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sorry for disturbing, but can you tell me that when should we use a straight horizontal line in the graph of E against x or V? thanks in advanceHi! I was just going through this when I stumbled upon this mistake.For 37, the correct answer is A. This is because the plate separation is increasing so the field strength would decrease. I'm sure it was just a slip. The rest is all correct, Masha'Allah.
The electric field strength between two parallel plates is always uniform as long as plate separation is unchanged. So you get a straight horizontal line in the graph of E against x, where x is the distance from one plate to any point between the two plates.sorry for disturbing, but can you tell me that when should we use a straight horizontal line in the graph of E against x or V? thanks in advance
2010 yes most of them r da same but 2011 about 60% same
can u please explain qs 34 a bit more???May/June 2008
1) C; kilo(k): 10^3
milli(m): 10^-3
Mega(M): 10^6
nano ( n):10^-9
2) D; Force = mass x acc. = kgms-²
work done = force x displacement = kgm²s-²
e.m.f. = work done/ charge = kgm²s-²/As
3) C; speed= distance/time = 100/10 = 10m/s
the approximate mass of an athlete can be estimated between 60 to 85 kg, if 80 kg is taken, by K.E.= 1/2 m V ² = 1/2 x 80 x 10 ² = 4000J
4) C; Here, R= V/I = 1/0.5 = 2 ohms.
ΔR/R = ΔV/V + ΔI/I
ΔR/2 = 0.05/1 + 0.o1/0.5
ΔR = ± 0.14
%age uncertainty = (0.14/2)x 100 = 7.0%
5) D; precise and not accurate result would be very different from 9.81 but within close range of one another.
6) C The case is similar to that when an object moves in a circular path. The speed remains same, but the direction keeps changing (direction at each instant is tangent to the circle). The acc. is because the velocity changes, by the changing direction.
7) B; Acc. of free fall on this planet would be, 9.81/6 = 1.635 m/s^2
W=mg
= 30 x 1.635
= 49.05N
8) C; At the top when it is released, it has a maximum acc. of 9.81 m/s^2. After that it keeps decreasing and finally when terminal velocity is reached, acc. becomes zero. The graph is often confused with that of D, but remember that shape is for a V-t graph.
9) D; The principle states that for a closed system (in which no external forces act) the momentum is always conserved.
10) D; Relative speed of approach = Relative speed of separation. Also, in an elastic collision, total kinetic energy is conserved.
11) A; F=ma
2000-R = 750 x 2
R = 500N
12) A;
The drag is a resisting force in fluids ( like air resistance in air). This is similar to friction.
Upthrust is only because of the pressure difference on top and bottom of an object when in a fluid.
Weight is the greatest of all.
13) A; Forces in equilibrium form a head to tail vector triangle (clockwise or anti-clockwise).
14) A; moment = force x distance
clockwise moment = anticlockwise moment
W x a + F x h = W x 2a
15) D; fact. the concept of ρgh.
16) A; for potential energy, vertical height is always taken, so distance would be s.
The field is towards the right side, so left is positive and the right negative, which means that the potential energy will decrease. energy = workdone, therefore, w.d. = F x s
17) C; linear momentum is Always conserved. B and D rejected.
The collision here is inelastic, so K.E. is not conserved. However, the Total energy is still conserved because of the law of conservation of energy.
18) B; The speed is constant, therefore the Kinetic energy is constt, which would give a straight line graph.
The ball is falling in a fluid of constt density. The potential energy varies linearly with the height. Ball is falling, h is decreasing so P.E. decreasing at a constt rate.
19) C; efficiency = (*useful* output/ total input) x 100
20) B; same old repeated question. fact.
A ignored cuz container's pressure is nowhere discussed.
C rejected bcuz the collisions are elastic.
D put down cuz the weight is downwards, and the statement is just wrong.
21) A; In the derivation of this formula, these formulae/principles are used, P = F/A, F=W, W=mg, m= ρv and v=area x height. Out of these, only mass= density x volume is given.
