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


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

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JEE Advanced Mock Test - 1 (Paper I) - Question 1

A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity ω. This can be considered as equivalent to a loop carrying a steady current . A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant .

The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is

Detailed Solution for JEE Advanced Mock Test - 1 (Paper I) - Question 1
Induced electric field due the change in magnetic flux is

Here

Hence magnitude of the induced electric field is

E = BR/2

JEE Advanced Mock Test - 1 (Paper I) - Question 2

A moving ball collides a pendulum bob as shown, the coefficient of restitution so that bob performs vertical circular motion about 0 is (g = 10 m/s2)

Detailed Solution for JEE Advanced Mock Test - 1 (Paper I) - Question 2

JEE Advanced Mock Test - 1 (Paper I) - Question 3

The mass of nucleus . z X A 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 mass m1 and m2 only if ( m1 + m2 ) < M . Also two light nuclei of massws m3 and m4 can undergo complete fusion 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:

Q. Which of the following statements is correct?

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

Q value of the reaction,

Hence α-decay not possible

So this reaction is also not possible

This reaction is possible

This reaction is also not possible.

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

Three simple harmonic motions in the same direction having the amplitudes a, 2a and 3a respectively are superposed. If each differ in phase from the next by 45o , then find out the correct options (Consider that all have the same frequency)

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

y1 = asinωt

y2 = 2asin(ωt + 45o)

y3 = 3asin(ωt + 90o)

Resultant =

*Multiple options can be correct
JEE Advanced Mock Test - 1 (Paper I) - Question 5

In the figure small block of mass m is kept on big block of mass M, then

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

Acceleration of M, a = (F / M)

*Multiple options can be correct
JEE Advanced Mock Test - 1 (Paper I) - Question 6

A fish looking upwards through the water seeing the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the surface, the radius of this circle in cm is

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

For total internal reflection,

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

A long, square wooden block is pivoted along one edge as shown. The block is in equilibrium when immersed in water to the depth shown.

Assuming friction in the pivot is negligible. Let the density of wood be 170α170α. Hence αα is (in SI unit, up to two significant digit).


Detailed Solution for JEE Advanced Mock Test - 1 (Paper I) - Question 7
Torque due to liquid pressure

=

= 360

Torque due to upthrust force

= (0.6 × 1.2 ) × 103 × 10 × 0.6 = 4320

Torque due to weight

= ρ × (1.2)2 × ×10 × 0.6 = 8.64 ρ

∑τ = 0

360 + 4320 - 8.64 ρ = 0

Hence α = 3

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

Two particles are projected horizontally in opposite directions from a point in a smooth inclined plane of inclination θ=60with the horizontal as shown in the figure. Find the separation between the particles in the inclined plane when their velocity becomes perpendicular to each other. v1 = 1 m/s, v2 = 3 m/s. Express your answer in the form of k/5 metre. Then, find the value of k.


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

v1 = v1i − g sin θt j

v2 =−v2 i − g sinθt j

v1.v2 = 0

or

or

or

The particle will be parallel to x - axis.

Separation between the particles will be

=

= 4/5

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

A plane monochromatic light wave falls normally on a diaphragm with two narrow slits separated by a distance d = 2.5 mm. A fringe pattern is formed on a screen placed at a distance l = 100 cm behind the diaphragm. By what distance in mm will these fringes be displaced when one of the slits is covered by a glass plate of thickness h = 10 μm?


Detailed Solution for JEE Advanced Mock Test - 1 (Paper I) - Question 9

Due to introduction of glass plate, width of fringe does not change.

Here, width of fringe,

According to problem,

D = l = 100 cm = 1m, d = 2.5 mm = 2.5 x 10-3 m

h = 10 x 10-6 m, n = 1.5

Putting the values, we get

Δy = 2×10−3

Δy = 2 mm.

In the diagram, upper slit is covered by glass plate.

So, fringe pattern on screen shifts upward.

Remarks : In this case, central bright fringe does not coincide with the centre of secreen.

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

If the switches S1, S2 and S3 in the figure are arranged such that current through the battery is minimum, then find the voltage across points A and B (in volt).


Detailed Solution for JEE Advanced Mock Test - 1 (Paper I) - Question 10

= 24/12

= 2A

VAB = 2 × 0.5 = 1 V

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

A parallel plate capacitor of capacitance 100 pF is to be constructed using paper sheet of 1 mm thickness as dielectric. If the dielectric constant of paper is 4.0, then the number of circular metal foils of diameter 2.0 cm is (N + 2). The value of N is


Detailed Solution for JEE Advanced Mock Test - 1 (Paper I) - Question 11

Let the number of sheets required be 'n'.

They will form (n - 1) capacitors in series. If K is the dielectric constant of dielectric, the capacitance is given by:

So,

(n - 1) = 9

n = 10

The number of circular metal foils of diameter 2.0 cm is 10.

N + 2 = 10 or N = 8

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

A square platform of side length 8 m is situated in x-z plane such that it is at 16 m from the x-axis and 8 m from the z-axis as shown in figure. A particle is projected with velocity)m/s relative to wind from origin and at the same instant the platform starts with acceleration . Wind is blowing with velocity  (g = 10 m/s2)

JEE Advanced Mock Test - 1 (Paper I) - Question 13

A siphon tube is discharging a liquid of specific gravity 0.9 from a reservoir as shown in the figure.

