FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Saturday 27th January, 2024) TIME : 3 : 00 PM to 6 : 00 PM
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. Considering only the principal values of inverse
trigonometric functions, the number of positive
real values of x satisfying
11
tan (x) tan (2x) 4
is :
(1) More than 2
(2) 1
(3) 2
(4) 0
Ans. (2)
Sol.
11
tan x tan 2x 4
; x > 0
11
tan 2x tan x
4
Taking tan both sides
1x
2x 1x
2
2x 3x 1 0
3 9 8 3 17
x88
Only possible
3 17
x8
2. Consider the function
f :(0,2) R
defined by
x2
f(x) 2x
and the function g(x) defined by
min{f(t)}, 0 t x and 0 x 1
g(x) 3x, 1 x 2
2
. Then
(1) g is continuous but not differentiable at x = 1
(2) g is not continuous for all
x (0,2)
(3) g is neither continuous nor differentiable at x = 1
(4) g is continuous and differentiable for all
x (0,2)
Ans. (1)
Sol.
f :(0,2) R
;
x2
f(x) 2x
12
f (x) 2x
f(x)
is decreasing in domain.
2
2x
f(x)
x20 x 1
2x
g(x) 3x 1 x 2
2
1 2O
g(x)
3. Let the image of the point (1, 0, 7) in the line
x y 1 z 2
1 2 3
be the point (, , ). Then
which one of the following points lies on the line
passing through (, , ) and making angles
2
3
and
3
4
with y-axis and z-axis respectively and an
acute angle with x-axis ?
(1)
1, 2,1 2
(2)
1,2,1 2
(3)
3,4,3 2 2
(4)
3, 4,3 2 2
Ans. (3)
Sol.
1
x y 1 z 2
L1 2 3
M( ,1 2 ,2 3 )
ˆ ˆ ˆ
PM ( 1)i (1 2 )j (3 5)k
PM
is perpendicular to line L1
PM.b 0
(
ˆ ˆ ˆ
b i 2j 3k
)
1 4 2 9 15 0
14 14 1
M (1,3,5)
Q 2M P
[M is midpoint of
P&Q
]
ˆ ˆ ˆ ˆ ˆ
Q 2i 6j 10k i 7k
ˆ ˆ ˆ
Q i 6j 3k
( , , ) (1,6,3)
Required line having direction cosine (l, m, n)
2 2 2 1 l m n
2
2
211
1
22
l
21
4
l
1
2
l
[Line make acute angle with x-axis]
Equation of line passing through (1, 6, 3) will be
1 1 1
ˆ ˆ ˆ ˆ ˆ ˆ
r (i 6j 3k) i j k
22 2
Option (3) satisfying for = 4
4. Let R be the interior region between the lines
3x y 1 0
and
x 2y 5 0
containing the
origin. The set of all values of a, for which the
points (a2, a + 1) lie in R, is :
(1)
1
( 3, 1) ,1
3
(2)
1
( 3,0) ,1
3
(3)
2
( 3,0) ,1
3
(4)
1
( 3, 1) ,1
3
Ans. (2)
Sol. P(a2, a + 1)
L1 = 3x – y + 1 = 0
Origin and P lies same side w.r.t. L1
L1(0) . L1(P) > 0
3(a2) – (a + 1) + 1 > 0
y
O(0,0)
L: x+2y–5=0
2
x
L: 3x–y+1=0
1
3a2 – a > 0
1
a ( ,0) ,
3
…………….(1)
Let L2 : x + 2y – 5 = 0
Origin and P lies same side w.r.t. L2
22
L (0).L (P) 0
2
a 2(a 1) 5 0
2
a 2a 3 0
(a 3)(a 1) 0
a ( 3,1)
…………….(2)
Intersection of (1) and (2)
1
a ( 3,0) ,1
3
5. The 20th term from the end of the progression
1 1 3 1
20,19 ,18 ,17 ,...., 129
4 2 4 4
is :-
(1) –118
(2) –110
(3) –115
(4) –100
Ans. (3)
Sol.
1 1 3 1
20,19 ,18 ,17 ,......, 129
4 2 4 4
This is A.P. with common difference
1
13
d1
44
11
129 ,..............,19 ,20
44
This is also A.P.
1
a 129 4
and
3
d4
Required term =
13
129 (20 1)
44
13
129 15 115
44
6. Let
1
f :R R
2
and
5
g:R R
2
be
defined as
2x 3
f(x) 2x 1
and
| x | 1
g(x) 2x 5
. Then
the domain of the function fog is :
(1)
5
R2
(2) R
(3)
7
R4
(4)
57
R,
24
Ans. (1)
Sol.
2x 3 1
f(x) ;x
2x 1 2
| x | 1 5
g(x) ,x
2x 5 2
Domain of f(g(x))
2g(x) 3
f(g(x)) 2g(x) 1
5
x2
and
| x | 1 1
2x 5 2
5
xR 2
and
xR
Domain will be
5
R2
7. For 0 < a < 1, the value of the integral
2
0
dx
1 2acosx a
is :
(1)
2
2
a
(2)
2
2
a
(3)
2
1a
(4)
2
1a
Ans. (3)
Sol.
2
0
dx
I ; 0 a 1
1 2acosx a
2
0
dx
I1 2acosx a
/2 2
2 2 2 2
0
2(1 a )
2I 2 dx
(1 a ) 4a cos x
/2 22
2 2 2 2
0
2(1 a ).sec x
I dx
(1 a ) .sec x 4a
/2 22
2 2 2 2 2
0
2.(1 a ).sec x
I dx
(1 a ) .tan x (1 a )
2
/2 2
2
2
022
2.sec x .dx
1a
I1a
tan x 1a
2
2
I0
(1 a ) 2
2
I1a
8. Let
x
g(x) 3f f(3 x)
3
and
f (x) 0
for all
x (0,3)
. If g is decreasing in (0, ) and
increasing in (, 3), then 8 is
(1) 24
(2) 0
(3) 18
(4) 20
Ans. (3)
Sol.
x
g(x) 3f f(3 x)
3
and
f (x) 0
x (0, 3)
f (x)
is increasing function
1x
g (x) 3 .f f (3 x)
33
x
f f (3 x)
3
If g is decreasing in (0, )
g (x) 0
x
f f (3 x) 0
3
x
f f (3 x)
3
x3x
3
9
x4
Therefore
9
4
Then
9
8 8 18
4
9. If
e
2
x0
3 sinx cosx log (1 x) 1
lim 3tan x 3
, then
2– is equal to :
(1) 2
(2) 7
(3) 5
(4) 1
Ans. (3)
Sol.
e
2
x0
3 sinx cosx log (1 x) 1
lim 3tan x 3
3 2 4 2 3
2
x0
x x x x x
3 x .... 1 .... x ...
