Structured Essay. Answers

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Godwin Anthony
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1 Structured Essay Answers 0. (a) AVρ...(0) (b) Momentum of air mass = mv = Av ρ v = Av ρ...(0) (c) (d) (e) (f) Force exerted on air /Force exerted on fan = F = Av ρ - 0 mv - mv t F = = Av Reaction forces acting on P, Q, R and S spheres are same. 4R = Mg ρ...(0) Mg R =...(0) 4 As a backward force is generated when the fan rotates reaction forces on P and S increase while reaction forces on Q and R decreases.(0) Anticlockwise moment = Clockwise moment by Mg acting on fan x Av ρ H = Mg...(0) (g) mgx Aρ H V =, 80 V = V = 3.6 = 0.56 m s -...(0) (h) When the fan rotates speeds at places with different radius on its wing differ from each other.. According to V = rw, V is heigher at places with heigher r while speed is lower at places with lower r. / Velocities of air particles pushed by the wing are different....(0) Applying Bernoulli theoram for points A and B on same flow line Pressve = ρ...(0) H + 0 = v + H + P

2 Pressure of air particles in front of the fan Pressure of air particles in front of the fan 5 P = P = Nm =... (0) 0 0.(a) (i) Marking to the right of C and marking eyeto the left of A (if both correct)....(0) (ii)...(0) No marks for these forms. (iii) A, It in necessary the adjusted telescope to adjest the colemotor or....(0) To get parallel rays from A, (b) (i) Adequate. Because only reflactions occur from the surfaces of the prism when levelling the prism table or No emergence takes place through the prism....(0) (ii)...(0)

3 (iii) X, Y and Z...(0) (c) (0) (d) (i) A A any one form...(0) (e) (ii) A = = A = = A = (0) n = Sin Sin A+D ( ) A ( ) n = Sin 0 ' ( ) Sin 60 ( ) Give marks for correct substitution if above A is wrong....(0) 03. (a) Stating the constant volume gas law correctly...(0) 0 (b) (i) Drawing and naming the stirer...(0) 3

4 (ii) Stire well and when temperature rises 0 C above the temperature to be messured, remove the burner, stire again and bringing to the temperature to be measured... (0) (c) (i) Cappilary tube...(0) (ii) To reduce the volume of air at room temperature in the tube outside the flask... (0) (d) (i) For sensitive variation of air pressure...(0) (ii) Not to have any temperature difference between water in the water bath and air in the flask...(0) (e) (i) C Pa C. 0 Pa...(0) (ii) 7 P = =.9 0 T (0) 7 P. 0 = =.9 0 T 373 P T P T 4...(0) = (a)...(0) (b) Y X...(0) (c)...(0) 4

5 (d) (i) (a) Increased current (b) Increased battery (or voltage)...(0) (ii) (a) Increased number of the coil (b) Increased the cross section area...(0) (e) Faraday Law I Faraday Law II...(0) (f) (i) The indigation of the galvonometer...(0) (ii) To flow the magnetic flox through aolevids...(0) (g) (i)...(0) (ii) N s Y s = Vp Np 500 = = 4 V...(0) 0 5

6 Essay Answers 0. (i) M = kg 4 =.5 0 kg...(0) (ii) mv - mu F = t = = 3 0 N...(0) 5 / mass = 3 0 kg...(0) (iii) mv P = = t 4...(0) 8 =.5 0 W...(0) (iv) F = ma 6 5 F = F = N...(0) if velocity of air current is v, m = 3 5 V = 75V 75V V-0 = (0) - V = 6 0 = 45m s...(0) (v) Not adequate. Should increase. As density of air decreases when going up, mass of air transmitted per second should be increased....(0) (vi) () Should be perpendicular to each other...(0) Because torque on the sketch due to rotation of the rotar acts in a plane perpendicular to the axis of the rotor...(0) () τ = FV 5 =. 0 = F 4...(0) 6

