Kvpy 2010 Class-XI solutions by Resonance

8/8/2019 Kvpy 2010 Class-XI solutions by Resonance

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QUESTIONS & SOLUTIONS OF

KISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) 2010TEST PAPER CLASS - XI

One-Mark Questions

MATHEMATICS

1. A student notices that the roots of the equation x2 + bx + a = 0 are each 1 less than the roots of the equation

x2 + ax + b = 0. Then a + b is :

(A) possibly any real number (B) 2

(C) 4 (D) 5

Ans. (C)

Sol. x2 + bx + a = 0

x2 + ax + b = 0

x = 1 x + 1 =

(x + 1)2 + a(x + 1) + b = 0

x2 + (a + 2)x + 1 + a + b = 0

x2 + bx + a = 0 and 1 + a + b = a b = 1

a = 3

_________

a + b = 4

2. If x, y are real numbers such that

1y

x

3 1y

x

3 = 24,

then the value of (x + y) / (x y) is :

(A) 0 (B) 1 (C) 2 (D) 3

Ans. (D)

Sol. 3x/y = t 3t 3

t= 24 8t = 3 24

t = a

x = 2y

yx

yx

y

y3 3 Ans. (D)

(TEST PAPER HELD ON 31-10-2010)


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3. The number of positive integers n in the set {1, 2, 3, ...., 100} for which the numbern......321

n.....321 2222

is an integer is :

(A) 33 (B) 34 (C) 50 (D) 100

Ans. (B)

Sol.)1n(n3

)1n2)(1n(n

2

3

1n2 = k n =

2

1k3

31 2

1k3 100 3 3k 201 1 k

3

120

1 k = 67

k is odd 34 (B)

4. The three different face diagonals of a cuboid (rectangular parallelopiped) have lengths 39, 40, 41. The length

of the main diagonal of the cuboid which joins a pair of opposite corners :

(A) 49 (B) 249 (C) 60 (D) 260

Ans. (A)

Sol. 2 + b2 = 392

b2 + h2 = 402

h2 + 2 = 412

adding

2(2 + b2 + h2) = 392 + 402 + 412

As =2

414039 222 =

2

)140(40)140( 222

=2

2)40(3 2 =

2

4802= 2401

= 49 Ans. (A)

5. The sides of a triangle ABC are positive integers. The smallest side has length . What of the following

statements is true ?

(A) The area of ABC is always a rational number.(B) The area of ABC is always an irrational number.

(C) The perimeter of ABC is an even integer.

(D) The information provided is not sufficient to conclude any of the statements A, B or C above.

Ans. (B)

Sol. It has to be isosceles triangle

=2

1 1

4

1x2 = 1x4

4

1 2

perimeter = 1 + 2x odd always irrational (B)


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6. Consider a square ABCD of Side 12 and let M, N be the midpoints of AB, CD respectively. Take a point P on

MN and let AP = r, PC = s. Then the area of the triangle whose sides are r, s, 12 is :

(A) 72 (B) 36 (C)2

rs(D)

4

rs

Ans. (B)

Sol. =2

1 12 6

= 36

Ans. (B)

7. A cow is tied to a corner (vertex) of a regular hexagonal fenced area of side a meters by a rope of length

5a / 2 meters in a grass field. (The cow cannot graze inside the fenced area.) What is the maximum possible

area of the grass field to which the cow has access to graze ?

(A) 5 a2 (B)2

5a2 (C) 6a2 (D) 3a2

Ans. ( ....)

Sol. Area =

4

a36

2

a5 22

=4

a36

4

a25 22

=4

a)3625(

2

Note : Options are not match with solution.

8. A closed conical vessel is filled with water fully and is placed with its vertex down. The water is let out at a

constant speed. After 21 minutes, it was found that the height of the water column is half of the original

height. How much more time in minutes does it empty the vessel ?

(A) 21 (B) 14 (C) 7 (D) 3

Ans. (D)


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Sol.dt

dv= c.

v = hr3

1 2 tan =h

r

=

3

h3 tan

dt

dv= h2 tan

dt

dh ct + k =

3

tanh3

when t = 0, h = H k =3

tanH3 ct =

3

(h3 H3) tan

when t = 21 , h =2

Hc = 36

H3

8

7

h = 0

tan)H(3

3= t

8

7H

6

3

3

24

7=

728

7

t

t = 24

more time in minutes does it empty the vessel is 3

9. I carried 1000 kg of watermelon in summer by train. In the beginning, the water content was 99%. By the time

I reached the destination, the water content had dropped to 98%. The reduction in the weight of the water-melon was:

(A) 10 kg (B) 50 kg (C) 100 kg (D) 500 kg

Ans. (D)

Sol. Water + solid = 1000

Water is 1000100

99 = 990

water evaporated is x.

so 100x1000

x990

= 98

99000 100 x = 96000 98x

1000 = 2x

x = 500


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10. A rectangle is divided into 16 sub-rectangles as in the figure; the number in each sub-rectangle represents

the area of that sub-rectangle. What is the area of the rectangle KLMN ?

