FULL TEST – I [ANSWERS, HINTS & SOLUTIONS FULL TEST –I]

other 11 Pages

Joint Entrance Examination (Main)

Miscellaneous

SRM

Contributed by

Sohail Rao Murthy

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

ANSWERS, HINTS & SOLUTIONS
FULL TEST –I
(Main)

S. No. PHYSICS CHEMISTRY MATHEMATICS
1.
A C C
2.
B A B
3.
D B B
4.
B D C
5.
C C C
6.
C B A
7.
A D A
8.
C B C
9.
B C B
10.
C D A
11.
B D C
12.
D A D
13.
C C C
14.
B D B
15.
B C D
16.
D C B
17.
C D A
18.
D B B
19.
D D B
20.
D C D
21.
D A C
22.
B A A
23.
A D C
24.
A D B
25.
A C B
26.
A C B
27.
D A B
28.
B A B
29.
D B B
30.
B B D

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P
P
h
h
y
y
s
s
i
i
c
c
s
s PART – I

SECTION – A

1. For full square about an axis passing through ‘O’ =
2
M
6


by symmetry for remaining portion it must be
2 2
3 M M
4 6 8
 

 
 
 

2. apply work-energy theorem

2 2
1 mg 1 mg
mgz K z K Fz 0
2 K 2 K
 
   
    
 
   
   
 
 

 z = 2F/K.

3. a = 2 + |t – 2|
for t  2
a = 2 – t + 2
a = 4 – t
dv = (4 – t)dt
v = 4t – t
2
/2
at t = 2, v = 6 m/s.
for t > 2
a = 2 + t – 2 = t

v t
6 2
dv tdv

 

v – 6 =
t
2
2
t /2
 
 

v =
2
t
4
2


at t = 4, v = 12 m/s.

10.
(a b) [a (a b)]
   
 
  
. Dot product is zero.

11. Say speed of boat is v w.r.t. water and speed of river is C. Then, distance travelled in ground
frame
= (c + v) 
1
2
hour + (v c) 
1
hour
2
= v  1 hour
= distance travelled by boat w.r.t. river.

12. In the shown frame the particle appears to be at rest.
 Net force on it must be zero. Therefore pseudo force must be
equal and opposite to the tension.’



P
O
m

L

13. Successive frequencies will differ by an amount v/2L

14. Linear magnification = v
2
/u
2

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15. Electric field between the capacitor plates =


2
1
0 0
2 2



 

E =
 
0 0 0
Q x x 1
Q 2x
2A 2A 2A

  
  

 Potential different E
d
=
 
0
d
Q 2x
2A
  



C

Q
+x


x


Q 2x
2C

 
 x =
Q
C
2
 

16. conservation of momentum

M 2gh m 2gh
 =MV
1
+ mV
2
(1)

1 2
V V
1
2 2gh

  (2)

2
V 3 2gh


2
2
V
h' 9h
2g
 

17.
1 2
1 1 1
C C C
 
=
0 0
x a b x
A A
 

 
 C =
0
A
a b



18. FBD of ‘B’ and ‘C’

a B

2g

T
a

C

3g

T

 T – 2g = 2a . . . (i)
and 3g – T = 3a . . . (ii)
T =
12g
5

for A

a

A

mg

T=2T

 T = mg
2 
12g
5
= mg  m = 4.8 kg.

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19. W =
/ 2
0
mgR
f.Rd
2

 
 

= 1 joule.

20. Coulombic force between them remains same.

21. Let us assume cylinder is no moving then
T + f
s
= mg sin 
T. R  f
s
R = 0
 f
s
=
mg 3
4
.
but (f
s
)
max
= N = mg cos 
= 0.4 . mg 
1 mg
2 5

 (f
s
) < (f
s
)
max
, our assumption is wrong. So, friction existing must be kinetic
f
k
= mg cos  = 0.4  mg 
1
2
=
mg
5

22. Circuit is forming a wheatstone bridge R
eq
= 2R
For maximum power transfer 2R = r.



R

R

R

R

R

4
R

r

23.  =
3
4

2
B
 =
2
d 3 dB
dt 4 dt

 
i =
2
3
R 4R




24. C =
0 0
RT 5RT
M 3M



dx = C. dt =
L 0
0
0
T T
5R
T x dt
3M L
 

 

 
 
 
 

L 0
2L 3M
t
5R
( T T )



A B
T
L
T
0
L
x
dx

25.
E
2
= E – ir
 i =
E
2r
. . . (i)
2E = i(3 + r) . . . (ii)
 r = 1 

26. dQ = dU + dW
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C = C
v
+
PdV
ndT
…(i)
Differentiating
2
TV constant


dV V
dT 2T
  …(ii)
PV = nRT …(iii)
solving eq. (i), (ii) and (iii)
C =
R
2

27. Equivalent circuit is

T T
90 V 20
2R R

  V = 50
0
C

P
R
R
T
V

2R
T
R
T
(90
0
C) (30
0
C)

C
C
h
h
e
e
m
m
i
i
s
s
t
t
r
r
y
y PART – II

SECTION – A

2. 2B atoms (ions are missing from every unit cell).

