Pioneer’s Class Sam… · If one of the roots of the quadratic equation x2 + mx + 24 = 0 is 1.5,...

Pioneer’s Sample Paper

Solution

10th CBSE Examination Centre: Pioneer Education, Sector – 40-D

General Instructions:

The question paper contains 90 multiple choice questions.

There are two sections in the question paper

Section: A– MATHEMATICS (1 to 45)

Section: B– SCIENCE (46 to 90)

Each right answer carries 4 marks and − 1 for every wrong answer.

The Maximum marks are 360 and Maximum Time 2.00 hrs.

Give your response in the OMR Sheet provided with the Question Paper.

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Section-A {Mathematics}

1. If one of the roots of the quadratic equation x2 + mx + 24 = 0 is 1.5, then what is the value of m?

(a) –22.5 (b) 16 (c) –10.5 (d) –17.5

Ans. (d)

Solution.

We know that the product of roots of a quadratic equation ax2 + bx + = 0 is c

a.

In the given equation, x2 + mx + 24 = 0, the product of the roots = 24

241 .

The question states that one of the root quadratic equation = 1.5.

If x1 and x2 are the roots of the given quadratic equation and let x1 = 1.5.

Therefore, x2 = 24

161

In the given equation, m is the co-efficient of the x term.

We know that the sum of the roots of the quadratic equation ax2 + bc + c = 0 is b m

ma 1

Sum of the roots = 16 + 1.5 = 17 = –17.5.

Therefore, the value of m = –17.5.

2. A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on

full ticket. One reserved first class ticket from Chennai to Trivandrum costs Rs. 216 and one full and one

half reserved first class tickets cost Rs. 327. What is the basic first class full fare and what is the

reservation charge?

(a) Rs. 105 and Rs. 6 (b) Rs. 216 and Rs12 (c) Rs. 210 and Rs. 12 (d) Rs. 210 and Rs. 6

Ans. (d)

Solution.

Let half of the full basic fare be Rs. X.

Therefore, full basic fare be Rs. 2X.

Let the reservation charge be Rs. Y per ticket.

Now, one full reservation ticket would cost 2X ( basic fare) +Y(reservation charge)

2X + Y = 216 …..(1)

The total basic fare for one half and one full ticket = X + 2X = 3X and the total reservation charge is 2Y.

Hence, 3X + 2Y = 327. ……(2)

Solving (1) and (2) we get,




X = 105 and Y = 6

Hence, the full basic fare is 2X = Rs. 210 and the reservation charge is Y = Rs. 6.

3. If 4x +6

y=15 and 6x –

8

y=14 then the value of ‘p’ will be (given y = px – 2)

(a) 2 (b) 4

3 (c)

2

3 (d) 3

Ans. (b)

4. If x2 – 3x + 1 =0, then the value of x2 + 2

1

xis

(a) 3 (b) 0 (c) 5 (d) 7

Ans. (d)

5. A man wishes to find the height of a flag post which stands on a horizontal plane; at a point on this

plane he finds the angle of elevation of the top of the flag post to be 450. On walking 30 metres towards

the town he finds the corresponding angle of elevation to be 600. Find the height of the flag post?

(a) 69 m (b) 70 m (c) 71 m (d) 74 m

Ans. (c)

6. A man on the top of a town, standing on the sea shore finds that a boat coming towards him takes 10

minutes for the angle of depression to change from 300 to 600. Find the time taken by the boat to reach

the shore from the position.

(a) 5 minutes (b) 10 minutes (c) 8 minutes (d) 15 minutes

Ans. (a)

7. From a point in the cricket ground, the angle of elevation of a vertical tower is found to be at a

distance of 200 m from the tower. On walking 125 m towards the tower the angle of elevation becomes

2 . Find the height of tower.

(a) 300 m (b) 100 m (c) 400 m (d) 200 m

Ans. (b)

8. The value of tan 10 tan 20 tan 30 …… tan 890 is

(a) 0 (b) 1 (c) –1 (d) 2

Ans. (b)

9. A man on the top of a vertical observation tower observes a car moving at a uniform speed coming

directly towards it. If it takes 12 minutes for the angle of depression to change from 300 to 450, how

soon after this will the car reach the observation tower. Give your answer correct to nearest seconds.




