KVPY_STREAM (SA) Class-XI Test Paper & Solutions

KISHORE VAIGYANIK PROTSAHAN YOJANA - 2012 Date : 30-10-2011

Duration : 3 Hours

Max. Marks : 100

STREAM - SA GENERAL INSTRUCTIONS •

The Test Booklet consists of 80 questions.

•

There are Two parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response.

MARKING SCHEME : PART-I : MATHEMATICS Question No. 1 to 15 consist of ONE (1) mark for each correct response. PHYSICS Question No. 16 to 30 consist of ONE (1) mark for each correct response. CHEMISTRY Question No. 31 to 45 consist of ONE (1) mark for each correct response. BIOLOGY Question No. 46 to 60 consist of ONE (1) mark for each correct response.

PART-II : MATHEMATICS Question No. 61 to 65 consist of TWO (2) marks for each correct response. PHYSICS Question No. 66 to 70 consist of TWO (2) marks for each correct response. CHEMISTRY Question No. 71 to 75 consist of TWO (2) marks for each correct response. BIOLOGY Question No. 76 to 80 consist of TWO (2) marks for each correct response.

KVPY QUESTION PAPER - STREAM (SA)

PART-I One Mark Questions MATHEMATICS 1.

Suppose a, b, c are three distinct real numbers. Let P(x) =

( x  b)( x  c ) ( x  c )( x  a) ( x  a)( x  b) + + (a  b)(a  c ) (b  c )(b  a) (c  a)(c  b)

When simplified, P(x) becomes (A) 1 Sol.

x 2  (a  b  c )(ab  bc  ca) (C) (a  b)(b  c )(c  a)

(B) x

(D) 0

(A) P(x) =

( x  b)( x  c ) ( x  c )( x  a) ( x  a)( x  b) + + (a  b)(a  c ) (b  c )(b  a) (c  a)(c  b)

Let

f(x) = P(x) – 1 f(a) = 1 + 0 + 0 – 1 = 0 f(b) = 0 + 1 + 0 – 1 = 0 f(c) = 0 + 0 + 1 – 1 = 0 f(x) is a polynomial of degree atmost 2, and also attains same value (i.e., 0) for 3 distinct values of x (i.e. a,b,c).  f(x) is an identity with only value equal to zero.  f(x) = 0  x R  P(x) = 1,  x R 2.

Let a, b, x, y be real numbers such that a2 + b2 = 81, x2 + y2 = 121 and ax + by = 99. Then the set of all possible values of ay – bx is  9 (A)  0,   11

Sol.

 9 (B)  0,   11 

b a = y  x

Aliter : a2 + b2 = 81 x2 + y2 = 121 ax + by = 99 (a2 + b2) (x2+ y2) = 81 × 121 (ax + by)2 = 992 (iv) – (v) (ay – bx)2 = 0  ay – bx = 0

Sol.

9  (D)  ,   11  

(C) Using cauchy schwartz's inequality (a2 + b2) (x2 + y2)  (ax + by)2 equality holds at

3.

(C) {0}

ay – bx = 0

......(i) ......(ii) ......(iii) ......(iv) ......(v)

1 1 1  a , x2  is  b , then x 3  3 x x2 x (A) a3 + a2 – 3a – 2 – b (B) a3 – a2 – 3a + 4 – b (A)

[(i) × (ii)] [squaring (iii)]

If x 

RESONANCE

(C) a3 – a2 + 3a – 6 – b (D) a3 + a2 + 3a – 16 – b

PAGE - 2


x

Then,

1 1  a and x 2  b x x3

2

1   x    a2 x 

x2 +

1

1  x   x 

x3 +

+2 = a2

x2

.........(i)

3

= a3

1

1   3 x   = a3 x  x

.........(ii)

3

add equation (1) and (2) 1   1  1  2   x  3    x 3  2   2  3 x   = a2 + a3 x x   x   

 3 1  b +  x  2  + 2 + 3a = a2 + a3 x  

x3 

4.

Sol.

5.

