FINAL JEE–MAIN EXAMINATION – APRIL, 2024
(Held On Thursday 04th April, 2024) TIME : 3 : 00 PM to 6 : 00 PM
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. If the function
x x x
ee
72 9 8 1 , x 0
fx 2 1 cosx
a log 2 log 3 , x 0
is continuous at x = 0, then the value of a2 is equal
to
(1) 968 (2) 1152
(3) 746 (4) 1250
Ans. (2)
Sol.
x0
lim f x a n2 n3
xx
x x x
n 0 x 0
8 1 9 1
72 9 8 1
lim lim
2 1 cosx 2 1 cosx
x x 2
n0
8 1 9 1 x
lim 2 1 cosx
x x 1 cosx
n8 n9 2 2 2 24 2 n2 n3
a =
24 2
, a2 = 576 × 2 = 1152
2. If > 0, let be the angle between the vectors
ˆˆˆ
a i j 3k
and
ˆˆ ˆ
b 3i j 2k
. If the vectors
ab
and
ab
are mutually perpendicular, then
the value of (14 cos )2 is equal to
(1) 25 (2) 20
(3) 50 (4) 40
Ans. (1)
Sol.
a b . a b 0
, 0
2
2
a b 0
1 + 2+ 9 = 9 + 1 + 4
= 2,
a b 3 6
cos a . b 14. 14
14cos = 3 – 8 = –5
(14 cos)2 = 25
3. Let C be a circle with radius
10
units and centre
at the origin. Let the line x + y = 2 intersects the
circle C at the points P and Q. Let MN be a chord
of C of length 2 unit and slope –1. Then, a distance
(in units) between the chord PQ and the chord MN
is
(1)
23
(2)
32
(3)
21
(4)
21
Ans. (2)
M
N
O
A
10
C : x2 + y2 = 10
AN =
MN 1
2
In OAN (ON)2 = (OA)2 + (AN)2
10 = (OA)2 + 1 OA = 3
Perpendicular distance of center from
0 0 2
PQ 2
2
Perpendicular distance between MN and
PQ OA 2
or
OA 2
32
or
32
4. Let a relation R on
be defined as :
(x1,y1) R(x2,y2) if and only if x1 < x2 or y1 < y2
Consider the two statements :
(I) R is reflexive but not symmetric.
(II) R is transitive
Then which one of the following is true ?
(1) Only (II) is correct.
(2) Only (I) is correct.
(3) Both (I) and (II) are correct.
(4) Neither (I) nor (II) is correct.
Ans. (2)
Sol. All ((x1y1), (x1,y1)) are in R where
x1, y1 N R is reflexive
((1,1), (2,3)) R but ((2,3), (1,1)) R
R is not symmetric
((2,4), (3,3)) R and ((3,3), (1,3)) R but ((2,4),
(1,3)) R
R is not transitive
5. Let three real numbers a,b,c be in arithmetic
progression and a + 1, b, c + 3 be in geometric
progression. If a > 10 and the arithmetic mean of
a,b and c is 8, then the cube of the geometric mean
of a,b and c is
(1) 120 (2) 312
(3) 316 (4) 128
Ans. (1)
Sol. 2b = a + c, b2 = (a + 1) (c + 3),
a b c 8 b 8,a c 16
3
64 = (a + 1) (19 – a) = 19 + 18a – a2
a2 – 18a – 45 = 0 (a – 15) (a + 3) = 0, (a > 10)
a = 15, c = 1, b = 8
((abc)1/3)3 = abc = 120
6. Let
12
A01
and B = I + adj(A) + (adj A)2+…+
(adj A)10. Then, the sum of all the elements of the
matrix B is :
(1) –110 (2) 22
(3) –88 (4) –124
Ans. (3)
Sol. Adj(A) =
12
01
(AdjA)2 =
14
01
|
|
10 1 20
AdjA 01
1 0 1 2 1 4 1 20
B ...
0 1 0 1 0 1 0 1
11 110
B0 11
sum of elements of B
= –88
7. The value of
2
22
2 2 2
1 2 2 3 ... 100 101
1 2 2 3 ... 100 101
is
(1)
306
305
(2)
305
301
(3)
32
31
(4)
31
30
Ans. (2)
Sol.
100 2
2
22
r1
2 2 2 100 2
r1
r r 1
1 2 2 3 ... 100 101
1 2 2 3 ... 100 101 r r 1
2
100 32
r1
100 2
32
r1
n n 1 2.n n 1 2n 1 n n 1
r 2r r 2 6 2
n n 1 n n 1 2n 1
rr
26
n n 1 n n 1 2. 2n 1 1
2 2 3
n n 1 n n 1 2n 1
2 2 3
;Put n = 100
100 101 2201 1 5185 305
23
100 101 201 5117 301
23
8. Let
x
t
0
f x t sin 1 e dt,x
.
Then
3
x0
fx
lim x
is equal to
(1)
1
6
(2)
1
6
(3)
2
3
(4)
2
3
Ans. (2)
Sol.
3
x0
fx
lim x
Using L Hopital Rule.
x
22
x 0 x 0
x sin 1 e
f ' x
lim lim
3x 3x
(Again L Hopital)
Using L.H. Rule
x x x x x
x0
sin 1 e e .e cos 1 e .e
lim 6
1
6
9. The area (in sq. units) of the region described by
{(x,y) : y2 < 2x, and y > 4x –1} is
(1)
11
32
(2)
8
9
(3)
11
12
(4)
9
32
Ans. (4)
Sol.
Q
1
P
y
x
0
–1/2
(0,–1)
y'
x'
(1/4,0
Shaded area
1
Right Left
1
2
x x dy
2
y 2x
y 4x 1 Solve
1
y 1,y 2
Shaded area
12
1
2
y 1 y dy
42
1
23
1
2
1 y y 9
y
4 2 6 32
10. The area (in sq. units) of the region
S z ; z 1 2; z z i z z 2,lm z 0
is
(1)
7
3
(2)
3
2
(3)
17
8
(4)
7
4
Ans. (2)
Sol. Put z = x + iy
22
z 1 2 x 1 y 4
…(1)
z z i z z 2 2x i 2iy 2
x y 1
…(2)
Im(z) > 0 y > 0 …(3)
(–3,0)
(1,0)
(5,0)
x–y=1
/4
Required area
= Area of semi-circle – area of sector A
2
12
22
3
2
11. If the value of the integral
1
x
1
cos x dx
13
is
2
.