(a) 16 mins 24 secs (b) 18 mins 20 secs (c) 20 mins 44 secs (d) 14 mins 38 secs

Ans. (a)

10. A cat saw a rat when it was exactly at the middle of a vertical tower. At that moment the angle of

elevation as observed by the cat was 30°. The cat ran a certain distance towards the tower to chose the

rat, which then an to the top of the tower. If the cat, which is now 20 m away from the foot of the tower,

finds the angle of elevation of the rat to be 60°, what distance cat run?

(a) 10 (b) 10 2 (c)10 3 (d) None of these

Ans. (b)

Solution.

Let the angle made by the foot of the ladder on the ground be ‘ ’ degree

We know that, ladder = 2 (Using of the Wall)

i.e., AC = 2 AB

In ABC ,

ABsin

AC

AB 1

sin2AB 2

030

11. If x = 2, then the value of x + x x x .... is

(a) 1 or 2 (b) 2 (c) 1 or 4 (d) 2

Ans. (c)

Let x + x x x .... y

2 2 2 2 ...... y

2 + y = y

2 2

y y 2

On solving above equation, we get

x = 1 or 4

A

B C




12. The area of the triangle formed by the points (k–1, k + 1), (k + 3, k –3) and (k + 1, k + 2) is

(a) 2 (b) 3 (c) 3k (d) none of these

Ans. (d)

Solution.

The points are (k–1, k + 1), (k + 3, k –3) and (k + 1, k + 2)

So, area of triangle

= 1

k 1 k 3 k 2 k 3 k 2 k 1 k 1 k 1 k 32

= 1

k 1 5 k 3 1 k 1 42

=1

5k 5 k 3 4k 42

=1

12 6sq.units2

13. If the end points of the diameter of a circle are (4, 6) and (8, 4), the radius of the circle is

(a) 2 5 units (b) 5 units (c) 10 units (d) 2 20 units

Ans. (b)

Solution.

Diameter of the circle = Distance between (4, 6) and (8, 4)

= 2 24 2 16 4 2 5 = Diameter

So, radius of the circle = 5 units

14. If the 3 consecutive vertices of a square are (–1, 4), (–3, 0) and (1, –2). Which of the following can be its

fourth vertex?

(a) (3, 2) (b) (2, 3) (c) (–3, 2) (d) (1, 2)

Ans. (a)

Solution.

Let the points are A(–1, 4), B(–3, 0), C(1, –2) and D(x, y).

The mid-point of

AC = 1 1 4 2

, 0,12 2

Mid-point of BD = x 3 y 0

,2 2

A (–1, 4) B (–3, 0)

C (1, –2) D (x, y)




x 3 y

0 x 3 and 12 2

y = 2.

Thus, (x, y) = (3, 2)

15. The triangle formed by joining the vertices (3, 2) (5, –3) and (–5, 4) will be a

(a) acute triangle (b) obtuse triangle (c) right triangle (d) isosceles triangle

Ans. (a)

Solution.

We have, (3, 2) (5, –3) and (–5, 4).

AB= 2 2 2 25 3 3 2 2 5 29

BC = 2 2

5 5 4 3 25 49 74

and CA = 2 2

5 3 4 2

= 64 4 70

BC2 < AB2 + CA2

So, ABC is an acute triangle.

16. If O is the centroid of the ABC, what is the length of median AD?

(a) 3 5

2

(b) 3 5

(c) 3 /2 (d) none of these

Ans. (a)

Solution.

We know, AO = 2 21 2 5 2/3 AD

( Centroid divides medians in 2 : 1 ratio)

AD = 3 5 /2

17. If the sum of 1st 9 terms of an AP is equal to the sum of 1st 13 terms, the sum of 1st 22 terms will be

respectively?

(a) 0 (b) 1 (c) –1 (d) 35

Ans. (a)

Solution.

