Download SECTION – A {Mathematics} 1. Find the sum of all natural numbers

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SECTION – A {Mathematics}
1.
Find the sum of all natural numbers between 250 and 1000 which are exactly divisible
by 3
(a) 866
(b) 867
(c) 868
(d) 869
Solution:
First natural no. between 250 & 1000
divisible by 3
= 252
Last Natural No. between 250 & 1000
Divisible by 3
= 999
252 + 255 + …………999
Let no. of terms = n
999 = 252 + (n – 1) 3
999 – 252 = (n – 1) 3
747
3
n 1
249 = n – 1
S
250
[2
2
252
n 250
250 1 3]
= 125[504 + 249 × 3]
= 125 [504 + 747] = 125 × 1251 = 156375
There is no option in the given question (in order to minimize fluke and guessing
probability )
2.
Find the value of k so that: 2k + 1, k2 +k + 1, 3k2 –3k +3 are in A.P.
(a) 2,3
(b) 1,-2
(c) 1,2
(d) 3,4
Solution:
2k + 1, k2 + k + 1, 3k2 – 3k + 3 are in A.P
(k2 + k + 1) – (2k + 1) = (3k2 – 3k + 3) – (k2 + k + 1)
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k2
+ k + 1 – 2k – 1 =
3k2
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– 3k + 3 –
k2
–k–1
k2 – k = 2k2 – 4k + 2
0 = 2k2 – 4k + 2 + k – k2
k2 – 3k + 2 = 0
k2 – 2k – k + 2 = 0
k(k – 2) – 1 (k – 2) = 0
k = 1, 2
Ans. (c)
3.
If the first, second and last term of an A.P are a, b and 2a respectively, then its sum is
(a)
ab
2(b a)
(b)
ab
(b a)
(c)
3ab
2(b a)
(d) none of these
Solution:
First term of A.P. =a
2nd term of A.P. = b
common difference = 2nd term – first term
d b a
last term of A.P. = 2a
a
n 1 d 2a
a
n 1 b a
n 1
a
b a
n 1
Sn
2a
n
2a
2
a
b a
n
b a a
b a
n
b
b a
n 1 d
b
2a
2 b a
b
b a
1 b a
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b
2a
2 b a
b b a
b a
b
2a a
2 b a
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b a
3ab
a b a
Ans. (c)
4.
Find the value of x of the given quadratic equation
(5 2 6)x
(a) 6,
2
3
(5 2 6)x
2
3
(b) 5,
6
=10
(c) 3,
5
(d) 2,
3
2
Solution:
5 2 6
x2 3
5 2 6
Let 5 2 6
y2 1
y
10
52
2 6
5 2 6
2
25 24
5 2 6
1
5 2 6
y
x2 3
5 2 6
1
y
x2 3
x2 3
5 2 6
5 2 6
5 2 6
Let 5 2 6
y
5 2 6
x2 3
1
5 2 6
x2 3
10
y
10
10
y 2 1 10y
y 2 10y 1 0
y
10
100 4
2
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y
10
96
5 2 6
2
5 2 6
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x2 3
5 2 6
x2 – 3 = 1
x2 = 4
x=
5 2 6
2
1
5 2 6
x2 3
x2 – 3 = – 1
x2 = 2
x
2
Ans. (d)
5.
Solve the given quadratic equation 6
6
(a) 1
(c) 3
(b) 2
6 ..............
(d) 4
Solution:
Let x = 6
x
6
6 ..............
6 x
x2 = 6 + x
x2 – x – 6 = 0
x2 – 3x + 2x – 6 = 0
x (x – 3) + 2 (x – 3) = 0
x = - 2, 3 (Negative value is rejected
Ans. x = 3
6.
Form a quadratic equation whose roots are given as: 3 + 5 , 3 – 5
(a) x2 + 6x - 4 = 0 (b) x2 – 6x + 4 = 0
(c) x2 – 6x – 4 = 0
(d) x2 +6x + 4 = 0
Solution:
x2 – 5x + P = 0
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S
3
5 3
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5
=6
P
3
5 3
5
9 5 4
x2 – 6x + 4 = 0
Ans. (b)
7.
If the root of the equation x2 px q 0 and x2 qx p 0 differ by the same quantity,
then p q k .find the value of k
(a) 1
(b)-1
(c)4
(d) -4
Solution:
x2 px q 0 , x2 qx p 0
square both the sides
2
2
2
2
4
4
p2 4q q2 4p
p2 q2 4q4p
p q p q
p q
4 p q
4
Ans. (d)
8.
