Physics (XII) Unit Test-04 PDF

Unit-4 UNIT TEST-04 Subject : Physics Class : XII Q.1 (2) Q.2 (1) Q.3 (1) Q.4 (1) Q.5 (3) Q.6 (1)...

Unit-4 UNIT TEST-04 Subject : Physics Class : XII Q.1 (2) Q.2 (1) Q.3 (1) Q.4 (1) Q.5 (3) Q.6 (1) Q.7 (2) Q.8 (1) Q.9 (2) Q.10 (3) Q.11 (3) Q.12 (1) Q.13 (3) Q.14 (4) Q.15 (2) Q.16 (2) Q.17 (1) Q.18 (4) Q.19 (4) Q. 20 (4) Q.21 (3) Q.22 (4) Q.23 (2) Q.24 (3) Q.25 (4) Q.26 (3) Q.27 (2) Q.28 (4) Q.29 (1) Q.30 (3) Q.31 (1) Q.32 (1) Q.33 (3) Q.34 (3) Q.35 (4) Q.36 (4) Q.37 (1) Q.38 (2) Q.39 (4) Q.40 (1) Q.41 (2) Q.42 (3) Q.43 (2) Q.44 (1) Q.45 (2) Q.46 (1) Q.47 (2) Q.48 (2) Q.49 (3) Q.50 (3) Q.1 (2) Q.6 (1) When  = 90° f = – 15 cm m = – 3 u=? v = + 3u 360 360 then   4 is an even number..  90 1 1 1 1 1 1      The number of images formed is given by v u f 3u u 15 n 360 1 360 1 4 1 3 1 1 1 (1  3) 1  90 =     3u u 15 3u 15 Q.2 (1) FG 4 IJ  15  u  u = 20 cm = H 3K Q.7 (2) vcos V vcos 25. Object vsin vsin Image. h0 = 7.5 cm f = 2 u = – 40 1 1 1 Q.3 (1)    v u f Height of man = 180 cm  Min. length of plane mirror for him to see his full 1 2 1 16  5 21   =  h v 25 40 200 200 length image = = 90 cm 2 200 5 m=–  21.  40 21 Q.4 (1) f = – 50 m=–2 h1 5 5 15 75   hI     1.78 v h0 21 21 2 42  2  v = 2u u Q.8 (1) 1 1 1 3 1     v u f 2u u Q.9 (2) 3f 3 4 u=  × – 50 = – 75 cm DBF = DF + DB water = 16 + 12 ×  32 2 2 3 Q.5 (3) Q.10 (3) |m| = 3 f 1  2  1 1  m= (a) f     1  R  R  f u  1  1 2  15 On value of focal length is there, number of images = 1 –3= 15 – u 1  2  1 1  – 15 – u = + 5 (b) f     1  R  R  1  1  1 2  u = –20 cm 1 UNIT TEST : Physics 1  3   1 1   3    1    cos = f 2  1   R1 R 2  2 2 Two values of focal length is there, so number of images are two   3 = cos1  2  =30 (c) As many reflections, that many images. Infinite 2   number of reflections will be there, so infinite number of images.  = 60° (d) Other side of lens have three different medium, so number of images are three Q.16 (2) I 1·5 II Q.11 (3) O O y > R 60cm 12cm 1  As cred  sin 1 R for I refracting surface 1 1·5 1 (1·5  1) c yellow  sin1   y v 60 12 solving, we get  As y > R v = 60 cm 1 1 1  for II refracting surface   sin  sin1 uy R v = + (60 – 24) = + 36 cm 1 1.5 1  1·5   c yellow     v 36 12 Solving, we get Q.12 (1) v = 12 cm Rainbow is formed due to dispersion of light where  distance from the centre is 12 + 12 = 24 cm all component clours got splitted into 7 colours. Q.17 (1) Q.13 (3) Lens Maker’s formula Angle of prism, A = 60º Angle of minimum deviation, ,m = 40º 1 1 1     –1    A  m 60º 40º f  R1 R 2  Angle of incidence, i   = 50 2 2 where, R 2  , R1  0.3m Q.14 (4) 1  5  1 1    –1 –  f  3  0.3   1 2 1    f 3 0.3 or f = 0.45 m i  r  90º Q.18 (4) r 1 + r2 = A If two thin lenses of focal lengths f1, f2 are placed in 2r = 60º  r = 30º contact coaxially, then equivalent focal length of com- sin 90º bination is  2 sin 30º 1 1 0 1 1 1 = f f ff = f f Q.15 (2) F 