3! 2! 4! 2 3 1
lim 3tan x 3
22
22
x0
1
(3 ) ( 1)x x .... x1
22
lim 3x tan x 3
3 0, 1 0
and
11
22
33
3, 1
2 2 3 5
10. If , are the roots of the equation,
2
x x 1 0
and
nn
n
S 2023 2024
, then
(1)
12 11 10
2S S S
(2)
12 11 10
S S S
(3)
11 12 10
2S S S
(4)
11 10 12
S S S
Ans. (2)
Sol.
2
x x 1 0
nn
n
S 2023 2024
n 1 n 1 n 2 n 2
n 1 n 2
S S 2023 2024 2023 2024
n 2 n 2
2023 [1 ] 2024 [1 ]
n 2 2 n 2 2
2023 [ ] 2024 [ ]
nn
2023 2024
n 1 n 2 n
S S S
Put n = 12
11 10 12
S S S
11. Let A and B be two finite sets with m and n
elements respectively. The total number of subsets
of the set A is 56 more than the total number of
subsets of B. Then the distance of the point P(m, n)
from the point Q(–2, –3) is
(1) 10
(2) 6
(3) 4
(4) 8
Ans. (1)
Sol.
mn
2 2 56
n m n 3
2 (2 1) 2 7
n3
22
and
mn
2 1 7
mn
n 3 and 2 8
n 3 and m n 3
n 3 and m 6
P(6,3) and Q(–2, –3)
22
PQ 8 6 100 10
Hence option (1) is correct
12. The values of , for which
33
122
11
10
33
2 3 3 1 0
, lie in the interval
(1) (–2, 1)
(2) (–3, 0)
(3)
33
,
22
(4) (0, 3)
Ans. (2)
Sol.
33
122
11
10
33
2 3 3 1 0
77
(2 3) (3 1) 0
66
77
(2 3). (3 1). 0
66
2
2 3 3 1 0
2
2 6 1 0
3 7 3 7
,
22
Hence option (2) is correct.
13. An urn contains 6 white and 9 black balls. Two
successive draws of 4 balls are made without
replacement. The probability, that the first draw
gives all white balls and the second draw gives all
black balls, is :
(1)
5
256
(2)
5
715
(3)
3
715
(4)
3
256
Ans. (3)
Sol.
69
44
15 11
44
CC3
C C 715
Hence option (3) is correct.
14. The integral
82
12 6 1 3 3
(x x )dx 1
(x 3x 1)tan x x
is
equal to :
(1)
1/3
13 3
e
1
log tan x C
x
(2)
1/2
13 3
e
1
log tan x C
x
(3)
13 3
e
1
log tan x C
x
(4)
3
13 3
e
1
log tan x C
x
Ans. (1)
Sol. I =
82
12 6 1 3 3
xx dx
1
(x 3x 1)tan x x
Let
13 3
1
tan x t
x
2
24
33
13
. 3x dx dt
x
1
1xx
66
12 6 4
x 3x 3
. dx dt
x 3x 1 x
1 dt 1
I ln | t | C
3 t 3
13 3
11
I ln tan x C
3x
1/3
13 3
1
I ln tan x C
x
Hence option (1) is correct
15. If
2
2tan 5sec 1
has exactly 7 solutions in
the interval
n
0, 2
, for the least value of
nN
then
n
k
k1
k
2
is equal to :
(1)
14
15
1(2 14)
2
(2)
15
14
1(2 15)
2
(3)
13
15
12
(4)
14
3
1(2 15)
2
Ans. (4)
Sol.
2
2tan 5sec 1 0
2
2sec 5sec 3 0
(2sec 1)(sec 3) 0
1
sec ,3
2
1
cos 2,3
1
cos 3
For 7 solutions n = 13
So,
13
k
k1
kS
2
(say)
2 3 13
1 2 3 13
S ....
2 2 2 2
2 3 13 14
1 1 1 12 13
S .....
2 2 2 2 2
13
13
14 13 13
1
1
S 1 13 2 1 13
2
. S 2.
1
2 2 2 2 2
12
16. The position vectors of the vertices A, B and C of a
triangle are
ˆ ˆ ˆ
2i 3j 3k
,
ˆ ˆ ˆ
2i 2j 3k
and
ˆ ˆ ˆ
i j 3k
respectively. Let l denotes the length of
the angle bisector AD of
BAC
where D is on the
line segment BC, then 2l2 equals :
(1) 49
(2) 42
(3) 50
(4) 45
Ans. (4)
Sol. AB = 5
AC = 5
A (2, –3, 3)
C
(–1, 1, 3)
B
(2,2,3) D
D is midpoint of BC
13
D , ,3
22
22
2
13
2 3 (3 3)
22
l
45
2
l
2l2 = 45
17. If y = y(x) is the solution curve of the differential
equation
22
(x 4)dy (y 3y)dx 0
,
3
x 2,y(4) 2
and the slope of the curve is never
zero, then the value of y(10) equals :
(1)
1/4
3
1 (8)
(2)
3
1 2 2
(3)
3
1 2 2
(4)
1/4
3
1 (8)
Ans. (1)
Sol.
