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JEE Advanced Practice Test- 6 - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Advanced Practice Test- 6

JEE Advanced Practice Test- 6 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2025 preparation. The JEE Advanced Practice Test- 6 questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Practice Test- 6 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Practice Test- 6 below.
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JEE Advanced Practice Test- 6 - Question 1

The coefficient of linear expansion for a certain metal varies with absolute  temperature as  α = aT. If Lo = 1 m is the initial length of the metal and the temperature of metal is changed from 27oC to 127oC, then final length is (where a = 2 × 10-5 K−2)  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 1


JEE Advanced Practice Test- 6 - Question 2

The radiations that can ionize the matter are

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JEE Advanced Practice Test- 6 - Question 3

A billiard player hits a billiard ball of mass m and radius r lying at rest on the billiard table. When the cue hits the ball horizontally with force F at a height (r + h) from the table top the ball rolls without slipping. h is given as  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 3


JEE Advanced Practice Test- 6 - Question 4

For c = 3a  = 2b, the magnetic field at point P is  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 4

Assuming z-axis to be coming out of the plane of the paper  


JEE Advanced Practice Test- 6 - Question 5

A conducting rod is rotated by means of strings in a uniform magnetic field with constant angular velocity as shown in the figure. Potential of points A, B and C are VA, VB and VC respectively. Then  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 5

Points A and C will rotate on the same circle.  Thus VA = VC and VA > VB.  

JEE Advanced Practice Test- 6 - Question 6

In the circuit shown (R = 1 Ω) current in branch CD is    

Detailed Solution for JEE Advanced Practice Test- 6 - Question 6

Let us split the circuit and then rearrange it as shown:  


Since this represents a balanced wheatstone bridge, current in CD branch is zero.  

JEE Advanced Practice Test- 6 - Question 7

A Π s haped wire frame with a light, smooth, sliding wire connected to its open end (width of open end is l) is dipped in a soap solution and then raised, such that a soap film forms between the slider and the closed end of the wire frame. If the wire frame is now placed in a vertical plane and a block of mass m is hung from the slider then for mass m to remain in equilibrium the surface tension of the soap film is  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 7


Force due to surface tension,
F = 2T × l  (since there are two free surfaces)
For equilibrium F = mg
⇒  T = mg/2l

JEE Advanced Practice Test- 6 - Question 8

An artificial satellite of mass m is moving in a circular orbit at a height equal to the radius R of the Earth. Suddenly due to internal explosion the satellite breaks into two parts of equal masses. One part of the satellite stops just after the explosion and then falls to the surface of the Earth. The increase in the mechanical energy of the system (satellite + Earth) due to explosion will be (take acceleration due to gravity on the surface of Earth as g) 

Detailed Solution for JEE Advanced Practice Test- 6 - Question 8

Conserving momentum during the explosion 
Increase in mechanical energy is ΔE = ΔK + ΔU = ΔK + 0  

*Multiple options can be correct
JEE Advanced Practice Test- 6 - Question 9

In position A kinetic energy of a particle is 60 J and potential energy is −20 J. In position B, kinetic energy is 100 J and potential energy is 40 J. Then in moving the particle from A to B  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 9

Work done by conservative forces  
= Ui − Uf = −20 − 40 = −60 J
Work done by external forces
=  Ef – Ei = (Kf + Uf) – (Ki + Ui) = 40 – 140 =  −100 J  
Work done by all the forces    
= Kf – Ki = 100 – 60 = 40 J  

*Multiple options can be correct
JEE Advanced Practice Test- 6 - Question 10

A metal rod is fixed in horizontal position and a force of magnitude F is applied as shown. If RA = force by wall A and RB = force by wall B, then   

Detailed Solution for JEE Advanced Practice Test- 6 - Question 10

F. B. D. of the two parts of the rod are as shown.  

Also, increase in length of 1 = decrease in length of 2  

*Multiple options can be correct
JEE Advanced Practice Test- 6 - Question 11

Two waves travelling in opposite directions produce a standing wave. The individual wave functions are given by y1 = 4 sin (3x – 2t) cm and y2 = 4 sin (3x + 2t) cm, where x and y are in centimeter. Now, select the correct statement(s) from the following.

