FULL TEST – I Paper-1 [ANSWERS, HINTS & SOLUTIONS]

other 11 Pages

Joint Entrance Examination (Advanced)

Miscellaneous

ZWM

Contributed by

Zara Wahid More

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ANSWERS, HINTS & SOLUTIONS
FULL TEST –I
(Paper-1)

Q.
No.
PHYSICS CHEMISTRY MATHEMATICS
1. B B B
2. B C A
3. A D A
4. B D A
5. A A C
6. A D B
7. C B A
8. D B C
9. B B A
10. A A A
11. D B C
12. B B D
13. B A C
14. A B A
15. B C C
16. C A C
17. A B A
18. C A B
19. C C C
1.
(A)  (r, s), (B)  (r, s),

(C)  (q, s), (D)  (p, s)
A → (q, r) B → (q, r)
C → (p, s) D → (p, s)
(A)  (s), (B)  (p),
(C)  (q), (D)  (r)
2.
(A)  (p, s), (B)  (p,
s),
(C)  (q, s), (D)  (p, s)
A → (r, s) B → (r, s)
C → (p, r) D → (p, q)
(A)  (q), (B)  (s),
(C)  (p), (D)  (r)
3.
(A)  (p, q, r, s),
(B)  (q, s) (C)  (q, s),
(D)  (p, q, r, s)
A → (q) B → (r)
C → (s) D → (p)
(A)  (q), (B)  (s),
(C)  (p), (D)  (r)

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P
P
h
h
y
y
s
s
i
i
c
c
s
s

PART – I

1. Now downward force on the right block is
more.

mg

T

T

mg

2. I
C
= I
0
+ M(OC)
2
= I
0
+ M(OB
2
+ BC
2
) = I
B
+ M(BC)
2

3. Potential difference across AC is zero as
I
AC
= 0
5  2I = 0
I = 2.5 A
Let the resistance of part BC be r
Applying KVL
10 + 5  2I  Ir  I = 0
2.5r = 7.5  r = 3 

G

E
1
= 10V

r
1
= 1
E
2
= 5V

r
2
= 2
A

B

C

As resistance of part AB = 9
 Length AC = 66.7 cm

5. Initial energy of electron = 2eV
Energy after formation of hydrogen atom in the ground state
= 13.6 eV.
Energy released = 2  (13.6) = 15.6 eV

hc
793
15.6
   Å

6. Threshold wavelength = 5000Å
Work function =
hc

= 2.48 eV
K.E. = eV = 3 eV

hC
2258
5.48
   Å

7.
2
V
P
R

 R =
2 2
2
V 300
900
P 100
  
 current
V 300
i
R 900
  =
1
A
3

Let L be the required inductance, then

500 1
z 3

or
2 2
2
500
(900) x

=
1
3

L
X 1200
 

1200 1200 1200
L 4H
150
W 2 f
2
   

 


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8. In steady state potential difference across each capacitor = E

9. H =
 
2
4 4
2
0 0
6t dt
E
dt
R 12

 
= 64 J

14. q
1
+ q
2
= 0
v
A
=
1 2
kq kq
kQ
R 2R 4R
 
v
C
=
1 2
kq kq
kQ
4R 4R 4R
 
v
A
= v
C

 q
1
= Q/3 and q
2
= Q/3

q
1
Q

q
2
B

A

C

15. v
A
= k
0
Q Q Q Q
3R 2R 12R 16 R

 
  
 

 

16. v
B
= k
0
Q Q Q 5Q
6R 2R 12R 48 R

 
  
 

 

17-19. When current is maximum
di
0
dt


 emf across L = 0 and potential difference across the capacitor will be same.
From conservation of charge
3CV + CV = 6CV
0
 CV
0

 V =
0
5CV
4

Loss in energy of capacitor = energy stored in inductor

 +

V

+ 

V

C

3C

S

L

 I
max
=
0
3V
3C
2 L

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

1.
Ag I AgI e
 
  

o
E 0.152 V


Ag e Ag
 
 
o
E 0.800 V
 
Ag I AgI
 
  E
cell
= 0. 952 V
At eqm,
o
SP
2.303RT 1
E
K
Ag I
 
 
   
   

SP SP
2.303RT
0.952 logK 0.059 logK
F

   
SP
logK 16.13
  
2.
2
A t
   

on differentiating the equation we get:

 
1
dA
2AdA dt or A
dt 2A 2

  
    
Hence order is -1.
3.
A
B
r
1
0.5
r 2


 
. As it lies in the range 0.414 to 0.732. AB has octahedral structure like that of NaCl.





