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30 Questions MCQ Test AMU B.Tech Entrance Exam Mock Tests - AMU B.Tech Entrance Exam Mock Test- 9

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AMU B.Tech Entrance Exam Mock Test- 9 - Question 1

In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26G and the dip angle is 60o. What is the magnetic field of the earth at this location

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 1

The earth's magnetic field is Be​ and its horizontal and vertical components are He​ and Hv​
cosθ= He​​/Be
∴cos60o= (​0.26×10−4​/ Be )T
⇒Be​=(​0.26×10−4)/ (½)​=0.52×10−4T=0.52G

AMU B.Tech Entrance Exam Mock Test- 9 - Question 2

A body of mass 0.5 kg travels on a straight line path with velocity v = (3x2 + 4) m/s. The net work done by the force during its displacement from x = 0 to x = 2 m is:

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 2

Solution:

  • Mass: 0.5 kg
  • Initial velocity at x = 0:
    • Formula: v = 3x2 + 4
    • Substitute: v = 3(0)2 + 4 = 4 m/s
  • Final velocity at x = 2:
    • Substitute: v = 3(2)2 + 4 = 16 m/s
  • Calculate the change in kinetic energy:
    • Initial kinetic energy: (1/2) × 0.5 × 42 = 4 J
    • Final kinetic energy: (1/2) × 0.5 × 162 = 64 J
  • Net work done by the force:
    • Work done = Change in kinetic energy = 64 J - 4 J = 60 J

Therefore, the net work done by the force is 60 J.

AMU B.Tech Entrance Exam Mock Test- 9 - Question 3

Planck's constant has the same unit as:

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 3
  • Force × distance = work done
  • The unit of work done is Joule. 
  • Therefore the unit of 
  • Force × distance × time = Joule × second
  • This is the same unit as Planck's constant.

So, the correct option is Force × distance × time

AMU B.Tech Entrance Exam Mock Test- 9 - Question 4

Select an incorrect alternative:
i. the radius of the nth orbit is proprtional to n2
ii. the total energy of the electron in the nth orbit is inversely proportional to n
iii. the angular momentum of the electron in nth orbit is an integral multiple of h/2π
iv. the magnitude of potential energy of the electron in any orbit is greater than its kinetic energy​

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 4

Statement i. Radius of Bohr's orbit of hydrogen atom is given by
r= n2h2​/4π2mKze2
or, r=(0.59A˚)(n2​/z)
So, from expression we found r∝n2
Hence the 1st statement is correct.
Statement ii.
 
We know that
En=-13.6 x z2/n2
So, En ∝1/n2
Hence the 2nd statement is wrong.
Statement iii.
Bohr defined these stable orbits in his second postulate. According to this postulate:

  • An electron revolves around the nucleus in orbits
  • The angular momentum of revolution is an integral multiple of h/2π – where Planck’s constant [h = 6.6 x 10-34 J-s].
  • Hence, the angular momentum (L) of the orbiting electron is: L = nh/2 π

 Hence the 3rd statement is correct.
Statement iv.
According to Bohr's theory
Angular momentum of electron in an orbit will be Integral multiple of (h/2π)
Magnitude of potential energy is twice of kinetic energy of electron in an orbit
∣P.E∣=2∣K.E∣
K.E=(13.6ev)( z2/n2)​
Hence, The 4th statement is correct.

AMU B.Tech Entrance Exam Mock Test- 9 - Question 5

In the circuit shown, if a conducting wire is connected between points A and B, the current in this wire will [2006]

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 5

Current will flow from B to A

Potential drop over the resistance CA will be
more due to higher value of resistance. So
potential at A will be less as compared with at
B. Hence, current will flow from B to A.

AMU B.Tech Entrance Exam Mock Test- 9 - Question 6

A particle of mass m moving with constant velocity v strikes another particle of same mass m but moving with the same velocity v in opposite direction stick together. The joint velocity after collision will be

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 6

Concept:

  1. Momentum: momentum is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.
  2. The unit of momentum (P) is kg m/s.
  3. Dimension: [MLT-1]
  4. Law of conservation of Momentum: A conservation law stating that the total linear momentum of a closed system remains constant through time, regardless of other possible changes within the system.
  5. P= P2
  6. m1 v1 = m2 v2
  7. Where, P1 = initial momentum of system, P2 = final momentum of system, m1 = mass of first object, v1 = velocity of first object, m= mass of second object and v2 = velocity of second object.

Calculation:
Given:  m1 = m kg,  m2 = m kg,  u= v m/s,  u2 ​=  -v m/s

Let the common velocity of the combined body be V m/s

Mass of combined body      M = m + m = 2m

Applying conservation of momentum:          

mv1 + m2 v2 = M V

mv + (-mv) = 2mV

0 = 2mV

V = 0 m/s
Hence the correct answer will be zero (0) m/s.

