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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
CODE : C
Sri Chaitanya Jr College
SRI CHAITANYA EDUCATIONAL INSTITUTIONS, INDIA. AP TELANGANA KARNATAKA TAMILNADU MAHARASHTRA DELHI RANCHI
AP EAMCET-2016 (Engineering) CODE-C MATHEMATICS 1.
In ABC ,if the sides a,b,c are in geometric progression and the largest angle exceeds the smallest angle by 600 ,then cos B 1)
Sol:
13 1 4
2)
1 13 4
3) 1
13 1 4
A B 600 , ac b2 4cos2 B 2cos B 3 0 cos B
2.
4)
13 1 4
R is equal to r2 r3
In a ABC if A 900 ,then cos 1 1) 900
2) 300
3) 600
4) 450
A 1 2R cos 1 600 2 2
Sol:
r2 r3 4R cos 2
3.
The Cartesian equation of the plane whose vector equation is 1 i 2 j 3 2 2 k , where are scalars,is
1) 2 x y 5 Sol:
2) 2 x y 5
3) 2 x z 5
4) 2 x z 5
x 1 , y 2 , z 3 2 2
2x z 5
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
4.
CODE : C
Sri Chaitanya Jr College
For three vectors p, q and r ,if r 3 p 4q and 2r p 3q , then 1) r 2 q and r , q have the same direction 2) r 2 q and r , q have opposite directions 3) r 2 q and r , q have opposite directions 4) r 2 q and r , q have the same direction
Sol:
By Eliminating p Then r
5.
13 q 5
r 2q
If a 2i 3 j 5k , b mi nj 12k , and a b 0 ,then m, n 24 36 , 5 5
24 36 , 5 5
1)
2)
24 36 , 5 5
3)
24 36 , 5 5
4)
2 3 5 24 36 m ,n m n 12 5 5
Sol:
a || b,
6.
If a 3, b 4 and the angle between a & b is 1200 ,then 4a 3b is equal to 1) 25
Sol:
2) 7 2
3) 13
4) 12
4a 3b 16 a 9 b 24 a b cos(1200 ) 144 144 24.3, 4 2
2
1 144 144 144 = 144 2
4a 4b 12
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
7.
CODE : C
Sri Chaitanya Jr College
If a , b , c are non-zero vectors such that a b c
1 b c a, c a and is the angle 3
between the vectors b , c then sin =
1)
Sol:
2 2 3
2)
a b c 1/ 3 b
1 3
3)
2 3
4)
2 3
ca
c a b c.b a 13 b c a c b cos
= cos
8.
1 b c 3
1 3
sin 1 1/ 9
2 2 3
If a b c 0 and atleast one of the scalars a,b,c is non zero then the vectors are 1) parallel
2) non coplanar
3) coplanar
4) mutually perpendicular
Sol:
coplanar
9.
If the mean of 10 observations is 50 and the sum of the squares of the deviations of the observations from the mean is 250 , then the coefficient of variation of those observations is 1) 25
Sol:
x
2) 50
4) 5
2 2 xi 1 50 , xi x 250 , 2 xi x 25 10 10
C.V (x) = 10.
3) 10
x
100
5 100 10 50
The variance of the first 50 even natural numbers is Page No : 3
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
1)
833 4
CODE : C
2) 833
Sri Chaitanya Jr College
3) 437
4)
437 4
Sol:
n2 1 2 d =833 12
11.
3 out of 6 vertices of a regular hexagon are chosen at a time at random .The probability that the triangle formed with these three vertices is an equilateral triangle, is 1)
1 2
2)
1 5
3)
1 10
4)
1 20
Sol:
n 1 PE n3 C3 10
12.
A speaks truth in 75% of the cases and B in 80% of the cases .Then the probability that their statements about an incident do not match is 1)
Sol:
7 20
3 20
3)
2 7
4)
5 7
P A B P A B
3 1 1 4 . . 4 5 4 5
13.
2)
7 20
If the mean and variance of a binomial distribution are 4 and 2 respectively , then the probability of 2 successes of that binomial variate X , is 1)
1 2
2)
219 256
3)
1 1 p ,q ,n 8 2 2
37 256
p x 2
4)
7 64
8
C2 7 28 64
Sol:
np 4, npq 2
14.
In a city 10 accidents take place in a span of 50 days. Assuming that the number of accidents follow the Poisson distribution, the probability that three or more accidents occur in a day is
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
e , 0.2 k! k 3
1)
k
CODE : C
e , 0.2 k k 3
2)
k
Sri Chaitanya Jr College
e k , 0.2 k! k 0
e , 0.2 k! k 0 3
3) 1
k
3
4)
Sol:
e k , 0.2 k! k 3
15.
Equation of the locus of the centroid of the triangle whose vertices are (a cos k, a sin k),
(b sin k,-b cos k) and (1,0) , where k is a parameter, is 1) 1 3x 9 y 2 a 2 b2
2) 3x 1 9 y 2 2a 2 2b2
3) 3x 1 3 y a 2 b2
4) 3x 1 3 y 3a 2 3b2
2
2
2
2
2
2
3x 1
3y = a sin k – b cos k
Sol:
3x-1=a cos k + b sin k
16.
If the coordinate axes are rotated through an angle
2
9 y 2 a 2 b2
about the origin , then the 6
transformed equation of 3x2 4 xy 3 y 2 0 is 1) 3 y 2 xy 0
2) x2 y 2 0
3) 3 y 2 xy 0
2
Sol:
4) 3 y 2 2 xy 0
2
3x y 3x y x 3 y x 3 y 3 4 0 2 2 2 2 3 y 2 xy 0
17.
If the lines x 3 y 9 0 , 4 x by 2 0 ,and 2 x y 4 0 are concurrent, then the equation of the line passing through the point (b,0) and concurrent with the given lines, is 1) 2x+y+10=0
2) 4x-7y+20=0
3) x-y+5=0
4) x-4y+5=0
Sol:
Here b = -5
P.O.C = (3,2)
By verification x-4y+5=0
18.
The midpoint of the line segment joining the centroid and the orthocenter of the triangle whose vertices are (a,b),(a,c) and (d,c) is 5a d b 5c , 6 6
1)
a 5d 5b c , 6 6
2)
3) a, c
4) a, 0 Page No : 5
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
Sol:
Sri Chaitanya Jr College
2a d b 2c G , 3 3 5a d b 5c , 6 6
O a, c
19.
CODE : C
Mid point is
The distance from the origin to the image of (1,1) with respect to the line x y 5 0 is 1) 7 2
2) 3 2
3) 6 2
4) 4 2
Sol:
Image =(-6,-6)
Distance = 6 2
20.
The equation of the pair of lines joining the origin to the points of intersection of x2 y 2 9 and x y 3 , is 1) x2 3 y 9 2
2) 3 y y 2 9 2
3) x2 y 2 9
4) xy 0
Sol:
x y x y 9 0 3
21.
The orthocenter of the triangle formed by the lines x+y=1 and 2 y 2 xy 6 x2 0 is
2
2
2
4 4
1) , 3 3 Sol:
xy = 0
4 4 3 3
3) ,
4) , 4 4
One altitude is 6x-9y=-4
Verification , 3 3
(Or) H Kl , Km When K 22.
2 2 3 3
2 2
2) , 3 3
n a b am 2hlm bl 2 2
Let L be the line joining the origin to the point of intersection of the lines represented by 2 x2 3xy 2 y 2 10 x 5 y 0 . If L is perpendicular to the line kx y 3 0 , then k=
1) Sol:
1 2
2)
1 2
3) 1
4)
1 3
Point of intersection (-1,2) Page No : 6
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
Slope L = - 2
23.