Then, a value of is
(1)
6
(2)
2
(3)
3
(4)
4
Ans. (2)
Sol. Let
1
x
1
cos x
I dx
13
…(I)
1
x
1
cos x
I dx
13
bb
aa
using f x dx f a b x dx
…(II)
Add (1) and (II)
11
10
2I cos x dx 2 cos x dx
sin 2
I given
2
12. Let
f x 3 x 2 4 x
be a real valued
function. If and are respectively the minimum
and the maximum values of f, then 2 + 22 is
equal to
(1) 44 (2) 42
(3) 24 (4) 38
Ans. (2)
Sol. f(x) =
3 x 2 4 x
x – 2 0 & 4 – x 0
x [2, 4]
Let x = 2sin2 + 4cos2
f(x) =
3 2 cos 2 sin
2 3 2 cos 2 sin 9 2 2
2 3 2 cos 2 sin 20
=
2
=
20
2 + 22 = 2 + 40 = 42
13. If the coefficients of x4, x5 and x6 in the expansion
of (1 + x)n are in the arithmetic progression, then
the maximum value of n is :
(1) 14 (2) 21
(3) 28 (4) 7
Ans. (1)
Sol. Coeff. of x4 = nC4
Coeff. of x5 = nC5
Coeff. of x6 = nC6
nC4, nC5, nC6 …. AP
2.nC5 = nC4 + nC6
n
n
6
4
nn
55
C
C
2CC
n
r
n
r1
Cn r 1
r
C
5 n 5
2n 4 6
12(n – 4) = 30 + n2 – 9n + 20
n2 – 21n + 98 = 0
(n – 14) (n – 7) = 0
nmax = 14 nmin = 7
14. Consider a hyperbola H having centre at the origin
and foci and the x-axis. Let C1 be the circle
touching the hyperbola H and having the centre at
the origin. Let C2 be the circle touching the
hyperbola H at its vertex and having the centre at
one of its foci. If areas (in sq. units) of C1 and C2
are 36 and 4, respectively, then the length (in
units) of latus rectum of H is
(1)
28
3
(2)
14
3
(3)
10
3
(4)
11
3
Ans. (1)
Sol. Let H :
22
22
xy1
ab
(b2 = a2(e2 – 1))
eqn of C1 = x2 + y2 = a2
Ar. = 36
a2 = 36
a = 6
Now radius of C2 can be a(e – 1) or a(e + 1)
for r = a(e – 1) for r = a(e + 1)
Ar. = 4 r2 = 4
a2(e – 1)2 = 4 a2(e + 1)2 = 4
36(e – 1)2 = 4 36(e + 1)2 = 4
e – 1 =
1
3
e + 1 =
1
3
e =
4
3
2
3
Not possible
b2 = 36
16 1
9
= 28
LR =
2
2b 2 28 28
a 6 3
15. If the mean of the following probability
distribution of a random variable X;
X 0 2 4 6 8
P X a 2a a b 2b 3b
is
46
9
, then the variance of the distribution is
(1)
581
81
(2)
566
81
(3)
173
27
(4)
151
27
Ans. (2)
Sol.
i
P1
a + 2a + a + b + 2b + 3b = 1
4a + 6b = 1 …. (I)
E(x) = mean =
46
9
ii
46
PX 9
4a + 4a + 4b + 12b + 24b =
46
9
8a + 40b =
46
9
4a + 20b =
23
9
…. (II)
Subtract (I) from (II) we get
b =
1
9
& a =
1
12
Variance = E(xi
2) – E(xi)2
E(xi
2) = 02 × 92 + 22 × 2a + 42(a + b) + 62(2b) + 82(3b)
= 24a + 280b
Put a =
1
12
b =
1
9
E(xi
2) = 2 +
280 298
99
2 = E(xi
2) – E(xi)2
2
298 46
99
2298 2116
9 81
566
81
16. Let PQ be a chord of the parabola y2 = 12x and the
midpoint of PQ be at (4,1). Then, which of the
following point lies on the line passing through the
points P and Q ?
(1) (3,–3) (2)
3, 16
2
(3)
(2, 9)
(4)
1, 20
2
Ans. (4)
Sol.
Q
(4,1
P
T = S1
y – 6(x + 4)
= 1 – 48
6x – y = 23
Option 4
1, 20
2
will satisfy
17. Given the inverse trigonometric function assumes
principal values only. Let x, y be any two real
numbers in [–1,1] such that
cos–1x – sin–1 y = ,
2
.
Then, the minimum value of x2 + y2 + 2xy sin is
(1) –1 (2) 0
(3)
1
2
(4)
1
2
Ans. (2)
Sol. cos–1x –
1
cos y
2
=
cos–1x + cos–1y =
2
3
, , 0,
2 2 2
1 2 2
cos xy 1 x 1 y 2
22
xy 1 x 1 y sin
(xy + sin) = (1 – x2)(1 – y2)
x2y2 + 2xysina + sin2a = 1 – x2 – y2 + x2y2
x2 + y2 + 2xy sin = 1 – sin2
x2 + y2 + 2xysin = cos2
Min. value of cos2 = 0
At
2
Option (2) is correct
18. Let y = y(x) be the solution of the differential
equation
(x2 + 4)2dy + (2x3y + 8xy – 2)dx = 0. If y(0) = 0,
then y(2) is equal to
(1)
8
(2)
16
(3) 2 (4)
32
Ans. (4)
Sol.
3
22
22
dy 2x 8x 2
y
dx x 4 x 4
22
2
dy 2x 2
y
dx x4 x4
IF =
2
2x dx
x4
e
IF = x2 + 4
y × (x2 + 4) =
2
2
2
2x4
x4
y(x2 + 4) =
22
dx
2x2
21
2x
y x 4 tan c
22
0 = 0 + c = c = 0
y(x2 + 4) = tan–1
x
2
y at x = 2
y(4 + 4) = tan–1(1)
y2 32
Option (4) is correct
19. Let
ˆ ˆ ˆ ˆ
ˆˆ
a i j k,b 2i 4j 5k
and
ˆˆˆ
c xi 2j 3k,x
. If
d
is the unit vector in the
direction of
bc
such that
a.d 1
, then
a b .c
is equal to
(1) 9 (2) 6
(3) 3 (4) 11
Ans. (4)
Sol.
d b c
a.d b.a c.a
1 = (1 + x + 5)
1 = (x + 6) ….(1)
d
= 1
1x6
b c 1
ˆˆˆ
x 2 i 6 j 2k 1
2((x + 2)2 + 62 + 22) = 1
x2 + 4x + 4 + 36 + 4 = (x + 6)2
x2 + 4x + 44 = x2 + 12x + 36
8x = 8, x = 1
1 1 1
2 4 5 a b .c
x 2 3
0 0 1
2 9 4
x 2 1 3
= 2 – 9(x – 2)
= 20 – 9x
at x = 1
20 – 9 = 11
Option 4 is correct
20. Let P the point of intersection of the lines
x 2 y 4 z 2
1 5 1
and
x 3 y 2 z 3
2 3 2
.
