PHYSICS (A-LEVEL)

PAPER I Marking Scheme

120marks (42%)

1.(a)(i)ke1 = ke2 + mg1M

(ii)T1 = k(e1 – y) = 5(e1 – y)1/2M

T2 = k(e2 + y) = 5(e2 + y)1/2M

(b)F = ma

T1 – T2 – mg = ma

k(e1 – y) - k(e2 + y) – mg = ma1M

1M

1M

(c)(i)Gravitational potential energy

(ii)T = 0.4s1M

1M

1M

m = 0.04kg1M

(d)(i)unchanged1M

(ii)increase1M

Total : 12M

2.(a)–Ve or negative1M

(b)1M

1M

(c)loss in electric PE/work done by E-field = qEr = mV0E/B1M

gain in KE = 1M

loss in electric PE/work done by E-field = gain in KE

1M

2M

(d) The speed Vs does not change. It is because the magnetic field does wok on

the changed particle.2M

Total: 8M

3.(a)=2.64mA1M

A1M

Vin = IBRB + VBE = 1.13V1M

(b)

1M

(c)2M

(d)(i) Ic = (6 – Vout)/RC1M

(ii) IE = (Vin–VBE)/RC

(iii) IC + IB = IE1M

IC = IE

1M

1M

(iv) before saturation.1M

Total: 14

4.(a)Alternating current in a magnetite field produces a force where direction and

magnitude varies at the same frequency as the a.c.1M

The wire is therefore undergoing forced vibration.1M

(b) (i) Increase the strength of the magnetic field1M

(ii) Increase the amplitude of the a.c.1M

(c)

2M

(d)–transverse Vs longitudinal1Ml

-stationary Vs travelling1M

-different speed /wavelength1M

Total: 9

5.(a)The disc cuts magnetic flux. Eddy current is induced in the disc. By Lenz’s

Law, the induced current tents to reduce the relative motion between the disc

and the magnet.2M

(b) lower1M

(c)

1M

(d)rate of cutting of magntic flux

2M

(e)(i) V + IR = 1.5 +2(2) = 5.5V2M

(ii)1M

(iii) Power output = Power input – Power loss

VI =

2M

(f)When the electromagnet is switch on, the disc cuts magnetic flux. Edding

current id induced.

Mechanical energy of the meter is converted to electrical energy and finally

converted to internal energy of the disc.

2M

Total: 14M

6.(a)I = P/V = 600/100m = 6mA1M

no. of e-1 per second = n = I/e = 3.75x10161M

(b)

2M

(c) Continuous background:

Some electrons are stepped by the target and their KE is converted directly to

x-ray.

The bombarding electron may be brought to rest in one or more collisions in

which thus give up energy gradually, thus emitted platters are if different

wavelengths.3M

Lines:

Some bombarding electrons have sufficient energy to knock out the inner

electrons of a target atom, leasing electrons of a in the shell, as electrons if

upper shells fill the vacancy, photons of definite frequencies are emitted.3M

Total:10

7.(a)(i)<P> = VrmsIrmscosA

60 = 220 * 05cosA1M

power factor = cosA = 0.5451M

(ii)A = cos-1(0.545) = 56.9o1M

(iii) 220 cosA = ImR

R = 220 * 0.545 / 0.5 = 240

220 sinA = IL

L = 1.17H1M

(b)(i)As the line switch is closed, gas discharge occurs inside the neon starter t

though temperature makes the bi-matallic strip bend towards the contact.

When the bi-matallic strip touches the contact, gasdischarge stops.

Temperature inside the starter decrease, as the bi-metallic strip restores to

its originalposition.

(ii)As the bi-metallic strip restores to its original position, the starter circuit

suffers from a sudden break. Large emf is induced due to the induction of

ballast.

(iii)It converts the u.v. light produced in the discharge process to visible light.

8.(a)

2M

(b) A2V2 = A1V11M

V2 = (1 x 40)/2 = 20ms-11M

Assumption: Water is incompressible.

(c)2M

(d)1M

1M

1M

= 780 Jkg-11M

(e)Power = W/t

=

= (780)(1000 x 1 x 40)1M

= 31.2MW1M

Total: 13M

9.(a) Hydrogen atoms in discharge tube are excited by E.H.T. to higher energy

states.1M

An atom in a higher energy state E1 will ‘fall’ to a lower energy state E2 by

emitting a photon whose wavelength corresponds to the energy difference

E1- E2.1M

(b)

(c)(i)1M

2M

color: green / blue/1M

(ii)The diffraction angles for 2nd order are larger than those for 1st order,

So error is smaller.1M

(d)

1M

1M

So, The bright line probably comes from the transition from n=4 to n=2.1M

Total: 13M

10.(a)(i)Consenation of Charge: Q = Q1 + Q2 ------(1)1M

Same p.d. across Co and C: ------(2)1M

(1) and (2) 2M

(ii)C2>C1Q2 = Q1M

(b)(i)Vout = Vin1M

(ii)original charge on unknown capacitor

= final charge on 10F capacitor

= 10F x 5V1M

= 50C1M

(c)(i)Consider Voltage gain:

Vout = Ao(Vin– Vout)

------(1)1M

Consider Ohm’s Law

------(2)1M

(1) and (2) ------(3)

As , 1M

As for discharging1M

(ii)time constant =(1 +Ao)CR

half-lift=In 2

= (1 + Ao)CRIn 21M

= (In 2)(1 + 105)(10x10-6)(2x106) = 385 hr1M

The is wrong rate of discharge can be neglected.1M

Total: 15M

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