(A) 20 (B) 30 (C) 40 (D) 50

Ans. (D)

Sol.

xy = 10 and yz = 4 z =5

x2

y

also

x

52 a = 12 a =

x30

and

x

30(b) = 15 b =

2

x

and

2

x(c) = 25 c =

x

50

Area = x

x

50= 50


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PHYSICS

11. A hollow pendulum bob filled with water has a small hole at the bottom through which water escapes at a

constant rate. Which of the following statements describes the variation of the time period (T) of the pendu-

lum as the water flows out ?

(A) T decreases first and then increases. (B) T increases first and then increases.

(C) T increases throughout. (D) T does not change.

Ans. (A)

Sol. T =g

2

First distance of com from suspension point will increase then decrease. So,

T .

12. A block of mass M rests on a rough horizontal table. A steadily increasing horizontal force is applied such

that the block starts to slide on the table without toppling. The force is continued even after sliding has

started. Assume the coefficients of static and kinetic friction between the table and the block to be equal.

The correct representation of the variation of the frictional forces, , exerted by the table on the block with

time t is given by :

(A) (B) (C) (D)

Ans. (A)

Sol. when sliding has started

till acceleration of block is zero F fk = maF f

s= 0

fs

= F


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13. A soldier with a machine gun, falling from an airplane gets detached from his parachute. He is able to resist

the downward acceleration if he shoots 40 bullets a second at the speed of 500 m/s. If the weight of a bullet

is 49 gm, what is the weight of the man with the gun ? Ignore resistance due to air and assume the

acceleration due to gravity g = 9.8 m/s2

(A) 50 kg (B) 75 kg (C) 100 kg (D) 125 kg

Ans. (C)

Sol. Mg = 40

1000

49

1

50

M =8.910

1054940

= 100 kg.

14. A planet of mass m is moving around a star of mass M and radius R in a circular orbit of radius r. The star

abruptly shrinks to half its radius without any loss of mass. What change will be there in the orbit of the

planet ?

(A) The planet will escape from the star. (B) The radius of the orbit will increase.

(C) The radius of the orbit will decrease. (D) The radius of the orbit will not change.

Ans. (D)

Sol.

2r

GMm=

r

mv2

No change because respectively between them will be from com to com distance.

15. Figure (a) below shows a Wheatstone bridge in which P, Q, R, S are fixed resistances, G is a galvanometer

and B is a battery. For this particular case the galvanometer shows zero deflection. Now, only the positions

of B and G are interchanged,. as shown in figure (b). The new deflection of the galvanometer.

(A) is to the left. (B) is to the right.

(C) is zero. (D) depends on the values of P, Q, R, S

Ans. (C)


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Sol.

Since, ig

= 0

PR = QS

Still it will be a balanced W.S.B.

So, again ig

= 0.

16. 12 positive charges of magnitude q are placed on a circle of radius R in a manner that they are equally

spaced. A charge +Q is placed at the centre. If one of the charges q is removed, then the force on Q is :

(A) zero

(B) 20R4

qQ

away from the position of the removed charge.

(C) 20R4

qQ11

away from the position of the removed charge.

(D) 20R4

qQ

towards the position of the removed charge.

Ans. (D)

Sol. 2R

KQq=

04

1

. 2R

Qq

Since initially net force on Q was zero by symmetry

So, 0FF 11mainingRe1

111mainingRe FF

So, 20 r

Qq.

4

1

towards the position of the removed charge.


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17. An electric heater consists of a nichrome coil and runs under 220 V, consuming 1 kW power. Part of its coil

burned out and it was reconnected after cutting off the burnt portion. The power it will consume now is :

(A) more than 1 kW. (B) less that 1 kW, but not zero.

(C) 1 kW. (D) 0 kW.

Ans. (A)

Sol. Pi =i

2

iR

V

Since, Rf< R

ikeeping V = constant

V = Vf

Pf=

f

2

R

V

Since, R P .

18. White light is split into a spectrum by a prism and it is seen on a screen. If we put another identical inverted

prism behind it in contact, what will be seen on the screen ?(A) Violet will appear where red was

(B) The spectrum will remain the same

(C) There will be no spectrum, but only the original light with no deviation.

