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JEE Advanced Mock Test - 1 (Paper II) - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Advanced Mock Test - 1 (Paper II)

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

Refractive index of an equilateral prism is √2

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

From

We can see that δm = 30o for μ = √2 and A = 60o

Further, at minimum deviation

∴ sin i1 = μ sin r1

= (√2) sin 30o

= 1/√2

∴ i1 = 45o

*Multiple options can be correct
JEE Advanced Mock Test - 1 (Paper II) - Question 2

There are three optical media, 1,2 and 3 with their refractive indices μ1 > μ2 > μ3 (TIR-total internal reflection)

Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 2
Total internal reflection takes place when ray of light travels from denser to rarer medium

Further,

Since,

θ12 > θ13

Smaller the value of critical angle more the chance of total internal reflection.

*Multiple options can be correct
JEE Advanced Mock Test - 1 (Paper II) - Question 3

A vernier callipers has 1 mm marks on the main scale. It has 20 equal divisions on the vernier scale which match with 16 main scale divisions. For this vernier calipers, the least count is

Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 3

Least count of vernier callipers

LC = 1MSD − 1VSD

= Smallest division on main scale / Number of divisions on vernier scale

20 divisions of vernier scale = 16 divisions of main scale

∴ 1VSD = 16 / 20mm = 0.8mm

LC = 1MSD − 1VSD

= 1mm − 0.8mm = 0.2mm

*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 4

One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dotted lines as shown in the graph below. If the work done along the solid line path is ws and that along the dotted line path is wd, then the ratio wd/ws is

(Round off up to 2 decimal places)


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

Solid line path work done (ws) is isothermal because PV is constant (Boyle's law) and dotted line (horizontal) path work done wd is isobaric. Work done in vertical line is zero as ΔV = 0.

Total work done on solid line path (ws) = 2.303 nRT log V2/V1

= 2.303 PV log V2/V1 = 2.303 x 4 x 0.5 log 5.5/0.5 = 4.8 L atm.

Total work done on dotted line path (wd) = PΔV

= 4 x (2 - 0.5) + 1(3 - 2) + 0.5 (5.5 - 3) = 6 + 1 + 1.25 = 8.25

So, ≈ 1.72.

*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 5

Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m2. The radii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N/m. Find the ratio nA/nB, where nA and nB are the number of moles of air in bubbles A and B, respectively. [Neglect the effect of gravity.]

(Round off upto 2 decimal places)


Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 5

Excess pressure above atmospheric pressure, due to surface tension in a bubble = 4T/r

Surrounding pressure, P0 = 8 N/m2

*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 6

Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T1 and T2, respectively. The maximum intensity in the emission spectrum of A is at 500 mm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that by B?


Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 6

According to Wien's displacement law,

According to Stefan Boltzmann law, rate of energy radiated by a black body is

[Here, A = 4πR2]

(Using (i))

= 9

*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 7

In the figure shown below, the maximum possible unknown resistance X (in Ω), that can be measured by the post office box is given by a × 10b Ω in scientific notation, then find the value of a−b?

(in this experiment, we take out only one plug in arm AB and only one plug in arm BC, but in arm AD we can take out many plugs).


Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 7

By Wheatstone bridge principle, we have P/Q = R/X

or X = RQ / P

Now

Xmax = 9000 × 1000 / 10

= 9 × 105

*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 8

A cylinder supports a piston of mass 5 kg5 kg and cross-sectional area 5×10−3 m2 enclosing a gas at 27°C. The gas is slowly heated to 77°C such that the piston rises by 0.1 m. The piston is now clamped at this new position and the gas is cooled down to its initial temperature. If atmospheric pressure is 1 atm, then the difference in heat supplied during heating and heat rejected during cooling is _____J.


Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 8

During heating, the process was isobaric, so, ΔW1 = PΔV

And internal energy, ΔU1 = nCvΔT = 50nCv

Thus, heat transfer, ΔQ1 = 50nCv + 55

During cooling, the process is isochoric, so, ΔW2 = 0 and ΔU2 = −50nCv

Thus, heat loss = ΔQ2 = −(ΔU2 + ΔW2) = 50nCv

∴ Heat supplied −- Heat rejected = 50nCv + 55 − 50nCv = 55 J

*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 9

Consider a frictionless plane of height h = 2 m m and angle α=30°. One end of an electric string is connected to point B, the other end is passed through an orifice at point D on the plane and connected to a body of mass m = 2 kg initially stationary at point A. This body comes to rest at point C at the bottom of the inclined plane. The line drawn from point B to point D is perpendicular to side AC. Knowing that the length of the string is unstretched state is equal to the length of segment BD. Determine the elastic constant in N m−1. Take g = 10 m s−2


Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 9
Using law of conservation of energy

*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 10

In figure all the pulleys and strings are massless, and all the surfaces are frictionless. A small block of mass m is placed on fixed wedge (take g = 10 ms-2):

The acceleration of m is____.


*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 11

In figure all the pulleys and strings are massless, and all the surfaces are frictionless. A small block of mass m is placed on fixed wedge (take g = 10 ms-2):

The acceleration of pulley P4 is _____.


JEE Advanced Mock Test - 1 (Paper II) - Question 12

Coefficient of mutual inductance for the given coils does not depend on

Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 12
The mutual inductance of two coils depends on their relative orientation, shape and size and also on medium in which the coils were placed.
JEE Advanced Mock Test - 1 (Paper II) - Question 13

Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.

The difference in maximum kinetic energy of photoelectrons from A and from B

Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 13

(13.6) − (3.2) = 10.2 eV

= constant

*Multiple options can be correct
JEE Advanced Mock Test - 1 (Paper II) - Question 14

Directions: The following question has four choices, out of which one or more is/are correct.

