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TS EAMCET 2016 Engineering Question Paper with Solutions
NARAYANA JUNIOR COLLEGE AP & TELANGANA TS-EAMCET-2016 (ENGINEERING)
MATHEMATICS 1.
2
If f(x) = x – 2x + 4 then the set of values of x satisfying f(x – 1) = f(x + 1) is 1) {–1}
2) {–1, 1}
3) {1}
4) {1, 2}
Key: 3 Hint: 2.
x 1
2
2
2 x 1 4 x 1 2 x 1 4 x 1
The number of real linear functions f(x) satisfying f(f(x)) = x + f(x) is 1) 0
2) 4
3) 5
4) 2
Key:4 Hint: Let f x ax b x 3.
ab 1 a a2
The remainder when 7n – 6n – 50 (n ) is divided by 36, is 1) 22
2) 23
3) 1
4) 21
Key: 2 Hint: Let f n 7 n 6n 50 Then f 1 49 hence the remainder is 23 4.
Consider the system of equations ax + by + cz = 2 bx + cy + az = 2 cx + ay + bz = 2 Where a, b, c are real numbers such that a + b + c = 0 1) has two solutions
2) is inconsistent
3) has unique solution
4) has infinitely many solutions
Key: 2 Hint: A AD 5.
Suppose A and B are two square matrices of same order. If A, B are symmetric matrices, then AB – BA is 1) a symmetric matrix
2) a skew symmetric
3) a scalar matrix
4) a triangular matrix
Key:2 T T T Hint: AB BA AB BA B T AT AT B T BA AB AB BA
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
6.
CODE–A
x 1 2 x 1 3x 1 If A x 2 x 1 3 x 1 x 1 then 3x 1
x 1
2x 1
1) –15
2)
1
A x dx 0
15 2
3) –30
4) –5
Key:2 6 x 3 6x 3 6x 3 Hint: By apply R1 R1 R2 R3 A x 2 x 1 3x 1 x 1 3x 1
x 1
2x 1
again C2 C2 C1 , C3 C3 C1 we get A x 6c 3 3x 2 1
7.
If z = x + y is a complex number such that z 3 a ib , then the value of 1) –1
2) –2
1 x y a b2 a b
3) 0
2
4) 2
Key: 2 3 3 2 Hint: Z a ib a 3 ib 3a 2 ib 3a ib X iY a3 3ab2 i 3a 2b b3
x y a 2 3b 2 , b 2 3a 2 a b 8.
The locus of z satisfying | z | + |z – 1| = 3 is 1) a circle
2) a pair of straight lines
3) an ellipse
4) a parabola
Key: 3 Hint: Compare with PA PB K We have AB K 9.
If the point z = (1 + i) (1 + 2i) (1 + 3i) ……. (1 + 10i) lies on a circle with centre at origin and radius r, then r2 = 1) 10 !
2) 2 × 3 × 4 × ………. × 10
3) 2 × 5 × 10 × ….. × 10!
4) 11!
Key: 3 2
Hint: z 0 2.5.10.....101 10.
The minimum value of | z – 1| + | z – 5 | is 1) 5
2) 4
3) 3
4) 2
Key: 2 Hint: put z 1, GE 4 x 5, GE 4 11.
The number of real roots of | x |2 – 5 | x | + 6 = 0 is 1) 2
2) 3
3) 4
4) 1
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
CODE–A
Key: 3 Hint: x 3, x 2 12.
If , are the roots of x2 – x + 1 = 0 then the quadratic equation whose roots are 2015, 2015 is 1) x2 – x + 1 = 0
2) x2 + x + 1 = 0
3) x2 + x – 1 = 0
4) x2 – x – 1 = 0
Key: 1 Hint: , 2
2015 2 , 2015 13.
If , , are roots of x3 – 5x + 4 = 0 then (3 + 3 + 3)2 = 1) 12
2) 13
3) 169
4) 144
Key:4 Hint: 0 , a 5, 4
3 3 3 3 2 2 2 0
3 3 3 3 12 14.
Suppose , , are roots of x3 + x2 + 2x + 3 = 0. If f(x) = 0 is a cube polynomial equation whose roots are + , + , + then f(x) = 1) x3 + 2x2 – 3x – 1
2) x3 + 2x2 – 3x + 1
3) x3 + 2x2 + 3x – 1
4) x3 + 2x2 + 3x + 1
Key: 3 Hint: y 1 1 x x 1 y
15.
Sub in GE
The number of 4 letter words that can be formed with the letters in the word EQUATION with at least one letter repeated is 1) 2400
2) 2408
3) 2416
4) 2432
3) 64
4) 60
Key: 3 Hint: n r n Pr 84 8 P4 16.
The number of divisors of 7! Is 1) 24
2) 72
Key: 4 Hint: 7! 1.2.3.4.5.6.8 2 4.32.5.7 No of devisors 5.3.2.2 60 2
17.
3
2 1 2 5 1 258 1 The sum of the series 1 ....... 3 8 3 6 5 3 6 9 8
1)
4 3 49
3
2)
49 4
3)
4 3 81
3
4)
81 4 Page 3
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
CODE–A
Key: 1 2
x P(P q) x Hint: G.E 1 P ..... q 2! q P=2, Q=3, X= 1 8
18.
= 1 x
p q
If Cr denotes the binomial coefficient nCr then (–1) C02 2C12 5C22 ...... 3n 1 Cn2 2n
1) 3n 2 Cn
3n 2 2 n 2) Cn 2
3) 5 3n 2 n Cn
3n 5 2 n 4) Cn 1 2
3) 2A + C
4) 2A + 2C
Key: 2 Hint: Put n 1 , verify 19.
x 1 A B C D E 2 3 4 B DE x x 2 x x x x x2 4
1) A + C
2) A – C
Key: 1 Hint: x 1 Ax 3 Bx 2 Cx D x 2 Ex 4
2 4 1 1 E A 1 , B , C , D 16 16 16 2 16 20.
2 4 If cos2 cos3 cos3 a cos 3, then a = 3 3 1)
1 4
2)
3 4
5 4
4)
7 4
3) 0
4)
1 2
3)
Key: 2 Hint: put 0 21.
cos13 sin13 1 cos13 sin13 cot148 1) 1
2) –1
Key: 3 Hint: tan(45 – 13) – tan 32 = 0 22.
x y If cos x + cos y + cos = 0 and si x + sin y + sin = 0, then cot 2 1) sin
2) cos
3) tan
4) cot
Key. 4 Hint: cos x + cos y = - cos , sin x + sin y = -sin using transformations. 23.
If f(x) = cos2 x + cos2 2x + cos2 3x, then the number of values of x [0, 2] for which f(x) = 1 is 1) 4
2) 6
3) 8
4) 10
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
CODE–A
Key: 4 Hint: sin 2 x cos 2 2x cos 2 3x
cos2 3x sin 2 x cos 2 2x 0 cos 4x cos 2x cos 2 2x 0 cos x 0 cos 2x 0 cos 3x 0 x 2n 1 x 2n 1 x 2n 1 , n z 2 4 6 24.
