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JEE Advanced Mock Test - 5 - 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 Mock Test - 5 - Question 1


JEE Advanced Mock Test - 5 - Question 2

The radiations that can ionize the matter are

JEE Advanced Mock Test - 5 - 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 Mock Test - 5 - Question 3


JEE Advanced Mock Test - 5 - Question 4

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

Detailed Solution for JEE Advanced Mock Test - 5 - Question 4

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


JEE Advanced Mock Test - 5 - 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 Mock Test - 5 - Question 5

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

JEE Advanced Mock Test - 5 - Question 6

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

Detailed Solution for JEE Advanced Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - Question 8

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

*Multiple options can be correct
JEE Advanced Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - Question 13

Equivalent circuit after completion of Event 1 is  


Equivalent circuit after completion of Event 2 is  

JEE Advanced Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - Question 18

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

JEE Advanced Mock Test - 5 - Question 19

The mechanism involved in the following conversion is 

JEE Advanced Mock Test - 5 - Question 20

The correct order of bond angle of the following compound is  

Detailed Solution for JEE Advanced Mock Test - 5 - Question 20

JEE Advanced Mock Test - 5 - 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 Mock Test - 5 - Question 21

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


Hence (C) is correct answer.   

JEE Advanced Mock Test - 5 - Question 22

The correct IUPAC name of compound given below is 

JEE Advanced Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - Question 24

Moles of gas 

JEE Advanced Mock Test - 5 - 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 Mock Test - 5 - 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 Mock Test - 5 - Question 26

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

Detailed Solution for JEE Advanced Mock Test - 5 - Question 26

MnO2 and Cr2O3 are reduced by aluminum. 

*Multiple options can be correct
JEE Advanced Mock Test - 5 - Question 27

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

Detailed Solution for JEE Advanced Mock Test - 5 - 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 Mock Test - 5 - Question 28

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

Detailed Solution for JEE Advanced Mock Test - 5 - 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 Mock Test - 5 - Question 29

Which of the following gives Cannizzaro’s reaction?  

Detailed Solution for JEE Advanced Mock Test - 5 - 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 Mock Test - 5 - Question 30

Which of the following have identical bond order?  

Detailed Solution for JEE Advanced Mock Test - 5 - Question 30

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

JEE Advanced Mock Test - 5 - Question 31

A reversible process is performed in such a way that the system passes from state 1 to state 2 by path-I and then from state 2 to state one by pathII as shown in figure 5. For the above process 

That the quantity  is not depends on the path of the reaction, but depends on initial and final state of the system. 

Q. 

The quantity in the above process is called 

JEE Advanced Mock Test - 5 - Question 32

A reversible process is performed in such a way that the system passes from state 1 to state 2 by path-I and then from state 2 to state one by pathII as shown in figure 5. For the above process 

That the quantity  is not depends on the path of the reaction, but depends on initial and final state of the system. 

Q.

In thermodynamics, a process is called reversible when:  

JEE Advanced Mock Test - 5 - Question 33

A reversible process is performed in such a way that the system passes from state 1 to state 2 by path-I and then from state 2 to state one by pathII as shown in figure 5. For the above process 

That the quantity  is not depends on the path of the reaction, but depends on initial and final state of the system. 

Q. 

Observe the given curve for a substance and identify the correct statement. 

Detailed Solution for JEE Advanced Mock Test - 5 - Question 33

Entropy follows the order of gas > liquid > solid 

JEE Advanced Mock Test - 5 - Question 34

There are some deposits of nitrates and phosphates in the earth’s crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions.  

Q. 

Select the correct statement among the following: 

Detailed Solution for JEE Advanced Mock Test - 5 - Question 34

Almost all metal nitrates are soluble in water 

JEE Advanced Mock Test - 5 - Question 35

There are some deposits of nitrates and phosphates in the earth’s crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions. 

