CONCEPT RECAPITULATION TEST - I Paper 2 [ANSWERS, HINTS & SOLUTIONS CRT–I] 2014

other 10 Pages

Joint Entrance Examination (Advanced)

Concept Recapitulation Test

SAS

Contributed by

Srinivasan Aayushman Sood

AITS-CRT-I(Paper-2)-PCM (Sol)-JEE (Adv)/14
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
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1

ANSWERS, HINTS & SOLUTIONS
CRT–I
(Paper-2)

Q. No. PHYSICS CHEMISTRY MATHEMATICS
1.
C A, B, D B
2.
B A, D A
3.
B B, C, D A
4.
B A, D D
5.
D A, D B, C, D
6.
C B, C, D A, B, C
7.
A A, B A, D
8.
C B, C A, B, C, D
9.
B. D C
10.
C C A
11.
B A C
12.
A B A
13.
C B A
14.
A B A
15.
A D A
16.
C C A
17.
B A C
18.
A B D
19.
C D B
20.
D D D

ALL INDIA

TEST SERIES

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P
P
h
h
y
y
s
s
i
i
c
c
s
s PART – I

SECTION – A

2. v
max
=   mg/k
F
av
=
 
max
mv
2mg
T 4



3. W =
2R
2
0
R
4S
p 4 r dr
r
 
 
 
 

= 4p
0

3
7R
3
+ (4)(4S) 
2
3R
2

4. Work done by all the forces on the block equal to change in kinetic energy.

6. No effect of ‘a’ and ‘g’ on time period of spring pendulum.

7.
3 1
1


= R
2 2
1 1 8R
1 3 9
 
 
 
 

2 1
1


= R
2 2
1 1 3R
1 2 4
 
 
 
 


3 1
2 1
27
32






8. Work done by friction
x
0
dx
F ds mgcos
cos
    

 



= mg x = 20 J





N

f

mg

dx
cos
ds
 

dx

dy

ds



9. Field inside the conductor is zero

10. Charge distribution is shown in figure.

Q

s
1
Q

Q

O

O

O

O

s
2

11.
0
( 3mg)/(m/ ) 2mg/(m/ )



 


0
3
2
  

12.
m
2m g
2
v
m/
 

 
 


=
5
g
2


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13-14. If the bike does not slip, path of both the wheel
must be parallel with its direction as shown in the
figure.
R
1
= /tan
R
cm
=
2
2
1
R
2
 

 
 

=
2
4 tan
2tan
 



Net friction force f =
2 2
2
cm
mv 2mv tan
R
4 tan


 


c.m.

R
2
R
cm


R
1

Rear
wheel

Front

wheel





15. p
0
A = ma  a = p
0
A/m

16. p
0
A = ma  m = m/5
 m  m = 4m/5

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

SECTION – A

2.
C C
CH
3
H CH
3
H
 C C
O
CH
3
H
CH
3
H
5
PCl
 C C
Cl CH
3
HH
CH
3
Cl
 
anti conformer

3. Thermal stability of nitrates of alkali metals increases down the group. MgO is basic; SnO is
amphoteric and B
2
O
3
is acidic oxides.

4. Cell reaction is:









2
Tl s Cu aq Tl aq Cu s
 
 



Reaction quotient
2
Tl
Q
Cu


 
 

 
 
and Nernst equation is
0
cell cell
0.0591
E E logQ; at 298 K
n
 
To increase E
cell,
Q should be decreased. Which is decreased by decreasing [Tl
+
] and increasing
[Cu
2+
].

5. Statement ‘B’ is true for weak electrolytes only and in case of ‘C’ osmosis never stops so dilution
on solution side continues and no limiting value of equivalent conductance of solution is obtained.

6.