1 2 1 2 1 2 sin  = Power for the combination is 3 sin /2 1 1 1 f1  f 2    P= = f f = ff 2 sin cos = 3 sin 2 F 1 2 1 2 2 2 2  Unit-4 Q.19 (4) Q.25 (4) Magnifying power when image is formed at near point Q.26 (3) f0  fe  50  5 M =  f 1  D  =  1    12 Monochromatic light means light containing single e   5  25  frequency or single wavelength or only one colour. Here bulb, candle, sun all are poly chromatic Q. 20 (4) Laser Monochromatic f0 m= Q.27 (2) f e ; f0 = 150cm n = m’ n  ' 5500 11 f 0 150     fe = = = 5 cm m  6000 12 m 30 Length of telescope Q.28 (4) l = f0 – fe = 150 – 5 = 145 cm Q.29 (1) Q.21 (3) Q.30 (3)  Wa  Q.31 (1) d Change in optical path diff x = (µ – 1)t  6000 10–10 1  Phase diff  = 2 x 180 d  d = 0.03 mm 2 20 = × 0.4 × 5 × 10–6 = Q.22 (4) 600 109 3     10  I0  a1  2 Ires = I0 cos2   = I0cos2    Ires = 1  2   3  4  I max  a2  a1  a 2 =  a1   a a =6 Q.32 (1)  a 1  I min 1 2 t( - 1) = n  2  n 4  6  10 7 t  7 a1  1 0.5 = 5 a2 t = 4.8 m  (1) Q.23 (2) Q.33 (3) d = 0.1 mm = 10–4m, D = 1.2 m, Let intensity of light coming from each slit of a coherent  = 6 mm = 6 × 10–3 m source is I. D As first slit has width 4 times the width of the second  slit, so d I1  4 I and I 2  I 1.2   6  10 –3     5  10–7 m  5000Å 10 –4   2 Imax ( I1  I 2 ) 2 4I  I 9        I 2 Imin I1  I 2 2 1 Q.24 (3) 4I   1 sin  =  d 2 Q.34 (3)  d = 2 = 2 × 6500 × 10–10 Fringe width   . Therefore,  and hence  decreases = 13000 × 10–10 = 1.3 mm 1.5 times when immersed in liquid. 3 UNIT TEST : Physics Q.35 (4) 2  sin 30º  1 sini2 Only transverse waves undergo polarisation. As sound waves are longitudinal in nature, so they can’t be i2 = 45º polarised  = i1 + i2 – A = 0º + 45º – 30º = 15º Q.36 (4) Q.41 (2) m2 = 1 µ1 = 1.5 R = – 5 cm u=–3 1 1.5 1  1.5 – =  v = – 2.5 cm v 3 5 Normal of the mirror won't change if the rotating axis is r to the plane of mirror. Q.42 (3)  i & r are the same. Q.43 (2) Q.37 (1) Direction of wave is perpendicular to the wavefront. u = –4f, O =6cm, I = ? 1 1 1 4 Q.44 (1) By mirror formula   v f  f v  4f 3 Also Q.45 (2) 4   f        v 3  IR = 4I cos2        2cm  2  O u ( 6 ) ( 4 f ) Im ax    Q.38 (2)  I max cos 2   is the critical angle. 2  2  = sin–1 (1/) = sin–1 (3/5)  or, sin = 3/5.  = 2 tan= 3/4 = r/4 or r = 3m.  r  x  4 xD 500  10 9 1 y=   = 1.25 × 10–4m d 4 1  10 3 4m  S Q.46 (1) Hence, the correct answer is option (2). Q.47 (2) Q.39 (4) Constructive interference occurs when the path differ- ence (S1P – S2P) is an integral multiple of . 1 C or S1P – S2P = n n where n = 0, 1, 2, 3,...... Cw ng   Cg n w Q.48 (2) 1.5  3 D or Cw = 2×10 × 8  2.25  108 m / s 4 Shift x = [(µ – 1) t] d Q.40 (1) Here x = 5  D    (µ – 1) t  D 30º  5 i1=0 r1=0 45º=i2  d  d r2=30  5 = (µ –) t 4  Unit-4 Q.50 (3) t   = (µ – 1) nD 5 y d (0.5)(6  10 –6 ) 2   2 = 1.6 102  5 0.14  103   = 6 × 10–7 m  = 5600Å   = 6000 Å Q.49 (3) D  d  'D '  d '  '  1         0.6mm '    0.4mm  1.5 5