The value of x which satisfies sin (cot–1 x) = cos(tan–1 (1 + x)) is 1)
1 2
2)
1 2
3) –1
4) 1
3) log tan 4 2
4) log tan 4 2
Key: 1 HINT: sin (cot-1 x ) = cos tan-1 (1 + x) 1 1 x
2
1 (1 x) 2 1
1 + x2 = (1 + x)2 + 1 2x + 1 = 0 x = 1/2 25.
For 0, ,sec h 1 cos 2 1) log tan 6 2
2) log tan 3 2
Key: 3 1 1 x 2 HINT: sech 1 x log x 26.
If ABC is such that A 90, B C, then 1)
1 3
2)
1 2
b2 c 2 sin B C b2 c2 3) 1
4)
3 2
Key: 3 HINT: Using sine rule 27.
In ABC, if 8R2 = a2 + b2 + c2, then the triangle is a 1) right angled triangle
2) equilateral triangle
3) scalene triangle
4) obtuse angled triangle
Key: 1 HINT: sin2 A + sin2 B + sin2C = 92. Right angle triangle 28.
In ABC, if 2R + r = r2, then B = Page 5
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
1)
3
2)
CODE–A
4
3)
6
4)
2
Key: 4 HINT: r2 – r = 4R sin2 B/L 29.
ABCDEF is a regular hexagon whose centre is O. Then AB AC AD AE AF is 1) 2AO 2) 3AO 3) 5AO 4) 6AO
Key: 4 HINT: AE BD
AF CD AD 2 AO 30.
ABCD is a parallelogram and P is the mid point of the side AD. The line BP meets the diagonal AC is Q. Then the ratio AQ : QC = 1) 1 : 2
2) 2 : 1
3) 1 : 3
4) 3 : 1
Key: 1 HINT: Take A be the origin AB b , AD d
31.
AP
d b d , AC 2 2
OQ
(b d ) d b , OQ Equate 1 1
The vectors 2i 3 j k , i 2 j 3k , 3i j 2k 1) are linearly dependent
2) are linearly independent
3) form sides of a triangle
4) are coplanar
Key: 2
2 3 1 HINT: 1 2 3 0 3 1 2 32.
a, b, c are three vectors such that a 1, b 2, c 3 and b, c are perpendicular. If projection of
b on a is the same as the projection of c on a , then a b c 1)
2)
2
7
3) 14
4)
21
Key: 3 HINT: b .c 0 &
b .a c .a |a | |a |
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
33.
If a, b, c are unit vectors satisfying the relation a b 3 c 0 , then the angle between
a and b is 1) 6 Key: 3 HINT: a b 3c S.O.B.S
34.
CODE–A
2)
4
3)
3
4)
a is perpendicular to both b and c . The angle between b and c is
2
2 . If a 2, b 3, c 4, 3
then c. a b 1) 18 3 2) 12 3 Key: 2 HINT: a || b c , c .(a b ) a .(b c ) . 35.
3) 8 3
4) 6 3
If the average of the first n numbers in the sequence 148, 146, 144,……. Is 125, then n = 1) 18
2) 24
3) 30
4) 36
Key: 2 HINT:
a a d ........ a ( n 1)d 125 n
a
( n 1) d 125 2
2a + (n – 1)d = 250 102 + (n – 1)2 = 148 36.
The standard deviation of a, a + d, a + 2d, …….. a + 2nd is 2) n2d
1) nd
3)
n n 1 d 3
4)
n n 3 d 3
Key: 3 HINT: x a nd
2 37.
1 ( xi x ) 2 n 1
1 1 1 Two events A and B are such that P A , P A | B and P B | A . Consider the 4 4 2 following statements:
I) P A | B
3 4
II) A and B are mutually exclusive
III) P(A | B) + P(A | B ) = 1 . Then 1) only (I) is correct
2) only (I) and (II) are correct
3) only (I) and (III) are correct
4) only (II) and (III) are correct
Key: 1
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
CODE–A
1 1 1 5 HINT: P ( A) , P ( B ) ,P ( A B ) , P ( A B ) 4 2 8 8
A 1 P( A B) 3 I) P 1 P( B) 4 B II) P ( A B ) 0 III) 38.
P( A B) P( A B ) 1 P( B) (B )
A five digit number is formed by the digits 1, 2, 3, 4, 5 with no digit being repeated. The probability that the number is divisible by 4, is 1)
1 5
2)
2 5
3)
3 5
4)
4 5
Key: 1 4
HINT: Required probability = 39.
5
P3 P4
When a pair of six faced fair dice are thrown, the probability that the sum of the numbers on the two dice is greater than 7, is 1)
1 3
2)
5 12
3)
1 2
4)
1 4
Key: 2 15 HINT: Required probability = 36
40.
x:
8
9
10
11
12
P( x ):
5 36
4 36
3 36
2 36
1 36
In a family with 4 children, the probability that there are at least two girls is 1)
1 2
2)
9 16
3)
3 4
4)
11 16
Key: 4 HINT: P(BBGG) + P(GGGB) + P(GGGG) 6 4 1 11 16 16 16 16 41. On an average nine out of 10 ships that have departed at A reach B safely. The probability that out of five ships that have departed at A at least four will reach B safely is 1) 14 (0.9)5 2) 1.4 (0.9)5 3) 0.14 (0.9)4 4) 1.4 (0.9)4 Key: 4 9 1 Hint : p , q , n 5 10 10 P X 4 P X 4 P X 5
42.
If A(5, –4) and B(7, 6) are points in a plane, then the set of all points P(x, y) in the plane such that AP : PB = 2 : 3 is
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
1) a circle Key: 1 Hint : P x, y
CODE–A
2) a hyperbola
3) an ellipse
4) a parabola
AP 2 PB 3
9 AP 2 4 PB 2 If the axes are rotated anticlockwise through an angle 90° then the equation x2 = 4ay is changed to the equation 1) y2 = 4ax 2) x2 = –4ay 3) y2 = –4ax 4) x2 = 4ay Key: 3 Hint : x y , y x
43.
44.
The combined equation of the straight lines of the form y = kx + 1 (where k is an integer) such that the point of intersection of each with the line 3x + 4y = 9 has an integer as its x-coordinate is 1) (y + x + 1) (y + 2x – 1) = 0 2) (y + x – 1) (y + 2x + 1) = 0 3) (y + x + 1) (y + 2x + 1) = 0 4) (y + x – 1) (y + 2x – 1) = 0 Key: 4
5 4k 3 45. A value of k such that the straight lines y – 3kx + 4 = 0 and (2k – 1) x – (8k – 1)y – 6 = 0 are perpendicular is 1 1 1) 2) 3) 1 4) 0 6 6 Key: 1 Hint : x
Hint : a1a2 b1b2 0 46.
The length of the segment of the straight line passing through (3, 3) and (7, 6) cut off by the coordinate axes is 1)
Key:
4 5
2)
5 4
Key:
7 4
4)
4 7
2
Hint : Find the line equation in the form 47.
