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SECTION – A {Mathematics}
1.
In a certain office, 1/3 of the worker are women1/2 of the women are married and 1/3
of the married women have children. If 3/4 of the men are married and 2/3 of the
married men have children, what part of the workers are without children?
(a) 5/18
(b) 4/9
(c) 11/18
(d) 17/36
Solution:
Let the total no. of workers = x
Women =
1
x
3
Man =
Married women =
1 x
2 3
2
x
3
x
6
Married women having children =
Married Men =
3 2
x
4 3
1 x
3 6
x
18
x
2
Married Men having children =
2 x
3 2
x
3
Total No. of workers having children
=
x
18
x
3
x 6x
18
7x
18
Total No. of workers without children
=x
7x
18
18x 7x
18
11x
18
Wor kers without children
Total no. of wor kers
=
11x
18x
11
18
Ans.(c)
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2.
x2 5x 6
Solve for x, 2
x 4x 4
3x 1
,x
5x 2
(a) -1
(b) -2
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2,
2
5
(c) -3
(d) –4
Solution:
x2 5x 6
x 2 4x 4
3x 1
5x 2
x2 3x 2x 6 3x 1
x2 2x 2x 4 5x 2
x 3 x 2
x 2 x 2
3x 1
5x 2
x 3 5x 2
3x 1 x 2
5x2 + 13x – 6 = 3x2 + 7x + 2
2x2 + 6x – 8 = 0
x2 + 3x –4 = 0
x2 + 4x – x – 4 = 0
(x + 4) (x – 1) = 0
3.
x = – 4, 1
A man leaves half of the property for his wife, one third for his son and remaining for
his daughter. If daughter’s share is Rs. 60,000, how much money did the man leave?
(a) Rs 3,60,000
(b) Rs 1,80,000
(c) Rs 1,20,000
(d) 1,60,000
Solution (c)
Let the total property = Rs x
Wife’s share = Rs
Son’s share = Rs
x
2
x
3
Daughter’s share (remaining of the property) = x
x
2
x
3
x
6
But Daughter’s share = Rs 60, 000
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x
60,000
6
x = Rs 3, 60, 000
Wife’s share =
Son’s share =
4.
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360000
= Rs 1, 80, 000
2
360000
= Rs 1, 20, 000
3
There are some benches in a classroom. If 4 students sit on each bench, then 3 benches
are left unoccupied. However, if 3 students sit in each bench, 3 students are left
standing. How many students are there in the class?
(a) 36
(b) 48
(c) 56
(d) 64
Solution: (b)
Let the no. of students = x
x
3
4
5.
x
3
x
4
x
48
x
1
3
3 1
4x 3x
12
4
A number consists of two digits. The digit at ten’s place is three time the digit at unit’s
place. The number formed by reversing the digits is 54 less than the original number.
Find the number?
(a) 91
(b) 92
(c) 93
(d) 94
Solution: (c)
Let the digit at units place be = x
Therefore digit at ten’s place = 3x
Number = 10(3x) + x = 31 x
New number by reversing the digits = 10x + 3x = 13x
According to given condition we get
31x – 13x = 54
18x = 54
x=3
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Digit at unit place = 3 digit at ten’s place = 3 × 3 = 9
Number = 93
6.
A and B can do a piece of work in 8 days which A alone can do in 12 days. In how many
days B can alone do the same work?
(a) 25 days
(b) 24 days
(c) 30 days
(d) 40 days
Solution:(b)
1
A
1
B
1
8
1 1
12 B
1
B
1
8
1 1
8 12
1 3 2
B 24
1 1
B 24
B 24 days
Ans. (b)
7.
