ts eamcet 2015 engineering question paper key solutions

8/18/2019 Ts Eamcet 2015 Engineering Question Paper Key Solutions

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01. If ( ) f x is a real function defined on [ ]1,1− , then the real function ( ) ( )5 4g x f x= + is defined

on the interval1) [ ]4,9− 2) [ ]1,9− 3) [ ]2,9− 4) [ ]3,9−

Key: 0/2Sol: 1 1 x− ≤ ≤

1 5 4 9 x⇒ − ≤ + ≤ 02. If : f N R→ is defined by ( )1 1 f = − and ( ) ( )1 3 2 f n f n+ = + for 1n > , then f is

1) one – one 2) onto 3) a constant function 4) ( ) 0 f n > for 1n > Key: 3Sol: ( ) ( ) ( )1 1& 1 3 2 f f n f n= − + = +

( ) ( ) ( )1 2 3 ..... f f f ⇒ = = = ⇒ Constant function.

03. The remainder of4 3 2

2 2 26n n n n− − + − when divided by 24 is1) 20 2) 21 3) 22 4) 23Key: 3Sol: 4 3 22 2 26n n n n− − + −

( )( ) ( )( )2 1 1 26n n n n= − − + − ( )( )( ) ( )2 1 1 48 22n n n n= − − + − +

( )24 22l= + ∴ Remainder = 22

04. ( )2 2 2

1 2 3

1 2 1 3 11 2 1 3 1

A x x x x x x x

= + + ++ + +

( )1

0 A x dx⇒ =∫

1) 0 2) 1 3) 2 4) 4Key: 1

Sol: ( )2

1 0 0

1 1 2

1 1 2

A x x

x

= + − −+ − −

2 2 12c c c→ − & 3 3 13c c c→ −

= 0 ( )1

0

0 A x dx∴ =∫

05. Let

2

2 4 3 2

2

1 1 2 3

3 1 2 1

5 1 2 3 4

x x x x

x x x ax bx cx dx c

x x x x

+ + + −− + − = + + + +

+ + + + be an identity in x. If a, b, c, d are

known, then the value of ‘e’ is1) 29 2) 24 3) 16 4) 9

Key: 1

Sol: ( )1 1 3

0 1 2 1 29

1 3 4

e A

−= = − − =


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06. The system of equations 4 2 5, 5 3 10,9 3 7 20 x y z x y z x y z+ + = − + = − + = has1) no solution 2) unique solution3) two solutions 4) infinite number of solutions

Key: 4Sol: By eliminating Y from equations,

We get 21 13 35 x z+ = 21 13 35 x z+ = ⇒ The given system has infinite number of solutions.

07. If 21, ,w w are the cube roots of unity and if 22 3α ω ω= + − then 3 212 48 3α α α+ + + = 1) – 63 2) – 62 3) – 61 4) – 60

Key: 4Sol: 24α ω= − +

24α ω+ = ( ) ( )33 24α ω+ =

3 212 48 3 60α α α⇒ + + + = − 08. If ,α β are the roots of 21 0 x x+ + = then the value of 4 4 4 4α β α β− −+ + =

1) 0 2) 1 3) – 1 4) 2Key: 1Sol: 2,α ω β ω= =

4 44 4

1.

α βα β

∴ + +

2 101ω ω= + + =

09. If ,α β are roots of the equation 2 4 8 0 x x− + = then for any 2 2, n nn N α β∈ + =

1) 2 12 cos2

n nπ+ 2) 32 cos2

n nπ 3) 3 12 cos2

n nπ+ 4) 32 cos4

n nπ

Key: 3Sol: 2 4 8 0 x x− + =

( )22 4 2 2 x x i− = − ⇒ = ± ( ) ( )2 22 2 2 22 1 2 1n nn n n ni iα β∴ + = + + −

3 12 .cos2

n nπ+= 10. If ,α β are the non-real cube roots of 2 then 6 6α β+ =

1) 8 2) 4 3) 2 4) 1Key: 1Sol: 1/3 1/3 22 , 2ω ω are non – real roots.

6 6 4 4 8α β∴ + = + = 11. Let α β≠ satisfy 2 21 6 , 1 6α α β β+ = + = . Then, the quadratic equation whose roots are

,

1 1

α β

α β+ + is

1) 28 8 1 0 x x+ + = 2) 28 8 1 0 x x− − = 3) 28 8 1 0 x x− + = 4) 28 8 1 0 x x+ − = Key: 2/3


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Sol: ,α β are roots of 2 6 1 0 x x− + = then 6, . 1α β α β+ = =

Required equation is( )( )

2 01 1 1 1

x xα β αβ

α β α β − + + = + + + +

2

8 8 1 0 x x⇒ − + = 12. The set of solutions of 2 5 4 0 x x− + < is

1) ( )4, 1− − 2) ( )1, 4 3) ( ) ( )1, 4 1, 4− − ∪ 4) ( )4, 4− Key: 3Sol: 2 5 4 0t x t t = ⇒ − + <

1 4t ⇒ < < 1 4 x⇒ < <

( ) ( )4, 1 1,4 x⇒ ∈ − − ∪

13. Let , ,α β γ be the roots of3

10 0 x x+ + = write 1 1 12 2 2, ,α β β γ γ α

α β γ γ α β+ + +

= = = . Then the value

of ( ) ( )3 3 3 2 2 21 1 1 1 1 1110α β γ α β γ + + − + + is

1)1

10 2)

15

3)3

10 4)

12

Key: 3

Sol: 1 1 11 1 1

, ,α β γ γ α β− − −= = =

Equation whose roots are 1 1 1, ,α β γ is3 210 1 0 x x− − =

3 21 1 010 10

x x⇒ − − = Synthetic division

3 3 31 1 1

3011000

α β γ + + =

2 2 21 1 1

1100

α β γ + + =

i.e., Required Expression =301 1 1 3

1000 10 100 10 − =

14. Suppose , ,α β γ are the roots of 3 2 2 0 x x x+ + + = . Then the value of2 2 2α β γ β γ α γ α β

γ α β + − + − + −

is

1)472

− 2) 472

3) 47− 4) 47Key: 1Sol: , ,α β γ are roots of 3 2 2 0 x x x+ + + =

Let2 3 1

3 y α β γ α β γ γ

γ γ γ + − + + − −= = = −

1 133

y y

γ γ − −+ = ⇒ = +


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∴ T.E is231 1 1

2 03 3 3 y y y

− − −+ + + = + + +

Constant term =47

2−

15. ( )10

405

0

r

r

C −

==∑

1) 41 305 5C C − 2) 41 306 6C C − 3) 41 305 5C C + 4) 41 6C Key: 2

Sol: ( )10

40 40 39 305 5 5 5

0

...r r

C C C C −

== + + +∑

41 306 6C C = −

16. The number of diagonals of a regular polygon is 35. Then the number of sides of polygon is

1) 12 2) 9 3) 10 4) 11Key: 3Sol: 2 35 10

n C n n− = ⇒ =

17.2 3

43 1 3 7 1 3 7 11 11 ...1! 6 2! 6 3! 6

x x× × × = + × + + + ⇒ =

1) 81 2) 54 3) 27 4) 8Key: 3

Sol: 3/4 3

42 11 273 3

x x− − = − ⇒ = =

18. If x is small so that2

x and higher powers of x may be neglected, then an approximate value

of( )

( )

31/5

4

21 1 15

3

2 3

x x

x

−− + −

− is

1) ( )1 1 78

x+ 2) ( )1 1 716

x− 3) 1 7 x− 4) ( )1 1 716

x+ Key: 4

Sol:

( ) ( )( )( ) ( )

3 15

44

21 1 15

1 2 1 3 131 716 1 6 163

2 12

x x x x

x x x

− − + − − + = = +− −

19. The coefficient of n x in the expansion of 215 6 x x− +

for 1 x < is

1) 1 11 1

2 3n n− −− 2) 2 2

1 12 3n n+ +

− 3) 1 11 1

2 3n n+ +− 4) 1 1

2 3n n−

Key: 3

Sol: 21 1 15 6 2 3 x x x x

−= +− + − −

( ) ( ) ( ) ( )1 11 1 1 11 / 2 1 / 3

2 1 / 2 3 1 / 3 2 3 x x

x x− −−= + = − − −− −


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Coefficient of 1 11 1

2 3n

n n x + += − 20. In a , ABC ∆ the value of A∠ is obtained from the equation 3cos 2 0 A+ = . The quadratic

equation, whose roots are sin A and tan A , is1) 23 5 5 0 x x+ − = 2) 26 5 5 0 x x− − = 3) 26 5 5 0 x x+ − = 4) 26 5 5 0 x x+ + =