1 x2

= a3 + a2 – 3a – b – 2

Let a, b, c, d be real numbers such that |a – b| = 2, |b – c| = 3, |c – d| = 4. Then the sum of all possible values of |a – d| is (A) 9 (B) 18 (C) 24 (D) 30 (B) |a – b| = 2, |b – c| = 3, |c – d| = 4 a–b=±2 b–c=±3 c–d=±4 possible values of (a – d) are ± 9, ± 5, ± 3 |a – d| = 9, 5, 3, 1 Sum of all possible values are 18 Below are four equations in x. Assume that 0 < r < 4. Which of the following equations has the largest solution for x ? x

x

r  (A) 5 1    9   

Sol.

r   (B) 51   9 17  

x

(C) 5 (1 + 2r)x = 9

1 (D) 5 1    9 r 

(B) Given 0 < r < 4 in all the obtain (Base)x =

9 5

the option having least base will give the largest x. r    is minimum for 0 < r < 4. So, in option B base  1  17  

RESONANCE

PAGE - 3


Aliter : (1 + x =

4 5



x log (1 + ) = log (1.8)



log (1.8) x = log (1   )

For x to be maximum log (1 + ) should be minimum =

r or 

2r

r 17

or

1 r

or

in 0 < r < 4

r is minimum. 17 6.

Let ABC be a triangle with B = 90°. Let AD be the bisector of A with D on BC. Suppose AC = 6 cm and the area of the triangle ADC is 10 cm2. Then the length of BD in cm is equal to (A)

Sol.

3 5

(B)

3 10

(C)

5 3

(D)

10 3

(D) Angle bisector theorem x y = 6 z

 

xz = 6y 20 = 6y



y=

6

x

2

10cm

10 cm 3

y

z

ALITER : Let ACB =  BC = 6 cos , AB = 6 sin  BD : CD = AB : AC = 6 sin  : 6 = sin  : 1 BD =

6 cos 

1  sin 

 sin 

 6 cos    1 CD =   1  sin  

.........(i)

.........(ii)

BD = CD sin  Now, area of ADC = 10 cm2

1 × 6(CD) sin = 10 2 CD sin = 10/3 cm

RESONANCE

PAGE - 4


7.

A piece of paper in the shape of a sector of a circle (see Fig.1) is rolled up to form a right-circular cone (see Fig. 2). The value of the angle  is

(A) Sol.

10 13

(B)

9 13

(C)

5 13

(D)

6 13

(A)

 

Slant height = 13 by  = r  2.5 = 13 





10 13

8.

In the adjoining figure AB = 12 cm, CD = 8 cm, BD = 20 cm; ABD = AEC = EDC = 90°. IF BE = x, then

Sol.

(A) x has two possible values whose difference is 4 (B) x has two possible values whose sum is 28 (C) x has only one value and x 12 (D) x cannot be determined with the given information (A) 20  x 12 = 8 x

x2 – 20x + 96 = 0 x = 8, 12 ALITER : 

A(0,12)

C(20,8) B(0,0)

E(x,0)

D(20,0)

= 4 52  1 = 4 26 20 2  4 2  (122 + x2) + (82 + (20 – x)2) = 202 + 42 2 2x – 40x + 400 + (122 – 202) + (82 – 42) = 0 2x2 – 40x + 144 + 12.4 = 0 AC =

RESONANCE

PAGE - 5


x2 – 20x + 72 + 24 = 0 x2 – 20x + 96 = 0 x = 12, 8 9.

Three circles each of radius 1 touch one another externally and they lie between two parallel lines. The minimum possible distance between the lines is (A) 2  3

Sol.

(B) 3  3



Sol.

11.

Sol.

(D) 2  1 3

(A)

AP = 2 sin 60° =

10.

(C) 4

3

d = 2 3

The number of distinct prime divisors of the number 5123 – 2533 – 2593 is (A) 4 (B) 5 (C) 6 (C) 5123 – (2533 + 2593) = 5123 – [(512) (2532 + 2592 – 253.259)] = 512 (5122 – ((512)2 – 3(253)(259)) = 512 (3.253 – 259) = 29. 3. 253 . 7.37 = 29. 3. (11) . (23). 7.37 So, number of distinct prime divisors are 6.

(D) 7

Consider an incomplete pyramid of balls on a square base having 18 layers; and having 13 balls on each side of the top layer. Then the total number N of balls in that pyramid satisfies (A) 9000 < N < 10000 (B) 8000 < N < 9000 (C) 7000 < N < 8000 (D) 10000 < N < 12000 (B) Top layer has (13 × 13) balls Simillary one layer below top layer will have (14 × 14) balls and we have 18 lesens to total number of ball N = (13)2 + (14)2 + .............+ (30)2 N=

30  31 61 12  13  25 – 6 6

N = 8805 12.

A man wants to reach a certain destination. One sixth of the total distance is muddy while half the distance is tar road. For the remaining distance he takes a boat. His speed of traveling in mud, in water on tar roaed is in the ratio 3 : 4 : 5. The ratio of the durations he requires to cross the patch of mud. stream and tar road is: (A)

1 4 5 : : 2 5 2

RESONANCE

(B) 3 : 8 : 15

(C) 10 : 15 : 18

(D) 1 : 2 : 3

PAGE - 6


Sol.