22) B; The metal stretches and then returns to its original shape. there is an extension, therefore, elastic behaviour. It follows the *same* curve when it contracts, therefore no plastic deformation.
23) A; The spring is being compressed, therefore there is a decrease in its length. The original length was 100mm. When the l is 70mm, the *change* in length = 30mm
so, the energy stored is area under the graph between 70 and 100 mm.
i.e. 1/2 x 30x10³ x 6 = 0.09J
24) C; the young modulus is same for a material.
25) B; p is not the height of the highest point on the wave so is not the amplitude. on x-axis is time, therefore q is time taken by one complete wave to pass through a point, time period.
26) D; I is inversely proportional to x² so doubling x means I would be divided by 4. I is directly proportional to A², so √(8²/4) = √16 = 4.0 μm
This cal also be simplified to this,
I1/ I2 = A1² / A2²
1/(1/4) = 8² /x²
4/8² = 1/x²
x²= 16
x= 4.0 μm
27) D; V= f λ where speed is c and wavelength is x+x+x+x =4x (as x is the distance between an antinode and a node, and one wave has three nodes and two antinodes).
28) B; d= 1/N
dsinθ = nλ
(1/N)sinθ = 3λ
29) A; λ = ax/D
re-arrange it to give, x = λD/a
this clearly shows that fringe spacing "x" is directly proportional to wavelength and inversely proportional to "a" (slit separation).
when λ is halved, x decreases, when a is doubled, x again is halved, so its like (3/2) /2 = 3/4 = 0.75 mm
30) A; the horizontal component of v is vcosθ
and u is in the same direction as vcosθ
re-arranging this gives u/cosθ = v
31) C;
E = V/d = F/Q
5000/0.8x10-² = E = 625000 N/C
now, Q=ne (where e is elementary charge)
625000 = F/5(e)
625000 x 5 x elementary charge = 5 x 10-¹³N
the drop has *gained* five electrons so is now negatively charged and will be attracted to the positive plate i.e. will experience an upward force.
32) B; P = I² x R
[(1/2)² x 2] / [1 x 1]
= 1/2
33) C; P= work done/ time
12 = w.d. / 50
w.d. = 600 J
v = work done/ charge
V= 600/100
V= 6.0V
34) C; R= ρL/A
which can also be written as R = 4ρL/πd²
the volume stays the same. L is increased four times, so the diameter is halved.
X has R1 = 4ρL/πd²
Y has R2 = 4ρ4L/π(d/2)² = 16ρL/(πd²/ 4)
ratio comes, (ignoring the constants) (16 x 4)/4 = 16.
35) C; e.m.f. = w.d./charge
w.d. = 12 x 4 = 48 J
P = w.d. /t
t = w.d. /P
t = 48/24 = 2 sec.
36) D;
basic concept of resistance of thermistor. More the resistance of a component, more will the voltage across it be. No other changes are made therefore those with LDRs rejected. when the temp. is reduced, resistance of thermistor increases and so does the p.d.
37) A; net resistance of R2 and voltmeter = 50kΩ
effecctive R of circuit = 150 kΩ
Total voltage across R2 and voltmeter is 1/3 rd of 6 V (same as ratio of resistances i.e. 50/150) = 2 V
V= IR
2= I x 100 x10³
I = 2 x 10-⁵ A
38) B; simple set up for a potentiometer.
39) C; estimations we're supposed to learn.
40) B; By beta decay, the proton nmbr increases by one and there is no change on the nucleon nmbr. After alpha emission, N decreases by 4 and Z decreases by 2.
On these graphs, x-axis is labelled Z and y-axis N.
thx
fr the question no.13 u need to multiply the radius and the weightso it will be
F is the torque on the lever. the distance between each forces of the couple is 1.2m.
for q.3, why is it nt 21?? (2 sig fig)
oooohhhh...he needed the torque, how stupid i am...thnk u very much
bt v have both the values to 2 sig fig dn y nt the ans *confused*
hey bro i cant get Q13....please explain thatt why to multiply raduis and not diameter with weight? :/
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