(D) There will be no spectrum, but the original light will be laterally displaced.

Ans. (D)

Sol.

the combination will behave as parallel slab so light get laterally displaced without any spectrum.

19. Two identical blocks of metal are at 20C and 80, respectively. The specific heat of the material of the two

blocks increases with temperature. Which of the following is true about the final temperature. Which of the

following is true about the final temperature Twhen the two blocks are brought into contact (assuming that

no heat is lost to the surroundings) ?

(A) Twill be 50C.

(B) Twill be more than 50C.

(C) Twill be less than 50C.

(D) T can be either more than or less than 50C depending on the precise variation of the specific heat with

temperature.

Ans. (B)

Sol. 20C 80C

S as T

d = m.s.d d = ms d

Since, average S of 80 is higher as 20C so average for so will be less as compared to 20.

So, Tf> 50C.


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20. A new temperature scale uses X as a unit of temperature, where the numerical value of the temperature tXin

this scale is related to the absolute temperature T by tX

= 3T + 300. If the specific heat of a material using this

unit is 1400 J kg1 X1 its specific heat in the S.I. system of units is :

(A) 4200 J kg1 K1 (B) 1400 J kg1 K1

(C) 466.7 J kg1 K1 (D) impossible to determine from the information provided

Ans. (A)

Sol. tx= 37 + 300 = ms

S =

.mS =

.m

Since, unit of is joule in both system

X T

m = m0kg m

0kg

= 0J

0.J

tx

T

Sx=

x0

0

tm = 1400 S

T= tm0

0

=

3

x0

0

tm

ST

= 3 1400 = 4200 J-kg1K1.

CHEMISTRY

21. The boiling points of 0.01 M aqueous solutions of sucrose NaCl and CaCl2would be :

(A) the same (B) highest for sucrose solution

(C) highest for NaCl solution (D) highest for CaCl2 solution

Ans. (D)

Sol. Aqueous solution containing more number of particles have more elevation in boiling point.

22. The correct electronic configuration for the ground state of silicon (atomic number 14) is :

(A) 1s2 2s2 2p6 3s2 3p2 (B) 1s2 2s2 2p6 3p4 (C) 1s2 2s2 2p4 3s2 3p4 (D) 1s2 2s2 2p6 3s1 3p3

Ans. (A)

Sol.14

Si : 1s2 2s2 2p6 3s2 3p2


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23. The molar mass of CaCO3is 100 g. The maximum amount of carbon dioxide that can be liberated on heating

25 g of CaCO3is :

(A) 11 g (B) 5.5 g (C) 22 g (D) 2.2 g

Ans. (A)

Sol. CaCO3(s)

CaO (s) + CO

2(g)

Number of mole

100

25

100

25

Amount of CO2

=

100

25 44 = 11 gram.

24. The atomic radii of the elements across the second period of the periodic table.

(A) decrease due to increase in atomic number (B) decrease due to increase in effective nuclear charge

(C) decrease due to increase in atomic weights (D) increase due to increase in the effective nuclear charge

Ans. (B)

Sol. As we move left to right in 2nd period atomic radii decreases due to increase in effective nuclear charge.

25. Among NH3, BCl

3, Cl

2and N

2, the compound that does NOT satisfy the octet rule is :

(A) NH3

(B) BCl3

(C) Cl2

(D) N2

Ans. (B)

Sol. In BCl3

octet rule is not satisfy.

Total number of 6 electrons in outermost shell of B after bonding.

26. The gas produce on heating MnO2with conc. HCl is

(A) Cl2

(B) H2

(C) O2

(D) O3

Ans. (A)

Sol. MnO2

+ 4HCl MnCl2

+ 2H2O + Cl

2

Cl2gas produces.


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27. The number of covalent bonds in C4H

7Br, is :

(A) 12 (B) 10 (C) 13 (D) 11

Ans. (A)

Sol. C4H

7Br

CH2

= CH CH2 CH

2 Br

HH||

BrCCCCH||||HHHH

Number of covalent bond = 12.

28. An aqueous solution of HCl has a pH of 2.0. When water is added to increase the pH to 5.0, the hydrogen ion

concentration :

(A) remains the same (B) decreases three-fold

(C) increases three-fold (D) decreases thousand-fold

Ans. (D)

Sol. pH = 2 [H+]i= 102 M

pH = 5 [H+]f= 105 M

i

f

]H[

]H[

=2

5

10

10

=

1000

1

So, H+ concentration decrease thousand fold.