S9 = S13




i.e., 9 13

2a 9 1 d 2a 13 1)d2 2

18a + 72 d = 26a + 156d

i.e., 8a = – 84d

i.e., 2a = –21d

Now, S22 = 22

1 2a 22 1 d 11 2a 11d2

= 11(2a – 2a) = 0

Shortcut Method

If Sp = Sq

Sp+q = S13

So, if S9 = S13

S9+13 = S22 = 0

18. In a AP series of 100 terms a1 + a50 + a51 + a100 = 40, so what is the sum of all the terms of the series ?

(a) 10000 (b) 5000 (c) 1000 (d) none of these

Ans. (a)

Solution.

As AP has 100 terms, we know

a1 + a100 = a2 + a99 = a3 + a98 = … and so on

So, a1 + a50 + a51 + a100 = 400

a1 + a100 = 1

2(400) = 200

Thus, S100 = 1 100a a100 100 100 10000

2

19. The ratio of the sum of 1st ‘P’ terms of two AP series is (4P +16) : (7P –17), what is the ratio of their 6th

terms?

(a) 1 (b) 15

43 (c)

59

160 (d) none of these

Ans. (a)

Solution.

p

p'

p /2 2a p 1 dS 4p 16

S 7p 17 p /2 2a' p 1 d'




=

2a p 1 d

2a' p 1 d'

Let P = 11

Given, 2a 10d 4 11 16 60

12a' 10d' 7 11 17 60

6

6

aa 5d1

a' 5d' a '

20. If a7 = 13a13, where a7 and a13 are the 7th and the 13th terms respectively and the common difference is

6, then 29th term of the same AP is

(a) 80 (b) 93 (c) –93 (d) none of these

Ans. (b)

Solution.

a7 = 13a13

a + 6d = 13(a + 12d)

12d = –25d

= –25 6 = –150 a = –75

a29 = a + (28)d = –75 + 28 6

= –75 + 168 = 93

21. In the fig. shown, PT and PAB are the tangent and the secant drawn to a circle. If PT = 12 cm and PB = 8

cm, then AB is equal to

(a) 16 cm (b) 10 cm

(c) 4 5 cm (d) 18 cm

Ans. (b)

Solution.

OA = OD = Radii

So, 2PT PA PB

(relation between PT tangent and secant PAB)

122 = PA 8 PA = 18

AB = PB – PA = 18 – 8 = 10 cm.

22. In the fig. shown, find the value of DEC.

(a) 1000 (b) 1200

(c) 800 (d) 600

A P

T

B




Ans. (c)

Solution.

ACD = Exterior angle of AOC = 400 + 200 = 600

Also, ACD = ABE = 600 (angle in the same arc)

Now, in ABE,

BAE + ABE + AEB = 1800

400 + 600 + AEB = 1800

Thus, AEB = DEC = 800 (vertically opposite angles)

23. PQRS is a rectangle with PQ = 16 cm and QR = 12 cm. If ABCD is a rhombus formed by joining the mid-

points of rectangle PQRS, what is the radius of the circle inscribed

in rhombus ABCD?

(a) 2.4 cm

(b) 4.8 cm

(c) 3.6 cm

(d) none of these

Ans. (b)

Solution.

OD = 8 cm and OA = 6 cm

AD = 10 cm (using Pythagoras theorem)

Let us draw OT AD.

Area of 1

AOD2

AO OD = 1

2 AD OT

1

2 6 8 =

1

2 10 OT

OT = 4.8 cm

O

A

B

C

D

S

P Q

R




24. Three identical circles of radius r cm each are placed inside an equilateral triangle. Find the length of

the side of the equilateral triangle in terms of r.

(a) 2r

(b) 3r

(c) 2r ( 2 +1)

(d) 2r ( 3 +1)

Ans. (d)

Solution.

Radius of each circle = r

We want to know the length of AB i.e., side of equilateral triangle in terms of r.

In AOD, we know OD AD (radius on tangent)

and DAO = 300 (equilateral triangle)

AOD = 600

As we know, in triangle having angles 900, 600 and 300 have opposite sides 2a, a 3 and a, respectively.

