Physics: Post your doubts here!

[funky brat]

M/J/09/4
Question 9b)
Please anyone.

The question mentions that there is an error of 2% in calibration so the error in decay can only be 8%.
I didn't get the exact answer but I tried

 

[Midnight dream]

The question mentions that there is an error of 2% in calibration so the error in decay can only be 8%.
I didn't get the exact answer but I tried

I think i get it.
Thank you!

 

[ahmedish]

w11 QP 41
Question 4)a)i) Mark scheme says its becuase theres zero field strength inside spheres. What does they are "conductors" mean and how does that prove it?

 

[ashcull14]

w11 QP 41
Question 4)a)i) Mark scheme says its becuase theres zero field strength inside spheres. What does they are "conductors" mean and how does that prove it?

free charges in a good conductor reside only on the surface... the free charge inside the conductor is zero. So the field in it is caused by charges on the surface. Since charges are of the same nature and distribution is uniform, the electric fields cancel each other..
also that electric flux =0 Furthermore, electric flux = electric field * area. Since area cannot be zero, electric field is zero

 

[ashcull14]

Thanks but I dont get this, somehow. :/

the diagram has two vertical bars carrying resistors ..the left one has 2 resistors and right one has 3
switch is connected to the left bar with 2 resistors and the negative terminal is connected to the top resistor in left bar ( encircled)
when switch is opened
for - terminal :- pd would b divided between both resistors and top resistor will hav a pd of 1.5 V so V- = 1.5V
for + terminal the current flows through the right bar with 3 resistors in series
as there is a total pd of 3V shown in the diag each resistor will receive 1 V
the positive terminal is connected to the bottom 2 resistors only not the top one in the right bar (encircled)
so therefore, the potential across V+ (positive terminal) would b 2*(1) = 2 V
thus, Vout = (V+ - V-) = 2- 1.5 = +0.5V

when the switch is closed:-
current flows through switch S not the resistor thus, V- receives 3V pd
and V+ remains 2 V as before
so Vout = 2-3 = -1V thus, output is negative

 

[funky brat]

the diagram has two vertical bars carrying resistors ..the left one has 2 resistors and right one has 3
switch is connected to the left bar with 2 resistors and the negative terminal is connected to the top resistor in left bar ( encircled)
when switch is opened
for - terminal :- pd would b divided between both resistors and top resistor will hav a pd of 1.5 V so V- = 1.5V
for + terminal the current flows through the right bar with 3 resistors in series
as there is a total pd of 3V shown in the diag each resistor will receive 1 V
the positive terminal is connected to the bottom 2 resistors only not the top one in the right bar (encircled)
so therefore, the potential across V+ (positive terminal) would b 2*(1) = 2 V
thus, Vout = (V+ - V-) = 2- 1.5 = +0.5V
View attachment 60337
when the switch is closed:-
current flows through switch S not the resistor thus, V- receives 3V pd
and V+ remains 2 V as before
so Vout = 2-3 = -1V thus, output is negative

THANKS I GOT IT NOW. :3

 

[ashcull14]

THANKS I GOT IT NOW. :3

no prob

 

[taimurahmed]

i cant understand the question

 

[ahmedish]

free charges in a good conductor reside only on the surface... the free charge inside the conductor is zero. So the field in it is caused by charges on the surface. Since charges are of the same nature and distribution is uniform, the electric fields cancel each other..
also that electric flux =0 Furthermore, electric flux = electric field * area. Since area cannot be zero, electric field is zero

And they act as point charges because the field from the further surface and the field from the closer surface, average to become as if its centred?

 

[ahmedish]

the diagram has two vertical bars carrying resistors ..the left one has 2 resistors and right one has 3
switch is connected to the left bar with 2 resistors and the negative terminal is connected to the top resistor in left bar ( encircled)
when switch is opened
for - terminal :- pd would b divided between both resistors and top resistor will hav a pd of 1.5 V so V- = 1.5V
for + terminal the current flows through the right bar with 3 resistors in series
as there is a total pd of 3V shown in the diag each resistor will receive 1 V
the positive terminal is connected to the bottom 2 resistors only not the top one in the right bar (encircled)
so therefore, the potential across V+ (positive terminal) would b 2*(1) = 2 V
thus, Vout = (V+ - V-) = 2- 1.5 = +0.5V
View attachment 60337
when the switch is closed:-
current flows through switch S not the resistor thus, V- receives 3V pd
and V+ remains 2 V as before
so Vout = 2-3 = -1V thus, output is negative

the diagram has two vertical bars carrying resistors ..the left one has 2 resistors and right one has 3
switch is connected to the left bar with 2 resistors and the negative terminal is connected to the top resistor in left bar ( encircled)
when switch is opened
for - terminal :- pd would b divided between both resistors and top resistor will hav a pd of 1.5 V so V- = 1.5V
for + terminal the current flows through the right bar with 3 resistors in series
as there is a total pd of 3V shown in the diag each resistor will receive 1 V
the positive terminal is connected to the bottom 2 resistors only not the top one in the right bar (encircled)
so therefore, the potential across V+ (positive terminal) would b 2*(1) = 2 V
thus, Vout = (V+ - V-) = 2- 1.5 = +0.5V
View attachment 60337
when the switch is closed:-
current flows through switch S not the resistor thus, V- receives 3V pd
and V+ remains 2 V as before
so Vout = 2-3 = -1V thus, output is negative

Does Physicsref have the diode orientation wrong for this question?
Red diode should light up when switch is open so when Vout is positive, the orientation, therefore should be forward biased, to let the current pass through, no?