22
(x 4)dy (y 3y)dx 0
22
dy dx
y 3y x 4
2
1 y (y 3) dx
dy
3 y(y 3) x 4
1 1 x 2
(ln | y 3| ln| y|) ln C
3 4 x 2
1 y 3 1 x 2
ln ln C
3 y 4 x 2
At
3
x 4, y 2
1
C ln3
4
1 y 3 1 x 2 1
ln ln ln(3)
3 y 4 x 2 4
At x = 10
1 y 3 1 2 1
ln ln ln(3)
3 y 4 3 4
3/4
y3
ln ln2
y
,
dy
x 2, 0
dx
as
3
y(4) y (0,3)
2
1/4
y 3 8 .y
1/4
3
y18
18. Let e1 be the eccentricity of the hyperbola
22
xy1
16 9
and e2 be the eccentricity of the ellipse
22
22
xy
1
ab
, a > b, which passes through the foci
of the hyperbola. If e1e2 = 1, then the length of the
chord of the ellipse parallel to the x-axis and
passing through (0, 2) is :
(1)
45
(2)
85
3
(3)
10 5
3
(4)
35
Ans. (3)
Sol.
22
xy
H: 1
16 9
1
5
e4
1 2 2
4
e e 1 e 5
Also, ellipse is passing through
( 5,0)
a = 5 and b = 3
22
xy
E: 1
25 9
Q P
(5,0)(–5,0)
(0, 2)
End point of chord are
55
,2
3
PQ
10 5
L3
19. Let
3!
4! !
4!
and
4!
5! !
5!
. Then :
(1)
N and N
(2)
N and N
(3)
N and N
(4)
N and N
Ans. (3)
Sol.
3! 4!
(4!)! (5!)!
,
(4!) (5!)
6 24
(24)! (120)!
,
(4!) (5!)
Let 24 distinct objects are divided into 6 groups of
4 objects in each group.
No. of ways of formation of group =
6
24! N
(4!) .6!
Similarly,
Let 120 distinct objects are divided into 24 groups
of 5 objects in each group.
No. of ways of formation of groups
=
24
(120)! N
(5!) .24!
20. Let the position vectors of the vertices A, B and C
of a triangle be
ˆ ˆ ˆ
2i 2j k
,
ˆ ˆ ˆ
i 2j 2k
and
ˆ ˆ ˆ
2i j 2k
respectively. Let l1, l2 and l3 be the
lengths of perpendiculars drawn from the ortho
center of the triangle on the sides AB, BC and CA
respectively, then
222
1 2 3
lll
equals :
(1)
1
5
(2)
1
2
(3)
1
4
(4)
1
3
Ans. (2)
Sol. ABC is equilateral
Orthocentre and centroid will be same
555
G , ,
333
A (2, 2, 1)
C
(2, 1, 2)
B
(1,2,2)
D
G
Mid-point of AB is
33
D ,2,
22
1
111
36 9 36
1 2 3
1
6
2 2 2
1 2 3
1
2
SECTION-B
21. The mean and standard deviation of 15
observations were found to be 12 and 3
respectively. On rechecking it was found that an
observation was read as 10 in place of 12. If µ and
2 denote the mean and variance of the correct
observations respectively, then
22
15( )
is
equal to ……………………..
Ans. (2521)
Sol. Let the incorrect mean be
and standard
deviation be
We have
ii
x12 x 180
15
As per given information correct
i
x
= 180–10+12
(correct mean) =
182
15
Also
22
ii
x144 3 x 2295
15
Correct
2
i
x
= 2295 – 100 + 144 = 2339
2 (correct variance) =
2339 182 182
15 15 15
Required value
22
15( )
182 182 182 2339 182 182
15 15 15 15 15 15 15
182 2339
15 15 15
= 2521
22. If the area of the region
2
{(x,y):0 y min{2x,6x x }}
is A, then 12 A
is equal to………………
Ans. (304)
Sol. We have
62
4
1
A 4 8 (6x x )dx
2
76
A3
12A = 304
23. Let A be a 2 × 2 real matrix and I be the identity
matrix of order 2. If the roots of the equation
|A xI| 0
be –1 and 3, then the sum of the
diagonal elements of the matrix A2 is……………
Ans. (10)
Sol. |A – xI| = 0
Roots are –1 and 3
Sum of roots = tr(A) = 2
Product of roots = |A| = –3
Let
ab
Acd
We have a + d = 2
ad – bc = –3
2
22
a b a b a bc ab bd
Ac d c d ac cd bc d
We need
22
a bc bc d
22
a 2bc d
2
(a d) 2ad 2bc
4 2(ad bc)
4 2( 3)
46
= 10
24. If the sum of squares of all real values of , for
which the lines 2x – y + 3 = 0, 6x + 3y + 1 = 0 and
x + 2y –2 = 0 do not form a triangle is p, then the
greatest integer less than or equal to p is ………
Ans. (32)
Sol.
2x y 3 0
6x 3y 1 0
x 2y 2 0
Will not form a if
x 2y 2 0
is concurrent
with
2x y 3 0
and
6x 3y 1 0
or parallel
to either of them so
Case-1: Concurrent lines
2 1 3 4
6 3 1 0 5
22
Case-2 : Parallel lines
6or 2
2 3 2
4 or 4
16
P 16 16 25
16
[P] 32 32
25
25. The coefficient of x2012 in the expansion of
2008 2 2007
(1 x) (1 x x )
is equal to
Ans. (0)
Sol.
2007 2 2007
(1 x)(1 x) (1 x x )
3 2007
(1 x)(1 x )
2007 2007 3
01
(1 x)( C C (x ) .......)
General term
r 2007 3r
r
(1 x)(( 1) C x )
r2007 3r r2007 3r 1
rr
( 1) C x ( 1) C x
3r = 2012
2012
r3
3r + 1 = 2012
3r = 2011
2011
r3
Hence there is no term containing x2012.
So coefficient of x2012 = 0
26. If the solution curve, of the differential equation
dy x y 2
dx x y
passing through the point (2, 1) is
2
1ee
y 1 1 y 1
tan log log | x 1|
x 1 x 1
,
then
5
is equal to
Ans. (11)
Sol.
dy x y 2
dx x y
x X h,y Y k
dY X Y
dX X Y
h k 2 0 h k 1
h k 0
Y vX
2
dv 1 v dv 1 v
vX
dX 1 v dX 1 v
2
1 v dX
dv
1 v X
12
1
tan v ln(1 v ) ln | X | C
2
As curve is passing through (2, 1)
2
1y 1 1 y 1
tan ln 1 ln | x 1|
x 1 2 x 1
1and 2
5 11
27. Let
x
e
0
1t
f(x) g(t)log dt
1t
, where g is a
continuous odd function.