3)
x y 1 and find a b
a2 b2
The equation of the pair of straight lines through the point (1, 1) and perpendicular to the pair of straight lines 3x2 – 8xy + 5y2 = 0 is 1) 5x2 + 8xy + 3y2 – 14x – 18y + 16 = 0
2) 5x2 + 8xy + 3y2 – 18x – 14y + 16 = 0
3) 5x2 – 8xy + 3y2 – 18x – 14y + 32 = 0
4) 5x2 – 8xy + 3y2 – 14x – 18y + 32 = 0
2 2
2
Hint : 5 x 1 8 x 1 y 1 3 y 1 0
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
48.
CODE–A
The combined equation of the three sides of a triangle is (x2 – y2) (2x + 3y – 6) = 0. If the point (0, ) lies in the interior of this triangle then 1) 2 0
2) 2 2
3) 0 2
4) 2
Key: 3 Hint : Given line equations are x = y, x = -y, 2x + 3y – 6 = 0 and draw the diagram 49.
The point where the line 4x – 3y + 7 = 0 touches the circle x2 + y2 – 6x + 4y – 12 = 0 is 1) (1, 1)
Key:
2) (1, –1)
3) (–1, 1)
4) (–1, –1)
3
Hint : Find the foot of perpendicular from centre 3, 2 to the line 4x – 3y + 7 = 0 50.
The normal to the circle given by x2 + y2 – 6x + 8y – 144 = 0 at (8, 8) meets the circle again at the point 1) (2, –16)
2) (2, 16)
3) (–2, 16)
4) (–2, –16)
Key: 4 Hint : If A(8, 8), C(3, -4) let other end B = 2C – A 51.
For all real values of k, the polar of the point (2k, k – 4) with respect to x2 + y2 – 4x – 6y + 1 = 0 passes through the point 1) (1, 1)
2) (1, –1)
3) (–3, 1)
4) (3, 1)
Key: 4 Hint : Find S1 = 0
L1 kL2 0 solve L1 = 0, L2 = 0 52.
If the circles x2 + y2 – 2x – 2y – 7 = 0 and 3(x2 + y2) – 8x + 29y = 0 are orthogonal then = 1) 4
Key:
2) 3
3) 2
4) 1
4
Hint : 2 gg 2 ff c c 53.
The radical centre of the circles x2 + y2 = 1, x2 + y2 – 2x – 3 = 0 and x2 + y2 – 2y – 3 = 0 is 1) (1, 1)
2) (1, –1)
3) (–1, 1)
4) (–1, –1)
Key: 4 Hint : Solve S S 0, S S 0 54. From a point (C, 0) three normals are drawn to the parabola y2 = x. Then 1 1 1 1 1 1) C 2) C 3) C 4) C 2 2 2 2 4 Key: 3 Hint : Equation of normal y = mx – 2am – am3 Substitute (c, 0) 55. The points of intersection of the parabolas y2 = 5x and x2 = 5y lie on the line 1) x + y = 10 2) x – 2y = 0 3) x – y = 0 4) 2x – y = 0 Key: 3 Hint : Solving y2 = 4x and x2 = 5y we get (0, 0) and (5, 5) Above two points passing through x – y = 0
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
56.
For the ellipse give by
x 3
2
25
CODE–A
y 2 16
2
1 , match the equations of the lines given in
List–I
List–II
i) The equation of the major axis
a) 3x = 34
ii) The equation of a directrix
b) y = 2
iii) The equation of a latus rectum
c) x + y = 9 d) x = 6 e) x = 3 f) 3y = 34
The correct matching is (i)
(ii)
(iii)
1)
(e)
(a)
(d)
2)
(b)
(f)
(e)
3)
(b)
(a)
(e)
4)
(b)
(a)
(d)
Key: 4 Hint : Equation of major axis y = 2; Equation of directrix 3x = 34 Equation of lotusrectum x = 6 57.
If S and S are the foci of the ellipse
x2 y 2 1 and if PSP is a focal chord with SP = 8 then SS 25 16
= 1) 4 + SP Key:
2) SP – 1
3) 4 + SP
4) SP – 1
1
Hint : SS 2ae = 6; SP S P 2a 10 ; S P 2 58.
Let A(2 sec , 3 tan) and B(2sec, 3tan) where
, be two points on the hyperbola 2
x2 y2 1 . If (, ) is the point of intersection of normals to the hyperbola at A and B, then = 4 9 1) Key:
13 3
2)
13 3
3)
3 13
4)
3 13
1
Hint : Solve equation of normals 59.
33 28 38 Points A(3, 2, 4), B , , and C(9, 8, 10) are given. The ratio in which B divides AC is 5 5 5 1) 5 : 3
2) 2 : 1
3) 1 : 3
4) 3 : 2
Key: 4 Hint : use x1 x : x x2
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
60.
CODE–A
C 1 2 If the angle between the lines whose direction cosines are , , and 21 21 21 3 6 3 , , is 2 then the value of C is 54 54 54 1) 6
Key:
2) 4
3) –4
4) 2
2
Hint : l1l2 m1m2 n1n2 = 0 61.
The image of the point 5, 2, 6 with respect to the plane x y z 9 is
7 2) , 1, 5 2
1) 3, 5, 2
7 2 10 3) , , 3 3 3
7 2 5 4) , , 3 3 3
3) e 4
4) e 2
Key: 3 Hint: Let (h,k,r) is image Then
h5 k 2 r 6 5 269 2 1 1 1 111
h5 k 2 r 6
8 8 7 h 5 3 3 3
8 2 2 3 3 8 10 r 6 3 3 7 2 10 h, k , r , , 3 3 3 k
x
x2 x 3 62. lim 2 x x x 2 1) Key: 4 x
Lt x2 x 3 x e Hint: Lt 2 x x x 2 2 e
63.
2) e x2 x 3 x 2 1 x x 2
e
2 x 1 Lt x 2 x x 2
x
p sin 1 sin x x , The values of p and q so that the function f x q , sin 2 x e sin 3 x , Is continuous at x 0 is 1 2 1) p , q e 2/ 3 2) p 0, q e 2 / 3 3) p , q e 2/ 3 3 3
x0 6 x0 0 x
6
2 4) p , q e 2 /3 3
Key: 4 Page 12
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
CODE–A
P
Hint: Lt f x Lt 1 sin x sin x e P x 0
x 0
sin 2 x
Lt f x Lt e sin3 x e 2/3
x 0
x 0
2 , q e 2/3 3 5cos x 12sin x dy If y tan 1 , then dx 12cos x 5sin x
Let e P e 2/3 q p 64.
1) 1
2) -1
3) -2
4)
1 2
Key: 2
5cos x 12sin x Hint: y tan 1 12cos x 5sin x 5 tan x y tan 1 12 5 1 tan x 12
y tan 1 tan x y x
dy 1 dx
1 sin x 1 sin x d tan 1 dx 1 sin x 1 sin x
65.