CODE : C
K
-2.-K= - 1
Sri Chaitanya Jr College
1 2
2 touches the line x+y-z=0 at (1,1) .Then the length of the
A circle S=0 with radius
tangent drawn from the point (1,2) to S=0 is 1) 1 Sol:
2)
3) 3
2
4) 2
Equation of the line at (1,1) is x+y-2=0 Slope = -1 Perpendicular line slope = 1
Centre 1 2.
1 1 1 1 ,1 2. ,1 2. (or) 1 2. 2 2 2 2
= (2,2) (or) (0,0) Equation of the circle x2 y 2 2 24.
Length of tangent 3
The normal drawn at P(-1,2) on the circle x2 y 2 2 x 2 y 3 0 meets the circle at another point Q. Then the coordinates of Q are 1) 3, 0
Sol:
3) 2, 0
4) 2, 0
P 1, 2 , Q x, y
Centre C = (1,1) 25.
2) 3, 0
Q x, y 3,0
C = midpoint of PQ
If the lines kx 2 y 4 0 and 5x-2y-4=0 are conjugate with respect to the circle x2 y 2 2 x 2 y 1 0 , then k=
1) 0 Sol:
2) 1
3) 2
4) 3
conjugate lines Page No : 7
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
CODE : C
Sri Chaitanya Jr College
r l1l2 m1m2 l1 g m1 f n1 l2 g m2 f n2 2
K=1
26.
The angle between the tangents drawn from the origin to the circle x2 y 2 4 x 6 y 4 0 is 5
5 12
1) tan 1 13 Sol:
tan
2
27.
12 5
3) tan 1
13 5
4) tan 1
r 3 s11 2
tan
2) tan 1
2 tan
2 12 tan 1 12 5 5 1 tan 2 2
If the angle between the circles x2 y 2 2 x 4 y c 0 and x2 y 2 4 x 2 y 4 0 is 600 then c is equal to 1)
Sol:
3 5 2
cos
2)
3)
9 5 2
4)
7 5 2
d 2 r12 r22 2r1r2
c2 7c 11 0
28.
6 5 2
c
7 5 2
A circle S cuts three circles x2 y 2 4 x 2 y 4 0, x2 y 2 2 x 4 y 1 0,
and x2 y 2 4 x 2 y 1 0 orthogonally. Then the radius of S is 1)
29 8
2)
28 11
3)
29 7
4)
29 5
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
Sol:
CODE : C
3 3
radical center C , 4 4 Radius = length of tangent from C to circle S11
29.
29 8
The distance between the vertex and the focus of the parabola x2 2 x 3 y 2 0 is 1)
Sol:
Sri Chaitanya Jr College
4 5
x 1
2) 2
3 4
3)
1 2
4)
5 6
3 y 1
Distance =|a|=3/4 30.
If x1 , y1 and x2 , y2 are the end points of a focal chord of the parabola y 2 5x, then 4x1 x2 y1 y2
1) 25 Sol:
2) 5
Sol:
4)
5 4
25 x1 x2 a 2 16
y1 y2 4a 2
31.
3) 0
25 4
4 x1 x2 y1 y2 0
The distance between the focii of the ellipse x 3cos , y 4sin is 1) 2 7
2) 7 2
x2 y 2 1 9 16
e
3)
7
4) 3 7
7 4
Distance between foci = 2be 2 7 32.
The equation of the latus recta of the ellipse 9 x2 25 y 2 36 x 50 y 164 0 are 1) x 4 0, x 2 0 2) x 6 0, x 2 0 3) x 6 0, x 2 0 4) x 4 0, x 5 0
Sol:
x 2 25
2
y 1 9
2
1
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
e
33.
4 x h ae 5
2 3
2 2 3
3)
8 9
4)
4 2 3
x2 y 2 1 9 4 c 2 a 2 m2 b 2
m
2 2 3
The harmonic conjugate of 2,3, 4 with respect to the point 3, 2, 2 , 6, 17, 4 is 1 1 1
1) , , 2 3 4 Sol:
x 6, x 2
2)
y mx 2
34.
Sri Chaitanya Jr College
The values of m for which the line y mx 2 becomes a tangent to the hyperbola 4 x2 9 y 2 36 is 1)
Sol:
CODE : C
18
18 5 4
4
2) , 5, 5 5
3) , , 5 4 5
18
4
4) , 5, 5 5
P divides AB in the ratio
l : m x1 x; x x2 3 2 : 2 6 1: 4 Harmonic conjugate Q divides AB in the ration l : m 1: 4
Q
35.
If a line makes angles , , and with the four diagonals of a cube, then the value of sin 2 sin 2 sin 2 sin 2 is 1)
Sol:
B 4 A (6, 17, 4) (12, 8,8) 4 18 25 4 18 , , , 5, 5 5 5 5 5 5 5
4 3
2)
8 3
3)
cos 2 cos 2 cos 2 cos 2
7 3
4)
5 3
4 3
sin 2 sin 2 sin 2 sin 2 4
4 8 3 3 Page No : 10
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
36.
Sri Chaitanya Jr College
If the plane 56 x 4 y 9 z 2016 meets the coordinate axes in A,B,C then the centroid of the triangle ABC is 1) 12,168, 224
Sol:
CODE : C
2) 12,168,112
3) 12,168,
224 3
4) 12, 168,
224 3
x y z 1 2016 2016 2016 56 4 9 224 2016 2016 2016 G of ABC , , 12,168, 3 3 56 3 4 3 9
37.
The value(s) of x for which the function 1 x , x 1 f x 1 x 2 x , 1 x 2 3 x , x2
fails to be continuous is (are) 1) 1 Sol:
2) 2
3) 3
4) All real numbers
f (1 ) f (1 ) , f is continues at x 1 f (2 ) f (2 ) f is not continues at x 2
f ( x) 3 x at x 2 f is continues at x 3 38.
6 x 3x 2 x 1 x 0 x2 Lt
1) loge 2 log e 3 Sol:
3 Lt
x 0
x
1 (2 x 1) x
2
2) loge 5
3) loge 6
4) 0
log3e .log e2
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
39.
x 2 bx c , x 1
Define f x
, x 1
x
1) 2 Sol:
CODE : C
. If f x is differentiable at x 1, then b c
2) 0
f 1 ( x) 2 x b ,
x 1
1
x 1
,
Sri Chaitanya Jr College
3) 1
4) 2
f is differentiable at x=1 f ' (1 ) f ' (1 ) 2 b 1 b 1 f is differentiable at x =1, f is continues at x =1 = f (1 ) f (1 ) 1 b c 1 b c 0 c b 1
b c 1 1 2 40.
If x a is a root of multiplicity two of a polynomial equation f x 0, then 1) f ' a f '' a 0 2) f '' a f a 0 3) f ' a 0 f '' a 4) f a f ' a 0; f '' a 0
Sol:
f ( x) ( x a ) 2 g ( x) f '( x) 2( x a) g ( x) ( x a)2 g '( x) f ''( x) 2 g ( x) 4( x a) g1 ( x) ( x a)2 g11 ( x)
f (a) f '(a) 0, f 11 (a) 0 41.
If y log 2 log 2 x , then 1)
Sol:
log e 2 x log e x
y
2)
dy dx
1 log e 2 x
x
3)
1 x loge x loge 2
4)
1 x log 2 x
2
log log x log log 2 log 2 Page No : 12
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
CODE : C
Sri Chaitanya Jr College
dy 1 dx x log x.log 2
42.