(C) Let distance is 6d

13.

Mud

:

Tar

:

Stream

Distance

d

:

3d

:

2d

Speed

3V

:

5V

:

4V

time

d 3V

:

3d 5V

:

2d 4V



10

:

18

:

15

A frog is presently located at the origin (0, 0) in the xy-plane. It always jumps from a point with integer coordinates to a point with integer coordinates moving a distance of 5 units in each jump. What is the minimum number of jumps required for the frog to go from (0, 0) to (0, 1) ? (A) 2

Sol.

(Note : order is changed in questions)

(B) 3

(C) 4

(D) 9

(B) 3 step

14.

A certain 12-hour digital clock displays the hour and the minute of a day. Due to a defect in the clock whenever the digit 1 is supposed to be displayed it displays 7. What fraction of the day will the clock show the correct time ? (A)

Sol.

1 2

(B)

5 8

(C)

3 4

(D)

5 6

(A) The clock will show the incorrect time (between 1 - 2, 10 - 11, 12 - 1 11 - 12 day and night both) Therefore incorrect time 8 × 60 = 480. Remaining 20 hour it will show the incorrect time 16 × 15 = 240 Total incorrect time = 240 + 480 = 720 the fraction day when the clock shows correct time is = 1 –

RESONANCE

720 1 1 =1– = 24  60 2 2

PAGE - 7


15.

Sol.

There are 30 questions in a multiple-choice test. A student gets 1 mark for each unattempted question, 0 mark for each wrong answer and 4 marks for each correct answer. A student answered x questions correctly and scored 60. Then the number of possible value of x is (A) 15 (B) 10 (C) 6 (D) 5 (C) Right 15 14 13 12 11 10

Wrong 15

Unattempted 0 4 8 12 16 20

6 cases only

PHYSICS 16.

Sol.

17.

A simple pendulum oscillates freely between points A and B.

We now put a peg (nail) at some point C as shown. As the pendulum moves from A to the right, the string will bend at C and the pendulum will go to its extreme point D. Ignoring friction, the point D. (A) will lie on the line AB (B) will lie above the line AB (C) will lie below the line AB (D) will coincide with B (A) By mechanical energy conservation KEi + Ui = KEf + Uf O + Ui = 0 + Uf Ui = Uf hi = hf So D will lie on line AB A small child tries to move a large rubber toy placed on the ground. The toy does not move but gets deformed 

under her pushing force (F) which is obliquely upward as shown . Then

(A)

Sol.



(B)

The resultant of the pushing force (F) , weight of the toy, normal force by the ground on the toy and the frictional force is zero. The normal force by the ground is equal and oipposite to the weight of the toy.

(C)

The pushing force (F) of the child is balanced by the equal and opposite frictional force

(D)

The pushing force (F) of hte child is ballanced by the total internal force in the toy generated due to deformation





(A) Since toy is not accelerating so net external force on toy is zero. So (A)

RESONANCE

PAGE - 8


18.

A juggler tosses a ball up in the air with initial speed u. At the instant it reaches its maximum height H, he tosses up a second ball with the same initial speed. The two balls will collide at a height. (A)

Sol.

H 4

H 2

(B)

(C)

3H 4

(D)

3 H 4

(C) |S1| + |S2| = H 1 2 1 gt + ut – gt 2 = H 2 2

ut = H .

....(I)

u2 2g

....(II)

H=

u

t = 2g S2 = ut – = 19.

Sol.

u 1 u2 3u2 1 2  g  gt = 4 2g 2 4g2 8g 2

3 H 4

On a horizontal frictional frozen lake, a girl (36 kg) and a box (9kg) are connected to each other by means of a rope. Initially they are 20 m apart. The girl exerts a horizontal force on the box, pulling it towards her. How far has the girl travelled when she meets the box ? (A) 10 m (B) Since there is no friction, the girl will not move (C) 16 m (D) 4m (D)

By concept of centre of mass 36x = 9(20–x) 36x = 180 – 9x 45x = 180 x = 4m 20.

Sol.

21.

Sol.

