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JEE Advanced Practice Test- 1 - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Advanced Practice Test- 1

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JEE Advanced Practice Test- 1 - Question 1

The switch in circuit shifts from 1 to 2 when VC > 2V/3 and goes back to 1 from 2 when VC < v/3.="" the="" voltmetre="" reads="" voltage="" as="" plotted.="" what="" is="" the="" period="" t="" of="" the="" wave="" form="" in="" terms="" of="" r="" and="" />

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

During time 't2', capacitor is discharging with the help of three resistors of resistance 'R' each.

Therefore, q = q0-3t/RC

[∵Q = CV]

V = V0e-3t/RC

As

t2 = (RC/3).loge2

During time 't1', capacitor is charging with the help of battery.

Therefore, q = q0(1 - e-3t/RC) or V = V0(1 - e-3t/RC)

As

t1 = (RC/3).loge2

T = t1 + t2 = (2RC/3).loge2

JEE Advanced Practice Test- 1 - Question 2

For a thermodynamic system, the pressure, volume and temperature are related as P = , where is a positive constant. The work done for this system in a constant pressure process, if the temperature changes from To to 2To, is

Detailed Solution for JEE Advanced Practice Test- 1 - Question 2

As PV = αT2

PdV + VdP = 2 αTdT

For constant pressure process, dP = 0

PdV = 2 α TdT

W = 3αT02

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JEE Advanced Practice Test- 1 - Question 3

One mole of a perfect gas expands isothermally to ten times its original volume.The change in entropy is :

Detailed Solution for JEE Advanced Practice Test- 1 - Question 3

We know ΔS = qrev / T

But

It is given n = 1,V2 / V1 = 10

qrev = 2.303 RT log 10 = 2.303 RT

Hence

= 2.303 R

JEE Advanced Practice Test- 1 - Question 4

Fe2+ and Fe3+ can be distinguished by :

Detailed Solution for JEE Advanced Practice Test- 1 - Question 4

Fe2+ + KSCN → No colouration

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 5

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

Detailed Solution for JEE Advanced Practice Test- 1 - Question 5

If charges are of opposite signs then the two fields are along the same direction. So, they cannot be zero. Hence, the charges should be of same sign.

Further, work done per unit charge by external force = change in potential energy

=

or

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 6

A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle θ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure. The block may remains stationary if (take g = 10 m s−2)

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

1 N × cos θ acts towards Q along the inclined plane, and mgsinθ acts towards P along inclined plane

At θ = 45o, mgsinθ = 1 × cosθ

At θ > 45o, mgsinθ > 1×cosθ (friction acts upward)

At θ < 45o, mgsinθ < 1×cosθ (friction acts downward)

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 7

There are four sphere of each with radius as a, b, c, d, each sphere is solid, insulating and uniformly charged. Variation of electric field with distance rr from the centre is given. Straight portion of curve c and d is overlapping and straight portion of curve a and b is overlapping.

Detailed Solution for JEE Advanced Practice Test- 1 - Question 7

Slope ∝ ρ

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 8

In the figure, a man of true mass M is standing on a weighing machine placed in a cabin. The cabin is joined by a string with a body of mass m. Assuming no friction, and negligible mass of cabin and weighing machine. Then

Detailed Solution for JEE Advanced Practice Test- 1 - Question 8

Mg − T = Ma ........(i)

T = ma ........(ii)

Solving (i) and (ii)

FBD of man

Mg − N = Ma

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 9

In the figure below, a closed loop is made of a U-shaped metallic wire of negligible resistance and a moving cross-bar of resistance R. The length of the cross-bar is L and has a mass m. It is initially placed at a distance h0 from the other end of the loop. The whole loop is placed in a uniform magnetic field of intensity Bo, as shown in the figure. Then,

Detailed Solution for JEE Advanced Practice Test- 1 - Question 9

On applying the Fleming's right-hand rule, the induced current is from left to right on the crossbar.

Current in the loop

Forces acting on the cross bar

Its weight mg acts downward.

Magnetic force B0iL acts in upward direction.