Key: 3

Sol: 2 5

cos ,sin3 3

A A−= =

R. E is 25 5 5

03 2 6

x x − − − =

26 5 5 0 x x⇒ + − = 21. If 2 4sin cos A θ θ= + , then for all values of , Aθ lies in the interval

1) [ ]1, 2 2) 3 , 14 3) 1 3,

2 4 4) 3 19,

2 16

KEY – 2HINT - 2 4cos A Sin θ θ= +

2cos 1θ ≤ 4 2cos cosθ θ≤ 1 A ≤ ----------- (1)

4 2cos 1 cos A θ θ= + − 4 2 1 1 1cos 2 os 1

2 4 4cθ θ= − + − +

=2

2 1 3 3cos2 4 4

Aθ − + ⇒ ≥

31

4 A≤ ≤

22. In a ∆ ABC,3

C π

∠ = , then 3 1a b c a c

− =+ + +

1)1

a b+ 2)

1b c+

3)1

2a b+ 4)

12b c+

KEY – 2

HINT - 01 1 3

60 ,C a c b c a b c∠ = + =+ + + + 3 1 1

a b c a c b c∴ − =+ + + +

23. The number of solutions of sec cos5 1 0 x x+ = in the interval [ ]0,2π is.1) 5 2) 8 3) 10 4) 12

KEY – 2HINT - sec cos5 1 0 x x+ = cos5 cos 0, cos 0 x x x+ = ≠ 2cos3 .cos 2 0 x x = cos3 0, cos 2 0 x x= = with cos 0 x ≠ 5 7 11 3 5 7

, , , , , , ,6 6 6 6 4 4 4 4π π π π π π π π

⇒


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24. If ( )1 11cos cot cot cos2

x− − =

, then a value of x is.

1)1

6

2)1

2

− 3)

2

6

4)2

6

−

KEY - 1

HINT - 11 1

cot cos2 5

A A− = ⇒ =

1

2cos cot

1

x x B B

x− = ⇒ =

−

2

1 1

5 61

x x

x= ⇒ = ±

−

25. If cos 2 199h x = , coth x =

1)5

3 11 2)5

6 11 3)7

3 11 4)10

3 11

KEY – 4

HINT - cosh 2 199 x =2

2

coth 1cosh 2

coth 1 x

x x

+= −

2 200coth198

x⇒ = 10

coth3 11

x =

26. The angles of a triangle ABC are in an arithmetic progression. The larger sides a , b satisfy the

relation 3 12

ba

< < , then the possible values of the smallest side are

1)2 24 3

2a b a

a± −

2)2 24 3

2a b a

b± −

3)2 24 3

2a b a

c± −

4)2 24 3

2a b a± −

KEY – 4

HINT - 2 23

1 4 3 0.2

bb a

a< < ⇒ − > Given 060 B =

2 2 2 02 cos 60b a c ac= + − 2 2 2 0c ac a b⇒ − + − =

2 24 32

a b aC ± −=

27. 2 2 2 21 2 3

1 1 1 1r r r r

+ + + =

1)2 2 2a b c+ +

∆ 2)

2 2 2

2

a b c+ +∆

3)2

2 2 2a b c∆

+ + 4) 2 2 2a b c

∆+ +

KEY – 2

HINT -( ) ( ) ( )2 2 22

2

s s a s b s c LHS

+ − + − + −= ∆

= ( )2 2 2 2

24 2s s a b c a b c− + + + + +

∆


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=2 2 2

2

a b c+ +∆

28. If in a ∆ ABC, 1 2 32 3r r r = = , then :b c = 1) 4 :3 2) 5 : 4 3) 2 :1 4) 3 : 2

KEY - 1HINT - 1 2 3lr mr nr = = Then : : : :a b c m n n l l m= + + +

: 4 : 3b c∴ = 29. P is the point of intersection of the diagonals of the parallelogram ABCD, If S is any point in

the space and SA SB SC SD SPλ+ + + = , then λ =1)2 2) 4 3) 6 4) 8

KEY – 2

Take S as origin &2 2

SA SC SB SDSP SP

+ += = , adding30. If M and N are the mid points of the sides BC and CD respectively of a parallelogram ABCD,

then AM AN + = 1)

43

AC 2)53

AC 3)32

AC 4)65

AC

KEY – 3Take A as origin

, , , AB b AC c AD d = = =

,2 2

b c c d AM AN

+ += =

2 32 2

c b d c AM AN + ++ = =

31. ABCD is a parallelogram and P is a point on the segment AD dividing it internally in the ratio3:1. The line BP meets diagonal AC in Q. Then : AQ QC = 1) 3 : 4 2) 4 :3 3) 3 : 2 4) 2 :3

KEY - 1HINT - A is origin , , , AB b AC c AD d = = =

34

AD d = , Q divides AC is the ratio 1: λ

1C AQ

λ= +

------------- (1 )

Q divides BP in the ratio K : 13

.41

b k d AQ

k

+= +

---------- (2)

equating 1 & 2 k λ = ratio 1: λ = 4 : 3

32. The position vectors of the vertices of ABC ∆ are ( )3 4 , 3 ,5i j k i j k i j k + − + + + + Respectively. The magnitude of the altitude from A onto the side BC is.1)

45

3 2)

55

3 3)

75

3 4)

85

3


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KEY – 1

HINT - AB BC

AD BC

×=

33. The shortest distance between the skew lines 3 4 2 1 7 2,1 2 1 1 3 2

x y z x y z− − + − + += = = =− is.

1) 6 2) 7 3) 3 5 4) 35KEY – 4

HINT - ( ) ( )3, 4, 2 , 1, 7, 2 A C = − = − − ( ) ( )1, 2,1 , 1,3, 2b d = − =

Shortest distance = AC b d

b d

×

34. If 2 3 5 , 3 4 5 5 3 2a i j k b i j k and c i j k = − + = − + = − −, then the volume of the parallelepipedwith co – terminus edges , ,a b b c c a+ + + is.1)1 2) 5 3) 8 4) 16

KEY – 4

HINT - V a b b c c a = + + +

= 2 a b c = 35. In a data the number i is repeated i times for 1, 2,.....,i n= . Then the mean of the data is .

1)2 1

6n +

2)2 1

4n +

3)2 1

3n +

4)2 1

2n +

KEY - 3

HINT - Mean1.1 2.2 3.3 .

1 2 3n n

xn

+ + + − − − += + + + −− −− +

( ) ( )

( )

1 2 12 16

1 32

n n nn

n n

+ ++= =+

36. Two teams A and B have the same mean and their coefficients of variation are 4, 2 respectively If, A Bσ σ are the standard deviations of teams A, B respectively then the relation between

them is.1) A Bσ σ= 2) 2 B Aσ σ= 3) 2 A Bσ σ= 4) 4 B Aσ σ=

KEY – 3

HINT - C.V of A = 100 A

x

σ ×

C.V of B = 100 B

x

σ × ( means are same)

37. If A and B are events such that ( ) ( ) ( )5 1 1, , ,6 4 3

P A B P A P B∪ = = = then A and B are .

1) mutually exclusive 2) independent3) exhaustive 4) exhaustive and independentKEY – 2


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HINT - ( ) ( ) ( ) ( )P A B P A P B P A B∪ = + − ∩

( ) 3 1 5 3 14 3 6 12 4

P A B∩ = + − = =

( ) ( ) ( )1. 4P A P B P A B= = ∩

38. If A and B are two events such that ( ) ( ) ( ) / 0.6, / 0.3, 0.1P A B P B A P A= = = then ( ) p A B∩ =(Here E is the complement of the event E)1)0.88 2)0.12 3)0.6 4)0.4

KEY – 1

HINT - ( ) ( )( )

/ 0.6P A B

P A BP B

∩= =

( ) ( )

( ) / 0.3P A B

P B A P A

∩= =

( ) ( )1P A B P A B∩ = − ∪ = ( ) ( ) ( )1 P A P B P A B− + − ∩

39. In a certain college, 4% of the men and 1% of the woman are taller than 1.8 meters. Also, 60%of the students are woman. If a student selected at random is found to be taller than 1.8meters, then the probability that the student being a woman is.

