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AP EAMCET Mock Test - 2 - JEE MCQ


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30 Questions MCQ Test AP EAMCET Mock Test Series - AP EAMCET Mock Test - 2

AP EAMCET Mock Test - 2 for JEE 2024 is part of AP EAMCET Mock Test Series preparation. The AP EAMCET Mock Test - 2 questions and answers have been prepared according to the JEE exam syllabus.The AP EAMCET Mock Test - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for AP EAMCET Mock Test - 2 below.
Solutions of AP EAMCET Mock Test - 2 questions in English are available as part of our AP EAMCET Mock Test Series for JEE & AP EAMCET Mock Test - 2 solutions in Hindi for AP EAMCET Mock Test Series course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt AP EAMCET Mock Test - 2 | 160 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study AP EAMCET Mock Test Series for JEE Exam | Download free PDF with solutions
AP EAMCET Mock Test - 2 - Question 1
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A voltmeter has a resistance of G ohm and range V volt. The value of resistance used in series to convert it into voltmeter of range nV volt is

AP EAMCET Mock Test - 2 - Question 2
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Light rays of wavelengths 6000 Å and of photon intensity 39 x 6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x 10-34  J − s ; Velocity of light = 3 x 108 m s-1 ]

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AP EAMCET Mock Test - 2 - Question 3
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Two spherical conductors A and B having equal radii and carrying equal charges on them, repel each other with a force F, when kept apart at some distance. A third spherical conductor C having same radius as that of A, but uncharged, is brought in contact with A, then brought in contact with B and finally removed away from both. The new force of repulsion between the conductors A and B is

AP EAMCET Mock Test - 2 - Question 4
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Neutron was discovered by
AP EAMCET Mock Test - 2 - Question 5
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In a Huygen's eye piece with an eyepiece of focal length F, the distance between the eye-piece and field lens should be
AP EAMCET Mock Test - 2 - Question 6
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Optical fibres uses the phenomenon of
AP EAMCET Mock Test - 2 - Question 7
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To a charged particle, which is moving with a constant initial velocity V → , a uniform magnetic field is applied in the direction of velocity; then
AP EAMCET Mock Test - 2 - Question 8
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In short wave communication waves of which of the following frequencies will be reflected back by the ionospheric layer having electron density 1011 per m-3?
Detailed Solution for AP EAMCET Mock Test - 2 - Question 8

N max = 1011 per m-3 

AP EAMCET Mock Test - 2 - Question 9
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When a wire of resistance R is stretched and its radius becomes r/2, then the new resistance will be
AP EAMCET Mock Test - 2 - Question 10
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Photoelectric effect proves the
AP EAMCET Mock Test - 2 - Question 11
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Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining centre to this point makes an angle θ with the axis of the dipole, then potential at the point P is given by
AP EAMCET Mock Test - 2 - Question 12
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Two charges 'Q' and 'Q' are kept at a distance 'd' .The charge 'q' is kept at the midpoint of line joining the charges. The magnitude of 'q' to keep the system in equilibrium is
AP EAMCET Mock Test - 2 - Question 13
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The function of the moderator in a nuclear reactor is
AP EAMCET Mock Test - 2 - Question 14
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In an n-p-n transistor, p-acts as a/an
AP EAMCET Mock Test - 2 - Question 15
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In an intrinsic - semiconductor the charge carriers responsible for electrical conduction are
AP EAMCET Mock Test - 2 - Question 16
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If f(3 − x) = f(x), then  is equal to
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AP EAMCET Mock Test - 2 - Question 17
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If X and Y are independent binomial variates  then 
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AP EAMCET Mock Test - 2 - Question 18
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is equal to
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AP EAMCET Mock Test - 2 - Question 19
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The solution of the differential equation
Detailed Solution for AP EAMCET Mock Test - 2 - Question 19

Now, differential equation is of type 
so assumey 

from (i) and (ii),

Solving equation (iii), we get

integrating both sides

Solving equation (iv), we get

integrating both sides, 

AP EAMCET Mock Test - 2 - Question 20
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A parabola is drawn with focus at one of the foci of the ellipse  where a > b, and directrix passing through the other focus and perpendicular to the major axis of the ellipse. If the latus rectum of the ellipse and that of the parabola are same, then the eccentricity of the ellipse is:
Detailed Solution for AP EAMCET Mock Test - 2 - Question 20

For ellipse, equation: 
Let focus be (ae, 0) and directrix be x + ae = 0


And, for parabola: y2 = 4aex

Length of latus rectum is 4ae.