If
2
/2 2
x
/2
x cosx
f(x) dx
1e
, then is
equal to………….
Ans. (2)
Sol.
x
0
1t
f(x) g(t)ln dt
1t
x
0
1t
f( x) g(t)ln dt
1t
x
0
1y
f( x) g( y)ln dy
1y
x
0
1y
g(y)ln dy
1y
(g is odd)
f( x) f(x) f
is also odd
Now,
/2 2
x
/2
x cosx
I f(x) dx
1e
…..(1)
/2 2x
x
/2
x e cosx
I f( x) dx
1e
……(2)
/2 /2
22
/2 0
2I x cosx dx 2 x cosxdx
/2
/2
2
00
I x sin x 2xsin xdx
2/2
0
2 xcosx cosxdx
4
2
22
2(0 1) 2 2
4 4 2
2
28. Consider a circle
22
(x ) (y ) 50
, where
,0
. If the circle touches the line y + x = 0 at
the point P, whose distance from the origin is
42
, then
2
()
is equal to…………….
Ans. (100)
Sol.
22
S:(x ) (y ) 50
CP = r
52
2
2
( ) 100
29. The lines
x 2 y z 7
2 2 16
and
x 3 y 2 z 2
4 3 1
intersect at the point P. If the
distance of P from the line
x 1 y 1 z 1
2 3 1
is l,
then 14l2 is equal to……………..
Ans. (108)
Sol.
x 2 y z 7
1 1 8
x 3 y 2 z 2 k
4 3 1
2 4k 3
3k 2
k 1, 1
8 7 k 2
P (1,1, 1)
P(1,1,–1)
(–1,1,1)
l
ˆ ˆ ˆ
2i 3j k
Projection of
ˆˆ
2i 2k
on
ˆ ˆ ˆ
2i 3j k
is
4 2 2
4 9 1 14
24 108
814 14
l
2
14 108l
30. Let the complex numbers and
1
lie on the
circles
2
0
| z z | 4
and
2
0
| z z | 16
respectively,
where
0
z 1 i
. Then, the value of
2
100| |
is……………
Ans. (20)
Sol.
2
0
| z z | 4
00
( z )( z ) 4
2
0 0 0
z z | z | 4
200
| | z z 2
………….(1)
2
0
| z z | 16
00
11
z z 16
2
00
(1 z )(1 z ) 16| |
22
0 0 0
1 z z | | | z | 16| |
2
00
1 z z 14| |
…………(2)
From (1) and (2)
2
5| | 1
2
100| | 20
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. The equation of state of a real gas is given by
, where P, V and T are
pressure. volume and temperature respectively and
R is the universal gas constant. The dimensions of
is similar to that of :
(1) PV
(2) P
(3) RT
(4) R
Ans. (2)
And [V] = [b]
32. Wheatstone bridge principle is used to measure the
specific resistance (S1) of given wire, having
length L, radius r. If X is the resistance of wire,
then specific resistance is : . If the
length of the wire gets doubled then the value of
specific resistance will be :
(1)
(2) 2S1
(3)
(4) S1
Ans. (4)
Sol. As specific resistance does not depends on
dimension of wire so, it will not change.
33. Given below are two statements : one is labelled as
Assertion (A) and the other is labelled as Reason (R ).
Assertion (A) : The angular speed of the moon in
its orbit about the earth is more than the angular
speed of the earth in its orbit about the sun.
Reason (R) : The moon takes less time to move
around the earth than the time taken by the earth to
move around the sun.
In the light of the above statements, choose the
most appropriate answer from the options given
below :
(1) (A) is correct but (R) is not correct
(2) Both (A) and (R) are correct and (R) is the
correct explanation of (A)
(3) Both (A) and (R) are correct but (R) is not the
correct explanation of (A)
(4) (A) is not correct but (R) is correct
Ans. (2)
Sol.
Tearth = 365 days 4 hour
34. Given below are two statements :
Statement (I) : The limiting force of static friction
depends on the area of contact and independent of
materials.
Statement (II) : The limiting force of kinetic
friction is independent of the area of contact and
depends on materials.
In the light of the above statements, choose the
most appropriate answer from the options given
below :
(1) Statement I is correct but Statement II is
incorrect
(2) Statement I is incorrect but Statement II is
correct
(3) Both Statement I and Statement II are incorrect
(4) Both Statement I and Statement II are correct
Ans. (2)
Sol. Co-efficient of friction depends on surface in
contact So, depends on material of object.
moon earth
T 27 days
moon
TT
21
S
2
1
S
4
1
L
r
SX 2
1
bV
aP
PV
22
2
V
a
P a PV
Sol.
2
2
b
a2
V
a
P V b RT
2
35. The truth table of the given circuit diagram is :
A
B
Y
(1)
A B Y
0 0 1
0 1 0
1 0 0
1 1 1
(2)
A B Y
0 0 0
011
1 0 1
110
(3)
A B Y
0 0 0
0 1 0
1 0 0
1 1 1
(4)
A B Y
0 0 1
011
1 0 1
110
Ans. (2)
Sol.
A
B
Y
AA.B
B
A
B
A.B
Y =
A B A B
This is XOR GATE
36. A current of 200 A deflects the coil of a moving
coil galvanometer through 600. The current to
cause deflection through
10
radian is :
(1) 30 A (2) 120 A
(3) 60 A (4) 180 A
Ans. (3)
Sol.
i
(angle of deflection)
2 2 2
11
ii
/10 3
i 200 A / 3 10
2
i 60 A
37. The atomic mass of 6C12 is 12.000000 u and that of
6C13 is 13.003354 u. The required energy to
remove a neutron from 6C13, if mass of neutron is
1.008665 u, will be :
(1) 62. 5 MeV (2) 6.25 MeV
(3) 4.95 MeV (4) 49.5 MeV
Ans. (3)
Sol.
13 12 1
6 6 0
C Energy C n
m = (12.000000 + 1.008665) – 13.003354
= – 0.00531 u
Energy required = 0.00531 × 931.5 MeV
= 4.95 MeV
38. A ball suspended by a thread swings in a vertical
plane so that its magnitude of acceleration in the
extreme position and lowest position are equal.