1)3

11 2)

511

3)6

11 4)

811

KEY – 1

( )P M = P ( selecting a man ) = 40100

( )P W = P ( selecting a woman )= 60100

( )P T = P( selecting a person tallen than 1.8 metres).

( ) ( ) ( )( ) ( ) ( ) ( )

. / /

. / . /

P W P T W P W T

P W P T W P M P T M = +

60 1 60 3

60 1 40 4 220 11

×= = =× + ×

40. X is a binomial variate with parameters 6n = and p. If ( ) ( )4 4 2P X P X = = = , then p is 6n =

1)12

2)13

3)14

4)16

KEY - 2n = 6

( ) ( )4 4 2P X P X = = =

4 2

4 2 2 44. 6 6 .C C p q p q= 2 24 p q=

2 1 p q p= = − 3 1 p =


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13

p =

41. The properties of coin showing head is p. 100 such coins are tossed. If the probability of 50

coins showing heads is same as the probability of 51 coins showing heads, then p =

(1)12

(2)49

100 (3)

51101

(4)50

101

KEY: 3

HINT:( ) ( )50 51P x P x= = =

( ) ( )50 49100 50 100 5150 511 1C P P C P P− = −

( ) ( )100 51100

50

1C P

PC

− =

( )100 51 1151

P P− +− =

51 51 50P P− = 51

101P⇒ =

42. The locus of the point P which is equidistant from 3 4 5 0 x y+ + = and 9 12 7 0 x y+ + = is

(1) a hyperbola (2) an ellipse (3) a parabola (4) a straight lineKEY: 4

HINT:1 1 1 13 4 5 9 12 7

25 81 144

x y x y+ + + += ±+

1 1 1 13 4 5 9 12 75 15

x y x y+ + + += ±

( )1 1 1 19 12 15 9 12 7 x y x y+ + = ± + +

( )1 1 1 19 12 15 9 12 7 x y x y⇒ + + = − + +

1 118 24 22 0 x y⇒ + + = Which represents a straight line

43. If the origin of a coordinate system is shifted to ( )2, 2− and then the coordinate system isrotated anticlockwise through an angle 045 , the point ( )1, 1P − in the original system has newcoordinates(1) ( )2, 2 2− (2) ( )0, 2 2− (3) ( )0, 2 2− − (4) ( )0, 2 2− +

KEY: 3

HINT:( ) ( )cos sin X x h y k θ θ= − + −

( ) ( )sin cosY x h y k θ θ= − − + −

( ) ( ), 2, 2h k = −

( ) ( ), 1, 1 x y = −

045θ =


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( ) ( )1 11 2 1 1 2 02 2

X ⇒ = + + − − =

( )2 1 11 1 2 2 21 2 2

Y = − + + − − = − −

( )0, 2 2− −

44. The combined equation of the straight lines passing through the point ( )4,3 and each line

making intercepts on the coordinate axes whose sum is -1 is(1) ( )( )3 2 6 2 2 0 x y x y− − − + = (2) ( )( )3 2 6 2 2 0 x y x y− + − + = (3) ( )( )3 2 6 2 2 0 x y x y− − − − = (4) ( ) ( )3 2 6 2 2 0 x y x y− + − − =

KEY: 1HINT: By verification the lines are

3 2 6 0, 2 2 0 x y x y− − = − + = Combined equation is

( )( )3 2 6 2 2 0 x y x y− − − + =

45. The value of 0k > such that the angle between the lines 4 7 0 x y− + = and 5 9 0kx y− − = is 045

, is

(1)253

(2)53

(3) 3 (4) 5

KEY: 3

HINT: 24 5

cos4516 1 25

k

k

+=+ +

2

1 4 5

2 17 25

k

k

+= +

By verification k = 346. An equation of a line whose segment between the coordinate axes is divided by the point

1 1,

2 3

in the ratio 2 :3 is

(1) 6 9 5 x y+ = (2) 9 6 5 x y+ = (3) 4 9 5 x y+ = (4) 9 4 5 x y+ = KEY: 3

HINT:( ) ( ) 1 1,0 , 0, , ,

2 3 A a B b P

2 3,

5 5a b

P = 1 1 2 3

, ,2 3 5 5

a b = 5 5

,4 9

a b= =

15 54 9

x y⇒ + =

4 9 5 x y+ =


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47. Two pairs of straight lines with combined equations 4 3 12 0 xy x y+ − − = and

3 4 12 0 xy x y− + − = form a square. Then the combined equation of its diagonals is(1) 2 22 0 x xy y x y− + + − = (2)

2 22 0 x xy y x y+ + + + = (3) 2 2 0 x y x y

− + − = (4) 2 2 0 x y x y

− + + =

KEY: 3

HINT:( ) ( )4 3 4 0 x y y+ − + =

( )( )3 4 0 x y− + =

And

( ) ( )3 4 3 0 x y y− + − =

( )( )4 3 0 x y+ − =

⇒

int po s are ( ) ( ) ( ) ( )4,3 , 3, 4 , 3,3 , 4, 4− − − −

( )73 47

y x−− = +

1 0 x y+ + =

( )7& 3 37

y x−− = −−

0 x y− = ( )( )1 0 x y x y⇒ − + + =

2 2 0 x y x y− + − = 48. The line x y k + = meets the pair of straight lines 2 2 2 4 2 0 x y x y+ − − + = in two points A and

B. If O is the origin and 090 AOB∠ = then the value of 1k > is(1) 5 (2) 4 (3) 3 (4) 2

KEY: 4

HINT:

22 2 2 4 2 0

x y x y x y x y x y

k k k + + + + − − + = 2 2 0coefficient of x coefficient of y+ =

1, 2k = 1, 2k k > =Q

49. The value of a, such that the power of the point ( )1,6 with respect to the circle2 2 4 6 0 x y x y a+ + − − = is -16, is

(1) 7 (2) 11 (3) 13 (4) 21KEY: 4

HINT:11 16S = −

1 36 4 36 16a⇒ + + − − = −

5 16a⇒ − = − 21a⇒ =

50. The area (in square units) of the triangle formed by the tangent, normal at ( )1, 3 to the circle2 2 4 x y+ = and the X-axis is

(1) 4 3 (2) 7 32

(3) 2 3 (4) 1 32

KEY: 3


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HINT:

21 12

ym

m∆ = +

3 1

3

2 3

∆ = − −

2 3= 51. If ( )4,2 and ( ), 3k − are conjugate points with respect to 2 2 5 8 6 0 x y x y+ − + + =, then k =

(1)283

(2)283

− (3) 328

(4)3

28−

KEY: 1

HINT:12 0S =

283

K ⇒ =

52. The length of the common chord of the two circles ( )2 2 2 x a y a− + = and ( )22 2 x y b b+ − = is

(1)2 2

ab

a b+ (2)

2 2

2ab

a b+ (3)

2 2

a b

a b

++

(4) 2 2a b+

KEY: 2HINT: Given circles are orthogonal

1 2

2 21 2

2r r d

r r =

+ 53. The equation of the circle passing through

( )1, 2 and the points of intersection of the circles

2 2 8 6 21 0 x y x y+ − − + = and 2 2 2 15 0 x y x+ − − = is(1) 2 2 6 2 9 0 x y x y+ + − + = (2) 2 2 6 2 9 0 x y x y+ − − + = (3) 2 2 6 4 9 0 x y x y+ − − + = (4) 2 2 6 4 9 0 x y x y+ − + + =

KEY: 3

HINT:( )2 2 8 6 21 6 0 x y x y x yλ+ − − + + + − =

( )sin 1,2 2 pas g through λ⇒ =

Required circle is

2 2

6 4 9 0 x y x y⇒ + − − + = 54. The equation of the parabola with focus ( )1, 1− and directrix 3 0 x y+ + = is(1) 2 2 10 2 2 5 0 x y x y xy+ − − − − = (2) 2 2 10 2 2 5 0 x y x y xy+ + − − − = (3) 2 2 10 2 2 5 0 x y x y xy+ + + − − = (4) 2 2 10 2 2 5 0 x y x y xy+ + + + − =

