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As physics p1 MCQS YEARLY ONLY.

chamiya

chamiya

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Thought blocker

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chamiya said:
Okay, so Young Modulus = (Force/Area)/(Extension/Original length). They have given Young modulus, Area and Original length. We calculate force by 70 x 9.81. So, extension of 1 wire = ((70 x 9.81) x 20)/(3.2 x 10^-6 x 2.1 x 10^11). There are 200 wires in all, so total extension is given by the previous answer divided by 200, 0.1 mm... C!
Click to expand...
Sorry, I got it as well :p
Ty ;)
 
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forever_chocoholic said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
how do we solve Q.8 ?
Click to expand...
P initial = 2mu - mu =mu
And for momentum to be conserved, P initial = P final
So check it for all question..
Option A)
-((2mu)/3)+(5mu/3) = 3mu/3 = mu so momentum is conserved Pi = Pf
Option B)
-((2mu)/6)+(2mu/3) = 1/3mu = Momentum is NOT conserved Pi =/(not equal) pf
Option C)
((2mu)/6)+((2mu)/3) = 6mu / 6 = mu so momentum is conserved Pi = Pf
Option D)
((2m+m)*u/3) = 3mu / 3 = mu so momentum is conserved Pi = Pf

Hence answer is B :)
 
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forever_chocoholic

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Mohammed Siddiq said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf

Q. 25 and 31
Click to expand...
Q.25:
astationary wave is formed
fundamental: f
first overtone: 3f
second overtone: 5f
and so on..
(odd numbers)

first higher note (first overtone) = 3*75= 225Hz
second higher note (second overtone) = 5*75= 375
therefore it is D

Q.31
Q =ne Q=It
therefore ne = It
re-arranging this gives:
n= It/e = 10 * 1/(1.6*10^-19) = 6.3*10^19
so C
 
forever_chocoholic

forever_chocoholic

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Thought blocker said:
P initial = 2mu - mu =mu
And for momentum to be conserved, P initial = P final
So check it for all question..
Option A)
-((2mu)/3)+(5mu/3) = 3mu/3 = mu so momentum is conserved Pi = Pf
Option B)
-((2mu)/6)+(2mu/3) = 1/3mu = Momentum is NOT conserved Pi =/(not equal) pf
Option C)
((2mu)/6)+((2mu)/3) = 6mu / 6 = mu so momentum is conserved Pi = Pf
Option D)
((2m+m)*u/3) = 3mu / 3 = mu so momentum is conserved Pi = Pf

Hence answer is B :)
Click to expand...

but momentum is conserved in elastic collision rite?
 
Mohammed Siddiq

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Thought blocker said:
25)
Use the formula (2n - 1)*75 i.e, for n = 1 we have frequency of 75Hz
we are asked for n = 2,3
n = 2 --> 3*75 = 225
n = 3 --> 5*75 = 375
Hence answer is D

31)
Q=ne
"n"denotes number of electrons
"e"=1.6x10^-19
sup
Q=IT
substitute both equations
IT=ne
or
I/e=n/t
10/1.6x10^-19
Hence answer is C
Click to expand...
Thank you so much! and can you also explain Q 27 from the same paper?
 
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forever_chocoholic said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
Q.21 and Q.22 pleaseeee!
btw for Q.22 i got x as 14...so do i take 13 or a 15? and why?
Click to expand...
21)
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.

22)
ultraviolet's wavelength lies in (4*10^-7) to 10^-8 longest wavelength is 10^-7. speed of all electromagnetic waves is 3*10^8 V=f*lamda.
lamda is wavelength
f is frequency of wave
v is speed
f=v/lamda=10^8/10^-7 .
do the maths u will add the powers it adds to 15 B is the answer
 
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Mohammed Siddiq said:
Thank you so much! and can you also explain Q 27 from the same paper?
Click to expand...
hmm.. let's see.. it is given : 500 lines in 1 mm, therefore, 1 line = 0.001/500 = 2 * 10^-6
the general equation to find the maximum no. of order, i.e the maximum no. of images = d/lamba = (2 * 10^-6)/(600*10^-9) = 3.33 so, the maximum no. or order is 3, but, you have to consider the other 3 images at position Z, hmm.. firstly, zero order is at position Y, so, that's one image, then there are three orders, i.e, three images at position X as well as at position Z, therefore, altogether there are 7 images, 3 images at position X, 1 image at position Y, and 3 images at position Z.. I hope you understand my explanations.. heheh.. :)
 
Hadi Murtaza

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forever_chocoholic said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
how do we solve Q.8 ?
Click to expand...
Q.8
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(2m)(u) + (m)(-u) = (2m)(v₁) + (m)(v₂)
2mu - mu = 2mv₁ + mv₂
mu = 2mv₁ + mv₂
u = 2v₁ + v₂ . . . . . (i)

e = 1 , as collision is elastic
e = (v₂ - v₁)/(u₁ - u₂)
1 = (v₂ - v₁)/(u₁ - u₂)
v₂ - v₁ = u₁ - u₂
v₂ - v₁ = u - (-u)
v₂ - v₁ = 2u
v₂ = 2u + v₁

Substitute in (i)
u = 2v₁ + (2u + v₁)
u = 3v₁ + 2u
3v₁ = -u
[ v₁ = -u/3 ]

v₂ = 2u + v₁
v₂ = 2u - u/3
[ v₂ = 5u/3 ]

So answer: A
 
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