The angle of intersection between the curves y 2 x2 a 2 2 and x2 y 2 a2 , is 1)
Sol:
3
2)
4
3)
6
4)
12
x2 y 2 a2 2 x2 y 2 a2 2 x 2 a 2 ( 2 1) x2
a ( 2 1) 2 2
= m1
x y
2 1 2 1 1 a 2 y 2 x2 a2 = a 2 2 2 2 1 x = m2 y 2 1
2 1 2 1
2 1 m m2 2 1 tan 1 1 m1m2 2 1 1 2 1 2
=
43.
If A 0, B 0 and A B 1)
Sol:
2 2 1 2 1 2 2 1 = 1 2 2 1 2 1
1 3
2)
3
, then the maximum value of tan A tan B is
1 3
3)
TanATanB . is maximum if A B
. Maximum of TanATanB
1 2
4) 3
6
1 3
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
44.
Sri Chaitanya Jr College
The equation of the common tangent drawn to the curves y 2 8x and xy 1 is 1) y 2 x 1
Sol:
CODE : C
2) 2 y x 6
3) y x 2
4) 3 y 8x 2
1 P t , be a point on xy 1 t Equation of Tangent at P is of
1 1 2 x ty 1 x x y 2t 2 t
C
y
x 2 Is Tangent to y 2 8 x 2 t t
a 2 2 1 t 3 t 1 2 m t 1/ t
Equation of C.T is y x 2 45.
Suppose f x x x 3 x 2 , x 1, 4. Then a value of c in 1, 4 satisfying f ' c 10 is 1) 2
Sol:
2)
5 2
3) 3
4)
7 2
f ( x) x3 x 2 6 x , f '( x) 3x2 2 x 6 , f '(c) 10 3c2 2c 6 10
c 23c 8 0 c 2 46.
If x3e5 x dx 1)
e5 x f x c, then f x 54
x3 3x 2 6 x 6 5 52 53 54
3) 52 x3 15x2 30 x 6 Sol:
3 3 5x x e dx x
2) 5x3 52 x2 53 x 6 4) 53 x3 75x2 30 x 6
e5 x e5 x e5 x 6e5 x e5 x 125 x3 75 x 2 30 x 6 c 3x 2 6x c 5 25 125 625 625
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
47.
Sol:
x
x2 2 x 2
2
CODE : C
dx
1)
x2 2 1 tan 1 x 1 c 2 x 2x 2 2
2)
3)
x2 2 1 tan 1 x 1 c 2 4 x 2x 2 2
4)
x
x 1 1 2
2
Sri Chaitanya Jr College
x2 2 1 tan 1 x 1 c 2 2 x 2x 2 2 2 x 1
1 tan 1 x 1 c x 2x 2 2 2
dx
x 1 tan
dx sec2 d 1 cos 2 sin 2 2 2 2
48.
tan 1 2 sec d sec4
1 sin 2 1 cos 2 d 2 x2 2 1 tan 1 x 1 c 2 4 x 2x 2 2
1 1 tan x 1 2
If log a 2 x2 dx h x c, then h x x
x
1) x log a 2 x 2 2 tan 1 a
2) x 2 log a 2 x 2 x a tan 1 a x
x
3) x log a 2 x 2 2 x 2a tan 1 a x 2 dx log a 2 x 2 .x
Sol:
log a
49.
For x 0, if
2
4) x 2 log a 2 x 2 2 x a 2 tan 1 a 1 x 2 x.x dx x log a 2 x 2 2 x 2a tan 1 2 a x a 2
log x dx 5
5 4 3 2 x A log x B log x C log x D log x E log x F constant, then
A B C D E F
1) 44
2) 42
3) 40
4) 36
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
Sol:
log x dx ,
CODE : C
Sri Chaitanya Jr College
put log x t dx e dt
5
t
t 5et dt et t 5 5t 4 20t 3 60t 2 120t 120 A B C D E F 44
50.
8a3 x2 and the curve y 2 is 4a x 4a 2
The area included between the parabola y
2
8
2) a 2 2 3
1) a 2 2 3
4
3) a 2 3
Sol:
2a 8a3 x2 4a 2 2 2 2 dx 2 a x 4 a 2 4a 3 0
51.
By the definition of the definite integral, the value of 1 1 1 lim ... 2 2 2 2 x 2 n 1 n 2 n n 1
1)
2)
n 1
Sol:
lim n
1
1 n r 2
1
2
0
2
1 x
dx
2
4
is equal to
3) 1
4) a 2 2 3
4
4)
6
2
x 4 4 dx 4 2 cos 2 x
52.
1)
8 3 5
2)
/4
Sol:
2 3 9
3)
4 2 3 9
4)
2 6 3
/4
x /4 dx dx 2 cos 2 x 2 cos 2 x /4 /4
02
1 dt 2 . . 3.tan 1 t 3 = 2 2 40 1 t 4 3 1 t 2 2 1 t 1
1
2
0
6 3
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
53.
CODE : C
Sri Chaitanya Jr College
The solution of the differential equation 1 y 2 x etan 1) x etan
1
3) 2 x etan
y
1
y
tan 1 y c
e2tan
1
y
2) x e2tan
c
4) x2etan
1
1
1
1
e tan
y
4e2tan
Sol:
54.
The solution of the differential equation 2 x 4 y 3
Sol:
e
1
y
dx 1 e tan y x dy 1 y 2 1 y2
tan 1 y
e tan y e tan x 1 y2
1
y
dy 2 x etan
1
y
1
y
c
y
1
dydx 0 is
y
e2tan
c
1
y
c
dy x 2 y 1 0 is dx
1) log 2 x 4 y 3 x 2 y c
2) log 2 2 x 4 y 3 2 x 2 y c
3) log 2 x 2 y 5 2 x y c
4) log 4 x 2 y 5 4 x 2 y c
dy x 2 y 1 , x 2 y t 1 dt / dx 2dy / dx dx 2 x 2y 3 1 dt t 1 1 2 dx 2t 3
2t 3
dx 4t 5dt 55.
The domain of the function f x log0.5 x ! 1) 0,1, 2,3,......
Sol: 56.
4 x 2 y c log 4 x 2 y 5
2) 1, 2,3,......
3) 0,
4) 0,1
f x is defined if x ! 0 and log0.5 x ! 0 x ! 1 x 0 or 1
If f x x 1 x 2 x 3 , 2 x 3, then f is 1) an onto function but not one-one
2) one-one function but not onto Page No : 17
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
CODE : C
3) a bijection
Sri Chaitanya Jr College
4) neither one-one nor onto
Sol:
If 2 x 3 , then x-1>0, x-2>0, x-3<0 f x x 1 x 2 x 3 x
57.
the greatest positive integer which divides
n 16 n 17 n 18 n 19 , for all positive
integers n, is 1) 6 Sol:
2) 24
3) 28
4) 20
Product of 4 consecutive integers is divisible by 4! a b c
58.
If a, b, c are distinct positive real numbers, then the value of the determinant b c a is c a b
1) < 0
2) > 0
4) 0
3) 0
a b c
Sol:
b c a 3abc a 3 b3 c3 negative c a b
59.
If x1 , x2 , x3 as well as y1 , y2 , y3 are in geometric progression with the same common ratio, then the points x1 , y1 , x2 , y2 , x3 , y3 are 1) vertices of an equilateral triangle
2) vertices of a right angled triangle
3) vertices of a right angled isosceles triangle 1 x1 rx1 2 y1 ry1
4) collinear x1 r 2 x1
Sol:
Area of triangle formed by given points
60.
The equations x y 2 z 4 , 3x y 4 z 6 , x y z 1 have
y1 r 2 y1
0
1) unique solution
2) infinitely many solution
3) no solution
4) two solutions
Page No : 18
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
Sol:
1 12 4 1 11 1 3 14 6
CODE : C
R2 R1 , R3 3R1
Sri Chaitanya Jr College
1 1 2 4 0 2 1 3 0 4 2 6
R1 2R1 R2 , R3 2R2
20 3 5 0 2 1 3 00 0 0 infinitely many solutions
61.