3 4 3 3 2 2
2
Pb O 4HNO 2Pb NO PbO 2H O
Before reaction 0.05 mole 2 mole
After reaction 1.8 mole 0.1mole 0.05 mole
residue
   
 




2
2
4
2Pb 8NaOH 2Na Pb OH

 
 
 

3 3 2
HNO NaOH NaNO H O
  
Moles of NaOH used to neutralise excess of HNO
3
= 1.8 mole
Moles of NaOH used with Pb(NO
3
)
2
= 0.4 mole
Total NaOH used = 2.2 mole.
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7. Coagulation value
1
Coagulating power

Less is the gold number, better is the protection power.
8. Maximum capacity or volumes of balloon
8
480 548.57 ml
7
  
Also, V
1
= 480 ml, T
1
= 278 K, n = 1 mole
(A) The balloon will burst at a minimum temperature (T
2
) when volume becomes 548.57 ml.
Using Charle’s law

1 2
1 2
V V
T T


2
480 548.57
278 T

 T
2
= 317.71 K
or T
2
= 44.71
o
C
Hence (A) is incorrect but (C) is correct.
(B) Pressure of the gas at 5
o
C having 1 mole and V = 480 ml
PV = nRT

480
P 1 0.0821 278
1000
   
 P = 47.5 atm
Hence (B) is also correct.
(D) Since V increase with increase in temperature from 5
o
C to 44.71
o
C, at which balloon bursts
and therefore pressure remains constant. Thus pressure of gas inside the balloon when it
bursts is 47.5 atm. Hence (D) is incorrect.
Hence (B, C) are correct.

Solution for the Q. No. 9 & 10
B is CO
2
(g)  the compound A is a ZnCO
3
and C is oxide of Zn, as it is white when cold and turn
yellow on heating. D is CaCO
3
which form Ca(HCO
3
)
2
which exist in solution only.

Solution for the Q. No. 11 & 12
Since ‘A’ react with NaOI, it must be
OH
OH
C
O
CH
2
Cl
other side products are
OH
O C
O
CH
2
Cl

and less likely to be
OH
OH
H
2
C
COCl

; B is
OH
OH
C
O
CH
2
NHMe

;

while ‘C’ is
OH
OH
CH
OH CH
2
NHMe

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13. The above paragraph shows that compounds (A to D) are

H
2
C
CH
2
CH
2
Cl
Cl
H
2
C
COOH
COOH
H
2
C
COOC
2
H
5
COOC
2
H
5
CH
3
COOH
(A)
(B)
(C)
(D)

In above problem concerned reaction is:

CH O
+ H
2
C
COOC
2
H
5
COOC
2
H
5

CH CH COOH
Cinnamic acid
(E)
 
 
 
3
i Py,
ii H O
iii





Benzaldehyde (D)

Hence (B) is correct.

14. Since compound (C) is CH
3
COOH. So,

CH
3
C OH
O
CH
3
C OH
O
4 10
2
P O
H O

CH
3
C
O
CCH
3
O
O
(F)
(ethanoic anhydride)

Hence (B) is correct.

15. (A) That is path independent e.g.
U q w;
  
U is state function, while q and w are path
functions.
(B) Work is a type of energy in transit, i.e. which appears only on surface/boundaries.
(C) w & q are not completely interconvertible.

16. (A) Work is path function.
(B) Initially there is no opposing force but opposing force increases gradually due to diffused gas.
(C) W = ab; (area inside the curve)
a = 3 litre, b = 1 atm.

17. Le-chateliers principle is applicable and for (R) P
total
=
2
CO
P
, which is equal to K
p
for the given
reaction.

20. Isodiaphers are those atoms for which (n-p) is equal. Isoesters are those species which contains
equal number of atoms as well as equal number of electrons.

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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. Normal vector to plane P
1
3
ˆ
k

Normal vector to plane P
2

ˆ ˆ ˆ
2i 4j 3k
 



ˆ ˆ
a 2i j
  


angle between
a

and vector
ˆ ˆ ˆ
i 2j 2k
  is given by
cos =




ˆ ˆ ˆ ˆ ˆ
2i j i 2j 2k
0
5 9
    



  =
2

.

2. f(x) = f(x)
Integrating log f(x) = x + k
f(x) = e
x+k

f(0) = 1
1 = e
0
.e
k

k = 0
 f(x) = e
x

g(x) = x  f(x) = x  e
x

 
1
x x
0
I e x e dx
 

1 1
x 2x
0 0
e x dx e dx
 
 

2
e 1
e (e 1)
2 2
    
2 2
3 e 3 e
2 2 2

   .