Which of the following can result in exactly 1 mol of K4[Fe(CN)6] stoichiometrically using the given compounds as the only source of carbon?

Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 14
Using POAC, 1 mol of C6H12O6 and 2 mol of Al4C3 have 6 mol of carbon atoms and hence, give 1 mol of K4[Fe(CN)6].

Hence, options (b) and (d) are correct.

*Multiple options can be correct
JEE Advanced Mock Test - 1 (Paper II) - Question 15

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

Cv and Cp denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then,

Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 15

For diatomic gas:

For monatomic gas:

Note: Cp - Cv = R (For all ideal gases)

Hence the options (a) and (c) are correct.

*Multiple options can be correct
JEE Advanced Mock Test - 1 (Paper II) - Question 16

Which of the following statements is/are correct?

Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 16

(b) is wrong because chemical adsorption first increases and then decreases with increase in temperature.

(d) is wrong because as a result of adsorption, there is a decrease in surface energy.

(a) and (c) are correct.

*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 17

Assume that water vapour behaves as an ideal gas and the volume occupied by the liquid water is negligible compared to the volume of the container. A sample of liquid water of mass 3 g is injected into an evacuated 76 L flask maintained at 320 K. At this temperature the vapour pressure of water is 32 mm of Hg. (1/0.0821 = 12.2)
Now give the answers to following questions:

What % of the water will be vaporized, when the system comes to equilibrium?


Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 17

*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 18

Assume that water vapour behaves as an ideal gas and the volume occupied by the liquid water is negligible compared to the volume of the container. A sample of liquid water of mass 3 g is injected into an evacuated 76 L flask maintained at 320 K. At this temperature the vapour pressure of water is 32 mm of Hg. (1/0.0821 = 12.2)
Now give the answers to following questions:

What should be the minimum volume (in litre) of the flask if no liquid water is to be present at equilibrium?


Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 18

*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 19

The Molecular Formula of compound (B) is CxHyOz. Then the value of x + y + z = ____


Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 19

Compound B is 
C6H10O1

*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 20

The Molecular mass (gm/mole) of compound (D) is :


Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 20

The compound (D) is 
C6H14O2

*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 21

Let each of the circles
S1 = x2 + y2 + 4y - 1 = 0,
S2 = x2 + y2 + 6x + y + 8 = 0,
S3 = x2 + y2 - 4x - 4y - 37 = 0
Touches the other two. Let P1, P2, P3 be the points of contacts of S1 and S2, S2 and S3, S3 and Srespectively and C1, C2, C3 be the centre of S1, S2, S3 respectively

The value of 


Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 21


*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 22

Let each of the circles
S1 = x2 + y2 + 4y - 1 = 0,
S2 = x2 + y2 + 6x + y + 8 = 0,
S3 = x2 + y2 - 4x - 4y - 37 = 0
Touches the other two. Let P1, P2, P3 be the points of contacts of S1 and S2, S2 and S3, S3 and Srespectively and C1, C2, C3 be the centre of S1, S2, S3 respectively

Length of direct common tangent for S1 and S3____.


Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 22


*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 23

Let f(x) and g(x) be two differentiable functions, defined as
f(x) = x2 + xg'(1) + g”(2) and g(x) = f(1)x2 + x f'(x) + f”(x).

The value of  is


*Answer can only contain numeric values
JEE Advanced Mock Test - 1 (Paper II) - Question 24

Let f(x) and g(x) be two differentiable functions, defined as
f(x) = x2 + xg'(1) + g”(2) and g(x) = f(1)x2 + x f'(x) + f”(x).

The value of  is ____


JEE Advanced Mock Test - 1 (Paper II) - Question 25

A dielectric slab of area A passes between the capacitor plates of area 2A with a constant speed v. The variation of current (i) through the circuit as function of time (t) can be qualitatively represented as

Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 25

JEE Advanced Mock Test - 1 (Paper II) - Question 26

Two identical cylinders have a hole of radius a (a<<R) at its bottom. A ball of radius R is kept on the hole and water is filled in the cylinder such that there is no water leakage from bottom. In case-1 water is filled upto height h and in second case it is filled upto height 2h. If F1 is force by liquid on sphere in case-1 and F2 is force by liquid on sphere in case-2 then.

Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 26

F1 = FB - (πa2) ρgh
F2 = FB - (πa2) ρgh

JEE Advanced Mock Test - 1 (Paper II) - Question 27

In a crystal structure of an element, atoms are present in fcc arrangement as well as intetrahedral voids. Which of the following relation between the atomic radius and the edge lengthis correct?

Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 27

Distance b/w corner and centre of tetrahedral void is 

JEE Advanced Mock Test - 1 (Paper II) - Question 28

Which one of the following reactions is not correct?

Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 28

Alcohols are less acidic than H2CO3 

JEE Advanced Mock Test - 1 (Paper II) - Question 29

From a pack of 52 playing cards; half of the cards are randomly removed without looking at them. From the remaining cards, 3 cards are drawn randomly. The probability that all are king.

Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 29

E1 “No card is king from removed cards”
E2 “1 card is king from removed cards”
E3 “2 card is king from removed cards”
E3 “3 card is king from removed cards”
E4 “4 card is king from removed cards”
E5 “5 card is king from removed cards”
F = 3 cards are drawn from pack those are kings.

JEE Advanced Mock Test - 1 (Paper II) - Question 30

  let f1(x) = f(x), fn(x) = f(fn-1(x)) for n ≥ 2, n ∈ N, then f2008 (x) + f2009 (x) =

Detailed Solution for JEE Advanced Mock Test - 1 (Paper II) - Question 30

f(f(x)) = x

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