29. Consider two sealed jars of equal volume. One contains 2 g of hydrogen at 200 K and the other contains 28

g of nitrogen at 400 K. The gases in the two jars will have :

(A) the same pressure. (B) the same average kinetic energy.

(C) the same number of molecules. (D) the same average molecular speed.

Ans. (C)

Sol. For 1st jar :

Number of moles of H2(g) =

2

2=1 mole.

Number of molecules of H2(g) = 6.02 1023.

For 2nd jar :

Number of moles of N2(g) =

28

28=1 mole.

Number of molecules of N2(g) = 6.02 1023.

So, both jar have same number of molecules.


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30. Identify the stereoisomeric pair from the following choices.

(A) CH3CH

2CH

2OH and CH

3CH

2OCH

3(B) CH

3CH

2CH

2Cl and CH

3CHClCH

3

(C) and (D) and

Ans. (C)

Sol. and

(cis) (trans)

cis and trans are stereoisomeric pair.

BIOLOGY

31. Which of the following is a water-borne disease ?

(A) Tuberculosis (B) Malaria (C) Chickenpox (D) Cholera

Ans. (D)

32. In has seminal work on genetics, Gregor Mendel described the physical traits in the pea plant as being

controlled by two 'factors'. What term is used to define these factors today ?

(A) Chromosomes (B) Genes (C) Alleles (D) Hybrids

Ans. (C)

33. A majority of the tree species of peninsular Indian origin fruit in the months of :

(A) April - May (B) August - September

(C) December - January (D) All months of the year

Ans. (A)

34. In frogs, body proportions do not change with their growth. A frog that is twice as long as another will be

heavier by approximately.(A) Two-fold (B) Four-foldSix (C) -fold (D) Eight-fold

Ans. (A)

35. Which of the following has the widest angle of binocular vision ?

(A) Rat (B) Duck (C) Eagle (D) Owl

Ans. (D)


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36. The two alleles of a locus which an offspring receives from the male and female gametes are situated on :

(A) Two different homologs of the same chromosome.

(B) Two different chromosomes.

(C) Sex chromosomes.

(D) A single chromosome.

Ans. (B)

37. Ants locate sucrose by :

(A) Using a strong sense of smell.

(B) Using a keen sense of vision.

(C) Physical contact with sucrose.

(D) Sensing the particular wavelength of light emitted / reflected by sucrose.

Ans. (A)

38. The interior of a cow-dung pile kept for a few days is quite warm. This is mostly because :

(A) Cellulose present in the dung is a good insulator.

(B) Bacterial metabolism inside the dung releases heat.

(C) Undigested material releases heat due to oxidation by air.

(D) Dung is dark and absorbs a lot of heat.

Ans. (B)

39. Which one of these is the correct path for a reflex action ?

(A) Receptor-Motor Neuron-Spinal Cord-Sensory Neuron-Effector.

(B) Effector-Sensory Neuron-Spinal Cord-Motor Neuron-Receptor.

(C) Receptor-Sensory Neuron-Spinal Cord-Motor Neuron-Effector.

(D) Sensory Neuron-Receptor-Motor Neuron-Spinal Cord-Effector.

Ans. (C)

40. Insectivorous plants digest insects to get an essential nutrient. Other plants generally get this nutrient from

the soil. What is this nutrient ?

(A) Oxygen (B) Nitrogen (C) Carbon dioxide (D) Phosphates

Ans. (B)


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3. (i) If there is natural number n relative prime with 10. Then show that there exist another natural number

m such that all digits are 1's and m is div. by 'n'.

(ii) Show that every natural number can represented as)110(10

acb

, where a, b, c N

Sol. from the question if m = 1111 or

(i) m = 111.11is always divisible by n = 11 which is coprime with 10

(ii) by choosing a = k 10b (10c 1) when k is any natural number we can option any natural number k.

The problem seen to have an error which may be due to memory retersion constraints.

PHYSICS

DISCLAIMER

The subjective problems are based memory rotations of our students hence they may differ slightly then

the original questions. The solutions presented here are in accordance with the given problemsstatement.

4. There is a smooth fixed concave surface. A particle is released from p. Find :

(i) PE as function of Q

(ii) KE as a function Q

(iii) time taken from P to Q

(iv) the reaction force at Q

Sol.

Assume to be spherical concave.

cos0

=R

)hR(

(i) P.E. = mg(R R cos) = mgR(1 cos)

(ii) = mgh mgR (1 cos) = kinetic energy


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(iii) m(g sin)R = (mR2) 2

2

dt

d

tPQ

=

4

161

g2

4

T

20

=

161

g2

20

(iv)

N mg =R

mV2

N = mg +R

mV2where V = gh2 .