OD = Radius = r

AD = r 3 (opposite to 600)

Now, in BO’E, we get

EB = r 3

Thus, AB = AD + DE + EB

= r 3 +2r+ r 3 = 2r( 3 1 )

25. If TA is a tangent to the circle at B, what is the value of x + y + z ?

(a) 560

(b) 1120

(c) 280

(d) 650

Ans. (a)

Solution.




x = PQB = 280

Also, OB TB

We get, PBO = 900 – x = 900 – 280 = 620

We get, OBP = OPB = 620

Now, in BPQ,

BPQ + PBQ + PQB = 1800

(620 + y) + (620 + z) + x = 1800

x + y + z = 1800–1240=560

26. A student cuts off two circles of radii r1 and r2 from a sheet of thick paper. He keeps these two in a

vertical position in contact with each other at a point P, Now, he keeps a rod of moderate thickness on

the top points A, B of the two circles. If r1 r2, then for any values of r1 and r2 the angle APB is equal to

(a) 30° (b) 60° (c) 90° (d) 120°.

Ans. (c)

Solution:

AT = TB also TB = TP

TAP = TPA also TBP = TPB

ATP = BTP = 900

TAP + TPB = 90°

Similarly, TBP +TPB= 90°

also, TAP + TPB +APB = 1800 APB=90°

27. If a and are roots of the equation x2+2x+ 1 = 0, then the equation whose roots are 1

and

1

is a

(a) x2+1

2x +

1

2 = 0 (b) x2+x +

1

2 = 0 (c) x2+

1

2x+ 1 = 0 (d) x2 + 2x +1 = 0.

Ans.(d)

Solution:

2 1

21 1 2 12 1 equation is x 2x 1 0

1

28. If a = sin4

, b = cos4

and c = –cosec4

, then the value of a3 + b3 + c3 is

(a) 3 2

2 (b) 1 (c) 0 (d)

3 2

2

A B

P

T




Ans.(d)

Solution:

1 1a sin Similarly b and c 2

4 2 2

3 3 3 1 1 3 2a b c 2 2

22 2 2 2

29. In the figure, O is the centre of circle; BCO = m°, BAC = n°. Then

(a) m + n = 90°

(b) m + n = 180°

(c) 2m + n = 180°

(d) m + 2n = 180°.

Ans. (a)

Solution.

Since OB = OC SCO = OBC = m

SOC = 180° - 2m also 180° - 2m = 2n

2m + 2n=180° m + n = 90°.

30. If 2x–1 + 2x+1 = 320, then x is equal to

(a) 6 (b) 8 (c) 5 (d) 7

Ans. (d)

Solution:

xx 1 x x x x 11 1 4 2 .5

2 .2 2 .2 2 2 2 2 .52 2 2

Thus 2x–1 . 5 = 320 x=7.

31. The solution of (25)x–2 = (125)2x –4 is

(a) 3/4 (b) 0 (c) 2 (d) –2

Ans. (c)

Solution:

(25)x–2 = (125)2x–4

(5)2x–4 = (5)6x–12 2x – 4 = 6x – 12 = x= 2.

O

n

m Q

A

B




32. In this figure the centre of the circle is O. AB BC, ADOE is a straight line, AP AD , and AB has a

length twice the radius. Then:

(a) AP2 = PB.AB (b) AP . DO PB . AD

(c) 2AB AD . DE (d) AB. AD OB . AO

Ans. (d)

Solution.

Since AB is tangent to the circle, we have 2AD/ AB AB/ AE, AB AD. AE.

But AE AD 2r AD AB ;

2 2AB AD AD AB AD AD . AB.

2 2AD AB – AD. AB AB AB.AD .

Since 2AP AD. AP AB AB– AP AB . PB.

33. Applied to a bill for Rs. 10,000 the difference between a discount of 40% and two successive discounts

of 36% and 4%, expressed in rupees, is:

(a) 0 (b) 144 (c) 256 (d) 400

Ans. (b)

Solution:

40% of Rs.10,000 is Rs. 4,000; 36% of Rs. 10,000 is Rs. 3,600; 4% of (Rs. 10,000 - Rs. 3,600) is Rs. 256.

