Title Series XI Sol..pdf ✅

1 Nepal's Largest Medical Entrance Preparation Centre CEE Based Model Examination [SET – A] Hints and Solutions Date: 2082-05- 7 VIBRANT MBBS ENTRANCE PREPARATION New Plaza, Putalisadak, (Beside Nepal Yatayat Stoppage). Kathmandu, Nepal, Website: www.vibrantmbbs.edu.np Face book:www.facebook.com/vibrantmbbs Contact Number:1-4526149/4540184 Physics 1. A 2. A tan = Q sin P + Qcos  tan90 ̊ = Qsin P + Qcos  1 0 = Qsin P + Qcos  cos = -P Q By question, | P + Q| = P  P 2 + Q2 = 2PQcos = P2  P 2 + Q2 + 2PQ    -P Q = P2  Q = 2P 3. D After 2 seconds,vx= ux=20cos30o =10 3 m/s and vy=20sin30o–10×2=–10m/s v = vx 2 + vy 2 = 20 m/s  KE = 1 2 m v 2 = 1 2 × 0.5 ×202 = 100 J 4. C In equilibrium state, frictional force on part lying on the table is equal to weight of hanging part.  M L (L–x)g = M L xg  x L–x  x = ( μ μ+1 ) L 5. A Since acceleration is constant So S= (u+v 2 )t = (6+16 2 )x2 = 22m MBBS BDS B. Sc. Nursing Paramedical/Applied Sciences
2 6. D 7. D At depth x inside earth, the value of g is g' = GM R 3 (R – x) = g R (R – x) Hence g' g = R – x R 1 4 = R – x R  x = 3 4 R OR Remember: x = n – 1 n R =     4 – 1 4 R = 3 4 R 8. D 9. B 10. A 11. A 12. B If V0 be the volume at 0°C then, at 27°C, V, = V0 ( 1 +  × 27) ........... (1) at °C, V' = V0 (1 +  × ) ................ (2) Dividing equation (2) by equation (1) V' V = 1 +  1 + 27  2 = 1 +  1 + 27  2 + 54 = 1 +   1 + 54 × 1 273 = 1 273        = 1 273°C 327 273 =  273  = 327°C 13. C 14. C 15. D 16. A 17. B 18. B ƒ = 1 -1 , and   1 λ 19. C 20. D fl =     g –1 g – 1 × 1 × fa =       3 2 –1 3 2 – 5 4 × 5 4 × 12 =     1 2 × 4 1 × 5 4 ×12 = 30 cm 21. B For achromatic combination of lens. f f' = – ' which is possible if  = 0, ' = 2 and f' = –2f 22. A 23. D 24. B v = 2(L2 – L1)f = 2 (52 – 17) 100 × 500 = 350 m/s 25. B 26. D 27. D