Physics: Post your doubts here!

[Taci12]

HELP NEEDED GUYS....!!!! AS LEVEL PHYSICS

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_1.pdf
CONFUSED IN QUESTIONS (37) PLEASE TELL ME IN DEATIL
(33) (31)(12) (24) (27) AND (28)

YOUR HELP WILL BE HUMBLY APPRECIATED..!!!!

Question 37:
e.m.f = sum of p.d
As e.m.f is constant, for p.d across thermistor to be maximum, p.d across LDR must be minimum.

for thermistor: decrease tem, resistance increases, p.d increases

for LDR ; increase light intensity, resistance decreases, p.d decreases

Answer is B

 

[sma786]

Anybody has any good notes for uncertainity ??

 

[Taci12]

Anybody has any good notes for uncertainity ??

Not exactly notes, but it may help for calculations:
1. if the given equation is a sum or subtraction, add absolute uncertainties
2. if given equation is a product or quotient (multiplication or division), add fractional uncertainties. In this case you would obtain the fractional uncertainty for the final answer. To obtain the absolute uncertainty, just multiply the fractional by the actual value of the answer.

absolute uncertainty is Δx and fractional uncertainty is Δx/x.

For values raised to powers, e.g x^5
the fractional uncertainty = fractional uncertainty of x x (power to which x is raised)
Δx^5/x^5 = 5 x (Δx/x)

 

[Alice123]

pls solv no 9..... http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf
http://www.xtremepapers.com/community/members/taci12.40527/
thanks

 

[Kandinsky]

pls solv no 9..... http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf
http://www.xtremepapers.com/community/members/taci12.40527/
thanks

The train decelerates uniformly between the yellow and reg lights. Hence, we can apply the equations of motion with uniform acceleration, especially this one:
2*a*x = v^2 - u^2.

the final velocity of the train v = 0. the initial is u. the acceleration is negative, hence we can rewrite it as:

2*a*x = u^2.
the deceleration doesn't change. hence, x is directly proportional to u^2. therefore, a 20% increase in u (making it 1.2u ; 1.2^2 = 1.44) will result in a 44% increase in x. The answer is C.

 

[Alice123]

The train decelerates uniformly between the yellow and reg lights. Hence, we can apply the equations of motion with uniform acceleration, especially this one:
2*a*x = v^2 - u^2.

the final velocity of the train v = 0. the initial is u. the acceleration is negative, hence we can rewrite it as:

2*a*x = u^2.
the deceleration doesn't change. hence, x is directly proportional to u^2. therefore, a 20% increase in u (making it 1.2u ; 1.2^2 = 1.44) will result in a 44% increase in x. The answer is C.

thanks!!!
no 22 also...........http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_11.pdf

 

[Kandinsky]

The force acting on the metal X is the same as the force acting on the metal Y(denote it as F) and according to Hooke's law: lambda(modulus of elasticity)/natural length = F/x.

for metal X: lambda = 3*F/x = 3 m*F/1.5 mm= 2000F
for metal Y: lambda = 1*F/x = 1m* F/1.0 mm= 1000F

hence the extension for the second case is:

for metal X: x = F*natural length/(lambda) = F*1 / 2000 F = 0.5 mm
for metal Y: x = F*3 / 1000 F = 3 mm

The total extension is 3.5 mm (the answer is B)

 

[Alice123]

no 32 wid explanation .......http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf

 

[queen of the legend]

no 32 wid explanation .......http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf

since electric field is acting downwards (directed from positive to negative) the negative charge on the rod will be directed towards the positive of field and vice versa ................this causes a resulatant torque (non-zero) on the rod but resultant force is zero so ans is C

 

[queen of the legend]

people please help me out with the following questions:

2009: q 13, 15 , 18 , 20

2010 : q 11, 27 , 33

2011: q 13, 14, 15, 19, 25, 27, question 34 : if R=(rho)L/A how do we relate L/A to volume V ? ......and for question 9 : one mass in motion collides to stationary mass and they stick together after collision ....so momentum after coliision should be 2mV and kinetic energy as KE= o.5 x 2m x V^2 ...which is equal to mv^2 ???!
please someone explain asap ......

will be v.grateful

 

[Alice123]

people please help me out with the following questions:

2009: q 13, 15 , 18 , 20

2010 : q 11, 27 , 33

2011: q 13, 14, 15, 19, 25, 27, question 34 : if R=(rho)L/A how do we relate L/A to volume V ? ......and for question 9 : one mass in motion collides to stationary mass and they stick together after collision ....so momentum after coliision should be 2mV and kinetic energy as KE= o.5 x 2m x V^2 ...which is equal to mv^2 ???!
please someone explain asap ......

will be v.grateful

2009
13. F*1.2o=900*0.20
hence F=150N
15. When tap is opened, for the water levels to be equal in both X and Y, height in both vessels must be h/2, and the mass also becomes half
potential energy lost=m/2*g*h/2=mgh/4
18. Initially pressure= h* Rho* g
When the flask is heated, the U tube rises through a distance h, so total height is 2h
Therefore, pressure = 2h* Rho *g
hope these will be useful

 

[Alice123]

people please help me out with the following questions:

2009: q 13, 15 , 18 , 20

2010 : q 11, 27 , 33

2011: q 13, 14, 15, 19, 25, 27, question 34 : if R=(rho)L/A how do we relate L/A to volume V ? ......and for question 9 : one mass in motion collides to stationary mass and they stick together after collision ....so momentum after coliision should be 2mV and kinetic energy as KE= o.5 x 2m x V^2 ...which is equal to mv^2 ???!
please someone explain asap ......

will be v.grateful

2010
11. 10-4=ma , m=2
a=3.0 ms^-2
27. Work done in a circular path=0
33.let the resistance of each resistor be 2 Ohms
effective resistance in circuit= 2 +[1/(1/2) + (1/2)]= 3 Ohms
power = I^2 x R
12= (I^2) x 3
I= 2 A (total current supplied by battery)

As resistance of R = resistance of Q,
current through R = current through Q
= 2/2
= 1 A

 

[queen of the legend]

2009
13. F*1.2o=900*0.20
hence F=150N
15. When tap is opened, for the water levels to be equal in both X and Y, height in both vessels must be h/2, and the mass also becomes half
potential energy lost=m/2*g*h/2=mgh/4
18. Initially pressure= h* Rho* g
When the flask is heated, the U tube rises through a distance h, so total height is 2h
Therefore, pressure = 2h* Rho *g
hope these will be useful

thank you a lot

 

[falcon678]

Not exactly notes, but it may help for calculations:
1. if the given equation is a sum or subtraction, add absolute uncertainties
2. if given equation is a product or quotient (multiplication or division), add fractional uncertainties. In this case you would obtain the fractional uncertainty for the final answer. To obtain the absolute uncertainty, just multiply the fractional by the actual value of the answer.

absolute uncertainty is Δx and fractional uncertainty is Δx/x.

For values raised to powers, e.g x^5
the fractional uncertainty = fractional uncertainty of x x (power to which x is raised)
Δx^5/x^5 = 5 x (Δx/x)

i wld like elaboration on tht please....

 

[Alice123]

people please help me out with the following questions:

2009: q 13, 15 , 18 , 20

2010 : q 11, 27 , 33

2011: q 13, 14, 15, 19, 25, 27, question 34 : if R=(rho)L/A how do we relate L/A to volume V ? ......and for question 9 : one mass in motion collides to stationary mass and they stick together after collision ....so momentum after coliision should be 2mV and kinetic energy as KE= o.5 x 2m x V^2 ...which is equal to mv^2 ???!
please someone explain asap ......

will be v.grateful

2011
for Q9, momentum before collision =mv
momentum after collision=0.5*2m* (v/2)^2=mv^2/4 (since the masses stick together, velocity becomes half)​

 

[queen of the legend]

2010
11. 10-4=ma , m=2
a=3.0 ms^-2
27. Work done in a circular path=0
33.let the resistance of each resistor be 2 Ohms
effective resistance in circuit= 2 +[1/(1/2) + (1/2)]= 3 Ohms
power = I^2 x R
12= (I^2) x 3
I= 2 A (total current supplied by battery)

As resistance of R = resistance of Q,
current through R = current through Q
= 2/2
= 1 A

for 2010 question 33 : can we take any number for the resistance ?

 

[Alice123]

2011​

for Q9, momentum before collision =mv​

momentum after collision=0.5*2m* (v/2)^2=mv^2/4 (since the masses stick together, velocity becomes half)​

14. a=10 ms^-2
v^2=u^2-2as
0=u^2-20s
s=u^2/20.........(i)

again,
v^2=2as /2( distance becomes half, as given in the qs)
s=v^2/10...........(ii)

equate (i) n (ii),
v^2/u^2=1/2
Therefore,
v/u=1/square root2

lemme knw if u hav prob in understanding

 

[Alice123]

for 2010 question 33 : can we take any number for the resistance ?

yup... its your choice

 

[Alice123]

2009
13. F*1.2o=900*0.20
hence F=150N
15. When tap is opened, for the water levels to be equal in both X and Y, height in both vessels must be h/2, and the mass also becomes half
potential energy lost=m/2*g*h/2=mgh/4
18. Initially pressure= h* Rho* g
When the flask is heated, the U tube rises through a distance h, so total height is 2h
Therefore, pressure = 2h* Rho *g
hope these will be useful

june 2011
15. sin theta=3/7
therefore theta=25.3 degrees
mg*sin(theta)-friction=ma
20sin(25.3)-5=2a
a=1.773
v^2=u^2+2as
v^2=2*1.773*7
v=4.9ms^-2
19. density(d)=m/v
m=d*v
mass of water=1000*1.5=1500kg
mass of alcohol=0.5*800=400kg
total mass=1900kg
density of the mixture=total mass/total volume=1900/2=950kgm^-3

 

[Alice123]

2011​

for Q9, momentum before collision =mv​

momentum after collision=0.5*2m* (v/2)^2=mv^2/4 (since the masses stick together, velocity becomes half)​

2011. no 26

use formula for diffraction:
dsin θ = nλ

In the question, N is given instead of d
d= 1/N


so (sin θ )/N= nλ
sin θ= nNλ

we take sin θ to be less than 1 for higher order images
so sin θ<1

nNλ<1
n<1/(Nλ)
n< 1/[(5oo x 1000) x (600 x 10^-9)]
n<3.33
n= 3
number of visible images on either side of Y =3
total number of images visible on both sides of Y=6

total number of visible images including straight-through image = 7


34.R=Rho*l/A
=(rho*V^1/3)/V^2/3=rho*V^-1/3