The angle () of thread deflection in the extreme
position will be :
(1)
1
tan 2
(2)
11
2tan 2
(3)
11
tan 2
(4)
11
2tan 5
Ans. (2)
Sol.
v
Loss in kinetic energy = Gain in potential energy
2
1mv mg 1 cos
2
2
v2g 1 cos
Acceleration at lowest point =
2
v
Acceleration at extreme point = gsin
Hence,
2
vgsin
sin 2 1 cos
1
11
tan 2tan
2 2 2
39. Three voltmeters, all having different internal
resistances are joined as shown in figure. When
some potential difference is applied across A and
B, their readings are V1, V2 and V3. Choose the
correct option.
V1V2
V3
A B
(1) V1 = V2 (2)
1 3 2
V V V
(3) V1 + V2 > V3 (4) V1 + V2 = V3
Ans. (4)
Sol. From KVL,
V1 + V2 – V3 = 0
V1 + V2 = V3
40. The total kinetic energy of 1 mole of oxygen at
27°C is :
[Use universal gas constant (R)= 8.31 J/mole K]
(1) 6845.5 J (2) 5942.0 J
(3) 6232.5 J (4) 5670.5J
Ans. (3)
Sol. Kinetic energy =
fnRT
2
51 8.31 300 J
2
= 6232.5 J
41. Given below are two statements : one is labelled as
Assertion(A) and the other is labelled as Reason
(R).
Assertion (A) : In Vernier calliper if positive zero
error exists, then while taking measurements, the
reading taken will be more than the actual reading.
Reason (R) : The zero error in Vernier Calliper
might have happened due to manufacturing defect
or due to rough handling.
In the light of the above statements, choose the
correct answer from the options given below :
(1) Both (A) and (R) are correct and (R) is the
correct explanation of (A)
(2) Both (A) and (R) are correct but (R) is not the
correct explanation of (A)
(3) (A) is true but (R) is false
(4) (A) is false but (R) is true
Ans. (2)
Sol. Assertion & Reason both are correct
Theory
42. Primary side of a transformer is connected to
230 V, 50 Hz supply. Turns ratio of primary to
secondary winding is 10 : 1. Load resistance
connected to secondary side is 46 . The power
consumed in it is :
(1) 12.5 W (2) 10.0 W
(3) 11.5 W (4) 12.0 W
Ans. (3)
Sol.
11
22
VN
VN
2
230 10
V1
2
V 23V
Power consumed =
2
2
V
R
23 23 11.5 W
46
43. During an adiabatic process, the pressure of a gas is
found to be proportional to the cube of its absolute
temperature. The ratio of
p
v
C
C
for the gas is :
(1)
5
3
(2)
3
2
(3)
7
5
(4)
9
7
Ans. (2)
Sol.
33
P T PT constant
PV const
nRT
P const
P
1
P T const
1
PT const
3
1
33
32
3
2
44. The threshold frequency of a metal with work
function 6.63 eV is :
(1)
15
16 10 Hz
(2)
12
16 10 Hz
(3)
12
1.6 10 Hz
(4)
15
1.6 10 Hz
Ans. (4)
Sol.
00
h
19 34 0
6.63 1.6 10 6.63 10
19
034
1.6 10
v10
15
0
v 1.6 10 Hz
45. Given below are two statements : one is labelled as
Assertion (A) and the other is labelled as Reason
(R)
Assertion (A) : The property of body, by virtue of
which it tends to regain its original shape when the
external force is removed, is Elasticity.
Reason (R) : The restoring force depends upon the
bonded inter atomic and inter molecular force of
solid.
In the light of the above statements, choose the
correct answer from the options given below :
(1) (A) is false but (R) is true
(2) (A) is true but (R) is false
(3) Both (A) and (R) are true and (R) is the correct
explanation (A)
(4) Both (A) and (R) are true but (R) is not the
correct explanation of (A)
Ans. (3 or 4)
Sol. Theory
46. When a polaroid sheet is rotated between two
crossed polaroids then the transmitted light
intensity will be maximum for a rotation of :
(1) 60° (2) 30°
(3) 90° (4) 45°
Ans. (4)
Sol. Let I0 be intensity of unpolarised light incident on
first polaroid.
I1 = Intensity of light transmitted from 1st polaroid
=
0
I
2
be the angle between 1st and 2nd polaroid
be the angle between 2nd and 3rd polaroid
0
90
(as 1st and 3rd polaroid are crossed)
0
90
I2 = Intensity from 2nd polaroid
22
0
21 I
I I cos cos
2
I3 = Intensity from 3rd polaroid
2
32
I I cos
22
31
I I cos cos
22
0
3I
I cos cos
2
90
22
0
3I
I cos sin
2
2
0
3I2sin cos
I22
2
0
3I
I sin 2
8
I3 will be maximum when sin 2 = 1
2 90
45
47. An object is placed in a medium of refractive
index 3. An electromagnetic wave of intensity
6 × 108 W/m2 falls normally on the object and it is
absorbed completely. The radiation pressure on
the object would be (speed of light in free space
= 3 × 108 m/s) :
(1) 36 Nm–2 (2) 18 Nm–2
(3) 6 Nm—2 (4) 2 Nm–2
Ans. (3)
Sol. Radiation pressure =
I
v
I
c
8
8
6 10 3
3 10
= 6 N/m2
48. Given below are two statements : one is labelled a
Assertion (A) and the other is labelled as
Reason(R)
Assertion (A) : Work done by electric field on
moving a positive charge on an equipotential
surface is always zero.
Reason (R) : Electric lines of forces are always
perpendicular to equipotential surfaces.
In the light of the above statements, choose the
most appropriate answer from the options given
below :
(1) Both (A) and (R) are correct but (R) is not the
correct explanation of (A)
(2) ((A) is correct but (R) is not correct
(3) (A) is not correct but (R) is correct
(4) Both (A) and (R) are correct and (R) is the
correct explanation of (A)
Ans. (4)
Sol. Electric line of force are always perpendicular to
equipotential surface so angle between farce and
displacement will always be 90°. So work done
equal to 0.
