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AP EAMCET Mock Test - 4 - JEE MCQ


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30 Questions MCQ Test AP EAMCET Mock Test Series - AP EAMCET Mock Test - 4

AP EAMCET Mock Test - 4 for JEE 2024 is part of AP EAMCET Mock Test Series preparation. The AP EAMCET Mock Test - 4 questions and answers have been prepared according to the JEE exam syllabus.The AP EAMCET Mock Test - 4 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for AP EAMCET Mock Test - 4 below.
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AP EAMCET Mock Test - 4 - Question 1

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 for AP EAMCET Mock Test - 4 - 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

A uniform metal chain is placed on a rough table, such that one end of it hangs down over the edge of the table. When one-third of its length hangs over the edge, the chain starts sliding. Then, the coefficient of static friction is

Detailed Solution for AP EAMCET Mock Test - 4 - Question 2

Weight of the hanging portion of the chain = mg/3
Weight of the part of the chain on the table = 2mg/3

For the hanging part of the chain,


For the part of the chain on the table, we have

and we have

From 1, 2, 3 and 4, we have

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AP EAMCET Mock Test - 4 - Question 3

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 for AP EAMCET Mock Test - 4 - Question 3

Let the velocity of ball I and ball III after collision be v1 and v2.
v− v= 0.5 x 10  ...(i)
mv+ mv= m x 10   ...(ii)
⇒   v+ 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 4

In the figure, the blocks A, B and C each of mass m have acceleration a1, a2 and a3, respectively. F1 and F2 are external forces of magnitude 2mg and mg respectively. Then

Detailed Solution for AP EAMCET Mock Test - 4 - Question 4

For A, T = f = 2mg

2mg − mg = ma1

∴ a1 = g

For B,

From the force diagram shown in the figure,

2mg − mg = 3ma2

a2 = g/3

For C,

∴ 3mg − mg = 2ma3

⇒ a3 = g

So, a1 = a3 > a2

AP EAMCET Mock Test - 4 - Question 5

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 for AP EAMCET Mock Test - 4 - Question 5




At t = 1 sec,

AP EAMCET Mock Test - 4 - Question 6

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 for AP EAMCET Mock Test - 4 - Question 6

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 7

A charged particle of mass 5 x 10-6  Kg is held stationary in space by placing it in an electric field of strength 106 N C-1 directed vertically downwards. The charge on the particle is g = 10 m s-2

AP EAMCET Mock Test - 4 - Question 8

The most stable ion is

AP EAMCET Mock Test - 4 - Question 9
A capacitor of 10 μF charged up to 250 volts is connected in parallel with another capacitor of 5 μF charged up to 100 volts. The common potential is
AP EAMCET Mock Test - 4 - Question 10
The energy stored in a 50 mH inductor carrying a current of 4 A will be
AP EAMCET Mock Test - 4 - Question 11
Depletion layer consists of
AP EAMCET Mock Test - 4 - Question 12

n small drops of same size are charged to V volt each. They coalesce to form a bigger drop. The potential of the bigger drop is

AP EAMCET Mock Test - 4 - Question 13

A ray of light is incident at 60o on a prism of refracting angle 30o . The emergent ray is at an angle 30o with the incident ray. The value of refractive index of the prism is

Detailed Solution for AP EAMCET Mock Test - 4 - Question 13

Angle of prism = 30o = A
Angle of incidence = i1 = 60o
Angle of emergence = i2 = ?
Angle of devation = D = 30o
∴ i2 = A + D - i2 = 30 + 30 - 60 = 0o
⇒ r2 = 0o
∴ r1 = A-r2 = 30 - 0 = 30o
∴ μ = sin i 1 sin r 1 = sin 60o /sin 30o = √ 3

AP EAMCET Mock Test - 4 - Question 14

Let A and B are two events and P(A′) = 0⋅3, P(B) = 0⋅4, P(A∩B′) = 0⋅5 then P(A∪B′) is:

Detailed Solution for AP EAMCET Mock Test - 4 - Question 14

Given that P(A′) = 0⋅3, P(B) = 0⋅4, P(A∩B′) = 0⋅5

P(B′) = 1 − P(B) = 1 − 0⋅4 = 0⋅6

P(A) = 1 − P(A′) = 1 − 0⋅3 = 0⋅7

P(A∪B′) = P(A) + (B′) − P(A∩B′)

= 0⋅7 + 0⋅6 − 0⋅5 = 0⋅8

AP EAMCET Mock Test - 4 - Question 15

If , then n = ........