When rod attains terminal velocity, resultant force on the rod is zero

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 10

Function x = A sin2 ωt + B cos2ωt + C sinωt cosωt represents SHM

Detailed Solution for JEE Advanced Practice Test- 1 - Question 10

We have, x = A sin2ωt + B cos2ωt + C sinωtcosωt = A/2 (1 - cos2ωt) + B/2(1 + cos2ωt) + C/2sin2ωt

For A = -B; C = 2B: x = B cos2ωt+ B sin2ωt; which is an SHM, whose amplitude is B√2.

For A = B, C = 0:

We get x = A, which in not an SHM.

For A = B, C = 2B:

x = B + B sin2ωt; which is an SHM, whose amplitude is B.

For any values of A, B and C (except C = 0):

;

which is an SHM.

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 11

In the circuit below, initially switch S1 is kept closed for a longer time and then switch S2 is closed. Which of the following statements is/are true?

Detailed Solution for JEE Advanced Practice Test- 1 - Question 11

When switch S1 is closed, the equation that describes the charging of the capacitor is given by:

Maximum charge on the capacitor, Q = CE.

Energy stored in the capacitor, when it is fully charged, is

When switch S2 is closed, according to the law of conservation of energy, the energy in the circuit cannot exceed 4.05 J, as the loss of energy is not considered; the energy in the circuit is 4.05 J.

Charge across the capacitor and current in the circuit are given by:

Magnitude of the rate of change of current is

At t = 0,

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 12

Directions: The question has four choices, out of which ONE or MORE may be correct.

A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat 'Q' flows only from left to right through the blocks. Then in steady state,

Detailed Solution for JEE Advanced Practice Test- 1 - Question 12

Using Fourier law,

Q = UA△T

QA = 2K/L

QB = 3K/4L

QC = K/L

QD = 5K/4L

QE = 6K/L

Also, heat flow across the system remains same.

∴ Option 1 is correct.

And since UE is maximum and heat flow across the system remains constant, option 3 is also correct.

Now,

∵ UC = K\L

∴ QC = KA△T/L

Also,

∵ UB + D = 5K/4L + 3K/4L + 4K/L

∴ QB + D = KA△T/L

∴ QC = QB + D

*Answer can only contain numeric values
JEE Advanced Practice Test- 1 - Question 13

An ideal gas (2 moles, diatomic) undergoes a thermodynamic process P.V-n = constant. For a change of temperature of 20 K, heat required is 200/3R (R is universal gas constant). The value of 'n' is


Detailed Solution for JEE Advanced Practice Test- 1 - Question 13

Let the heat capacity for the polytropic process be C.

Here, μ is the number of moles.

Hence,

For the polytropic process,

1 + n = 6

⇒ n = 5

*Answer can only contain numeric values
JEE Advanced Practice Test- 1 - Question 14

If wavelength associated with an electron revolving in nth orbit of a singly ionised He-atom is 14.976 Å, then the value of 'n' is


Detailed Solution for JEE Advanced Practice Test- 1 - Question 14

Using the quantum condition: 2π = nλ

*Answer can only contain numeric values
JEE Advanced Practice Test- 1 - Question 15

A bullet loses 1/x4 of its velocity in passing through a given plank of energy dissipating medium. If after passing through 8 such planks, the bullet comes to rest, the value of 'x' is


Detailed Solution for JEE Advanced Practice Test- 1 - Question 15

Putting x4 = N, so the bullet loses the velocity by v/n on passing through each plank.

As the total number of planks = 8, therefore

change in K.E. in passing through 8 planks is 100%.

Solving this, we get:

N = 15.5

or approximately N = 16

x4 = 16

x = 2

*Answer can only contain numeric values
JEE Advanced Practice Test- 1 - Question 16

A string having length 0.5 m fixed at one end and other end is connected to a block of mass m = 2 kg as shown in figure. The string is set into vibrations which is represented by y = 4sin(πx/5)cos(50πt), where x and y are in cm and t is in second. The number of anti-nodes between point A and the fixed pulley is 2n, find n.


Detailed Solution for JEE Advanced Practice Test- 1 - Question 16

M × 5 × 10−2 = 0.5

m = 10

2n =10

n = 5

*Answer can only contain numeric values
JEE Advanced Practice Test- 1 - Question 17

Two particles A and B of masses mm and 2m are connected through a rod of length l. The system lies on a smooth horizontal surface. An impulse of magnitude mv to the particle B is imparted through impact horizontally but perpendicular to the rod. The angular velocity of the rod just after impact is v / nl. Find 2n.