3. Only KI produces
I

ions in aq. solution.

6. Time period
3
3
n
z


3 3
1 1 2
3 3
2
2 1
T n z
;
T
n z



z
1
= z
2
for same element.
Hence,
3
1
3
2
T
1 1
T 27
3
 

7. From K
eq
value; % of -D-glucose is 64.3.

9. Leaving group is at  position to anion stabilizing group (i.e.
C
O
group.)

11.
O
OH
H


OH
OH
/Methyl
Shift


OH
OH

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12. All others are used for this purpose and they form cyclic acetal or cyclic thioacetals at
C O

group.

17. Bauxite is (Al
2
O
3
.2H
2
O)
Magnetite is (Fe
3
O
4
)
Bornite is (Cu
3
FeS
3
)
Cerussite is (PbCO
3
)

21.
NH Ac
X =
NH
SO
3
H
Ac
Y =
HN
NO
2
SO
3
H
Ac
Z =
NH
2
NO
2
Product P =

22.
 
2
4
NiCl

 dsp
2
, square planer and diamagnetic

 
2
3
4
Cu NH

 
 
 dsp
2
, square planer and paramagnetic

 
4
Ni CO
 
 
 is sp
3
, tetrahedral but diamagnetic

25.
2 2
CaF CaF
  
(solution) -
2
H O


= 4.05  10
–3
S/m

4 3
m
Conc 2.025 10 mol / dm


  


3 11 3
sp
K 4C 3.32 10 M

  

27. Heat supplied at constant pressure in the range of temperature, T is H and change in internal
energy is U hence
 
U 1 1
0.71 approx
H 1.4

  
 

28. (A) 2KClO
3
+ I
2


2KIO
3
+ Cl
2

(B) 5NaBr + NaBrO
3
+ 6HCl

6NaCl + 3Br
2
+ 3H
2
O
(C) CaOCl
2
+ 2NaI + 2HCl

I
2
+ CaCl
2
+ H
2
O + NaCl
(D) NaCl + H
2
SO
4


NaHSO
4
+ HCl

29. A and C possesses plane of symmetry. While D possesses centre of symmetry.

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M
M
a
a
t
t
h
h
e
e
m
m
a
a
t
t
i
i
c
c
s
s PART – III

SECTION – A

1.
n n
3 2
C C
  n = 5

2.
 
1/ 2
PA PB
PA PB
2

 
|PA| + |PB| > 2|PT| =
2 3

Maximum length occurs when PAB passes through centre
i.e. |PA| + |PB| = 4 (Maximum)
So, range is

2 3, 4



3. Length of tangent is =
2
1
z i 4
 

=
2
3 2i 4 9 4 4 3
     

4. Required number of sentences =
 
6 6 6
0 1 2
7!
C C C
2!
  

5. The given line can be written as y = x
Hence reflection is 3 + 2i

6.
 
n 1 n n n n
1 1 1 1
a a a 1 a a 1

  
 


1 2 100 101
1 1 1 1
S .....
a a a a
    
=
1 101 101
1 1 1
2
a a a
  
Since, a
101
> 1  [S] = 1

7. Height of the cone =
5
3

Radius of cone =
2
2
5 8
9 11
3 3
 
 
 
 

Volume =
2
8 5
11
3 3
   

   
   
=
3520
27


8. 4! = 24
5! = 24 × 5
6! = 24 × 5 × 6
1! + 2! + 3! = 9
So, 1! + 2! + 3! + ….. 100! = 9 + 24p
  = 9

9. |z| = O × O – I × I = –I
2

det(|z|) = |–I
2
| = (–1)
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10.
 