Rs. 3,600 + Rs. 256 = Rs. 3,856; the difference is Rs. 4,000 - Rs. 3,856 = Rs. 144; or two successive

discounts of 36% and 4% are equivalent to one discount of 38.56%.

A

B

E O D

C

P




34. Triangle PAB is formed by three tangents to circle O and APB = 40°; then angle AOB equals:

(a) 45° (b) 50° (c) 55° (d) 70°

Ans. (d)

Solution:

P = 40°; PAB + PBA = 180° – 40° = 140°.

TAS=180°– PAB; RBS= 180° – PBA

TAS + RBS = 360° – 140° = 220°.

Since OA and OS bisect angles TAS and RBS, respectively

OAS+ OBS=1

2(220°)= 1 1 0

AOB= 180° – 1 1 0 ° = 70°.

The number of degrees in AOB is independent of tangent ASB.

35. If x = 1 1 1 1 ......... , then :

(a) x =1 (b) 0 < x <1 (c) 1 < x < 2 (d) x is infinite

Ans. (c)

Solution:

x = x 1 , x2 = 1 + x, x2 – x – 1=0, and x 1.62 1 < x < 2.

36. The area of the largest triangle that can be inscribed in a semi-circle whose radius r is:

(a) r2 (b) r3 (c) 2r2 (d)2r3

Ans. (a)

Solution:

Of all triangles inscribable in a semi-circle, with the diameter as base, the one with the greatest area is

the one with the largest altitude (the radius); that is, the isosceles triangle. Thus the area is

1

2. 2r . r = r2

P

O

A T

R

B

S 40o




37. John ordered 4 pairs of black socks and some additional pairs of blue socks. The price of the black socks

per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of

the two colours had been interchanged. This increased the bill by 50% .The ratio of the number of pairs

of black socks to the number of pairs of blue socks in the original order was:

(a) 4 : 1 (b) 2 : 1 (c) 1 : 4 (d) 1 : 2

Ans. (c)

Solution:

Let x be the number of pairs of blue socks, and let y be the price of a pair of blue socks. Then 1y is the

price of a pair of black socks and

x × 2y + 4 × y =3

2 (4 –2y + x–y). Divide by y and solve for x.

x= 16; 4 : x = 4 : 16 = 1 : 4.

38. If the radius of a circle is increased 100%, the area is increased:

(a) 100% (b) 200% (c) 300% (d) 400%

Ans. (c)

Solution:

Let r be the original radius; then 2r = new radius, 2r =original area, 4 2r = new area.

Percent increase in area=2 2

2

4 r r

r

×100=300.

39. A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped

to each side. The sum of these perpendiculars is:

(a) least when the point is the centre of gravity of the triangle

(b) greater than the altitude of the triangle

(c) equal to the altitude of the triangle

(d) one-half the sum of the sides of the triangle




Ans. (c)

Solution.

Let P be an arbitrary point in the equilateral triangle ABC with sides of length s, and denote the

perpendicular segments by pa, pb, pc. Then Area ABC = Area APB + Area BPC + Area CPA =

=1

2 (spa + spb + spc )=

1

2 s(pa + pb + pc)

Also, Area ABC = 1

2 s h, where h is the length of the altitude of ABC. Therefore, h = pa + pb + pc and this

sum does not depend on the location of P.

40. Tom, Dick and Harry started out on a 100 km/hr journey. Tom and Harry went by automobile at the

rate of 25 km/hr, while Dick walked at the rate of 5 km/hr. After a certain distance, Harry got off and

walked on at 5 km/hr, while Tom went back for Dick and got him to the destination at the same time

that Harry arrived. The number of hours required for the trip was:

(a) 5 (b) 6 (c) 7 (d) 8

Ans. (d)

Solution:

Let t1, t2, t3 be the number of hours, respectively, that the car travels forward, back to pick up Dick, then

forward to the destination. Then we may write

25 . t1 – 25 . t2 + 25 . t3 = 100 for car

5 . t1 + 5 . t2 + 25 . t3 = 100 for Dick

25 . t1 + 5 . t2 + 5 . t3 = 100 for Harry.

