Kvpy 2010 Class-XI solutions by Resonance

(TEST PAPER HELD ON 31-10-2010)

QUESTIONS & SOLUTIONS OF KISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) 2010 TEST 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.

Sol.

(C)

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

2.

b=–1 a=–3 _________ a+b=–4

If x, y are real numbers such that

3

x 1 y

–

3

x 1 y

= 24,

then the value of (x + y) / (x – y) is : (A) 0 (B) 1 Ans. Sol.



(D) 3

(D)

3x/y = t 3t –

xy x–y

(C) 2



t = 24 3

3y y

RESONANCE



8t = 3 × 24



t=a x = 2y



3

Ans. (D)

Page # 1

3.

The number of positive integers n in the set {1, 2, 3, ...., 100} for which the number is an integer is : (A) 33 Ans.

Sol.

(B) 34

(C) 50

(D) 100

(B)

n(n  1)(2n  1) 2 3n(n  1) 31 

3k – 1  100 2



2n  1 =k 3



n=

3k – 1 2



3  3k 201



1 k 

20 1 3

1  k = 67 k is odd 34

4.

Ans.

(B) 49 2

(D) 60 2

(A)

=

392  402  412 = 2

=

3( 40)2  2 = 2

= 49

( 40 – 1)2  402  ( 40  1)2 2

4802 = 2

2401

Ans. (A)

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.

Sol.

(C) 60

2 + b2 = 392 b2 + h2 = 402 h2 + 2 = 412 adding 2(2 + b2 + h2) = 392 + 402 + 412 As

5.

(B)

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

Sol.

12  2 2  3 2  .....  n 2 1  2  3  ......  n

(B)

It has to be isosceles triangle =

1 1 1 2 4x 2 – 1 ×1 x – = 4 2 4

perimeter = 1 + 2x  odd

RESONANCE

always irrational

(B) Page # 2

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 Ans.

Sol.

=

(B) 36

(C)

rs 2

(D)

rs 4

(B)

1 × 12 × 6 2

= 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 Ans.

Sol.

(B)

5 a2 2

(C) 6a2

(D) 3a2

( ....)

2  3a 2   5a     Area =  2  – 6 4     

=

25a 2 6 3 a 2 – 4 4

= (25  – 6 3 )

a2 4

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)

RESONANCE

Page # 3

Sol.

dv = c. dt v=

=

1 2 r h 3

tan  =

r h

 3 h tan  3

dv dh = h2 tan  dt dt

when

t = 0, h = H

when

t = 21 , h =



ct + k =

k=



H 2

c=

h3 tan  3

 6

3

H3 tan  3



ct =

 3 (h – H3) tan  3

–7  H3   8 

h=0

  3 – 7     t (–H3 ) tan  =  3 H   8  3 6 7 7 = t 24 8  72 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 watermelon was: (A) 10 kg (B) 50 kg (C) 100 kg (D) 500 kg Ans.

Sol.

(D)

Water + solid = 1000 Water is

99  1000 = 990 100

water evaporated is x. so

 990 – x   100 = 98  1000 – x 

99000 – 100 x = 96000 – 98x 1000 = 2x x = 500

RESONANCE

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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 Ans.

(B) 30

(C) 40

(D) 50

(D)

Sol.

xy = 10 and yz = 4



 2x z= y  5

2  also  x  a = 12 5 



a=

30 x

and

 30    (b) = 15   x 

b=

x 2

and

x   (c) = 25 2



c=

50 x



 50   = 50 Area = x   x 

RESONANCE

Page # 5

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 pendulum 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.

Sol.

(A)

 T = 2 g

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

 12.

T.

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, ƒ, exerted by the table on the block with time t is given by :

(A)

Ans.

(B)

(C)

(D)

(A)

Sol.

when sliding has started till acceleration of block is zero F – fs = 0 fs = F

RESONANCE

F – fk = ma

Page # 6

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.

Sol.

 49  50  × Mg = 40 ×   1000  1

M=

14.

(C)

40  49  5  10 = 100 kg. 10  9.8

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.

GMm r

2

=

mv 2 r

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. (C) is zero. Ans.

(B) is to the right. (D) depends on the values of P, Q, R, S

(C)

RESONANCE

Page # 7

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 qQ

(B) 4 R2 away from the position of the removed charge. 0 11qQ

(C) 4 R2 away from the position of the removed charge. 0 qQ

(D) 4 R2 towards the position of the removed charge. 0 Ans.

Sol.

KQq R

2

(D)

1 Qq = 4 . 2 0 R

Since initially net force on Q was zero by symmetry   F1  FRe maining 11  0 So,

  FRe maining 11  F1 1 Qq So, 4 . 2 towards the position of the removed charge. 0 r

RESONANCE

Page # 8

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.

Sol.

(A) Pi =

Vi2 Ri

Since, Rf < Ri keeping V = constant V = Vf Pf =

V2 Rf

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.

Sol.

Two identical blocks of metal are at 20ºC and 80º, 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 Tƒ when the two blocks are brought into contact (assuming that no heat is lost to the surroundings) ? (A) Tƒ will be 50ºC. (B) Tƒ will be more than 50ºC. (C) Tƒ will be less than 50ºC. (D) Tƒ can be either more than or less than 50ºC depending on the precise variation of the specific heat with temperature. Ans. 20ºC

(B) 80ºC

S  as T  d = m.s.d d = ms d Since, average S of 80º is higher as 20ºC so average  for so will be less as compared to 20º. So, Tf > 50ºC.

