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As physics p1 MCQS YEARLY ONLY.

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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
 
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q27 plzzzz?
i know its a simple question but.........didnt get it
POSTED IT THOUSAND TIMES :p
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.
 
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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
 
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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?
 

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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
 
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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
 
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