This system of simultaneous equations is equivalent to the system

t1 – t2 + t4 = 4

t1 + t2 + 5t3 = 20

5t1 + t2 + t3 = 20

whose solution is t1 = 3, t2 = 2, t3 = 3.

Hence t1 + t2 + t3 = 8 = total number of hours.

A

B C

Pc

P Pa

s

Pb




Section-B {Science}

41. Focal length of a convex lens will be maximum for –

(a) blue light (b) yellow light (c) green light (d) red light

Ans. (d)

42. If a convex lens of focal length 80 cm and a concave lens of focal length 50 cm are combined together,

what will be their resulting power –

(a) +6.5 D (b) –6.55 D (c) +7.5 D (d) –0.75 D

Ans. (d)

43. When a mirror is rotated through an angle θ , the refracted ray from it turns through an angle of –

(a) θ (b) θ/2 (c) 2θ (d) 0

Ans. (c)

44. A ray of light strikes a transparent surface from air at angle θ . If the angle between the reflected and

refracted ray is a right angle, the refractive index of the other surface is given by –

(a) μ 1/ tanθ (b) 2μ tan θ (c) μ sin θ (d) μ tan θ

Ans. (d)

45. A point source of light B, placed at a distance L in front of the centre of a mirror of width d, hangs

vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance

2L from it as shown. The greatest distance over which he can see the image of the light source in the

mirror is –

(a) d/2 (b) d (c) 2d (d) 3d

Ans. (d)




46. A ray of light passes through four transparent media with refractive indices 1 2μ , μ , 3μ , and 4μ as shown

in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray

AB, we must have:

(a) 1 2μ μ (b) 2 3μ μ (c) 3 4μ μ (d) 4 1μ μ

Ans. (d)

47. An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the

figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h,

he can see the lower end of the rod. Then the refractive index of the liquid is –

(A) 5/2 (b) 5/2 (c) 3/2 (d) 3/2

Ans. (b)

48. A ray of light is incident at the glass-water interface at angle i, it emerges finally parallel to the surface

of water, then the value of gμ would be :

(a) (4/3) sin i (b) 1/sin i (c) 4/3 (d) i

Ans. (b)

49. A ray of light is incident on the surface of separation of a medium with the velocity of light at an angle

450 and is refracted in the medium at an angle 300. What will be the velocity of light in the medium –

(a) 1.96 108 m/s (b) 2.12 108 m/s (c) 3.18 108 m/s (d) 3.33 108 m/s

Ans. (b)




50. A fish looking up through the water sees the outside world contained in a circular horizon. If the

refractive index of water is 4/3 and the fish is 12 cm. Below the surface, the radius of this circle in

cm is –

(a) 36 5 (b) 4 5 (c) 36 7 (d) 36 / 7

Ans. (d)

51. What will be refractive index of glass for total internal reflection –

(a) 3 1

2

(b)

5 1

2

(c)

2 1

2

(d)

3

2

Ans. (d)

52. A light ray is incident perpendicularly to one face to a 900 prism and is totally internally reflected at the

glass-air interface. If the angle of reflection is 450, we conclude that the refractive index n –

(a) 1

n2

(b) n 2 (c) 1

n2

(d) n 2

Ans. (b)

53. Which of the prism is used to see infrared spectrum of light –

(a) Rock salt (b) Nicol (c) Flint (d) Crown

Ans. (a)




54. Myopia is due to –

(a) elongation of eye ball (b) irregular change in focal length

(c) shortening of eye ball. (D) older age

Ans. (a)

55. Rainbow is formed due to a combination of –

(a) Refraction and absorption (b) Dispersion and focusing

(c) Refraction and scattering (d) Dispersion and total internal reflection

Ans. (d)

56. A person using a lens as a simple microscope sees an –

(a) inverted virtual image (b) inverted real magnified image

(c) upright virtual image (d) upright real magnified image

Ans. (c)

57. Blue colour of sky is due to phenomenon of –

(a) Reflection (b) Refraction (c) Scattering (d) Dispersion

Ans. (c)

58. An endoscope is employed by a physician to view the internal parts of a body organ.

It is based on the principle of –

(a) refraction (b) reflection (c) total internal reflection (d) dispersion

Ans. (c)