Determine the ratio in which the line y –x + 2 = 0 divides the line segment joining the
points (3, –1) and (8,9)
(a) 1:3
(b)1:2
(c)1:4
(d) 2:3
Solution:
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P
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8k 3 9k 1
,
k 1 k 1
Since point P lies on the line
y–x+2=0
9k 1
k 1
8k 3
k 1
9k 1
8k 3
k 1
2 0
2 k 1
0
9k – 1 – 8k – 3 + 2k + 2 = 0
3k – 2 = 0
k=
2
3
Ratio 2 : 3
Ans. (d)
9.
What point on x-axis is equidistant from the points A (1, 3) and B (2, -5)?
(a)
16
,0
2
(b)
19
,0
2
(c)
18
,0
2
(d)
21
,0
2
Solution:
Let the required point on A(1, 3)
x-axis is P(x, 0) B(2, – 5)
PA = PB
x 1
2
0 3
2
x 2
2
x 5
2
Square both sides
x 1
2
9
x 2
2
25
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x2
+ 1 – 2x + 9 =
x2
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+ 4 – 4 x + 25
4x – 2x + 10 = 29
2x = 19
19
2
x
Ans. (b)
10. The coordinate of centroid of a triangle are ( 3,2 ) and two of its vertices are (2 3, 1)
and (2 3,5) Find the third vertex of the triangle.
3, 2
(a)
(b)
3,2
(c)
3, 2
(d)
3,2
Solution:
x1 x 2 x3 y 1 y 2 y 3
,
Centroid =
3
3
Let the coordinates of the third vertex = (x, y)
x 2 3 2 3 y 1 5
,
3
3
3,2
x 4 3
3
y 4
2
3
3,
x 4 3 3 3, y 4 6
x
3, y
Ans.
3, 2
2
Ans. (a)
11. Find the third vertex of a triangle whose centroid is (1, 4) and two vertices are (–1, 4)
and (3, –1).
(a)(1,9)
(b) (1,6)
(c) (1,5)
(d) (1,8)
Solution:
Two vertices (–1, 4) & (3, –1)
centroid = (3, –1)
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when the three vertices of the
are (x1, y2), (x2, y2) & (x3, y3)
x1 x 2 x3 y 1 y 2 y 3
,
3
3
Coordinates of centroid are =
Let the third vertices of the
1,4
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be (x, y).
x 1 3 y 4 1
,
3
3
x 1 3
y 4 1
1 &
4
3
3
x 2 3 &
x 1
y 3
4
3
y 3 12
y 9
Ans.(a) (1, 9)
12. Find the value of x such that PQ = QR, where P, Q and R are (6,1), (1,3) and
(x, 8) respectively.
(a) –1
(b) –2
(c) –3
(d) –4
Solution:
PQ = QR
6 1
P(6, 1) Q(1, 3)
2
1 3
2
1 x
R(x, 8)
2
3 8
2
square both the sides
25 + 4 = (1 – x)2 + 25
4 = 1 + x2 – 2x
x2 – 2x = 3
x2 – 2x – 3 =0
x2 – 3x + x – 3 =0
x(x –3) + 1(x–3)=0
x = –1, 3
Ans. (a)
13. If pth of an AP is q and qth term is p, then its nth term is
(a) (p +q –n)
(b) (p –q +n)
(c) (p +q +n)
(d) (p –q –n)
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Solution:
A + (p – 1) D = q
..(1)
A + (q – 1) D = p
..(2)
(1) – (2)
(p – q) D = q – p
D=–1
A + (p – 1) D = q
A + (p – 1) (– 1) = q
A–p+1=q
A=p+q–1
Now, nth term = A + (n – 1) D
= (p + q – 1) + (n – 1) (– 1)
=p+q–1–n+1
=p+q–n
Ans. (a)
14. The value of a such that x2 – 11x + a =0 and x2–14x + 2a = 0 may have a common root is
(a) 0
(b) 12
(c) 24
(d) 32
Solution:
Let the common root is
2
11
a 0 ............ 1
2
14
2a 0 ........ 2
Solving by cross multiplication method
2
22a 14a
a 2a
1
14 11
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2
8a
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1
3
a
2
a
,
3 8a
a
2
8
2
a
3
a2
9
8
2
8
a
3
8a
3
a2
3
8a
a2
8a 0
3
a
a
8
3
a
a 24
3
a a 24
0
0
0
a 0, 24
Ans. (a), (c)
15. The points (a2, 0) (0, b2) and (1, 1) are collinear if
(a) 1
(b)2
(c)3
1
a2
1
b2
k . find the value of k
(d)4
Solution:
a2 ,0 0,b2 1,1
Since the three points are collinear