A diagonal of a rectangle is inclined to one side of the rectangle at 350. The acute angle
between the diagonals is
(a) 900
(b) 950
(c) 70o
(d)1300
Solution:(c)
Ext. angle = sum of two opposite angles
8.
x = 350 + 350 = 700
Through points A, B, C of the ABC , lines are drawn parallel to the sides BC, CA and AB,
respectively. These lines form PQR , then BC = K QR. Then the value of K is
(a) 1
(b)2
(c)1/2
(d) ¼
Solution: (c)
Given : In the Fig, BC  QR, AC  PR and AB  QP.
Since QC  AB and QA  BC,
ABCQ is a parallelogram.
(opposite sides are  )
BC = AQ
(opposite sides of  gm) (1)
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Similarly BCAR is parallelogram and,
BC = AR
(2)
Adding Eqs. (1) and (2), we get
2 BC = AQ + AR
2 BC = QR
BC
9.
1
QR .
2
Which of the following is not true for a parallelogram?
(a) opposite sides are equal
(b) opposite angles are equal
(c) diagonals bisect each other
(d) opposite angles are bisected by the diagonals
Solution:(c)
Angles will be bisected only when all sides are equal
10. The line joining mid-points of the diagonals of a trapezium which is parallel to the
parallel sides of the trapezium is equal to ______________ of the difference of the parallel
sides.
(a) double
(b)half
(c) thrice
(d) one third
Solution: (b)
Given Trapezium ABCD Fig , in which M and N are the mid-points of the diagonals AC
and BD.
To Prove : MN  AB
and
MN
1
AB CD
2
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Construction : Join C to N and produce it to meet AB at E.
Proof : In the
CDN and the
BNE,
DN = NB
DNC =
BNE
CDN =
NBE
CDN
BNE
(vert. opposite angles)
(alternate angles of  sides)
CD = BE
and NC = NE
Now in the
ACE, M and N are the mid-points of CA and CE respectively.
MN
1
AE and MN  AE
2
MN
1
AB BE and MN  AB
2
MN
1
AB CD and MN  AB
2
11. ABC is triangle, right angled at B and P is mid-point of AC. Then PB = PA =AC= k. find
the value of k.
(a)
1
2
(b)
1
3
(c)
1
4
(d)
1
8
Solution: (a)
Given : In the ABC , P is the mid-point of AC and
To Prove : PB
AP
B = 900.
1
AC.
2
Construction: Draw PK  BC
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Proof :
2
ABC
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1800
(the sum of the interior angles of  sides)
But
900
ABC
2
But
900
2 1800
1
1
900
Now in the PAK and the PBK ,
1
2
(each 900)
PK = PK
(common)
AK = BK (P is the mid-point of AC and PK  BC)
PAK
PKB
PA = PB
But
PA = PC
Now AC = PA + PC
PA
1
AC
2
PB
PA
AC = PA + PA = 2PA
1
AC
2
12. In a quadrilateral ABCD,
(a) 600
A
(b) 800
C is 2 times B
D . If
(c) 1200
A 400 , then B
(d) None of these
Solution: (d)
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A+
C=2(
400 +
Also
B+
C=2
A+
2
B+2
3(
B+
B+
D)
B+2
B+
C+
D+
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D
D = 3600
B+
D = 3600
D) = 3600
D = 1200
Now using Eqn (1)
400 +
C=2(
400 +
B+
D)
C = 2 × 1200
C = 240 – 40 = 2000
We cannot determine the angle B
13. AD, the median of ABC divides it into two
(a) Right triangles
(b) Congruent triangles
(c) Triangles of equal area
(d) Two equilateral triangles
Solution:(c)
Triangles of equal area
14. The mid-point of the sides of a triangle along with any one of the vertices as the fourth
point make a parallelogram whose area is equal to
(a) ar ABC
(b)
1
ar ABC
4
(c)
1
ar ABC
3
(d)
1
ar ABC
2
Solution:(d)
Area of  gm BDEF
area  gm CEFD
area of  gm AFDE
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area ABC
2
15. The figure obtained by joining the mid-points of the adjacent sides of a rectangle of
sides 18 cm and 8 cm is :
(a) a rectangle of area 72 cm2
(b) a square of area 72 cm2
(c) a trapezium of area 72 cm2
(d) a rhombus of area 72 cm2
Solution:(d)
Area of rectangles = 18 × 8 = 144 cm2
Area of rhombus PQRS =
=
1
area ABCD
2
1
× 144 = 72 cm2
2
16. In ABC, AD is the median, such that area of the ADC 10cm2 and BC = 5 cm. Find the
altitude from the vertex A to the side BC.