Then, the shortest distance of P from the line
4x = 2y = z is
(1)
5 14
7
(2)
14
7
(3)
3 14
7
(4)
6 14
7
Ans. (3)
Q
L3
L1
P
L2
1
x 2 y 4 z 2
L1 5 1
P 2,5 4, 2
2
x 3 y 2 z 3
L2 3 2
P 2 3,3 2,2 3
+ 2 = 2 + 3 3 + 2 = 5 + 4
= 2 + 1 3 = 5 + 2
3 = 5(2 + 1) + 2
3 = 10 + 7
= –1 = –1
Both satisfies (P)
P(1,–1,1)
3
x y z
L1 / 4 1 / 2 1
3
x y z
Lk
1 2 4
Coordinates of Q(k,2k,4k)
DR’s of PQ = <k–1, 2k + 1, 4k – 1>
PQ to L3
(k – 1) + 2(2k + 1) + 4(4k – 1) = 0
k – 1 + 4k + 2 + 16k – 4 = 0
1
k7
1 2 4
Q , ,
7 7 7
2 2 2
1 2 4
PQ 1 1 1
7 7 7
36 81 9 126
49 49 49 7
3 14
PQ 7
Option-3 will satisfy
SECTION-B
21. Let S = {sin22 : (sin4 + cos4)x2 + (sin2)x +
(sin6 + cos6) = 0 has real roots}. If and be
the smallest and largest elements of the set S,
respectively, then 3(( – 2)2 + ( – 1)2) equals…..
Ans. (4)
Sol.
2
22
sin 2 3
D sin 2 – 4 1 – 1 – sin 2
24
224
53
sin 2 – 4 1 – sin 2 sin 2
48
42
3
D – sin 2 6sin 2 – 4 0
2
42
3sin 2 –12sin 2 8 0
2
212 12 –12.8 12 4 3 6 2 3
sin 2 6 6 3
22
sin 2 2 3
, but sin2 2 [0, 1]
2
2– 3
, = 1
24
– 2 ,
3
2
–1 0
2
2
3 – 2 –1 4
22. If
52
e
3x
cosec xdx cot x cos ecx cos ec x log tan C
22
where and C is constant of integration ,
then the value of 8( + ) equals …..
Ans. (1)
Sol.
32
cosec x cosec xdx I
By applying integration by parts
32
I – cot xcosec x cot x –3cosec x cot xcosecx dx
3 3 2
I – cot xcosec x – 3 cosec x cosec x – 1 dx
33
I – cot xcosec x – 3I 3 cosec xdx
let
32
1
I cosec xdx –cosecxcot x – cot xcosecxdx
2
1
I –cosecxcotx– cosec x – 1 cosecxdx
1
x
2I –cosecxcotx + n tan 2
1
1 1 x
I – cos ecx cot x n tan
2 2 2
33 3 x
4I – cot xcosec x – cosecxcot x n tan 4c
2 2 2
2
1 3 3 x
I – cosecxcot x cosec x n tan c
4 2 8 2
1,
4
3
8
81
23. Let f : be a thrice differentiable function
such that f(0) = 0, f(1) = 1, f(2) = –1, f(3) = 2 and
f(4) = –2. Then, the minimum number of zeros of
(3f' f'' + ff"') (x) is …..
Ans. (5)
Sol.
2'
3f 'f '' ff ''' x ff '' f ' (x)
2'
ff '' f ' (x) ff ' x
''
3f 'f '' f ''' (x) f(x) f '(x)
1
2
3
4
min. roots of f(x) 4
min. roots of f '(x) 3
min. roots of
f(x) f '(x)
7
min. roots of f(x) f '(x) '' 5
24. Consider the function f : defined by
f(x) =
2
2x
1 9x
. If the composition of
10
2
10 times
2x
f, f o f o f o...o f x
1 9 x
, then the
value of
31
is equal to …..
Ans. (1024)
Sol.
2 2 2 2
2f(x) 4x
f f x
1 9f x 1 9x 9.2 x
3 2 3
22 2 2 4
2
2
2 x / 1 9x 2 x
f f f(x)
2x 1 9x 1 2 2
1 9 1 2 1 9x
By observation
10
2 20
2 4 18
2
2 – 1 2 – 1
1 2 2 .... 2 1 3
2 –1
20 10
3 1 2 3 1 2 1024
25. Let A be a 2 × 2 symmetric matrix such that
13
A17
and the determinant of A be 1.
If A–1 = A + I, where I is an identity matrix of
order 2 × 2, then + equals …..
Ans. (5)
Sol. Let
ab
Abd
2
a b 1 3 ,ad – b 1
b d 1 7
a + b = 3, b + d = 7, (3 – b) (7 – b) –b2 = 1
21 – 10b = 1 b = 2, a = 1, d = 5
12
A,
25
–1 5 –2
A–2 1
–1
A A I
5 –2 2
–2 1 2 5
= –1, = 6
5
26. There are 4 men and 5 women in Group A, and 5
men and 4 women in Group B. If 4 persons are
selected from each group, then the number of ways
of selecting 4 men and 4 women is …..
Ans. (5626)
Sol.
From
Group A
From
Group B
Ways of selection
4M
4W
44
44
C C 1
3M 1W
1M 3W
4 5 5 4
3 1 1 3
C C C C 400
2M 2W
2M 2W
4 5 5 4
2 2 2 2
C C C C 3600
1M 3W
3M 1W
4 5 5 4
1 3 3 1
C C C C 1600
4W
4M
55
44
C C 25
Total
5626
Ans. 5626
27. In a tournament, a team plays 10 matches with
probabilities of winning and losing each match as
1
3
and
2
3
respectively. Let x be the number of
matches that the team wins, and y be the number of
matches that team loses. If the probability
P(|x – y| < 2) is p, then 39p equals……
Ans. (8288)
Sol.