A B
a 2 r r 2 1 2 6 pm
 
     
Volume = a
3
= (6 pm)
3
= 216 pm
3

4.
     
2
2
BaF s Ba aq 2F aq
 




2
sp
2
sp
2
K
K Ba F F
Ba
  

   
  
   
 
 

Again
2
2
sp
K 2 Ba F
 
   
 
   

sp
2
K
F
2 Ba


 
 
 
 
 

5.
3
Co

1s
2
2s
2
2p
6
3s
2
3p
6
3d
6

6
3d
4s
4p
4d
3
sp hybridisation

6. Bond order of
2 2
N ,N ,NO ,NO,CN
  
and CN are 3, 2.5, 3, 2.5, 3, 2.5 respectively. Higher is bond
order smaller is bond length. Bond order of CO and CO
+
are 3 and 3.5.
7.
2
Cl /h
4 3
CH CH Cl

 
MgCl
2
Br /h

Br
KCN

C N
3
H O


COOH

8.
3 2
CH CH O


O
Br
H C
6
H
5

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
C
6
H
5
O
Br
C
6
H
5
O
Br

O CH
3
C
6
H
5

9.
3 2
O /Zn H O

O
O

OH
O
H
O
O
H
KOH
Cannizzaro reaction

 
Y
K
+

14.
   
o
4 2 2
1 3
H H CaSO . H O s H H O g
2 2
   
  
   
   



1
4 2
H CaSO .2H O s 833 kJ mol

  
 

= + 484 kJ for 1 kg
15.
o o o
H H S
    

= 17920 J mol
-1

o
p
G 2.303RTlogK
  

 
2
3
o
4
2
p H O
G
logK 7.22 10 p
2.303RT


  

2
3
H O
p 8.1 10 atm

 
16.
2
H O p
p 1, K 1
 

o
G 0
 
at eqm.

o
o o
o
H
H T S T 380 K
S

     


= 107
o
C
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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

1. x
2
+ y
2
– 2x – 2y – 1 = 0
x
2
+ y
2
– 1 = 0
Common chord is –2x – 2y = 0  2x + 2y = 0
y = –x
Points of intersection of circles are
1 1
,
2 2
 

 
 


2.
4
k 1
8
2
k
10 2 5
P(E)
8 7 14
C


  



3. The line can be written as y = mx and curve as x
2
+ y
2
= 4
Let, C(h, k) be a point on the circles and


A 3, 1
be
given point then,
h 2 3
3

 


h 3 2 3
  

k 2
m
3

 

 k = 3m – 2
Now, this point (h, k) lies on the circle



3, 1 A

C(h, k)

B

2

(

, m

)

1






(0, 0)

y = mx


 
 
2
2
3 2 3 3m 2 4
     

2 2 2
9 12 12 3 9m 4 12m 4
         






2 2
9 1 m 12 3 m 12 0
      





2 2
3 1 m 4 3 m 4 0
      

 
 
 
2
2
16 3 m 4 3 1 m 4 0
    

 
 
2
2
3 m 3 1 m 0
   

2 2
3 m 2 3m 3 3m 0
    

2
2 3m 2m 0
 

2
2m 2 3m 0
 



m 0, 3


4. Suppose m is an integer root of x
4
– ax
3
– bx
2
– cx – d = 0 as d  0
 m  0
(I) m > 0
m
4
– am
3
– bm
2
– cm = d
 d = m(m
3
– am
2
– bm – c)
 d  m
Also, m
4
– am
3
= bm
2
+ cm + d
m
3
(m – a) = bm
2
+ cm + d
 m > a  contradiction
(II) m < 0
m = –n
 n
4
+ an
3
– bn
2
+ cn – d = 0
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n
4
+ n
2
(an – b) + (cn – d) > 0
So, contradiction
Hence, equation has no integral solution

