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PAPER 1 phisics Multiple choise doubts

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Could anyone explain the questions on O/N 2002 . Q 7,11,18,19,35 and 36. full explanation would be nice please
 
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for q.7 we will count da waves
that is coming about 8.5 waves
then we will divide it by 60 ms
becx for 1 box it is 10 ms so we've got 6 boxes so 60 ms
and so we get 141.7 waves/sec
as they've askd approximate so itx B
140 waves/sec
in other words 140 Hz
 
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for q.11 da formula ov elastic collision is applied
that is;

Relative speed of approach = relative speed of departure
Ua - Ub = Vb - Va

then according to da directions v put da signs lyk:
(+ Ua) - (- Ub) = (+ Vb) - (+ Va)

so finally v get:
Ua + Ub = Vb - Va

that is part A
 
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for q.18 we'll use two formulas;
firstly: P.E = m.g.h

so the energy is coming
P.E = (1.3 x 10^9) x (10) x (2)
P.E = 2.6 x 10^10

now we will calculate power by formula;
P = E/t
as da tym is 1 day so we will convert it to sec to get
t = 24 x 60 x 60
t = 86400 s

now putting in da values
P = (2.6 x 10^10) / (86400)
P = 300.9 kW
so da ans is D
 
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for q.19;
initially da twig has P.E.
as mass is unknown so we will take it as m
so initially:
P.E. = m.g.h
P.E.= (200).(10).(m)
P.E. = 2000m (J)

as 40% is lost as heat so we will subtract it lyk:
first calculate 40% of 2000m
its = 2000m x 40/100 = 800m

as dis 800m is lost so we will subtract it from 2000m
so we will get 1200m (J)

now as dis much energy is converted to kinetic energy so
1200m = (0.5).m.v^2 -(formula for kinetic energy)
m will cancel out so we will get
v = 49 m/s
so da ans is C
 
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obr*** said:
for q.11 da formula ov elastic collision is applied
that is;

Relative speed of approach = relative speed of departure
Ua - Ub = Vb - Va

then according to da directions v put da signs lyk:
(+ Ua) - (- Ub) = (+ Vb) - (+ Va)

so finally v get:
Ua + Ub = Vb - Va

that is part A

this question i still dont understand, why is it ( Ua - Ub = Vb - Va) is this the equation for the elastic collision??
 
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for q.36
we will find the potentials or voltages at X and at Y
then we will subtract them

at X the potential is
2 - (2/3) = 4/3 (V)

at Y the potential is
2 - (4/3) = 2/3 (V)

nw we will subtract Y from X
(4/3) - (2/3) = 2/3 (V)

so our ans is A
 
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1992 said:
obr*** said:
for q.11 da formula ov elastic collision is applied
that is;

Relative speed of approach = relative speed of departure
Ua - Ub = Vb - Va

then according to da directions v put da signs lyk:
(+ Ua) - (- Ub) = (+ Vb) - (+ Va)

so finally v get:
Ua + Ub = Vb - Va

that is part A

this question i still dont understand, why is it ( Ua - Ub = Vb - Va) is this the equation for the elastic collision??

yap!! dis is the equation for elastic collision-Remember IT!!!
 
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for q.35
as we increase the resistance of varible resistor
so the voltage across it will also increase
this means dat the voltage across XY will decrease
(as they r in series and voltage divides proportionally b/w the resistors according to their resistances)
so v r left wd either option C or D

now as the voltage across XY decreases the voltage across XN will also decreases
for zero deflection we have to increase the value of voltage to the same value as it was before
we will get that value near Y
becux near X the voltage will further decrease as length is directly proportional to pd across it
so our ans is D
 
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obr*** said:
for q.36
we will find the potentials or voltages at X and at Y
then we will subtract them

at X the potential is
2 - (2/3) = 4/3 (V)

at Y the potential is
2 - (4/3) = 2/3 (V)

nw we will subtract Y from X
(4/3) - (2/3) = 2/3 (V)

so our ans is A

At X, whre do u get 2/3 ??
And at Y, 4/3?
 
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obr*** said:
for q.35


now as the voltage across XY decreases the voltage across XN will also decreases
for zero deflection we have to increase the value of voltage to the same value as it was before
we will get that value near Y
becux near X the voltage will further decrease as length is directly proportional to pd across it
so our ans is D

which formula shows that lenght is proportional to pd??
 
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the formula is..

R=$L/A

r=resistance
$=resistivity
L=length
A=area
now...R=V/I
so accordig to the equation.....V is directly propotional to the length
 
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1992 said:
obr*** said:
for q.36
we will find the potentials or voltages at X and at Y
then we will subtract them

at X the potential is
2 - (2/3) = 4/3 (V)

at Y the potential is
2 - (4/3) = 2/3 (V)

nw we will subtract Y from X
(4/3) - (2/3) = 2/3 (V)

so our ans is A

At X, whre do u get 2/3 ??
And at Y, 4/3?


The emf of the battery is 2 V
in paralell the voltage will remain same so pd across each branch is 2 V

there are 3 resistors connected in series in each branch
we will calculate the voltage dat each resistor will use

to do this we will first calculate the total resistance in the whole ckt
Total R = 7.5 ohms

Now we will calculate the value for total current
I = V/R
I = 2/7.5
I = 0.267 A

Now we will calculate current in each branch
As the resistance in each branch is the same -i.e- 15 ohms
so current will divide equally in both branches
in 1 branch;
I = 0.267/2
I = 0.133 A

Now we will calculate the voltage each resistor will take
V = IR
V = 0.133 x 5
V = 2/3 V

So at X the Voltage will be
V = 2 - 2/3
V = 4/3 V

and at Y
V = 2 - (2/3) - (2/3)
V = 2/3 V

so to get the pd across X and Y
we will subtract voltage at X from Voltage at Y

so we get A as the answer

I hope this extensive explanation helps!!! :)
 
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Btw i cant help u on other Qs
at least not today as i hav math paper tomoro
sorry!
 
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sanakhalil said:
the formula is..

R=$L/A

r=resistance
$=resistivity
L=length
A=area
now...R=V/I
so accordig to the equation.....V is directly propotional to the length


Or we can also use;

LAC/LAB = VAC/Vtotal

this shows dat length is directly prop to voltage across it
 
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can any one help me in oct/nov 2006 Q 21 please???????
i need the full explanations with the answer please. :| :| :|
 
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plz any one help me i realy want 2 know how 2 get it plzzzzzzzzz i need help
 
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