3. n (S) =
64
C
3

Let ‘E’ be the even of selecting 3 squares which from the letter ‘L’
No ways of selecting squares consisting of 4 unit squares =
7
C
1

7
C
1
= 49
Each square with 4 unit squares forms 4L – shapes consisting of 3 unit squares
 n (E) = 7  7  4 = 196
 p (E) =
64
3
196
C

4. x
2
+ 9y
2
+ 25z
2
= 15yz + 5 zx + 3xy
(x)
2
+ (3y)
2
+ (5z)
2
 (x) (3y)  (3y) (5z)  (x) (5z) = 0

2 2 2
1
2(x) 2(3y) 2(5z) 2(x)(3y) 2(3y)(5z) 2(x)(5z)
2
 
    
 

=
2 2 2
1
(x 3y) (3y 5z) (x 5z) 0
2
 
     
 

x  3y = 0, 3y  5z = 0, x  5z = 0

1 1
x 3y
 and
5 1
3y z

and
1 1
5z x


1 1 1 5 2
x z 3y 3y y
   


1 1 1
, ,
x y z
are in A.P. and x, y, z are in H.P.
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5. When z = n + 1 we can choose x, y from {1, 2, …, n}
 when z = n + 1, x, y can be chosen in n
2
ways and z = n, x, y can be chosen in (n – 1)
2
ways
and so on
 n
2
+ (n – 1)
2
+ … + 1
2
=
1
6
n(n + 1)(2n + 1)
ways of choosing triplets
Alternatively triplets with x = y < z, x < y < z, y < x < z can be chosen in

n + 1
C
2
,
n + 1
C
3
,
n + 1
C
3
ways.
There are

n + 1
C
2
+ 2(
n + 1
C
3
) =
n + 2
C
2
+
n + 1
C
3
= 2(
n + 2
C
3
) –
n + 1
C
2
.

6. z = ze
i

… (1)
z = ze

i

… (2)
 zz = z
2
 z, z, z are in G.P.

2 2
z z
2cos2
z z
 
   
  
   
   

 z
2
+ z
2
= 2z
2
cos 2
z + z = 2z cos .

7.
2 2
c
3
1 1
3 4

= c  c = 6, 1.

8. cos =
1 1
x
2 x
 

 
 
and cos =
1 1
y
2 y
 

 
 

since xy > 0  x +
1
x
 2 or  – 2
y +
1
y
 2 or  – 2
 cos = 1, cos = 1 …(1)
or
cos = –1, cos = –1 …(2)
 cos cos = 1
(cos + cos)
2
= 4
  +  is an even multiple of .
 sin( +  + ) = sin(2n + ) = sin
Also, sin = sin = 0

9. IAB =
2

, IAC =
2


i
2 1 4 1
2
2 1 4 1
z z z z
e
| z z | | z z |


 

 

i
3 1 4 1
2
3 1 2 1
z z z z
e
| z z | | z z |


 

 

 
 
 
2
3 12 1 4 1
2
2 1 3 1
4 1
z z
z z z z
e
| z z || z z |
| z z |


 

 


(z
1
) A

C (z
3

B (z
2
)

I (z
4
)
D
z z
1 2
2

 
 
 

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




 
 
3 1
2 1
2 2
4 1
z zz z
AB AC
z z IA




=
2
AB AC
IA AB
   
   
   

10.




 
2
3 12 1
2
4 1
z zz z
AD AC
2
IA AD
z z

   

   
   

(Since AB = 2AD)



4 1
z z

2
(1 + cos ) sec  =




3 1
2 1
z z
z z 

11. Equation of the tangent is

x y
sec tan 1
a b
   

Normal is ax cos  + by cot  = a
2
+ b
2

The normal at P meets the co-ordinate axes at
2 2
a b
G
sec , 0
a
 


 
 
and
2 2
a b
g
0, tan
a
 


 
 

 PG
2
=
 
2
2 2
2
a b
btan 0
sec asec
a
 


 
  
 
 

PG
2
=
 
2
2 2 2 2
2
b
b sec a tan
a
  

12. When tan  = 0
PG =
2
b
a

13. When 1 all rounder and 10 players from bowlers and batsman play number of ways
=
4
C
1

14
C
10

when 1 wicket keeper and 10 players from bowlers and batsman play number of ways
=
2
C
1