AMU B.Tech Entrance Exam Mock Test- 9 - Question 7

A silver ball of radius 4.8 cm is suspended by a thread in the vacuum chamber. UV light of wavelength 200 nm is incident on the ball for some times during which a total energy of 1 × 10–7 J falls on the surface. Assuming on an average one out of 103 photons incident is able to eject electron. The potential on sphere will be

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 7

AMU B.Tech Entrance Exam Mock Test- 9 - Question 8

At resonance the impedance of RLC circuit is

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 8

Resonance is a condition in LCR circuit where the inductive reactance ( XL) and capacitive reactance (Xc) is equal.
Impedance at resonance is equal to the Resistance of the resistor because the other two cancel themselves off according to the below formula -
Z2=R2+(XL2-Xc2)
As XL = Xc ,
Z = R
 

AMU B.Tech Entrance Exam Mock Test- 9 - Question 9

Equal masses of three liquids of specific heats C1, C2 and C3 at temperatures t1, tand t3 respectively are mixed. If there is no change of state, the temperature of the mixture is

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 9

For the composite system, energy conservation yields no net energy flow in or out of the system.
Let final temperature be T
Then, heat absorbed by A+heat absorbed by B+heat absorbed by C=0

Here in the question m1=m2=m3

AMU B.Tech Entrance Exam Mock Test- 9 - Question 10

The restoring force in a simple harmonic motion is _________ in magnitude when the particle is instantaneously at rest.

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 10

The restoring force in a simple harmonic motion is maximum in magnitude when the particle is instantaneously at rest because in SHM object’s tendency is to return to mean position and here particle is instantaneously at rest after that instant restoring force will be max to bring particle to mean position.

AMU B.Tech Entrance Exam Mock Test- 9 - Question 11

The correct statement about the following reaction is : 

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 11

It is Benzil-benxilic acid rearrangement reaction.

AMU B.Tech Entrance Exam Mock Test- 9 - Question 12

Which of the following reactions produces an alkene?

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 12

Option a) Diazene(N2H2) is a hydrogenating agent. So, there will be no reaction.
Option b) Al2O3+CrO3 acts as a dehydrogenation catalyst and so an alkene is formed. (Here 1-propene is formed)
Option c) This reaction is Wolff Kishner reaction. Here, acetone would be converted to alkane.
Option d) Zn/CH3COOH substitutes Cl with H and an alkane is formed.

AMU B.Tech Entrance Exam Mock Test- 9 - Question 13

For the reaction in equilibrium, A B

Thus, K is

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 13

As the reaction is in equilbrium, therefore
-d[A]/dt = d[B] /dt
⇒2.3 × 106 [A] = k [B]     -- (1)
 as given in question [B] /[A] = 4 × 108
so [A] / [B] = 1/ 4 ×10-8

Using equation (1), we get
⇒ 2.3 × 106 . [A] /[B] = k

Substituting value of  [A] / [B], we get
⇒  2.3 × 106 / 4 × 108 = k
Or k = 5.8 × 10-3 /sec¹

AMU B.Tech Entrance Exam Mock Test- 9 - Question 14

xSo2 + yMnO-4 + zH2O → Mn2+ + SO42- + H+ . In this, three values of x, y, z are

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 14

AMU B.Tech Entrance Exam Mock Test- 9 - Question 15

An ideal gas expands in volume from 1 × 10–3m3 to 1 × 10–2m3 at 300K against a constant pressure of 1 × 105Nm–2. The work done is

[AIEEE 2004]

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 15

AMU B.Tech Entrance Exam Mock Test- 9 - Question 16

 obtained by chlorination of n-butane, will be [2001]

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 16

Chlorination of n-butane taken place via free radical formation i.e., 

This will lead to racemisation i.e.racemic form.

AMU B.Tech Entrance Exam Mock Test- 9 - Question 17

Direction: In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A) : Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97°C) and propanone (boiling point 56°C).

Reason (R) : Liquids with a difference of more than 20°C in their boiling points can be separated by simple distillation.

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 17
Simple distillation technique is used for separating components that have a difference of more than 20C in their boiling points. Thus, the statement 'simple distillation can help in separating a mixture of propan-1-ol (boiling point 97C) and propanone (boiling point 56C)' is correct. Thus, the assertion is correct.
AMU B.Tech Entrance Exam Mock Test- 9 - Question 18

The equivalent conductance of  solution ofa weak monobasic acid is 8.0 mhos cm2 and at infinite dilution is 400 mhos cm2. The dissociation constant of this acid is: [2009]

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 18

Degree of dissociation

= 1.25 x 10-5

AMU B.Tech Entrance Exam Mock Test- 9 - Question 19

Ethylene oxide, when treated with Grignard reagent, yields

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 19

Ethylene oxide, when treated with Grignard reagent, gives primary alcohol.

AMU B.Tech Entrance Exam Mock Test- 9 - Question 20

Complete hydrolysis of XeF4 and XeF6 results in the formation of:

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 20

Complete hydrolysis of XeF4 and XeF6 results in the formation of XeO3.
6XeF4 + 12 H2O ----> 2XeO3 + 4Xe + 3O2 + 24HF
XeF6 + 3H2O ----> XeO3 + 6HF

AMU B.Tech Entrance Exam Mock Test- 9 - Question 21

On mixing certain alkane with chlorine and irradiating it with ultraviolet light, it forms only one monochloro alkane. The alkane could be

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 21

Neopentane gives only one monochloro product.

