Chemistry and Physics AS paper 12 MCQS

[bogus]

help needed in nov 2002 q 35 and mj 02 q 28 plzzz! physicssss syed1995 ?

 

[Sheikh Nahiyan]

help needed in nov 2002 q 35 and mj 02 q 28 plzzz! ?

Couldnt find the Mj paper.
The answer for the Oct-nov 02 question is:
Option 1 is correct as bond length increase down the group and so the dissociation increases and bond energy increases.
Option 2, I am unsure about it :\
Option 3 is also correct as trend in reduction increases down the group

I think the answer is A

 

[mannat0033]

you can post your doubts anytime. I believed all of us here are willing to help.

i have a big problem in resolving the vectors and finding the resultant and all . mainly in mechanic i need help .

 

[h4rriet]

i have a big problem in resolving the vectors and finding the resultant and all . mainly in mechanic i need help .

You'll find this video helpful, hopefully:

.

 

[bogus]

Couldnt find the Mj paper.
The answer for the Oct-nov 02 question is:
Option 1 is correct as bond length increase down the group and so the dissociation increases and bond energy increases.
Option 2, I am unsure about it :\
Option 3 is also correct as trend in reduction increases down the group

I think the answer is A

lol physics not chem ... thanks anyway

 

[Sheikh Nahiyan]

lol physics not chem ... thanks anyway

Achha .Sorry jeee You didnt mention the subject y'know

 

[Sheikh Nahiyan]

Physics:
Oct/Nov'11 Paper 11. Question: 12,20,24,26,27,28,29,30.

Please explain these questions to me as I didnt understand even after seeing the mark scheme. I know I have problem in my concept. Can anyone please help me out?

 

[syed1995]

help needed in nov 2002 q 35 and mj 02 q 28 plzzz! physicssss syed1995 ?

35 D...

cuz if the resistance of Variable resistor is increased.. It will have more PD dropped on the resistor and less PD on the wire.

And Zero deflection occurs when The PD across both points is zero (SAME PD means no current flows).. so that means we have to increase the resistance of the wire.. for that we would have to move the pointer further towards Y. And it will be nearer to Y.

MJ 02 Q28

The signal is decreasing meaning it's going towards a dark fringe. And for dark fringe path difference needs to be Lambda/2 .. and Path difference = the distance from one fringe - distance from other fringe..

S2 Distance - S1 distance = Lambda/2

so the answer according to me would be C.


Let me know if the answers are correct.. and if you got the explanation .. since I have the unsolved past papers with no MS.

 

[mannat0033]

You'll find this video helpful, hopefully:

.

thanks alot
it actually helped.

 

[ahmed abdulla]

ans. is B ie 1&2 ..but how

 

[Tina Yuzuki]

Physics:
Oct/Nov'11 Paper 11. Question: 12,20,24,26,27,28,29,30.

Please explain these questions to me as I didnt understand even after seeing the mark scheme. I know I have problem in my concept. Can anyone please help me out?

Qn 12-it says the ice rink is frictionless so there shouldnt be any resistive forces and therefore momentum remains constant (no change in momentum). after it collides with the wall, it rebounds in the opposite direction so Velocity is negative. In addition to that since the collision is inelastic, energy(kinetic) is dissipated or lost and therefore Velocity decreases resulting is a smaller momentum. Negative velocity does not mean there is less velocity, just a CHANGE IN DIRECTION.

Qn 20
-It says that the mass moves from P to Q and it losses Potential Energy so according to our knowledge of gravity, the direction should be --> (moving closer towards = lost in PE and vice versa). The magnitude of acceleration can be found with 2 equations. E(lost)=MGH which is E=MGX and F=MA. Modify the equation F=MA into A=F/M. add displacement to both sides so you get AX=FX/M. sub E=FX and you get AX=E/M and move X over to get A=E/MX. This equation is correct as if you sub MGX=E you will get A=G which is viable.

