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As physics p1 MCQS YEARLY ONLY.

Thought blocker

Thought blocker

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33)Seriously, this one is a doozy :) the wording is not very nice, to be honest, and it makes the question rather confusing.
What it actually asks for (as far as I can tell) are the variables required to find the different between OPEN circuit voltage and CLOSED circuit voltage. In other words, the difference between the potential difference measured when the circuit is OPEN and when it is fully connected with an external resistor.
(I thought it spoke about the normal decrease in voltage with running duration, which can be expected as a battery runs down. My mistake!)

A Google Images search for "internal external resistance" provides some very clear images, and i'm using the attached one for reference.
img635-jpg.38354

When the circuit is open, no current flows through the battery, and the initial reading the equivalent to the EMF of the battery; no energy is lost per second to heat in the internal resistance, so it is equal to the EMF.
Let's write this down as V.

Once the circuit is connected as above, a current indeed does flow through all components, and a potential difference is maintained across each too. However, since the battery consists of an internal resistance (this internal resistance is NEVER separate from the battery; it can be considered as such, but the battery consists both of the cells and the internal resistance), there is a drop in potential across this internal resistance before the current even leaves the battery. Thus, the potential difference across the terminals of the battery drop.

Applying Kirchoff's Second Law while moving from B to A, what we get is (V is still the EMF, r(internal) is the internal resistance):

(Potential at A) + V - i * r(internal) = (Potential at B)

All this says is that the potential at A is one thing, it changes by so and so amount, as a result of which is becomes another value, the Potential at B.

So, since the Potential Difference is the (Potential at B) - (Potential at A), we have:

PD = V - i * r(internal)

Thus, the drop in potential is

V - (V- i * r(internal)) = i * r(internal)

The change is negative, but the decrease is positive, so the signs change there. Therefore, all you need to find the drop in potential are the final current, and the internal resistance of the battery = C
 
forever_chocoholic

forever_chocoholic

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Thought blocker said:
9) Get the time from eqn s = ut + 0.5at^(2)
S= 1.25, u = 0, a = 9.8, t = ?
t = 0.5
Now as v = d/t
V = 10/0.5 = 20 m/s :)

12)upthrust is always due to a difference in PRESSURE

17)First take the any random value of X that is half than Y, Say X = 2 and Y = 4 hence they are the respective velocities of X and Y
Now same way, for MASS Take Y as 2 and X as 4
Now for X, m = 4 and V = 2
---> 0.5*4*4 = 8
Now for Y, m =2 and V = 4
---> 0.5*2*16 = 16
Hence it is shown, that,
Car X has half the kinetic energy of car Y.

18)W.D = F*d Where F = force and d = perpendicular distance
so force is 50 times 9.8 that is 490 and d is 1.6
W.D = 490*1.6 = 784 that is near to 780

25) The image of a wave has been given, and so has the direction it is traveling in.

Note that since this is a transverse wave and the disturbances are traveling from left to right, the particles at any point are vibrating in a direction perpendicular to that propagation.
In other words, if you focused on any one segment of the rope and that one segment alone (let's say by painting it a different color from that of the rope), then you would find it does not experienceany sideways displacement; it only moves up, then down, then up, then down, etc.

At the same time, the waveform shifts towards the right, since that is the direction in which it is traveling.
So just imagine the waveform displayed in the question shift to the right; imagine each part of the waveform progress, and you would then see that P is moving downwards; since the trough just before P has to travel to the right, P has to "fall" into that trough, and if it "falls" into that trough, it is moving downwards.

Of course, this eliminates all the options in the question, leaving behind only A, the correct answer, but it is worth discussing the motion of Q:

At the instant of the displayed image, Q is a segment at the crest of the waveform; it is debatable that after some time, Q has to fall towards the equilibrium position, but taking a closer look at the wave motion (and seeing that we can estimate the motion of any one segment as Simple Harmonic Motion, SHM) it turns out that Q would be stationary.