KEY: 1

HINT:SP PM =

( ) ( )2 2 1 1 31 11 1

x y x y

+ +− + + =+

2 2

10 2 2 5 0 x y x y xy+ − − − − =


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55. If P is a point on the parabola 2 8 y x= and A is the point ( )1,0 , then the locus of the mid point

of the line segment AP is

(1) 21

42

y x = −

(2) ( )2 2 2 1 y x= + (3) 2 12

y x= − (4) 2 2 1 y x= +

KEY: 1

HINT:( ) ( )22 , 4 1,0P t t A

( )2

1 1

2 1 4 0int , ,

2 2t t

Mid po x y + += =

212 12

xt

−⇒ =

221&

4 y

t =

221 11 1

2 1 14

4 2 2

y x y x

− ⇒ = ⇒ = −

56. For the ellipse

2 2

125 16 x y+ = , a list of lines given in List-I are to be matched with their equations

given in List-IIList-I List-IIi) Directrix corresponding to the focus ( )3,0− a) 4 y = ii) Tangent at the vertex ( )0, 4 b) 3 25 x = iii) Latus rectum through ( )3,0 c) 3 x =

d) 4 0 y + = e) 3 0 x+ = f) 3 25 0 x+ =

The correct matching(i) (ii) (iii)

1) (b) (a) (e)2) (f) (a) (c)3) (b) (d) (c)4) (f) (a) (e)KEY: 2

HINT:5, 4a b= =

35

e =

( )3,0S = ±

i)

a x

e= −

5

35

x −=

3 25 0 x+ = ii) Tangent at (0, 4) is y = 4iii) Latus rectum x = ae


15/39

x = 3

57. The centre of the ellipse( ) ( )2 23 1

19 16

x y x y+ − − ++ = is

(1)( )

1, 2−

(2)( )1, 2

− (3)

( )1, 2

− − (4)

( )1, 2

KEY: 4HINT: On solving x + y – 3 = 0

x – y + 1 = 0(x,y) = (1,2)

58. The product of lengths of perpendiculars from any point on the hyperbola 2 2 16 x y− = to itsasymptotes is

(1) 2 (2) 4 (3) 8 (4) 16KEY: 3

HINT:

2 2

2 28

a bP

a b= =

+ 59. ( ) ( )4,3,5 , 0, 2, 2 A B − and ( )3,2,1C are three points. The coordinates of the point in which thebisector of BAC ∠ meets the side BC is

(1)15 4 11

, ,8 8 8

(2)12 2 10

, ,7 7 7

(3)9 2 7

, ,5 5 5

(4)3 3

,0,2 2

KEY: 1

HINT:: 5 :3 AB AC =

3 5

8

B C D

+=

15 4 11, ,

8 8 8 D =

60. If the extremities of a diagonal of a square are ( )1,2,3 and ( )2, 3,5− , then its side is of length

(1) 6 (2) 15 (3) 15 (4) 3KEY: 3

HINT:2 2 2 x x AC + =

22 30 x =

2 15 x = 15 x =

61. A plane meets the coordinate axes in P,Q,R respectively. If the centroid of ∆PQR is 1 11, ,2 3

,

then the equation of the plane is(1) 2 4 3 5 x y z+ + = (2) 2 3 3 x y z+ + = (3) 4 6 5 x y z+ + = (4) 2 2 6 3 x y z− + =

KEY: 2

HINT:

1 1, , 1, ,3 3 3 2 3

a b c =


16/39

2

13 3 1 x y z+ + =

2 3 3 x y z+ + =

62.1/

0lim tan

4

x

x x

π→

+ =

(1) 2e (2) e (3) 3/ 2e (4) 1e−

KEY: 1

HINT: It is in the form1∞

0

1lim tan 14 x x xe

π→

+ − apply LH rule.

2

0limsec

24 x x

e eπ

→ + =

63. The value that should be assigned to ( )0 f so that the function ( ) ( )cot1 x f x x= + is continuous at0, x = is

(1) e (2) 1 (3) 2 (4) 1e−

KEY: 1

HINT: ( ) ( )00 lim x f f x→= ( ) ( )cot

00 lim 1

x

x f x

→= + 1∞− − −

( )

0lim cot 1 1

x x x

e →+ −

0lim 1tan x

x xe e→= =

64. If3 3

1 12 2 4

3 4 4tan tan

1 3 1 6 x x x x

y x x x

− − − −= + − − + then

dydx

=

(1) 22

1 x+ (2) 2

41 x+

(3) 26

1 x+ (4) 2

71 x+

KEY: 4

HINT: Puttan x θ=

,( ) ( )1 1tan tan 3 tan tan 4 y θ θ− −= +

17 7 tan y Y xθ −= ⇒ =

2

71

dydx x

= +

65. If 2 22

x y t t

+ = + and 4 4 2 24

x y t t

+ = + , then 3 dy x ydx

=

(1) 1− (2) 2− (3) y x

(4) xy


17/39

KEY: 2

HINT: Squaring on both sides

4 4 2 2 22

42 4 x y x y t

t + + = + +

2 2 2 x y⇒ =

differentiate

3 2dy y dy

x ydx x dx

− = ⇒ = −

66. If1

1

y x

y

−=+

, then ( )2

2

3 11

yd y dy x

dx dx y

++ + =

(1) 2 y− (2) 0 (3) y− (4) y

KEY: 2

HINT:

1 111

y x x y x y

− −= ⇒ = ++

Differentiate two times w.r.t to x,

we get( ) // / 3 11 0

y x y y

y

++ + =

67. If the lines 4 y x b= − + are tangents to the curve 1 y x

= , then b = (1) 4± (2) 2± (3) 1± (4) 8±

KEY: 1HINT: Applying the condition for tangency.

4b = ±

68. An equilateral triangle is of side 10units. In measuring the side, an error of 0.05 units ismade. Then the percentage error in the area of the triangle is

(1) 5 (2) 4 (3) 1 (4) 0.5 KEY: 3

HINT:10, 0.05a aδ= =

234

A a= apply logs and differentiate

0.05100 2 100 2 100 1

10 A a

A aδ δ × = × × = × × =

69. Define ( ) ( )

( )0 1

2 1 2

x x f x

x x

≤ ≤= − ≤ ≤ Then Rolle’s theorem in not applicable to ( ) f x

because(1) ( ) f x is not defined everywhere on [ ]0,2 (2) ( ) f x is not continuous on [ ]0,2 (3) ( ) f x is not differentiable on ( )1, 2 (4) ( ) f x is not differentiable on ( )0, 2


18/39

KEY: 4

HINT:( ) f x

is not differentiable at1 x =

( ) f x∴ is not differentiable on ( )0, 2

70. Two particles P and Q located at the points with coordinates ( ) ( )3 3, 16 3 , 1, 6 6P t t t Q t t t − − + − − aremoving in a plane. The minimum distance between them in their motion is

(1) 1 (2) 5 (3) 169 (4) 49

KEY: 1

HINT:( )21 10 3PQ t = + −

applying first derivative test

310

t = , min.distance

1PQ =

71.22

xdx

x+ =−∫

(1) 1 22sin 42 x

x c− + − + (2) 1 2cos 4

2 x

x c− − − +

(3) 1 2sin 42 x

x c− − − + (4) 1 22sin 4

2 x

x c− − − +

KEY: 4

HINT: Multiply Nr and Dr with2 x+

2 2 2 22

2 2

dx xdx

x x⇒ +

− −∫ ∫

72. ( ) ( ) ( ) ( )1 21tan log 12

x x xe e dx f x e c f x− − = − + + ⇒ =∫ (1) ( )1tan x x xe e e− −− (2) ( )2 1tan x x x e e− −+

(3) ( )1

tan x x

e e− −

− (4) ( )1

tan x x

x e e− −

− KEY: 4

HINT: put

xe t =, apply integration by parts.