2)
1) 1
1 2
3)
1 2
4) -1
Key: 3 Hint:
1 sin x 1 sin x d tan 1 dx 1 sin x 1 sin x
1 cos x 1 cos x d 2 2 tan 1 dx 1 cos x 1 cos x 2 2 x x cos sin d 4 2 4 2 tan 1 x dx cos sin x 4 2 4 2 x d tan 1 tan dx 4 4 2
d tan 1 dx
x 1 tan 2 2
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
66.
CODE–A
If y a cos sin 2 x b sin sin 2 x , then y '' 2 tan 2 x y ' 1) 0
2) 4 cos 2 2x y
3) 4 cos 2 2x y
4) cos 2 2x y
Key: 3 Hint: y1 a sin sin 2 x cos 2 x 2 b cos sin 2 x cos 2 x 2 y11 4a cos 2 2 x cos sin 2 x sin 2 x sin sin 2 x 4b sin sin 2 x cos 2 x sin 2 x cos sin 2 x y11 2 tan 2 x y1
4 cos 2 2x y
The length of the segment of the tangent line to the curve x a cos3 t , y a sin 3 t , at any point on the curve cut off by the coordinate axes is 1) 4a 2) a 3) a 2 4) 2a Key: 2 Hint: x a cos3 t , y a sin 3 t dx dy 3a cos 2 t sin t 3a sin 2 t cos t dt dt dx sin t dt cos t sin t Tangent is y a sin 3 t x a cos3 t cos t x sin t y cos t a sin t cos t x y 1 AB a 2 cos2 t a 2 sin 2 t a a cos t a sin t 68. The area of the triangle formed by the positive x-axis, the tangent and normal to the curve
67.
x 2 y 2 16a 2 at the point 2 2a, 2 2 is
1) a 2 2) 16 a 2 3) 4 a 2 Key: 4 dy Hint: 1 m dx y2 1 8a 2 8a 2 1 1 2 8a 2 Req. area = 1 m 2 m 2 2 1 69. Define f x sin x sin x , 0 x 2 . Then f is 2 3 1) increasing in , 2 2 2) decreasing in 0, and increasing in , 2 2 3) increasing in 0, and decreasing in , 2 2 4) increasing in 0, and decreasing in , 4 4 Key: 3
4) 8 a 2
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TS EAMCET 2016 Engineering Question Paper with Solutions
EAMCET QP | KEY & SOLUTIONS
CODE–A
1 sin x sin x 2 0 x sin x sin x
Hint: f x
1 2sin x 2 f x sin x f x
f 1 x cos x f 1 x 0 cos x 0 Increa sin g 0, Decreasing , 2 2 70.
The smallest value of the constant m 0 for which f x 9mx 1 1)
1 9
2)
1 16
3)
1 36
1 0 for all x 0 , is x 1 4) 81
Key: 3
1 0 9mx2 x 1 0 x 1 0 1 36 m 0 m 36 2 x 1 dx x4 7 x2 1 x2 1 1 1) tan 1 c 3 3x
Hint: f x 9mx 1
71.
3)
x2 1 1 tan 1 c 3 x
x2 1 2) tan 1 c x x2 1 1 4) tan 1 c 3 3x
Key: 1
x2 1 dx Hint: 4 x 7 x2 1 1 1 2 1 x t 2 x 1 dx x x 7 2 x dt t2 2 7 2 dt 1 1 1 t 1 x 1 Tan Tan c t2 9 3 3 3 3x 72.
x3 1 x2
dx
3/ 2 x 1 x2 c 3 3/ 2 2 1 x 2 1 x 2 c 3
3/ 2 2 1 x2 c 3 1/ 2 1 1 x 2 1 x 2 c 3
1) 1 x 2
2) 1 x 2
3) x 2
4) x 2
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CODE–A
Key: 3 Hint:1 x 2 t 2 2 xdx 2tdt xdx tdt x 3 dx
1 x2 t 2 1 dt
t2 2 t 1 dt dx
cos x 4 cos x 2
73.
sec x 2 1 log c 2 sec x 4
1)
1 2 log cos x 4 c sin 2
2)
3)
sec x 4 1 log c sin 2 sec x 2
4) log
sec x 4 c sec x 2
Key: 3 Hint:
dx
cos x 4 cos x 2 sin x 4 x 2 1 sin 2 cos x b cos x 2
1 Tan x y Tan x z dx sin 2 sin x 4 1 log sin 2 sin x 2 2x 2 x 2 4 x 5 dx
74.
1)
x2 4 x 5 log x x 2 4 x 5 c
2) log
x2 4 x 5 x2 4 x 5 c
3)
x2 4 x 5 6log x 2 x2 4 x 5 c
4) 2 x 2 4 x 5 6log x 2 x2 4 x 5 c
Key: 4 Hint:
2x 2 x2 4x 5 2x 4 x2 4x 5
dx 6
2 x 2 4 x 5 b
2x 4 6 dx x2 4x 5 dx x2 4 x 5 dx
x 2
2 x 2 4 x 5 b log
2
9
x 2
x2 4 x 5
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/4
75.
0
1)
CODE–A
sin x cos x dx 7 9 sin 2 x
log 3 4
2)
log 3 36
3)
log 7 12
1 log 7e 24
4)
log 7 24
Key: 4 /4
sin x cos dx 0 7 9 sin 2 x Let sin x cos x t cos x sin x dx dt
Hint:
0
0
0 1 dt dt dt 1 7 9 1 t 2 1 16 9t 2 9 1 4 2 2 t 3
4 t 1 1 log 3 4 93 4 t 2 3 3
0
1 1 log1 log 24 7
2)
2
3)
1
/4
76.
0
tan x cot x dx
1)
2
3 2
4)
Key: 1 /4
Hint:
Tanx cot x dx
0 /4
sin x cos x sin x cos x 0 Let sin x cos x t cos x sin x dx dt
0
1
dt 1 t2 2
2 sin 1 t
0 1
2 0 2 2 77. If the area bounded by the curves y ax 2 and x ay 2 , a 0 is 3 sq. units, then the value of a is 2 1 1) 2) 3) 1 4) 4 3 3 Key: 2
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CODE–A
Hint: y ax 2 , x ay 2 1 1 x 2 y, y 2 x a a 1 1 16 4a 4a 3 3 1 1 2 3 a 3a 3 78. Let p IR , then the differential equation of the family of curves y x e px , where , are arbitrary constants, is 2) y '' 2 py ' p 2 y 0 1) y '' 4 py ' p 2 y 0 3) y '' 2 py ' p 2 y 0 4) y '' 2 py ' p 2 y 0 Key: 2 Hint: y x e px y1 x pe px e px
y1 py e px y11 py1 pe px y11 py1 p y1 py
y11 2 py1 p 2 y 0 79.
