03’ AL Physics: Structural Questions Marking Scheme83-04)+MC ans(80-00)/… · 03’ AL...

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03’ AL Physics/Structural Questions/Marking/P.1 03’ AL Physics: Structural Questions Marking Scheme 1. (a) (External) force required F ext = T - F r = 2mg – 0.6mg (away from center) = 1.4mg (N) (b) (i) F = mω 2 r For ω min : (2mg – 0.6mg) = ) 1 . 0 ( 2 min ω m ω min = 11.8 rad s -1 For ω max : (2mg + 0.6mg) = ) 1 . 0 ( 2 max ω m ω max = 16.1 rad s -1 (ii) unchanged. (c) By conservation of energy τθ = ) ( 2 1 2 min 2 max ω - ω I (1.3 × 10 -3 )(2π × n) = ) 8 . 11 1 . 16 )( 10 2 ( 2 1 2 2 3 - × - (n = no. of rev.) n = 14.7 (rev) 2. (a) (i) With normal incidence, the path difference between the 2 rays is too large. The interference fringes are due to the interference of the rays from - bottom surface of the glass sheet and - upper surface of the plastic sheet (ii) The ray from the upper surface of the plastic plate suffers fixed-end reflection and this introduces a phase difference of π to the reflected ray. The physical path difference at A is zero. The rays from the upper surface of plastic plate and rays from the lower surface of glass sheet are exactly out of phase. They interfere each other destructively and produce a dark fringe. (iii) Consider the dark fringe at B, Path difference = 2t = 10λ t = 2 ) 10 600 ( 10 9 - × = 3.0 × 10 -6 m = 3.0 μm m Reaction Tension (T) Weight Friction (Fr) (F ext )

Transcript of 03’ AL Physics: Structural Questions Marking Scheme83-04)+MC ans(80-00)/… · 03’ AL...

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03’ AL Physics/Structural Questions/Marking/P.1

03’ AL Physics: Structural Questions Marking Scheme

1. (a) (External) force required Fext = T - Fr = 2mg – 0.6mg (away from center) = 1.4mg (N) (b) (i) F = mω2r For ωmin: (2mg – 0.6mg) = )1.0(2

minωm ωmin = 11.8 rad s-1

For ωmax: (2mg + 0.6mg) = )1.0(2

maxωm ωmax = 16.1 rad s-1 (ii) unchanged. (c) By conservation of energy

τθ = )(21 2

min2max ω−ωI

(1.3 × 10-3)(2π × n) = )8.111.16)(102(21 223 −× − (n = no. of rev.)

n = 14.7 (rev) 2. (a) (i) With normal incidence, the path difference between the 2 rays is too large. The interference fringes are due to the interference of the rays from - bottom surface of the glass sheet and - upper surface of the plastic sheet (ii) The ray from the upper surface of the plastic plate suffers fixed-end reflection and this introduces a

phase difference of π to the reflected ray. The physical path difference at A is zero. The rays from the upper surface of plastic plate and rays from the lower surface of glass sheet are

exactly out of phase. They interfere each other destructively and produce a dark fringe. (iii) Consider the dark fringe at B, Path difference = 2t = 10λ

t = 2

)10600(10 9−×

= 3.0 × 10-6 m = 3.0 µm

m

Reaction

Tension (T)

Weight

Friction (Fr)

(Fext )

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(iv) Fringe separation ≈ 10

107.1 2−×

= 1.7 × 10-3 m (b) - number of fringes increases / fringe separation decreases - a bright fringe is observed at A instead

3. (a) f1 = fc

vc −

f2 = 1fvcc+

= fvcvc

+−

∆f = ff −2 = fvc

vcvc+

−−−

≈ fcv2

(Θ v << c)

(b) (i) f = 5.0 × 106 Hz, ∆f = 1.5 × 103 Hz, c = 1.5 × 103 ms-1

∆f = c

fv θcos2

1.5 × 103 = 3

6

105.130cos)100.5(2

×°× v

v = 0.26 ms-1

volume flow rate : 4

2dv

π× = 4

)105.1(26.0

23−×π×

= 4.6 × 10-7 m3s-1

(ii) ∆f = c

fv θcos2

1.8 × 103 = 3

6

105.130cos)100.5(2

×°× v

v1 = 0.31 ms-1 v1A1 = 4.6 × 10-7

A1 = 31.0106.4 7×

= 1.48 × 10-6 m2

A = 4

2dπ = 23)105.1(

4−×π

= 1.77 × 10-6 m2 % change in cross-sectionl area = 17% 4. (a) (i) Yes. The motion brings about the change of flux linkage through the coil, by Lenz’s law an

induced e.m.f./current results and it would oppose the turning motion. (ii) No. The coil would remain at that position as there is no change in magnetic flux linkage through

the coil. (b) (i) B = 0.1 + 0.3 × t

Alternatively:

As ∆f ∝ v and v ∝ A1

1ff

∆∆

= 1vv

= AA1

3

3

108.1

105.1

××

= AA1

A1 = A65

% change in A = 17%

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(ii) dtdB

= 0.3

ε = dtNBAd )(− =

dtdB

NA−

ε = -(100)(π × 0.052)(0.3) And ε = IR (100)(π × 0.052)(0.3) = I (10)

I = 0.024 A (iii) 5. (a) A typical op amp has a very high input resistance (~ 1012 Ω), it therefore draws a minute current from

the circuit supplying its inputs. It has a very low output resistance (~ 100 Ω) which account for its very small loss in transferring the output voltage.