73. 4 xe dx− =∫ (1) 1

4tan 4

2

x xe e c−

− + − + (2) 1

42 4 4 tan

2

x x ee c−

−− − +

(3) 14

2 4 4cot2

x x ee c−

−− − + (4) ( )14 4 tan 4 x xe e c−− − − +


19/39

KEY: 2

HINT: Put

24 xe t − =

74. If ( ) ( )125

log 4 5 tan4 5

xdx a x x b x k

x x2 −+ = + + + ++ +∫ +constant then ( ), ,a b k =

(1)1

,3,22

(2)1

,1, 22

(3)1

,3,12

b (4) ( )1,3,2

KEY: 1

HINT:

( )( )2

2 4 612 4 5

xdx

x x

+ + + +

∫ ( )2 11 log 4 5 3 tan 2

2 x x x k −= + + + + +

75.1

0

11

xdx

x− =+∫

(1) 12π − (2) 1

2π + (3) 1π − (4) 3

2π

KEY: 1

HINT: Rationalising

76. /2

4 40

16 sin cossin cos

x x xdx

x x

π

=+∫

(1)2

4π

(2)2

2π

(3) 2π (4) 22π

KEY: 3

HINT:

( ) ( )0 0

a a

f x dx f a x dx= −∫ ∫

77. The area of region bounded by the curves 29 y x= and 25 4 y x= + (in square units) is

(1) 64 (2)643

(3)323

(4)163

KEY: 4

HINT: Required area,( )

12 2

0

2 5 4 9 A x x dx = + − ∫ 78. The differential equation of the family of curves

1, y ax

a= + where 0a ≠ is an arbitrary constant,

has the degree(1) 4 (2) 3 (3) 1 (4) 2


20/39

KEY: 4

HINT:

/ y a= put ‘a’ value in

1 y ax

a= +

79. The solution of the differential equation / / 2 y x xy xe y−= + is

(1) / 0 y xe In cx+ = (2) / y xe x c− = + (3) / y xe In cx= (4) / 2 y xe In cx=

KEY: 4

HINT: put y vx=

80. Match the differential equations in list I to their Integrating factors in List II

List I List IIDifferential equation Integrating factor

(i) ( )3 2 21 3dy x x y xdx

+ + = (a) 3 x

(ii) 2 63dy

x xy xdx

+ = (b) ( )23 1 x +

(iii) ( ) ( )23 2 3 21 6 1dy x x x y xdx

+ + + = (c ) ( )21 x2 +

(iv) ( )1 4dy x xy Inxdx

2 + + = (d) 2 1 x +

(e) ( )1/33 1 x + (f) ( )

1/23 1 x + The correct match is:

(i) (ii) (iii) (iv)(1) d a b c(2) e a b c(3) e b c f(4) e a c d

KEY: 2

HINT: (i)( )

2

3 1/331. 1 x

dx x I F e x+∫ = = +

(ii)

33.

dx x I F e x∫ = =

(iii)

( ) ( )2

3

621 3. 1

xdx

x I F e x

+∫ = = +

(iv)

( )

( )2

421 2

. 1

xdx

x

I F e x+∫

= = +


21/39

PHYSICS

81. Match the following:A B

a) Rocket propulsion e) Bernoulli’s principle in fluid dynamicsb) Aeroplane f) Total internal reflection of lightc) Optical fibers g) Newton’s laws of motiond) Fusion test reactor h) Magnetic confinement of plasma

f) Photoelectric effectThe correct match is: (a) (b) (c) (d)1) g f e h2) g e f i3) i e f g4) g e f h

Key: 4Sol: Conceptual

82. Force F is given by the equation X

F Linear density

= . Then dimensions of X are

1) 2 0 2 M L T − 2) 0 0 1 M L T − 3) 2 2 L T − 4) 0 2 2 M L T −

Key: 1

Sol: X

F

Linear Density=

2 M MLT X L

− × =

2 2 X M T −∴ = 83. The displacement of a particle moving in a straight line is given by the expression

3 2 x At Bt Ct D= + + + in meters, where ‘t’ is in seconds and A, B, C and D are constants. Theratio between the initial acceleration and initial velocity is

1) 2C B

2) 2 BC

3) 2C 4)2C B

Key: 2Sol: 3 2 x At Bt Ct D= + + +

23 2V At Bt C = + + At 0t V C = ⇒ =

6 2a tA B= +

At 0 2t a B= ⇒ =

2 BC

∴


22/39

84. A, B, C are points in a vertical line such that AB = BC. If a body falls freely from rest at A, and

1t and 2t are times taken to travel distances AB and BC, then ratio ( )2 1 / t t is

1) 2 1+ 2) 2 1− 3) 2 2 4) 12 1+

Key: 2 & 4

Sol: 2112

AB S gt = =

( )21 21

22

S g t t = +

( )221 1 21 1

22 2

gt g t t × = +

1 1 22t t t = + ( )1 22 1t t − =

2

1

2 1t t

= −

85. Sum of magnitude of two forces is 25 N. The resultant of these forces is normal to the smallerforce and has a magnitude of 10 N. Then the two forces are1) 14.5 N, 10.5 N 2) 16 N, 9 N 3) 13 N, 12 N 4) 20 N, 5 N

Key: 1

Sol: x

10

25 x−

( )2 2 225 10 x x− = +

14.5 & 10.5

86. A body of mass ‘m’ thrown up vertically with velocity 1v reaches a maximum height 1h in 1t

seconds. Another body of mass 2 m is projected with a velocity 2v at an angle θ . The second

body reaches a maximum height 2h in time 2t seconds. If 1 22t t = , ration 12

hh

is

1) 1: 2 2) 4: 1 3) 1: 1 4) 3: 2Key: 2Sol: 1 22t t =

1 22 sinv vg g

θ×=

1 22 sinv v θ= 2

1 12 2

2 2

22 sin

h v gh g v θ= ×

A

B

C

s

s

1t

2t


23/39

Substitute 1' 'v in above

1

2

4 :1hh∴ =

87. Hammer of mass M strikes a nail of mass ‘m’ with a velocity 20 m/s into a fixed wall. The nailpenetrates into the wall to a depth of 1 cm. The average resistance of the wall to thepenetration of the nail is

1)2

310 M

M m × +

2)2

42 10 M

M m×+

3) 22 10 M m

M + × 4)

2210

M M m

×+

Key: 2Sol: ( )20 M M m V × = +

20 M V

M m×= +

( )21

12w M m v f = × + = × 2

42 10 M

f M m

∴ = ×+

88. A body of mass 10kg is acted upon by a force given by equation ( )23 30F t = − Newtons. Theinitial velocity of the body is 10 m/s. The velocity of the body after 5secs. is1) 4.5 m/s 2) 6 m/s 3) 7.5 m/s 4) 5 m/s

Key: 3

Sol: . dv

F ma m dtt = =

. . M dv F dt =

( )2

1

2

0

3 30v t

v

M dv t dt = −∫ ∫

( )3

2 1

330

3t

M v v t − = −

( )210 10 125 30 5v − = − ×

2 7.5v =

89. A ball (initially at rest) is released from the top of a tower. The ratio of work done by the forceof gravity in the first, second and third seconds is1) 1 : 3 : 5 2) 1 : 4 : 16 3) 1 : 9 : 25 4) 1 : 2 : 3

Key: 1Sol: .w f s=

w s∝

1 2 3: : 1 :3 :5s s s =

90. A body of mass 2.4 kg is subjected to a force which varies with distance as shown in figure. Thebody starts from rest at 0 x = . Its velocity at 9 x m= is


24/39

0 3 6 9

20

xinmeters→

Force N

1) 5 3 / secm 2) 20 3 / secm 3) 10 / secm 4) 40 / secm

Key: 3

Sol: 21 1 12 2 2

bh Lb bh mv+ + =

21 1 13 20 20 3 3 20 2.42 2 2

v× × + × + × × = × × 110v ms−∴ =

91. The moment of inertia of a solid cylinder of mass M, length 2 R and radius R about an axispassing through the centre of mass and perpendicular to the axis of the cylinder is 1 I and about

an axis passing through one end of the cylinder and perpendicular to the axis of cylinder is 2l ,then

1) 2 1 I I < 2) 22 1 I I MR− = 3) 21

1912

I I

= 4) 21

76

I I

=

Key: 2

Sol: 2 2

12

12 4

ml mR I and l R= + =

2 2

2 23 4ml mR

I and l R= + = 2

2 1 I I MR− = 92. A body of mass 1kg, initially at rest explodes and breaks into three parts. The masses of the

parts are in the ratio 1 : 1 : 3. The two pieces of equal mass fly off perpendicular to each otherwith a speed of 30 m/s each. The velocity of the heavier part in m/s is1) 10 2 2) 6 3) 3 4) 6 2