According to the question,

AP EAMCET Mock Test - 2 - Question 21
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If α, β, γ​​​​​​​ are the roots of the equation 2x3 - 3x2 + 6x + 1 = 0, then α2 + β2​​​​​​​ + γ2​​​​​​​ is equal to
Detailed Solution for AP EAMCET Mock Test - 2 - Question 21


Sum of the roots = 3/2
Sum of product (two at a time) of roots = 6/2 = 3
Required Answer = (3/2)2 - 2 x 3
= -15/4
(Note : Answer is negative, It means there must be 2 roots which are imaginary. This is because the sum of square of real number can never be negative.
Imaginary roots are always in pair. )
Correct answer is Option (a)

AP EAMCET Mock Test - 2 - Question 22
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The number of common tangents to the circles x2 + y2 + 2x + 8y − 25 = 0 and x2 + y2 − 4x − 10y + 19 = 0 is
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The equation of circle is x2 + y2 + 2gx + 2fy + c = 0 with centre (-g, -f) ......(1)
Comparing the equation x2 + y2 + 2x + 8y − 25 = 0 with (1), we get,
g = 1 and f = 4
∴ Centre of first circle C1 (−1, −4).

Comparing the equation x2 + y2 − 4x − 10y + 19 = 0 with (1), we get
g = −2 and f = −5
∴ Centre of second circle C2 (2, 5).

 ⇒ It means circles intersecting each other.
Hence, number of common tangents = 2

AP EAMCET Mock Test - 2 - Question 23
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AP EAMCET Mock Test - 2 - Question 24
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If a2 + 2b = 7, b2 + 4c = −7 and c2 + 6a = −14, then the value of (a2 + b2 + c2) is
Detailed Solution for AP EAMCET Mock Test - 2 - Question 24

We have,

a2 + 2b = 7   …(i)

b2 + 4c = −7   …(ii)

c2 + 6a = −14   …(iii)

Adding equations (i), (ii) and (iii), we have

a2 + 2b + b2 + 4c + c2 + 6a = 7 − 7 − 14

⇒ a2 + 6a + b2 + 2b + c2 + 4c = −14

⇒ (a2 + 6a + 9) + (b2 + 2b + 1) + (c2 + 4c + 4) = −14 + 9 + 1 + 4

⇒ (a2 + 6a + 9) + (b2 + 2b + 1) + (c2 + 4c + 4) = 0

⇒ (a + 3)2 + (b + 1)2 + (c + 2)2 = 0

It is possible only when a = −3, b = −1 and c = −2.

So,

a2 + b2 + c2

= (−3)2 + (−1)2 + (−2)2 = 14

AP EAMCET Mock Test - 2 - Question 25
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If f : R → R is defined by , where [x] is the greatest integer not exceeding x, then

is equal to
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AP EAMCET Mock Test - 2 - Question 26
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If g(x) = max(y2 − xy) (where 0 ≤ y ≤ 1), then the minimum value of g(x) (for real x) is:
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AP EAMCET Mock Test - 2 - Question 27
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For what value of 'a' will the lines represented by the equation ax2 - 6xy + 9y2 + 3x - 9y - 4 = 0 be perpendicular?
Detailed Solution for AP EAMCET Mock Test - 2 - Question 27

ax2 - 6xy + 9y2 + 3x - 9y - 4 = 0
This is a homogeneous second degree equation.
It represents a pair of perpendicular lines if
a + b = 0, where a is coefficient of x2 and b is coefficient of y2
Here, b = 9
a = a
⇒ a + 9 = 0
Or, a = -9

AP EAMCET Mock Test - 2 - Question 28
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If , then  is equal to
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AP EAMCET Mock Test - 2 - Question 29
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The area between the parabolas x2 = y/4 and x2 = 9y and the straight line y = 2 is
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Given curves x2 = y/4 and x2 = 9y are the parabolas whose equations can be written as 

Also, given y = 2
Now, shaded portion shows the required area which is symmetric.

AP EAMCET Mock Test - 2 - Question 30
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If the mean and coefficient of variation are 12 and 45% respectively, then its standard deviation is
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