The locus of the point representing the complex number z for which z 3 z 3 15 is 2
1) a circle Sol:
2) a parabola
3) a straight line
4) an ellipse
3) 2
4) -2
2
x iy 3 x iy 3 15 2
x 3
2
2
y 2 x 3 y 2 15 2
12 x 15
1 i 2014 1 i
a straight line
2016
62.
1) -2i
2) 2i
Sol:
1 i 1 i
63.
If z1 1, z2 2, z3 3 and 9 z1 z2 4 z1 z3 z2 z3 12, then the value of z1 z2 z3 is 1) 3
Sol:
2014
1 i = i 2
2014
2i 2i
2) 4
3) 8
4) 2
| 9 z1 z2 4 z1 z3 z2 z3 | 12 | z3 z3 z1 z2 z2 z2 z1 z3 z1 z1 z2 z3 | 12 Page No : 19
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
CODE : C
| z1 z2 z3 | | z3 z2 z1 | 12
64.
| z1 z2 z3 | 2
If 1, z1 , z2 ,......zn1 are the nth roots of unity, then 1 z1 1 z2 .... 1 zn1 1) 0
Sol:
Sri Chaitanya Jr College
2) n - 1
3) n
4) 1
xn 1 0
x 1 x z1 x z2 x zn1 xn 1 x z1 x z2 x z3 x zn1 1 z1 1 z2 1 z3 1 zn1 xLt1 65.
2
1)
Sol:
If 124 2 x 24 3
12
13 12
4 2 x2
12 2 3
3 x2 2
14 5
3)
12 13
4)
5 14
3 14 3x 2 2 x 2 5
The product and sum of the roots of the equation x2 5 x 24 0 are respectively 1) -64, 0
Sol:
xn 1 x 1
, then x =
2)
2x2 4
66.
3 x2 2
xn 1 x 1
2) -24, 5
3) 5, -24
4) 0, 72
If x 0, x2 5x 24 0
x 8 x 3 0 x 8
If x<0 x2 5x 24 0
x 8 x 3 0
X = -8
Sum of roots = 8-8 = 0
Product of roots = 8(-8)=-64 Page No : 20
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
67.
Sri Chaitanya Jr College
The number of real roots of the equation x5 3x3 4 x 30 0 is 1) 1
Sol:
CODE : C
2) 2
3) 3
4) 5
f 1 x 5x4 9 x2 4 0 for all x R
f(x) = 0 has only one real root 68.
If the coefficients of the equation whose roots are k times the roots of the equation
x3
1 2 1 1 x x 0, are integers then a possible value of k is 4 16 144
1) 3 Sol: 69.
2) 12
3) 9
4) 4
x f 0 k 12 k
The sum of all 4-digit numbers that can be formed using the digits 2,3,4,5,6 without repetition, is 1) 533820
Sol:
2) 532280
3) 533280
4) 532380
4P3 2 3 4 5 6 1111
=533280 70.
If a set A has 5 elements, then the number of ways of selecting two subsets P and Q from A such that P and Q are mutually disjoint, is 1) 64
2) 128
3) 243
Sol:
35 243
71.
The coefficient of x 4 in the expansion (1 x x2 x3 )4 is 1) 31
Sol:
1 x
2) 30 4
3) 25
4) 729
4) -14
1 x
2 4
Page No : 21
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
CODE : C
Coefficient of x 4C0 .4C2 4C2 .4C1 4C4 .4C0 4
72.
Sri Chaitanya Jr College
=31
If the middle term in the expansion of (1 x)2 n is the greatest term, then x lies in the interval
n n 1 1) , n 1 n Sol:
n 1 n 2) , n n 1
3) (n 2, n)
4) (n 1, n)
Tn is the Greatest term | Tn1 || Tn || Tn1 | n n 1 x , n 1 n
73.
To find the coefficient of x 4 in the expansion of
3x the interval in which the ( x 2)( x 1)
expansion is valid is 1) 2 x
Sol:
1 1 2) x 2 2
3) 1 x 1
4) x
3x 6 3 x 2 x 1 x 2 x 1 1
1 x 3 1 3 1 x 2
x 1 and | x | 1 2
x 1,1
74.
If (1 tan )(1 tan 4 ) 2, 0, ,then 16 1)
20
2)
30
3)
40
4)
60 Page No : 22
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
Sol:
4
75.
If cos
1) cot
Sol:
2
Sri Chaitanya Jr College
20
cos cos , then one of the values of tan is 1 cos cos 2
tan
2) tan tan
2
2
3) tan cot 2 2
4) tan 2
2
tan 2
2
cos 1 1 cos 1 cos cos 1 1 cos 1 cos tan
76.
4
CODE : C
2
tan
2
.cot
2
The value of the expression
1 1 sin 2 3 sin 2 cot cot is 3 4 2 2 2 cos(2 2 ) tan 4 1) 0
Sol:
cos sin
3) sin 2
2) 1 2
cos sin cos 2 cos sin
2
4) sin 2
1 cos 2sin cos 2 4 sin
1 cos2 sin 2
77.
if
1 sin ,cos and tan are in geometric progression, then the solution set of is 6
1) 2n 6 Sol:
2) 2n 3
3) n (1)n 3
4) n 3
1 sin cos2 sin . 6 cos 6cos3 cos2 1 0 Page No : 23
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
cos cos
78.
3
2n
3
x
2) x y
3) x 0 y
4) x y
2 tan 4 2 1 tan 5
1 cos 1 y sin 2 5 2
79.
Sri Chaitanya Jr College
4 1 if x sin(2 tan 1 2) and y sin tan 1 , then 3 2 1) x y
Sol:
CODE : C
x y
5 If cosh( x) , then cosh(3x) 4 1)
61 16
2)
63 16
3)
65 16
4)
61 63
65 16
Sol:
cosh 3x 4cosh 3 x 3cosh x
80.
B A C B C C A A B In ABC, if x tan and z tan then tan , y tan tan tan 2 2 2 2 2 2
( x y z) 1) xyz
Sol:
2) xyz
3) 2xyz
4)
1 xyz 2
1 x 1 y 1 z . . 1 1 x 1 y 1 z 1 s1 s2 s3 1 s1 s2 s3
s1 s3 x y z xyz
Page No : 24
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
CODE : C
Sri Chaitanya Jr College
PHYSICS 81.
When a long straight uniform rod is connected across an ideal cell, the drift velocity of electrons in it is v. If a uniform hole is made along the axis of the rod and the same battery is used, then the drift velocity of electrons becomes 1) v
Sol:
vd
82.
2) >v
3)
4) zero
I neA
v v RneA l neA A
Independent of „A‟ Hence (1)
In a meter bridge experiment, when a nichrome wire is in the right gap, the balancing length is 60cm. When the nichrome wire is uniformly stretched to increase its length by 20% and again connected in the right gap, the new balancing length is nearly. 1) 61cm
Sol:
2) 31cm
3) 51cm
4) 41cm
RL 60 3 RR 40 2 RL l 1.44 RR 100 l 1.44 RR 3 100 l RR 2 l
100 l 2 1.44 0.96 l 3 100 l 0.96l 1.96l 100
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
l
83.
CODE : C
Sri Chaitanya Jr College
100 51cm 1.96
A loop of flexible conducting wire lies in a magnetic field of 2.0 T with its plane perpendicular to the filed . The length of the wire is 1m. When a current of 1.1A is passed through the loop, it opens into a circle , then the tension developed in the wire is 1) 0.15N
Sol:
T
2) 0.25N
3) 0.35N
4) 0.45N
Bil 2
0.35N
84.