Key: 1Sol:

3 32P M V

=

3

32 30

5 5 M M

v× × = ×

3 10 2v = 93. A particle of mass 4kg is executing S.H.M. Its displacement is given by the equation

[ ]8cos 100 / 4Y t cmπ= + . Its maximum kinetic energy is1) 128 J 2) 64 J 3) 16 J 4) 32 J

Key: 1

Sol: 2 2max1

.2

K E mw A=

4 41 4 10 64 102

−= × × × ×


25/39

=128 Joules

94. Infinite number of spheres, each of mass m are placed on the X-axis at distances 1, 2, 4, 8, 16,…. Meters from origin. The magnitude of the gravitational field at the origin is

1)2

3Gm 2)

4

3Gm 3) Gm 4) 6Gm

Key: 2

Sol: 1 1 1

...1 4 16

I GM = + + +

43

GM I =

95. When a force 1F is applied on a metallic wire, the length of the wire is 1 L . If a force 2F is

applied on the same wire, the length of the wire is 2 L . The original length of the wire L is

1) 1 1 2 2

1 2

L F L F

F F

+

+ 2) 2 1

1 2

L L

F F

−

+ 3) 1 2 2 1

1 2

F L F L

F F

−

− 4) 1 1 2 2

1 2

F L F L

F F

−

−

Key: 3Sol: e f ∝

1 1

2 2

L L F L L F

− =−

1 2 2 1

1 2

F L F L L

F F −

∴ = −

96. 1000 spherical drops of water each 810 m− in diameter coalesce to form one large sphericaldrop. The amount of energy liberated in this process in Joules is(Surface tension of the water is 0.075 N/m)

1) 1510.75 10π −× 2) 156.75 10π −× 3) 158.65 10π −× 4) 153.88 10π −× Key: 2

Sol: ( )2 2/34W r T n nπ= −

( )( )16

2/33104 0.075 1000 104

W π−

= × × −

15

6.75 10W J π −

= × 97. A thermos flask contains 250g of coffee at 090 C . To this 20 g of milk at 05 C is added. After

equilibrium is established, the temperature of the liquid is(Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as

01.00 / cal g C )

1) 03.23 C 2) 03.17 C 3) 083.7 C 4) 037.8 C Key: 3

Sol: 1 1 1 2 2 21 1 2 2

m s t m s t t

m s m s+= +

and 1 2s s=

250 90 20 5 83.7250 20

t × + ×= =+


26/39

98. A copper rod of length 75 cm and an iron rod of length 125 cm are joined together end to end.

Both are of circular cross section with diameter 2 cm. The free ends of the copper and iron aremaintained at 0100 C and 00 C respectively. The surfaces of the bars are insulated thermally.The temperature of the copper- iron junction is

1) 0100 C 2) 00 C 3) 093 C 4) 050 C Key: 3

Sol: ( ) ( )1 1 2 2

1 2

K A K A

l l

θ θ θ θ− −=

( ) ( )386.4 100 48.46 075 125

θ θ− −= 093 C θ∴ =

99. 1 g of water at 0100 C is completely converted into steam at 0100 C . 1g of steam occupies avolume of 1650 cc . (Neglect the volume of 1 g of water at 0100 C ). At the pressure of 5 210 / N m ,

latent heat of steam is 540 / cal g (1 Calorie = 4.2 Joules). The increase in the internal energy inJoules is1) 2310 2) 2103 3) 1650 4) 2150

Key: 2Sol: dQ du dw= +

mL du pdv= + 2103du J =

100. R.M.S. velocity of oxygen molecules at N.T.P is 0.5 / km s . The R.M.S velocity for the hydrogenmolecule at N.T.P is1) 4 / km s 2) 2 / km s 3) 3 / km s 4) 1 / km s

Key: 2

Sol: RMS RT

V M

γ =

1V

M ∝

1 1

2 2

V M V M

=

2 2 / V km s= 101 A thin wire of 99cm is fixed at both ends as shown in figure. The wire is kept under a tension

and is divided into three segments of lengths 1 2,l l and 3l as shown in figure. When the wire ismade to vibrate, the segments vibrate respectively with their fundamental frequencies in theratio 1 : 2 : 3 . Then the lenghts 1 2 3, ,l l l of the segments respectively are ( in cm )

1) 27, 54, 18 2) 18, 27, 54 3) 54, 27, 18 4) 27, 9, 14KEY - 3

HINT - 1 2 31 2 3

1 1 1: :l l l

n n n= = =

6 : 3 : 2=


27/39

1

699 54

11l = × =

2

399 27

11l = × =

32 99 18

11l = × =

102. Three thin lenses are combined by placing them in contact with each other to get moremagnification in an optical instrument . Each lens has a focal length of 3 cm. If the lensdistance of distinct vision is taken as 25 cm. The total magnification of the lens combination innormal adjustment is1) 9 2) 26 3) 300 4) 3

KEY - 2

HINT -1 2 3

1 1 1 1F f f f

= + + 1 3 13F

= =

1 D M f

= +

103. A convex lens of glass ( )1.45gµ = has a focal length g f in air. The lens is immersed in a liquidof refractive index ( )gµ 1.3. The ratio of the / liquid g f f is1) 3.9 2) 0.23 3) 0.43 4) 0.39

KEY – 1

HINT -( )1

1

g L

ga

L

u f u f

u

−=

−

104. Through a narrow slit of width 2mm, diffraction pattern is formed on a screen kept at adistance 2 m from the slit. The wavelength of the light used is 6330 A and falls normal to theslit and screen. Then the distance between the two minima on either side of the central,maximum is1) 12.7 mm 2) 1.27 mm 3) 2.532 mm 4) 25.3 mmKEY- 2

HINT - Dx nγ = yd n D

γ = Distance = 2 y

105. Charges ‘Q’ are placed at the ends of a diagonal of a square and charges ‘q’ are placed at theother two corners. The condition for the net electric force on ‘Q’ to be zero is

1) 2 2Q q= − , q being – ve 2)2q

Q = − , q being – ve

3) 2 2Q q= , q being – ve 4) 2Q q= , q being – ve

KEY - 1

HINT - 0net F = 2

0 0 2

1 12 .

4 4 2net Q

F q

πε πε= +

106. In the arrangement of capacitors shown in the figure, if each capicator is 9 PF then the effective

capacitance between the points A and B is


28/39

1) 10 PF 2) 15 PF 3) 20 PF 4) 5PF KEY - 2HINT - 1 2c and c are in parallel

2 3c and c are in series

Resultant of 2 3c and c is in parallel with 4c 107. A battery of the emf 18 V and internal resistance of 3Ωand another battery of emf 10 V and internal

resistance of 1Ω are connnected as shown in figure. Then the voltmetre reading is1) 10 V 2) 12 V 3) 16 V 4) 8 V

KEY - 2HINT - V E ir = −

108. A wire of Aluminium and a wire of Germanium are cooled to a temperature of 077 K . Then1) Resistance of each of them decreases2) Resistance of each of them Increases3) Resistance of Aluminium wire increases and that of Germenium wire decreases4) Resistance of Aluminium wire decreases and that of Germenium wire increases

KEY - 4HINT - With decrease in temp resistance of metals decrease and semi conductor increases

109. A voltmeter of 250 mV range having a resistance of10Ω

is converted into an ammeter of 250 mArange. The value of necessary shunt is ( nearly )1) 2Ω 2) 0.1Ω 3) 1Ω 4) 10Ω

KEY - 3

HINT -V

K a

= 1

GS

iig

= −

110. A circular loop and a square loop are formed from two wires of same length and cross-section. Samecurrent is passed through them. Then the ratio of their dipole moment is

1) 4 2)2

π

3) 2 4)4

π

KEY - 4HINT - 1 2 A A=


29/39

111. At a certain place a magnet makes 30 oscillations per minute. At another place where the magnetic field

is doubled its time period will be

1) 2 sec 2) 2 sec 3) 4 sec 4)12

sec

KEY - 1HINT -

1T α

β

112. A small square loop of wire of side ' 'l is placed inside a large square loop of side L. ( ) L l> if the loopsare coplanar and their centres co-incide, the mutual induction of the system is directly proportional to