x1 y 2 y3
a2 b2 1
x2 y 3 y 1
01 0
x3 y 1 y 2
1 0 b2
0
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2 2
ab
a
2
2
0 b
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0
a2 b2 a2b2
Dividing both sides by a b2
a2
a2b2
1
a2
b2
a2b2
1
b2
a2b2
a2b2
1
k 1
Ans. (a)
16. If the pth, qth and rth terms of an A.P are a, b, and c respectively, then:
(b–c) p + (c–a)q +(a–b)r = k. Find the value of k
(a) 0
(b) 1
(c) 2
(d) 3
Solution :
A + (p –1) D = a
A + (q –1) D = b
A + (r –1) D = c
(b – c)p + (c – a) q + (a – b)r = k
[{A + (q –1) D} – {A + (p –1) }] p + [{ A + (r –1) D} – { A + (p –1) D}] r
= [(q –r)D] p + [(r –p) D]q + [(p –q) D] r
= D[pq –pr +rq –pq + pr –qr]
=0
Ans. (A)
17. If the sum of p, q and r terms of an A.P be a, b and c respectively. Then:
a
b
c
(q r)
(r p) (p q) k
p
q
r
(a) 0
(b) 1
(c) 2
(d) 3
Solution :
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p
2A
2
p 1 D
a
q
2A
2
q 1 D
b
r
2A
2
r 1 D
a
q r
p
Now,
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c
b
r p
q
c
p q
r
=
p
2A
2p
=
1
2A
2
=
1
2A q r r p p q
2
p 1 D p r
q
2A
2q
1
2A
2
p 1 D q r
q 1 D r p
q 1 D r p
r
2A
2r
1
2A
2
r 1 D p q
r 1 D p q
D pq pr q r qr pq r p rp rq p q
=0
Ans. (A)
18. The digits of a positive integer, having three digits are in A.P and their sum is 15. The
number obtained by reversing the digits is 594 less than the original number. Find the
number.
(a) 850
(b)851
(c) 852
(d) 853
Solution :
Look at the options only 3rd option is satisfying all (C) the condition as per question
19. Find the value of x in the given quadratic equation 51+x +51–x = 26.
(a) 1
(b) 2
(c) 3
(d) 4
Solution :
55+x + 51–x = 26
51
.
5x
51
+ x
5
26
Let 5x = y
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5y +
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5
26
y
5y2 – 26y + 5 = 0
5y2 – 25y +–y + 5 =0
5y(y–5) –1(y –5) =0
y=
5x =
1
,5
5
1
5
x = –1
5x =5
x=1
Ans. (a)
20. The co-ordinates of A, B, C are (6, 3), (3, 5) and (4, –2) respectively and P is a point
(X, Y), Then
area PBC
area ABC
(a) 6
X Y 2
. The value of k is :
k
(b) 7
(c) 8
(d) 9
Solution :
A(6, 3), B(3, 5), C(4, –2)
area of PBC
areaof ABC
P (x, y)
x y 2
k
1
3 2 y 4 y 5 x 5 2
2
1
6 7 3 2 3 4 3 5
2
1 6 3y 4y 20 7x
42 15 81
7x y 26
19
x y 2
k
x y 2
k
x y 2
k
We cannot find the value of k from this sequence.
There is no option in the given question (in order to minimize fluke and guessing
probability )
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21. The coordinates of two points A and B are (–1, 4) and (5, 1) respectively Find the
coordinates of the point P which lie on extended line AB such that it is three times as
far from B as from A.
(a) (–4,–11/2)
(b) (4,11/2)
(c) (4,–11/2)
(d) (-4,11/2)
Solution :
PA = 3P B
PA
PB
3
1
AB PB 3
PB
1
AB PB 3PB
AB 2PB
AB 2
PB 1
5=
2X
1
2 1
2X 1
3
5
15 1 2x
2x 16
x 8
1=
2y 4
2 1
3 2y 4
3 – 4 =2y
y=
1
2
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8,
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1
2
PB = 3PB
1
2xx 1 5
2 1
3 2x 5
8 2x
x
4=
4
2y 1 1
2 1
12 2y 1
11 2y
y
11
2
4,
11
2
Ans. (d)
22. Find the area of triangle formed by joining the mid points of the sides of triangle
whose vertices are (0,–1), (2, 1) and (0, 3). Find the ratio of this triangle with given
triangle.