A charge q is spread uniformly over an isolated ring of radius ‘R’. The ring is rotated about its natural axis with an angular velocity ‘ ’. Magnetic dipole moment of the ring is 1)
Sol:
qR 2 2
2)
q R 2
3) qR 2
4)
q 2R
M iA
q 2
2 R
q R 2 2
85.
A magnetic dipole of moment 2.5Am2 is free to rotate about a vertical axis passing through its centre. It is released from East-West direction . Its kinetic energy at the moment it takes North-South position is BH 3 105 T . 1) 50J
Sol:
2) 100J
3) 175J
4) 75J
K .E MB cos 2 cos 1 2.5 3 105 75 106 Page No : 26
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
86.
CODE : C
Sri Chaitanya Jr College
A branch of a circuit is shown in the figure. If current is decreasing at the rate of 103 As 1 , the potential difference between A and B is + A
1) 1V Sol:
2A
2) 5V
7
4V
3) 10V
9mH
B
4) 2V
VAB 14 4 9 103 103 0 VAB 1V
87.
The natural frequency of an LC circuit is 125kHZz. When the capacitor is totally filled with a dielectric material, the natural frequency decreases by 25kHz. Dielectric constant of the material is nearly. 1) 3.33
Sol:
f
2) 2.12
4) 1.91
1 2 LC
125 k 100
88.
3) 1.56
2
5 k k 1.56 4
Choose the correct sequence of the radiation sources in increasing order of the wavelength of electromagnetic waves produced by them. 1) X-ray tube, Magnetron valve, Radio active source, Sodium lamp 2) Radio active source , X-ray tube, Sodium lamp, Magnetron valve 3) X-ray tube, Magnetron valve, Sodium lamp, Radio active source 4) Magnetron valve, Sodium lamp, X-ray tube, Radio active source
Sol:
Theory Page No : 27
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
89.
CODE : C
Sri Chaitanya Jr College
A photo sensitive metallic surface emits electrons when X-rays of wavelength ‘ ’ fall on it. The de Broglie wavelength of the emitted electrons is (Neglect the work function of the surface, m is mass of the electron. h-Planck’s constant c-velocity of light) 1)
Sol:
90.
2mc h
2)
h 2mkE
1
h 2mc h 2m
hc
3)
mc h
4)
h mc
h 2mc
An electron in a hydrogen atom undergoes a transition from a higher energy level to a lower energy level. The incorrect statement of the following is 1) Kinetic energy of the electron increases. 2) Velocity of the electron increases 3) Angular momentum of the electron remains constant. 4) Wavelength of de-Broglie wave associated with the motion of electron decreases.
Sol:
Theory
91.
The radius of germanium ( Ge) nuclide is measured to be twice the radius of 94 Be . The number of nucleons in Ge will be 1) 72
2) 73
3) 74
4) 75
1/3
Sol:
R1 A1 R2 A2 23
A1 A2
23
A1 9
A1 72 Page No : 28
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
92.
CODE : C
Sri Chaitanya Jr College
For a common-emitter transistor amplifier, the current gain is 60. If the emitter current is 6.6mA then its base current is 1) 6.492mA
Sol:
2) 0.108mA
3) 4.208mA
4) 0.343mA
Ic 60 IB
I c 60I B
But I B I E IC I B 6.6 60I B
61I B 6.6 I B 0.108mA
93.
If a transmitting antenna of height 105m is placed on a hill, then its coverage area is 1) 4224km2
2) 3264km2
Sol:
A 2Rh
94.
Match the list-I with list-II
3) 6400 km2
List-I
List-II
A)Boltzmann constant
I) ML0T 0
B)Coefficient of viscosity
1 1 II) ML T
C)Water equivalent
3 1 III) MLT K
4) 4864 km2
2 2 1 D) Coefficient of thermal conductivity IV) ML T K
Sol:
1) A-III,B-I,C-II,D-IV
2)A-III,B-II,C-I,D-IV
3) A-IV,B-II,C-I,D-III
4) A-IV,B-I,C-II,D-III
Theory Page No : 29
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
95.
CODE : C
Sri Chaitanya Jr College
Two trains , which are moving along different tracks in opposite directions are put on the same track by mistake. On noticing the mistake, when the trains are 300m apart the drivers start slowing down the trains . The graphs given below show decrease in their velocities as function of time. The separation between the trains when both have stopped is V(ms-1)
V(ms-1) 40 20
8 0 Train-I
10
t(s)
t(s) 20
1) 120m
2) 20m
u1 40 a1 4
3) 60m
Train-II
4) 280m
u2 20 10 a2 4
Sol: u12 s1 2a1
1600 8
s1 200
u22 s2 2a2
400 5
s2 =80
Reaming distance = 20 m
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
96.
CODE : C
Sri Chaitanya Jr College
A point object moves along an arc of a circle of radius ‘R’ . Its velocity depends upon the distance covered ‘S’ as V K S where ‘K’ is a constant. If ‘ ’ is the angle between the total acceleration and tangential acceleration , then 1) tan
Sol:
at
2) tan
S 2R
dv v ds
d k s k s ds
3) tan
S 2R
4) tan
k sk
1
k2 = 2
2S R
dv dt
dv ds ds dt
97.
S R
2 s
A body projected from the ground reaches a point ‘X’ in its path after 3 seconds and from there it reaches the ground after further 6 seconds. The vertical distance of the point ‘X’ from the ground is ( acceleration due to gravity 10ms 2 )
Sol:
98.
1) 30m
2) 60m
3) 80m
2u 9 g
u 45m / s
1 h ut gt 2 2
45(3) 5 3
2
4) 90m
= 90 m
A particle of mass ‘m’ is suspended from a ceiling through a string of length ‘L’ . If the particle moves in a horizontal circle of radius ‘r’ as shown in the figure, then the speed of the particle is
0
r
1) r
g L2 r 2
2) g
r L2 r 2
L
m
3) r
g L r2 2
4) g
r L r2 2
Page No : 31
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
Sol:
T sin
mv r
CODE : C
Sri Chaitanya Jr College
2
T cos mg
v2 L2 r 2 rg
v2 tan rg
99.
r
v 2
r2g L r 2
2
vr
g L r2 2
A particle is placed at rest inside a hollow hemisphere of radius ‘R’. The co-efficient of friction between the particle and the hemisphere is
1 . The maximum height upto 3
which the particle can remain stationary is 1)
Sol:
R 2
2) 1
3 R 2
3)
3 R 2
4)
3R 8
1 R h 1 2 1 3 h 1 R 2
100.
A 1kg ball moving with a speed of 6ms 1 collides head-on with a 0.5kg ball moving in the opposite direction with a speed of 9 ms 1 . If the co-efficient of restitution is
1 , the energy 3
lost in the collision is 1) 303.4J Sol:
KE
2) 66.7J
3) 33.3J
4) 67.8J
m1m2 2 1 e2 u1 u2 2 m1 m2
1 1 2 2 1 1 15 9 1 2 1 2
1 28 225 2 69
8 225 33.33 54 Page No : 32
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
101.
CODE : C
Sri Chaitanya Jr College
A ball is thrown vertically down from a height of 40m from the ground with an initial velocity ‘v’. The ball hits the ground , loses
1rd of its total mechanical energy and rebounds 3
back to the same height. If the acceleration due to gravity is 10ms 2 , the value of ‘v’ is 1) 5ms 1 Sol:
102.