1
P(W) 3
2
P(L) 3
x = number of matches that team wins
y = number of matches that team loses
x – y 2
and x + y = 10
x – y 0,1,2
x, y N
Case-I :
x – y 0 x y
x y 10 x 5 y
55
10
5
1
P x – y 0 C 33
Case-II :
x – y 1 x – y 1
x = y + 1
x = y –1
x y 10
x y 10
2y = 9
2y = 11
Not possible
Not possible
Case-III :
x – y 2 x – y 2
x – y 2 OR x y –2
x y 10 x y 10
x 6,y 4 x 4,y 6
64
10
6
12
P x – y 2 C 33
46
10
4
12
C33
5 4 6
10 10 10
5 6 4
10 10 10
2 2 2
p C C C
3 3 3
9 10 5 10 4 10 6
5 6 4
1
3 p C 2 C 2 C 2
3
= 8288
28. Consider a triangle ABC having the vertices
A(1,2), B() and C() and angles
ABC 6
and
2
BAC 3
. If the points B and C lie on the
line y = x + 4, then 2 + 2 is equal to …..
Ans. (14)
Sol.
y=x+4
m=1
/6
/6
2/3
A(1,2)
B(,)
C(,)
Equation of line passes through point A(1, 2)
which makes angle
6
from y = x + 4 is
1 tan 6
y 2 (x 1)
1 tan 6
31
y 2 (x 1)
31
y 2 (2 3)(x 1) y 2 (2 3)(x 1)
solve with y x 4 solve with y x 4
x 2 (2 3)x 2 3 x 2 (2 3)x 2 3
4 3 4 3
xx
1 3 1 3
Θ
22
22 4 3 4 3
1 3 1 3
22
14
29. Consider a line L passing through the points
P(1,2,1) and Q(2,1,–1). If the mirror image of the
point A(2,2,2) in the line L is (), then
+ + 6 is equal to …..
Ans. (6)
C
B(,,)
DR's of Line L –1 : 1 : 2
DR's of AB – 2 : – 2 : – 2
ABar L 2 – + – 2 + 2 – 4 = 0
2 + – = 4 ...(1)
Let C is mid-point of AB
2 2 2
C,,
2 2 2
DR's of PC =
2
::
2 2 2
line L | | PC
2K(let)
2 2 4
= – 2K
= 2K + 2
= 4K
use in (1)
1
K6
value of + + 6 = 24K + 2 = 6
30. Let y = y(x) be the solution of the differential
equation (x + y + 2)2 dx = dy, y(0) = –2. Let the
maximum and minimum values of the function
y = y(x) in
0, 3
be and , respectively. If
22
3 3, ,
, then + equals
…..
Ans. (31)
Sol.
2
dy (x y 2)
dx
...(1), y(0) = –2
Let x + y + 2 = v
dy dv
1dx dx
from (1)
2
dv 1v
dx
2
dv dx
1v
tan–1(v) = x + C
tan–1(x + y + 2) = x + C
at x = 0 y = – 2 C = 0
tan–1(x + y + 2) = x
y = tanx – x – 2
f(x) = tanx – x – 2, x
0, 3
f '(x) = sec2x – 1 > 0 f(x)
fmin = f(0) = –2 =
fmax =
f 3 2
3
3
now (3 + )2 + 2 = +
3
(3 + )2 + 2 = (3
3
–6)2 + 4
3
= 67 –
36 3
= 67 and = –36 = 31
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. The translational degrees of freedom (ft) and
rotational degrees of freedom (fr) of CH4 molecule
are :
(1) ft = 2 and fr = 2 (2) ft = 3 and fr = 3
(3) ft = 3 and fr = 2 (4) ft = 2 and fr = 3
Ans. (2)
Sol. Since CH4 is polyatomic Non-Linear
D.O.F of CH4
T. DOF = 3
R DOF = 3
32. A cyclist starts from the point P of a circular
ground of radius 2 km and travels along its
circumference to the point S. The displacement of
a cyclist is :
P
Q
S
0
R
(1) 6 km (2)
8
km
(3) 4 km (4) 8 km
Ans. (2)
Sol.
P
Q
S
R
R
Displacement
Displacement =
R2
=
2 2 8
km
33. The magnetic moment of a bar magnet is 0.5 Am2.
It is suspended in a uniform magnetic field of
8 × 10–2 T. The work done in rotating it from its
most stable to most unstable position is :
(1) 16 × 10–2 J (2) 8 × 10–2 J
(3) 4 × 10–2 J (4) Zero
Ans. (2)
Sol. At stable equilibrium
U = –mB cos 0° = –mB
At unstable equilibrium
U = –mB cos 180° = + mB
W = U
W.D. = 2 mB
= 2 (0.5) 8 × 10–2 = 8 × 10–2 J
34. Which of the diode circuit shows correct biasing
used for the measurement of dynamic resistance of
p-n junction diode :
(1)
D4
R
(2)
D2
R
5V
(3)
D3
R
(4)
D1
R
5V
Ans. (2)
Sol. Diode should be in forward biased to calculate
dynamic resistance
Hence correct answer would be 2.
35. Arrange the following in the ascending order of
wavelength :
(A) Gamma rays (1) (B) x-ray (2)
(C) Infrared waves (3) (D) Microwaves (4)
Choose the most appropriate answer from the
options given below :
(1) 4 < 3 < 1 < 2 (2) 4 < 3 < 2 < 1
(3) < 2 < < 4 (4) 2 < 1 < 4 < 3
Ans. (3)
Sol. < 2 < < 4
36. Identify the logic gate given in the circuit :
A
B
Y
(1) NAND - gate (2) OR - gate
(3) AND gate (4) NOR gate
Ans. (2)
Sol.
Y A.B
By De-Morgan Law
Y A B
Y = A + B
Hence OR gate
37. The width of one of the two slits in a Young's
double slit experiment is 4 times that of the other
slit. The ratio of the maximum of the minimum
intensity in the interference pattern is :
(1) 9 :1 (2) 16 : 1
(3) 1 : 1 (4) 4 : 1
Ans. (1)
Sol. Since, Intensity width of slit ()
so, I1 = I, I2 = 4I
Imin =
2
12
I I I
2
max 1 2
I I I 9I
max
min
I9I 9
I I 1
38. Correct formula for height of a satellite from earths
surface is :
(1)
1/2
22
T R g R
4
(2)
1/3
22
2
T R g R
4
(3)
1/3
22
2
TR R
4g
(4)
1/3
22
2
TR R
4
Ans. (2)
Sol.
