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JEE Advanced Mock Test - 4 (Paper II) - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Advanced Mock Test - 4 (Paper II)

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*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 1

A driver in a stationary car blows a horn which produces monochromatic sound waves of frequency 1000 Hz normally towards a reflecting wall. The wall approaches the car with a speed of 3.3 m s−1.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 1

The frequency received by the wall is

The reflected frequency will be same as the frequency received by the wall. Since the wall is moving towards the driver, the frequency heard by the driver will be,

Now the percentage increase,

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 2

A vertical glass capillary tube, open at both ends, contains some water. Which of the following shapes may not be taken by the water in the tube ?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 2
The two free liquid surface must provide a net upward force due to surface tension to balances the weight of the liquid column.

Usually, the water contact angle is smaller than 90o with glass surface and so the surface will be hydrophilic. Shape of water surface on top will be concave. If gravity is taken into account, water will take the shape as shown in option D.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 3

Monochromatic light of wavelength 4000 A∘. is incident on an isolated neutral metal sphere of work function 2 eV. Choose the correct statements(s).

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 3

KEmax = hυ − W

= hc/λ - W

= (12400/4000 - 2)eV

= 1.1 eV

Emitted electron can have kinetic energy from 0 to 1.1 eV.

After sometime the metal plate gets '+'charge and the electrong cant overcome the potential so photoelectric effect stops

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 4

A long current carrying straight wire having current I1 = 104 ampere is placed at the centre of an another current carrying loop having current I2 = 104 ampere. Straight wire is perpendicular to the plane of the loop as shown in the figure. The torque (in N m) acting on the loop about its center is 10x. Find the value of x. (Radius R = 1 m)


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 4

Net torque acting,

=

= 2 × 10−7 × 104 × 104 × (1) = 20 N m

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 5

During an adiabatic process, the pressure of a gas is found to be proportional to cube of its absolute temperature, Cv / Cp is


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 5
Given in the question that during an adiabatic process, the pressure P of a gas is found to be proportional to cube of its absolute temperature T,

∴ P ∝ T3

⇒ P = kT3

⇒PT-3 = k ...(1)

where kk is the constant of proportionality.

Also, during adiabatic process, pressure and temperature are related as

P1−γTγ = constant

where γ = CP / CV

Comparing equations (1) and (2), we get

Hence, Cv/Cp = 1/γ = 2/3 = 00.67

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 6

A part of circuit in a steady state along with the currents flowing in the branches, the values of resistances, etc. is shown in the figure. Calculate the energy (in mJ) stored in the capacitor C (4μF).

(Round off up to 2 decimal places)


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 6
Using Kirchhoff's first law at junction a and b, we have found the current in other wires of the circuit on which currents were not shown.

Now, to calculate the energy stored in the capacitor, we will have to first find the potential difference Vab across it.

∴ Va - 3 x 5 - 3 x 1 + 3 x 2 = Vb

∴ Va - Vb = Vab = 12 V

= ½(4 x 10-6)(12)2 J = 0.29 mJ

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 7

The figure below shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance 2 Ω/m. The position of the conducting rod at t = 0 is shown. A time-dependent magnetic field B = 4t T is switched on at t = 0.

The rod moves towards right at 10 cm/s. The current (in A) in the loop at t = 3 s due to induced emf is

(Round off up to 2 decimal places)


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 7

Induced emf, e = Blv = 12 × 0.30 × 0.10 = 0.36 V

Resistance of U-shaped rail = 2 W, as its length is 100 cm

Induced current, I = e/R = 0.36/2 = 0.18 A

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 8

A sample of air is initially at 533 K and 700 N/m2, and occupies a volume of 0.028 m3. The air is expanded at constant pressure to three times. A polytropic process with n = 1.5 is then carried out followed by an isothermal process which completes the cycle.