(a) 6cm
(b) 7cm
(c)8cm
(d)9cm
Solution:
Here area of the
ADC = 10 cm2
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Area of the ABC = 2 × 10 = 20 cm2
(Median bisects the triangle in two parts of equal areas)
BC = 5 cm
ar ABC
Again
20
h
1
base × altitude
2
1
5 h
2
2 20
= 8 cm.
5
Altitude = 8 cm
Ans.(c)
17. BD is one of the diagonals of a quadrilateral ABCD. AM and CN are the perpendicular
from A and C, respectively, on BD. Then area (Quad. ABCD) = k × BD (AM + CN). Find
the value of k
(a)
1
2
(b)
1
3
(c)
1
4
(d)
1
8
Solution: (a)
ABCD is a quadrilateral
AM
BD and CN
BD.
Area of quad. ABCD = ar
Area ABCD
ABD
ar
BCD
1
1
BD AM
BD CN
2
2
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1
BD AM CN
2
18. In figure ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and
EQ drawn parallel to AD meets CD produced at Q. then ar ABCDE
k ar APQ . Find
the value of k.
(a)
1
2
(b)
1
4
(c)
1
(d)
1
3
Solution:
Since BP|| AC
Ar AVCP
Ar ABC ...... 1
Also EQ || AD
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Ar ADQ
Ar AED
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............ 2
(1) + (2)
Ar ACP
Ar ADQ
Ar ABC
Ar AED
Adding Ar ACP on both the sides
Ar ACP
Ar ADQ
Ar ACD
Ar APQ
Ar ABCDE
Ar ABC
Ar AED
Ar ACD
k 1
Ans. (c)
19. The area of a trapezium is ____________the product of its height and the sum of the
parallel sides.
(a)
1
2
(b)
1
3
(c)
1
4
(d)
1
8
Solution: (a)
Given : Trapezium ABCD in which AB  DC. Let AB = a, DC = b and distance between the
parallel sides is DL = BN = h.
To Prove : ar(ABCD) =
1
a b h
2
Construction : Join BD.
Proof : Since BD is the diagonal of ABCD.
ar ABCD
=
ar
ABD
ar
BCD
1
1
AB DL
DC NB
2
2
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=
1
1
a h
b h
2
2
=
1
h a b
2
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20. In figure CD || AE and CY || BA. Then, ar CBX
(a)
1
2
(b)
1
4
k ar AXY . so the value of k is
(c) 1
(d)
1
3
Solution:
Since CY||BA
Ar CAY
Ar BCY
Ar CAY
Ar CXY
Ar AXY
Ar CBX
Ar BCY
Ar CXY
k 1
Ans. (c)
21. A pharmacist needs to strengthen a 15% alcoholic solution to one of 32% alcohol. How
much pure alcohol should be added to 800 ml of 15% solution?
(a) 600 ml
(b) 800 ml
(c) 400ml
(d) 200 ml
Solution: (d)
Quantity of pure alcohol in 800 ml of 15% solution
= 15% of 800
=
15
× 800 = 120 ml.
100
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Suppose that x ml of pure alcohol is added to the 15% solution to get 32% alcoholic
solution.
New volume of the solution = (800 + x) ml.
Quantity of pure alcohol in (800 + x) ml of 32% solution = (120 + x) ml,
32% of (800 + x) = (120 + x)
32
800 x
100
120 x
32(800 + x) = 100(120 + x)
25600 + 32x = 12000 + 100x
68x = 13600
x = 200 ml
200 ml of pure alcohol should be added.