1)l

L 2)

2l L

3) 2l

L 4)

2

2

l L

KEY - 2HINT D Mi= BA Mi= 113 In a circuit L , C and R are connected in series with an alternating voltage source of f. When current in

the circuit leads the voltage by 045 , the value of C

1)( )

12 2 f fL Rπ π +

2)( )

12 2 f fR Lπ π +

3)( )1

2 f R Lπ + 4)

11

2 f R L

π +

KEY – 1

HINT - tan C L X X

Rϕ

−= c

114 Suppose that the electric flux inside a parallel plate capacitor changes at a rate of 147 10× units / sec,then the magnetic induction field density at any point inside the capacitor is

Area of the plate of the Capacitor =2

1m Permitivity of free space = 12 2 28.8 10 Nm c− −× Permitivity of free space = 74 10 / Tesla m Ampπ −× 1) 37.79 10 T −× 2) 50.779 10 T −× 3) 48.85 10 T −× 4) 1288.5 10 T −×

KEY – 3

HINT - 21 d

BC dT

ϕ= 115 If an electron has an energy such that its De Broglie wave length is 5500A, hten the energy value of that

electron is ( )34 316.6 10 9.1 10ch Js m kg− −= × = ×

1)20

8 10 J −

× 2)10

8 10 J −

× 3) 8 J 4)25

8 10 J −

× KEY – 4HINT -

h p

λ = 2

2P

KE m

= 116 The following statements are given about hydrogen atom

A) The wave length of the spectral lines of Lyman series are greater than the wavelength of the seccondspectral line of Balmer seriesB) The orbits correspond to circular standing waves in which the circumference of the orbit equals awhole numbver of wavelengths1) A is False , B is True 2) A is True , B is False3) A is False , B is True 4) A is True , B is True

KEY – 1


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HINT – Conceptual

117 A radioactive nucleus can decay by two different processes. The half lives of the first and seccond decayprocesses are 35 10× and 510 years respectively. Then the effective half – life of the nucleus is1) 5105 10× yrs 2) 4762 yrs 3) 410 yrs 4) 47 . 6 yrs

KEY – 2

HINT - 1 2

1 2

TT T

T T = +

118 In ahalf wave rectifier the AC input source of frequency 50Hz is used. The fundamental frequency ofthe output is1) 50 Hz 2) 150 Hz 3) 200 Hz 4) 75 Hz

KEY – 1HINT - CONCEPTUAL119 If en and hn are electron and hole concentrations in the extrinsic semiconductor and in is electron

concentration in an intrensic semiconductor then,

1) e ih

n nn =

2) ( )e h in n n+ = 3) ( ) 2e h in n n− = 4) 2e h in n n=

KEY – 4HINT - 21 nn nen= 120 A carrier wave of peak voltage 12 volts is used to transmit a signal. If the modulation index is 75% the

peak voltage of the modulating signal is1) 18 V 2) 22 V 3) 16 V 4) 28 V

KEY – 3

HINTS- 0C V V m=

CHEMISTRY

121. The number of radial nodes present in 3p orbital is

1) 0 2) 1 3) 2 4)3

KEY:- 2Hint :- number of radial nodes = ( )1n l− − 122. The radiation with maximum frequency is

1) X –rays 2) Radio waves 3) UV rays 4) IR raysKEY:- 1Hint :- Theoretical

123. The equation used to represent the electron gain enthalpy is1) ( ) ( ) X g e X g− −+ → 2) ( ) ( ) X s e X g− −+ → 3) ( ) ( ) X g X g e+ −→ + 4) ( ) ( ) X s X g e+ −→ +

KEY:- 1Hint :- Theoretical ( elements must be in gaseous state)124. An element in +2 oxidation state has 24 electrons. The atomic number of the element

and the number of unpaired electrons present in it respectively are1) 24, 4 2) 26, 4 3) 24, 2 4) 26, 5

KEY:- 2

Hint :- Element has 26 electrons that is Fe. It contains 4 unpaired electrons 125. Number of bonding electron pairs and number of lone pairs of electrons in 3 4 5ClF ,SF ,BrFrespectively are


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1) 3, 2; 4, 2; 5, 2 2) 3, 1; 4, 1; 5, 2 3) 3, 1; 4, 2; 5,1 4) 3, 2; 4, 1; 5, 1

KEY:- 4Hint :- Theoretical126. What is the bond order of 2N ?

1) 3 2) 4 3) 2 4) 1

KEY:- 1

Hint :- bond order = 10 4 32 2

b a N N − −= =

127. Match the following :

List – I List – II

(A) Viscosity (I) Critical temperature(B) Ideal gas behavior (II) Isobar

(C) Liquefaction of gases (III) Compressibility factor

(D) Charles’ law (IV) Kg 2S −

(V) Kg 1 2m S − −

The correct answer is :(A) (B) (C) (D)

1) (IV) (III) (I) (II)2) (V) (III) (I) (II)3) (V) (III) (II) (I)4) (IV) (III) (II) (I)

KEY:- 2Hint :- Theoretical 128. The most probable speed of 2O molecules at T(K) is

1)4

RT

π 2)

RT

16p 3)

RT

16 4)

3RT

32

KEY:- 3

Hint :- 2 232 16

RT RT RT MPV

M = = =

129. According to significant figure convention the result obtained by adding 12.11, 18.0 and1.012 is1) 31.12 2) 31.1 3) 31 4) 31.122

KEY:- 2

Hint :- Added value is 31.122 can be rounded to 31.1130. An organic compound having C, H and O has 13.13% H, 52.14% C. Its molar mass is

46.068 g. What are its empirical and molecular formulae?


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1) 2 6 4 12 2C H O, C H O 2) 3 2 6 2CH O, C H O

3) 2 6 2 6C H O, C H O 4) 2 6 2 3 9 4C H O , C H O

KEY:- 3Hint :- EF = 2 6C H O here n = 1 ( ) MF n EF ∴ =

MF = 1 ( )2 6 2 6C H O C H O=

131. Which one of the following is not a state function ?

1) Internal energy 2) Work 3) Entropy 4) Free energy

KEY:- 2

Hint :- Theoretical (work is a path function)

132. When one mole of A and one mole of B were heated in a one litre flask at T(K), 0.5moles of C was formed in the equilibrium

A+B C+D

The equilibrium constant CK is

1) 0.25 2) 0.5 3) 1 4) 2

KEY:- 3Hint :-

( ) ( ) ( ) ( )g g gg A B C D+ +ƒ

Initial mole 1 1 0 0At equilibrium 1-0.5 1-0.5 0.5 0.5

[ ] [ ][ ][ ] 0.5 0.5

10.5 0.5C

C DK

A B×= = =×

133. If the solubility of ( )3 4 2Ca PO in water is ‘X’ mol1 L− , its solubility product in 5 5mol L− is

1)5

6 X 2)5

36 X 3)5

64 X 4)5

108 X KEY:- 4Hint :- ( ). . x y x yspK x y s

+= Here solubility = XFor ( )3 4 2 ; 3& 2Ca po x y= =

( )3 23 2 53 .2 . 108spK X X += =

134. Which one of the following is not a method to remove permanent hardness of water?1) Clark’s method 2) Calgon method3) Ion-exchange method 4) Synthetic resins method

KEY:- 1Hint :- Theoretical


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135. White metal is an alloy of

1) Na, Mg 2) Na, Pb 3) Li, Mg 4) Li, PbKEY:- 4Hint :- Theoretical 136. Which one of the following elements does not form triiodide on reacting with iodine?

1) B 2) Tl 3) Al 4) GaKEY:- 2Hint :- Theoretical ; Inertpair effect

Tl in +3 state unstable 137. The buffer system which helps to maintain the pH of blood between 7.26 to 7.42 is

1) 2 3 3 / H CO HCO− 2) 4 4 / NH OH NH Cl

3) 3 3 / CH COOH CH COO− 4) 3 4CH COONH

KEY:- 1

Hint :- Theoretical 138. Municipal sewage BOD values (ppm) are________1) 1 5− 2) 100 4000− 3) 50 90− 4) 20 40−

KEY:- 2Hint :- Theoretical 139. The two bonds N = O and N – O in 3 2 H CNO are of same bond length due to_____

1) Inductive effect 2) Hyperconjugation3) Electromeric effect 4) Resonance effect

KEY:- 4Hint :- Theoretical

140. Assertion (A) : Reaction of 1-butene with HBr gives 1-bromobutane as major product.

Reason (R) : Addition of hydrogen halides to alkenes proceeds according toMarkovnikov’s rule.