The solution of the differential equation 3 xy ' 3 y x2 y 2
1/ 2
0 , satisfying the condition
y 1 1 is y 1) 3cos 1 ln x x y 3) 3cos 1 2ln x x Key: 1 Hint: 3 xy 3 y x y 1
3x
2
1 2 2
y 2) 3cos ln x x y 4) 3sin 1 ln x x
0
dy 3 y x2 y 2 dx
dy 3 y x 2 y 2 dx 3x Let y Vx dy dv V x dx dx
V x
dv 3Vx x 2 V 2 x 2 dx 3x
3V 1 V 2 3
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CODE–A
dv 1V 2 dx 3 dv dx 3 x 1V 2 x
3sin 1 V log x log c 1
y c 3sin 1 log x x Let x 1, y 1 y 3 1 3sin 1 log x 2 x y 3 sin 1 log x x 2 y 3cos 1 log x x 80.
The solution of the differential equation y '
1) x e y y c Key: 1 1 Hint: y1 y e x dy ey dx 1 xe y dx 1 xe y e y x y dy e dx s x e y dy 1dy I .F e e y
2) y e y x c
1 , is e x 3) x e y y c y
4) x y e y c
x.e y e y .e y dy
xe y y c x e y y c
PHYSICS 81.
Electron microscope is based on the principle 1) Photoelectric effect
2) Wave nature of electron
3) Superconductivity
4) Laws of electromagnetic induction
Key: 2 HINTS :
h . mv
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EAMCET QP | KEY & SOLUTIONS
82.
CODE–A
Force is given by the expression F A cos Bx C cos Dt where x is displacement and t is D time. The dimension of is same as that of B
1) Velocity
2) Velocity gradient
3) Angular velocity
4) Angular momentum
Key: 1 HINTS : Dt 1 Bx
D x L LT 1 velocity B t T 83.
A car accelerates from rest with 2 m/s2 on a straight line path and then comes to rest after applying brakes. Total distance travelled by the car is 100 m in 20 seconds. Then the maximum velocity attained by the car is 1) 10 m/s
2) 20 m/s
3) 15 m/s
4) 5 m/s
Key: 1 HINTS : S
1 2 t 2
1 2 100 20 20 2 2 2 4 2 / 3
Vmax
84.
t
2 2 2 20 2 20 3 3 10 m / sec 2 8 2 3 3
A body is falling freely from a point A at a certain height from the ground and passes through points B, C and D (vertically as show below) so that BC = CD. The time taken by the particle to move from B to C is 2 seconds and from C to D 1 second. Time taken to move from A to B in seconds is
1) 0.6
2) 0.5
3) 0.2
4) 0.4
Key: 2 HINTS : x
1 2 gt 2
x BC
1 2 g t 2 2
x 2 BC
1 2 g t 3 2
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BC
CODE–A
1 1 2 2 2 g t 3 t 2 g t 2 t 2 2 2
On solving t = 0.5 Sec. 85.
A particle moves from (1, 0, 3) to the point (-3, 4, 5), when a force F iˆ 5kˆ acts on it. Amount of work done in Joules is 1) 14
2) 10
3) 6
4) 15
Key: 3 HINTS : F iˆ 0 ˆj 5kˆ S 4iˆ 4 ˆj 2kˆ
W F . S 1 4 0 4 5 2 4 10 6J 86.
A particle is projected with velocity 2 gh and at an angle 600 to be horizontal so that it just clears two walls of equal height ‘h’ which are at a distance 2h from each other. The time taken by the particle to travel between these two walls is 1) 2
2h g
2)
h 2g
3) 2
h g
4)
h g
Key: 3 HINTS : 2h 2 gh cos 60 t
1 h h g t t 2 2 g 87.
A body of mass 20 kg is moving on a rough horizontal plane. A block of mass 3kg is connected to the 20 kg mass by a string of negligible mass through a smooth pulley as shown in the figure. The tension in the string is 27 N. The coefficient of kinetic friction between the heavier mass and the surface is g 10m / s 2 .
1) 0.025
2) 0.035
3) 0.35
4) 0.25
Key: 2 HINTS : 30 27 3a a 1 m / sec2
27 K 20 10 20 1
K 88.
7 0.035 200
Two masses m1 and m2 are placed on a smooth horizontal surface and are connected by a string of negligible mass. A horizontal force F is applied on the mass m2 as show in the figure. The tension in the string is
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m1 1) F m1 m2
2)
CODE–A
m 3) 1 F m2
m2 F m1 m2
4)
m2 F m1
Key: 1 m1 F m1 m2
HINTS : T m1 a 89.
A body of mass 3 kg moving with a velocity 2iˆ 3 ˆj 3kˆ m / s collides with another body of mass 4 kg moving with a velocity
3iˆ 2 ˆj 3kˆ m / s . The two bodies stick together after
collision. The velocity of the composite body is 1)
1 ˆ 4i 6 ˆj 3kˆ 7
3)
1 ˆ 6i 4 ˆj 6kˆ 7
2)
1 18iˆ 17 ˆj 3kˆ 7
4)
1 ˆ ˆ 9i 8 j 6kˆ 7
Key: 2 HINTS : V 90.
m1V1 m2V2 18iˆ 17 ˆj 3kˆ = 7 m1 m2
A simple pendulum of length L carries a bob of mass m. When the bob is at its lowest position, it is given the minimum horizontal speed necessary for it to move in a vertical circle about the point of suspension. When the string is horizontal the net force on the bob is 1) 10 mg
2)
5 mg
3) 4 mg
4) 1mg
Key: 1
1 1 m 5rg m v 2 mgr 2 2 2 5 v rg rg 2 2 3 v2 rg v 2 3rg 2 2 mv 2 Fx Fc 3mg r Fy mg
HINTS :
FNex Fx2 Fy2 mg 10
91.
A system of two particle is having masses m1 and m 2 . If the particle of mass m1 is pushed towards the center of mass of particles through a distance d, by what distance the particle if mass m 2 should be moved so as to keep the center of mass of particles at the original position?
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CODE–A
Key:- 3 Hint:- m1x1 m 2 x 2 1) 92.
m1 d m1 m 2
2) d
3)
m1 d m2
4)
m2 d m1
a thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its center and perpendicular to its plane with an angular velocity . 1 Another disc of same thickness and radius but mass M is placed gently on the first disc co8 axially. The angular velocity of the system is now 1)
9 8
2)
5 9
3)
1 3
4)
2 9
Key:- 1 Hint:- I11 I1 I 2 2 93.
93. 9 kg solution is poured into a glass U-tube as shown in the figure below. The tube's inner
m and the solution oscillates freely up and down about its position of 5 equilibrium (x = 0). The period of oscillation in seconds is (1 m3 of solution has a mass IA = 900 kg, g = 10 m/s', Ignore frictional and surface tension effects), 1) 0.1 Diameter is 2
2) 10
3) 4) 1 Key:- 1
Hint:- m 2 x pxA 94.
The bodies of masses 100 kg and 8100 kg are held at a distance of 1 m. The gravitational field at a point on the line joining them is zero. The gravitational potential at that point in J/kg is G 6.67 1011 N.m 2 / Kg 2 1) 6.67 10 7
2) 6.67 10 10
3) 13.34 10 7
4) 6.67 10 9
Key:- 1 Hint:- x
95.
d m2 1 m1
An elastic spring of unstretched length L and force constant K is stretched by a small length x. It is further stretched by another small length y. Work done during the second stretching is 1)
ky x 2y 2
2)
k 2x y 2
3) ky x 2y
4)
ky 2x y 2
Key:- 4
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Hint:- w 1 w2
1 2 kx 2
1 2 k x y 2
w 2 w1
96.
