# Physics: Post your doubts here!

#### A.ELWY 7

hey guys these are short AS physics notes...all the credit goes to KEITAK from coloumbia, he posted them as with A2 notes in another thread, so i thought of putting them here as this thread have more views.

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#### A.ELWY 7

and these are for the A2 guys....all my thanx to Keitak from coloumbia

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#### unique840

the activity A0 is stated with 2% error. we hav to get the time for 10% ka error. means original se 8% error. if original is A0, then after 8% error, it will be 0.92A0. we will consider the expression: A = A0* e^(-lambda*time)
lambda will be 0.693/half life
lambda = 0.693/(5.27*365*24*3600)
A = 0.92 A0
so the equation will be:
0.92A0 = A0 * e^(-lambda*time)
A0 will be cancelled
0.92 = e^ (-lambda*time)
ln (0.92) = -lambda*time

#### Mustehssun Iqbal

the activity A0 is stated with 2% error. we hav to get the time for 10% ka error. means original se 8% error. if original is A0, then after 8% error, it will be 0.92A0. we will consider the expression: A = A0* e^(-lambda*time)
lambda will be 0.693/half life
lambda = 0.693/(5.27*365*24*3600)
A = 0.92 A0
so the equation will be:
0.92A0 = A0 * e^(-lambda*time)
A0 will be cancelled
0.92 = e^ (-lambda*time)
ln (0.92) = -lambda*time
thanks! btw I was asking about Q.6 and it got mistyped. Can you solve that one too, about electromagnetism??
And thanks!

#### unique840

which part of q6?

#### unique840

part b and c??
bi) reading on balance is increasing which means force on balance is acting downwards. according to newton's third law, every action has an equal and opposite reaction. so the wire will have force upwards with the same magnitude. the direction of current in from X to Y and the force is upwards. according to fleming's rule, magnetic field is from ryt to left so pole P is north cox magnetic field is directed from north to south

#### unique840

part b and c??
bii) B = F/(IL)
I and L is given. force is mg
m is given.
B = mg/IL

#### Mustehssun Iqbal

Thanks!
And how do we know that the force due to current-carrying wire is acting upwards??

#### ABDSyed

Any one know how to Take Uncertainty when dealing with Log

#### XPFMember

XPRS Moderator
Assalamoaalikum wr wb!
some doubts in Nov 2005 Paper 4 Physics...
· Q:3 I’m sorta confused with the calculation...part b ...and I don’t understand why they give the ans to C to 2s.f when they’re penalising 2s.f in part b?
· Q:4 c Why speed max when displacement is zero? :s
· Q:5b(ii) why when speed is reduced, deflection is larger? :s
· Q:6 b I don’t understand

#### XPFMember

XPRS Moderator
Thanks A lot Bro
JazzakAllah
waeyyakum..
and sorry..forgot to say salam..assalamoalaikum

#### sumaiyarox:)

Question 2b plzzz...

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#### smzimran

Assalamoaalikum wr wb!
some doubts in Nov 2005 Paper 4 Physics...
· Q:3 I’m sorta confused with the calculation...part b ...and I don’t understand why they give the ans to C to 2s.f when they’re penalising 2s.f in part b?
· Q:4 c Why speed max when displacement is zero? :s
· Q:5b(ii) why when speed is reduced, deflection is larger? :s
· Q:6 b I don’t understand
AoA,
Q3:
We know that when a liquid turns into gas, work is done by the liquid
In this case, they are asking about work done on the system, so this must mean - (work done by the system)
(i)
W = - p(deltaV)
W = - 1.03 × 10^5 [(2.96 × 10^-2) - (1.87 × 10^-5 )]
W = - 3050 J

(ii)
Heating we know it is given Q = 4.05 × 10^4 J

(iii) U = Q + W
U = 4.05 × 10^4 + (- 3050)
U = 3750 J
I dont think the calculation is much tricky
And, I dont get their approach of penalising 2 s.f in (b) but not in (c)
What i think is because in (b) they have given all data values to 3 s.f so they want the answer to atleast 3 s.f
In (c) they have not given any values, so they accept s.f below 3,
This is what i think altough im not sure

#### smzimran

Assalamoaalikum wr wb!
some doubts in Nov 2005 Paper 4 Physics...
· Q:3 I’m sorta confused with the calculation...part b ...and I don’t understand why they give the ans to C to 2s.f when they’re penalising 2s.f in part b?
· Q:4 c Why speed max when displacement is zero? :s
· Q:5b(ii) why when speed is reduced, deflection is larger? :s
· Q:6 b I don’t understand
AoA,
Q4:
Speed is max when displacement is zero,
Consider the ball thrown in the air and down again case (which you must have studied)
At max. height the ball has max. gravitational potential energy (P.E) and zero kinetic energy (K.E) meaning zero speed
Similarly, when the ball falls and touches the ground,
P.E is zero as the height is zero and by law of conservation of energy the whole P.E at top is converted into max. K.E at bottom, so speed is max at ground (at zero displacement)

Same applies in S.H.M:
At amplitude, speed is zero (K.E is zero) because the object has max P.E as it is at max height and at momentary rest.
At zero displacement, speed has increased to maximum value as whole of P.E at amplitude has been converted into K.E
That is why speed is taken max at displacement = 0

#### sumaiyarox:)

Q2? AND
how to find the magnitude in Q 6A 1?

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#### confused123

Q2? AND
how to find the magnitude in Q 6A 1?
Electric field stregth = Voltage / Distance between the plates
350 / 2.5 into 10^-2 = 1.4 into 10^4 N/C

#### smzimran

Assalamoaalikum wr wb!
some doubts in Nov 2005 Paper 4 Physics...
· Q:3 I’m sorta confused with the calculation...part b ...and I don’t understand why they give the ans to C to 2s.f when they’re penalising 2s.f in part b?
· Q:4 c Why speed max when displacement is zero? :s
· Q:5b(ii) why when speed is reduced, deflection is larger? :s
· Q:6 b I don’t understand
AoA,
Q5
When an electron is moving fast in a magnetic field, it experiences the magnetic force for a shorter duration of time thus is deflected less!
When it is passing slowly (with smaller value of speed), it experiences the magnetic force for a longer duration of time thus is deflected more!

Imagine you have to pass a patch of quicksand;
If you run through you will go through easily and will not be much drenched in the sand
If you walk slowly, you will be going down and down in the sand

^ Just came across my mind, hope it helps

Q6(b)
You just have to derive the expression for magnetic flux linkage (Φ) :
We know that
Φ = BAsinθ
For N number of lines,
Φ = NBAsinθ
Done
Small derivation so worth 2 marks only