1 2 2
f x f
1 x x 1 x
 
  
 
   
….. (1)
Now replace x by
1
1 x

we get
 
1 1 1 2
f f 1 2 1 x 2 1 2x
1 x x x x
     
        
     
      
….. (2)
Replace x by
1
1
x

we get
 
1 2x
f 1 f x 2x
x x 1
 
   
 
  
….. (3)
From equation (1) – (2), we get
Now,
 
1 2
f x f 1 2x
x 1 x
 
   
 
  
….. (4)
From equation (3) + (4), we get

Again,
 
2x 2
2f x
x 1 1 x
 
 

 
x 1
f x
x 1



which is a one–one function

11. x = –a is the only asymptote to the given curve

a
2
a
A 2 ydx 3 a

  


N

(
–
a, 0)B

M

(a, 0)

12. M = e
y
, N = x · e
y
+ 2y
Now,
dM
dy
(keeping x constant) = e
y

=
dN
dx
(keeping y constant)
Hence, the given equation is exact
Now,
   
y y
f e dx g y x e g y
    


Differentiating w.r.t y we get,
y y
dg
x e N x e 2y
dy
     
i.e.
dg
2y
dy

Integrating g(y) = y
2
+ c
1
 f = x·e
y
+ y
2
+ c
1

So, desired solution is x·e
y
+ y
2
= 0

13.


b v b b c
   



4v b v b b c
   

v b c
 

v b sin c
 

2 v sin 3
 

3
sin
2
 
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1
cos
2
 


1
4v 2 1 b b c
2
 
    
 
 

 
1
v b b c
4
  

14. Area under curve and axis =
 
1
2
0
2
1 x dx
3
 

at point (, 1 – 
2
) tangent is
[y – (1 – 
2
)] = –2(x – )
 2x + y = 
2
+ 1 
2
1
P , 0
2
 
 

 

 
, Q  (0, 
2
+ 1)
Hence,


2
1 1
Q 2, 1 4x x
  





2 2 2
1 1 1 1 1 1
A 1 2x x 1 4x x 1 3x x
 
        
 

1
1
dA
3 2x
dx
   
1
3
x
2
 

15.
, 0
4

 

 
 
lies on directrix and centre be (x
1
, 0). Equation of one of the tangent is
x y 0
4

  

and chord is
x 0
4

 
, centre is (x
1
, 0)

1 1
x x
4 4
1
2
 
 
 
2
2
1
x
x
16 2
 
  =
2
2
1
2x x
8

  

2
2
1
3 x
x 0
2 16
 
  

1 2
3
x x
2

   Midpoint 
3
, 0
4

 
 
 

16. 2x + y – 1 is the axis and (1, –1) is the point of intersection with the tangent at the vertex, length
of latus rectum is
5

So, coordinates can be (h, k)

5 1 1 3 5
h 1 1 ,
4 4 4 4
5
 
 
    
 
 
 
 


5 2 1 1 3
k 1 1 ,
4 2 2 2
5
 
 
        
 
 
 
 

3 1
,
4 2
 

 
 
and
5 3
,
4 2
 

 
 
are required points

17. For the common tangent m
2
sin
2
 + cos
2
 = m
2
+ 1
m
2
(sin
2
 – 1) = 1 – cos
2


2
2
2
1 cos
m
sin 1
 

 
=
2
2
2
cos
cot
sin

   


Clearly no such m is possible (when
4 2
 
 
  
 
 
)

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18. Let midpoint is (h, k) then chord is
h
y x 2h
k
   and if it is a tangent then
2
h
x 4 x 2h
k
 
  
 
 

D = 0

2
h
k
2
 
(Parabola)

19.
2 2 2 2
1 1 2 2 1 1 1 2
           


1 1 2 2
1
     

20.
 
 


 
2
n n 1
x / 2
0
x
g n e d x
2
 



=
 
 


1
2 x / 2
0
x
n n 1 e dx
2
 


=
 
1
2 x / 2
0
x
n n 1 e dx
2
 
 
 
 


= n
2
+ n + 1[4 – 2e
1/2
]
So, minimum value is
12 6 e


21. f(0) < f(1) now if suppose
 f(n – 1) < f(n)

   
f n 1 6 f n 6
   

 f(n) < f(n + 1)
Also, f(0) < 3, let f(n) < 3

 
f n 6 3 6
  

 f(n + 1) < 3
So, f(n) is bounded above 3

22. Any plane containing the line is
1 2
p p 0
  
. It’s distance (d) from origin
 
2
1
a a'
 

 

.
Let
2
2
2
2 1
d
A B C
   
  
   
where
2
A a' , B 2 aa'
 
 
and
2
C a


.







2
A 1 B 2 C 1 0
           
. As
R
 
,
D 0




2
4 A C B
p
0
q
4AC B
 
   

. So,
max
p
d
q
 .

23. We have f(x + a) 
1
2
and so f(x) 
1
2
 x  R.
Lets set, g(x) = f(x) –
1
2
we have g(x)  0  x  R

 
 
 
2
1
g x a g x
4
  

 
 
2 2
1
g x a g x
4
   =
   
2
1
g x 2a g x a
4
   

Page 10

FULL TEST – I [ANSWERS, HINTS & SOLUTIONS FULL TEST –I]

FULL TEST – I [ANSWERS, HINTS & SOLUTIONS FULL TEST –I]

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