CODE–A
ky 2x y 2
A soap bubble of radius 1.0 cm is formed inside another soap bubble of radius 2.0 cm. The radius of an another soap bubble which has the same pressure difference as that between the inside of ihe smaller and outside of large soap bubble, in meters is 1) 6.67 103
2) 3.34 10 3
3) 2.23 10 3
4) 4.5 10 3
Key:- 1 Hint:- R 97.
R1R 2 R1 R 2
A slab of stone area 3600 cm2 and thickness 10 cm is exposed on the lower surface to steam at 100°C. A block of ice at 0°C rests on upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of the stone in js1 m 1 k 1 is (Latent heat of ice
3.36 105 J / Kg ) 1) 12.0
2) 10.5
3) 1.02
4) 1.24
Key:- 4 Hint:98.
mL kA t x
The surface of a black body is at a temperature 727°C and its cross section is 1 m2. Heat radiated from this surface in one minute in Joules is (Stefan's constant = 5.7 x 10-8 W/m2/k4) 1) 34.2 105
2) 2.5 105
3) 3.42 105
4) 2.5 10 6
Key:- 1 Hint:99.
E eT 4 Axt
Two moles of a gas is expanded to double its volume by two different processes. One is isobaric and the other is isothermal. If w1 and w2 are the works done respectively, then 1) w 2
w1 ln 2
2) w1 = w2
3) w1 = w2 ln2
4) w12 w 2 ln 2
Key:- 4 Hint:- Isobaric w1 p 2v v v w 2 nRT log e 2 v1
2v = pv log e v
w 2 w1 log e 2 Page 24
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CODE–A
100. Uranium has two isotopes of masses 235 and 238 units. If both of them are present in Uranium hexafluoride gas, find the percentage ratio of difference in rms velocities of two isotopes to the rms velocity of heavier isotope. 1) 1.64
2) 0.064
3)0.64
4)6.4
Key:- 3 Hint:- v rms
1 M
101. A source of frequency 340 Hz is kept above a vertical cylindrical tube closed at lower end. The length of the is 120 cm. Water is slowly poured in just enough to produce resonance. Then the minimum height (velocity of sound =340m/s) of the water level in the tube for that resonance is, 1) 0.75 m
2) 0.25
3) 0.95
4) 0.45
Key: 4 Hint : l0
v 340 0.25 25cm 4 f 4 340
h 120 3l0 45cm 102. A thin convex lens of focal length ‘f’ made of crown glass is immersed in a liquid of refractive index 1 1 c where c is the refractive index of the crown glass. The convex now is 1) A convex lens of longer focal length
2) A convex of shorter focal length
3) A divergent lens
4) A convex of focal length c l f
Key: 3 Hint : Conceptual 103. Two convex lenses of focal lengths f1 and f2 form images with magnification m1 and m2, when used individually for an object kept at the same distance from the lenses. Then f1/f2 is 1)
m1 1 m1
2)
m2 1 m2
m1 1 m2 m2 1 m1
3)
m2 1 m1 m1 1 m2
4)
m2 1 m2 m1 1 m1
Key: 2 Hint : u
m1 1 f1 m2 1 f 2 m1
m2
f1 m1 1 m2 f 2 m2 1 m1
104. With the help of a telescope that has an objective of diameter 200 cm, it is proved that light of 0
wavelengths of the order of 6400 A coming from a star be easily resolved. Then the limit of resolution is 1) 39 108 deg
2) 39 108 rad
3) 19.5 108 rad
4) 19.5 108 deg
Key: 2 Hint:
1.22 d
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CODE–A
105. Two charged identical metal spheres A and B repel each other with a force of 3 105 N. Another identical uncharged sphere C is touched with sphere A and then it is placed midway between A and B. Then the magnitude of net force on C is 1) 110 5 N
2) 3 105 N
3) 2 105
4) 5 105 N
Key: 2
kq 2 / r 2 3 105 Hint : FB FC q / 2.q q / 2.q / 2 k 2 r / 2 2 r / 2 3 105 1 1 F 2 1
106. The electrostatic potential inside a charged is given as V=Ar2 +B, where r is the distance from the center of the sphere; A and B are constants. Then the charge density in the sphere is 1)16A 0
2)-6A0
3)20A 0
4)-15A0
Key: 2
3E 0 3 2 Ar 0 qr r 6 A 0 = 3 r 3 0 R r
Hint: E k .
107. Three unequal resistances are connected in parallel. Two of these resistances are in the ratio 1:2. The equivalent resistance of these three connected in parallel is 1 . What is the highest resistance value among these three resistances if no resistance is fractional ? 1)10
2)8
3)15
4)6
Key: 4
1 1 1 1 Hint: 1 R1 2 R1 R3 1
3 1 2R1 R3
Verify options, R1 2, R2 3, R3 6 108. Two electric resistors have equal values of resistance R. Each can be operated with a power of 320 watts (w) at 220 volts. If the two resistors are connected in series to a 110 volts electric supply, then the power generated in each resistor is 1)90 watts
2)80 watts
3) 60 watts
4) 20 watts
Key: 2 Hint : PS
v2 P0 P0 nPS P0 2 n 2R
109. A current of 1 A is flowing along the sides of an equilateral triangle of side 4.5 102 . The magnetic field at the centroid of the triangle is 0 4 107 H / m 1) 4 105 T
2) 2 105 T
3) 4 104 T
4) 2 104 T
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CODE–A
Key: 2 Hint: B0 3Beach side =
30 sin 1 sin 2 4 r
110. A charged particle (charge=q: mass=m) is rotating in a circle of radius ‘R’ with uniform speed ‘V’. Ratio of its magnetic moment to the angular momentum (L) is 1)
q 2m
2)
q m
3)
q 4m
4)
2q m
Key: 1
niA q L mvr 2m 111. Two small magnets have their masses and lengths in the ratio 1:2. The maximum torques experienced by them in a uniform magnetic field are the same. For small oscillations, the ratio of their time periods is 1 1 1 1) 2) 3) 4) 2 2 2 2 2 2 Key: 1 Hint:
Hint: T 2
I I m 2t 2 MBH
112. Two coils have mutual inductance 0.005H. The current changes in the first coil according the equation I I 0 sin t , where I 0 10 A and 100 rad S 1 . The maximum value of emf in the second coil is 1) 5 2) 5 3) 0.5 4) Key: 2 di Hint: e M dt F