Nitrates are mainly produced for use as fertilizers in agriculture because of their 

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 - Question 36

The degree of unsaturation in the product of following reaction is  


Detailed Solution for JEE Advanced Mock Test - 5 - Question 36


*Answer can only contain numeric values
JEE Advanced Mock Test - 5 - Question 37

Which chlorine in the following compound is most reactive towards nucleophilic (weak base) substitution reaction? 


Detailed Solution for JEE Advanced Mock Test - 5 - Question 37

Chlorine (3) gives IPSO substitution reaction more easily because of –R effect of nitro group.

JEE Advanced Mock Test - 5 - Question 38

Range of the function, 

, (where sgn (⋅) denotes the signum function and [⋅] greatest integer function, {.} is a fractional function) 

Detailed Solution for JEE Advanced Mock Test - 5 - Question 38

JEE Advanced Mock Test - 5 - Question 39

Let the circles S1 ≡ x2 + y2 – 4x – 8y + 4 = 0 and S2 be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 is  

Detailed Solution for JEE Advanced Mock Test - 5 - Question 39

Centre of circle S1 = (2, 4)  
Centre of circle S2 = (4, 2)  
Radius of circle S1 = radius of circle S2 = 4
∴ equation of circle S2  
(x – 4)2 + (y – 2)2 = 16  
⇒ x2 + y2 – 8x – 4y + 4 = 0  . .  . (i)
Equation of circle touching y = x at (1, 1) can be taken as  
(x – 1)2 + (y – 1)2 + λ(x – y) = 0
or, x2 + y2  + x (λ – 2) + y(– λ – 2) + 2 = 0  . . . (ii)  
As this is orthogonal to S2

JEE Advanced Mock Test - 5 - Question 40

Let the line x + y =  meets the lines y = xtanθ; θ = 0,  at A, B, C and D respectively. If AB, BC and CD are in A.P., then α =   

Detailed Solution for JEE Advanced Mock Test - 5 - Question 40


Since AB, B C and CD are in A.P. and let their lengths be taken as a – d, a and a + d respectively. 

JEE Advanced Mock Test - 5 - Question 41

Three parallel chords of a circle have lengths 2, 3, 4 and subtend angles α, β, α + β at the centre respectively (given α + β < π), then cos α is equal to 

Detailed Solution for JEE Advanced Mock Test - 5 - Question 41


Using the property that equal chords subtends equal angles at centre of circle, then problem can be converted to the diagram in adjoining figure.
AB = 4, AC = 2, BC = 3
Apply Cosine rule in ΔABC,

JEE Advanced Mock Test - 5 - Question 42

If α, β , γ are the roots of the equation x3 = x2 + 1, then the equation whose roots are  α2 + β3 + γ4, β2 + γ3 +  α4,  γ2 + α3 + β4 is 

Detailed Solution for JEE Advanced Mock Test - 5 - Question 42


JEE Advanced Mock Test - 5 - Question 43

Find the sum of first n terms, whose nth term is  

Detailed Solution for JEE Advanced Mock Test - 5 - Question 43


JEE Advanced Mock Test - 5 - Question 44

Detailed Solution for JEE Advanced Mock Test - 5 - Question 44



JEE Advanced Mock Test - 5 - Question 45

 be a point and be a line. If PQ is the distance of plane from point P measured along a line inclined at an angle of 60o with the L and is minimum, then PQ is equal to

Detailed Solution for JEE Advanced Mock Test - 5 - Question 45

Required distance PQ is perpendicular distance of plane from P (as angle between line and plane is 30o)  

*Multiple options can be correct
JEE Advanced Mock Test - 5 - Question 46

All functions f: R → R which obey the equation

Detailed Solution for JEE Advanced Mock Test - 5 - Question 46

Put x = 0  
⇒  f(y2) = f2(0) + y2 (1)  
Taking y = 0 in (1), we get  
f(0) = f2(0)    
⇒ f(0) = 0   or  f(0) = 1  
Now, taking y = x in the given equation  
f(0) = f2(x)  –  2xf(x) + x2    (2)  
Case I: If f(0) = 0, then f(x) = x.  
Case II: If f(0) = 1, then f(x) = x ± 1
hence f(x) = x + 1 (only)  

