Kvpy 2010 Class-XI solutions by Resonance

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. (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

1 y x 3  – y 1 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 y – x y x  y y 3  3 Ans. (D)

3. The number of positive integers n in the set {1, 2, 3, ...., 100} for which the number n ... 3 2 1 n ... 3 2 12 2 2 2         is an integer is : (A) 33 (B) 34 (C) 50 (D) 100 Ans. (B) Sol. ) 1 n ( n 3 ) 1 n 2 )( 1 n ( n    2  3 1 n 2  = k  n = 2 1 – k 3 31  2 1 – k 3  100  3  3k 201  1 k  3 1 20 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) 49 2 (C) 60 (D) 60 2 Ans. (A) Sol. 2 + b2 = 392 b2 _{+ h}2 _{= 40}2 h2 ₊ 2 = 412 adding 2(2 + b2 + h2) = 392 + 402 + 412 As = 2 41 40 392 2 2 = 2 ) 1 40 ( 40 ) 1 – 40 ( 2 2  2 = 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.

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 5 a2 (C) 6a2 (D) 3a2 Ans. ( ....) Sol. Area = _              4 a 3 6 – 2 a 5 2 2 = 4 a 3 6 – 4 a 25 2 2 = 4 a ) 3 6 – 25 ( 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 ?

Sol. dt dv = c. v = r h 3 1 2  tan  = h r = 3  h3 _tan  dt dv = h2 tan  dt dh  ct + k = 3 tan h3   when t = 0, h = H  k = 3 tan H3    ct = 3  (h3 – H3) tan  when t = 21 , h = 2 H c = ₃ 6  H3 _      8 7 – h = 0   tan ) H (– 3 3 = H –₈7 t 6 3 3               24 7 = 72 8 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 1000 100 99  = 990 water evaporated is x. so _x 100 – 1000 x – 990       = 98

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 x 2 y   also       x 5 2 a = 12  a = x 30 and       x 30 (b) = 15  b = 2 x and       2 x (c) = 25  c = x 50  Area = x       x 50 = 50

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 = 2 _g

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, ƒ, 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 = ma

F – fs = 0

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 . 9 10 10 5 49 40     = 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. 2 r 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

Sol.

Since, i_g = 0 PR = QS

Still it will be a balanced W.S.B. So, again i_g = 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) 2 0R 4

qQ

 away from the position of the removed charge.

(C) 2 0R 4

qQ 11

 away from the position of the removed charge.

(D) 2 0R 4

qQ

 towards the position of the removed charge.

Ans. (D) Sol. ₂ R KQq = 0 4 1  .R2 Qq

Since initially net force on Q was zero by symmetry So, F₁F_Remaining110   1 11 maining Re F F  So, 2 0 r Qq . 4 1

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. P_i = i 2 i R V

Since, R_f < R_i keeping V = constant V = V_f P_f = 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 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 T

ƒ when the two blocks are brought into contact (assuming that

no heat is lost to the surroundings) ? (A) T

ƒ will be 50ºC.

(B) T

ƒ will be more than 50ºC.

(C) T

ƒ will be less than 50ºC.

(D) T

ƒ can be either more than or less than 50ºC depending on the precise variation of the specific heat with

temperature.

Ans. (B)

Sol. 20ºC 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.

20. A new temperature scale uses X as a unit of temperature, where the numerical value of the temperature t_X in this scale is related to the absolute temperature T by t_X = 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. (A) Sol. t_x = 37 + 300  = ms  S =    . m S =   . m Since, unit of  is joule in both system

X T m = m₀kg m₀kg  = 0J 0.J tx T S_x = x 0 0 t m   = 1400 S_T = t m₀ 0   = 3 x 0 0 t m          S_T = 3 × 1400 = 4200 J-kg–1_K–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) 1s}2 _2s2 _2p6 _3p4 _{(C) 1s}2 _2s2 _2p4 _3s2 _3p4 _{(D) 1s}2 _2s2 _2p6 _3s1 _3p3 Ans. (A)

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       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, BCl₃, Cl₂ and N₂, 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 MnO₂ with conc. HCl is

(A) Cl₂ (B) H₂ (C) O₂ (D) O₃

Ans. (A)

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

27. The number of covalent bonds in C₄H₇Br, is : (A) 12 (B) 10 (C) 13 (D) 11 Ans. (A) Sol. C₄H₇Br CH₂ = CH – CH2– CH2– Br H H | | Br – C – C – C C – H | | | | H H 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. (D) Sol. pH = 2  [H+]_i = 10–2 _M pH = 5  [H+]_f = 10–5 _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 :}

30. Identify the stereoisomeric pair from the following choices.

(A) CH₃CH₂CH₂OH and CH₃CH₂OCH₃ (B) CH₃CH₂CH₂Cl and CH₃CHClCH₃

(C) and (D) and

Ans. (C)

Sol. and

(cis) (trans) cis and trans are stereoisomeric pair.

BIOLOGY

(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

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

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

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

) 1 10 ( 10 a c b  , where a, b, c  N

Sol. 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.

(iii) m(g sin)R = (mR2) 2 2 dt d  t_PQ = 4 16 1 g 2 4 T 2 0         _     = _       _   16 1 g 2 2 0  (iv) N – mg = R mV2 N = mg + R mV2 where V = 2gh. 5. Given: R_A << R << R_V Restimated =  V S_R = |R – Rest| (i) Calculate SR_P (ii) Calculate SR_Q

Sol. (i) SR_P R =  V V =            A V V _R R R RR ' V V A A V R R R R R RR RR R S P      R_P =  V = V V R R RR  + RA (ii) RP = A V V _R R R R . R R _         Since, R_A < < R < < R_V R_est = _{R R} R ~– R RR A V V   P R  = 0 (iii) RQ V = V A V A R R R R ) R R (     R_estimated = V A V A R R R R ) R R (    RQ = R – V A V A R R R R ) R R (    = V A V A V V A 2 R R R R R RR RR RR R       = V A V A A 2 R R R R R RR R    

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 1 2  V = 3 20  (ii) f_LM = –5 cm 3 20 + d = 10 d = 10 – 3 20 = 3 10 cm (iii)

(A) Ist _{reference with lens}

V = 3 0 2  (B) Then mirror,

(C) Again by lens, X_im = –X0 M 40 3 V 1   = 10 1  40 3 10 1 V 1    = 40 3 4  V = 7 40  cm

It means right of lens at a distance 7 40

cm.

CHEMISTRY

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.

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.

8. A copper ore was treated with HNO₃ to form Cu(NO₃)₂, Which on further reaction with K, forms Cu2, which

decomposes to form Cu₂2 and 2, 2 was titrated with Na2S2O3.

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

(a) a Cu + b HNO₃ c Cu(NO3)2 + d NO + e H2O

(b) f Cu2 g Cu22 + h 2

(c) i Na₂S₂O₃ + j 2 k Na2S4O6 +  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(NO3)2 + 4 NO + 3 H2O

(b) 2 Cu₂ Cu₂₂ + ₂ (c) 2 Na₂S₂O₃ + 2 Na2S4O6 + 2 Na (ii) Mole of 2 = ₂₅₄ 54 . 2 = 0.01 Mole of Cu2 = 2 × mole of 2 = 0.02 Mole of Cu2 = 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 C₁₂H₂₂O₁₁ + 12 O₂ 12 CO₂ + 11 H₂O 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 CO₂ evolved during the process.

Sol. (i) 18 . 4 2500 kJ = 598 kJ

(ii) Mole of sucrose required = _{6 3} 10 10 6 . 5 598    = 0.106 wt. of sucrose required = 0.101 × 342 = 36.52 gm