The efficiency of the cycle (in %) is:


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 8

= -58.4 J

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 9

A sonometer wire resonates with a given turning fork forming standing waves with five antinodes between two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by M, the wire resonates with the same tuning fork forming three antinodes for the same positions of bridges. The value of M (in kg) is


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 9

As per the given condition,

Or M = 25 kg

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 10

If two structures of the same cross-sectional area but different numerical apertures NA1 and NA2 (NA2 < NA1) are joined longitudinally, the numerical aperture of the combined structure is ____.

1. 
2. NA1 + NA2
3. NA1
4. NA2
(Enter your answer as per respective numbering 1, 2, 3, 4 given above.)


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 10

The numerical aperture is limited by the second slab. [smaller NA]
∵ sin i2 < NA2
sin i1 > sin i2
∴ NA of combination = smaller NA
Hence, the numerical aperture of the combined structure must be NA2.

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 11

For two structures namely S1 with n1 = √45/4 and n2 = 3/2 and S2 with n1 = 8/5 and n2 = 7/5, and taking the refractive index of water to be 4/3 and that of air to be 1, consider the following statements:

  1. NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive index 16/3√15.
  2. NA of S1 immersed in liquid of refractive index 6/√15 is the same as that of S2 immersed in water.
  3. NA of S1 placed in air is the same as that of S2 immersed in liquid of refractive index 4/√15.
  4. NA of S1 placed in air is the same as that of S2 placed in water.

How many of the above given statements are correct?


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 11






∴ option (3) is correct.

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 12

If the piston is pushed at a speed of 5 mms–1, the air comes out of the nozzle with a speed of ____ m/s.


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 12

Av = constant ⇒ 5 mms1 × (20 mm)2 = v × (1 mm)2
⇒ v = 2 ms-1

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 13

If the density of air is pa and that of the liquid is pl, then for a given piston speed, the rate (volume per unit time) at which the liquid is sprayed will be proportional to _____.

(Enter your answer as per respective numbering 1, 2, 3, 4 given above.)


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 13

JEE Advanced Mock Test - 4 (Paper II) - Question 14

If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 14

On a system of particles if,

No other conclusions can be drawn.

JEE Advanced Mock Test - 4 (Paper II) - Question 15

When a potential difference of 103V is applied between A and B , a charge of 0.75 mC is stored in the system of capacitors. The value of C is (μF)

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 15

q = CeqV

0.75C + 1.5 = C + 1

0.5 = 0.25 C

⇒ C = 2

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 16

The decreasing order of hyperconjugative effect is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 16
Hyperconjugation effect depends on the number of alpha-H-atoms.

Both options (a) and (b) are arranged in the order of decreasing number of alpha-hydrogen atoms and hence, in the decreasing order of the hyperconjugating effect.

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 17

Enthalpy of dissociation of acetic acid obtained from Experiment 2 is _____. (kJ mol-1)


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 17

Let the heat capacity of insulated beaker be C.
Mass of aq. content = 100 + 100 = 200 g
Total heat capacity = C + 200 x 4.2 x ΔT
0.1 x 5.7 x 10-3
5.7 x 100 = C + 200 x 4.2 x 5.7
5,700 = C + 200 x 4.2 x 5.7
5,700 = C + 4,788
C = 5,700 - 4,788 = 912 J
msΔT = 912/5.7 = 160 JK-1
In Experiment 2:
nCH3COOH = 0.2, nNaOH = 0.1
Heat evolved = 200 x 4.2 x 5.6 + 160 x 5.6 = 5,600 J
Only 0.1 mol of CH3COOH undergoes neutralisation.
Enthalpy of dissociation for 0.1 mole of CH3COOH = 5,700 - 5,600 = 100 J
ΔH(dissociation) for CH3COOH = 1000 J/mole = 1

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 18

The pH of the solution after Experiment 2 is k × 10-1. What will be the value of k?


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 18

[CH3COOH] = 0.5
[CH3COONa] = 0.5
pH = -log(2 × 10-3) + log10(0.5/0.5) = 4.7
= 47 × 10-1
So, k = 47

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 19

Compound X is

(Mark the compound as per their respective numbering given in the question.)


