Suggested Solution to 2010 JC2 Physics Prelims H2 P1
Section A (MCQs)1. / Ans: B
The mass of an average person is about 60 kg and it consists of mainly water. Molar mass of water
is 18 g = 0.018 kg.
60 kg consist of 60/0.018 = 3.3 x 103 moles.
Hence no. of molecules = 3.3 x 103 x 6.02 x 1023 = 2 x 1027 molecules.
2. / Ans: A
Units of (a/Vm2) is equivalent to units of pressure.
Hence units of a = Pa x (m3. mol-1)2 = Pa m6 mol-2
Units of b = units of Vm = m3 mol-1
3. / Ans: A
Time/s / 0 / 1.0 / 2.0 / 3.0 / 4.0 / 5.0 / 6.0
Speed/m s-1 / 0 / 1.0 / 2.0 / 3.0 / 3.5 / 4.0 / 4.5
Acc/m s-2 / 1.0 / 1.0 / 1.0 / 0.5 / 0.5 / 0.5
For the first 3 s, the acceleration is constant and equal to 1.0 m s-2.
For t = 3 to 6 s, the acceleration is constant and equal to 0.5 m s-2.
The slope in option A gives the best possible way to attain this acceleration.
4. / Ans: D
sy= uyt + ½ gt2
180 = 0 + ½ (9.81)t2
t = 6.06 s
sx = uxt
2000 = ux(6.06)
ux = 330 m s-1
5. / Ans: A
F = kx
mg sin θ = kx
x =
6. / Ans: D
Taking moments about B:
W (1.5) = R(0.90) + FA(2.0)
FA =
Since net force in the vertical plane = 0
12000 = 5850 + FB
FB = 6150 N
7. / Ans: B
Action and reaction pair must act on different bodies.
8. / Ans: A
Change in momentum,
Final velocity,
9. / Ans: B
Using
10. / Ans: B
By conservation of energy,
11. / Ans: C
Tension is greatest at the bottom of the circular path.
Fnet = T – mg
v = d / t = 2πrf = 2π (0.50)(1.9) = 5.969 m s-1
T = Fnet + mg =
12. / Ans: B
mg = L cos θ =>
Fnet =
= 3.0 m s-1
13. / Ans: C
For each moon,
The mass of Ekapluto is
Using the data of Moon A, 3.0 1023 kg
14. / Ans: D
The minimum distance between the moons, d = (3.0 2.0) 108 = 1.0 108 m
The maximum gravitational force between moons A and B,
4.0 1014 N
15. / Ans: C
The total energy E of a mass-spring system is
where k is the spring constant, is the angular frequency, and x0 is the amplitude.
Option A: E will be halved.
Option B: E will increase by 4 times.
Option D: E will increase by 8 times.
Option C: According to , doubling m does not affect k or x0. So E does not change. It is not advisable to use here, because will also change as we vary m (recall that 2 = k/m).
16. / Ans: B
Sticking the chewing gum on the back of the mirror does not cause damping, but it changes the natural frequency of the mirror. As a result, resonance will occur (“mirror shakes a lot”) at a different speed.
17. / Ans: C
“Heat is added… at a constant rate” means that the power P supplied is constant.
Energy needed to melt the substance of mass m, Qf = Ptf = mLf
Energy needed to vaporize the substance, Qv = Ptv = mLv
So,
18. / Ans: C
For an ideal gas, pVT
So, .
Also, the average KE of a molecule of an ideal gas =
So,
19. / Ans: C
If the wave is travelling to the left, its next position can be represented by the dotted line shown above. X can be seen to be moving from equilibrium to a displacement in the positive direction. This matches the displacement-time graph of X shown in Fig 19.2.
20. / Ans: C
By quick calculation of the 4 options given, it can be shown that only option C will meet the condition for constructive interference to occur (i.e. path difference = nλ, where n is an integer):
When D is moved 1.0 m further away from S1,
Path difference = = 2.0 = nλ, where n is 1.
21. / Ans: D
d sin θ = nλ
d = 4.45 x 10-6 m
22. / Ans: D
Electric field strength is zeroimplies, not .
23. / Ans: C
24. / Ans: A
25. / Ans: B
26. / Ans: C
I = 5 mA, Vacross diode= 0.80 V (from graph)
VR = (5x10-3)x 50 = 0.25 V
Vsupply = Vacross diode + VR
= 0.80 + 0.25
= 1.05 V
27. / Ans: D
When R increases, VLM drops, the potential drop per unit length across LM decreases, a longer length is needed to balance the e.m.f of the standard cell.
28. / Ans: B
29. / Ans: D
,E =E0 sin t
When the speed of rotation of the coil is doubled, is doubled, is doubledand E0 also doubled.
30. / Ans: D
Using Fleming’s Left Hand rule, the wire experiences a force to the East.
By Newton’s 3rd Law, the magnet will experience a force to the West.
31. / Ans: B
Input power = output power(Assume 100% efficiency)
IpVp = IsVs
Primary current, Ip = Is
=
= 0.75 A
32. / Ans: A
The equation is of the form V = V0 sint
Vrms =
=
= 2.1 V
Frequency, f =
=
= 60 Hz
33. / Ans: A
is a constant, depending on the type of metal. When the frequency of the incident photons is increased, Kmax will increase.
Constant intensity,
The area illuminated remains unchanged, so the increase in f will be accompanied by a decrease in (n/t), i.e. the no. of incident photons per unit time will decrease. As a result, the rate of ejected electrons will decrease.
34. / Ans: B
Let E32 = E =
The other possible transitions are:
E21 = 2E = 2= and E31 = 3E = 3=
These transitions will produce the wavelengths and .
35. / Ans: A
In general, the x-rays of the continuous spectrum are produced when the bombarding electrons lose their energy as they impact the metal target. To decrease the cutoff wavelength, more energetic x-ray photons must be produced. In other words, the energy of the bombarding electrons must be increased, and the most direct method is to increase the electric p.d. used to accelerate the electrons.
36. / Ans: C
A donor atom contributes an extra electron to the conduction band easily, because its energy level is close to the conduction band.
37. / Ans: A
38. / Ans: A
39. / Ans: B
Rutherford’s -particle scattering experiment deduced that the nucleus in an atom is very small, positive and heavy.
40. / Ans: D
Actual count-rate of the source, C0 = 90-10 = 80 min-1
After 12 hours, actual count-rate,
Observed count-rate is 57 + 10 = 67 min-1
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