Detailed Solution for AP EAMCET Mock Test - 4 - Question 15
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 16

In a triangle ABC, if  then the projection of the vector  is equal to:

Detailed Solution for AP EAMCET Mock Test - 4 - Question 16


Projection of 

AP EAMCET Mock Test - 4 - Question 17

If  and g′(1)=2, g(1)=1 , then  is equal to

Detailed Solution for AP EAMCET Mock Test - 4 - Question 17

AP EAMCET Mock Test - 4 - Question 18

Which of the following diagrams correctly represents intersection of sets A and B?

Detailed Solution for AP EAMCET Mock Test - 4 - Question 18

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

AP EAMCET Mock Test - 4 - Question 19

The sum of 

Detailed Solution for AP EAMCET Mock Test - 4 - Question 19

For the case when the number n is not a positive integer the binomial theorem becomes, for -1 < x < 1,

AP EAMCET Mock Test - 4 - Question 20

The median of a grouped data is 39.8. The lower limit of the median class is 35 and the frequency is 10. The cumulative frequency of the preceding class of the median class is 34 and the total number of observations is 80. The class size of the grouped data is

Detailed Solution for AP EAMCET Mock Test - 4 - Question 20

l = lower boundary point of median class
c.f. = Cumulative frequency of the class preceding the median class
f = Frequency of the median class
n = total number of values or observations
h = class length of median class

AP EAMCET Mock Test - 4 - Question 21

If a random variable X follows the Binomial distribution B(33, p) such that 3P(X = 0) = P(X = 1), then the value of  is equal to:

Detailed Solution for AP EAMCET Mock Test - 4 - Question 21

3P(X = 0) = P(X = 1)

⇒ 1 - P = 11P
⇒ P = 1/12

⇒ 113 - 11 = 1320

AP EAMCET Mock Test - 4 - Question 22

The solution of the differential equation ydx - xdy = xy dx is ...

Detailed Solution for AP EAMCET Mock Test - 4 - Question 22
We have differential equation ydx - xdy = xydx

On integrating both sides, we get

log(x/y) = x ⇒ x/y = ex

⇒ x = yex

AP EAMCET Mock Test - 4 - Question 23

Detailed Solution for AP EAMCET Mock Test - 4 - Question 23

Since, we know that cos2θ + sin2θ = 1

Put cosx = t, −sinx dx = dtcos,

AP EAMCET Mock Test - 4 - Question 24

Five men and four women are to be seated on 9 adjacent seats in a cinema hall in such a way that no two women sit together. Find the number of ways in which they can be arranged.

Detailed Solution for AP EAMCET Mock Test - 4 - Question 24

* M1 * M2 * M3 * M4 * M5 *
Five men can be seated in 5! ways.
The women can occupy any of the places marked *.
Thus, the women can be seated in 6P4 = 360 ways.
 Total number of ways = 5! × 360 = 43200

AP EAMCET Mock Test - 4 - Question 25

 is equal to

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


AP EAMCET Mock Test - 4 - Question 26

The line y = mx intersects the circle x2 + y2 − 2x − 2y = 0 and x2 + y2 + 6x − 8y = 0 at points A and B (points being other than origin). The range of m such that origin divides AB internally is

Detailed Solution for AP EAMCET Mock Test - 4 - Question 26

   

Let C1: (x − 1)2 + (y − 1)2 = 2

C2: (x + 3)2 + (y − 4)2 = 52
Both the circle pass through the origin.

Hence, tangents at the origin (using T = 0) to C1 and C2 are x + y = 0 and 3x − 4y = 0, respectively.

The slope of the tangents are −1 and 3/4 respectively.

Thus, if  −1 < m < 3/4, then origin divides AB internally.

AP EAMCET Mock Test - 4 - Question 27

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 for AP EAMCET Mock Test - 4 - Question 27

We know that

AP EAMCET Mock Test - 4 - Question 28

The area (in sq. units) of the region bounded by y ≤ x+ 3x, 0 ≤ y ≤ 4 and 0 ≤ x ≤ 3 is

Detailed Solution for AP EAMCET Mock Test - 4 - Question 28

The area of the region ODC is

So, the area of the given region OABC is equal to
Area of OABD – Area of OCD

AP EAMCET Mock Test - 4 - Question 29

The first, second and last terms of an AP are a, b and 2a, respectively. Then, its sum is

Detailed Solution for AP EAMCET Mock Test - 4 - Question 29

First term = a
Common difference = (b - a)
Last term = 2a
=> Tn = 2a = a + (n - 1)(b - a)

AP EAMCET Mock Test - 4 - Question 30

If a2x4+b2y4=c4, then the maximum value of xy is

Detailed Solution for AP EAMCET Mock Test - 4 - Question 30

Therefore, maximum value of xy is 

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