2) 10ms 1
3) 15ms 1
4) 20 ms 1
21 2 mv mgh mgh 32
21 2 v gh gh , 32
1 2 2 v gh gh , 3 3
u2 2 gh gh 3 3
u 2 gh 3 3
u gh
40 10
u 20
Three identical uniform thin metal rods form the three sides of an equilateral triangle. If the moment of inertia of the system of these three rods about an axis passing through the centroid of the triangle and perpendicular to the plane of the triangle is ‘n’ times the moment of inertia of one rod separately about an axis passing through the centre of the rod and perpendicular to its length , the value of ‘n’ is 1) 3
Sol:
2) 6
3) 9
4) 12
moment of inertia about an axis passing through centre and perpendicular to rod
l 2 3
ml 2 ml 2 M.I about central 12 12
MI
System MI 3
ml 2 I1 12
2ml 2 12
2ml 2 ml 2 6 6I 12 12
n6
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
103.
CODE : C
Sri Chaitanya Jr College
Two smooth and similar right angled prisms are arranged on a smooth horizontal plane as shown in the figure. The lower prism has a mass ‘3’ times the upper prism. The prisms are held in an initial position as shown and are then released . As the upper prism touches the horizontal plane, the distance moved by the lower prism is b
a
1) a-b Sol:
2)
a b 3
3)
4)
a b 4
3mx m a b) x x
3x a b x 4x a b
104.
ba 2
a b 4
A particle is executing simple harmonic motion with an amplitude of 2m. The difference in the magnitudes of its maximum acceleration and maximum velocity is 4. The time period of its oscillation and its velocity when it is 1m away
from the mean position are
respectively. 1) 2s, 2 3ms 1 Sol :
2)
7 s, 4 3ms 1 22
3)
22 s, 2 3ms 1 7
4)
44 s, 4 3ms 1 7
A 2m
2 A A 4
2
2 4 ,
2 2 0 ,
2 1 2 0 , 2 , T
2 2 2 0 , 2 2 22 S, 2 7
V A2 42
2 4 1 2 3m / s Page No : 34
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
105.
CODE : C
Sri Chaitanya Jr College
Two bodies of masses ‘m’ and ‘9m’ are placed at a distance ‘r’. The gravitational potential at a point on the line joining them, where gravitational field is zero, is ( G is universal gravitational constant) 1)
Sol:
106.
14Gm r
x
2)
d m2 1 m1
16Gm r
3)
12Gm r
4)
8Gm r
r r 4 9m 1 m
r x r
r 3r 4 4
V V1 V2
G m 4 G 93m 4 16Gm r 3r r
When a load of 80N is suspended from a string , its length is 101mm. If a load of 100N is suspended , its length is 102mm. If a load of 160N is suspended from it, then length of the string is ( Assume the area of cross-section unchanged)
Sol:
1) 15.5cm
2) 13.5cm
Natural length (l)=
l1T2 l2T1 T2 T1
l
3) 16.5cm
4) 10.5cm
101 100 5 102 80 4 505 408 97cm 20
80 4 l3 97 8 105mm 160 l3 97
=10.5cm
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A.P EAMCET (Engineering)
107.
CODE : C
Sri Chaitanya Jr College
A sphere of material of relative density 8 has a concentric spherical cavity and just sinks in water. If the radius of the sphere is 2cm , then the volume of the cavity is 1)
Sol:
76 3 cm 3
2)
79 3 cm 3
3)
82 3 cm 3
4)
88 3 cm 3
88 3 cm 3
mg FB
4 R3 r 3 d g 4 R3d g B w 3 3 8 R3 r3 1 1 3 1 R3 r 3 R 8 r3 7 7 3 r 3 2 7 3 R 8 8 4 3
4 22 7 3 7
Volume of cavity r 3 108.
A hunter fired a metallic bullet of mass ‘m’ kg from a gun towards an obstacle and it just melts when it is stopped by the obstacle . The initial temperature of the bullet is 300K. If
1 th of heat is absorbed by the obstacle , then the minimum velocity of the bullet is 4 [ Melting point of bullet =600K, 1 o 1 Specific heat of bullet =0.03 cal g C ,
1 Latent heat of fusion of bullet 6cal g
1) 410ms 1 Sol:
2) 260ms 1
31 2 mv ms mL 42
V2
3) 460ms 1
4) 310ms 1
3 2 v 0.03 4200 300 6 4200 8
8 0.01 4200 300 2 4200 3
V 2 40 4200 V 16800 410m / s
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
109.
CODE : C
Sri Chaitanya Jr College
0
‘M’ kg of water ‘t’ C is divided into two parts so that one part of mass ‘m’ kg when converted into ice at 00C would release enough heat to vapourise the other part, then
m is M
equal to [Specific heat of water = 1 cal g-1 0C-1 Latent heat of fusion of ice = 80 cal g-1, Latent heat of steam = 540 cal g-1] 1) 640 – t Sol:
2)
720 t 640
3)
640 t 720
4)
640 t 720
m 80 m 1 t M m 1 100 t 540 M m m 80 mt M m 100 M m t 540M m 540 80m mt M 100 m 100 4t mt 540M 540M 640M 100m 640M Mt
720m 640 t M
110.
m 640 t M 720
A diatomic gas 1.4 does 300J work when it is expanded isobarically. The heat given to the gas in this process is 1) 1050J
Sol:
2) 950J
3) 600J
4) 550J
dW 1 1 dQ r 300 1 1 dQ 1.4
dQ
300 1.4 300 75 14 0.4 4
1050J Page No : 37
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
111.
CODE : C
Sri Chaitanya Jr College
When the absolute temperature of the source of a Carnot heat engine is increased by 25%, its efficiency increases by 80%. The new efficiency of the engine is 1) 12%
Sol:
2) 24%
3) 48%
4) 36%
5 100 T2 125 9 100 125 T2
900 9T2 500 4T2 n 1
112.
80 9 100 36 % 125 25
A cylinder of fixed capacity 67.2 litres contains helium gas at STP. The amount of heat needed to rise the temperature of the gas in the cylinder by 200C is (R = 8.31 Jmol-1K-1) 1) 748J
Sol:
3) 1000J
4) 500J
dQ ncv dT 3
113.
2) 374J
831 3 20 2
748J
For a certain organ pipe, three successive resonance frequencies are observed at 425, 595 and 765Hz, respectively. The length of the pipe is (speed of sound in air = 340 ms-1) 1) 0.5m
Sol:
2) 1m
3) 1.5m
4) 2m
5V 425 4l 5 340 425 4l
1700 1700l l 1m
114.
A student holds a tuning fork oscillating at 170Hz. He walks towards a wall at a constant speed of 2ms-1. The beat frequency observed by the student between the tuning fork and its echo is (Velocity of sound = 342 ms-1)
Sol:
1) 2.5Hz
2) 3Hz
3) 1Hz
V V0 n1 n V V0
340 2 170 172 2 HZ 340 2
4) 2hz
Page No : 38
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
115.
CODE : C
Sri Chaitanya Jr College
An infinitely long rod lies along the axis of a coneave mirror of focal length ‘f’ the nearer end of the rod is at a distance u, (u>f) from the mirror. It’s image will have a length 1)
Sol:
uf u f
2)
uf u f
3)
f2 u f
4)
f2 u f
1 1 1 u v f
1 1 1 u v f 1 1 1 v u f
uf v f u
f2 u f
uf f u f
L v f
0
116.
In Young’s double slit experiment, red light of wavelength 6000 A is used and the nth bright fringe is obtained at a point ‘P’ on the screen. Keeping the same setting, the source 0
of light is replaced by green light of wavelength 5000 A and now (n+1)th bright fringe is obtained at the point P on the screen. The value of ‘n’ is 1) 4 Sol:
n
2) 5
3) 6
4) 3
6000 D n 1 5000 D d d
6n 5n 5 n5
117.