22. In a right angled triangle the ratio of the acute angles is 1: 2. Then the smallest angle is
(a) 300
(b) 400
(c) 900
(d) 550
Solution:
Let the acute angles are x & 2x
Now, x + 2x + 90 = 180
3x 90
x 300
Ans.(a)
23. The perimeter of a rectangular plot is 120 m. If the length of the plot is decreased by
5m and breadth is increased by 5 m; the area in increased by 75 sq m. Find the length
(x) and breadth (y) of the rectangular plot.
(a) x = 60, y = 30 (b) x = 40, y = 20
(c) x = 50, y = 25
(d) x = 80, y = 40
Solution: (d)
Let the length of rectangle = x m.
and breadth = y m, and area of rectangle = xy.
Perimeter = 2 (x + y) = 120
x + y = 60
…(i)
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After change
New length = (x – 5) m
New breadth = (y + 5) m
Therefore increased area = 75 m2
(x – 5) (y + 5) – xy = 75
xy – 5y + 5x – 25 – xy = 75
5x – 5y = 100
x – y = 20
..(ii)
Solving Eq. (i) and Eq. (ii), we get x = 40 m and y = 20 m.
24. A number consists of two digits. The digit at unit’s place is two-times the digit at ten’s
place. The numbers, formed by reversing the digits, is 36 more than the original
number. Find the originals number.
(a) 24
(b) 48
(c) 36
(d) 12
Solution:
Let the digit at place =x
digit at unit place = 2x
number = 10(x) + 2x = 12x
Reverse no. = 10(2x) + x= 21x
21x = 36 + 12x
9x = 36
x 4
number = 21x = 12×4= 48
Ans.(b)
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25. The angles of a triangle are 5(x –1), 3(2x –5) and 9x. Name the triangle.
(a) isosceles triangle
(b) equilateral triangle
(c) right angled triangle
(d) isosceles right angled triangle
Solution:
We know that sum of three angles of a triangle is 1800. Three angles are
5(x – 1), 3(2x – 5) and 9x.
5(x – 1) + 3(2x – 5) + 9x = 1800
5x – 5 + 6x – 15 + 9x = 1800
20x – 20 = 1800
20x = 180 + 20 = 200
x = 10
Three angles are
5(10 – 1)0, 3(2. 10 – 5)0, (9.10)0
or 450, 450, 900
It is a isosceles right angled triangle.
26. BD is one of the diagonals of a quadrilateral ABCD, AM and CN are perpendiculars from
A and C respectively, on BD. Then, (quad. ABCD) = k BD (AM +CN). Find the value of k.
(a)
1
2
(b)
1
4
(c) 2
(d)
1
3
Solution:
Ar Quad. ABCD
ar
ABD
ar
BCD
1
1
BD AM
BD CN
2
2
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BD AM CN
2
Ans.(a)
27. If
x 5 x 3
6
4
x 1
, find y when y = –7x.
9
(a) 1
(b)2
(c) 3
(d) 4
Solution: (a)
x 5
6
x 3
4
6 x 5
36
x 1
9
9 x 3
36
4 x 1
36
6x + 30 – 9x – 27 = 4x + 4
– 3x + 3 = 4x + 4
1
7
x
7x = – 1
1
=1
7
Now y = – 7x = – 7.
28. A thief steals a car at 2:30 PM and drives it at 60 km/h. The theft is discovered at 3 PM
and the owner sets off in another car at 75 km/h. When will he overtake the thief?
(a) 4 : 30 PM
(b) 4 : 45 PM
(c) 5 PM
(d) 5:15 PM
Solution:
Let they meet after t hrs after 3 pm
Speed =
Dis tance
Time
Distance will be the same in both t Distance = Speed × Time
60
60 t
T
1
2
60
75 t
75 t
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60t + 30 = 75 t
30 = 15t
t
2 hrs
29. The medians of a ABC intersect at G. then AGB s
(a)
1
2
(b)
1
3
(c)
ABC . Find the value of s.