49. A heavy iron bar of weight 12 kg is having its one
end on the ground and the other on the shoulder of
a man. The rod makes an angle 60° with the
horizontal, the weight experienced by the man is :
(1) 6 kg
(2) 12 kg
(3) 3 kg
(4)
6 3 kg
Ans. (3)
Sol.
N1
N2
60° 120
Of
Torque about O = 0
2
L
120 cos60 N L 0
2
N2 = 30 N
50. A bullet is fired into a fixed target looses one third
of its velocity after travelling 4 cm. It penetrates
further D × 10–3 m before coming to rest. The
value of D is :
(1) 2
(2) 5
(3) 3
(4) 4
Ans. (Bonus)
Sol. v2 – u2 = 2aS
222
2u u 2 a 4 10
3
222
4u u 2a 4 10
9
22
5u 2a 4 10
9
…(1)
2
2u
0 2 a x
3
2
4u 2ax
9
…(2)
(1) /(2)
2
5 4 10
4x
2
16
x 10
5
2
x 3 2 10 m
3
x 32 10 m
Note : Since no option is matching, Question
should be bonus.
SECTION-B
51. The magnetic field at the centre of a wire loop
formed by two semicircular wires of radii R1 = 2 m
and R2 = 4m carrying current I = 4A as per figure
given below is × 10–7 T. The value of is
______. (Centre O is common for all segments)
OR1
R2
Ans. (3.00)
Sol.
OR1
R2
00
21
ii
2R 2 2R 2
00
21
ii
4R 4R
77
4 10 4 4 10 4
4 4 4 2
77
3 10 10
3
52. Two charges of –4 C and +4 C are placed at the
points A(1, 0, 4)m and B(2, –1, 5) m located in an
electric field
ˆ
E 0.20 i V / cm
. The magnitude of
the torque acting on the dipole is
5
8 10 Nm
,
Where = _____.
Ans. (2.00)
Sol.
AB
(1, 0, 4) (2, -1, 5)
-4 C4 C
pE
pq
VV
E 0.2 20
cm m
ˆ ˆ ˆ
p 4 i j k
ˆ ˆ ˆ
4i 4j 4k C m
6
ˆ ˆ ˆ ˆ
4i 4j 4k 20i 10 Nm
55
ˆˆ
8k 8j 10 8 2 10
2
53. A closed organ pipe 150 cm long gives 7 beats per
second with an open organ pipe of length 350 cm,
both vibrating in fundamental mode. The velocity
of sound is ________ m/s.
Ans. (294.00)
Sol.
150 cm
closed
pipe open
pipe
350 cm
c1
v
f4
o2
v
f2
c0
f f 7
vv
7
4 150 2 350
vv
7
600 cm 700 cm
vv
7
6m 7m
1
v7
42
v 42 7
=294 m/s
54. A body falling under gravity covers two points A
and B separated by 80 m in 2s. The distance of
upper point A from the starting point is ________
m (use g = 10 ms–2)
Ans. (45.00)
Sol.
{
O
S
start
u = 0
A
B
80m
v1
From
AB
2
11
80 v t 10t
2
2
11
80 2v 10 2
2
1
80 2v 20
1
60 2v
v1 = 30 m/s
From O to A
22
v u 2gS
2
30 0 2 10 S
900 = 20 S
S = 45 m
55. The reading of pressure metre attached with a
closed pipe is 4.5 × 104 N/m2. On opening the
valve, water starts flowing and the reading of
pressure metre falls to 2.0 × 104 N/m2. The
velocity of water is found to be
Vm / s
. The
value of V is ________
Ans. (50)
Sol. Change in pressure
2
1v
2
4 4 3 2
1
4.5 10 2.0 10 10 v
2
4 3 2
1
2.5 10 10 v
2
v2 = 50
v 50
Velocity of water =
V
50
= V = 50
56. A ring and a solid sphere roll down the same
inclined plane without slipping. They start from
rest. The radii of both bodies are identical and the
ratio of their kinetic energies is
7
x
where x is
________.
Ans. (7.00)
Sol. In pure rolling work done by friction is zero.
Hence potential energy is converted into kinetic
energy. Since initially the ring and the sphere have
same potential energy, finally they will have same
kinetic energy too.
Ratio of kinetic energies = 1
71 x 7
x
57. A parallel beam of monochromatic light of
wavelength 5000 •Å is incident normally on a
single narrow slit of width 0.001 mm. The light is
focused by convex lens on screen, placed on its
focal plane. The first minima will be formed for
the angle of diffraction of ________ (degree).
Ans. (30.00)
Sol. For first minima
asin
10
6
5000 10 1
sin a2
1 10
30
58. The electric potential at the surface of an atomic
nucleus (z = 50) of radius 9 × 10–13 cm is
________ × 106 V.
Ans. (8.00)
Sol. Potential
kQ k.Ze
RR
9 19
13 2
9 10 50 1.6 10
9 10 10
6
8 10 V
59. If Rydberg’s constant is R, the longest wavelength
of radiation in Paschen series will be
7R
, where
= ________.
Ans. (144.00)
Sol. Longest wavelength corresponds to transition
between n = 3 and n = 4
22
22
1 1 1 1
RZ RZ 9 16
34
2
7RZ
9 16
144
7R
for Z = 1
144
60. A series LCR circuit with
3
100 10
L mH, C F
and R = 10 , is connected across an ac source of
220 V, 50 Hz supply. The power factor of the
circuit would be ________.
Ans. (1.00)
Sol.
c3
1
X 10
C2 50 10
3
L100
X L 2 50 10
10
CL
XX
,Hence, circuit is in resonance
power factor
RR
1
ZR
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. The order of relative stability of the contributing
structure is:
CH =CH–
2 C–H
I
O
CH – CH = C – H
2II
O
CH – CH = C – H
2III
O
Choose the correct answer from the options given
below:
(1) I > II > III
(2) II > I > III
(3) I = II = III
(4) III > II > I
Ans. (1)
Sol. I > II > III, since neutral resonating structures are
more stable than charged resonating structure. II > III,
since stability of structure with –ve charge on more
electronegative atom is higher.