(a) 1:4
(b)4:1
(c) 2:1
(d)1:2
Solution :
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Area of DEF
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1
12 1
2
11 0
00 2
1
01 3
2
23 1
0 1 1
1
1 1 1
2
Area of ABC
=
1
8 4
2
Areaof DEF
Areaof ABC
1
4
Ans.(a)
23. Form a quadratic equation whose roots are given as: 3,
(a)
3x2 4x
3 0
3 0
(c) 3x2 4x
Solution :
1 3 1 1
5= 3
3
3
3
1
1
P= 3.
3
x2 – 5x + P =0
x2 –
1
3
(b) 3x2 4x
3 0
(d) 3x2 4x
3 0
4
x 1 0
3
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3x
2
4x
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3 0
Ans. (c)
24. x
12
12
12........
(a)1
(b) 2
(c)3
(d)4
Solution :
x 12
12
12
x
12 x
square both the sides
x2 = 12 +x
x2 – x –12 =0
x2 –4x +3x –12 =0
x(x–4) +3(x –4) =0
(x–4) (x +3)=0
x = –3, 4
Ans. (d)
25. If Sr denotes the sum of the first r terms of an A.P. Then, S3n : (S2n – Sn) is
(a) n
(b) 3n
(c) 3
(d) none of these
Solution :
Sn
n
2a
2
Sn
S2n Sn
n 1 d
3n
2a
2
2n
2a
2
3n
2a
2
3n 1 d
3n 1 d
n
2a
2
n 1 d
3n 1 d
n
4a 2d 2n 1
2
2a
n 1 d
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3n
2a
2
3n 1 d
n
4a 2d 2n 1
2
3 2a
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2a
n 1 d
3n 1 d
2a 2d 2n 1
n 1 d
6a 3d 3n 1
2a d 4n 2 n 1
6a 9nd 3d
2a d 3n 1
6a 9nd 3d
2a 3nd d
3 2a 3nd d
2a 3nd d
3
Ans.(c)
26. If 7 times of 7th term of an AP is equal to 11 times the 11th term. then find the value of
18th term
(a)1
(b)0
(c) 2
(d)3
Solution :
7[A + (7 –1) D] = 11[ A + (11 – 1) D]
7A +42 D = 11A + 110 D
42D –110D = 4A
–68D = 4A
4A +68D=0
2A + 34D =0
A +17D =0
A + (8 –1)D=0
18th
Ans. (b)
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27. If , are roots of the equation ax2 + bx + c = 0 find the value of : 3 + 3
3abc b3
3abc b3
3abc b3
3abc b3
(a)
(b)
(c)
(d)
a3
a3
a3
a3
Solution :
ax2 + bx + c =0
b
c
,
a
a
3
3
2
=
3
b
=
a
2
b
a
b b2
=
a a2
3
2
2
c
a
3c
a
b b2 3ac
=
a
a2
b3 3acb 3abc b3
=
a3
a3
Ans. (d)
28. If the roots of the equation p(q r)x2 q(r p)x r(p q) 0 are equal, then
1 1
p r
N
.
q
then the value of N is:
(a) 0
(b) 1
(c) 2
(d) 3
Solution :
p(q –r)x2 + q(r –p) x +r(p –q) =0
D =0
q r p
2
4pr q r p q
0
q2 (r –p)2 – 4pr(pq –q2 –pr + qr) =0
q2(r 2 + p2 –2pr) – 4p2rq + 4q2pr + 4p2r2 – 4p2rq + 4q2pr +4p2r2 – 4pqr2 =0
q2r2 + p2q2 + 2prq2 +(2pr)2 – 4p2rq –4pqr2 =0
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(qr + pq –
2pr)2
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=0
qr + pq =2pr
1 1
p r
2
q
N 2
Ans. (c)
29. If the point (a, 0) (0,b) and (1,1) are collinear then
(a)1
(b) 2
1 1
k find the value of k
a b
(c)3
(d)4
Solution :
(a, 0), (0, b) (1, 1)
x 1(y2 –y3) + x2(y3 – y1) x3(y1 – y2) =0
a(b –1) + 0(1 –0) + 1(0 –b) =0
ab –a –b =0
a + b =ab
1 1
1
a b
Ans. (a)
30. In an A.P if Sn = n2p and Sm = m2 p , m≠ n, show that Sp = k p2
(a)1
(b) 2
(c) 3
(d) 4
Solution :
Sn = n2p
n
2A
2
n 1 D
2A +(n –1)D = 2np
n2p
…..(1)
……(2)
(1) –(2)
(n –m) D =2np –2mp
D=
2p n m
n m
D = 2p
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2A +(n–1) D = 2np
2A + (n –1)2p = 2np
A + np –p = np = A p
Sp =
=
p
2A
2
p
2p
2
p 1 D
p 1 2p
p
= ,2p 1 p 1
2
=p2 . p = p3 = p . p2
k p Ans.