Two charges each of charge + 10 c are kept on Y-axis at y = -a and y = +a respectively. Another point charge -20 c is placed at the origin and given a small displacement x(x<
1)
3.6x N a2
2)
2.4x 2 N a
3)
3.6x N a3
4)
4.8x N a2
Page No : 39
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
Sol:
CODE : C
Sri Chaitanya Jr College
Fnet 2F cos 12 x 9 200 10 2 9 10 a2 x2 a2 x2
10 C
a
20 C
x
36 101
118.
x
a
2
x2
3
2
3.6x a3
Three identical charges, each 2 C lie at the vertices of a right angled triangle as shown in the figure. Forces on the charge at B due to the charges at A and C respectively are F1 and F2. The angle between their resultant force and F2 is A
3m
B
9 16
1) tan 1
4m
9
2) tan 1 7
C
16 9
3) tan 1
7
4) tan 1 9
Page No : 40
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
CODE : C
Sri Chaitanya Jr College
A 3m
F2
B
C
4m
Fnet
F1
Sol: tan
F1 F2
kq1q2 tan 9 kq1q2 16
tan
119.
16 9
The figure shows equipotential surface concentric at ‘O’. The magnitude of electric field at a distance ‘r’ meters from ‘O’ is 20V 30V 60V
20
cm
O 10cm 30cm
1)
Sol : 60
9 Vm1 2 r
2)
16 1 Vm r2
3)
2 Vm1 2 r
4)
6 Vm1 2 r
kq 10 102
kq 6
E
kq r2
E
6 r2 Page No : 41
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
120.
CODE : C
Sri Chaitanya Jr College
ur A region contains a uniform electric field E 10 i 30 j Vm1 . A and B are two points in the
field at (1,2,0) m and (2,1,3) m respectively. The work done when a charge of 0.8C moves from A to B in a parabolic path is 1) 8J Sol:
2) 80J
3) 40J
4) 16J
dv E.dr
10i 30 j . i j 3k 20
w qv
=0.8 (20)
= 16J
CHEMISTRY 121.
Sol:
Match the following List – 1
List – 2
(magnetic property)
( substance)
A) Ferromagnetic
1) O2
B) Anti ferro magnetic
2) CrO2
C) Ferri magnetic
3) MnO
D) Para magnetic
4) C6 H 6
1) A – 3, B – 2, C – 4, D – 1
2) A – 2, B – 3, C – 4, D – 1
3) A – 1, B – 3, C – 5, D – 4
4) A – 4, B – 2, C – 3, D - 5
Key: 2 Conceptual
Page No : 42
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
122.
CODE : C
Sri Chaitanya Jr College
The vapour pressures of pure benzene and toluene are 160 and 60 mm Hg respectively. The mole fraction of benzene in vpour phase in contact with equimolar solution of benzene and toluene is 1) 0.073
Sol:
2) 0.027
3) 0.27
4) 0.73
YA PA0 Key: 4 X A PTotal PTotal PA0 . X A PB0 .X B
1 1 PTotal 160 60 110 mm 2 2
123.
YA 160 1 110 2
YA 0.73
6g of a non volatile, non electrolyte X dissolved in 100 g of water freezes at 0.930 C . The molar mass of x in g mol 1 is K f of H 2O 1.86K Kg mol 1 1) 60
Sol:
2) 140
3) 180
4) 120
Key: 4 Tf K f .m T f 0.930 C 0.93 1.86
124.
6 1000 M 100
M 120
The products obtained at the cathode and anode respectively during the electrolysis of aqueous K 2 SO4 solution using platinum electrodes are 1) O2 , H 2
Sol:
2) H 2 , O2
3) H 2 , SO2
4) K , SO2
Key:2 K2 SO4 2K SO42 H 2O H OH
Cathode :- 2H 2e H 2 Anode:- 4OH 4e 2H 2O O2
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
125.
CODE : C
Sri Chaitanya Jr College
The slope of the graph drawn between in k and 1/T as per Arrhenius equation gives the value (R= gas constant, Ea = Activation energy) 1)
Sol:
R Ea
2)
Key:3 K Ae ln K
Ea R
3)
Ea R
4)
R Ea
Ea RT
Ea ln A RT
y mx c slope
126.
Ea R
which is not the correct statement in respect of chemisorption? 1) Highly specific adsorption
2) Irreversible adsorption
3) Multilayered adsorption
4) High enthalpy of adsorption
Sol:
Key: 3 Conceptual
127.
Which of the following is carbonate ore? 1) Cuprite
2) Siderite
3) Zincite
Sol:
Key: 2 Siderite is PbCO3
128.
Which one of the following statements is not correct?
4) Bauxite
1) O3 is used as germicide 2) In O3 , O O bond length is identical with that of molecular oxygen 3) O3 is an oxidising agent Sol:
4) The shape of O3 molecule is angular
Key: 2 Ozone exhibit resonance
Page No : 44
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
129.
CODE : C
Sri Chaitanya Jr College
Which of the following reactions does not take place? 1) F2 2Cl 2F Cl2
2) Br2 2I 2Br I 2
3) Cl2 2Br 2Cl Br2
4) Cl2 2F 2Cl F2
Sol:
Key: 4 Less reactive halogen cannot displace more reactive halogen
130.
Which of the following statements regarding sulphur is not correct? 1) At about 1000K, it maninly consists of S 2 molecules 2) The oxidation state of sulphur is never less than +4 in its compounds 3) S 2 molecule is paramagnetic 4) Rhombic sulphur is readily soluble in CS2
Sol:
Key: 2 Conceptual
131.
Which of the following reactions does not involve, liberation of oxygen ? 1) XeF4 H 2O 2) XeF4 O2 F2 3) XeF2 H 2O
Sol:
4) XeF6 H 2O
Key: 4 XeF6 H 2O XeOF4 2HF XeF6 2H 2O XeO2 F2 4HF XeF6 3H 2O XeO3 6HF
132.
Select the correct IUPAC name of [Co( NH3 )5 (CO3 )]Cl 1) penta ammonia carbonate cobalt (III) chloride 2) pentammine carbonate cobalt chloride‟ 3) pentammine carbonato cobalt (III) chloride 4) Cobalt (III) pentammine carbonate chloride
Sol:
Key: 3 Conceptual
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
133.
CODE : C
Sri Chaitanya Jr College
Which of the following characteristics of the transition metails is associated with their catalytic activity ? 1) Colour of hydrated ions
2) Diamagnetic behaviour
3) paramagnetic behaviour
4) variable oxidation states
Sol:
Key: 4 Conceptual
134.
Observe the following polymers PHBV
Nylon 2 – nylon 6
Glyptal
Bakelite
(A)
(B)
(C)
(D)
Biodegradable polymer(s) from the above is /are 1) (C)
2) (A), (B)
Sol:
Key: 2 Conceptual
135.
Observe the following statements i) Sucrose has glycosidic linkage
3) (D)
4) (C), (D)
2) Cellulose is present in both plants and animals
iii) lactose contains D- galactose and D-glucose units The correct statements are 1) (i), (ii), (iii)
2) (i), (ii)
3) (ii), (iii)
Sol:
Key: 4 Cellulose cannot be present in animals
136.
identify the antioxidant used in foods
Sol:
4) (i), (iii)
1) Aspartame
2) Sodium benzoate
3) Ortho-sulpho benzimide
4) Butylated hydroxy toluene
Key: 4 Conceptual
Page No : 46
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
CODE : C
Sri Chaitanya Jr College
CH 3
Cl dry ether
2 Na CH 3Cl
2NaCl
137. This reaction is known as 1) Wurtz – Fitting reaction
2) Wurtz reaction
3) Fitting reaction
4) Friedel – crafts reaction
Sol:
Key: 1 Conceptual
138.
what is Z in the following sequence of reactions ? H 2O Mg 2-methyl-2-bromo propane X Z dryether
Sol:
139.