M
m
R
h
2
2
GMm mv
Rh
Rh
2
GM v
Rh
….(1)
v = (R + h)
v =
2
RhT
…….(2)
2
GM g
R
GM = gR2 ……..(3)
Put value from (2) & (3) in eq. (1)
2
22
gR 2
Rh
R h T
22 3
2
T R g Rh
2
1/3
22
2
T R g Rh
2
39. Match List I with List II
List–I
List–II
A.
Purely
capacitive
circuit
I.
V
I
90°
B.
Purely
inductive
circuit
II.
V
I
C.
LCR
series at
resonance
III.
I
V
D.
LCR
series
circuit
IV.
V
I
90°
Choose the correct answer from the options given
below :
(1) A-I, B-IV, C-III, D-II
(2) A-IV, B-I, C-III, D-II
(3) A-IV, B-I, C-II, D-III
(4) A-I, B-IV, C-II, D-III
Ans. (4)
Sol. A – V lags by 90° from I hence option (I) is
correct.
B – V lead by 90° from I hence option (IV) is
correct
C – In LCR resonance XL = XC. Hence circuit is
purely resistive so option (II) is correct
D – In LCR series V is at some angle from I hence
(III) is correct
Hence option (4) is correct.
40. Given below are two statements :
Statement I : The contact angle between a solid
and a liquid is a property of the material of the
solid and liquid as well.
Statement II : The rise of a liquid in a capillary
tube does not depend on the inner radius of the
tube.
In the light of the above statements, choose the
correct answer from the options given below :
(1) Both Statement I and Statement II are false
(2) Statement I is false but Statement II is true.
(3) Statement I is true but Statement II is false.
(4) Both Statement I and Statement II are true.
Ans. (3)
Sol. Statement I is correct as we know contact angle
depends on cohesine and adhesive forces.
Statement II is incorrect because height of liquid is
given by h =
C
2T cos
gr
where r is radius of
Tube (assuming length of capillary is sufficient)
Hence option (3) is correct.
41. A body of m kg slides from rest along the curve of
vertical circle from point A to B in friction less
path. The velocity of the body at B is :
A
(Vertical Circle)
B
45°
(given, R = 14 m, g = 10 m/s2 and
2
= 1.4)
(1) 19.8 m/s (2) 21.9 m/s
(3) 16.7 m/s (4) 10.6 m/s
Ans. (2)
Sol.
R
Apply W.E.T. from A to B
Wmg = KB – KA
mg ×
RR
2
=
2
B
1mv 0
2
A
v 0 rest
2
B
21 1
mgR mv
2
2
B
2 2 1
gR v
2
B
10 14 2 2.4 v
1.4
21.9 = vB
Hence option (2) is correct
42. An electric bulb rated 50 W – 200 V is connected
across a 100 V supply. The power dissipation of
the bulb is :
(1) 12.5 W (2) 25 W
(3) 50 W (4) 100 W
Ans. (1)
Sol. Rated power & voltage gives resistance
2
2200
V 40000
RP 50 50
R = 800
22
applied
V100
PR 800
P = 12.5 watt
Hence option 1 is correct.
43. A 2 kg brick begins to slide over a surface which is
inclined at an angle of 45° with respect to
horizontal axis. The co-efficient of static friction
between their surfaces is :
(1) 1 (2)
1
3
(3) 0.5 (4) 1.7
Ans. (1)
Sol.
45°
fL
N
mgsin45
mgcos45
mg sin 45 = fL
mg cos 45 = N
fL = µsN
µs = tan 45 = 1
or
tan = µs ( is angle of repose)
tan 45 = µs = 1
correct option (1)
44. In simple harmonic motion, the total mechanical
energy of given system is E. If mass of oscillating
particle P is doubled then the new energy of the
system for same amplitude is :
K
m
P
(1)
E
2
(2) E
(3)
E2
(4) 2E
Ans. (2)
Sol. T.E. =
2
1kA
2
since A is same T.E. will be same
correct option (2)
45. Given below are two statements : one is labelled as
Assertion A and the other is labelled as Reason R.
Assertion A : Number of photons increases with
increase in frequency of light.
Reason R : Maximum kinetic energy of emitted
electrons increases with the frequency of incident
radiation.
In the light of the above statements, choose the
most appropriate answer from the options given
below :
(1) Both A and R are correct and R is NOT the
correct explanation of A.
(2) A is correct but R is not correct.
(3) Both A and R are correct and R is the correct
explanation of A.
(4) A is not correct but R is correct.
Ans. (4)
Sol. Intensity of light I =
nh
A
Here n is no. of photons per unit time.
IA
nh
so on increasing frequency , n decreases
taking intensity constant.
kmax = h –
So on increasing , kinetic energy increases.
46. According to Bohr's theory, the moment of
momentum of an electron revolving in 4th orbit of
hydrogen atom is :
(1)
h
8
(2)
h
(3)
h
2
(4)
h
2
Ans. (3)
Sol. Moment of momentum is
rP
L r mv
L = mvr =
nh 4h 2h
22
47. A sample of gas at temperature T is adiabatically
expanded to double its volume. Adiabatic constant
for the gas is = 3/2. The work done by the gas in
the process is : (µ = 1 mole)
(1)
RT 2 2
(2)
RT 1 2 2
(3)
RT 2 2 1
(4)
RT 2 2
Ans. (4)
Sol.
nR T
W1
1
1
f
TV cons tan t T 2V
Tf =
1/2
1T
T22
T
RT 21
2
W 2RT
32
12
RT 2 2
48. A charge q is placed at the center of one of the
surface of a cube. The flux linked with the cube
is :-
(1)
0
q
4
(2)
0
q
2
(3)
0
q
8
(4) Zero
Ans. (2)
Sol. From
q
0
q
2
0
q
2
49. Applying the principle of homogeneity of
dimensions, determine which one is correct.
where T is time period, G is gravitational constant,
M is mass, r is radius of orbit.
(1)
2
2
2
4r
TGM
(2)
2 2 3
T 4 r
(3)
23
24r
TGM
(4)
22
24r
TGM
Ans. (3)
Sol. According to principle of homogeneity dimension
of LHS should be equal to dimensions of RHS so
option (3) is correct.
23
24r
TGM
3
2
1 3 2
L
TM L T M
(Dimension of G is
1 3 2
M L T
)
3
22
32
L
TT
LT
50. A 90 kg body placed at 2R distance from surface
of earth experiences gravitational pull of :
(R = Radius of earth, g = 10 ms–2)
(1) 300 N (2) 225 N
(3) 120 N (4) 100 N
Ans. (4)
Sol. Value of g =
2
s
h
g1R
=
2s
s
g
g 1 2 9
Here gs = gravitational acceleration at surface
Force = mg = 90 ×
s
g
9
= 100 N
SECTION-B
51. The displacement of a particle executing SHM is
given by x = 10 sin
tm
3
. The time period
of motion is 3.14 s. The velocity of the particle at
t = 0 is ______ m/s.