The following three objects (1) a metal tray, (2) a block of wood, and (3) a wooden cap are left in a closed room overnight. Next day the temperature of each is recorded as T1 , T2 and T3 respectively. The likely situation is (A) T1 = T2 = T3 (B) T3 > T2 > T1 (C) T3 = T2 > T1 (D) T3 > T2 = T1 (A) All three will be in thermal equilibrium with air of room. so temperature of the three will be same We sit in the room with windows open. Then (A) Air pressure on the floor of the room equals the atmospheric pressure but the air pressure on the ceiling is negligible (B) Air pressure is nearly the same on the floor, the walls and ceiling (C) Air pressure on the floor equals the weight of the air coloumn inside the room (from floor to ceiling) per unit area (D) Air pressure on the walls is zero since the weight of air acts downward (B) Pressure of gas is same everywhere in the vessel.

RESONANCE

PAGE - 9


22.

Sol.

23.

Sol.

A girl standing at point P on a beach wishes to reach a point Q in the sea as quickly as possible. She can run at 6 kmh–1 on the beach and swim at 4 km-h–1 in the sea. She should take the path

(A) PAQ (B) PBQ (C) PCQ (C) To travel from P to Q in minimum time, she should travel on path PCQ.

Light enters an isosceles right triangular prism at normal incidence through face AB and undergoes total internal reflection at face BC as shown below :

The minimum value of the refractive index of the prifsm is close to : (A) 1.10 (B) 1.55 (C) 1.42 (C) i = 45°  C C = 45° for minimum   sin 45 = 1  =

24.

Sol.

(D) PDQ

(D) 1.72

2 = 1.42

A convex lens is used to form an image of an object on a screen. If the upper half of the lens is blackened so that it becomes opaque. Then : (A) Only half of the image will be visible (B) the image position shifts towards the lens (C) the image position shifts away from the lens (D) the brightness of the image reduces (D)

In this case only half part of lens is used to form the image so intensity will reduce 25.

Sol.

A cylindrical copper rod has length L and resistance R. If it is melted and formed into another rod of length 2L. the resistance will be : (A) R (B) 2R (C) 4R (D) 8R (C) R=

 A

R=

   A 

RESONANCE

PAGE - 10


R=

 2 V

R’ =

(2 )2 = 4R V

26.

Two charges +Q and _2Q are located at points A and B on a horizontal line as shown below :

Sol.

The electric field is zero at a point which is located at a finite distance : (A) On the perpendicular bisector of AB (B) left of A on the line (C) between A and B on the line (D) right of B on the line (B)

E1 = Electric field due to +Q E2 = Electric field due to –2Q There resultant is 0 at this point

27.

Sol.

A 750 W motor drives a pump which lifts 300 litres of water per minute to a height of 6 meters. The efficiency of the motor is nearly (take acceleration due to gravity to be 10 m/s2) (A) 30% (B) 40% (C) 50% (D) 20% (B) mgh / t  100 750

=

300  10  6 60 =  100 750

=

300  100 750

= 40% 28.

Figure below shows a portion of an electric circuit with the currents in ampreres and their directions. The magnitude and direction of the current in the portion PQ is :

(A) 0A

RESONANCE

(B) 3A from P to Q

(C) 4A from Q to P

(D) 6A from Q to P

PAGE - 11


Sol.

(D)

6A from Q to P 29.

A nucleus of lead pb 314 emits two electrons followed by an alpha particle. The resulting nucleus will have 82

Sol.

(A) 82 protons and 128 neutrons (C) 82 protons and 130 neutrons (A)

(B) 80 protons and 130 neutrons (D) 78 protons and 134 neutrons

210  2e–1 + He 24 + X82 pb 314 82

So 82 proton and 128 Neutron 30.

Sol.

The number of air molecules in a (5m × 5m × 4m) room at standard temperature and pressure is of the order of (A) 6 × 1023 (B) 3 × 1024 (C) 3 × 1027 (D) 6 × 1030 (C) PV = NKT 105 × 100 = N × 1.38 × 10–23 × 273 N  3 × 1027

CHEMISTRY 31.

Sol.

32.

Two balloons A and B containing 0.2 mole and 0.1 mole of helium at room temperature and 2.0 atm. respectively, are connected. When equilibrium is established, the final pressure of He in the system is (A) 0.1 atm (B)1.5 atm (C) 0.5 atm (D) 2.0 atm (D) Since pressures of the gases are same in both the containers. So, the final pressure will not change In the following set of aromatic compounds

(i)

Sol.

(ii)

(iii)

(iv)

the correct order of reactivity toward friedel-crafts alkylations is (A) i > ii > iv (B) ii > iv > iii > i (C) iv > iii >i (D) iii > i > iv > ii (C) Reativity towards Friedel-Crafts alkylation is proportional to electron density in the benzene ring.

RESONANCE

PAGE - 12


33.

Sol.