There is no option in the given question (in order to minimize fluke and guessing
probability)
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SECTION – B {PHYSICS}
31. Velocity of light in medium 1 is 2.4×107 m/s and velocity of light in medium 2 is 1.8×
107 m/s then the refractive index of medium 2 with respect to medium 1 is
(a) 3
(b) 4
4
(c) 1
3
3
(d) 1
4
Solution: (b)
1
n2
V1
V2
2.4 107 4
=
1.8 107 3
Ans. (b)
32. If the a convex lens of refractive index n1 is placed in a medium of refractive index n2
such that n2 > n1 the lens will behave as a
(a) convex lens itself
(b) diverging lens
(c) concave mirror
(d)convex mirror
Solution: (b)
33. A concave mirror forms an erect image twice the size of an object. The object distance
from the mirror is:
(a) f/2
(b) 2f
(c) 3f/2
(d) 2f/3
Solution: (a)
Since
h2
h1
v
u
Since h2 = 2h1
Therefore v = 2u
According to mirror formula
1
2u
1
u
1
v
1
u
1
f
1
f
Therefore u =
f
2
34. A point source of light B is placed at a distance L in front of the centre of a mirror of
width d hung vertically on a wall. A man walks in front of the mirror Along a line
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parallel to a mirror at a distance 2L from it, as shown in fig. find the greatest distance
over which he can see the image of the light source from the mirror.
(a) d
(b) d/2
(c) 2d
(d) 3d
Solution:
From geometry, CD = d = C’D’
BD = DP = d/2; CB = QC = d/2
P’Q’ = C’D’ + P’D’ + Q’C’
ND
DP
ND'
D'P'
L
d/2
2L
D'P'
D'P' d
Similarly, Q’C’ = d
Hence, distance through which the man can observe the image
= Q’C’ + C’D’ + D’P’ = d +d + d = 3d
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35. Two plane mirrors AB and CD each of length 2m are arranged parallel to each other
3 m apart and a ray of light is incident at A as shown in figure. How much refraction
does the ray of light undergo?
(a) 4
(b) 2
(c) 3
(d) 1
Solution: (c)
DC
AD
tan300
DC
DC
AD tan 300
3
1
3
DC = 1 m
Now no. of images = 2 + 1
=3
36. A glass lens has a focal length 5 cm in air. In water its focal length would be
(a) 5 cm
(b) More than 5 cm but finite
(c) infinite
(d) Less than 5 cm
Solution:(b)
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37. In the above equation what is the angle of deviation of the incident ray.
(a) 300
(b) 400
(c)900
(d) 1200
Solution:
Since the instant ray came out with same angle (450) Therefore angle of deviation is
900.
Ans.(c)
38. Consider two convex lens of same focal length f separated by a distance 2f. If a parallel
rays of light falls on one convex lens as shown in figure then the final image will be
formed at
(a) at F
(b) at infinity
(c) at 2f
(d) none of these
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Solution : (b)
Since the distance between the two lenses is 2f their for the focus of the two lenses will
coincide each other thus first of all ray will focus at point f and then it will move to
infinity when passes through second lens.