7 F = N...(0) (3) If volocity of air currents is V, m = 3 v 3v v- = (0) - v = 00 m s...(0) 5 (0) (i) Power Area - - MT T = = MT L -3...(0) (ii) (iii) P = ρvi = ML LT MT ML T = M L T = ML T P ML T A = = = L πρvf ML LT T L = L...(0) P = ρvi = Substitution...(0)...(0) =.97x0-5 Nm -...(0) -5 ρ.97 0 A = = πρvf Substitution...(0) - =.07 0 m...(0) (iv) (iv) For threshold of pain, P = 9.7 N m 5 A =.07 x 0 m F = P A = = N...(0)...(0)...(0) 7

8 (v) (a) M B I f r r...(0)...(0) (b) For the case, reaching m v+v0 f = f v 350 f = 680 = 700 Hz (0) For the case, reaching B v f = f v-v5 340 = (0) 330 =7. Hz...(0) or V+V V-Vs = = 7. Hz 0 f = f =

9 (03) (a) (i) (ii) P = π + hρs X = =. 0 N m V π = t 8η 0. = (0.5 0 ) (P - 0 ) X π( 0 ) (. 0 -P X) 8η (0)...(0) P =.08 0 N m X (0) (iii) = 3 (0.5 0 ) ( ) t (0) t = 30 s...(0) (b) (i) πr σg a = πr (σ-ρ)g 3 3 σ-ρ a = g ρ...(0) = 0 = 5 m s (0) (ii) V = r 0 ( σ - ρ) 9η g...(0) ( ) 0 = (0.5 0 ) = 0.56 m s - = 9...(0) (iii).5 5 = t 9 t =.7 s 9... (0)

10 (C) (), F+u = m g+t...(), F+u+T = mg... () () - () T = mg - m s -T T = mg - m g...(0)...(0) T = π r σg - π(r -r )σg 3 T = π r σ g 3...(0) T = T = =.56 0 N -5...(0) - V 0 = 0.7 N s...(0) 5 0

11 (04) i) Varying negative potential of W...(0) ii) (). Air resistance or Air friction... (0) (). Influence of current... (0) iii) downwards because relatively +ve potential exists on lower plate or because electric field between plates is upwards...(0) iv) 0Hz...(0) v) any sinosoidal form...(0) (no need to have initial position from 0) vi) (). Mv = Qv x 9 x 0-3 x (4 x 0 7 ) =.6 x 0-9 v... (0) V = 45 V... (0) (). S = ut 4cm 4 x 0 - = 4 x 0 7 t... (0) mm t = 0-9 S s = ut + / at x 0-3 = 0 + / x a x (0) a = 4 x 0 5 ms - F=ma, F = 9 x 0-3 x 4 x (0) = 36 x 0-6 N 36 x x 0 3 F = Eq, E = =... (0).6 x E = x 0 3

12 > E = v V p = 9 x0 3 x 4 x (0) d 0.4 V p = 90 v... (0) vii) No magnetic field is generated around those charged particles when cathode rays travel.... (0) 5 (5)(A) i) a) l = 000 m r =.5 x 0-3 m p =.5 x 0-8 Ω m resistance of one cable R = pl A.5 x 0-8 x 0 3 =... (0) /7 (.5 x 0-3 ) R = 3.53 Ω resistance of 54 cables = = Ω... (0) b) resistance of aluminium cables = Ω resistance of steal cables = 4 Ω R = (0) or stating that equivalant resistance is closer to smaller resistance steel cables have no considerable affect...(0) c) As elastic modeulus of steel is grater than that of aluminium it has the ability to withstand tention... (0) ii) a) resistance of pump motor = 40 Ω R = 600 Ω V P = 36 Ω...(0) > 40 V M 36 Ω V = R Ω

13 p.d across the motor = 40 x 36 V = 7.37 V...(0) 38 b) Working power of pump = (7.37) 36...(0) = W = 436 W... (0) c) p.d required for the pump to have a power of.5 kw according to p = V /R V =.5 x 0 3 x 36 V = 3.38 V... (0) If R is the combined resistance in both wires required to supply the above p.d. V = IR R =.8 Ω ( R + 36 ) = (0) the resistance should have in one cable = 0.59 Ω Ω 0.6 Ω... (0) d) if the starting carent in I 40 = I x 36 I = = 6.67 A... (0) e) Using a starter switch... (0) ( Then the starting current reduce because a resistor is connected in series with the motor.... (0) 5 3