5.

Given: RA


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Sol.

(i)PR

S R =

V

V =

A

V

V RRR

RR'

V

VAAVR

RR

RRRRRRRS

P

RP

=V =

V

V

RRRR + RA

(ii)PR

= AV

V RRR

R.RR

Since, RA

< < R < < RV

Rest

= R~RRR

RRA

V

V

PR

= 0

(iii) RQ

V =VA

VA

RRR

R)RR(

Restimated

=VA

VA

RRR

R)RR(

RQ

= R VA

VA

RRR

R)RR(

=

VA

VAVVA2

RRR

RRRRRRRRR

=VA

VAA2

RRR

RRRRR

= AV

A

V

2

RR

RR

R

R

RQ

~ RA .

(iii)VV

VAA2

RRR

RRRRR

= 0

After solving, R = VARR .


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6. An object is placed. 20 cm from a concave lens of focal length 10 cm :

(i) find the position of the image

(ii) find the position of another concave mirror of focal length 5 cm when is to be placed. right side of above

concave lens such that final image coincides with the object.

(iii) If a plane mirror is placed in place of concave mirror at the same position, then find the position of final

image.

Sol.

(i)10

1

20

1

V

1

20

1

10

1

V

1 =

20

12

V =3

20

(ii) fLM = 5 cm

3

20+ d = 10

d = 10 3

20=

3

10cm

(iii)

(A) Ist reference with lens

V =302

(B) Then mirror,


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(C) Again by lens,

Xim

=X0 M

40

3

V

1

=

10

1

40

3

10

1

V

1 =

40

34

V =7

40cm

It means right of lens at a distance7

40cm.

CHEMISTRY

DISCLAIMER

The subjective problems are based memory rotations of our students hence they may differ slightly thenthe original questions. The solutions presented here are in accordance with the given problems

statement.

7. There are four bottles 1,2,3,4 containing following compounds.

Bottle-1 : ; Bottle-2 :

Bottle-3 : ; Bottle-4 :

Identify the bottle :

(I) (A) whose content does not react with 1N HCl or 1N NaOH.

(B) whose content reacts with 1N NaOH only.

(C) whose content reacts with 1N NaOH and 1N HCl both.

(D) whose content reacts with 1N HCl only.

(II) The bottle whose content is highly soluble in distilled water.

Sol. (I) (A) Bottle-3 does not react with HCl or NaOH.

(B) Bottle-2 reacts only with NaOH.

(C) Bottle-4 reacts both NaOH or HCl both.

(D) Bottle-1 reacts with HCl only.

(II) Bottle-4 is highly soluble in distilled water due to zwitter ion formation.


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8. A copper ore was treated with HNO3to form Cu(NO

3)

2, Which on further reaction with K, forms Cu

2, which

decomposes to form Cu2

2and

2,

2was titrated with Na

2S

2O

3.

(I) Balance the reactions involved (find out values of a to )

(a) a Cu + b HNO3 c Cu(NO

3)

2+ d NO + e H

2O

(b) f Cu2 g Cu

2

2+ h

2

(c) i Na2S

2O

3+ j

2 k Na

2S

4O

6+ Na

(II) 2.54 g of 2 was evolved : Find the percentage purity of copper in 2 g of ore.

Sol. (i) Balanced equation are :

(a) 3 Cu + 8 HNO3 3 Cu(NO

3)

2+ 4 NO + 3 H

2O

(b) 2 Cu2Cu

2

2+

2

(c) 2 Na2S

2O

3+

2 Na

2S

4O

6+ 2 Na

(ii) Mole of 2

=254

54.2= 0.01

Mole of Cu2= 2 mole of

2= 0.02

Mole of Cu2= mole of Cu = 0.02

wt. = 0.02 6.25 = 1.27 g

% purity =2

27.1 100 = 63.5%

9. Human being 2500 K cal of energy per day

C12

H22

O11

+ 12 O2 12 CO

2+ 11 H

2O

H = 5.6 106 J / mol

(i) Human being required ..................... kJ of energy per day.

(ii) Amount of sucrose required per day and volume of CO2evolved during the process.

Sol. (i)18.4

2500kJ = 598 kJ

(ii) Mole of sucrose required = 36 10106.5

598

= 0.106

wt. of sucrose required = 0.101 342 = 36.52 gm

Kvpy 2010 Class-XI solutions by Resonance

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Transcript of Kvpy 2010 Class-XI solutions by Resonance