JEE Advanced Mock Test - 5 - Question 47

If polynomials f(x), g(x) and h(x), are such that 

Q.

f(x) =

Detailed Solution for JEE Advanced Mock Test - 5 - Question 47


The function G (x) = max .{−1, 3x + 2} agrees with F(x) for x ≤0 , but

JEE Advanced Mock Test - 5 - Question 48

If polynomials f(x), g(x) and h(x), are such that

Q. 

g(0) =

Detailed Solution for JEE Advanced Mock Test - 5 - Question 48


The function G (x) = max .{−1, 3x + 2} agrees with F(x) for x ≤0 , but

JEE Advanced Mock Test - 5 - Question 49

If polynomials f(x), g(x) and h(x), are such that  

Q. 

h' (0) =  

Detailed Solution for JEE Advanced Mock Test - 5 - Question 49


The function G (x) = max .{−1, 3x + 2} agrees with F(x) for x ≤0 , but

JEE Advanced Mock Test - 5 - Question 50

P, Q are two points on the rectangular hyperbola (x – 1)(y – 2) = c2, O is the centre of hyperbola, also tangent at P is perpendicular to OQ and meets OQ at N such that (OQ)(ON) = 4  

Q. 

One of the equation of directrix of the hyperbola is

Detailed Solution for JEE Advanced Mock Test - 5 - Question 50


so equation of hyperbola is (x – 1)(y – 2) = 2  
Its foci are (−1, 0), (3, 4) and equation of directrix is x + y – 5 = 0, x + y – 7 = 0 is the equation of latus rectum of ellipse and tangents at end points of latus rectum always intersects on corresponding directrix.   

JEE Advanced Mock Test - 5 - Question 51

P, Q are two points on the rectangular hyperbola (x – 1)(y – 2) = c2, O is the centre of hyperbola, also tangent at P is perpendicular to OQ and meets OQ at N such that (OQ)(ON)=4 

Q.

An ellipse confocal with the hyperbola and with eccentricity equal to  is intersected by the line x + y – 7 = 0 at A and B, then intersection point of tangents at A and B will lie on

Detailed Solution for JEE Advanced Mock Test - 5 - Question 51


so equation of hyperbola is (x – 1)(y – 2) = 2  
Its foci are (−1, 0), (3, 4) and equation of directrix is x + y – 5 = 0, x + y – 7 = 0 is the equation of latus rectum of ellipse and tangents at end points of latus rectum always intersects on corresponding directrix.   

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 - Question 52

‘A’ tosses a coin. If it shows a tail, ‘A’ is asked to roll a fair die and A’s score is the number that die shows. If the coin shows a head, ‘A’ is asked to toss five more coins and A’s score is that total number of heads shown including the first coin. If A tells you only that his score is 3, and if ‘P’ is the probability that ‘A’ rolled a die, then [1/P] is equal to  (where [ ] is the greatest integer function) 


Detailed Solution for JEE Advanced Mock Test - 5 - Question 52

Let us define the events as follows:  
E1 : A rolls the die  
E2 : A tosses five more coins  
E : A’s score is 3  


*Answer can only contain numeric values
JEE Advanced Mock Test - 5 - Question 53

The number of solution of the curve sin x = cos y and the circle x2 + y2 = 1 is/are 


Detailed Solution for JEE Advanced Mock Test - 5 - Question 53

Equation of circle in parametric form
x = cos θ, y = sin θ
So, we have to solve the equation sin (cos θ) = cos (sin θ)   (1) 

Which is not satisfied by any n = 0, ±1, ± 2, ......
∴  sin x = cos y does not intersect the circle x2 + y2 = 1.  

*Answer can only contain numeric values
JEE Advanced Mock Test - 5 - Question 54


Detailed Solution for JEE Advanced Mock Test - 5 - Question 54

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