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JEE Exam  >  JEE Test  >  AP EAMCET Mock Test Series  >  AP EAMCET Mock Test - 4 - JEE MCQ

AP EAMCET Mock Test - 4 Free Online Test 2026


Full Mock Test & Solutions: AP EAMCET Mock Test - 4 (160 Questions)

You can boost your JEE 2026 exam preparation with this AP EAMCET Mock Test - 4 (available with detailed solutions).. This mock test has been designed with the analysis of important topics, recent trends of the exam, and previous year questions of the last 3-years. All the questions have been designed to mirror the official pattern of JEE 2026 exam, helping you build speed, accuracy as per the actual exam.

Mock Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 180 minutes
  • - Total Questions: 160
  • - Analysis: Detailed Solutions & Performance Insights
  • - Sections covered: Physics, Chemistry, Mathematics

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AP EAMCET Mock Test - 4 - Question 1
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Find the elongation of the steel and brass wire in the fig. unloaded length of steel wire = 1.5 m, unloaded length of brass wire 0.3 cm, diameter of each wire = 0.3 cm. Young's modulus for steel = 20 × 1010 Pa and that for brass = 9.0 × 1010 Pa.

Detailed Solution: Question 1

For steel wire: Y = 20 × 1010 Pa; L = 1.5 m ; d = 0.3 cm = 0.3 × 10−2m
Therefore, area of cross-section of the steel wire,

The stretching force for the steel wire,
F = 4.0 + 6.0 = 10 kgf = 10 × 9.8 N
If l is extension in the steel wire, then

For brass wire: Y = 9.0 × 1010 Pa ; L = 1m;

The stretching force for the brass wire,

F = 6 kgf = 6 × 9.8 N

AP EAMCET Mock Test - 4 - Question 2
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When sounded together, a column of air and a tuning fork produce 4 beats per second. The tuning fork gives the lower note. The temperature of air is 15°C. When the temperature falls to 10°C, the two produce 3 beats per second. Find the frequency of the fork.

Detailed Solution: Question 2

Let v = Frequency of the tuning fork
v1 = Frequency of the air column
At 15°C, v1 = v + 4
At 10°C, v1 = v + 3
We know, velocity u = v
 u15 = (v + 4)
and u10 = (v + 3)
 
Also,  …(2)
From (1) and (2), we get

Using binomial expansion and neglecting terms, we get


Or, 1 + 
Or, (v + 3) = 113.2
 v = 110.2 Hz  110 Hz

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AP EAMCET Mock Test - 4 - Question 3
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A particle moves in xy plane according to the law x = a sin ωt and y = a(1 - cos ωt), where a and w are constant. The particle traces

Detailed Solution: Question 3

x = a sin t ...(i)
y = a(1 - cos t) ...(ii)
x2 = a2 sin2 ωt = a2(1 - cos2 ωt)
x2 = a2 - a2 cos2 t
a2 cos2 ωt = a2 - x2 ...(iii)
y = a(1 - cos t) = a - a cos t
a - y = a cos t
(a - y)2 = (a cos t)2
a2 + y2 - 2ay = a2 cos2 ωt ...(iv)
Put the value from equation (iii),
a2 + y2 - 2ay = a2 - x2
x2 + y2 - 2ay = 0
This is the equation of a circle.

AP EAMCET Mock Test - 4 - Question 4
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A semiconductor has an electron concentration of 8×1013 per cm3 and a hole concentration of 5 x 1012 per cm3. The electron mobility is 25000 cm2 V−1 s−1 
 and the hole mobility is 100 cm2 V. s−1 . Then,

Detailed Solution: Question 4

AP EAMCET Mock Test - 4 - Question 5
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A 2 μF, capacitor C1 is charged to a voltage 100 V and a 4 μF capacitor C2 is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?

Detailed Solution: Question 5

Loss in energy when two capacitors are connected in parallel is


AP EAMCET Mock Test - 4 - Question 6
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Figure shows, a glass prism ABC (refractive index 1.5), immersed in water (refractive index 43). A ray of light incident normally on face AB. If it is totally reflected at face AC then

Detailed Solution: Question 6

The ray falls normally on AB is refracted undeviated into the prism and incident on face AC at an angle of incidence i=θ.

So, for total reflection at AC face,

AP EAMCET Mock Test - 4 - Question 7
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A hollow sphere of mass M and radius R is initially at rest on a horizontal rough surface. It moves under the action of a constant horizontal force F as shown in the figure.