14
C
10

when 1 all rounder 1 wicket keeper and 9 from batsmen and bowlers play number of ways
=
4
C
1

2
C
1

14
C
9

when all eleven players play from bowlers and batsmen then the number of ways
=
14
C
11

 Total number of selections =
4
C
1

14
C
10
+
2
C
1

14
C
10
+
4
C
1

2
C
1

14
C
9
+
14
C
11
.

14. If 2 batsmen don’t want to play then the rest of 10 players can be selected from 17 other players
number of selection =
17
C
10

If the particular bowler doesn’t play then number of selection =
19
C
11

 Total number of selection =
17
C
10
+
19
C
11

15. Slope of tangent at any point P(x, y) is
dy
y
dx



dy
dx
= ky  y = e
k(x – 1)

Since a(y – 1) + (x – 1) = 0 is normal at (1, 1)  k = 1
Hence f(x) = y = e
a(x – 1)

16. Area bounded =
 
1
a
x 1
0
1 x
dx
1 e
a a

 
  
 
 

=
a
1
1
a e
a
2

 
 
 
 
sq. units.

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17. (P) Limit is easily reducible to
2
1
k
cosec
2

 
 
 
by L’Hospital rule which exist if
2
k
2
 1.
(Q) kx
2
+ (3 – 2k)x – 6 = (kx + 3)(kx – 2)
4 
3
k

 5  –
3
4
 k  –
3
5
.
(R) (2k + 1, k – 1) is an interior point
(2k + 1)
2
+ (k – 1)
2
– 2(2k + 1) – 4(k – 1) – 4 < 0
0 < k <
6
5
….. (1)
Centre (1, 2) and point (2k + 1, k – 1) must lie on opposite side of chord x + y – z = 0
k <
2
3
….. (2)
0 < k <
2
3

(T) x > –
5
2
, – 4 < x < 4  x 
5
, 4
2
 

 
 

2
5
16 x
log
2x 5
 

 

 
 1

2
1
16 x
5
2x 5




 x  (– , – 9)  [– 1, )
 x  [– 1, 4]

18. I =
2
tx
0
e dt




x
t
= u, then dx =
du
t

I =
2
u
0
du
e
t



=
2
u
0
1
e du
t 2 t






2
x
e dx




= 2
2
x
0
e dx


 

.

19. r
1
r
2
+ r
2
r
3
+ r
3
r
1
=
2
1 2 3
r r r
s 144
r
 
4(r
2
+ r
3
) + r
2
r
3
= 2r
2
r
3
= 144
 r
2
r
3
= 72, r
2
+ r
3
= 18  r
2
= 6, r
3
= 12

1
r s a 1
a 6
r s 2

   

Similarly b = 8, c = 10
 ABC is right angle triangle.
Smallest angle is
1
3
sin
4


 =
1
2
 6  8 = 24 sq. units
R = 5
|r
2
– r
3
| = 6
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AITS-CRT-I(Paper-2)-PCM (Sol)-JEE (Adv)/14
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
website: www.fiitjee.com
10

20. (P) Equation of normal is y = tx + 2at+ at
3
at P(t)
It intersect the curve again at point Q(t
1
) on the parabola such that

2
2
t t
t
  

Again slope of OP is
OP
2
M
t

Also, slope of OQ is
OQ
1
2
M
t

Since
OP OQ
1
4
M M 1
tt
   
 tt
1
= 4

2
t t 4
t
 
   
 
 

 t
2
= 2
(Q) P(1, 2), Q(4, 4), R(16, 8)
Now, ar(PQR) = 6 sq. units
(R) Equation of normal from any point P(am
2
, 2m) is
y = mx  2am  am
3

It passes through
11 1
,
4 4
 
 
 

 4m
3
+ 8m  11m + 1 = 0
 4m
3
 3m + 1 = 0
Now, f(m) = 4m
3
 3m
 f(m) = 12m
2
 3 = 0

1
m
2


Since
1 1
f f 0
2 2
   
 
   
   
has 3 normals are possible.
(T) Since, normal at P(t
1
) if meets the curve again at (t
2
), then

2 1
1
2
t t
t
  

Such that here normal at P(1) meets the curve again at Q(t)
 t = 1 
1
3
2
 

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