The set of principal (n). azimuthal (l) and magnetic (m 1) quantum numbers that is not allowed for the electron in H-atom is (A) n = 3, l = 1, m  = -1 (B) n = 3, l = 0, m  = 0 (D) n = 2, l = 1, m = 0 (D) n = 2, l = 2, m  = -1 (D) n=2  = 0, 1 0  m = 0,  1  1  

34.

Sol.

At 298 K, assuming ideal behaviour, the average kinetic energy of a deuterium molecule is : (A) two times that of a hydrogen molecule (B) four times that of a hydrogen molecule (C) half of that of a hydrogen molecule (D) same as that of a hydrogen molecule (D) Average Kinetic Energy depends only on temperature (KE)avg =

3kT per molecule 2

35.

An isolated box, equally partitioned contains two ideal gases A and B as shown

Sol.

When the partition is removed, the gases mix. The changes in enthalpy (H) and entropy (S) in the process, respectively, are : (A) zero, positive (B) zero, negative (C) positive, zero (D) negative, zero (A) Ideal gas H = 0 S > 0 (randomness increases)

36. Sol.

The gas produced from thermal decomposition of (NH4)2Cr2O7 is : (A) oxygen (B) nitric oxide (C) ammonia (D)

(D) nitrogen

 (NH4)2 Cr2O7  Cr2O3 + N2 + 4H2O 37.

The solubility curve of KNO3 in water is shown below.

The amount of KNO3 that dissolves in 50 g of water at 40°C is closest to : (A) 100 g (B) 150 g (C) 200 g (D) 50 g

RESONANCE

PAGE - 13


Sol.

(A) According to the graph solubility 40° is approx. 200 g per 100 ml. For 50 ml, amount is 100 g approx.

38.

A compound that shows positive iodoform test is : (A) 2-pentanone (B) 3-pentanone (C) 3-pentanol (A)

Sol.

Aldehyde, ketones with acetyl group CH3 –

(D) 1-pentanol

– show Iodoform test.

After 2 hours the amount of a certain radioactive substance reduces to 1/16th of the original amount (the decay process follows first-order kinetics). The half-life of the radioactive substance is : (A) 15 min (B) 30 min (C) 45 min (D) 60 min (B)

39.

Sol.

1 2

40.

n



1 n, number of half-lives = 4 = 2 hrs. half-life = 30 min. 16

Ans.

In the conversion of a zinc ore to zinc metal, the process of roasting involves. (A) ZnCO3 ZnO (B) ZnO ZnSO4 (C) ZnS ZnO (D) ZnS ZnSO4 (C)

Sol.

2ZnS + 3O2  2ZnO + 2SO2

41.

The number of P-H bond(s) in H3PO2, H3PO3 and H3PO4, respectively, is : (A) 2, 0, 1 (B) 1, 1, 1 (C) 2, 0, 0 (D) 2, 1, 0 (D)

Sol.

42.

Sol.

43.

Sol.

When chlorine gas is passed through an aqueous solution of KBr, the solution turns orange brown due to the formation of : (A) KCl (B) HCl (C) HBr (D) Br2 (D) Cl2 + 2KBr 2KCl + Br2 reddish brown Among

the compound which is not aromatic is : (A) i (B) ii (C) iii (D) iv (B) (i) and (iv) are hetro aromatic and the resonance form of azulene (iii) is aromatic.

(ii) Is nonaromatic.

RESONANCE

PAGE - 14


44.

Among the following compounds :

(i)

Sol.

45.

(ii)

2, 3–dimethylhexane is : (A) i (B) ii (B) Simple nomenclature of alkane.

(iii)

(iv)

(C) iii

(D) iv

The major product formed in the reaction :

Product is :

Sol.

(A) i (B) ii (C) iii (D) iv (C) The given reaction is SN2 which occurs at sp3 carbon with good leaving group.

BIOLOGY 46.

Sol.

47.

Sol.

48.

Sol.

49.

Sol.

If parents have free ear lobes and the offspring has attached ear lobes, then the parents must be (A) homozygous (B) heterozygous (B) co-dominant (D) nullizygous (B) heterozygous During meiosis there is (A) one round of DNA replication and one division (B) two round of DNA replication and one division (C) two round of DNA replication and two division (D) one round of DNA replication and two division (D) one round of DNA replication and two division Blood clotting involves the conversion of (A) prothrombin to thromboplastin (C) fibrinogen to fibrin (C) fibrinogen to fibrin The gall bladder is involved in (A) synthesizing bile (C) degrading bile (B) Storing and secreting bile

RESONANCE

(B) thromboplastin to prothrombin (D) fibrin to fibrinogen

(B) storing and secreting bile (D) producing insulin

PAGE - 15


50.

