EJC 2022 GCE A Level H3 Physics 9814 Suggested Solutions

©EJC 2022 9814/H2 PHYSICS 2022 PHYSICS SUGGESTED MARK SCHEME 9814 October/November 2022 1(a) It is given in the formula list that the moment of inertia of rod about one of its ends is 21 3 ML=I uniform rod so CG is in middle of rod ( ) ( ) 2 2 sin2 1 sin23 sin 3 sin 2 1.5 3 2 L Mg L Lg L g L Mg ML k τθ α θα θα θ = =   = =  =  = I 1(b) by conservation of energy 2 2 2 1loss in GPE 2 1 cos22 s 3 co 3g L L MLMg τθ ω θω θω = =  =   = I

2 ©EJC 2022 9749/H2 PHYSICS 2022 2(a) the line integral of B.ds around any closed path equals μ0I, where I is the total steady current passing through any surface bounded by the closed path 0.dB s µ=∫ I 2(b)(i) in conductor ( ) 1 2 11 area J J rπ = = I since there is an opposite current in metal braid, ( ) ( ) ( ) 2 22 32 2 11 22 32 2 1 122 32 J r r J r r r J r r r π π π =− − − = − =− − I 2(b)(i)1. insulator no current flow ( ) 2 11 1 1 1 . 2 2 dc c c B JB s r r r JB µ µπ π µ =  =  =  ∫I  2(b)(i)2. ( ) ( ) 2 11 2 2 1 1 . 2 2 di i i B s r r JB JB r µ µπ π µ = = = −   ∫I  Note: at plastic jacket, no net current flow B = 0 2(b)(iii)

3 ©EJC 2022 9749/H2 PHYSICS 2022 2(c) When a current passes through a standard electrical transmission cable, the magnetic flux density in the vicinity of the cable will be nonzero. If high frequency signals are passing through, the rapidly changing magnetic field will lead to EM waves being generated. Hence there will be energy loss. Note: Compare with the coaxial cable discussed in the previous sections. By design, the magnetic field outside a coaxial cable is zero.

4 ©EJC 2022 9749/H2 PHYSICS 2022 3(a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 max max max 2 max 0 2 si m cos sin 2 n2 d cos 2d2 d 0d co 0 s sin 2 si 2 150 9. n 2 2 cos sin cos is when 0 29 45 2290 m 2.28 9 k1 x y x x x s ut tu u s gs us g R RR Ru ut gst g t g t g u u R R g R θ θ θ θ θ θθ θ θ θθ θ θ θ θ = = =  = − = −  =    = = ° = ° = = = = = = = =

5 ©EJC 2022 9749/H2 PHYSICS 2022 3(b)(i), (ii), (iii), (iv) ( ) cos cos Ru t t u R θ θ = = tan h Rφ = ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 2 2 2 2 22 22 2 2 2 2 2 tan 2 cos tan2 cos 1 cos22cos 1 cos2 sin1c t os2 1 cos sin 2 sin 2 sin cos 2 cos an tan tan tan tan ta 2o n cs g t g hu t hu t RRu gRR u gR u uR u t g uu h R g R R g u g u g θ θ θ θθ θ θ θθ φθ θθ φθ θ θφθ θ θ φ   + = − −= − = −+ =−+  +== = + = ++  =− + = +  +  ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) 2 2 2 2 1 sin21 w cos2 1 cos2cos 2cos 1 cos21 cos2 sin2 2cotan tan hen 0, equation reduces to: 2 0 s 1 cos sin2 sin2 u R u g g u g θθθ θθ θφ θθ θ φ θθ φ θ  +   = + +   +  = + + + + = + =

