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For Q37 (Attachement 1)
Ans: Option A
For QQ35 (Attachement 2)
Ans: Option C
Ans: Option A
For QQ35 (Attachement 2)
Ans: Option C
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Hooke's law is F=k*x.
its A. Since pipe open at both ends 1st Harmonic occurs where 0.5lamda=L. 2nd Harmonic occurs where lamda=Length. Likewise 3rd Harmonic occurs where 1.5lamda=Length and finally the fourth Harmonic where 2lamda=Length. This gives u 50 cm so Ahelpaaa
View attachment 64128 help ANS is B
Thank youuu,i suck at physics lol
what about 17 tho
Can anybody give me please The Answer of Cambridge International AS A Level Physics Coursebook 3rd Edition EXAM-STYLE QUESTIONS?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)
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.
thankyou but can you please tell which formula is p1v1=p2v2 and what do v1 and v2 signify here?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
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.thankyou but can you please tell which formula is p1v1=p2v2 and what do v1 and v2 signify here?
thankyou so muchSince 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!
Resistance of wire [imath]R = \frac{ρL}{A}[/imath]
wont we subtract 0.94 from 1D as it says DECREASED BY 94%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!
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