5. (I)
2
1
x x
2
 

2
1 1
x 1
2 4
 
  
 
 


1 3
x
2 2
 

(II) x
2
– 3x + 2

(III)
1
x 1 x 1
2
   

 set x can be


1 3
,
4 4
 
   
 
 

1
+
1
–
1
+

1 3
2 2


1 3
2 2


+ – +
1 2

+ + – – +

1
+
1
–

1
–
1
+
1
+

1

2

1 3
2 2


1 3
2 2


2

–1 1
–2x

1
2x
2
 


1
x
4
 

1
2x 1
2
  


3
x
4
 

6. Required area,
/ 2
0
dx
A 4 y dt
dt

 


=
 
/ 2
2
0
a
4 asint cos t dt
sint

 
 
 


=
/ 2
2 2 2
0
4a cos t dt a

 


x


x

y

(0, a)

0

(0, –a)

y


7. Distance from origin,
2 2 2
D x y z
   where P(x, y, z) is any point on the curve
D
2
= x
2
+ y
2
+ z
2
= x
2
+ y
2
+
2
xy

2
2xy 4
xy
 

 D = 2 and occurs at point(s)


1, 1, 2
and


1, 1, 2
  ,


1, 1, 2
 ,


1, -1, 2
 

8. Differentiating w.r.t. ‘r’, 2r dr = 2a cos 2 d,
2
r
a
sin2




2
r
rdr cos2 d
sin2
  



d
r tan2
dr

  

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So, differential equation of orthogonal trajectory,
d
r cot 2
dr

   

Solving
dr
tan2 d
r
   
 ln(cos 2) = 2 ln r + k  r
2
= c cos 2

9. Let






3 2
f x x bx cx 1. f 0 1 0, f 1 b c 0
         

So,


1,0
   . So,
 


1 1 2
2tan cosec tan 2sin sec
 
   

 
1 1 1 1
2
1 2sin 1
2tan tan 2 tan tan sin
sin sin
1 sin
   
 

     
    
     
 
 
 
     
 

2
2

 
   
 
 



as sin 0
 

10.
2
1
1 z

is continuous every where except where 1 + z
2
= 0
 z = i so when |z| < 1 then above points are excluded so f(z) is continuous

11. Continuity : for any point z,




0 0
0 0
z z z z
lim f z lim x x f z
 
  
. Non differentiable : for any point z,
 




z 0 z 0
f z z f z x
f ' z lim lim
z x i y
   
   
 
   
now
x 0
y 0
x
lim 0
x i y
 
 


  
and
y 0
x 0
x
lim 1
x i y
 
 


  

12.
 
1
1
2
m
y cos msin x
1 x

 


 
2 1
1
1 x y mcos msin x

 



2 2
2 1
1 x y xy m y 0
   

 
 


 
 
2 2
n 2 n 1 n n 1 n n
n n 1
1 x y n 2x y 2 y xy n.1.y m y 0
2
  
 

        
 
 

Simplifying we get,


 


2 2 2
n 2 n 1 n
1 x y 2n 1 y n m y 0
 
     

13.


 
3
2
3
dy 3at 2 t
dt
1 t




 
3
2
3
1
t
dx
2
6a
dt
1 t
 

 
 



Now
dy
dx
  at t  (2)
1/3

14.-16. Here,
v 2 K
P P P
 
  
so
v
P

will make acute angle with all the vectors from
2 3 k
P , P , ....., P
  

P
n
P
n–1
P
n–2
P
k
P
m

P
3
P
2
P
1

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So,


v 2 3 k 1
OP OP OP ..... OP 0

    
   

Again, if n is odd  n = 2k – 1

1 2k 1 k
OP OP OP

 
  

Now
k
OP

will make acute angle with all the vectors from
1 2 2k 1
OP , OP ..... OP

  



k 1 2 k k 1 2k 1 k k
OP OP OP ..... OP OP ..... OP OP OP 1
 
         
      