Which one of the following colors is the LEAST useful for plant life ? (A) red (B) blue (C) green (C) Green

Sol.

51. Ans. 52.

At rest the volume of air that moves in and out per breath is called (A) resting volume (B) vital capacity (C) lung capacity (D)

(C) violet

(D) tidal volume

How many sex chromosomes does a normal human inherit from father ? (A) 1 (B) 2 (C) 23 (D) 46 (A) 1

Sol.

53.

In the 16th century, sailors who travelled long distances had diseases related to malnutrition, because they were not able to eat fresh vegetables and fruits for months at a time scurvy is a result of deficiency of (A) carbohydrates (B) proteins (C) vitamin C (D) vitamin D (C) vitamin C

Sol.

54.

The following structure is NOT found in plant cells (A) vacuole (B) nucleus (C) centriole (C) centriole

Sol.

55.

The cell that transfers information about pain to the brain is called a (A) neuron (B) blastocyst (C) histoblast (A) Neuron

Sol.

56.

(D) vitamin D

(D) vitamins

The presence of nutrient sin the food can be tested. Benedict's's test is used to detect. (A) sucrose (B) glucose (C) fatty acid (D) vitamins (B) Glucose

Sol. 57.

Several mineral such as iron, iodine, calcium and phosphorous are important nutrients. Iodine is found in (A) thyroxine (B) adrenaline (C) insulin (D) testosterone (A) thyroxine

Sol. 58.

The principle upon which a lactometer works is (A) viscosity (B) density (C) surface tension (B) Density

Sol. 59.

Ans. 60.

Ans.

Mammalian liver cells will swell when kept in (A) hypertonic solution (C) isotonic solution (B) Hypotonic solutions

(D) presence of protein

(B) hypotonic solutions (C) isothermal solutions

The form of cancer called 'carcinoma' is associated with (A) lymph cells (B) mesodermal cells (C) blood cells (D) epithelial cells (D) Epithelial cells

RESONANCE

PAGE - 16


PART-II Two Mark Questions MATHEMATICS 61.

Let f(x) = ax2 + bx + c, where a, b, c are integers. Suppose f(1) = 0, 40 < f(6) < 50, 60 < f(7) < 70, and 1000t < f(50) < 1000 (t + 1) for some integer t. Then the value of t is (A) 2

Sol.

(B) 3

(C) 4

(D) 5 or more

(C) f(x) = ax2 + bx + c, given,

f(1) = 0

f(1) = 0 a+b+c=0

and

40 < f(6) < 50 40 < 36a + 6b + c < 50 40 < 35a + 5b < 50 8 < 7a + b < 10

7a + b = integer = 9 and

.......(1)

60 < f(7) < 70 60 < 49a + 7b + c < 70

60 < 48a + 6b < 70 10 < 8a + b < 11.6 8a + b = 11

.......(2)

solving equation (1) and (2) a = 2, b = –5, so c = 3 f(x) = 2x2 – 5x + 3 f(50) = 4753 t=4 62.

The expression 22  1 22  1

32  1



32  1



42  1 42  1

 .......... ... 

( 2011) 2  1 ( 2011)2  1

lies in the interval

Sol.

(A) (2010, 2010

1 ) 2

1 1   , 2011   (B)  2011  2011 2012  

(C) (2011, 2011

1 ) 2

(D) (2012, 2012

1 ) 2

(C) We can write the expression as 2011 r 2

 r 2

2

1

r 1

=

  2  1      r  1 r  1  

=

1 1     1    r  1 r  1

RESONANCE

PAGE - 17


Putting r = 2, 3, ............, 2011 = 2010 + 1 +

= 2011 +

1 1 1   2 2012 2011

1  1 1    2  2011 2012 

1 ) 2

this lies between (2011, 2011

63.

The diameter of one of the bases of a truncated cone is 100 mm. If the diameter of this base is increased by 21% such that it still remains a truncated cone with the height and the other base unchanged, the volume also increases by 21%. The radius the other base (in mm) is (A) 65

Sol.

(B) 55

(C) 45

(D) 35

(B) Let initially 2 bases have radii 5 & r. and finally bases have radii (1.21 × 5) & r. Ratios of volumes =



V2  1.21 V1

V2 =

h ((6.05)2 + 6.05 r + r2) 3

V1 =

h 2 (5 + 5r + r2) 3

V2 (6.05)2  6.05 r  r 2  1.21   1.21 V1 5 2  5r  r 2 36.6025 + 6.05 r + r2 = 30.25 + 6.05r + 1.21 r2 .21r2 = 6.3525 r2 =

r= 64.