1
4
(d)
1
8
Solution: (b)
Given : A triangle ABC in which the medians AD, BE and CF intersect at the point G.
ABC
To Prove : ar
3 ar
AGB
Proof : In the DABC,
ar
ADB
ar
ADC
(1)
ar
BAE
ar
BCE
(2)
ar
BCF
ar
ACF
(3)
(A median divides the triangles into two parts of equal area).
In the DBCG, DC is a median.
ar
GBD
ar
GDC
(4)
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Subtracting Eq. (4) from Eq. (1)
ar
ABD
ar
GBD
ar
ar
ABG
ar
AGC
Similarly ar
ABG
ar
ADC
ar
GDC
(5)
BGC
… (6)
Comparing Eq. (5) and Eq. (6)
As
ar
ar
AGC
ar
BGC
ABC
ar
ABG
ar
= 3 ar
or
ar
AGB
AGB
ACG
ar
BGC
AGB
1/3 ar
ABC
30. In the figure AD and BE are perpendicular in a line l. C is the mid-point of AB. Then.
DC
k . then find the value of k.
CE
(a)1
(b)2
(c)3
(d)4
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Solution:
Given points A and B are on the same side of the line l, AD
l and BE
l. C is the
mid-point of AB.
To Prove : CD = CE
Construction : Draw a line CF through C, which is parallel to AD.
Proof:
ADF +
CFD = 1800
ADF = 900
CFD 900
(given)
CFE 900
Since AD  CF  BE
and AB and DE are transversal and C is the mid-point of AB, therefore, F is the midpoint of DE or DF = FE.
Now in the
CDF and the
DF = FE
CEF,
[proved above]
CF = CF
CFD =
CDF
DC = EC.
CFE
[proved above] (each 900)
CFE
(proved)
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SECTION – B {Physics}
31. The apparent weight of an object on the surface of the moon, if the mass of the object
and liquid displaced are X and Y respectively, on earth is.
(a)
X Y
g
(b)
g
X Y
6
(c) 6g X Y
(d)
6
X Y
g
Solution: (b)
The weight of the body on the surface of earth = X g
The weight of liquid displaced is = Yg
The apparent weight is on earth is Xg – Yg
Now the value of g on the surface of earth is
surface of the moon is
1
of g therefore apparent weight on the
6
g
X Y
6
32. The work done by the centripetal force F when the body complete half the rotation
around the circle of radius R is
(a) 2 RF
(b) 2RF
(c) RF
(d) Zero
Solution:(d)
Since centripetal force is perpendicular to the displaced therefore
W = Fs cos 900
=0
33. The sudden fall in pressure due to a rise in humidity indicates
(a) a cyclone
(b) rainfall
(c) a dust storm
(d) dry weather
Solution:(a)
34. A balloon filled with air is weighted (W) so that it just floats in water as shown in the
figure. When it is further pushed by a short distance in water, it will
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(a) Come back to its original position
(b) Stay at the depth where it stands submerged
(c) Sink to the bottom
(d) Sink down a little further but will not reach the bottom
Solution:(a)
35. If the kinetic energy of a body increases by 300%, Calculate the percentage increase in
momentum
(a) 100%
(b) 150%
(c) 200%
(d) 50%
Solution:(a)
Let kinetic energy be E1
Therefore,
E2 = E1 300% of E1
E2 = 4E1
Now,
P1
P2
P1
P2
E1 2m
E2 2m
1
2
P2 2P1
Now, increase in linear momentum given by
2P 1 P1
100 =100%
P1
36. The diagrams represent four measuring cylinders containing liquids. The mass and
volume of the liquid in each cylinder are stated. Which two measuring cylinders could
contain an identical liquid?
