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JEE Advanced Practice Test- 7 National Level Test Series MCQs & solutions


MCQ Practice Test & Solutions: JEE Advanced Practice Test- 7 (54 Questions)

You can prepare effectively for JEE National Level Test Series for JEE Advanced 2026 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "JEE Advanced Practice Test- 7". These 54 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 180 minutes
  • - Number of Questions: 54

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JEE Advanced Practice Test- 7 - Question 1

A tank of uniform cross-section is completely filled with ice. The height of ice is H and its mass is m. When the entire ice melts, the work done by gravity is  

ice = 0.9 gm/cc, ρwater = 1 g/cc and g represents acceleration due to gravity) 

Detailed Solution: Question 1

H height of frozen water becomes 0.9 H height of liquid on melting.  
Thus Ui = m g(H/2) = 0.5 mgH,  Uf = mg(0.9H/2) = 0.45 mgH  
WG = Ui – Uf = 0.5mgH − 0.45mgH = 0.05 mgH 

JEE Advanced Practice Test- 7 - Question 2

A hydrogen like species having atomic number Z = 2, in ground state, is excited by means of electromagnetic radiation of frequency 1.315 × 1016 Hz. How many spectral lines will be observed in the emission spectrum? 

(Planck’s constant h = 4.14 × 10-15 eVs) 

Detailed Solution: Question 2

E = hν = 4.14 × 10-15 × 1.315 × 1016 = 54.4 eV  
Energy of the given species in ground state is 

Ionization energy of the given species is = |E1| = 54.4 eV  
Thus the electron will get ionized and therefore become free.
So no spectral lines will be observed. 

JEE Advanced Practice Test- 7 - Question 3

Velocity of a point on the equator of a rotating spherical planet is v. The angular velocity of the planet is such that the apparent value of acceleration due to gravity ‘g’ at the equator is half of that at the poles. The escape velocity of a particle from the surface of the planet is  

Detailed Solution: Question 3

Apparent value of g at pole is equal to

JEE Advanced Practice Test- 7 - Question 4

A positive charge q is projected from origin with a velocity along positive x-axis in a region having uniform magnetic field directed towards negative y-axis. If T is the time period of circular motion then the velocity vector of charge q at some instant t where 

Detailed Solution: Question 4


The charge executes circular motion in xz plane with its center lying on the negative z-axis. At time t where the charge has completed (3/4)th of the circle and so its velocity vector is pointing towards 1st quadrant in x-z plane.  
⇒ vx is positive, vz is positive and vy is zero.  

JEE Advanced Practice Test- 7 - Question 5

Four charges q1, q2, q3 and q4 are placed at the positions as shown in the figure, given q1 +q2+ q3 + q4 = 0 . The electric field on z-axis  

Detailed Solution: Question 5

As q1 +q2 + q3 + q4 = 0 , z axis is an equipotential line. Thus electric field is either zero or perpendicular to the z-axis. When q2 = q4 and q3 = q1 electric field is zero at origin and in all other cases it is perpendicular to z-axis. 

JEE Advanced Practice Test- 7 - Question 6

For the circuit as shown in the figure q1 and q2 be the charges on 3μF and 6μF capacitors respectively, then 

Detailed Solution: Question 6

At any time t the current distribution in the circuit will be as shown. If i2 is the current going out of upper loop through 3 μF capacitor then current i2 must also come back to this through 6μF capacitor to make the current in that loop i1.Thus the current through both the capacitors will be same at all time instants. Thus they will have equal charge. 

*Answer can only contain numeric values
JEE Advanced Practice Test- 7 - Question 7

A ball is released from position A and drops 10 m before striking a smooth incline. The coefficient of .  If the time taken by the ball to strike the incline again is t then find the value of t2 [in (second)2]. (g = 10 m/s2


Detailed Solution: Question 7


Just after collision component of velocity perpendicular to the incline is

Component of g perpendicular to incline is  

∴  time after which the ball strikes the incline again is  

*Answer can only contain numeric values
JEE Advanced Practice Test- 7 - Question 8

Two identical beads P and Q of mass 1 kg each are connected by an inextensible massless string and they can slide along the two arms AB and BC of a rigid smooth wire frame in vertical plane. If the system is released from rest and vQ is the speed of bead Q when they have both moved by a distance of 0.1 m then find the value of (in m/s). (g = 10 m/s2


Detailed Solution: Question 8


Velocity of P and Q along the string should be same.  