RESONANCE

Page # 9

20.

A new temperature scale uses X as a unit of temperature, where the numerical value of the temperature tX in 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 kg–1 X–1 its specific heat in the S.I. system of units is : (A) 4200 J kg–1 K–1 (B) 1400 J kg–1 K–1 (C) 466.7 J kg–1 K–1 (D) impossible to determine from the information provided Ans.

Sol.

(A)

tx = 37 + 300 S=

 = ms 

 m.

S=

 m.

Since, unit of  is joule in both system X T m = m0kg m0kg  = 0J 0.J tx T 3

0 Sx = m t = 1400 0 x

0  0 ST = m t =  0  m 0 t x

   

ST = 3 × 1400 = 4200 J-kg–1K–1.

CHEMISTRY 21.

The boiling points of 0.01 M aqueous solutions of sucrose NaCl and CaCl2 would 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.

Sol.

14

(A)

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

RESONANCE

Page # 10

23.

The molar mass of CaCO3 is 100 g. The maximum amount of carbon dioxide that can be liberated on heating 25 g of CaCO3 is : (A) 11 g (B) 5.5 g (C) 22 g (D) 2.2 g Ans.

(A) 

Sol.

CaCO3 (s)  CaO (s) + CO2 (g) Number of mole

 25     100 

 25     100 

 25   × 44 = 11 gram. Amount of CO2 =   100 

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 Ans.

(D) increase due to increase in the effective nuclear charge

(B)

Sol.

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

25.

Among NH3, BCl3, Cl2 and N2, the compound that does NOT satisfy the octet rule is : (A) NH3 (B) BCl3 (C) Cl2 (D) N2 Ans.

Sol.

(B)

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 MnO2 with conc. HCl is (A) Cl2 (B) H2 (C) O2 Ans.

Sol.

(D) O3

(A)

MnO2 + 4HCl  MnCl2 + 2H2O + Cl2 Cl2 gas produces.

RESONANCE

Page # 11

27.

The number of covalent bonds in C4H7Br, is : (A) 12 (B) 10 Ans.

Sol.

(C) 13

(D) 11

(A)

C4H7Br CH2 = CH – CH2 – CH2 – Br H H H H | | | | H – C  C – C – C – Br | | H H

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.

Sol.

(D)

pH = 2 pH = 5

[H ] f 

[H ]i

 

=

10 5 10

2

[H+]i = 10–2 M [H+]f = 10–5 M

 1   =   1000 

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.

Sol.

(C)

For 1st jar : Number of moles of H2 (g) =

2 =1 mole. 2

Number of molecules of H2 (g) = 6.02 × 1023. For 2nd jar : Number of moles of N2 (g) =

28 =1 mole. 28

Number of molecules of N2 (g) = 6.02 × 1023. So, both jar have same number of molecules.

RESONANCE

Page # 12

30.

Identify the stereoisomeric pair from the following choices. (A) CH3CH2CH2OH and CH3CH2OCH3 (B) CH3CH2CH2Cl and CH3CHClCH3

(C)

Ans.

and

(D)

and

(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 Ans.

32.

(A)

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.

35.

(C)

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.

34.

(D)

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.

33.

(D) Cholera

(A)

Which of the following has the widest angle of binocular vision ? (A) Rat (B) Duck (C) Eagle Ans.

(D) Owl

(D)

RESONANCE

Page # 13

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.

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.

38.

(B)

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.

40.

(A)

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.

39.

(B)

(C)

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)

RESONANCE

Page # 14

SUBJECTIVE QUESTIONS 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 problems statement.

MATHEMATICS 1.

Leela and madan collectea their CD's and sold them such that each C.D. was sold at the same amount as number of C.D's. From this amount colluted leela juist borrowed Rs. 10, then madan borrowed Rs. 10, they borrowed money alternately until amount less than Rs. 10 was left for Madan to borrow. Find has much was left for Madan to borrowed at the end.

Sol.

Let total amount is n2 total borrowed amount = (2u + 1) 10 n2 – (2u + 1) 10 < 10 True for. n = 6 u = 1 so the left amount = 6

2.

In ABC, DE is drawn parallel to BC, D on A and B on E. Such that ADE has area 3. Now BE and DC are joined to get intersection point P. DPE = 1 sq. unit. Find area of ABC ?

Sol.

Let  ADE is equilateral and D is mid point of AB and E is mid point of AC [given condition is true for above assumption]  Area of quad. ADPE = Area of quad. DPFB = Area of quad. EPFC  Area of  ABC = 12 sq. units

RESONANCE

Page # 15

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

a

Sol.

b

10 (10 c  1)

, where a, b, c  N

from the question if m = 1111 or (i) m = 111.11 is 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 problems statement. 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.

cos0 =

(R  h) R

(i) P.E. = mg(R – R cos) = mgR(1 – cos) (ii) = mgh – mgR (1 – cos) = kinetic energy

RESONANCE

Page # 16

d2 

(iii) m(g sin)R = (mR2)

dt 2 2

tPQ = T  2 4

   1 0 g  16 4

   

2  0     1  = 2 g 16  

(iv)

N – mg =

mV 2 R

N = mg +

mV 2 R

where V =

2gh .

5.

Given: RA