 

Now,


k 1 2 2k 1 k 1 2k 1
OP OP OP ..... OP OP OP ..... OP
 
      
      

=
1 2 n
OP OP ..... OP
  
  

Again,
V

makes acute angle with all the vectors
1 2 7
OP , OP , ....... OP
  

So,
1 2 7
V S OP + OP +.......+ OP 1
  
   


V S 1
 



17. Eigen values are roots of the equation A – X = 0  (A – I)X = 0
 |A – I| = 0

8 4
0
2 2
  

 

 ( – 4)( – 6) = 0
  = 4, 6

18. When  = 6

1 1
2 2
x x
8 6 4 2 4 0
x x
2 2 6 2 4 0
  
   
     
 
   
     
 
     
   

 2x
1
– 4x
2
= 0  x
1
= 2x
2


2
X C
1
 

 
 

19. |A – I| = 0

1 0 1
1 2 1 0
2 2 3
  
  
 

 
3
– 6
2
+ 11 – 6 = 0   = 1, 2, 3,
For  = 3

1
X C 1
2
 
 

 
 

 
which is orthogonal

SECTION – B

1. (A) Assume sphere as, x
2
+ y
2
+ z
2
+ 2ux + 2vy + 2wz = 0
Now P, Q, R are (–2u, 0, 0), (0, –2v, 0), (0, 0, –2w)
So, equation of plane is
x y z
1
2u 2v 2w
  
  

Since it passes through (1, 2, 3)
1 2 3
1
2u 2v 2w
  
  

If centre is (x, y, z)  (–u, –v, –w)
 locus is
1 2 3
2
x y z
  

(B) Assume equation of sphere as, x
2
+ y
2
+ 2ux + 2vy + 2wz = 0
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It passes through (, 0, 0), (0, , 0), (0, 0 )

r
u , v= ,
2 2 2
 
     

Radius = 1 
2 2 2
1
2 2 2
  
     
  
     
     


2
+ 
2
+ 
2
= 4 if (x, y, z) are coordinates of centroid
 x , y , z
3 3 3
  
  

2 2 2
4
x y z
9
  

(C) P  (, 0, 0), Q  (0, , 0), OPQ = 15º
 tan 15º =


….. (1)
Now if sphere is made with PQ as diameter (x – )x + (y – )y + z
2
= 0
 x
2
+ y
2
+ z
2
= ax + by ….. (2)
A plane through PQ parallel to z–axis is

x y
1
 
 
….. (3)
Using (1), (2), (3) we get x
2
+ y
2
+ z
2
= (x + y)
x y
 

 
 
 

 z
2
= xy(tan 15º + cot 15º) 
2
z
2xy 0
2
 

(D) Using family of planes any plane through line of intersection is x + y – 6 + (x – 2z – 3) = 0
Now, its distance from centre of sphere is radius
 
2
2
6 3
3
1 1 4
  

    
,  = 1,
1
2


 Equation is x + 2y + 2z = 9; 2x + y – 2z = 9

2. (A) Let p = cos x sin y cos z. As,
y z
2

 
, sin (y – z)  0

   
2
1 1
p cosz sin x y sin x y cos z
2 2
 
    
 

As, sin (x – y)  0 and sin (x + y) = cos z

 
1 1 2 3
p 1 cos2z 1 cos
4 4 6 8
 
 
    
 
 

(B) The equation can be written as
2
1
8 2 7
3u 8u 3 0 u
6
 
     
2
8 2 7
u
6
 

Clearly, u
1
, u
2
are negative so
1 2
x , x
2

  

  < x
1
+ x
2
< 2 as cot x tan x = 1 = cot x
cot x
2

 

 
 
=
3
cotx cot x
2

 
 
 
 


1 2
3
x x
2

  and another pair,
 
1 2 1 2
7
x' x' x' , x' 2
2

     

 x
1
+
1
x'
+ x
2
+
2
x'
= 5
Page 10

FULL TEST – I Paper-1 [ANSWERS, HINTS & SOLUTIONS]

FULL TEST – I Paper-1 [ANSWERS, HINTS & SOLUTIONS]

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