www.eenadupratibha.net MODEL EAMCET (MEDICAL) PHYSICS - SOLUTIONS − i + 3^ j 82. A = 2 ^ t e − − A B =^ i +^ j − Scalar component of A along − − A.B − B = Ax = A cos θ =  θ n . a − B − Ax h b ( ti) ( a ) r up B − − A .B − − ^ Vector component of A along B = Ax B =  − − − B − B  B A .B B =  B2 d a = (2 + 3)(i + j) 5  =  (i^ +^j). 2 2 n e ( ) ( ) e . w () w w 83. y = x tan θ x 2 1 -  = 2 (tan 45°) 1 -  R 6 2 = 2(1)  3 4 u1 84.  =  v1 1 t e n 4 = m 3 u 4 . a h 1 ∴ v1 =  v2 - v1 e=  u1 } u1 v2 -  1 4  =  2 u1 u1 u1 = 2v2 -  2 3 u1 or v2 =  4 b i t m1u1 = m1v1 + m2v2 ra p u n e e . ww d a u1 m1u1 = m1  + m2  4 4 m1 3m2 m1 =  +  4 4 3 4 3 4  m1 =  m2 ⇒ m1 = m2 86. Contact force is independent of friction. w 3u1 ( ) L 87. m1 = m 0,  2 ( ) L m2 = m , 0 2 www.eenadupratibha.net www.eenadupratibha.net L m3 = m L,  2 ( ) mL 0 +  + mL 2 m1x1 + m2x2 + m3x3 xc =  m1 + m2 + m3 L =  2 =  3m L L m +0+ m m1y1 + m2y2 + m3y3 2 2 L yc =  =  =  m1 + m2 + m3 3m 3   2 L2  L 13L2 √  13 L 2 2 r = xc + yc =  +  =  =  4 9 36 6 √ √ t e n . a h b i √   88. VBottom = √  2gH = √ 2g(3R) = √ 6gR  ∴n=6 But VBottom = √ ngR t a r p u d a n I 2 2 TBottom = THor : TTop = (n + 1)mg : (n - 2)mg : (n - 5)mg = 7 : 4 : 1 e e . w w 1 1    mv2 2 k2 r2ω2 Translational KE 2 r2 2 89.  =  =  =  =  Rotational KE 1 2 1 1 k2  Iω  Iω2 w k2 r . a h 1 2  =  2 ( ) b i t 1 k2 Body having  =  is solid cylinder. r2 2 ( a r p) ) 1 90. W =  K x22 - x12 2 ( t e n u d a 1 =  (400) 225 × 10-4 - 25 × 10-4 2 1 =  × 400 × 200 × 10-4 = 4 J. 2 n e e . ww 91. 2l 3 w l 3 O θ B A h Loss of energy = Work done to overcome friction l mgh = µmg cos θ  3 ( ) l or mg l sin θ = µmg cos θ  3 ( or µ = 3 tan θ. www.eenadupratibha.net ) www.eenadupratibha.net 92. In artificial satellite, there will be no gravity. Then water will have spherical shape due to Surface Tension, since density is constant, volume remains constant. 4 3 t e 4Π 3 ∴  Π r3 =  ⇒ r = 1 cm n . a Surface area of sphere = 4Πr2 = 4Π. h b i Fnet = F2 - F1 = a(v22 - v21)ρ 93. v1 = a(2g∆h)ρ t a r = 0.01 × 2 × 10 × 1 × 103 v2 = 200 N p u d 94. At steady state QAO = QOB + QOC a n 3K(100 - θ) = 2K (θ - 50) + K(θ - 0) e e . w w 300 - 3θ = 2θ - 100 + θ 100°C A 400 = 6θ 3K O 95. vrms = √  3RT  = M = √ 2gR ve √     3KNAT √  = M √  3KT  . a h b i t m a r p vrms = Ve  3KT   m = √ 2gR u d a n e e . ww 3KT  m = 2gR w t e n K 200° or θ =  C. 3 w ∴ 2K 50°C B 2mgR T=  3K 96. W = JQ 60% Pt = Jms (∆θ) 3 5  Pt = Jms∆θ www.eenadupratibha.net 0°C C www.eenadupratibha.net 3 Pt 3 5000 × 840 ∆θ =   =  ×  = 210 °C 5 Jms 5 (1)(30)(400) 1 97. Slope of V-T curve ∝  Pressure t e Volume while going from A to B slope B n . a A decreases and pressure increases. h b i θA t a r 98. θB Temp. p u d 1  1 1 1   8  8 8 8 T A  T 6  T T -A  6  12 12 A -A   0 2 2 3 T T T 5T a) For  oscillation from mean, t =  +  +  =  8 12 6 6 12 a n e e . w w w t e n 3 T T T 4T T b) For  oscillation from extreme, t =  +  +  =  =  8 6 12 12 12 3 . a h 5 c) For  oscillation from mean, 8 b i t T T 7T t=  +  =  2 12 12 a r p 5 T T 4T 2T d) For  oscillation from extreme, t =  +  =  =  8 2 6 6 3 ( a) du n( ) e .e 3R 99. H = n cv (dT) = (1)  10 = 15 R 2 7R H' = n cp (dT) = (2)  5 = 35 R 2 w w w H' 35 R 7 =  =  H 15 R 3 www.eenadupratibha.net www.eenadupratibha.net 10 sinΠx 20  cos 96 Πt 101. y = 2A sin kx cos ωt = 2Π Π λ ∴ k =  =  or  = length of each loop = 20 cm λ 20 2 x = 15 cm lies in 1st loop ↑ ↓ x = 25 cm lies in 2nd loop. Phase difference for particles in successive loops = 180°. ( v + vο 102. n' = n  v ) () b i t a r v + 300 = 2v or v = 300 ms-1. p u d µg - 1  µg  -1 µl a n e t e n Pair 1.5 - 1 0.5 × 1.6  =  =  = -8 Pliq 1.5 -0.1  -1 1.6 Pair -8D =  = 1D. Pliq =  -8 -8 e . w w w 1 1 1 104.  =  -  f v u 1 1 =  +  -40 ∞ -1 =  or f = -40 cm 40 . a h b i t a r p u d a n e e . ww 100 100 P =  in cm =  = -2.5 D. f -40 βο 105. βm =  µ w n . a h Here vο = gt = 300 ms-1 v + 300 ∴ 2000 = 1000 v fliq 103.  = fair t e Width of 10 fringes in air ∴ Width 10 fringes in liquid =  µ 3 cm =  = 2 cm 1.5 Central fringe will not shift. Hence Y coordinates are 0, 2 cm. www.eenadupratibha.net www.eenadupratibha.net 106. i sin i sin r0 µO =  rο rE t e sin i sin rE µE =  n . a Here i is same h b i O E sin r0 < sin rE ∴ µO > rE. t a r( ) 107. W = MB (cos θ1 - cos θ2) p u d () 1 W1 = MB (cos 0° - cos 60°) = MB  2 1 W2 = MB (cos 60° - cos 90°) = MB  2 a n e e . w w ∴ W2 = W1 = 8 × 10-5 J. 108. w BH 60° r ' BH BV BV i) For True Dip, Tan θ =  = 1 BH ii) When rotated by 60°, BV For apparent dip, Tan θ' =  ' BH t e n . a h b i t BV Tan θ Tan θ' =  =  = 2 BH Cos 60° Cos 60° θ' = Tan-1 (2). a r p u d a 109. At steady state, no current flows through Capacitor Branch. E 3 3 V, 0.5 Ω i =  =  = 2 amp R + r 1 + 0.5 n e e . ww VCD = VCE = VAB = iR = 2 volt Charge on capacitor w D C Q = CV = (1 µF) 2 Volt = 2 µc. A 4Ω 1 µF 1Ω www.eenadupratibha.net E B www.eenadupratibha.net 110. ← d → y x -Q1 = -2 A t e Q2 = 8 n . a d 1 x=  =  Q2 5  +1 Q1 h b i d 1 y=  =  Q2 3  -1 Q1 1 1 3+5 8 x+y=  +  =  =  5 3 15 15 t a r p u d 111. Brown - 1 a n Black - 0 Green - 5 e e . w w Resistance = 10 × 105 = 1 × 106 Ω t e n Tolerance for gold = 5%. w i1 . a h 112. i2 b i t i2 a r p u d a 1 1 1 i1 : i2 : i3 =  :  :  = a1 : a2 : a3 = 3 : 4 : 5 R3 R1 R2 n e e . ww i1 i2 F12 ∝  r1 i2 i3 F23 ∝  r2 Here F12 = F23 w i1 i2 i2 i3 ∴ =  r r 1 2 r1 i1 3 or  =  =  r2 i3 5 www.eenadupratibha.net www.eenadupratibha.net 113. i a b R L di Va - Vb = L  + iR dt When current increasing L(+1) + 2 R = 8 t e → (1) When current is drcreasing L(-1) + 2 R = 4 → (2) n . a (1) + (2) gives 4R = 12 or R = 3 Ω then L = 2 H.  114. Z = √ R2 + (XL Xc)2 ∼ Z2 = R2 + (XL ∼ Xc)2 400 = 100 + (XL ∼ Xc)2  = 10  ∴ XL ∼ Xc = √300 √3  XL∼ Xc 10 √ 3 Tan φ =  =  t a r p u d a n R e e . w w 10 h b i  = √3 ∴ φ = 60°. t e n 115. KE = Kmax = E - W → (1) When enters normally into magnetic field, w . a h B2e2r2 KE =  → (2) 2m b i t B2e2r2 2m ∴  =E-W 2m(E - W) or r2 =  B2e2  √ 2m (E - W) r =  Be a r p u d a n e e . w( w) 1 1 1 =  =  116. For X:  2n 16 24 ∴ Number of Half life periods in 8 Hours = 4 w T1/2 for X = 2 1 1 1 =  =  For Y:  2n 2256 28 ∴ Number of Half life periods in 8 Hours = 8 ∴ T1/2 ( ) www.eenadupratibha.net for Y = 1. www.eenadupratibha.net 117. Truth Table for given logic gate Input Output A B Y 0 0 0 1 0 0 0 1 0 1 1 1 t e n . a h b i t a r It corresponds to AND gate. 119. When Amplifiers are connected in series p u d A = (A1)(A2) = (10)(20) = 200 Voutput a n ∴  = Voltage amplification = 200 Vinput e e . w w Vout put = 200(0.01) = 2. 