{4} = 0.8836(\frac{πd^{2}}{4})[/imath] = 0.8836A

Volume of the wire is cross-sectional area x length,
[imath]V = AL[/imath]
And since the volume of the wire is unchanged,
[imath]AL=A_{s}L_{s}[/imath]
and the diameter is reduced to 94%, the length has to be
[imath](\frac{πd^{2}}{4})L=(\frac{π(0.94d^{2})}{4})(L_{s})[/imath]
[imath]L_{s}=\frac{L}{0.94^{2}}[/imath]

So, the length of the stretched wire is [imath]\frac{L}{0.94^{2}}[/imath] or [imath]1.132L[/imath]

Resistance of wire [imath]R = \frac{ρL}{A}[/imath]

So Resistance of stretched wire
[imath]R_{s} = \frac{ρL_{s}}{A_{s}} = \frac{ρ(1.132L)}{0.8836A} = 1.28\frac{ρL}{A} = 1.28R[/imath]

Hope this explains!

 

[redox233]

tysm

 

[redox233]

Resistance of wire [imath]R = \frac{ρL}{A}[/imath]
And cross sectional area [imath]A = \frac{πd^{2}}{4}[/imath]

When the wire is stretched, the diameter is reduced to 94%.
This causes the area of stretched wire to be
[imath]A_{s} = \frac{π(0.94d)^{2}}{4} = 0.8836(\frac{πd^{2}}{4})[/imath] = 0.8836A

Volume of the wire is cross-sectional area x length,
[imath]V = AL[/imath]
And since the volume of the wire is unchanged,
[imath]AL=A_{s}L_{s}[/imath]
and the diameter is reduced to 94%, the length has to be
[imath](\frac{πd^{2}}{4})L=(\frac{π(0.94d^{2})}{4})(L_{s})[/imath]
[imath]L_{s}=\frac{L}{0.94^{2}}[/imath]

So, the length of the stretched wire is [imath]\frac{L}{0.94^{2}}[/imath] or [imath]1.132L[/imath]

Resistance of wire [imath]R = \frac{ρL}{A}[/imath]

So Resistance of stretched wire
[imath]R_{s} = \frac{ρL_{s}}{A_{s}} = \frac{ρ(1.132L)}{0.8836A} = 1.28\frac{ρL}{A} = 1.28R[/imath]

Hope this explains!

wont we subtract 0.94 from 1D as it says DECREASED BY 94%

 

[PlanetMaster]

wont


wont we subtract 0.94 from 1D as it says DECREASED BY 94%

It says decreases to 94%!
If it were decreases by 94%, then we would subtract 0.94 from 1.

 

[redox233]

CAN SOMEONE EXPLAIN THIS QUESTION?

 

[redox233]

AND THIS QUESTION AS WELL THANKYOU. HOW WILL WE FIND THE AMPLITUDE IN THIS QUESTION?

 

[PlanetMaster]

View attachment 70518
CAN SOMEONE EXPLAIN THIS QUESTION?

You need to take intersections of both waves where they are going in the same direction.
From the graph, these can be either 0 to 0.5 or it can be 0.5 to 0.8.

We can then use ratios to calculate the phase difference
[imath]\frac{\Delta t}{T} = \frac{\Delta \phi}{360^{\circ}}[/imath]

Since T is 0.8 i.e one full cycle, using [imath]\Delta t = 0.3[/imath] gives us [imath]\Delta \phi = 135^{\circ}[/imath] and using [imath]\Delta t = 0.5[/imath] gives us [imath]\Delta \phi = 225^{\circ}[/imath]

So the answer is C i.e 135 which is also 360-225 i.e. 135

 

[PlanetMaster]

View attachment 70537

AND THIS QUESTION AS WELL THANKYOU. HOW WILL WE FIND THE AMPLITUDE IN THIS QUESTION?

During one oscillation, point A moves a distance of 80mm as given in the question.
And during this one oscillation, point A would return to its original position. Any point would move a distance = 4 time the amplitude during one oscillation, .

Remember, as defined above, the displacement is up and down while the wave travels to the right (in this case). Amplitude is the maximum displacement form the equilibrium position.

Consider the point A for example, which is originally on the minima. During one oscillation, it will move
1. a distance a from its current position at A to the equilibrium position,
2. another distance a from the equilibrium position to reach the maxima (crest),
3. another distance a to move from the maxima (crest) back to reach the equilibrium position again,
4. and finally, another distance a from the equilibrium position to return to its original position at A.

Thus, amplitude = 80/4 = 20mm

 

[redox233]

Sorry to disturb you again and again but can you pls tell that how the graphs are intersecting at 0.5 and 0.8

You need to take intersections of both waves where they are going in the same direction.
From the graph, these can be either 0 to 0.5 or it can be 0.5 to 0.8.

We can then use ratios to calculate the phase difference
[imath]\frac{\Delta t}{T} = \frac{\Delta \phi}{360^{\circ}}[/imath]

Since T is 0.8 i.e one full cycle, using [imath]\Delta t = 0.3[/imath] gives us [imath]\Delta \phi = 135^{\circ}[/imath] and using [imath]\Delta t = 0.5[/imath] gives us [imath]\Delta \phi = 225^{\circ}[/imath]

So the answer is C i.e 135 which is also 360-225 i.e. 13