CLASS XII [MAIN] ANSWER KEY WITH SOLUTION TARGET IIT-JEE PHYSICS MATHEMATICS CHEMISTRY DATE : 24-07-2016 [ All Batch ] SECTION – A 1. A 2. C 3. B 4. D 5. D 6. A 7. D 8. A 9. D 10. C 11. C 12. B 13. D 14. B 15. C 16. B 17. B 18. C 19. B 20. C 21. C 22. C 23. D 24. A 25. A 26. A 27. B 28. B 29. C 30. D SECTION – A 1. B 2. A 3. A 4. A 5. C 6. B 7. B 8. D 9. B 10. C 11. C 12. D 13. D 14. A 15. D 16. A 17. B 18. B 19. A 20. D 21. C 22. B 23. D 24. C 25. D 26. B 27. A 28. B 29. C 30. A 394 - Rajeev Gandhi Nagar Kota, Ph. No. 93141-87482, 0744-2209671 www. motioniitjee.com , [email protected]1 SECTION – A 1. A 2. C 3. A 4. D 5. B 6. A 7. B 8. C 9. D 10. A 11. C 12. A 13. B 14. D 15. A 16. D 17. D 18. C 19. B 20. D 21. C 22. A 23. D 24. D 25. D 26. C 27. A 28. B 29. B 30. B
12
Embed
TARGET IIT-JEEerp.motioniitjee.com/images/041d65de-15e8-4f23-b77... · class xii [main] answer key with solution target iit-jee physics mathematics chemistry date : 24-07-2016 [ all
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
E = kq/r2 outside a uniformly charged sphereand also for a point charge.
2. A
R
M R m
dx
x
F = 2R
2R
mGM dxR
x
= 2
GMm2R
3. AThe given circuit is equivalent to
As 10 × 4 = 5 × 8 this is balancedWheatstone network
Therefore R = 189)810()45(
= 6 ohm
4. A
Use C = dA0
A = 0
dC
5. CThe gravitational field at any point on thering due to the sphere is equal to the fielddue to single particle of mass M Placed atthe centre of the sphere. Thus, the force onthe ring due to the sphere is also equal tothe force on it by particle of mass M placedat this point. By Newton's third law it is equalto the force on the particle by the ring. Nowthe gravitational field due to the ring at adistance d = 3 a on its axis is given as
10. CFringe width . Therefore, and hence decreases 1.5 times when immersed in liquid.The distance between central maxima and10th maxima is 3 cm in vacuum. Whenimmersed in liquid it will reduce to 2 cm.Position of central maxima will not changewhile 10th maxima will be obtained at y= 4cm.
11. C
i = 50
20 RPotential drop across R = Potential dropacross AB
50 .R
20 R = 30 R = 30
12. DIf d sin = ( - 1)t, central fringe is ob-tained at O.If d sin > ( - 1)t, central fringe is obtainedabove O andif d sin < ( - 1)t, central fringe is obtainedbelow O. (D)
12. DAt earth's equator effective value of gravityis
geq = gs – 2 Re
If geff at equator to be zero, we havegs – 2 Re = 0
or = e
e
Rg
14. A
22 r16kq
)r4(kqE
q
y
x
2r
(0, 0)
+ + +++++
+++++
+++++
+ + + q
– ––––
––––
(4r, 0)
15. DAfter charging, isolatedSo Q = constant d increasing C
V C1
V increasing
16. AWe can observer that 10, 10 and 20are shorted and bypassed, so no current will
flow in then and I = 90V45
= 0.5 A.
+
–A
20
9010 1045V
Alternative :
+–
A90
10 10
20
–
–
I
–
A5.090
V45I
17. B
Shift 1 7 1 1 4 1
5t . D t . D DS
d d d
t = 8 × 10–3 mm = 8 m
18. BFrom symmetry flux through each point ofthe sphere is same.
21. Cproduct A is obtained by OMDM (i.e. additionof HOH acc. to morkovnikov’s without rear-rangement.)product B is obtained by HBO (Hydroborationoxidation) (i.e. final product is addition ofHOH acc. to antimorkovnikov’s without rear-rangement.)