From conservation of mechanical energy, decrease in potential energy of Q = increase in kinetic energy of both  


*Answer can only contain numeric values
JEE Advanced Practice Test- 7 - Question 9

A solid uniform sphere rotating about its axis with kinetic energy Eo is gently placed on a rough horizontal plane. The coefficient of friction on the plane varies from point to point. After some time, the sphere begins pure rolling with total kinetic energy equal to E. Then find the value of 


Detailed Solution: Question 9

From conservation of angular momentum about point of contact  


*Answer can only contain numeric values
JEE Advanced Practice Test- 7 - Question 10

A variable voltage V = 2t is applied across an inductor of inductance L = 2H as shown in figure. Then find the rate at which magnetic potential energy stored in the inductor is increasing at t = 1 s (in J/s). Take the current through the inductor at t = 0 as zero. 


Detailed Solution: Question 10


Potential energy stored in the inductor is  

JEE Advanced Practice Test- 7 - Question 11

A calorimeter of mass m contains an equal mass of water in it. The temperature of water and calorimeter is t2. A block of ice of mass m and temperature t3 < 0oC is gently dropped into the calorimeter. Let C1, C2 and C3 be the specific heats of calorimeter, water and ice respectively and L be the latent heat of fusion of ice.  

Q. 

The whole mixture in the calorimeter becomes ice if  

Detailed Solution: Question 11

For whole mixture to become ice tf ≤ 0oC  
Let tf = −to  then  
From heat lost = heat gained we get

JEE Advanced Practice Test- 7 - Question 12

A calorimeter of mass m contains an equal mass of water in it. The temperature of water and calorimeter is t2. A block of ice of mass m and temperature t3 < 0oC is gently dropped into the calorimeter. Let C1, C2 and C3 be the specific heats of calorimeter, water and ice respectively and L be the latent heat of fusion of ice.  

Q. 

The whole mixture in the calorimeter becomes water if  

Detailed Solution: Question 12

For whole mixture to convert to water tf ≥ 0oC  
Let tf = to then  
From heat lost = heat gained we get
 

JEE Advanced Practice Test- 7 - Question 13

A calorimeter of mass m contains an equal mass of water in it. The temperature of water and calorimeter is t2. A block of ice of mass m and temperature t3 < 0oC is gently dropped into the calorimeter. Let C1, C2 and C3 be the specific heats of calorimeter, water and ice respectively and L be the latent heat of fusion of ice.

Q. 

Water equivalent of calorimeter is  

Detailed Solution: Question 13

Let water equivalent of calorimeter be mo then  moC2 = mC1 ⇒ 

JEE Advanced Practice Test- 7 - Question 14

A parallel plate capacitor has its plate horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown. The distance between the plates is d1. Now the capacitor is connected with an electric source having voltage V. The separation between the plates changes to d2 at equilibrium. The mass of lower plate is ‘m’ and cross-sectional area of each plate is A.

Q. 

The spring constant k is  

Detailed Solution: Question 14

When capacitor was uncharged, for equilibrium of lower plate of mass m   kxo = mg   where xo is the compression in the spring.   When voltage source is connected and lower plate again reaches equilibrium then compression in the spring is xo – (d1 – d2).  
Electric force between capacitor plates is 

JEE Advanced Practice Test- 7 - Question 15

A parallel plate capacitor has its plate horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown. The distance between the plates is d1. Now the capacitor is connected with an electric source having voltage V. The separation between the plates changes to d2 at equilibrium. The mass of lower plate is ‘m’ and cross-sectional area of each plate is A.

Q. 

The maximum voltage Vm for a given k for which an equilibrium exists is  

Detailed Solution: Question 15



JEE Advanced Practice Test- 7 - Question 16

A parallel plate capacitor has its plate horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown. The distance between the plates is d1. Now the capacitor is connected with an electric source having voltage V. The separation between the plates changes to d2 at equilibrium. The mass of lower plate is ‘m’ and cross-sectional area of each plate is A.

Q. 

When lower plate is slightly displaced about equilibrium position, time period T of small oscillations is  

Detailed Solution: Question 16

Let the lower plate be displaced upward by small distance x, then FBD of the plate is as shown. 