Detailed Solution for JEE Advanced Practice Test- 6 - Question 11

y = y1 +  y2  = (8 sin 3x. cos 2t) = Axcos 2t  
Where Ax = 8  sin 3x
∴  Nodes are formed where 3x = 0, π, 2π, ….. etc.  

*Multiple options can be correct
JEE Advanced Practice Test- 6 - Question 12

A point source S is placed anywhere in between two converging mirrors having focal lengths f and 2f, respectively as shown. The value of d for which only single image may be formed is /are 

Detailed Solution for JEE Advanced Practice Test- 6 - Question 12

Only one image will be formed if the object is either at the common focus or common centre of curvature of both the mirrors. The ray diagrams are as shown. 

JEE Advanced Practice Test- 6 - Question 13

In the circuit shown, initially keys K1, K2 and K3 are all open. Now certain events as described below are performed successively on the circuit which involve closing and opening of the keys K1, K2 and K3.  

Event 1: Keys K1 and K2 are closed and K3 is left open.  
Event 2: After a long time K2 is opened and K3 is closed.  
Event 3: After a long time K1 is opened and K2 is closed. 

Q. 

The time constants for charging of capacitor (completion of Event 1) and rise of current through the inductor (completion of Event 2) respectively are  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 13

Equivalent circuit after completion of Event 1 is  


Equivalent circuit after completion of Event 2 is  

JEE Advanced Practice Test- 6 - Question 14

In the circuit shown, initially keys K1, K2 and K3 are all open. Now certain events as described below are performed successively on the circuit which involve closing and opening of the keys K1, K2 and K3.  

Event 1: Keys K1 and K2 are closed and K3 is left open.  
Event 2: After a long time K2 is opened and K3 is closed.  
Event 3: After a long time K1 is opened and K2 is closed. 

Q. 

Find the maximum charge that can come on the capacitor after the completion of all the three events  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 14


Steady state charge on capacitor is  

Steady state current (I) through the inductor is  

Thus the capacitor again starts charging.  
Charge on the capacitor will reach maximum value when current through the inductor becomes zero.  
From conservation of energy between initial state (when i = 5 A, q = 40 μC) and final state (when i = 0, q = Qmax) we get

JEE Advanced Practice Test- 6 - Question 15

In the circuit shown, initially keys K1, K2 and K3 are all open. Now certain events as described below are performed successively on the circuit which involve closing and opening of the keys K1, K2 and K3.  

Event 1: Keys K1 and K2 are closed and K3 is left open.  
Event 2: After a long time K2 is opened and K3 is closed.  
Event 3: After a long time K1 is opened and K2 is closed. 

Q. 

Find the maximum current that can pass through the inductor after the completion of all the three events 

Detailed Solution for JEE Advanced Practice Test- 6 - Question 15

After completion of Event 3

The capacitor again starts charging, the current through inductor starts decreasing, becomes zero, changes direction and again starts increasing and goes beyond the value of 5 A.
Current through the inductor will reach maximum value when charge on the capacitor has become zero.
From conservation of energy between initial state (when i = 5 A, q = 40 μC) and final state (when i = Imax, q = 0) we get

JEE Advanced Practice Test- 6 - Question 16

A liquid flowing from a vertical pipe has a very definite shape as it flows from the pipe. To get the equation for this shape assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed vo and the radius of the cross section of the liquid stream is ro.  

Q. 

An equation for the speed v of the liquid as a function of the distance y it has fallen, is 

Detailed Solution for JEE Advanced Practice Test- 6 - Question 16

Applying Bernoulli’s equation between the initial crosssection and the cross-section at a distance ‘y’ below the pipe we get,  

Since both cross-sections are in atmosphere, the pressure there is atmospheric pressure,  

JEE Advanced Practice Test- 6 - Question 17

A liquid flowing from a vertical pipe has a very definite shape as it flows from the pipe. To get the equation for this shape assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed vo and the radius of the cross section of the liquid stream is ro.  