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 19

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 20

The major compound Y is

(Mark the compound as per their respective numbering given in the question.)


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 20

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 21

The value of r is .  Find the value of n.


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 21

P ≡ (at2, 2at), F ≡ (a, 0)

Applying condition of collinearity, we get

Since QR ∥ PK,

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 22

If st = 1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is  . Find the value of m.


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 22

S = 1/t

Equation of normal at S is x/t + y = 2a/t + a/t3 ............ (1)
Equation of tangent at P is ty = x + at2 ............ (2)
Solving Eq. (1) and (2) simultaneously, we get,

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 23

The value of   is nπ.  Find the value of n.


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 23

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 24

The value of  is


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 24






JEE Advanced Mock Test - 4 (Paper II) - Question 25

A sphere of radius R carries a positive charge whose volume density depends only on separation r from the sphere's centre. It is according to the formula where p0 is a constant.
Which of the following is/are true?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 25


Consider the spherical shell of radius 'r' and thickness 'dr'. Let the charge enclosed by the spherical shell be dq.

Total charge enclosed within the sphere of radius r,

Using Gauss's law, Electric field at any point inside the sphere is

For electric field outside the sphere, total charge enclosed within the sphere (put r = R),

Using Gauss's law,
Electric field outside the sphere is

At the centre of the sphere, E = 0

JEE Advanced Mock Test - 4 (Paper II) - Question 26

The mass of a nucleus is less than the sum of the masses of (A – Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m1 and m2 only if (m1 + m2) < M. Also, two light nuclei of masses m3 and m4 can undergo complete fusing and form a heavy nucleus of mass M' only if (m3 + m4) > M'. The masses of some neutral atoms are given in the table below:

The kinetic energy (in keV) of the alpha particle, when the nucleus   at rest undergoes alpha decay, is:

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 26

Q value of the reaction
Q = Δmc² = (209.982876 - 4.002603 - 205.97455)c²
Q = 5.422 MeV
Using conservation of linear momentum
If K1 is the kinetic energy of the alpha particle and K2 is the kinetic energy of the recoil nucleus, then
2 × 4 × K1 = 2 × 206 × K2
K1 = (103/2)K2 ----1
K1 + K2 = 5.422 MeV ----2
From 1 and 2
K1= 5.319 MeV or 5319 kev 

JEE Advanced Mock Test - 4 (Paper II) - Question 27

Find which of the following compounds can have mass ratio of C : H : O as 6 : 1 : 24.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 27

Given mass ratio of C : H : O is 6 : 1 : 24; so the molar ratio will be 6/12 : 1/1 : 24/16 = 1 : 2 : 3.
HO-(C=O)-OH has molar ratio 1 : 2 : 3.
Hence, option (1) is correct.

JEE Advanced Mock Test - 4 (Paper II) - Question 28

Which of the following represents Schotten-Baumann reaction?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 28

The following reaction is called Schotten-Baumann reaction.

JEE Advanced Mock Test - 4 (Paper II) - Question 29

The sum of the series 

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 29

Given

This is an infinite series of the form:

which resembles the expansion of the standard logarithmic function:

Comparing this with our series, we observe that it matches the Maclaurin series expansion of tan-1x when x = a/n. Thus, the given series sums to:

To determine the final sum, we use the known identity:

By setting x = a/n  and considering its logarithmic expansion, we obtain:

JEE Advanced Mock Test - 4 (Paper II) - Question 30

If the number of terms in the expansion of (x - 2y + 3z)n is 45, then what is the value of n?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 30

Since the number of terms in (x1 + x2 + x3 + ... + xm)n is n + m - 1Cm-1, the number of terms in (x - 2y + 3z)n, where m = 3, is n + 3 - 1C3-1 - 1.
(n + 1)(n + 2)/2
We must have (n + 1)(n + 2)/2 = 45
⇒ (n + 1)(n + 2) = 90
⇒ n = 8

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