120. For electron revolving in circular orbit t e n KE = x w . a h PE = -2x TE = -x b i t Here -x = -3.4 ∴ KE = x + 3.4 eV a r p PE = -2x = -6.8 eV. u d a n e e . ww w www.eenadupratibha.net www.eenadupratibha.net MODEL EAMCET (MEDICAL) CHEMISTY - SOLUTIONS 20 121. E = E0 + Ek; Given Ek = E ×  100 t e 80 ∴ E0 = E ×  100 n . a h b i λ 80 4 5 5 E 0 =  =  =  ; λ0 = λ ×  = 4000 ×  = 5000 A° E λ0 100 5 4 4 t a r 122. Palladium; atomic no: 46; Electronic configuration 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10  p u d 126. PbO2 is oxidising agent: PbO2 + 4 HCl → PbCl2 + Cl2 + 2 H2O PbO is basic oxide. 2 PbO + 4 HCl → 2 PbCl2 + 2 H2O a n 129. As molecular weight of the gas decreases, the curve shifts towards right hand side i.e., higher speed. e e . w w √ VCH tX 4  = VX tCH4 w 130.  MX 360 × 10  ;  = MCH4 120 × 15 √ t e n  MX  16 . a h √  MX  = 2; MX = 2 × 2 × 16 = 64, X = SO2 16 131. Pi = Pi°. Xi(l) ; Pi = Ptotal. Xi(g); b i t a r p Ptotal 500 5 XB(l) More Volatile is B.  =  =  =  XB(g) PB 600 6 u d a 1000w = Kf. m.i = Kf  i exp gmwW n e e . ww 132. ∆Tf ∆Tfexp gmw.W 0.96 60 × 30 i =  .  =  ×  1000 w 4.8 1000 × 6 Kf = 0.6 w (1 - 0.6)2 1- i x =  . n =  = 0.8; P = 100 α = 80 n- 1 2- 1 0.06 [P] 0.06 [R] 133. Ecell = E°cell -  log  ; Ecell - E°cell =  log  n [R] n [P] www.eenadupratibha.net www.eenadupratibha.net 0.06 1 0.06 1)  log  =  log 1022 = 0.22 6 (10-2)2 (10-3)6 6 t e 0.06 1 0.06 2)  log  =  log 108 = 0.24 2 2 (10-2)(10-3)2 0.06 (10-3)2 3)  log  = 0 6 (10-2)3 n . a h b i 0.06 (10-3)2 4)  log  = 0.06. 6 (10-2)6 t a r 134. For compound CXHY p u d Y ∆HComb + ∆fH = X. HCO +  HH2O 2 2 a n 6 ∆Hf C6H6 = 6 (-394) +  (-285) - (-3267) = + 48 KJ mol-1. 2 135. e e . w w Answer: 10 : 1. t e n 136. Stage I Calculation of KC w 2 HI (g) . a h H2 (g) + I2 (g) i X X X X r - +  +  2 4 4  X X X    e 2 4 4 b i t a r p u d a X X .  [H2] [I2] 4 4 1 Kc =  =  =  [HI]2 4 X X  .  2 2 n e e . ww w www.eenadupratibha.net www.eenadupratibha.net Stage II: Mole of HI added 'n' 2 HI (g) H2 (g) + I2 (g ) t e X X X  +n   2 4 4 X X X r -  +  + 2 4 4  X X   e n 2 2 n . a i 1 Kc =  4 h b i t a r X X  .  2 2 =  n2 X X n2 =  .  . 4 = X2 2 2 p u d a n e e . w w n=X t e n 137. For zero order reaction unit of rate and rate constant same; mol L-1 S-1. w A B ohv thv n ( atoms in hcp) n 2n 138. 4 4n . a h b i t 3 : 3n 4 n: n 3 4n 2 ∴  =  (2n) 3 3 a r p u d a n e e . ww 2 atoms 'A' are in  of thv. 3 VO (STP) 0.56 2 149. Vol. Strength =  =  = 0.112 5 VH2O2 w 11.2 V ≡ 34 g L-1; 0.112 V = 0.34 g L-1 = 340 mg L-1 = 340 ppm. www.eenadupratibha.net www.eenadupratibha.net t e n . a h b i t a r p u d a n e e . w w t e n w . a h b i t a r p u d a n e e . ww w www.eenadupratibha.net