The frictional force between the sphere and the surface is

Detailed Solution: Question 7


Let a and α be the linear and angular accelerations of the sphere respectively.
For translational motion,
F + f = Ma ...(1)
The magnitude of the net torque acting on the sphere = FR - fR.
Hence, for rotational motion the equation is

  • FR - fR = Iα = Ia/R (∵ a = αR)

For a hollow sphere, I = 2/3 MR2. Hence
FR - fR = 2/3 MR2 ×  MRa
F - f = 2/3 Ma ...(2)
From Eqs. (1) and (2) we get f = .
Hence the correct choice is (4).

AP EAMCET Mock Test - 4 - Question 8
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The escape velocity from the earth is 11 km s−1. The escape velocity from a planet having twice the radius and same mean density as that of the earth is

Detailed Solution: Question 8

Escape velocity,

∴ve∝R if ρ=constant.

Since the planet is having double radius in comparision to earth, therefore escape velocity becomes twice ie, 22 km s−1 .

AP EAMCET Mock Test - 4 - Question 9
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A force of 1200 N acts on a 0.5 kg steel ball as result of collision lasting 25 ms. If the force is in a direction opposite to the initial velocity of 14 m s−1 then the final speed of the steel ball would be

Detailed Solution: Question 9

Initial velocity of steel ball u=14 m s−1
Mass of steel ball m=0.5
Force acting on the steel ball from opposite direction.

F = −1200 N
Time duration for collision,

t = 25 ms = 25 x 10−3 s
Let final velocity of ball=v
Applying Newton's II law of motion to steel ball ie,

[−ve sign shows that now ball will move in the direction of force]

AP EAMCET Mock Test - 4 - Question 10
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Three identical balls, ball I, ball II and ball III are placed on a smooth floor on a straight line at a separation of 10 m
between them as shown in the figure. Initially, the balls are stationary. Ball I is given a velocity of 10 m s−1 towards ball II. The collision between ball I and II is inelastic with coefficient of restitution 0.5 but collision between ball II
and III is perfectly elastic. What is the time interval between two consecutive collisions between ball I and II ?

Detailed Solution: Question 10

Let the velocity of ball I and ball III after collision be v1 and v2.
v2 − v1 = 0.5 x 10  ...(i)
mv2 + mv1 = m x 10   ...(ii)
⇒   v2 + v1=10
Solving equations (i) and (ii),

v1=2.5 m s−1 
v2=7.5 m s−1

Ball II after moving 10 m collides with ball III elastically and stops. But ball I moves towards ball II. Time taken between two consecutive collisions,

AP EAMCET Mock Test - 4 - Question 11
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A particle of mass 5 kg moving in the x−y plane has its potential energy given by U = (−7x+24y) Joule, x and y
being in meter. Initially at t =0 ,the particle is located at the origin (0, 0) and moving with a velocity of . Find the magnitude of force acting on the particle.

Detailed Solution: Question 11


Hence, magnitude of force is 25 N

AP EAMCET Mock Test - 4 - Question 12
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A charge Q is placed at a distance a/2 above the centre of a square surface of side length a. The electric flux through the square surface due to the charge would be?

Detailed Solution: Question 12


Charged particle can be considered at centre of a cube of side a, and given surface represents its one face. So, flux 

AP EAMCET Mock Test - 4 - Question 13
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A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of water in the capillary tube is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water, that will rise in this tube, is

Detailed Solution: Question 13

Mass of water in first tube is

Now, surface tension 
where h' is the height to which water rises in the second tube and r' its radius.
Since r' = 2r, h' = h/2.
Therefore, the mass of water in the second capillary tube is

= 

AP EAMCET Mock Test - 4 - Question 14
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If a = 18√v (where 'a' and 'v' are acceleration and velocity at any instant, respectively), then the acceleration when the time t = 1 second is

Detailed Solution: Question 14




At t = 1 sec,

AP EAMCET Mock Test - 4 - Question 15
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The following figure shows a spherical Gaussian surface and a charge distribution (magnitude of all the given point charges is different). When calculating the flux of electric field through the Gaussian surface, the electric field will be due to

Detailed Solution: Question 15

The electric flux is given by the surface integral. Here, the electric field E is due to charge inside the Gaussian surface only. Hence, the correct option is (d).