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(a) W and X
(b) W and Y
Medical and Non - Medical Classes
(c) X and Y
(d) X and Z
Solution: (d)
Since the density of both liquid are same therefore option (d) should be correct.
37. What is the power of pump which takes 10 seconds to lift 100 kg of water to a water
tank situated at a height of 10 m? Take g = 10m/s2
(a) 1 kw
(b) 2 kw
(c) 3 kw
(d) 4 kw
Solution:
Power =
=
Work done
time
mgh 100 10 10
=
time
10
= 1000 w = 1 kw
38. A boy of mass 40kg sits in a swing by a rope of 6 m long. A person pulls the swing to a
side so that the rope makes an angle of 600 with the vertical. What is the gain in
potential energy of the boy?
(a) 2200 J
(b) 5012 J
(c) 2455 J
(d) 1176 J
Solution:(d)
As shown in Fig, on pulling the swing, from B to A, the height through which the body is
raised,
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h = BA’ = OB – OA
= 6 6 cos 600
6 6
1
2
3m
Gain in potential energy of the boy = mgh
= 40 × 9.8 × 3 = 1176 J
39. A car and a truck are moving on the road with equal kinetic energy. Breaks are applied
to both of them which produce equal retarding forces in them. Both will cover distance
S1 and S2 respectively so which of the following is correct?
(a) S1 > S2
(b) S1 = S2
(c) S1 < S2
(d) S1
S2
Solution: (b)
40. An electric motor of power 100 W is used to drive the stirrer in a water bath. If 50 % if
the energy supplied to the motor is spent in stirring the water, calculate the work done
on the waster in one minute.
(a) 2000 J
(b) 3000 J
(c) 3330 J
(d) 3456 J
Solution:(b)
Given that :
Power used in stirring the water, P = 50% of power of motor
=
50
100 W
100
50 W
t = one minute = 60 s
Hence, work done on water, W = P t
= 50 × 60 J = 3000 J
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SECTION – C {CHEMISTRY}
41. How many years in it would take to spend one avogadro’s number of rupees at a rate
10 lakh of rupees in one second?
(a) 1.9089×1025 year
(b) 1.9089×109 year
(c) 1.9089×1010 year
(d) 1.9089×1012 year
Solution: (c)
Number of rupees spent in one second = 106
Number of rupees spent in one year
= 106 × 60 × 60 × 24 × 365
Avogadro’s number of rupees will be spent in
=
106
6.02 1023
60 60 24 365
= 19.089 × 109 year
= 1.9089 × 1010 year
42. Which one of the following will have largest number of atoms?
(a) 1 g Au(s)
(b) 1 g Na(s)
(c) 1 g Li(s)
(d) 1g Cl2 (g)
Solution: (c)
1 g of Li will have largest number of atoms because it has lowest atomic mass.
43. The number of oxygen atoms present in 0.25 moles of magnesium perchlorate.
(a) 4 N
(b) 2 N
(c) 6 N
(d) 8 N
Solution:(b)
1 mole of magnesium per chlorate Mg(ClO4 )2 = 8 mol of oxygen atom
0.25 mol of magnesium per chlorate = 0.25 × 8 = 2
44. Suppose the chemists had chosen 1029 as the number of particles in a mole. What
would be the molecular mass of oxygen gas?