Detailed Solution for JEE Advanced Practice Test- 1 - Question 17

From momentum conservation, Velocity of B just after impact = v/2

Angular velocity, ω = v / 2l

So, n = 2

*Answer can only contain numeric values
JEE Advanced Practice Test- 1 - Question 18

The friction coefficient between the horizontal surface and each of the blocks shown in the figure is 0.2. The collision between the blocks is perfectly elastic. Find the separation (in cm) between them when they come to rest.

Take g=10 m s-2.


Detailed Solution for JEE Advanced Practice Test- 1 - Question 18

Velocity of first block before collision,

By conservation of momentum,

Also v'2−v'1 = v1 for elastic collision

It gives V'2 = 0.4m/s

v'1 = −0.2m/s

Now distance moved after collision

∴ s = s1 + s2 = 0.05m = 5 cm

JEE Advanced Practice Test- 1 - Question 19

Which of the following is the correct order of stability of the given compounds?

Detailed Solution for JEE Advanced Practice Test- 1 - Question 19

(I) ⇒ Aromatic 6πe system in conjugation

(II) ⇒ Non - aromatic, 8πe system but ring is not planar, so not anti - aromatic but non - aromatic.See

(III) ⇒ anti - aromatic 8πe- in conjugation

Stability order: Aromatic > Non - aromatic > Anti - aromatic.

(I) > (II) > (III).

JEE Advanced Practice Test- 1 - Question 20

When chlorine water is added to an aqueous solution of sodium halide in the presence of chloroform, a violet colouration is obtained. When more of chlorine water is added, the violet colour disappears and solution becomes colourless. This confirms that sodium halide is :

Detailed Solution for JEE Advanced Practice Test- 1 - Question 20

2NaI + Cl2 ⟶ 2NaCl + I2(violet colouration)

JEE Advanced Practice Test- 1 - Question 21

During the dehydration of alcohols, the hydrogen is removed from the carbon atom having a lesser number of hydrogen atoms to yield the most stable alkene as a major product. Identify the most stable alkene formed during the following conversion:

Detailed Solution for JEE Advanced Practice Test- 1 - Question 21

The reaction proceeds through the formation of a carbocation intermediate, which re-arranges by a hydride shift followed by ring expansion in order to reduce the Bayer strain.

Thus, the major product formed is 1,2-dimethyl cyclohexene.

JEE Advanced Practice Test- 1 - Question 22

Directions: The question is Based on the following paragraph. This question has FOUR options, out of which ONLY ONE is correct.

A tertiary alcohol H upon acid catalysed dehydration gives a product I. Ozonolysis of I leads to compounds J and K. Compound J upon reaction with KOH gives benzyl alcohol and a compound L, whereas K on reaction with KOH gives only M.

Compound H is formed by the reaction of

Detailed Solution for JEE Advanced Practice Test- 1 - Question 22

Step 1:

Step 2:

Steo 4:

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 23

Which of the following substances exhibit tautomerism

Detailed Solution for JEE Advanced Practice Test- 1 - Question 23
Tautomerism is spontaneous interconversion of two isomeric forms with different functional groups. The prerequisite for this is the presence of the C=O, C☰N or N=O in the usual cases and an alpha H atom. In case of keto enol tautomerism, the keto form is more stable. Enols can be formed by acid or base catalysis from the ketone and are extensively used in making C-C single bonds in organic synthesis. Compound in (1), (3) and (4) exhibit tautomerism as shown below.

Hence, option (a), (c) and (d) are correct.

Benzoquinone is highly stable due to conjugation and does not exhibit tautomerism.

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 24

Observe the following sequence of reactions and select the option(s) that is/are true in this context.

Detailed Solution for JEE Advanced Practice Test- 1 - Question 24

Option (a) is incorrect as cold alkaline solution of KMnO4 would produce vicinal diols.

Option (b) is correct as KMnO4 is stronger oxidising agent and in the acidic medium, it oxidises the alkene by breaking the carbon-carbon double bond and replacing it with two carbon-oxygen double bonds.

Option (c) is correct as the process is ozonolysis by ozone, which is arrested at the aldehyde stage by adding zinc water.

