As physics p1 MCQS YEARLY ONLY.

[Thought blocker]

Now post new Doubts

 

[Catherine7532]

in q27 why is the polarity of the charge negative??

 

[Thought blocker]

in q27 why is the polarity of the charge negative??

27)
look, for you to determine the polarity you just see,weight of body should act downwards
therefore electric field should act upwards
body should be attracted upwards(towards positive pole)means it is negatively charged
.now since body is kept at rest..electric force=weight
electric force = qE and weight = mg qE = mg if u make q divide by m (which is the charge to mass ratio) you are left with g divided by E
hence answer is B

 

[Catherine7532]

thanx a lot

 

[Catherine7532]

q27 plzzzz?
i know its a simple question but.........didnt get it

 

[Thought blocker]

q27 plzzzz?
i know its a simple question but.........didnt get it

POSTED IT THOUSAND TIMES
27)
Let the distance between the double slit and the screen be 1m initially.
When the distance is increased BY 2m the NEW distance is 3m.
Using the formula:
Wavelength= (fringe seperation x distance between the double slits)/ distance between the screen and the slits.
Wavelenght=600nm=600x10^-9m.
Fringe seperation= 3mm= 3x10^-3m
Distance between screen and slits= 3m (1+2)
Distance between the double slit=?
Put these values in the formula, the final answer is 6x10^-4m =0.6mm
Hence the answer is B.

 

[Catherine7532]

thanks but i asked for q27. it doesnt even need explanation but i would be glad if u could give one.

 

[Thought blocker]

thanks but i asked for q27. it doesnt even need explanation but i would be glad if u could give one.

oh sorry.
27)
imagine that you are the one who the electric field comes from..the direction of the field will be from you to the computer..and then the electron passes from your left arm to your right through the field..so because you are the source of the field you r +vely charged so you will attract the electron towards you and o the answer is D

 

[Catherine7532]

in q2 why is the answer C and not D??
in Q31 why is the ans D and not C?
IN q38 how do we know whether the speed of the electron will increase or decrease?

 

[Thought blocker]

in q2 why is the answer C and not D??
in Q31 why is the ans D and not C?
IN q38 how do we know whether the speed of the electron will increase or decrease?

8702 ?? CIE ?

 

[Thought blocker]

in q2 why is the answer C and not D??
in Q31 why is the ans D and not C?
IN q38 how do we know whether the speed of the electron will increase or decrease?

Suchal Riaz
sagar65265

 

[Suchal Riaz]

Suchal Riaz
sagar65265

better do chemistry papers today

 

[Thought blocker]

better do chemistry papers today

But solve this ._.

 

[hellangel1]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_12.pdf
Q28, 32 and 34 anyone please!

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_12.pdf
Q28, 32 and 34 anyone please!

28)
28 the charged particle moves with a displacement perpendicular to the direction of the force
when the force and the displacement are perpendicular to each other, the workdone is 0
F cos (90) x distance = 0

32)
Where R will be lowest Ammeter readings would be highest
D is answer

34)
All the power in the battery is dissipated in the resistors respectively.
12 = I^2R + (I/2)^2R + (I/2)^2R <the current in the parallel resistors is halved as they are identical)
12 = I^2R + (I^2/4)R + (I^2/4)R
12 = 1/2(2I^2R + I^2R)
24 = 3 I^2R
8 = I^2R

the power dissipated in R is (I^2/4)R
1/4 of 8 is 2W

 

[Zepudee]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_12.pdf

Q15 (B) ?

 

[hellangel1]

28)
28 the charged particle moves with a displacement perpendicular to the direction of the force
when the force and the displacement are perpendicular to each other, the workdone is 0
F cos (90) x distance = 0

32)
Where R will be lowest Ammeter readings would be highest
D is answer

34)
All the power in the battery is dissipated in the resistors respectively.
12 = I^2R + (I/2)^2R + (I/2)^2R <the current in the parallel resistors is halved as they are identical)
12 = I^2R + (I^2/4)R + (I^2/4)R
12 = 1/2(2I^2R + I^2R)
24 = 3 I^2R
8 = I^2R

the power dissipated in R is (I^2/4)R
1/4 of 8 is 2W

Thanku so much! God bless you

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_12.pdf

Q15 (B) ?

Impulse = m(v - u)
Impulse = 20 000(30 - 6)
Impulse = 480 000 Ns

Force = Impulse/Time
Force = 480 000/300
Force = 1600 N

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_12.pdf

Q15 (B) ?

Change of momentum = m(v-u)
-------------------------> = 20,000(30 -6)
-------------------------> = 480,000 Ns
Rate of change of momentum = [m(v-u)]/(t)
so 480,000 / 300 = 1600 N
so B

 

[Thought blocker]

in q2 why is the answer C and not D??
in Q31 why is the ans D and not C?
IN q38 how do we know whether the speed of the electron will increase or decrease?

ZaqZainab