10 3 100 113. A capacitance of mH and a resistance of 10 are and an inductance of 2 connected in series with an AC voltage source of 220V, 50Hz. The phase angle of the circuit is 1) 600 Key: 3
2) 300
3) 450
4) 900
XL XC R 114. Two equations are given below Q (A) E.d A ; (B) B.d A 0 0 Hint: Tan
1) (A) - Ampere’s law 2) (A) – Gauss law for electric fields 3) (A) – Faraday law 4) Both (A) and (B) represent Faraday law
(B) – Gauss law for electricity (B) – Gauss law for magnetic fields (B) – Gauss law for electric fields
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CODE–A
Key: 2 Hint: Conceptual 115. A charged particle is accelerated from rest through a certain potential difference. The de Broglie wavelength is 1 when it is accelerated through V1 and is 2 when accelerated through V2 . The ratio 1 / 2 is 1) V13/ 2 : V23/ 2
2) V21/ 2 : V11/ 2
3) V11/ 2 : V21/ 2
4) V12 : V22
Key: 2 Hint:
h 2meV 0
116. If the first line of Lymann series has a wavelength 1215.4 A , the first line of Balmer series is approximately 0
1) 4864 A Key: 3 Hint:
0
2) 1025.5 A
0
3) 6563 A
0
4) 6400 A
1 1 1 RZ 2 2 2 n1 n2
117. A certain radioactive element disintegrates with a decay constant of 7.9 1010 / sec . At a given instant of time, if the activity of the sample is equal to 55.3 1011 disintegration/sec, then number of nuclei at that instant of time 1) 7.0 1021 2) 4.27 1013 3) 4.27 103 4) 6 1023 Key: 1 Hint: R N 118. The change in current through a junction diode is 1.2 mA when the forward bias voltage is changed by 0.6V. The dynamic resistance is 1) 500 2) 300 3) 150 4) 250 Key: 1 V Hint: R i 119. A semiconductor has equal electron and hole concentration of 2 108 m3 . On doping with a certain impurity, the electron concentration increases to 4 1010 m3 , then the new hole concentration of the semiconductor is 1) 106 m3 2) 108 m3 3) 1010 m3 4) 1012 m3 Key: 1 Hint: ne nh ni 2 4 1010 nh 4 1016
6
nh 106 m3 120. A message single of 12kHz and peak voltage 20V is used to modulate a carrier wave of frequency 12MHz and peak voltage 30V. Then the modulation index is 1) 0.32 2) 6.7 3) 0.67 4) 67 Key: 3 E 1 m Hint: mix Emin 1 m
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CODE–A
CHEMISTRY 121. Assertion (A) : Atoms with completely filled and half filled subshells are stable. Reason (R) : Completely filled and half filled subshells have symmetrical distribution of electrons and have maximum exchange energy. The Correct answer is 1) (A) and (R) are correct, (R) is the correct explanation of (A) 2) (A) and (R) are correct, (R) is not correct explanation of (A) 3) (A) is correct, but (R) is not correct 4) (A) is not correct, but (R) is correct Key: 1 122. The element with the electronic configuration 1s2 2s2 2p6 3s2 3p6 3d10 4s1 is 1) Cu
2) Ca
3) Cr
4) Co
Key:1 Cu z 29 : 1s 2 2s 2 sp 6 3s 2 3 p 6 4s1 3d 10 123. Among the following, the isoelectronic species is/are i) O 2 , F , Na , Mg 2
ii) Na , Mg , Al 3 , F iii) N 3 , O 2 , F , Ne
1) (i) & (ii)
2) (i) , (ii) & (iii)
3) (ii) & (iii)
4) (i) & (iii)
Key:4
i O 2 , F , Na , Mg 2 ii N 3 , O 2 , F , Ne
contains 10 electrons each
contains 10 electrons each
124. What is the atomic number of the element with symbol Uus? 1) 117
2) 116
3) 115
4)114
Key: 1 Hint: Uus = ununseptium =117 125. Match the following List – I
List – II
(A) PCl3 (B) BF3 (C) ClF3 (D) XeF4 A
B
I) Square planar II) T-shape III) Trigonal-pyramidal IV) see-saw V) Trigonal planar C
D
A
B
C
D
1) (IV) (II) (I) (III)
2) (III) (V) (II)
(IV)
3) (III) (V) (II) (I)
4) (II) (IV) (III) (V)
Key:3 Hint: As per VSEPR theory 126. The order of covalent character of KF, Kl, KCl is 1) KCl < KF < KI
2) KI < KCl < KF
3) KF < KI < KCl
4) KF < KCI < KI Page 29
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CODE–A
Key:4 Hint: Anion size increases covalent character increases ( As per Fajan’s rules ) 127) If the kinetic energyin J, of CH4 ( molar mass = 16g mol-1) at T(K) is X, the kinetic energy in J, of O2 ( molar mass =32 g mol-1 ) at the same temperature is 1) X
3) X2
2) 2X
4) X/2
Key:1 Hint: If temp same, their kinetic energies are also same 128) The given figure shows the Maxwell distribution of molecular speeds of a gas at three different temperatures T1, T2 and T3. The correct order of temperature is :
1) T1 > T2 > T3
2) T1 > T3 > T2
3) T3 > T2 > T1
4) T2 > T3 > T1
Key:4 Hint: If temp increases, the peak of the curve shifted to right side, because velocity of molecules increases 129) In Haber’s process 50.0 g of N2(g) of H2(g) are mixed to produce NH3(g). What is the number of moles of NH3(g) formed ? 1) 3.33
2) 2.36
3) 2.01
4) 5.36
Key: 1 Hint: N 2 g 3H 2 g
1 1.786
2NH 3 g
2
3
5
--
(excess (Limiting reagent ) regent ) Products are formed as per limiting reagents H2 NH 3 3
2
5
5 10 2 3.33 2 3
130) The following reaction occurs in acidic medium KMnO4 + 8H+ + 5e– K+ + Mn2+ + 4H2O What is the equivalent weight of KMnO4 ? ( Molecular weight of KMnO4 =158 ) 1) 79.0
2) 31.6
3) 158.0
4) 39.5 Page 30
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CODE–A
Key: 2 Hint:
EKMnO4
158 mw 31.6 changeinO. N 5
131. Given that N2(g) + 3H2(g) 2NH3(g), rH = -92 kJ, the standard molar enthalpy of formation in kJ mol-1 of NH3(g) is 1) -92
2) +46
3) +92
4) -46
Key: 4 HINT:
H 2
132. Which one of the following is correct ? 1) The equilibrium constant (Kc) is independent of temperature 2) The value of Kc is independent of initial concentrations of reactants and products 3) At equilibrium, the rate of the forward reaction is twice the rate of the backward reaction 4) The equilibrium constant (Kc) for the reaction Ni(s) + 4CO(g) Ni(CO)4(g) is,
