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Physics: Post your doubts here!

usama321

usama321

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Atif Aziz said:
B Part, in detail please, explain any formulas used :/
Click to expand...
The wave from S1 has to travel 100cm. Using Pythagoras theorem, S2 will have to travel 128.1 cm.There is a path difference of 28.1 cm

Using v =f lambda, we know that
lambda= 330/1000 to 330/4000 Thus
8.25cm <= lambda <= 33 cm



Now, we know that a minima is formed when S2 is at its trough, while S1 at its crest. As both waves have the same frequencies, they will have travelled the same amount of distance in their respective first 100cm. However, in the next 28.1 cm, S2 must have travelled at least half, 1.5, 2.5 and so on cycles to cause destructive interference and thus a minima. We will check this with different possible wavelengths.

28.1 = 0.5 lamda
lamda = 56 cm. This is outside our range

28.1 = 1.5 lamda
lamda = 18.73. This is in our range

28.1 = 2.5 lamda
lamda = 11.24 cm This is in our range
28.1/3.5 = 8.02 Outside our range
answer = 2
 
mehria

mehria

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Mohammed salik said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf

In Both variants ..
Q 7 b (ii) and (iii) Plz !! :)
Click to expand...

Variant 1
we know that Kirchoff's law states 2 information
first=> the sum of current entering a junction = the sum of current leaving the junction
second=> the sum of e.m.f= the sum of potential difference

Using the Second Law
the sum of e.m.f=the sum of potential difference
for b(ii)
-E2=-I2R + (-I3R)
-E2=-R(I2 + I3)
the negative signs will cancel out eachother so our ans wll be:- E2= R(I2+I3)

(iii)
E1-E2=I1R+ I1R - I2R
E1-E2 = 2I1R - I2R

Focus on the direction of current n the junctions of cell for the given the loop...

Variant 2
this is solved the same way as the one in Variant 1... all u have to focus is on the directions... n for a given loop only focus on that loop n ignore the rest of the loops...

b (ii) in this part we can see that the direction of the current is the same as the path of the loop so across R, I1 will flow n as the E2's n I3's direction is opposite then they will be negative...
E1-E2= I1R - I3R

(iii) -E2=-I3R - I4R - I4R
E2=I3R + 2I4R
 
M

Mohammed salik

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mehria said:
Variant 1
we know that Kirchoff's law states 2 information
first=> the sum of current entering a junction = the sum of current leaving the junction
second=> the sum of e.m.f= the sum of potential difference

Using the Second Law
the sum of e.m.f=the sum of potential difference
for b(ii)
-E2=-I2R + (-I3R)
-E2=-R(I2 + I3)
the negative signs will cancel out eachother so our ans wll be:- E2= R(I2+I3)

(iii)
E1-E2=I1R+ I1R - I2R
E1-E2 = 2I1R - I2R

Focus on the direction of current n the junctions of cell for the given the loop...

Variant 2
this is solved the same way as the one in Variant 1... all u have to focus is on the directions... n for a given loop only focus on that loop n ignore the rest of the loops...

b (ii) in this part we can see that the direction of the current is the same as the path of the loop so across R, I1 will flow n as the E2's n I3's direction is opposite then they will be negative...
E1-E2= I1R - I3R

(iii) -E2=-I3R - I4R - I4R
E2=I3R + 2I4R
Click to expand...
Thanx A Million :)
This really Helped!
 
mehria

mehria

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Mohammed salik said:
Just One Thing.
Could YOU Plz Elaborate V1 Part (iii). (What about I3 and How do we know E1-E2 or E2-E1!?)
Click to expand...
n as i told u..for a given junction u only have to focus on that junction n ignore the rest of the junctions...
take that junction as a separate circuit n solve ur question
n ths part we dnt have to make I3 a part of our ans as it is a part of the junction BY ... n we r not asked abt anythng related to ths junction...
 
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Mohammed salik

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mehria said:
n as i told u..for a given junction u only have to focus on that junction n ignore the rest of the junctions...
take that junction as a separate circuit n solve ur question
n ths part we dnt have to make I3 a part of our ans as it is a part of the junction BY ... n we r not asked abt anythng related to ths junction...
Click to expand...
Ohhh .. I missed That Part.. FeEL Realy Stupid :oops:..!
Anyways.. Thanks For ur Time!!:)
 
MemoryMatrix 21

MemoryMatrix 21

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faizay warsi said:
can someone provide the short notes on Xray n MRI .....? :(
Click to expand...
X-Ray crystallography
Tells us how the atoms are arranged in a crystal of the compound by determining the
positions of all atomic nuclei/atoms except that of hydrogen atom. Hydrogen atoms in a
structure cannot be detected by this technique because there is insufficient electrons / electron
density / electron cloud around the H atom. structures of complex molecules such as
enzymes, derived from X-ray crystallography can help explain their biochemical behaviour
such as their 3D shape and hence their active sites. It is also called X-ray diffraction.
It uses a solid crystal or powder as its sample.
Principle;
The X-rays are diffracted by the electron clouds around the individual atoms in the structure.
Interference patterns relate to the position and electron density of atoms
 
MemoryMatrix 21

MemoryMatrix 21

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faizay warsi said:
can someone provide the short notes on Xray n MRI .....? :(
Click to expand...

Nuclear Magnetic Resonance (NMR) spectroscopy
Tells us;
The number of hydrogen atoms in each chemical environment in a molecule (from the
integration trace and the chemical shift (δ) values), and
The number of their nearest neighbours (from the splitting patterns).
If a particular type of proton has n nearest neighbours, its peak is split into (n+1) lines.
When a spinning 1H nucleus is put in an external magnetic field the nuclei may either align
itself with the external field or against (i.e. opposed to) the external field thus;
In the absence of an applied field, the spin states of a given nucleus are of equal energy,
however, the spin states in an external magnetic field are no longer of equivalent energy and
the two spin states occupy two different energy levels;
The phenomenon of NMR occurs when nuclei aligned with the applied field absorb energy
and change their spin orientation with respect to the field.
Quiz: Explain with reference to energy states how 1H NMR can supply information about
the structure of molecules.
The sample is dissolved in a solvent that contains no protons, e.g.CCl4 or heavy water, D2O,
in which all H atoms are deuterium isotope 2H. A drop of tetramethylsilane (TMS), (CH3)4Si,
is then added to act as an internal reference.
N/B;
 TMS is used as a references because it is volatile, inert and produces a strong singlet peak at
a higher field than most 1H absorptions in organic molecules, so its peak does not interfere
with the other peaks.
 The TMS peak is always ignored when determining the structures of compounds, no matter
how high it is!
An NMR spectrum consists of a graph of absorbance against frequency. The frequency scale
is measured in units of chemical shift (symbol δ). The chemical shift of a proton is the
difference between its absorption frequency and that of TMS, measured in part per million
 
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