6.3525 .21 11 cm. = 55 mm 2

Two friends A and B are 30 km apart and they start simultaneously on motorcycles to meet each other. The speed of A is 3 times that of B. The distance between them decreases at the rate of 2 km per minute. Ten minutes after they start A's vehicle breaks down and A stops and waits for B to arrive. After how much time (in minutes) A started riding, does B meet A ? (A) 15

Sol.

(B) 20

(C) 25

(D) 30

(D) Let speed of B = V km/hr. Let speed of A = 3V km/hr. Given 4r = 2 × 60 km/hr  V = 30 km/hr Distance covered by then after 10 min. = 2 × 10 = 20 km So, remaining distance = (30 – 20) km = 10 km. Time table by B to cover 10 km =

10 = 20 min 30

Total Time 20 + 10 = 30

RESONANCE

PAGE - 18


65.

Sol.

There taps A, B, C fill up a tank independently in 10 hr, 20 hr, 30 hr, respectively. Initially the tank is empty and exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour. What is the minimum number of hours required to fill the tank ? (A) 8 (B) 9 (C) 10 (D) 11 (A) A B C 10 hr 20 hr 30 hr Exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour. So, first we say, A and B are open for 1 hour, then B & C and then C & A. 1  1  1   1  1  1 22      +   +   = 60  10 20   20 30   30 10 

First then

second then

Third then

 22   In three hours the tank will be filled   60 

th

part

1 9   1    Now, for minimum time, the rest tank must be filled with A and B taps.   10 20 60   38   So, the rest   60 

th

Part of tank will take 5 hour more

So, the tank will be filled in 8th hour.

PHYSICS 66.

An object with uniform desity  is attached to a spring that is known to stretch linearly with applied force as shown below

When the spring object system is immersed in a liquid of density 1 as shown in the figure, the spring stretches by an amount x1 ( > 1). When the experiment is repeated in a liquid of density 2 < 1 . the spring strethces by an amound x2. Neglecting any buoyant force on the spring, the density of the object is : (A)   Sol.

1x1  2 x 2 x1  x 2

(B)  

1x 2  2 x1 x 2  x1

(C)  

1x 2  2 x1 x1  x 2

(D)  

1x1  2 x 2 x1  x 2

(B)

RESONANCE

PAGE - 19


kx1 + 1vg = vg kx1

v = g   g 1 kx2 + 2vg = vg kx 2

v = g   g 2 kx1 kx 2  (  1)g (  2 )g

x1 – x1 = x2 – x2 x1 – x2) = x1 – x2 = 67.

Sol.

2x1  1x 2 x1  x 2

A body of 0.5 kg moves along the positive x - axis under the influence of a varying force F (in Newtons) as shown below :

If the speed of the object at x = 4m is 3.16 ms–1 then its speed at x = 8 m is : (A) 3.16 ms–1 (B) 9.3 ms–1 (C) 8 ms–1 (D) 6.8 ms–1 (B) 8

1

2

1

2

 F.dx  2 mv f  2 mvi 4

1 1 1 ×3×8– × 1.5 × 4 = × 0.5 (vf2 – 3.16)2 2 2 2

24 – 6 = 0.5 (vf2 – 3.162) 36 = vf2 – 3.162 vf  6.8 m/s 68.

In a thermally isolated system. Two boxes filled with an ideal gas are connected by a valve. When the valve is in closed position, states of the box 1 and 2. respectively, are (1 atm, V, T) and (0.5 atm, 4V, T). When the valve is opened, the final pressure of the system is approximately. (A) 0.5 atm (B) 0.6 atm (C) 0.75 atm (D) 1.0 atm

RESONANCE

PAGE - 20


Sol.

(B)

vi = vf 5 5 5 ×1 × V + × 0.5 × 4v = P 5V 2 2 2

3V = 5PV P= 69.

A student sees the top edge and the bottom centre C of a pool simultaneously from an angle  above the horizontal as shown in the figure. The refractive index of water which fills up to the top edge of the pool is 4/ 3. If h/x = 7/4 then cos  is :

(A) Sol.

3 = 0.6 5

2 7

(B)

8 3 45

(C)

8 3 53

(D)

8 21

(C)

sin(90 – ) = sinr 4

cos = 3 

x/2 h2 

RESONANCE

x2 4

PAGE - 21


cos =

2 3

1 h2

1  2 4 x

=

4

8

2

= 3 53 70.

Sol.