AP EAMCET Mock Test - 4 - Question 16
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If an unchanged capacitor is charged by connecting it to a battery, then the amount of energy lost as heat is

AP EAMCET Mock Test - 4 - Question 17
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Acetaldehyde cannot exhibit

AP EAMCET Mock Test - 4 - Question 18
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Two wires of same metal have the same length but their cross-sections are in the ratio 3:1. They are joined in series. The resistance of the thicker wire is 10 Ω. The total resistance of the combination will be

AP EAMCET Mock Test - 4 - Question 19
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The number of molecules of ATP produced in the lipid metabolism of a molecule of palmitic acid is

AP EAMCET Mock Test - 4 - Question 20
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Tollen's reagent is

AP EAMCET Mock Test - 4 - Question 21
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If , then n = ........

Detailed Solution: Question 21

We have,

⇒ 3 + 5 + 7 … + (2n + 1) = 440

⇒ n/2[2 x 3 + (n-1)(2) = 440

⇒ n(3 + n - 1) = 440

⇒ n(n + 2) = 440

⇒ n = 20

AP EAMCET Mock Test - 4 - Question 22
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The distance between 4x + 2y + 4z- 16 = 0 and 4x + 2y + 4z + 5 = 0 is 

Detailed Solution: Question 22

The distance between 4x + 2y + 4z- 16 = 0 and 4x + 2y + 4z + 5 = 0 is 

AP EAMCET Mock Test - 4 - Question 23
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Which of the following diagrams correctly represents intersection of sets A and B?

Detailed Solution: Question 23

Option (1) represents set A.
Option (2) represents set A - B.
∴ Only option (3) represents set A ∩ B.

AP EAMCET Mock Test - 4 - Question 24
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For LP.P, maximize z = 4x, + 2x2 subject to 3x1 + 2x2 ≥ 9, x1 - x2 ≤ 3, x1 ≥ 0, x2 ≥ 0 has...

Detailed Solution: Question 24

We have, maximise z = 4x1 + 2x2

Subject to contracts, 3x1 + 2x2 ≥ 9, x1 - x2 ≤ 3, x1 ≥ 0, x2 ≥ 0

On taking given constraints as equation, we get the following graphs

Here, we get feasible region is unbounded.

AP EAMCET Mock Test - 4 - Question 25
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Detailed Solution: Question 25

Since, we know that cos2θ + sin2θ = 1

Put cosx = t, −sinx dx = dtcos,

AP EAMCET Mock Test - 4 - Question 26
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The number of distinct real roots of the equation tan2 2x + 2tan2x tan3x − 1 = 0, in the interval  is:

Detailed Solution: Question 26

The given equation is tan2 2x + 2tan2x tan3x = 1

So the given equation becomes

To find the solutions in the given interval, put n=0,1,2,3

Hence, four values in the interval 

AP EAMCET Mock Test - 4 - Question 27
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If 2sec2α = tanβ + cotβ, then one of the values of (α + β) {where, (α + β) is not an odd multiple of π/2} is

Detailed Solution: Question 27

Given, 2sec2α = tanβ + cotβ


Taking +ve sign, we have

AP EAMCET Mock Test - 4 - Question 28
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 is equal to

Detailed Solution: Question 28


AP EAMCET Mock Test - 4 - Question 29
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If x and y are deviations from arithmetic mean, r=0.8, Σxy=60 , σy=2.5 and Σx2=90, then number of items in the series is,

Detailed Solution: Question 29

We know that

AP EAMCET Mock Test - 4 - Question 30
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For any non zero vector, a,b,c

a.[(b + c) x (a + b + c)] = ...

Detailed Solution: Question 30

We have, a.[(b + c) x (a + b + c)]

= a.[(b x a ) + (b x c) + (c x a) + (c x b)]

= a[(b x a) + ( b x c) + (c x a) - ( b x c)]

= a[(b x a) + (c x a)]

=[a b a] + [a c a] = 0 + 0 = 0

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About this JEE Mock Test

This mock test contains 160 questions designed to replicate the pattern, difficulty level and time pressure of the actual JEE exam. It is part of AP EAMCET Mock Test Series on EduRev, covering all the topics that are being asked as per the current JEE syllabus. Attempting this test under timed conditions will help you assess your exam readiness, identify subject-wise weak areas, and improve your speed and accuracy before the actual exam.

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