Download Aiits 2016 Hct Vi Jeem Jeea Main Solutions Solutions PDF

TitleAiits 2016 Hct Vi Jeem Jeea Main Solutions Solutions
TagsPhysics Physics & Mathematics Trigonometric Functions Ellipse
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Page 1

AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16


FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
website: www.fiitjee.com

1







ANSWERS, HINTS & SOLUTIONS


HALF COURSE TEST-VI

(Main)






Q. No. PHYSICS CHEMISTRY MATHEMATICS
1. A B A

2. B A A
3. B D B
4. B D A

5. B B D
6. C B A

7. AA D B
8. D D B
9. C B B
10. B B B

11. D B B
12. B D D

13. B B B
14. B D C
15. A B A

16. C B C
17. C B A
18. D B B

19. D D D
20. B B C
21. D C B

22. C C A
23. C C B
24. C A B

25. A A A
26. C A A
27. C C B

28. C A


D
29. C D B

30. B C A


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Page 2

AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16


FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
website: www.fiitjee.com

2

PPhhyyssiiccss PART – I




SECTION – A

1. Friction force on block A, fK = N = 10 N (opposite to v)

 acceleration of A =
10N

2.5
4kg

 m/s2 (opposite to v)

Time taken by the plank to come at rest,


2

10 m / s
t

2.5 m / s
 = 4 sec (using, v = u + at)

In 4s distance moved by the plank SA = 20 m (using v
2 = u2 + 2as).

To reach the other end block will have to move by 20 m + 40 m = 60 m. As B is moving with
constant speed


60 m / s

T 6sec
10 m / s

 


2. Reference frame is at the edge of the plane with the leftward direction as the x-axis. Let x0 be the

distance of the cart at t = 0
mg  T = ma
T = Ma

 a =
mg

M m
= 2.8 m/s2

Since, M has an initial velocity to the left, a is in fact retardation for M.
 V = 7 + (2.8)  5 = 7 m/s
i.e. after 5 sec the cart will have the same speed but in opposite direction.

x = x0 + 7  5 +
1
2

(2.8)  5

x = x0 + 35  35 = x0
Let the cart go by S to the left from its initial position
02  72 = 2(2.8)  S
 S = 8.75 m
 Distance covered in up and down journey = 8.75 + 8.75 = 17.5 m

3. Let the pressures be P1 and P2 of the spheres having initial temperatures 0C and 20C

respectively then.

Initially P1 = P2 
1

2

n
n

293
273



If 1 2P ',P ' be the pressures after rasing temperature then
1

1 2
2

P ' 283 293
1 P ' P '

P ' 303 273
    

thus the mercury pellet will be displaced towards the sphere at higher temp.

4. No eddy current form in case (i) & (ii) and hence motion is not opposed.

7. 3 2 3 4(1.2 10 10 10 10 ) (10 10 ) F       
= Buoyant force – weight
 F = 20 N in down ward direction


8. Kinetic energy is maximum at centre.

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7

CChheemmiissttrryy PART – II




SECTION – A

4. 2 2 6 1Na 1s 2s 2p 3s
3s1 orbitals has n = 3, 0
Number of radial nodes = n  – 1 = 3 – 0 – 1 = 2


6. 2 2 4 2
4 4 2 4

O O CH O

CH CH O CH

r P M n 16
r P M n 32

  =
3 16 16 3
32 2 32 4 2

 


7. Half life t1/2 = 140 days

Number of half lives in total time =
280 140

3
140




R = R0(1/2)
n

 3000 = R0(1/2)
3, R0 = 24000 dps


9.




11. Cl2
5 8

n 1.63
0.082 300


 




Mass of  1(s)Al 0.05 10 20 20 6 1.13 13.56 gm
      

So mass of (s)Al dissociated = 0.5 moles
So eq is established

(s) 2(g) 3(s)
3

Al Cl AlCl
2

 


3

1.63 0.5
2

  0.5

   
3

3 32
2 2p Cl2

0.88 0.082 300
K P 2.7

8


        



(Volume of the vessel is taken 8 lit approx even after eq is established).