[Ni(CO)4 ] . [CO]
Key: 2 HINT: Conceptual 133. pH of an aqueous solution of NH4Cl is 1) 7
2) >7
3) <7
4) 1
Key: 3 HINT: NH 4 Cl Due to atomic hydrolysis 134. What is the change in the oxidation state of Mn, in the reaction of MnO4- with H2O2 in acidic medium ? 1) 7 4
2) 6 4
3) 7 2
4) 6 2
Key: 3 7
HINT: Mn O4 Mn 2 135. Which one of the following will not give flame test ? 1) Ca
2) Ba
3) Sr
4) Be
3) Al
4) Ga
Key: 4 HINT: Conceptual 136. Which one of the following forms a basic oxide ? 1) B
2) Tl
Key: 2 HINT: Metal oxides are basic
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TS EAMCET 2016 Engineering Question Paper with Solutions
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137. The gas produced by the passage of air over hot coke is 1) Carbon monoxide 2) Carbon dioxide 3) Producer gas 4) Water gas Key: 3 HINT: Conceptual 138. In environmental chemistry the medium which is affected by a pollutant is called the 1) Sink 2) Slag 3) Solvent 4) Receptor Key: 4 HINT: Conceptual 139. The hybridization of each carbon in the following compound is 1) sp3 sp2 sp3 sp 2) sp3 sp3 sp2 sp 3) sp3 sp sp3 sp2 4) sp3 sp2 sp sp3 Key: 1 HINT: Conceptual 140. The product Z of the following reaction is 2 HBr Z H 2 CC CH 1) H3CCH2CHBr2 2) H3CCBr2CH3 3) H3CCHBrCH2Br 4) BrCH2CH2CH2Br Key: 2 HINT: Markownicouf’s rule 141. Identify X and Y in the following reaction sequence O3 Zn X Y CH 3 2 CO CH 2O Zn H 2O X 1)(CH3)2 CHCH3 | Br 2)(CH3)2CHCH2Br 3)(CH3)2CBrCH2Br 4)(CH3)2 CHCHBr2 Key: 3 Hint : x = (CH3)2CBrCH2Br
Y CH3CH = CHCH3
CH3CH=CHCH3 (CH3)2C =CH2 (CH3)2C =CH2 Zinc O3 (CH3)2C =CH2 CH 3 2 CO CH 2O Zn H 2O
142. The packing efficiency of simple cubic (sc), body centred cubic (bcc) and cubic close packing (ccp) ;attics follow the order 1)bcc
experimental depression in freezing point van’t Hoff factor
0.025 0.0125 2
144. The molality of an aqueous dilute solution containing non-volatile solute is 0.1 m. What is the boiling temperature (in °C) of solution ? (Boiling point elevation constant, Kb=0.52 kg mol-1K; boiling temperature of water = 100°C). 1)100.0052
2)100.052
3)100.0
4)100.52
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Key: 2 Hint : Tb kb .m
T T0 kb .m ; T = T0 + kb . m; T0 100 0.1 0.52 100.052 145. Which one of the following is the correct plot of m (in s cm2 mol-1) and
KCI solution ? y m ; x c
c (in mol/L)1/2) for
1)
2)
3)
4)
Key: 2 Hint : Conceptual 146. For the reaction 5 Br aq 6 H aq BrO3 aq 3Br2 aq 3H 2O l if,
BrO3 t
0.01m L-1 min-1,,
1) 0.01
Br2 t
in mol min-1 is
2) 0.3
3) 0.03
4) 0.005
2) Soap lather
3)Butter
4)Vanishing Cream
2)Cu2O,FeO
3)Cu2S,FeS
4)Cu2S,FeO
Key: 3 Hint :
BrO3 1 Br 1 Br2 5 t t 3 t
BrO3 t
Br2 t
=3
1 Br2 3 t BrO3 t
= 3(0.01) = 0.03
147. Which one of the following is an emulsion ? 1) Milk Key: 1 and 4 Hint : Conceptual 148. Copper matte contains 1)Cu2O,Cu2S
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Key: 3 Hint : Conceptual 149. X reacts with dilute nitric acid to form ‘laughing gas’. What is X ? 1)Cu
2)P4
3)S8
4)Zn
Key: 4 Hint : Conceptual 150. Xenon reacts with fluorine at 873 K and 7 bar to form XeF4 . In this reaction the ratio of Xenon and fluorine required is 1)1:5
2)10:1
3)1:3
4) 5:1
Key: 1 Hint : Conceptual 151. Which of the following metal ions has a calculated magnetic moment value of 24 B.M. ? 1) Mn2 2) Fe2 3) Fe3 40 Co2 Key: 2 Hint: 4 = unpaired e is Fe2 Ar 4s 0 3d 6 = 4 4 2 24 152. Which one of the following does not exhibit geometrical isomerism? 1) Octahedral complex with formula MX 2 L4 2) Square planar complex with formula MX 2 L2 3) Tetrahedral complex with formula MABXL 4) Octahedral complex with formula MX 2 L L 2 Key: 3 Hint: Tetrahedral complex of MAB L Only exhibits geometrical 153. The P Dispersity Index (PDI) of a polymer is ( M w = weight average molecular mass and M n = number average molecular mass 1) The product of M n and M w
2) The sum of M n and M w
3) The difference between M w and M n Key: 4
4) The ratio between M w and M n
Mw Mn 154. Hormone that maintains the blood glucose level within the limit is 1) Thyroxine 2) Insulin 3) Testosterone Key: 2 Hint:Blood glucose level = insulin
Hint: PDI =
4) Epinephrine
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155. Chloroxylenol is an example of 1) Antiseptic 2) Antipyretic 3) Analgesic Key: 1 Hint:Chloroxylenol is Antiseptic 156. Which one of the following has highest boiling point ? 1) H 3CCH 2CH 2 CH 2Cl
4) Transquilizer
2) H 3C 2 CHCH 2Cl
3) H 3C 3 CCl
4) H 3CCH 2 C HCH 3 |
Cl
Key: 1 Hint: Straight chain compounds have more surface area. Therefore has high BP
H 157. X Y Aspirin H 3CCOOH
Identify X and Y from the following X Y CO2 H 1) H 3CCOCl
2)
H 3CCO 2 O
CO2 H
OH
3)
H 3CCO2 H
CO2 H
OH
4)
CO2 H
H 3CCOCH 3
OH
Key: 2 Hint: COOH
COOH OH
CH 3CO 2 O
O CO CH 3 CH 3CO
2 158. R CN R CHO ii H O i SnCl HCl 3
What is the name of the above reaction?
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EAMCET QP | KEY & SOLUTIONS
1) Rosenmund Key: 3 Hint: Stephen’s reaction
CODE–A
2) Williamson
3) Stephen
4) Kolbe
SnCl2 HCl R CN R CHO H 3O
159.
CH 2CH 3
CH 2CH 2CH 3
4 Y ii H O i KMnO KOH /
4 Z ii H O i KMnO KOH /
3
3
What are the structures of Y and Z? Y Z CH 2CO2 H CH 2CH 2 CO2 H 1)
2)
CO2 H
CH 2 CO2 H
3)
COCH 3
CH 2CO2 H
4)
CO2 H
CO2 H
Key: 4 Hint: CH 2CH 3
CH 2 CH 2 CH 3
COOH
COOH
Y
Z
160. Which is the strongest base among the following? 1)
H 3CNH 2
2)
NH 2
3)
NHCH 3
4)
CH 3 N CH 3
Key: 1 Hint: Aliphatic amines are strong bases
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