AMU B.Tech Entrance Exam Mock Test- 9 - Question 22

Statement Type
This section is based on Statement I and Statement II. Select the correct answer from the codes given below.

Statement I: N2 is less reactive than P4.
Statement II: Nitrogen has more electron gain enthalpy than phosphorus.

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 22

P — P single bond (213 kJ mol-1)  in Pmolecule is much weaker than N ≡ N triple bond (941.4 kJ mol-1) in N2

Thus, Statement I is correct and Statement II is incorrect.

AMU B.Tech Entrance Exam Mock Test- 9 - Question 23

The reciprocal of the distance between two points, one on each of the lines
(x - 2)/3 = (y - 4)/2 = (z - 5)/5 and (x - 1)/2 = (y - 2)/3 = (z - 3)/4, Then which of the following is FALSE.

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 23

The shortest distance (SD)

= 1 / √78

So, 1 / SD = √78

AMU B.Tech Entrance Exam Mock Test- 9 - Question 24

A person has to catch a train. To catch train, from his home he can take a taxi or take rickshaw or walk by foot with respective probabilities 1/2,1/3,1/6. Probability of him catching train when he takes rickshaw from his home is half that of when the takes the taxi and probability of catching the train when he walked by foot is 1/4 th that of when he takes rickshaw. He finally reached the train, the probability he walked by his foot to catch the train, is

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 24

T : took a taxi
R : Took a rickshaw
W : Walked by foot
P(T) = 1/2, P(R) = 1/3, P(W) = 1/6
C : Catching the train

AMU B.Tech Entrance Exam Mock Test- 9 - Question 25

If  then a2 + b2 is

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 25


Taking magnitude and squaring on both sides, we get

AMU B.Tech Entrance Exam Mock Test- 9 - Question 26

Calculate the least whole number, which when subtracted from both the terms of the ratio 5 : 6 gives a ratio less than 17 : 22.

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 26

Given:

Initial ratio = 5 ∶ 6

Final ratio should be less than 17 ∶ 22

Calculation:

Let the least whole number that is needed to be subtracted be a.

According to the question,

(5 - a)/(6 - a) < 17/22

⇒ 5 × 22 - 22a < 17 × 6 - 17a 

⇒ 110 - 22a < 102 - 17a 

⇒ 110 - 102 < - 17a + 22a 

⇒ 8 < 5a 

⇒ 8/5 = 1.6 < a 

∴ The least whole number must be 2.

AMU B.Tech Entrance Exam Mock Test- 9 - Question 27

Let A be a vertex of the ellipse S ≡ (x² / 4) + (y² / 9) - 1 = 0 and F be a focus of the ellipse S' ≡ (x² / 9) + (y² / 4) - 1 = 0. Let P be a point on the major axis of the ellipse S' = 0, which divides OF in the ratio 2:1 (O is the origin). If the length of the chord of the ellipse S = 0 through A and P is , then k = ? ​

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 27

We have, S ≡ (x² / 4) + (y² / 9) - 1 = 0

∴ A(0, ±3) (vertex =(0, ±b)

Again,

S' ≡ (x² / 9) + (y² / 4) - 1 = 0

∴  F(√±5, 0) (Focus =(±ae, 0)

Again P is a point on the major axis of the ellipse S′ = 0, which divides of in the ratio 2:1.

Now, equation of line passing through A and P is given by

Now, intersection point of line in Eq. (i) and S = 0 are (0,3) and .

∴ Length of chord

∴ k = 7.

*Answer can only contain numeric values
AMU B.Tech Entrance Exam Mock Test- 9 - Question 28

Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.

Q. The points (1, 3) and (5, 1) are two opposites vertices of a rectangle and the other two vertices lie on the line y – 2x + C = 0. Then the value of C is


Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 28

Since the diagonals of a rectangle bisect each other, so the point

lies on y = 2x - C, which gives C = 4

AMU B.Tech Entrance Exam Mock Test- 9 - Question 29

 

If  = x + 4,x ≠ −5/3,2/3 and ∫f(x)dx = Ax + Bln|3x−2| + C, then 3B − A =

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 29

We have,

Put,

Put, 3x − 2 = t ⇒ x = t + 2 / 3

AMU B.Tech Entrance Exam Mock Test- 9 - Question 30

If A and B are two square matrices such that B = –A–1 BA, then (A+B)2 is equal to

Detailed Solution for AMU B.Tech Entrance Exam Mock Test- 9 - Question 30

B = –A–1 BA
AB = A (–A–1BA)

AB = –BA
AB + BA = 0
Now, (A + B)2 = A2 + B2 + AB + BA
(A + B)2 = A2 + B2, (∵AB + BA = 0)

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