(b) (i) Comparator. (ii) Vo = Ao(V2 – V1) 13 = 12

510 VV −

12 VV − = 1.3 × 10-4 V or 130 µV

I/A

0 t/s

0.024

-0.032

5 10 14

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(iii) (I)

(II)

(c) (i) A feedback resistor.

(ii) Vo = )k 10

(1

21f R

VVR +

Ω− (resistance in kΩ)

i.e. –1 = )2

101

(1

f RR

−+−

and +3 = )1

101

(1

f RR +−−

On solving, R1 = 25 kΩ Rf = 50 kΩ 6. (a)

(b) There is no current supply to voltmeter or CRO in this experiment. or Induced charge on the probe of voltmeter and CRO cannot be eliminated.

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(c) The flame produces ion-pairs which eventually neutralize the induced charge on the needle so that it is electrically neutral.

The electric field is undisturbed and the electric potential is unaffected by the presence of the flame probe there.

(d) Unchanged. The electric potential is constant (as it is on an equipotential surface). (e) (i)

Distance from center of ball r/cm 10 11 12 15 17.5 20 Electric potential V/V 1000 900 800 600 500 400

r1

/cm-1 0.100 0.091 0.083 0.067 0.057 0.050

Axes labeled with appropriate scales Points correctly plotted Correct graph (ii) Induced charges on the bench surface and the experimenter (nearby earthed object) reduced the

electric potential. (iii) Work done = (1000 – 900) q = 100 q (J)

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E = rV

∆∆− =

210)1011(

1000900−×−

−−

= 104 Vm-1 7. (a)

(b) y = 2

21

sin gttu −θ

For y= 0, t = g

u θsin2 ……….(1)

x = u cos θ t ……….(2)

Sub. (1) into (2): R = )sin2

(cosg

uu

θθ

= g

u θ2sin2

For R’ = R, sin (180° - 2θ’) = sin (2 × 40°) 2θ’ = 100° θ’ = 50° (c) (i) By conservation of energy,

2)10(21

m = )6.1)(10(21 2 mmvc +

2cv = 68

vC = 8.25 ms-1 (ii) v2 = u2 + 2as 02 = (10 sin 40°)2 – 2(10)yD yD = 2.07 m (d) (i)

(ii) Angular speed decreases (due to air resistance).

Alternatively: v = u + at For vy = -u sin θ at E -u sin θ = u sin θ - gt

t = g

u θsin2

& R = u cos θ t

Alternatively: v2 = u2 + 2as vy

2 = (10 sin 40°)2 – 2(10)(1.6) = 9.32 m2s-2 vy = 3.05 ms-1 & vx = 10 cos 40° = 7.66 ms-1

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(iii)

Due to the spinning of the ball, the speed of air flow on side A is reduced compared to that on the

opposite side. By Bernoulli’s principle, a pressure force results as the pressure on side A is greater than that on the opposite side.

8. (a) (i) α-particle. The range in air is 60 mm. (ii) (I) No. of rectangle under the graph = 18 Each rectangle represents 10 × 100 = 104 ion-pairs. Total no. of ion-pairs produced by the radiation = 18.0 × 104 = 1.8 × 105 (II) Total energy of the radiation = 1.8 × 105 × (5 × 10-18) = 9.0 × 10-13 J

= MeV 10

1

106.1

100.9619

13

×××

= 5.6 (MeV) (iii) More ion-pairs are produced at the end of the track as the α-particle travels slower so that its

interaction with the air molecules is more effective.

(b) (i) No. of atoms of X in the sample = 233

1002.60.234

101 ××× −

= 2.56 × 1018 Minimum no. of radiation = 2.56 × 1018 × 4 = 1.03 × 1019

(ii) A = A0e-λt and t1/2 =

λ2ln

= 24 days

For activity to decrease by 10%, 0.9 = e-λt

ln 0.9 = tt 2/1

2ln− (t1/2 = 24 days)

t = 3.6 days or 3.1 × 105 s 9. (a) (i) Avoid the formation of bubbles adhering to the fishing line. (Accept any other reasonable answer)

(ii) ρ = Vm

= 6

3

10)0.56.7(

10756.2−

×−×

= 1100 kg m-3

No. of ion pairs = Area under the curve

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(iii) %100×ρρ∆

= %100)( ×∆+∆VV

mm

≈ %1006.21.0 × (or

6.22.0=∆

VV

, ans. = 8%)

= 4 % (b) (i) To ensure that the elastic limit is not exceeded/elastic deformation/the line returns to its original

length. (Accept any other reasonable answers)

(ii) Slope of the graph: 34.7

5.105.26−

= 3.64 cm kg-1 = 3.64 × 10-2 m kg-1

E = leAF

//

= 24 )100.6(

4

00.21−×π

gslope

= 1.9 × 109 (Nm-2)

(iii) AF

=

4)100.6(

101524−×π

×

= 5.3 × 108 N m-2

(c) (i) v = ρE

= 1100

109.1 9×

= 1300 (ms-1)

∴ t = vd

= 130010

= 0.0075 s or 7.5 ms

(ii) Breaking stress (AF

) remains unchanged, i.e. 5.3 × 108 Nm-2

Therefore 4kg 15

= 2m

m = 1.875 kg

0