14 B) a) i) V pq = 5 V... (0) ii) I = 0... (0) b) i) V pq = 0V... (0) ii) I = x 0 3 = 8 x 0 - A or 80 ma...(0) c) V AB = 5 V i. V PQ = 7.5 V... (0) ii. I = 0... (0) V AB = 60 V i. V PQ = 0 V... (0) µ ii. I = x x 0 3 = 8 x x 0 - = 40 µ A... (0) d) 60-0 R = 0.5 k Ω I = = 80 ma 0 I = R min 3...(0) = R min min 0 R = 50 Ω...(0) 4

15 e) i) V BE = 0.7 V... (0) ii) I = = MA 4.3 x 0 3 x 0-3 A... (0) iii) V R = 5.0 = 5 volt... (0) iv) I = x (0) = 3 x 0 - A v) Vo 5 t... (0) 5 (6) (A) When the rate of heat supply from the coil to the metal cube and the rate of heat loss from the metal cube to the environment become equal, the temperature of the cube is constant. if the rate of heat loss is Q / t i) Q/t = k (θ θ 0 ) θ room temperature θ 0 k temperature of environment - cooling constant at 50 0 C 4 = K (50-30) 4 = k x 0... (0) k = 6/5 5

16 when applied for 35 0 C (q/t) 35 = 6/5 (35-30) = 6 W... (0) when applied for 40 0 C (q/t) 40 = 6/5 (40-30) = W... (0) when applied for 45 0 C (q/t) 45 = 6/5 (45-30) = 8 W... (0) ii) marking...(0) graph... (0) iii) Rate of heat loss from a heated object by convection is directly proportional to the excess temperature between the temperature of the object and the temperature of environment..... (0) (This law is true for larger excess temperatures when cooling under forced convection and smaller excess temperature when cooling under natural convection) iv) Heat supplied for metal cube = 4 x 3 x 60 = 430J... (0) in 3 minits Heat absorbed by metal cube = M.C. θ =. S (40-30)... (0) Loss of heat to the environment 0+ = 3 60 during this time interval... (0) = 8 60 = 080 J Heat given = Heat loss + Heat obtained by the cube S = S = - - S = 70 J kg C... (0) 6

17 v) a A a B a a a a Rate of heat loss by cube θ = t A Q t = m s A t A d d d 3 θ = a ρ s d t A k 6 a ( θ θ r ) = Q 3 dθ = a = k 6a ρ S ( θ θr )... () t A dt A θ... (0) Rate of heat loss by cube B Q dθ = m = s t B dt B = 8a 3 dθ ρ s dt = B k 6 4a ( θ θr ) Q 3 dθ = 8a = k 4a ρ S ( θ θr )... () t B dt B dθ () dt A = () dθ 4 8 dt B...(0) dθ dθ : = : dt dt A B...(0) 5 7

18 (B) i) If maximum wavelength of electrons emited from metal surface is λ max λ max C = f ϕ = hf 0 0 ii) Area of light beam = λ (0-3 ) = λ x 0-6 m Power density of light beam λ hc = = φ max 9-7 = m = 653 A 0 iii) Accordingly as the maximum wavelength that can eject electrons is 653A 0, electrons are not emitted when 7000 A 0 light falls on metal surface are emitted only by radiation of wavelength 6000 A 0. iv) Energy of 6000 A 0 light = = π (0 ) = π 0 m -3-6 For 6000 A 0 light concentration = = -6 π 0 π 00 = = 3.8 w m 7-0 = π Number of electrones relased v) Charge corresponding to e n flux = = = = 7.7 C vi) As lesser energy is used by photons to remove surface electrons, energy of produced electrons is heigher. As heigher energy is used to remove electrons inside the surface, energy of produced electrons is lower. 8

Fig. 2.1 I =... A [2] Suggest why it would be impossible for overhead cables carrying an alternating current to float in the Earth s magnetic field.

Fig. 2.1 I =... A [2] Suggest why it would be impossible for overhead cables carrying an alternating current to float in the Earth s magnetic field. 1 (a) Fig. 2.1 shows a horizontal current-carrying wire placed in a uniform magnetic field. I region of uniform magnetic field wire Fig. 2.1 The magnetic field of flux density 0.070 T is at right angles