(a) 5.25×103
(b) 5.32× 10-3
(c) 5.25×102
(d) 5.25×10–2
Solution:(b)
According to present concept of mole,
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6.022 × 1023 molecules of oxygen weigh = 32 g
Mass of 1 molecule of oxygen =
32
6.022 1023
= 5.32 × 10–23 g
If the mole contained 1020 particles
The mass of one mole i.e., 1020 particles = 5.32 × 10–23 × 1020 = 5.32 × 10–3 g
Molecular mass of oxygen = 5.32 × 10–3 g
45. Which has the highest mass:
(a) 1 g atom of C
(b) ½ mole of CH4
(c) 10 mL of water
(d) 3.011×1023 atom of oxygen
Solution:(a)
46. What is the mass of 1 u in grams
(a) 1.99×10–23
(b) 6.022×10–23
(c) 1.66×10–24
(d) 1.99×1023
Solution:(c)
1u
1
th of the mass of carbon atom
12
Let us calculate the mass of a carbon atom 6.022 × 1023 atoms of carbon = 12 g
Mass of one carbon atom =
12
6.022 1023
= 1.99 × 10–23 g
1u
1
× 1.99 × 10–23 g = 1.66 × 10–24 g
12
47. 5.6 litre of oxygen at NTP is equivalent to :
(a) 1 mole
(b) ½ mole
(c) ¼ mole
(d) 1/8 mole
Solution:(c)
22.4 litre O2 at STP = 1 mole.
5.6 litre O2 at STP =
5.6
22.4
1
mole
4
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48. How many electrons are present in 16 g of CH4
(a) 6.023×10–23
(b) 3.023×1023
(c) 6.023×1023
(d) 6.022×1022
Solution:(c)
1 molecule of CH4 = 6 + 4 = 10 electrons
16 g of CH4 contains 10 × 6.022 × 1023 = 6.022 × 1024
49. From 200 mg of CO2, 1021 molecules are removed. How many moles of CO2 are left?
(a) 2.88×10–3
(b) 1.88×10–3
(c) 288×10–3
(d) 2.88×10–5
Solution: (a)
Gram-molecular mass of CO2 = 44 g
Mass of 1021 molecules of CO2 =
44
6.02 1023
1021 0.073g
Mass of CO2 left = (0.2 – 0.073) = 0.127 g
Number of moles of CO2 left =
0.127
= 2.88 × 10–3
44
50. Which of these weights most?
(a) 32 g of oxygen
(c) 0.5 mole of Fe,
(b) 2 g atom of hydrogen
(d) 3.01×1023 atoms of carbon.
Solution: (a)
(i) 32 g of oxygen weighs most.
(ii) 2 g atom of H2 = 2g
(iii) 0.5 mole of Fe = 0.5 × 56 = 28 g (iv) 3.01 × 1023 atoms of C =
1
12 6 g
2
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SECTION – D {Biology}
51. The ability of nitrogen fixation is found in
(a) monerans only
(b) both monerans and protistans
(c) protistans only
(d) fungi only
Solution: (a)
52. Find the incorrect pair
(a) Leech – phylum Annelida
(b) Octopus – phylum Mollusca
(c) Fasciola – phylum platyhelminthes
(d) starfish – phylum Chordata
Solution: (d)
53. Which of the following plant group bears naked seeds?
(a) pteridophyta
(b) bryophyta
(c)gymnospermae
(d) angiospermae
Solution: (c)
54. Which of the following is a monocot?
(a) Carrot
(b) wheat
(c) mango
(d) mustard
Solution: (b)
55. Presence of chitinous exoskeleton is an identifying feature of :
(a) Corals
(b) Molluscs
(c) Coelenterates
(d) Arthropods
Solution: (d)
56. In Whittaker’s classification unicellular organism are grounded under
(a) Protista
(b) Porifera
(c) Fungi
(d) Protozoa
(c) Jelly fish
(d) Sponges
Solution: (a)
57. Radial symmetry is best seen in:
(a) Fished
(b) Star fish
Solution: (b)
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58. Which of the following is parasitic protozoan?
(a) Amoeba
(b) Entamoeba
(c) Euglena
(d) paramecium
Solution: (b)
59. A plant body not differentiated into root, stem and leaves is termed as
(a) thallus
(b) Mycelium
(c) hyphae
(d) herb
Solution: (a)
60. Male Ascaris can be distinguish from female ones by
(a) Curved posterior
(b) Round shape end
(c) presence of penial setae
(d) Both (a) and (c)
Solution: (d)
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