It's just like the motion of a pendulum; at any extreme, the rate of change of it's position - it's velocity - is zero, but due to the forces acting on the pendulum (in THIS case, the force is the tension in the string) this changes slowly to make it change position. So that's most likely why the velocity of Q is zero at the instant shown.

31)I=V/R AND R=pL/A So you can write it as I= V/pL/A ; A is pi*r^(2)
I = [V/pL/pie *(1 * 10^-3)^2 / V/pL/pie * (0.5 * 10^-3)^2]
substitue u will reach to a point where it will be 1/0.25=4
Click to expand...


Thanks!!!!!! :):):)
 
forever_chocoholic

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Thought blocker said:
33)Seriously, this one is a doozy :) the wording is not very nice, to be honest, and it makes the question rather confusing.
What it actually asks for (as far as I can tell) are the variables required to find the different between OPEN circuit voltage and CLOSED circuit voltage. In other words, the difference between the potential difference measured when the circuit is OPEN and when it is fully connected with an external resistor.
(I thought it spoke about the normal decrease in voltage with running duration, which can be expected as a battery runs down. My mistake!)

A Google Images search for "internal external resistance" provides some very clear images, and i'm using the attached one for reference.
img635-jpg.38354

When the circuit is open, no current flows through the battery, and the initial reading the equivalent to the EMF of the battery; no energy is lost per second to heat in the internal resistance, so it is equal to the EMF.
Let's write this down as V.

Once the circuit is connected as above, a current indeed does flow through all components, and a potential difference is maintained across each too. However, since the battery consists of an internal resistance (this internal resistance is NEVER separate from the battery; it can be considered as such, but the battery consists both of the cells and the internal resistance), there is a drop in potential across this internal resistance before the current even leaves the battery. Thus, the potential difference across the terminals of the battery drop.

Applying Kirchoff's Second Law while moving from B to A, what we get is (V is still the EMF, r(internal) is the internal resistance):

(Potential at A) + V - i * r(internal) = (Potential at B)

All this says is that the potential at A is one thing, it changes by so and so amount, as a result of which is becomes another value, the Potential at B.

So, since the Potential Difference is the (Potential at B) - (Potential at A), we have:

PD = V - i * r(internal)

Thus, the drop in potential is

V - (V- i * r(internal)) = i * r(internal)

The change is negative, but the decrease is positive, so the signs change there. Therefore, all you need to find the drop in potential are the final current, and the internal resistance of the battery = C
Click to expand...

thanks!!!!=)
 
Thought blocker

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Thought blocker

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forever_chocoholic said:
v=0 so acceleration is max rite?
n if v=max acceleration is 0?
is this way correct?
Click to expand...
Yes, a valid reason...
Here other way too, but all thing is same, D is only the correct :¬
it isnt B because option B states that the displacement is always zero. it is untrue as the displacement is just "instantaneously" zero.. when the wave will move forward there will be non-zero displacement.
there is maximum acceleration and maximum force at maximum points.. so option D is correct
 
forever_chocoholic

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Thought blocker said:
Yes, a valid reason...
Here other way too, but all thing is same, D is only the correct :¬
it isnt B because option B states that the displacement is always zero. it is untrue as the displacement is just "instantaneously" zero.. when the wave will move forward there will be non-zero displacement.
there is maximum acceleration and maximum force at maximum points.. so option D is correct
Click to expand...

thanks a lot!!! =)
 
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forever_chocoholic said:
13 :(
going in order...
same here.. didnt do chemistry
Click to expand...
Solve 5 papers per day, Our paper is of 1 hour, so... take out 5 hours for paper, do 2 in morning 1 in afternoon and 2 in eve... (replace at instead of in :p I am vored to write again :p) So by this still you have loads of time to revise your weak part of physics, YOu'll fininsh it by 3 days ;)
 
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