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AS Physics P1 MCQs Preparation Thread.

confused123

confused123

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leosco1995 said:
3. Intensity = energy/(area * time) (also given in the question)
= kgm²s¯² / s¯1 * m2
= kgs¯3
Hence the correct answer is D.

8. You should know how to “convert” displacement/acceleration/velocity graphs to one another. Anyway, here is how this should be:

At the first stage, the speed is constant. This means that the object’s displacement is increasing at a uniform rate. At the second stage, the speed is decreasing, this logically means that the displacement is increasing, but at a decreasing rate. This means that eventually, at S there will be a straight horizontal line. The only graph that shows these 2 stages correctly is C.

28.

y = λd / a = 6.0 * 10-9 / a (this is for the original wavelength of 600nm)
ay = 6.0 * 10-9

Now we want to find the new value of d,
Y = (4.0 * 10-9 * d) / a

Substitute a from the old equation, a and y will cancel out and the value of d turns out to be 1.5m.

31. The best way to solve this is to see each option, and find out which one is correct.
A is wrong because the word force shouldn't be there, it should be replaced with energy.
B is wrong because the word energy should be replaced with power.

Now from C and D, use the equation P = I2R.
P = IV
V = P/I

This is the ratio of power dissipated to current, so C is correct.

34. C. You have to learn these graphics (diode, filament lamp, ohmic conductor and thermistor/semi-conductor type material). Google them if you don't already know them.

37. The easiest way to solve these questions is by using some fake values of “I” entering the parallel junction. You know that the current is divided into the inverse proportional of the resistances in parallel, so that means if say, you have 6 A entering the junction, 4 A would go the 2 Ohm resistor and 2 A to the 6 Ohm resistor in the 1st parallel combination (lets call this “block 1”).

In “block 2”, 6 A is also entering the junction, and 3 A is shared equally between the resistors. I1 is 4 A and I2 is 2 A, and I1 > I2. Since you also have the currents entering the resistors, you can easily find out the values of V too and compare them.
Click to expand...
i didn't get 31, B and C means almost same i guess, B is energy dissipated, nd C is energy dissipated per unit time. how does dat change the P.D description :/
 
Peter Check

Peter Check

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leosco1995 said:
I am not sure about the 1st one, because I also got the answer wrong the first time (I did the same thing as you). I guess it is because the wire is undergoing elastic deformation from point X and Y so it doesn't have the same gradient as from 0 to X. Someone correct me if I'm wrong, though.

The 2nd one is just simple harmonic/wave motion. At point B, the speed is 0, but it is at the highest point of its motion, not lowest. Let me know if you are confused about this stuff.
Click to expand...
For the 2nd one. how do we know if it is at the highest or lowest point of the motion?? Isnt the velocity like you said zero? So how does it even have any motion to begin with!
 
L

leosco1995

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confused123 said:
i didn't get 31, B and C means almost same i guess, B is energy dissipated, nd C is energy dissipated per unit time. how does dat change the P.D description :/
Click to expand...
Where does it say the word energy dissipated per unit time? It doesn't say that in option B. It just says "the ratio of energy dissipated between the points to the current"
 
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leosco1995

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Peter Check said:
For the 2nd one. how do we know if it is at the highest or lowest point of the motion?? Isnt the velocity like you said zero? So how does it even have any motion to begin with!
Click to expand...
Yes.. the velocity is zero but they are talking about the lowest point of motion, i.e. the lowest position of the mass where it will be going down fast and then slowing to zero speed, not going upwards which is what's happening in B. Look:

H
.
.
.
E
.
.
.
L

Let H be the maximum position, E be the equilibrium position and L be the lowest position. From 0 to A, the mass moves from L to E. (because at A the speed is maximum so it crosses the central position, and you know A is upwards velocity because of the graph labeling). From A to B, the mass goes from E to H, which is clearly not the highest position. From B to C, the mass goes back to the equilibrium position. And from C to D, the mass moves down with negative velocity again.
 
R

raamish

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leosco1995 said:
Weight = upthrust + drag

Upthrust is always the smallest force, then it's drag and then weight.
Click to expand...
ok so if an object is falling in water will upthrust still be lower than drag and weight would be the biggest force. the object is not floating.
 
SalmanslK

SalmanslK

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i love my country said:
OH here is the link ....sorry(y)
Click to expand...

Q8 => as there is air resistance acceleration will decrease at first until the weight=Air resistance; Terminal Velocity.

Q9 => initial speed is equal the horizontal component speed of projectile motion as vertical component is ZERO.

Horiz Comp.=Horizontal Distanc/Time

for time we know for vertical displacement : s=ut+ 0.5 .a .t^2
1.25=0+4.9t^2

0.255=t^ t=0.5 s

thus Horiz Comp.=Horizontal Distanc/Time= 10/0.5= 20m/s
 
xxfarhaxx

xxfarhaxx

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Silent Hunter said:
Asalaoalikum wr wb

anybody?
Click to expand...
16) first find the velocity using KE=1/2 mv^2..
u will find v= 30ms^-1
then use the equation F=ma to find the acceleration of the car =F/m=a
6000/1000 = 6 ms^-2
and the final part to find the distance travelled use the equation V^2= u^2 +2as
since v (final velocity =0), substitue 3o ms^-1 for u, and 6 ms^-2 for a..
calculate n u wil gt= 30^2/ 12= 75 m-----> ans B
 
Silent Hunter

Silent Hunter

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xxfarhaxx said:
16) first find the velocity using KE=1/2 mv^2..
u will find v= 30ms^-1
then use the equation F=ma to find the acceleration of the car =F/m=a
6000/1000 = 6 ms^-2
and the final part to find the distance travelled use the equation V^2= u^2 +2as
since v (final velocity =0), substitue 3o ms^-1 for u, and 6 ms^-2 for a..
calculate n u wil gt= 30^2/ 12= 75 m-----> ans B
Click to expand...


thanks alot :)
 
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