Option (d) is incorrect as decarboxylation, with soda lime, would produce ethane. Ethene is obtained by Kolbe's electrolysis. Hence, 'S' represents the process of electrolysis and not a reagent.

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 25

A crystalline substance composed of Ba, Ti and O crystallises in a perovskite structure. This structure may be described as a cubic lattice, with barium ion occupying corners of the unit cell and oxide ions occupying the face centres of the unit cell. Which of the following statements is/are true?

Detailed Solution for JEE Advanced Practice Test- 1 - Question 25

1) Number of Ba ions per unit cell = 1/8 x 8 = 1

Number of Ti ions per unit cell = 1 x 1 = 1

Number of O ions per unit cell = 1/2 x 6 = 3

Hence, the empirical formula of the substance should be BaTiO3.

Thus, (a) is correct.

2) Titanium ions can be assumed to occupy a hole in the lattice formed by barium and oxide ions. The titanium ions occupy the body centres of the face centred cubic and this is one of the octahedral holes in the fcc lattice.

So, (b) is also true.

3) The octahedral holes at the centres of unit cell constitute just one-fourth of all the octahedral holes in a face centred cubic lattice. Thus, (c) is also true.

4) An octahedral hole at the centre of a unit cell, occupied by titanium ion, has 6 nearest neighbour oxide ions. The other octahedral holes located at the centres of the edges of unit cells have 6 nearest neighbours each, as in case with any octahedral hole, but 2 of the 6 neighbours are barium ions (at unit cell corners terminating the given edge) and 4 oxide ions. The proximity of two cations Ba2+ and Ti4+ would be electrostatically unfavourable. So, titanium cannot occupy all of the octahedral holes and restrictions as stated above exist.

So, (d) is wrong.

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 26

Directions: The question has four choices, out of which ONE or MORE may be correct.

Extraction of metal from the ore cassiterite involves

Detailed Solution for JEE Advanced Practice Test- 1 - Question 26

Tin is obtained by reducing the ore cassiterite with coal in a reverberatory furnace. Limestone is added to produce a slag with the impurities, which can be separated.

SnO2 + 2C → Sn + 2CO

Crude tin so obtained is contaminated with iron, copper, lead and other metals. It is, therefore, remelted on an inclined furnace. The process is called liquation.

The easily fusible tin melts away and the less fusible impurities are left behind. Molten tin is finally stirred with green poles of wood in contact with air. In this process, any remaining metal impurities are oxidised forming a scum, which rises to the surface and is removed. This process is called poling.

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 27

Amongst the following, the correct statements(s) is/are

Detailed Solution for JEE Advanced Practice Test- 1 - Question 27

In NO (15 electrons), the last electron is unpaired in

NO + (BO = 3.0) & O2+(BO = 2.5)

In OF+ (Electrons = 16), 2 unpaired electrons in

In Ne2+ (Electrons = 19), 1 unpaired electron in

ie.

π -bond has inter-nuclear axis coincident with the nodal plane.

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 28

Given is the arrangement of atoms in crystallographic plane.

Which plane correctly represent(s) the adjacent drawn structure ?

Detailed Solution for JEE Advanced Practice Test- 1 - Question 28

Hence, answer ​is Face plane in FCC and ​Body diagonal plane in BCC.

Body diagonal plane in FCC the atom at body centre will not be touching four corner atoms. Face plane in BCC shall not contain the atom at body centre.

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 29

In the following reaction, the product(s) formed is(are):

Detailed Solution for JEE Advanced Practice Test- 1 - Question 29
The above given reaction is Reimer-Tiemann reaction where a strong base reacts with chloroform to generate reactive species called as dichlorocarbene. The reaction is used for the ortho-formylation of phenols.

The dichlorocarbene formed acts as an electrophile and attacks the para methyl phenol molecule at ortho and para position which are the sites where the negative charge is localised on the ring.

There are two products formed out of which the para substitution product is minor due to steric hindrance caused by the methyl group while the ortho substitution product is the major as the site has no effect of steric hindrance.

*Multiple options can be correct
JEE Advanced Practice Test- 1 - Question 30

Which of the following is a cyclic oxo-acid?

Detailed Solution for JEE Advanced Practice Test- 1 - Question 30

H3P3O9 is a cyclic tri-metaphosphoric acid.

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