Ans. (10)
Sol. Given,
T = 3.14 =
2
= 2 rad/s
x 10 sin t 3
dx
v 10 cos t
dt 3
at t = 0
v 10 cos 3
=
1
10 2 2
[using = 2 rad/s]
v = 10 m/s
52. A bus moving along a straight highway with speed
of 72 km/h is brought to halt within 4s after
applying the brakes. The distance travelled by the
bus during this time (Assume the retardation is
uniform) is _______m.
Ans. (40)
Sol. Initial velocity = u = 72 km/h = 20 m/s
v = u + at
0 = 20 + a × 4
a = –5 m/s2
v2 – u2 = 2as
02 – 202 = 2(–5).s
s = 40 m
53. A parallel plate capacitor of capacitance 12.5 pF is
charged by a battery connected between its plates
to potential difference of 12.0 V. The battery is
now disconnected and a dielectric slab (r = 6) is
inserted between the plates. The change in its
potential energy after inserting the dielectric slab is
_______× 10–12 J.
Ans. (750)
Sol. Before inserting dielectric capacitance is given
C0 = 12.5 pF and charge on the capacitor Q = C0V
After inserting dielectric capacitance will become
rC0.
Change in potential energy of the capacitor
= Ei – Ef
=
22
if
QQ
2C 2C
=
2
0r
Q1
1
2C
2
0
0r
CV 1
1
2C
2
0
r
11
C V 1
2
Using C0 = 12.5 pF, V = 12 V, r = 6
2
11
12.5 12 1
26
2
15
12.5 12
26
= 750 pJ = 750 × 10–12J
54. In a system two particles of masses m1 = 3kg and
m2 = 2kg are placed at certain distance from each
other. The particle of mass m1 is moved towards
the center of mass of the system through a distance
2cm. In order to keep the center of mass of the
system at the original position, the particle of mass
m2 should move towards the center of mass by the
distance ____ cm.
Ans. (3)
Sol.
m1=3kg
m2=2kg
2cm
x
1 1 2 2
C.O.M.
12
m x m x
Xmm
3 2 2 x
032
x = 3 cm
55. The disintegration energy Q for the nuclear fission
of
235 140 94
U Ce Zr n
is ____MeV.
Given atomic masses of
235 140
U : 235.0439u; Ce;139.9054u
,
94 Zr : 93.9063u;n :1.0086u
,
Value of c2 = 931 MeV/u.
Ans. (208)
Sol.
235 140 94
U Ce Zr n
Disintegration energy
Q = (mR – mP).c2
mR = 235.0439 u
mP = 139.9054u + 93.9063u + 1.0086 u
= 234.8203u
Q = (235.0439u – 234.8203u)c2
= 0.2236 c2
= 0.2236 × 931
Q = 208.1716
56. A light ray is incident on a glass slab of thickness
43
cm and refractive index
2
. The angle of
incidence is equal to the critical angle for the glass
slab with air. The lateral displacement of ray after
passing through glass slab is ____cm.
(Given sin 15° = 0.25)
Ans. (2)
Sol.
t
µ
r
i
i = C
i = sin–1
1
i = 45º
and according to snell's law
1 sin 45° =
2 sin r
r = 30°
Lateral displacement
t sin i r
cos r
4 3 sin15
cos30
= 2cm
57. A rod of length 60 cm rotates with a uniform
angular velocity 20 rad s–1 about its perpendicular
bisector, in a uniform magnetic field 0.5 T. The
direction of magnetic field is parallel to the axis of
rotation. The potential difference between the two
ends of the rod is ____V.
Ans. (0)
Sol.
B
O
B
A
2
0A
B
VV 2
2
0B
B
VV 2
AB
VV
AB
V V 0
58. Two wires A and B are made up of the same
material and have the same mass. Wire A has
radius of 2.0 mm and wire B has radius of 4.0 mm.
The resistance of wire B is 2. The resistance of
wire A is _____.
Ans. (32)
Sol.
2
V
RAA
24
A B B
24
B A A
R A r
R A r
4
3
A
3
R4 10
2 2 10
RA = 32 .
59. Two parallel long current carrying wire separated
by a distance 2r are shown in the figure. The ratio
of magnetic field at A to the magnetic field
produced at C is
x
7
. The value of x is ___.
2I
A
I
r
r
r
O
2r
C
r
Ans. (5)
Sol.
0
00
A
2i
i 5 i
B2 r 2 3r 6 r
000
C
2i i 7 i
B2 r 2 3r 6 r
A
C
B5
B7
x = 5
60. Mercury is filled in a tube of radius 2 cm up to a
height of 30 cm. The force exerted by mercury on
the bottom of the tube is ___N.
(Given, atmospheric pressure = 105 Nm–2, density
of mercury = 1.36 × 104 kg m–3, g = 10 ms–2,
=
22
7
)
Ans. (177)
Sol.
0m
F P A ghA
= 105 ×
2
2
22 2 10
7
2
4 2 2
22
1.36 10 10 30 10 2 10
7
F = 51.29 + 125.71 = 177 N
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. The equilibrium constant for the reaction
3 2 2
1
SO g SO g O g
2
is KC = 4.9 × 10–2. The value of KC for the reaction
given below is
2 2 3
2SO g O g 2SO g
is
(1) 4.9 (2) 41.6
(3) 49 (4) 416
Ans. (4)
Sol.
22
C2
C
11
K' K 4.9 10
K'C = 416.49
62. Find out the major product formed from the
following reaction. [Me: –CH3]
Br
Br
(1)
NMe2
NMe2
(2)
NMe2
NMe2
(3)
NMe2
NMe2
(4)
NMe2
NMe2
Ans. (2)
Sol.
Br
Br
Me2NH
. .
SN2
NMe2
Br
. .
N
Me
Me
Me2NH
. .
SN2
NMe2
. .
NMe2
H
–H+
NMe2
NMe2
–H+
The above mechanism valid for both cis and trans
isomers. So the products are same for both cis and
trans isomers.
63. When MnO2 and H2SO4 is added to a salt (A), the
greenish yellow gas liberated as salt (A) is :
(1) NaBr (2) CaI2
(3) KNO3 (4) NH4Cl
Ans. (4)
Sol.
