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AS Physics P1 MCQs Preparation Thread.

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I = k A^2
k=I/A^2 (step 1)
at other position of meter...2I=k(x^2)
2I= put the value of k u took in first * x^2
2I= (I/A^2) * x^2
now solve for x .....u will get x=square root 2 * A
hope u get it....:)

Thnx a lot!! :)
 
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calculating the mass of the liquid,
it is (70 - 20) +/- (1+1)
= 50 +/- 2 [uncertainties are added because it is addition we are doing]

now density is mass / volume that means in this absolute uncertainties of mass and volume will be added

so
absolute uncertainty in density = absolute uncertainty in mass + absolute uncertainty in volume
absolute uncertainty in density = ( 2/50 ) + ( 0.6/10 )
absolute uncertainty in density = 0.10

uncertainty = absolute uncertainty * real value
uncertainty = 0.10 * 5 = 0.5

So, B is the answer!



Thank you so much for this! :) i always find uncertainty questions confusing :/
 
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Q4- substitute symbols by units. m^3/P = QS^2
then substitute P and Q using the options. Both sides should cancel eachother out
Q20- first find magnitude equation. E=F*d and F=ma so E=mad. a=E/md and as d is X here a=E/mx
then find direction. energy decreases towards Q means it is attracted towards Q so direction is to the right.
 
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first mcq 20: step 1 Eof wire P=E of wire Q
then put Fl/Ae= Fl/Ae
put values of all nd solve u vl get Ep/Eq = 4/1 hence ans is D
now 25: d sintheta =n lamda
sin theta / lamda = n/d
d=1/N (where N = number of lines on grating)
so, n/(1/N)
n* N/1
hence gradient= nN
ans = A
now 40: B^11 -5 + He^4-2 ----------> N^14-7 + n ^1-0
i hope u get this thing i really cudnt understand how to write this....:p
 
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Can someone explain O/N 2003 #4 for me please

The percentage uncertainty is found by the formula- uncertainty/total value *100
since the reading is being taken twice in this question...one at 40 degrees and the other at 100 degrees...the uncertainty is doubled due to two readings being taken and the total value is 60 degrees here.
0.5*2/60*100 =1.7% D
 
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