Q. 

The expression for the radius of the cross-section of the liquid stream at a distance ‘y’ below the pipe is  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 17

Applying equation of continuity between the initial cross-section and the cross section at a distance ‘y’ below the pipe, we get  

Using the expression for ‘v’ from above question, we get 

JEE Advanced Practice Test- 6 - Question 18

The correct order of acidic strength of the compounds, 1 – 3 is 

JEE Advanced Practice Test- 6 - Question 19

The mechanism involved in the following conversion is 

JEE Advanced Practice Test- 6 - Question 20

The correct order of bond angle of the following compound is  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 20

JEE Advanced Practice Test- 6 - Question 21

The rate law for one of the mechanism of the pyrolysis of CH3CHO at 520oC and 0.2 bar is 

The overall activation energy Ea in terms of rate law is: 

Detailed Solution for JEE Advanced Practice Test- 6 - Question 21

From Arrhenius equation, K = AeEa/RT
From rate equation given in question


Hence (C) is correct answer.   

JEE Advanced Practice Test- 6 - Question 22

The correct IUPAC name of compound given below is 

JEE Advanced Practice Test- 6 - Question 23

The stopping potential (V0) of photosensitive surface varies with the frequency of incident radiation as shown in the following figure.  

The work function of photosensitive surface. 

Detailed Solution for JEE Advanced Practice Test- 6 - Question 23


From photoelectric effect we have 

From equation (ii) and figure given in question, we have  v0 = 4×1015Hz
The work function of photosensitive surface is given by the relation: 

JEE Advanced Practice Test- 6 - Question 24

5.6 litre of an unknown gas at NTP requires 12.5 calorie to raise its temperature by 10oC at constant volume. 

Detailed Solution for JEE Advanced Practice Test- 6 - Question 24

Moles of gas 

JEE Advanced Practice Test- 6 - Question 25

The value of (PVm)P→0 of a real gas is independent of the nature of gas because  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 25

At very low pressure the molar volume of gas is very high therefore; Vm >>>b , and molecular attraction are also negligible. Thus gas behaves as an ideal gas.  
Hence (A) is correct answer. 

*Multiple options can be correct
JEE Advanced Practice Test- 6 - Question 26

Which of the following oxide cannot be reduced by carbon reduction?  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 26

MnO2 and Cr2O3 are reduced by aluminum. 

*Multiple options can be correct
JEE Advanced Practice Test- 6 - Question 27

Which of the following can show both optical and geometrical isomeris?  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 27

[Co(NH3)3(SO4)(NO2)Cl] forms four geometrical isomers (two cis and two trans) one of the cis-isomer is optically active.


Cis isomer is optically active due absence of plane of symmetry

*Multiple options can be correct
JEE Advanced Practice Test- 6 - Question 28

Which of the following is (are) correct statement (s)?

Detailed Solution for JEE Advanced Practice Test- 6 - Question 28

A is correct because nitrogen in +3 oxidation state.
HOF is unstable but acts as very strong oxidizing agent.
moles of CH3COONa = 150 x 0.1 = 15  
moles of HCl = 50 x 0.1 = 5
moles of CH3COONa unreacted = 10 
moles of weak acid CH3COOH = 5  
hence solution is an acidic mixture.  
(D) is correct because pH of the salts = (pKa1+pKa2)/2 

*Multiple options can be correct
JEE Advanced Practice Test- 6 - Question 29

Which of the following gives Cannizzaro’s reaction?  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 29

The presence of electron donating group decreases the reactivity of benzaldehyde derivatives towards Cannizzaro’s reaction. 

*Multiple options can be correct
JEE Advanced Practice Test- 6 - Question 30

Which of the following have identical bond order?  

Detailed Solution for JEE Advanced Practice Test- 6 - Question 30

Bond ord er of , SO2 and C6H6 is 1.5, and bond order of O2 is 2. 

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