62. Which among the following halide/s will not show
SN1 reaction:
(A) H2C = CH – CH2Cl
(B) CH3 – CH = CH – Cl
(C)
CH –Cl
2
(D)
CH3
H C
3
CH Cl–
Choose the most appropriate answer from the
options given below:
(1) (A), (B) and (D) only
(2) (A) and (B) only
(3) (B) and (C) only
(4) (B) only
Ans. (4)
Sol. Since
3
CH CH CH
is very unstable, CH3–CH =
CH – Cl cannot give
N1
S
reaction.
63. Which of the following statements is not correct
about rusting of iron?
(1) Coating of iron surface by tin prevents rusting,
even if the tin coating is peeling off.
(2) When pH lies above 9 or 10, rusting of iron
does not take place.
(3) Dissolved acidic oxides SO2, NO2 in water act
as catalyst in the process of rusting.
(4) Rusting of iron is envisaged as setting up of
electrochemical cell on the surface of iron
object.
Ans. (1)
Sol. As tin coating is peeled off, then iron is exposed to
atmosphere.
64. Given below are two statements:
Statement (I) : In the Lanthanoids, the formation
of Ce+4 is favoured by its noble gas configuration.
Statement (II) : Ce+4 is a strong oxidant reverting
to the common +3 state.
In the light of the above statements, choose the
most appropriate answer from the options given
below:
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are true
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are false
Ans. (2)
Sol. Statement (1) is true, Ce+4 has noble gas electronic
configuration.
Statement (2) is also true due to high reduction
potential for Ce4+/Ce3+ (+1.74V), and stability of
Ce3+, Ce4+ acts as strong oxidizing agent.
65. Choose the correct option having all the elements
with d10 electronic configuration from the following:
(1) 27Co, 28Ni, 26Fe, 24Cr
(2) 29Cu, 30Zn, 48Cd, 47Ag
(3) 46Pd, 28Ni, 26Fe, 24Cr
(4) 28Ni, 24Cr, 26Fe, 29Cu
Ans. (2)
Sol. [Cr] = [Ar]4s1 3d5
[Cd] = [Kr]5s24d10
[Cu] = [Ar]4s13d10
[Ag] = [Kr]5s14d10
[Zn] = [Ar]4s23d10
66. Phenolic group can be identified by a positive:
(1) Phthalein dye test
(2) Lucas test
(3) Tollen’s test
(4) Carbylamine test
Ans. (1)
Sol. Carbylamine Test-Identification of primary amines
Lucas Test - Differentiation between 1°, 2° and 3°
alcohols
Tollen's Test - Identification of Aldehydes
Phthalein Dye Test - Identification of phenols
67. The molecular formula of second homologue in the
homologous series of mono carboxylic acids is
_________.
(1) C3H6O2
(2) C2H4O2
(3) CH2O
(4) C2H2O2
Ans. (2)
Sol. HCOOH, CH3COOH
Second homologue
68. The technique used for purification of steam
volatile water immiscible substance is:
(1) Fractional distillation
(2) Fractional distillation under reduced pressure
(3) Distillation
(4) Steam distillation
Ans. (4)
Sol. Steam distillation is used for those liquids which
are insoluble in water, containing non-volatile
impurities and are steam volatile.
69. The final product A, formed in the following
reaction sequence is:
Ph–CH=CH2A
(i) BH 3
22,
(ii) H O OHOH
(iii) HBr
(iv) Mg, ether, then HCHO/H O
3
+
A
(1)
2 2 3
Ph CH CH CH
(2)
3
3
Ph CH CH
|
CH
(3)
3
2
Ph CH CH
|
CH OH
(4)
222
Ph CH CH CH OH
Ans. (4)
Sol.
PhCH = CH2B H /H O ,OH
2 6 2 2
–
PhCH2CH2OH
PhCH CH OH + HBr
2 2 PhCH CH Br + H O
2 2 2
Mg/dry ether
(S )2
PhCH CH CH OH
2 2 2
(I) H –C–H
O
(II) H O
3+PhCH CH
2 2MgBr
(SNNGP)
70. Match List-I with List-II.
List – I List – II
(Reaction) (Reagent(s))
(A)
OH OH COOH
(I) Na2Cr2O7, H2SO4
(B)
OH OH CHO
(II) (i) NaOH (ii) CH3Cl
(C)
OH O
O
(III) (i) NaOH, CHCl3
(ii) NaOH (iii) HCl
(D)
OH
OCH3
(IV) (i) NaOH (ii) CO2
(iii) HCl
Choose the correct answer from the options given
below:
(1) (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(2) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
(3) (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
(4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Ans. (4)
Sol. (A)
Kolbe Schmidt Reaction
(B)
Reimer Tiemann Reaction
(C)
Oxidation of phenol to p-benzoquinone
(D)
PhOH + NaOH
H2O + PhO–
PhO–+ CH3– Cl
PhOCH3 + Cl–
71. Major product formed in the following reaction is a
mixture of:
O C
CH3
CH3
CH3
HI Major product
(1)
I
and (CH ) CI
3 3
(2)
I
and (CH ) COH
3 3
(3)
OH
and (CH ) COH
3 3
(4)
OH
and CH –C–I
3
CH3
CH3
Ans. (4)
Sol.
O
+ H – I
O
H
+ I–
+
OH
++I–I
72. Bond line formula of HOCH(CN)2 is:
(1)
H
CCN
CNHO
(2)
HO CH C N
C N
(3)
HO CH CN
CN
(4)
CN CN
OH
Ans. (4)
Sol. CH (OH) (CN)2 is
OH
CH
CN CN
or
OH
CNNC
73. Given below are two statements:
Statement (I) : Oxygen being the first member of
group 16 exhibits only –2 oxidation state.
Statement (II) : Down the group 16 stability of +4
oxidation state decreases and +6 oxidation state
increases.
In the light of the above statements, choose the
most appropriate answer from the options given
below:
(1) Statement I is correct but Statement II is incorrect
(2) Both Statement I and Statement II are correct
(3) Both Statement I and Statement II are incorrect
(4) Statement I is incorrect but Statement II is correct
Ans. (3)
Sol. Statement-I: Oxygen can have oxidation state from
–2 to +2, so statement I is incorrect
Statement- II: On moving down the group stability
of +4 oxidation state increases whereas stability of
+6 oxidation state decreases down the group,
according to inert pair effect.
So both statements are wrong.