39. A air bubble in water will act like a
(a) Concave mirror
(b) Convex lens
(c) Convex mirror
(d) Concave lens
Solution:(d)
40. If camera a has f= 4.5 m lens and camera B has f =2.8m lens, and the diameters of both
the lenses are equal, then
(a) A is better for photographing fast moving objects
(b) Pictures taken by B will always be sharper
(c) Pictures taken by A will always be sharper
(d) B is better for photographing fast moving objects
Solution:(c)
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SECTION – C {Chemistry}
41. In CH3 CH2 OH the bond that undergoes heterolytic cleavage most readily is (a)C–C
(b)C–O
(c)C–H
(d)O–H
Solution: (b)
42. 'Drinking alcohol' is very harmful and it ruins the health 'Drinking alcohol' stands for(a) drinking methyl alcohol
(b) drinking ethyl alcohol
(c) drinking propyl alcohol
(d) drinking isopropyl alcohol
Solution: (b)
drinking ethyl alcohol
43. Which of the following is a soap (a) Sodium alkylsulphonate
(b) Sodium chloride
(c) Sodium sulphonate
(d) Sodium stearate
Solution: (d) Sodium stearate (C17H35 COONa)
44. Acidic hydrogen is present in (a) Ethyne
(b) Ethene
(c) Ethane
(d) Benzene
Solution: (a)
Ethyne (CH CH)
RC CH
RC C– + H+
45. An organic compound 'A' with molecular formula C2H4O2 turns blue litmus to red and
gives brisk effervescence with sodium hydrogen carbonate. Identify the compound (a) Methanoic acid
(b) Ethanoic acid
(c) Propanoic acid
(d) Butanoic acid
Solution: (b)
Ethanoic acid (CH3COOH)
46. Power alcohol contains (a) 50% petrol and 50 % ethanol
(b) 75% petrol and 25 % ethane
(c) 25% petrol and 75% ethane
(d) 70% petrol and 30% ethane
Solution:(b)
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47. Glacial acetic acid is a(a) frozen acetic acid
(b) 5-8% of solution of acetic acid in water
(c) mixture of acetic acid and alcohol
(d) None of these
Solution:(a)
48. A & B both compounds give H2 gas with solution. If A & B react in presence of acid
catalyst then they from ethyl acetate. Thus, A & B would be (a) CH3COOH, CH3OH
(b) HCOOH, CH3COOH
(c) CH3COOH, C2H5OH (d) C3H7COOH, C3H7OH
Solution :(a)
49. The IUPAC name for
(a) 1, 1-Dimethyl-1-1,3-butanediol
(b) 2-methyl-2,4-pentanediol
(c) 4-methyl-2,4-pentanediol
(d) 1,3,3-Trimethyl1-1,3-propanediol
Solution :(b)
50. The number of 40 carbon atoms in 2, 3, 4, 4-tetramethyl pentane is (a) 1
(b) 2
(c) 3
(d)4
Solution:(b)
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SECTION – D {Biology}
51. Mucosal lining of uterus is
(a) Mesometrium
(b) Endometrium
(c) Epimetrium
(d) Epidermis
Solution: (b)
52. In spirogyra, asexual reproduction takes place by
(a) breaking up the filaments into smaller bits
(b) division of a cell into two cells
(c) division of a cell into many cells
(d) formation of young cells from older cells
Solution: (a)
53. Jasmine is multiplied vegetatively through
(a) Stem cutting
(b) Leaves
(c) root cutting
(d) Layering
Solution: (d)
54. Out of the four sides I, II, III, IV whose details are shown, which one should be focused
under the microscope for showing budding in yeast?
(a) I
(b) II
(c) III
(d) IV
Solution: (c)
55. Seed is formed from
(a) Unfertilized ovary
(c) Fertilized ovule
(b) Fertilized ovary
(d) Unfertilized ovule
Solution: (c)
56. Characters transmitted from parents to offspring are present in
(a) Cytoplasm
(b) Ribosome
(c) Golgi bodies
(d) Genes
Solution :(d)
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57. Gestation period in humans is
(a) 270 days
(b) 290 days
(c) 200 days
(d) 245 days
Solution :(a)
58. The slide of Amoeba showing binary fission is characteristic by
(a) Rounded Amoeba with elongated nucleus
(b) Amoeba having pseudopodia and rounded nucleus
(c) Amoeba constructed in the middle with round nucleus
(d) Amoeba constructed in the middle with elongated nucleus
Solution :(d)
59. Which of the following diagrams shows Yeast cell undergoing budding
(a)I
(b) II
(c) III
(d) IV
Solution :(d)
60. A surgical technique of reversible but permanent contraception in females is
(a) Vasectomy
(b) Tubectomy
(c) Norplant
(d) IUCD
Solution :(b)
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