The correct answer is

1) (A) and (R) are correct (R) is the correct explanation of (A)

2) (A) and (R) are correct but (R) is not the correct explanation of (A)

3) (A) is correct but (R) is not correct4) (A) is not correct but (R) is correct

KEY:- 4Hint :- Butene-1+HBr →2-Bromo butane ∴Assertion is wrong

141. The product (Z) of the following reaction is

3

3

CH Cl Anhydrous AlCl →

1)

Cl

2) Cl

3CH

3)

3CH

4)

Cl

Cl


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KEY: 2HINT:

33

Anh y AlClCH Cl+ − →

3CH

142. An example of covalent solid is(1) MgO (2) Mg (3) SiC (4) 2CaF

KEY: 3HINT: Silicon carbide is covalent solid

143. What is the weight (in g) of 2 3 Na CO (molar mass =106) present in 250ml of its 0.2M solution?(1) 0.53 (2) 5.3 (3) 1.06 (4) 10.6

KEY: 2HINT:

1000.

wt M

GM wt vol inml= ×

10000.2

106 250wt = ×

106 0.25.3

4wt

×= =

144. An aqueous dilute solution containing non-voltile solute boils at 100.052 0 c What is the molalityof solution? ( 10.52 . . ;bK kg mol K

−= boiling temperature of water 0100 c= )(1) 0.1m (2) 0.01m (3) 0.001m (4) 1.0m

KEY: 1

HINT:.b bT K m∆ = 0.b b bT T T ∆ =

3

2

0.052 52 100.1

0.52 52 10b

b

T m

K

−

−∆ ×= = = =×

145. A lead storage battery is discharged. During the charging of this battery, the reaction that occursat anode is

(1) ( ) ( ) ( )24 42PbSO s e Pb s SO aq− −+ → + (2) ( ) ( ) ( ) ( ) ( )24 2 2 42 4 2PbSO s H O l PbO s SO aq H aq e− + −+ → + + + (3) ( ) ( ) ( )2 24 4PbSO s Pb aq SO aq+ −→ + (4) ( ) ( ) ( ) ( ) ( )24 2 2 42 2 2PbSO s H O l e PbO s SO aq H aq− − ++ + → + +

KEY: 2HINT:At anode ( ) ( ) ( ) ( ) ( )

22 44 22 4 2s l s aq aqPbSO H O PbO SO H e

− + −+ → + + +


35/39

146. For the reaction ( ) ( ) ( ) ( ) ( )3 2 25 6 3 3 Br aq BrO aq H aq Br aq H O l− − ++ + → + if

1 10.05 min , Br

molLt

−− − ∆ − =∆

3 1 1min BrO

inmolL ist

−− − ∆ − ∆

(1) 0.005 (2) 0.05 (3) 0.5 (4) 0.01 KEY: 4HINT:

315

d Br d BrO

dt dt

− − − = −

30.01

d BrO

dt

− =

147. Which one of the following is used in the hardening of leather ?(1) Light sensitive silver bromide in gelatin (2) Sodium lauryl suphate(3) Alum (4) Tannim

KEY: 4HINT: Tannin

148. German silver contains which of the following metals?(1) Cu, Zn (2) Fe, Zn (3) Zn, Fe, Ni (4) Cu,Zn,Ni

KEY: 4

HINT:The composition of German silver , ,Cu Zn Ni−

149. The key step in the manufacturing of 2 4 H SO by contact process is

(1) Absorption of 3SO in 2 4 H SO to give oleum(2) Dilution of oleum with water(3) Burning of sulphur in air to generate 2SO

(4) Catalytic oxidation of 2SO with 2O to give 3SO

KEY: 4HINT: Conceptual

150. Ammonia on reaction with chlorine forms an explosive 3 NCl . What is the mole ratio of 3 NH and

2Cl required for this reaction?(1) 8 : 3 (2) 1 : 1 (3) 1 : 3 (4) 10 : 1

KEY: 3HINT:

3 2 33 3 NH Cl NCl HCl+ → + 151. Which one of the following lanthanide ions does not exhibit paramagnetism?

(1) 3 Lu + (2) 3Ce + (3) 3 Eu + (4) 3Yb + KEY: 1

HINT:3 14 0 04 5 6 Lu f d s+ →

152. The increasing order of field strength of ligands is

(1) 3 2 NH H O Cl CO CN − −

< < < < (2) 2 3Cl H O NH CN CO− −

< < < < (3) 3 2Cl CO NH Cl H O− −< < < < (4) 3 2CN CO NH Cl H O− −< < < < KEY: 2


36/39

HINT:

2 3Cl H O NH CN CO− −< < < <

153. Identify condensation homopolymer from the following

1)

2CH

OH

2CH

OH

2CH

2)

2 2OCH CH O CO− − − CO ][ n

3) ( )2 5 nCO CH NH − − − −

s

4) ( ) ( )2 26 4 n NH CH NH CO CH CO − − − − − − −

KEY: 3HINT:

( )2 5 nCO CH NH − − − −

154. Identify the nucleoside from the following

1)

O Base

H

H H

H

2 HOH C

OH OH 2)

O Base

H

H H

H

2O P O H C − − −

OH OH

O

O

3)

N

N N

N

H

2 NH

4)

OOH

H

H H

H

2 HO H C −

OH OH

KEY: 1HINT:


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O

Base

H

H H

H

2 HOH C

OH OH

155. Which one of the following is the correct structrure of sulphapyridine ?

1)

2 2SO NH

2 NH 2)

2 H N S

O

O

NH −

N

3)

N N OH

4)

2 NH NH

KEY: 2HINT:

2 H N S

O

O

NH − N

156. Identify Z in the following reaction

2 N H

21 . , , 2 7 3 2782 .

Na NO H C l K K l

− →

1)

2 NO

2)

Cl

3)

I

4) I

2 NH

KEY: 3HINT:

2

273 278

Na NO HC l

K

+− → 2 N C l

+ − K I

→

I 2 N C l+ −


38/39

157. Which intermediate is formed in the Reimer-Tiemann reaction?(1) Aldehyde (2) Carbocation(3) Carbanion (4) Substituted benzal chloride

KEY: 4HINT:

OH

3CHCl NaOH + →

ONa −

2CHCl

158. Which one of the following is an acetal ?

1)

R CH −

'OR

'OR 2)

R CH −

OH

'OR 3)

C

OH

'OR

R

R 4)

C

'OR

'OR

R

R

KEY: 1HINT:

1C O R O H = + − − → C R

H

OR

OH

1 R OH − → C

H

1OR

1OR

R

159.2 5 2 2

3 2 2P O H O SOCl H CCH CO H X Y Z ∆ → → → identitfy X,Y and Z

X Y Z(1) 2 2 H C CHCO H = 2 2 HOH CCHOHCO H 2 HOH CCHOHCOCl (2) ( )3 2 2 H CCH CO O 3 2 2 H CCH CO H 2ClCCH COCl (3) ( )3 2 H CCO O 3 2 H CCO H 2ClCH COCl (4) ( )3 2 2 H CCH CO O 3 2 H CCO H 3 H CCOCl

KEY: 2HINT:


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2 5 / 3 2

P OCH CH C OH ∆− − − → O3 2 H C CH C − − OH

O

3 2 H C H C C − −O

O

23 22

H O CH CH COOH → − −

2SOCl

3 2CH CH C Cl− − −

O

160. 3 2 2 2 3 24 2 2 H CCONH Br NaOH Y Na CO NaBr H O+ + → + + + what is Y in the reaction?

(1) 3 2 2 H CCH NH (2) 3 2 H CNH (3) 3 H CCOBr (4) 2 HCONH KEY: 2

HINT:

3 2 2 3 2 2 3 24 2 2CH C NH Br NaOH CH NH NaBr Na CO H O− − + + → − + + +

O

ts eamcet 2015 engineering question paper key solutions

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