=

2 3

1 49 4  16 16

3 53

In the following circuit, the 1 resistor dissipates power P. If the resistor is replaced by 9. the power dissipated in it is

(A) P (A)

(B) 3P

(C) 9P

(D) P/3

Pi = i2R 2

100  10   1  16  4 

P= 

Pf = i2R 2

100 100  10  9  =   9  =P 12  12 16  12 

CHEMISTRY 71.

Sol.

An aqueous buffer is prepared by adding 100 ml of 0.1 mol 1–1 acetic acid to 50 ml of 0.2 mol 1–1 of sodium acetate. If pKa of acetic acid is 4.76, the pH of the buffer is : (A) 4.26 (B) 5.76 (C) 3.76 (D) 4.76 (A) 100 mL 0.1 M CH3COOH + 50 mL 0.4 M CH3COONa

CH COO  = 1 3



Ch3COOH

pH = pKa + log 1 = pKa = 4.76

RESONANCE

PAGE - 22


72.

Sol.

The maximum number of stgructural isomers possible for the hydrocarbon having the molecular formuka C4H6, is : (A) 12 (B) 3 (C) 9 (D) 5 (C) 9-structural isomers are possible. CH3CH2CCH, CH3 –CC–CH3 CH2 =CH–CH=CH2 , CH2 =C=CH–CH3

73.

In the following reaction sequence, X and Y, respectively, are : X



Y

OH



(A) H2O2 ; LiAlH4 (C) C6H5 COOOH ; Zn/Hg HCl (B)

Sol.

74.

(B) C6H5 COOOH ; LiAlH4 (D) Alkaline KMnO4 ; LiAlH4

Among (i) [Co(NH3)6]Cl3, (ii) [Ni(NH3)6]Cl2 , (iii) [Cr(H2O6)]Cl3, (iv) [Fe(H2O)6]Cl2 the complex which is diamagnetic is : (A) i (B) ii (C) iii (D) iv (A) Co+3 : 3d6 45° strong field ligand (NH3) Ni+2 : 3d8 45° SFL (NH3) Cr+3 : 3d3 45° WFL (H2O) Fe+2 : 3d6 45° WFL (H2O) Co+3 will be diamagnetic (i)

Sol.

75.

At 783 K in the reaction H2 (g) + I2 (g) 2Hl(g), the molar concentrations (mol 1–1) of H2, I2 and HI at some instant of time are 0.1, 0.2 and 0.4, respectively. If the equilibrium constant is 46 at the same temperature, then as the reaction proceeds. (A) the amount of HI will increase (B) the amount of HI will decrease (C) the amount of H2 and I2 will increases (D) the amount of H2 and I2 will not change (A)

Sol.

Reaction quotrient Q =

[H] 2 (0.4)2 = =8 [H2 ] [ 2 ] 0.1 0.2

Q
BIOLOGY 76.

Ans.

You removed four fresh tobacco leaves of similar size and age. Leave "leaf 1" as it is , smear " leaf 2" with vaseline on the upper surface, "leaf 3" on the lower surface and "leaf 4" on both the surface . Hang the leaves for a few hours and you observe that leaf 1 wilts the most, leaf 2 has wilted. leaf 3 wilted less than leaf 2 and leaf 4 remains fares. Which of the following conclusion is most logical ? (A) tobacco leaf has more stomata on the upper surface (B) tobacco leaf has more stomata on the lower surface (C) stomata are equally distributed in upper and lower surface (D) no conclusion on stomatal distribution can be drawn from this experiment (B)

RESONANCE

PAGE - 23


77.

Ans. 78.

Ans. 79.

Ans. 80.

Ans.

Vestigial organs such as the appendix exist because (A) they had and important function during development which is not needed in the adult (B) they have a redundant role to play if an organ with similar functions fails. (C) nature cannot get rid of structures that have already formed (D) they were inherited from an evolutionary ancestor in which they were functional (D) Mendel showed that unit factors, now called alleles, exhibit a dominant recessive relationship. In a monohybrid cross. the .............................trait disappears in the first filial generation (A) dominant (B) co-dominant (C) recessive (D) semi-dominant (C) If a man with an X-linked dominant disease has six sons with a woman having a normal complement of genes, then the sons will (A) not show any symptoms of the disease (B) show strong symptoms of the disease (C) three will show a disease symptom, while three will not (D) five will show a disease symptom, while one will no (A) In evolutionary terms, an Indian school boy is more closely related to (A) an Indian frog (B) an American snake (C) a Chinese horse (D) an African shark (C)

RESONANCE

PAGE - 24

KVPY_STREAM (SA) Class-XI Test Paper &amp; Solutions

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