13.

B

OH

OH OH

OH
2

CH2

CH2

OH

OH
 B

O

O O

O

H2C

H2C

CH2

CH2
24H O




14. (i)
2 2 3 2 2Na S O 2HCl 2NaH SO H O S    

(ii)
3 2 2 3 2 4 6KI 2Na S O KI 2NaI Na S O   

(iii)
 

3 2 2 3 2 2 3 3
white

2AgNO Na S O Ag S O 2NaNO  

2H O 2 2 4Ag S H SO
Black

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8

15. Stability of intermediate carbocation increases rate of reaction.

16. CH3

Cl

2

3

NH
NH

CH3

2NH

CH3

NH2

H

CH3

NH2

18.

O O

(Aromatic)



19.

4KMnO
H



O

O

O

OH

C

O

OH



21.  2 3 2 32Mg C 2H O Mg OH CH C CH    


24.
 
 

4
b 10

4

NH Cl
pOH pK log

NH OH
 

0.1
4.7 log 5.7

0.01
  

pH = 14 – 5.7 = 8.3

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13


1

4 2 3
1 sinx

 



1

sinx 1
4 2 3

 





(4 2 3) 2 3 3

sinx 1
4 4 2


   

 x
3


 or
2
3


.


20. Adding and subtracting the given relation,
we get 3 2(m n) acos 3acos sin       32 sinsin.cos3 aa 

3a(cos sin )   

and similarly 3(m n) a (cos sin )    

Thus, 2/3 2/3(m n) (m n)  

2/3 2 2a {cos sin ) (cos sin ) }       

2/3 2 2 2/3a {2(cos sin )} 2a     .


21. We have
n sinA

sinA nsinB
1 sinB

  



A B A B
2cos sinn 1 sinA sinB 2 2

A B A Bn 1 sin A sinB 2sin cos
2 2

 
 

  
  




A B A B

tan cot
2 2
 




n 1 A B A B

tan tan
n 1 2 2
   

    
.


22. We have equal roots, therefore 2 4   .
Now second equation 2x x 12 0    has a root 2, so put x 2  4 2 12 0     4 

Hence from 2 4   , we have
16

4
4

  

 ( , ) (4,4)   .

23. Let the roots are  and n

Sum of roots,
b

n
a

     
b

a(n 1)
  


.....(i)

and product,
c

.n
a

    2
c

na
  ....(ii)

From (i) and (ii), we get


2

b c
a(n 1) na

 
   


2

2 2
b c

naa (n 1)





 2 2nb ac(n 1)  .
Note: Students should remember this question as a fact.


24. We have 3 i (a ib)(c id)   

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ac bd 3   and ad bc 1 

Now tan-1
1b dtan

a c
      

   

1 1 1
b d

bc ad 1a ctan tan tan
b d ac bd 31 .
a c

  

 
    

           
 

n ,n I
6


   


25. We know that a alog m log n  m n or m n , according as a 1 or 0 a 1  . Hence for

z x iy 

(1/3) (1/3)log | z 1| log | z 1| | z 1|     | z 1| 
1

0 1
3

 
  

 


 | x iy 1| | x iy 1|    

 2 2 2 2(x 1) y (x 1) y    
 4x 0 x 0 Re(z) 0    

26. 2n 1 2n 1x y  is always contain equal odd power. So it is always divisible by x y .


27. Check through option, the condition
nn 1

n!
2
 

 
 

is true for n 1 .


28. H dtan  and H h dtan  


60 tan

60 h tan



 


60tan 60tan

h
tan
  

 





60sin( )

h
sin

cos cos
cos

  



 



 x cos sin   .





H=60m



d

h





29. o 2
2

h
tan45 1 T A h

T A
   




Hence 120 h 30 h h 45     m.



45° 45°
A B 30 T2 T1

T2 T1

h h

120 m


30. x 1 1 1 .....    to  

We have x 1 x 
 2 2x 1 x x x 1 0     


1 1 4 1 5

x
2 2

  
 

As x 0 , we get
1 5

x
2




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