59. The asexual process replaced by the sexual method is known as –

(a) Semigany (b) Amphimixis (c) Apospory (d) Apomixis

Ans. (d)

60. Gemmule formation in sponges is helpful in –

(a) Parthenogenesis (b) Sexual reproduction

(c) Only dissemination (d) Asexual reproduction

Ans. (d)

61. Ovulation in mammals is caused by –

(a) FSH and TSH (b) FSH and LH (c) FSH and LTH (d) LTH and LH

Ans. (b)

62. If both ovaries are removed from a rat, then which hormone is decreased in blood –

(a) Oxytocin (b) Estrogen (c) Prolactin (d) Gonadotrophic

Ans. (b)




63. Which part of the ovary in mammals acts as an endocrine gland after ovulation –

(a) Vitelline membrane (b) Graffian follicles (c) Stroma (d) Germinal epithelium

Ans. (b)

64. Tunica albuginea is the covering around

(a) ovary (b) testis (c) kidney (d) heart

Ans. (b)

65. Acrosome is made up of –

(a) mitochondria (b) centrioles (c) golgi bodies (d) ribosomes

Ans. (c)

66. Example of External fertilization is -

(a) fish and frog (b) frog and monkey (c) dog and goat (d) goat and fish

Ans. (a)

67. Development of embryo in human body takes place in –

(a) ovary (b) oviduct (c) uterus (d) ostium

Ans. (c)

68. Which of the following statements is not true for ethane –

(a) It can be chlorinated with chlorine (b) It can be catalytically hydrogenated

(c) When oxidised produces CO2 and H2O (d) It is a homologous of isobutane

Ans. (b)

69. Which gives CH4 when treated with water –

(a) Silicon carbide (b) Calcium carbide (c) Aluminium carbide (d) Iron carbide

Ans. (c)

70. Water gas is –

(a) CO + CO2 (b) CO + N2 (c) CO + H2 (d) CO + N2 + H2

Ans. (c)

71. Which one gives only one monosubstitution product on chlorination –

(a) n-pentane (b) Neopentane (c) Isopentane (d) n-butane

Ans. (b)

72. Main constituent of marsh gas is –

(a) C2H2 (b) CH4 (c) H2S (d) CO

Ans. (b)




73. Ethyl bromide gives ethylene when reacted with –

(a) Ethyl alcohol (b) Dilute H2SO4 (c) Aqueous KOH (d) Alcoholic KOH

Ans. (d)

74. Which is the most reactive hydrocarbon in the following –

(a) Ethane (b) Ethyne (c) Ethene (d) Methane

Ans. (b)

75. Which of the following gases is used for welding –

(a) Methane (b) Ethane (c) Acetylene (d) Ethene

Ans. (c)

76. The IUPAC name of the compound having the formula (CH3)3CCH=CH2 is –

(a) 3, 3, 3-trimethyl-1-propane (b) 1, 1, 1-trimethyl-1-1-butene

(c) 3,3-dimethyl-1-butene (d) 1,1-dimethyl-1,3-butene

Ans. (c)

77. On moving from left to right across a period in the table the metallic character –

(a) Increases (b) Decreases

(c) Remains constant (d) first increases and then decreases

Ans. (b)

78. The size of the following species increases in the order –

(a) Mg2+ < Na+ < F– < Al (b) F– < Al < Na+ > Mg2+

(c) Al < Mg < F– < Na+ (d) Na+ < Al < F– < Mg2+

Ans. (a)

79. The set representing the correct order of first ionisation potential is –

(a) K > Na > Li (b) Be > Mg > Ca (c) B > C > N (d) Ge > Si > C

Ans. (b)

80. Which one of the following oxides is neutral –

(a) CO (b) SnO2 (c) ZnO (d) SiO2

Ans. (a)

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Pioneer’s Class Sam… · If one of the roots of the quadratic equation x2 + mx + 24 = 0 is 1.5,...

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Transcript of Pioneer’s Class Sam… · If one of the roots of the quadratic equation x2 + mx + 24 = 0 is 1.5,...