 

[ahmedish]

The question mentions that there is an error of 2% in calibration so the error in decay can only be 8%.
I didn't get the exact answer but I tried

1) How did we get delta N/ N? I solved
A= delta N / delta T
decay constant = (From A= lambda*N) so lambda = delta N / (delta N* delta T)
is this wrong????
2) why would it have to decay by 8%? I get the percentage error is added, so its 10% - 2%, but I dont get why this 8% is the percentage of decay

 

[ahmedish]

HOW DO I SORT MY SIGNIFICANT FIGURES
LIKE IN JUNE 09 9B, HE HAS DAYS IN 2 SF
WHY?
Do I lose marks if I write it to 3 s.f? 231 vs 230
mark scheme always rounds up or down im not sure when to round and when to give exact answer

 

[ashcull14]

Does Physicsref have the diode orientation wrong for this question?
Red diode should light up when switch is open so when Vout is positive, the orientation, therefore should be forward biased, to let the current pass through, no?

the que states that red LED lights up when the door is opened
when the door is opened as stated in the first part Vout is positive...so current flows from v+
...and for V- ...the switch is closed ...door is closed so the current flows from switch through v-

 

[ashcull14]

1) How did we get delta N/ N? I solved
A= delta N / delta T
decay constant = (From A= lambda*N) so lambda = delta N / (delta N* delta T)
is this wrong????
2) why would it have to decay by 8%? I get the percentage error is added, so its 10% - 2%, but I dont get why this 8% is the percentage of decay

λN = dN/dt
dN / N = -λdt
t=0 when N=N and t=t when N= No
nw apply limits of N and No across dN/N = limits across -λ of o and t ...and solve

 

[ahmedish]

the que states that red LED lights up when the door is opened
when the door is opened as stated in the first part Vout is positive...so current flows from ground to ouput
...and for V- ...the switch is closed ...door is closed so the current flows from switch to the ground

Don't we use conventional current here?

λN = dN/dt
dN / N = -λdt
bro, what?
t=0 when N=N and t=t when N= No
nw apply limits of N and No across dN/N = limits across -λ of o and t ...and solve

bro, what?

 

[ahmedish]

λN = dN/dt
dN / N = -λdt
t=0 when N=N and t=t when N= No
nw apply limits of N and No across dN/N = limits across -λ of o and t ...and solve

i know thats how you get the N=N0*e(lambda*t) but that doesnt explain why A= dN/N

 

[ashcull14]

Don't we use conventional current here?

bro, what?

the current flows from the right vertical bar carrying three resistors in a row as switch is only connected to the left bar with two resistors ...track the path of the current ...when v = + ...and ull understand tht it will pass through the red LED..i didnt mean to say tht it is generated frm the ground .. i was talking abput the pathway

 

[ashcull14]

i know thats how you get the N=N0*e(lambda*t) but that doesnt explain why A= dN/N

its not A = dN / N
as A= λN......a is the activity...as activity is the rate at which the isotope decays .....it is = dN / dt....so λN = dN / dt

 

[funky brat]

1) How did we get delta N/ N? I solved
A= delta N / delta T
decay constant = (From A= lambda*N) so lambda = delta N / (delta N* delta T)
is this wrong????
2) why would it have to decay by 8%? I get the percentage error is added, so its 10% - 2%, but I dont get why this 8% is the percentage of decay

I think the question mentioned sth like there's already an error of 2% in calibration and then it asked the maximum possible error which ruled out the possibility of using -2% as the error and the max error had to be 10% in total therefore only 8% could be the possible error in decay. Get it?
Also it is derived like this:
rate of decay is directly propotional to the no of nuclei
delta n/ delta t= N where N is no nuclei
delta n/ delta t= -lambda*N where lambda is the decay constant
and then make lambda the subject to get (delta n/ N) / T

 

[ahmedish]

I think the question mentioned sth like there's already an error of 2% in calibration and then it asked the maximum possible error which ruled out the possibility of using -2% as the error and the max error had to be 10% in total therefore only 8% could be the possible error in decay. Get it?
Also it is derived like this:
rate of decay is directly propotional to the no of nuclei
delta n/ delta t= N where N is no nuclei
delta n/ delta t= -lambda*N where lambda is the decay constant
and then make lambda the subject to get (delta n/ N) / T

Yeah man I agree with that! But in the mark scheme its written lambda = (delta N)/ (N)
whereas I think its (delta N)/( N* delta T)