74. Identify from the following species in which d2sp3
hybridization is shown by central atom:
(1) [Co(NH3)6]3+
(2) BrF5
(3) [Pt(Cl)4]2–
(4) SF6
Ans. (1)
Sol. [Co(NH3)6]+3 – d2sp3 hybridization
BrF5 – sp3d2 hybridization
[PtCl4]-2 – dsp2 hybridization
SF6 – sp3d2 hybridization
75. Identify B formed in the reaction.
24
Cl (CH ) Cl
3
excessNH NaOH
A
2
B H O NaCl
(1)
NH
(2)
2 2 2
4
H N CH NH
(3)
3 2 4 3
ClNH (CH ) NH Cl
(4)
N
N
Ans. (2)
Sol.
76. The quantity which changes with temperature is:
(1) Molarity
(2) Mass percentage
(3) Molality
(4) Mole fraction
Ans. (1)
Sol.
Molesof solute
Molarity Volumeof solution
Since volume depends on temperature, molarity
will change upon change in temperature.
77. Which structure of protein remains intact after
coagulation of egg white on boiling?
(1) Primary
(2) Tertiary
(3) Secondary
(4) Quaternary
Ans. (1)
Sol. Boiling an egg causes denaturation of its protein
resulting in loss of its quarternary, tertiary and
secondary structures.
78. Which of the following cannot function as an
oxidising agent?
(1) N3–
(2)
2
4
SO
(3)
3
BrO
(4)
4
MnO
Ans. (1)
Sol. In N3– ion 'N' is present in its lowest possible
oxidation state, hence it cannot be reduced further
because of which it cannot act as an oxidizing
agent.
79. The incorrect statement regarding conformations
of ethane is:
(1) Ethane has infinite number of conformations
(2) The dihedral angle in staggered conformation is
60°
(3) Eclipsed conformation is the most stable
conformation.
(4) The conformations of ethane are inter-
convertible to one-another.
Ans. (3)
Sol. Eclipsed conformation is the least stable
conformation of ethane.
80. Identity the incorrect pair from the following:
(1) Photography - AgBr
(2) Polythene preparation – TiCl4, Al(CH3)3
(3) Haber process - Iron
(4) Wacker process – Pt Cl2
Ans. (4)
Sol. The catalyst used in Wacker's process is PdCl2
SECTION-B
81. Total number of ions from the following with
noble gas configuration is _________.
Sr2+ (Z = 38), Cs+ (Z = 55), La2+ (Z = 57) Pb2+
(Z = 82), Yb2+ (Z = 70) and Fe2+ (Z = 26)
Ans. (2)
Sol. Noble gas configuration = ns2 np6
[Sr2+] = [Kr]
[Cs+] = [Xe]
[Yb2+] = [Xe] 4f14
[La2+] = [Xe] 5d1
[Pb2+] = [Xe] 4f14 5d10 6s2
[Fe2+] = [Ar] 3d6
82. The number of non-polar molecules from the
following is ___________
HF, H2O, SO2, H2, CO2, CH4, NH3, HCl, CHCl3, BF3
Ans. (4)
Sol. The non-polar molecules are CO2, H2, CH4 and
BF3
83. Time required for completion of 99.9% of a First
order reaction is ________ times of half life (t1/2)
of the reaction.
Ans. (10)
Sol.
3
99.9%
1/ 2
2.303 a 100
log
tlog10 3
k a x 100 99.9 10
2.303
t log2 log 2 0.3
log2
k
84. The Spin only magnetic moment value of square
planar complex [Pt(NH3)2Cl(NH2CH3)]Cl is
________ B.M. (Nearest integer)
(Given atomic number for Pt = 78)
Ans. (0)
Sol. Pt2+ (d8)
22
Pt dsp
hybridization and have no unpaired e–s.
Magnetic moment = 0
85. For a certain thermochemical reaction
MN
at
T = 400 K,
H 77.2
kJ mol–1, S = 122 JK–1,
log equilibrium constant (logK) is –______ × 10–1.
Ans. (37)
Sol.
G H T S
3
77.2 10 400 122 28400 J
G 2.303RTlogK
28400 2.303 8.314 400 logK
1
logK 3.708 37.08 10
86. Volume of 3 M NaOH (formula weight 40 g mol–1)
which can be prepared from 84 g of NaOH is
________ × 10–1 dm3.
Ans. (7)
Sol.
1
NaOH
sol
84 / 40
n
M 3 V 0.7L 7 10 L
V (in L) V
87. 1 mole of PbS is oxidised by “X” moles of O3 to
get “Y” moles of O2. X + Y = __________
Ans. (8)
Sol. PbS + 4O3PbSO4 + 4O2
x = 4, y = 4
88. The hydrogen electrode is dipped in a solution of
pH = 3 at 25°C. The potential of the electrode will
be –________ × 10–2 V.
2.303RT 0.059V
F
Ans. (18)
Sol.
2
aq.
2H 2e H (g)
2
cell
H
0
cell 2
P
0.059
E E log
2H
= 0- 0.059 × 3 = - 0.177 volts. = -17.7 × 10-2 V.
89. 9.3 g of aniline is subjected to reaction with excess
of acetic anhydride to prepare acetanilide. The
mass of acetanilide produced if the reaction is
100% completed is _________ × 10–1 g.
(Given molar mass in g mol–1 N : 14, O : 16, C :
12, H : 1)
Ans. (135)
Sol.
6 5 2 3 3
AnilineMM 93
OO
| | | |
C H NH CH C O C CH
6 5 3 3
Ace tan ilide MM 135
O
||
C H NH C CH CH COOH
Ace tan ilide Aniline
nn
m 9.3
135 93
m 13.5g
90. Total number of compounds with Chiral carbon
atoms from following is ________.
O O
O
3 2 2
CH CH CH(NO ) COOH
3 2 2 3
CH CH CHBr CH CH
3 2 2
CH CH(I) CH NO
3 2 2
CH CH CH(OH) CH OH
325
CH CH CH(I) C H
|
I
Ans. (5)
Sol. Chiral carbons are marked by.
O
*,CH CH –CH–COOH
3 2
NO2
*I
*,
NO2
, ,
I
*
I
*
OH
*OH