Qn 24
-As you know strain energy is area under graph of force against extension therefore the energy that the rubber band will have will be area X+Y when it is stretched to e. So in order for the rubber to have an extension e, an equivalent work has to be done so minimum energy required will be the area under the graph. Do not get confused with the graph of Stress against Strain that is different

Qn 27
-Fact. Memorise the wavelenghts of the wave spectra RMIVUXG . Lengths(1 x 10 to the power of) 1,-1,-5,-6,-7,-9,-10 p.s these are only an approximate value as there are a range for each em wave.

Qn 29
-In order for stationary waves to be formed, there must be two coherent waves of same frequency and wavelength moving in opposite directions that meet and interfere causing a resultant wave. So P is possible as the wave is reflected at the end of the tube. Q is a little bit special. It is an effect of the boundary conditions imposed at the end of the tube(the fact that pressure at the end of a tube must be equal to atmospheric pressure at the other end forming a boundary and therefore enabling the wave to reflect)

haha these are my answers and may not necessarily be 100% correct but hopefully it make a little sense.

 

[mannat0033]

ans. is B ie 1&2 ..but how

when CL reacts with mg it forms MgCl2 and when it reacts with Na , NaCl is formed so the number of moles reacting with X which is Mg is twice the number of moles reacting with Y which is Na
but i dont how it applies to H2 and kBr

 

[ahmed abdulla]

when CL reacts with mg it forms MgCl2 and when it reacts with Na , NaCl is formed so the number of moles reacting with X which is Mg is twice the number of moles reacting with Y which is Na
but i dont how it applies to H2 and kBr

same doubt

 

[mannat0033]

same doubt

im trying to slove . but the moles of Cl reacting with H2 and Kbr are the same . its so confusing .

 

[Tina Yuzuki]

im trying to slove . but the moles of Cl reacting with H2 and Kbr are the same . its so confusing .

Mg + Cl2 -> MgCl and 2Na + Cl2 -> 2NaCl,
H2 + Cl2 -> 2HCl and Cl2 + 2KBr -> 2KCl + Br2 ??

 

[ahmed abdulla]

Mg + Cl2 -> MgCl and 2Na + Cl2 -> 2NaCl,
H2 + Cl2 -> 2HCl and Cl2 + 2KBr -> 2KCl + Br2 ??

in the question they said one mole of x and y
from where the 2Na and 2Kbr

 

[mannat0033]

Mg + Cl2 -> MgCl and 2Na + Cl2 -> 2NaCl,
H2 + Cl2 -> 2HCl and Cl2 + 2KBr -> 2KCl + Br2 ??

Mg+Cl2 -> MgCl2 not MgCl

 

[bogus]

35 D...

cuz if the resistance of Variable resistor is increased.. It will have more PD dropped on the resistor and less PD on the wire.

And Zero deflection occurs when The PD across both points is zero (SAME PD means no current flows).. so that means we have to increase the resistance of the wire.. for that we would have to move the pointer further towards Y. And it will be nearer to Y.

MJ 02 Q28

The signal is decreasing meaning it's going towards a dark fringe. And for dark fringe path difference needs to be Lambda/2 .. and Path difference = the distance from one fringe - distance from other fringe..

S2 Distance - S1 distance = Lambda/2

so the answer according to me would be C.


Let me know if the answers are correct.. and if you got the explanation .. since I have the unsolved past papers with no MS.

june 07 mcq 36 plzzzz

 

[syed1995]

june 07 mcq 36 plzzzz

This is a series circuit.. So current is constant and we will use the formula P=I^2R

PD= 4/6 * 3
PD= 2V

Total Power = 0.5^2*6
= 1.5W

Power Dissipated at the internal resistor = 0.5^2*2
=0.5W

Output Power=Total Power - Power Dissipated at Internal Resistor
= 1.5-0.5
=1 Watts

so C answer?

 

[syed1995]

ans. is B ie 1&2 ..but how

1.) 2Mg + 2Cl2 --> 2MgCl2 , 2Na + Cl2 --> 2NaCl Yes 2:1
2.) H2 + Cl2 --> 2HCl , KBr + 1/2 Cl2 --> KCl Yes 2:1
3.) No. cuz the reaction produces NaClO and NaClO3 the Cl amount in not in the ratio of 1:2.