Net restoring force acting on the lower plate is 






Comparing with standard equation F = −Kx for SHM, we get

JEE Advanced Practice Test- 7 - Question 17

A certain weak acid has a dissociation constant 1.0 x 10-4. The equilibrium constant for its reaction with strong base is  

Detailed Solution: Question 17

JEE Advanced Practice Test- 7 - Question 18

The major product formed in the following reaction is 

Detailed Solution: Question 18

JEE Advanced Practice Test- 7 - Question 19

Identify the major product of the following reaction: 

Detailed Solution: Question 19

This is nucleophilic addition followed by dehydration. 

JEE Advanced Practice Test- 7 - Question 20

The oxidation state of molybdenum in [(η7-tropylium)Mo(CO)3]+ is 

Detailed Solution: Question 20

Tropylim cation has one positive charges and CO is neutral ligand. 

JEE Advanced Practice Test- 7 - Question 21

In metal-olefin interaction, the extent of increase in metal ⎯→ olefin π-back-donation would 

*Answer can only contain numeric values
JEE Advanced Practice Test- 7 - Question 22

The highest oxidation state of an element in the following compound that behaves as an acid in H2SO4 is  

AcOH, HNO2,  HNO3, H2O, HClO, HClO4


Detailed Solution: Question 22

HClO4 is stronger acid than H2SO4 

*Answer can only contain numeric values
JEE Advanced Practice Test- 7 - Question 23

How many unpaired electrons are present in O2 molecule? 


*Answer can only contain numeric values
JEE Advanced Practice Test- 7 - Question 24

One mole of Pb3O4 is separately reacted with excess of HCl and HNO3. The difference in moles of HCl and HNO3 is 


Detailed Solution: Question 24

Pb3O4 is a mixture of PbO2 and PbO. Only PbO reacted with HNO3 since Pb4+ is a very good oxidizing agent. HCl is a good reducing agent that can reduces Pb4+ into Pb2+.  
Pb3O4 + 8HCl ⎯⎯→ 3PbCl2 + Cl2 + 4H2O  
Pb3O4 + 4HNO3 ⎯⎯→ 2Pb(NO3)2 + PbO2 + 2H2

*Answer can only contain numeric values
JEE Advanced Practice Test- 7 - Question 25

How many moles of phenyl hydrazine are used in the formation of osazone from glucose? 


Detailed Solution: Question 25


*Answer can only contain numeric values
JEE Advanced Practice Test- 7 - Question 26

How many S – S bonds are present in (SO3)3


Detailed Solution: Question 26

JEE Advanced Practice Test- 7 - Question 27

If a concentrated solution of copper sulphate is placed at the bottom of a beaker of water or that of a dilute solution of copper sulphate is carefully poured over it there will be a two distinct layer visible. However, after some time the boundaries will disappear. This property is called diffusion. If we now consider a solution which is separated from the pure solvent by a semipermeable membrane then the solvent particles move from the pure solvent region through the SPM to the solution region. This phenomenon is called osmosis.



Q. 

for an indefinitely dilute boundary then dμ1 is 

JEE Advanced Practice Test- 7 - Question 28

If a concentrated solution of copper sulphate is placed at the bottom of a beaker of water or that of a dilute solution of copper sulphate is carefully poured over it there will be a two distinct layer visible. However, after some time the boundaries will disappear. This property is called diffusion. If we now consider a solution which is separated from the pure solvent by a semipermeable membrane then the solvent particles move from the pure solvent region through the SPM to the solution region. This phenomenon is called osmosis.



Q. 

Which of the following solution has highest osmotic pressure? 

JEE Advanced Practice Test- 7 - Question 29

If a concentrated solution of copper sulphate is placed at the bottom of a beaker of water or that of a dilute solution of copper sulphate is carefully poured over it there will be a two distinct layer visible. However, after some time the boundaries will disappear. This property is called diffusion. If we now consider a solution which is separated from the pure solvent by a semipermeable membrane then the solvent particles move from the pure solvent region through the SPM to the solution region. This phenomenon is called osmosis.



Q. 

Which of the following is correct?

JEE Advanced Practice Test- 7 - Question 30

Thionyl chloride can be synthesized by chlorinating SO2 using PCl5. Thionyl chloride is used to prepare anhydrous ferric chloride starting from its hexahydrated salt. Alternatively, the anhydrous ferric chloride can also be prepared from hexahydrated salt by treating with 2,2-dimethoxypropane.

Q. 

Consider the following reaction 


The compound X is 

Detailed Solution: Question 30

SO2 + PCl5 ⎯→ SOCl2 + POCl3 

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