4 2 2 4 4
4 2 4 2 2
greenish
yellow
solution
2NH Cl MnO 2H SO MnSO
(NH ) SO 2H O Cl
64. The correct statement/s about Hydrogen bonding
is/are :
A. Hydrogen bonding exists when H is covalently
bonded to the highly electro negative atom.
B. Intermolecular H bonding is present in o-nitro
phenol
C. Intramolecular H bonding is present in HF.
D. The magnitude of H bonding depends on the
physical state of the compound.
E. H-bonding has powerful effect on the structure
and properties of compounds.
Choose the correct answer from the options given
below :
(1) A only (2) A, D, E only
(3) A, B, D only (4) A, B, C only
Ans. (2)
Sol. (A) Generally hydrogen bonding exists when H is
covalently bonded to the highly
electronegative atom like F, O, N.
(B) Intramolecular H bonding is present in
N
O
O
O
H
(C) Intermolecular Hydrogen bonding is present
in HF
(D) The magnitude has Hydrogen bonding in solid
state is greater than liquid state.
(E) Hydrogen bonding has powerfull effect on the
structure & properties of compound like
melting point, boiling point, density etc.
65.
“A”
H
“B”
O–Na+
H
O
O
O
O
In the above chemical reaction sequence “A’’
and “B” respectively are :
(1) O3, Zn/H2O and NaOH(alc.) / I2
(2) H2O, H+ and NaOH(alc.) / I2
(3) H2O, H+ and KMnO4
(4) O3, Zn/H2O and KMnO4
Ans. (1)
Sol.
,
O
Na
O
O3
ZnH2O
“A”
O
H
NaOH
(alc.)
“B”
I2
O
O
H
+ CHI3
66. Common name of Benzene-1, 2-diol is
(1) quinol (2) resorcinol
(3) catechol (4) o-cresol
Ans. (3)
Sol.
OH
OH
IUPAC name : Benzene-1,2-diol
Common name : catechol
67. CH3 – CH2 – CH2 – Br + NaOH
25
C H OH
Product 'A'
H2O
H+
Product "B"
Product "C"
Diborane
H2O/H2O2/–OH
Product A
Consider the above reactions, identify product B
and product C.
(1) B = C = 2-Propanol
(2) B = 2-Propanol C = 1-Propanol
(3) B = 1-Propanol C = 2-Propanol
(4) B = C = 1-Propanol
Ans. (2)
Sol.
OH
CH3–CH2–CH2–Br + NaOH
C2H5OH
CH3–CH=CH2
H2O
H+
CH3–CH–CH3
2-Propanol
[B]
Product
[A]
B2H6
H2O/H2O2/OH
CH3–CH2–CH2–OH
[C]
1-Propanol
68. The adsorbent used in adsorption chromatography
is/are
A. silica gel B. alumina
C. quick lime D. magnesia
Choose the most appropriate answer from the
options given below :
(1) B only (2) C and D only
(3) A and B only (4) A only
Ans. (3)
Sol. The most common polar and acidic support used is
adsorption chromatography is silica. The surface
silanol groups on their supported to adsorb polar
compound and work particularly well for basic
substances. Alumina is the example of polar and
basic adsorbent that is used in adsorption
chromatography.
69.
Br
KOH alc
major product "P"
Product P is
(1)
(2)
(3)
(4)
Ans. (2)
Sol.
Br
Alc.KO
H
(Major)
70. Correct order of stability of carbanion is
a
b
c
d
(1) c > b > d > a (2) a > b > c > d
(3) d > a > c > b (4) d > c > b > a
Ans. (4)
Sol. As we know compound (d) is aromatic and the
compound (a) is anti-aromatic. Hence compound
(d) is most stable and compound (a) is least stable
among these in compound (b) and (c) carbon atom
of that positive charge is sp3 hybridised they on the
basis of angle strain theory compound (c) is more
stable than compound (b).
>
>
>
. .
71. The correct order of the first ionization enthalpy is
(1) Al > Ga > Tl (2) Ga > Al > B
(3) B > Al > Ga (4) Tl > Ga > Al
Ans. (4)
Sol. (i) due to lanthanide contraction T has more I.E.
as compared to Ga and A
(ii) due to scandide contraction Ga has more I.E. as
compared to A
72. If an iron (III) complex with the formula
–
3xy
Fe NH CN
has no electron in its eg
orbital, then the value of x + y is
(1) 5 (2) 6
(3) 3 (4) 4
Ans. (2)
Sol. Complex is
III
3 2 4
[Fe(NH ) (CN) ]
x = 2
y = 4
so x + y = 6
73. Fuel cell, using hydrogen and oxygen as fuels,
A. has been used in spaceship
B. has as efficiency of 40% to produce electricity
C. uses aluminium as catalysts
D. is eco-friendly
E. is actually a type of Galvanic cell only
(1) A,B,C only (2) A,B,D only
(3) A,B,D,E only (4) A,D,E only
Ans. (4)
Sol. Fuel cell is used in spaceship and it is type of
galvanic cell.
74. Choose the Incorrect Statement about Dalton's
Atomic Theory
(1) Compounds are formed when atoms of
different elements combine in any ratio
(2) All the atoms of a given element have
identical properties including identical mass
(3) Matter consists of indivisible atoms
(4) Chemical reactions involve recorganization of
atoms
Ans. (1)
Sol. In compound atoms of different elements combine
in fixed ratio by mass.
75. Match List I with List II
LIST I LIST II
A.
a- Glucose and a-Galactose I. Functional isomers
B.
a- Glucose and b-Glucose II. Homologous
C.
a- Glucose and a-Fructose III. Anomers
D.
a- Glucose and a-Ribose IV. Epimers
Choose the correct answer from the options given
below:
(1) A-III, B-IV, C-II, D-I
(2) A-III, B-IV, C-I, D-II
(3) A-IV, B-III, C-I, D-II
(4) A-IV, B-III, C-II, D-I
Ans. (3)
Sol. Based on biomolecules theory and structure of
these named compounds –
(A) a-Glucose and a-Galactose (IV) Epimers.
(B) a-Glucose and b-Glucose (III) Anomers
(C) a-Glucose and a-Fructose (I) Functional isomers
(D) a-Glucose and a-Ribose (II) Homologous
76. Given below are two statements:
Statement I : The correct order of first ionization
enthalpy values of Li, Na, F and Cl is Na < Li < Cl < F.
Statement II : The correct order of negative
electron gain enthalpy values of Li, Na, F and Cl is
Na < Li < F < Cl
In the light of the above statements, choose the
correct answer from the options given below :
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is false but Statement II is true
(4) Statement I is true but Statement II is false
Ans. (1)
Sol..