1) propane
2) 2-methyl propene
3) 2 – methyl propane
4) 2 – methyl butane
CH3 |
CH3 |
CH3
CH3
2 CH 3 CH CH 3 Key: 3 CH 3 C| Br Mg CH 3 C| MgBr |
H O
CH3
In which of the following reactions the product is not correct ? LiAlH 4 CH3CH 2OH 1) CH3CHO
Zn Hg CH 3 CH CH 3 2) CH 3COCH 3 HCl | OH
( i ) H 2 N NH 2 3) CH3CH 2CHO CH3CH 2CH3 ( ii ) KOH , ethylene glycol /
KMnO4 CH3CH 2COOH 4) CH3CH 2CHO
Sol:
Key: 2 It is not clemenson reduction product
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
140.
CODE : C
Sri Chaitanya Jr College
identify the name of the following reactions.
CH 3
CHO H O
Cr O2Cl
3 2 A CS 2
1) Gatterman – Koch reaction
2) Gatterman reaction
3) Stephen reaction
4) Etard reaction
Sol:
Key: 4 Conceptual
141.
What is C in the following sequence of reactions ? PCl3 KCN hydrolysis CH3OH A B C
1) CH3CH 2OH
2) CH 3CHO
3) CH3COOH
4) HOCH 2 CH 2OH
Sol:
PCl HO KCN Key: 3 CH3OH CH3Cl CH3CN CH3COOH
142.
The order of basic strength of the following in aqueous solution is
3
1) C6 H 5 NH 2
3
(CH 3 )3 N
NH 3
CH 3 NH 2
(CH3 )2 NH
1) 4 1 5 3 2 2) 2 5 4 3 1 3) 5 4 2 3 1 4) 4 3 5 2 1 Sol:
Key: 3 Conceptual
143.
Yellow dye can be prepared by a coupling reaction of benzene diazonium chloride in acid medium with X. Identify X from the following 1) Aniline
Sol:
2) Phenol
3) Cumene
4) Benzene
Key: 1 aniline yellow is produced
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
144.
CODE : C
Sri Chaitanya Jr College
In which of the following the product of uncertainity in velocity and uncertainity in position of a micro particle of mass ‘m’ is not less than 1) h
3 m
2)
h m 3
3)
h 1 4 m
4)
h m 4
h 4 m
Sol:
Key: 3 x.p
145.
An element has [ Ar ]3d 1 configuration in its +2 oxidation state. Its position in the periodic table is 1) period-3, group-3
2) period -3,group -7
3) period-4,group -3
4) period -3, group -9
Sol:
Key: 3 Conceptual
146.
In which of the following molecules all bond lengths are not equal ? 1) SF6
2) PCl5
3) BCl3
4) CCl4
Sol:
Key: 2 Conceptual
147.
In which of the following molecules maximum number of lone pairs is present on the central atom ? 1) NH 3
2) H 2O
3) CIF3
4) XeF2
Sol:
Key: 4 XeF2 contains 3 lonepairs on Xenon
148.
Which one of the following is the kinetic energy of of a gaseous mixture containing 3g of hydrogen and 80g of oxygen at temperature T(K) ? 1) 3 RT
Sol:
2) 6 RT 3 2
Key: 2 K .E nRT
3) 4 RT
4) 8 RT
n 6
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
149.
CODE : C
Sri Chaitanya Jr College
A,B,C and D are four different gases with critical temperatures 304.1,154.3,405.5 and 126.0 K respectively, While cooling the gas which gets liquefied first is 1) B
2) A
3) D
4) C
Sol:
Key: 4 Higher the critical temperature easier the liquification
150.
40 ml of x M KMnO4 solution is required to react completely with 200 ml of 0.02 M oxalic acid solution in acidic medium. The value of x is 1) 0.04
Sol:
Key: 1 V1 N1 V2 N2
151.
Given that
2) 0.01
3) 0.03
4) 0.02
C( g ) O2( g ) CO2( g ) ; H 0 xkJ 2CO( g ) O2( g ) 2CO2( g ) ; H 0 ykJ 1)
y 2x 3
2)
y 2x 2
3)
2x y 2
4)
x y 2
Sol:
Key: 2 Conceptual
152.
At 400 K, in a 1.0 L vessel N 2O4 is allowed to attain equilibrium
2 NO2( g ) N2O4( g ) At equilibrium the total pressure is 600 mm Hg, when 20% of N 2O4 is dissociated, The K p value for the reaction is 1) 50 Sol:
2) 100
3) 150
4) 200
Key: 2 N2O4 ƒ 2 NO2 Initial 1mol
0 mol
K P
2 PNO 2
PN2O4
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
CODE : C
Sri Chaitanya Jr College 2
Equi
1 0.2 0.8 mol
153.
0.4 600 1.2 100 KP 0.8 600 1.2
0.4
0.4 mol
In which of the following salts only cationic hydrolysis is involved? 1) CH3COONH 4
2) CH3COONa
3) NH 4Cl
Sol:
Key: 3 NH 4Cl is the salt of weak base and strong acid
154.
Calgon is 1) Na2 HPO4
2) Na3 PO4
Sol:
Key: 3 Conceptual
155.
Consider the following statements
3) Na6 P6O18
4) Na2 SO4
4) NaH 2 PO4
I) CS ion is more highly hydrated than other alkali metal ions II) Among the alkali metals, only lithium forms a stable nitride by direct combination with nitrogen III) Among alkali metal Li, Na, K, Rb, the metal, Rb has the highest melting point IV) Among alkali metals Li, Na, K, Rb only Li forms peroxide when heated with oxygen 1)I
2) II
3) III
4) IV
Sol:
Key: 2 6Li N2 2Li3 N
156.
Assertion (A) : AlCl3 exists as a dimer through halogen birdged bonds. Reasson (R): AlCl3 gets stability by accepting electrons from the bridged halogen. 1) Both (A) and (R ) are true and ( R) is the correct explanations of (A ) 2) Both (A) and (R ) are true but ( R) is not the correct explanations of (A ) 3) (A) is true, but (R ) is not true.
Sol:
4) (A) is not true, but (R ) is true.
Key: 1 Conceptual Page No : 51
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AP EAMCET 2016 Solutions by Sri Chaitanya
A.P EAMCET (Engineering)
157.
CODE : C
Sri Chaitanya Jr College
Which of the following causes “Bule baby syndrome” 1) High concentration of lead in drinking water 2) High concentration of sulphates in drinking water 3) High concentration of nitrates in drinking water 4) High concentration of copper in drinking water
Sol:
Key: 3 Conceptual
158.
Which of the following belong to the homologous series of C5 H8O2 N ? 1) C6 H10O3 N
2) C6 H8O2 N2
3) C6 H10O2 N2
4) C6 H10O2 N
Sol:
Key: 4 Conceptual
159.
In Dumans method , 0.3 g of an organic compound gave 45 ml. of nitrogen at STP. The percentage of nitrogen is 1) 16.9
2) 18.7
3) 23.2
4) 29.6
V 45 45 18.7 8W 8 0.3 2.4
Sol:
Key: 2 % N
160.
The IUPAC name of
CH3 2 CH CH CH CH CH CH CH 3 C2 H 5
Sol:
1) 2,7 - dimethyl -3,5-nonadiene
2) 2-7-dimethyl-2- ehtylheptadiene
3) 2- methyl-7-ethyl1-3, 5-octadiene
4) 1,1-dimethyl -6-ethyl -2,4- heptadiene
Key: 1 Conceptual
Page No : 52
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