6 ©EJC 2022 9749/H2 PHYSICS 2022 3(b)(v) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 21 2 21 2 2 tan tan d1 1 0 tan cos2d2 2 tan cos2 1 sin 2 tan 1 sin 2 tan 2 1 sin 2 cos2 sin2 sin2 sin22 1tan2 tan tan tan2 tan2 R R u g u g u g u g u g u g φφθ θ φθ θθ φθ θ φθ φ πφθ φ φ − − +     =    = + = ++  +  = ++    = ++ −   = +  ( ) ( ) ( ) ( )( ) ( ) 22 22 2 1 sin 22 1 cos 2 dset n 0d cos 2 0 2 2 42 ta 2 tan 2 u g R u g πθφ φ π θφ φ θφ θ θφ πθφ πφθ   +−  +  = +−   + = − = −= −= = −

7 ©EJC 2022 9749/H2 PHYSICS 2022 4(a)(i) each spring originally in 4.1 is subject to tension of T Mg= In 4.2, each spring at equilibrium is subject to 1 2 Mg but is being stretched by additional tension provided by hand. So will move up 4(a)(ii) In Fig. 4.2 at equilibrium, strings will be slack so effectively is 2 springs in parallel ( ) 4.1, eff 4.1, eff 00 0 4.2, eff 10 10 0 00 0 Fig. 4.1 : 12 2 222 2 Fig. 4.2 : 2 2 2 2 2 2 2 2 2 3 22 2 L Mg kk kk kLMg L k Mg L k kk LMg k LL LM k Mg k Mg g Mg kk L Mg kk L Mg k =  = =−= −  = − =   − = + = ++ − = −  + +    = + +−   + +  − = l l l l l l ll l

8 ©EJC 2022 9749/H2 PHYSICS 2022 4(b)(i) ( ) 10 20 21 1 1 00 0 0 2 2 5 222 3 22 33 4 14 3 14 4 22 8 4 2 2 6 4 M M k g Lk Mg k Lg kL LMg k L Mg k L L Mg Lk Mg Mg k Mgk = −  = − −  − = = ++  = − = = = = l l ll l l 4(b)(ii) 0 0 10 0 0 2 2 2 2 2 3 2 3 22 3 2 2 2 Mg Mg L k Mg k L Mg Mg k kk Mg MgL kk MgL k MgL Mg L L = = + + + + = − = −  +  = − = =    = + l l l l l

9 ©EJC 2022 9749/H2 PHYSICS 2022 5(a)(i) consider moment of inertia 2 2m L=  I consider restoring torque ( ) 2 22 22 22 2 22 2 2 2 2 4 2 2 k mL mL mL T mL T k k T τθα θα ωθ ω π π π = = = = =   = = =  I I I 5(a)(ii) ( ) 23 330 50 2 m 2 air gap 10 0 0.17 6 10 1 Mmr r r −− − ++ ×× = = += + 5(a)(iii) consider sector 3 1 1 1 11 1 10 41 0.00456 rad9000 41 0.1 0.01 0.00014 rad9000 4.1 1.8 0.0046 0.0002 rad 4.1 0.9 0.00013642 s sr sr sr s r r r s θ θ θ θ θθ θ − = =   = = + = ×= = = ∆ ∆∆= ∆∆ ∆+ + ≈  = ±     0.90 ± 0.005 m θ1 4.1 ± 0.1 mm

10 ©EJC 2022 9749/H2 PHYSICS 2022 5(a)(iii) 2 1 2 22 122 22 1 2 2 2 g g k LF GMm r GMmkL r GMmL r G F mL T Lr MT τθα θ π θ πθ = = = = = = = I 5(b)(i) consider sector 2 2 2 102 0.013 r ad 0.745 2 0.0065 rad 1 0.372 5 2 5 .6 1 sr s r θ θ θ −×= = = ° = ° = = = 5(b)(ii) 10 min 22 sec is three quarters of a period ( ) ( ) ( ) ( )( ) ( ) ( ) 2222 2 22 11 3 1 2 4 3 2 1.8 3 10 60 22 622 s4 2488622 s 3 0.0065 2488 3 6.58 10 k 0.12 m g s 76 158 Lr MT T T G ππθ − −− = += =   = = = = × 12.00 m θ2 15.6 cm 2θ2