1
(i) Na < Li < Cl < F
I.E. in kJ/mol 496 520 1256 1681
eg
(ii) Na < Li < F < Cl
H in kJ/mol –53 –60 –328 –349
77. For a strong electrolyte, a plot of molar conductivity
against (concentration)1/2 is a straight line, with a
negative slope, the correct unit for the slope is
(1) S cm2 mol–3/2 L1/2 (2) S cm2 mol–1 L1/2
(3) S cm2 mol–3/2 L (4) S cm2 mol–3/2 L–1/2
Ans. (1)
Sol.
o
mm
AC
Units of
AC
= S cm2 mole–1
Uits of A = S cm2 mole–3/2 L1/2
78. A first row transition metal in its +2 oxidation state
has a spin-only magnetic moment value of 3.86 BM.
The atomic number of the metal is
(1) 25 (2) 26
(3) 22 (4) 23
Ans. (4)
Sol. 22Ti+2 [Ar]3d2
23V+2 [Ar]3d3
25Mn+2 [Ar]3d5
26Fe+2 [Ar]3d6
79. The number of unpaired d-electrons in
[Co(H2O)6]3+ is______
(1) 4 (2) 2
(3) 0 (4) 1
Ans. (3)
Sol. [Co(H2O)6]+3
eg
t2g
d6
Co+3
No unpaired electrons
80. The number of species from the following that
have pyramidal geometry around the central atom
is________
2 2 2 2
2 3 4 3 2 7
S O ,SO ,SO ,S O
(1) 4 (2) 3
(3) 1 (4) 2
Ans. (3)
Sol.
S
O–
O–
O
Pyramidal
S
O–
O–
O
··
S
S
O–
O–
O
O
S
O–
O
O
O
S
O
O–
O
,
,
tetrahedral with respect to central atom
SECTION-B
81. The maximum number of orbitals which can be
identified with n = 4 and ml = 0 is_____
Ans. (4)
Sol.
n = 4 ,
4s
4p
4d
4f
m
1
1
1
1
So answer is 4.
82. Number of compounds/species from the following
with non-zero dipole moment is_____
BeCl2, BCl3, NF3, XeF4, CCl4, H2O H2S, HBr,
CO2, H2, HCl
Ans. (5)
Sol.
3 2 2
(µ 0)
Polar molecule: NF ,H O,H S,HBr,HCl
2 3 4 4 2 2
(µ 0)
NonPolar molecule: BeCl ,BCl ,XeF ,CCl ,CO ,H
So answer is 5.
83. Three moles of an ideal gas are compressed
isothermally from 60 L to 20 L using constant
pressure of 5 atm. Heat exchange Q for the
compression is – ____ Lit. atm.
Ans. (200)
Sol. As isothermal U = 0
and process is irreversible
Q = –W = – [– Pext (V2 – V1)]
Q = 5 (20 – 60) = – 200 atm-L
84. From 6.55 g of aniline, the maximum amount of
acetanilide that can be prepared will be___×10–1 g.
Ans. (95)
Sol.
NH2
+
CH3 – C – Cl
O
NH – C – CH3
O
93 g aniline form 135 gm acetanlide
so 6.55 g anilne form
135 6.55 9.5
93
95 × 10–1
85. Consider the following reaction, the rate
expression of which is given below
A + B C
rate = k [A]1/2 [B]1/2
The reaction is initiated by taking 1M
concentration A and B each. If the rate constant (k)
is 4.6 × 10–2 s–1, then the time taken for A to
become 0.1 M is_____sec. (nearest integer)
Ans. (50)
Sol.
2.303 1
K log
t 0.1
4.6 × 10–2 =
2.303
t
t = 50 sec.
86. Phthalimide is made to undergo following
sequence of reactions.
Phthalimide ' P'
Total number of bonds present in product 'P'
is/are
Ans. (8)
(i)KOH
(ii)Benzylchloride
Sol.
O
. .
K+OH–
N – H
O
O
. .
N K
O
CH2 – Cl
SN2
–KCl
O
. .
N – CH2
O
(P)
(Phthalimide)
Total number of -bonds present in product P is 8
87. The total number of 'sigma' and 'Pi' bonds in
2-oxohex-4-ynoic acid is____.
Ans. (18)
Sol.
O
HO–C–C–CH2–C C–CH3
O
2-Oxohex-4-ynoic acid
Number of -bonds = 14
Number of -bonds = 4
= 18
88. A first row transition metal with highest enthalpy
of atomisation, upon reaction with oxygen at high
temperature forms oxides of formula M2On (where
n = 3,4,5). The 'spin-only' magnetic moment value
of the amphoteric oxide from the above oxides
is___ BM (near integer)
(Given atomic number : Sc : 21, Ti : 22, V : 23,
Cr : 24, Mn : 25, Fe : 26, Co : 27, Ni : 28 ,Cu : 29,
Zn : 30)
Ans. (0)
Sol. 'V' has highest enthalpy of atomisation (515 kJ/mol)
among first row transition elements.
V2O5
Here 'V' is in +5 oxidation state
V+5 1s2 2s2 2p6 3s2 3p6 (no unpaired electrons)
89. 2.7 Kg of each of water and acetic acid are mixed,
The freezing point of the solution will be –x °C.
Consider the acetic acid does not dimerise in
water, nor dissociates in water x = ______(nearest
integer)
[Given : Molar mass of water = 18 g mol–1,
acetic acid = 60 g mol–1]
f
K
2
HO
: 1.86 K kg mol–1
f
K
acetic acid : 3.90 K kg mol–1
freezing point : H2O = 273 K, acetic acid = 290 K]
Ans. (31)
Sol. As moles of water > moles of CH3COOH
water is solvent.
T°F – (TF)S = KF × M
0 – (TF)S = 1.86 ×
2700 / 60
2700 /1000
(TF)S = –31°C.
90. Vanillin compound obtained from vanilla beans,
has total sum of oxygen atoms and electrons is__
Ans. (11)
Sol. Vanillin compound is an organic compound
molecular formula C8H8O3. It is a phenolic aldehyde.
Its functional compounds include aldehyde, hydroxyl
and ether. It is the primary component of the extract
of the vanilla beans.
OCH3
OH
O
H
Total sum of oxygen atoms and -electrons is 3 + 8 = 11
Total number of oxygen atoms = 3
Total number of -bonds = 4
Total number of -electrons = 8