JEE Advanced Mock Test - 1 (Paper I) - Question 14

Figure shows three concentric thin spherical shells A, B and C of radii R, 2R, and 3R, respectively. The shell B is earthed, and A and C are given charges q and 2q, respectively. If the charge on surfaces 1, 2, 3 and 4 are q1, q2, q3, and q4 respectively, then match the following list:

JEE Advanced Mock Test - 1 (Paper I) - Question 15

A bird in air is diving vertically over a tank with speed 5 cm s-1. Base of tank is silvered. A fish inthe tank is rising upward along the same line with speed 2 cm s-1. The water level is falling atrate of 2 cm s-1. (Take: μwater = 4/3)

JEE Advanced Mock Test - 1 (Paper I) - Question 16

Two aliphatic aldehydes P and Q react in the presence of aqueous K2CO3 to give compound R, which upon treatment with HCN provides compound S. On acidification and heating, S gives the product shown below.

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

JEE Advanced Mock Test - 1 (Paper I) - Question 17

The hydrogen-like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light, the ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state energy of the hydrogen atom.

The orbital S2 and the energy of the state S1 in units of the hydrogen atom ground state energy respectively are;

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

The spherically symmetric state S1 of Li2+ has one radial node 2s. Upon absorbing light, the ion gets excited to state S2, which also has one radial node.

The energy of electron in S2 is same as that of H-atom in its ground state.

, where E1 is the energy of H-atom in the ground state

But, E1 = En

Thus, n = 3

Energy state S2 of Li2+ having one radial node is 3p.

Energy State

JEE Advanced Mock Test - 1 (Paper I) - Question 18

The correct set of the products obtained in the following reactions

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

A) When alkyl cyanide undergoes reduction reaction then primary amine is produced.

B) When alkyl cyanide is treated with grignard reagent then ketone is produced as a major product.

C) When alkyl cyanide undergoes hydrolysis then primary amine is produced.

D) when alkyl cyanide undergoes reaction with nitrous acid then alcohol is produced.

JEE Advanced Mock Test - 1 (Paper I) - Question 19

Which starting material should be used to produce the compound shown below ?

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

The product is having 8 sided cyclic ring

Reductive ozonolysis of alkenes breaks-up

C = C bonds completely to produce carbonyl carbon centres

The remaining compounds will not give the required product.

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

Graph shows variation of internal energy U with density ρ of one mole of an ideal monoatomic gas. Process AB is a part of rectangular hyperbola. Find work done in the process (in Joules)


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

U.ρ. = Constant
T/V = constant
P = constant
W = nR T .....(i)
n(Cv)ΔT = 3

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

Two blocks of masses m1 = 5kg and m2 = 2 kg are connected by threads which pass over the pulleys as shown in the figure. The threads are massless, and the pulleys are massless and smooth. The blocks can move only in the vertical direction. If the acceleration of m1, if expressed in simplest from, is equal to  calculate 'n'. (Take g = 10 m/s2)


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

How many maximum numbers of cyclic isomer(s) is/are possible with formula C5H10? (consider stereoisomers)


JEE Advanced Mock Test - 1 (Paper I) - Question 23

Match the following:

JEE Advanced Mock Test - 1 (Paper I) - Question 24

Match the following:

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

Match the following:

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

Match the following:

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

Match the following:

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

(P) CS1 and CS2 divided the triangle into three of equal area so that S1 and S2 are points of trisection of AB. Thus S1 = (0, 0), S2 = (2, -3). Equation of CS1 is y - x = 0. Slope of CS2 = -4. The line through (0, 0) drawn parallel to CS2 is Required equation is y2 + 3xy - 4x2 = 0
Required equation y2 + 3xy - 4x2 = 0
⇒ λ+ μ = 3 + 4 = 7
(Q) The required area is bounded by the lines y = ±x and the parabola x2 = - (y - 2) having the vertex at
(0,2) and passing through (±1, 1). By summary about the y-axis,



Whose solution is x(1 - v2) = c. When x = 2, v = 1/2 so that c = 3/2
The equation of the curve is 
This is a rectangular hyperbola with eccentricity √2.
∴ e6 = 8

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

Match the following:

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

(P):Equation of the plane A(x → 1) + B(y → 1) + C(z → 1) = 0
Since the lines is perpendicular to the plane (1)
∴ 3(x → 1) + 0(y → 1) + 4(z → 1) = 0
3x + 0y + 4z → 7 = 0
Distance from (0, 0, 0)



 

(R) If x is replaced by - x in the given equation, then
- x f( - x) + (1 + x) f(x) = x2 - x + 1
subtracting the two equations we get
f(- x) = f(x) + 2x, subtracting the value of f(- x) in the given equation we get
xf(x) + (1 - x) (f(x) + 2x) = x+ x + 1 and thus

(S) Using the notation from the above diagram and the conditions from the problem one obtains:


And consequently y = 3r. On the other hand the area of
the trapezoid ABCD is 4, thus (r + x + r + y)r =4.
Substituting for x = r/3 and y = 3r,
we get r = √3/2 ⇒ 4r2 = 3

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

Match the following:

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



Comparing coefficient of x9 both sides, we get

(R) Let P be origin and position vectors of point A, B, C be 

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

Match the following:

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

(P) The equation of tangents to hyperbola having slope m are

 are eccentricities of a hyperbola and its conjugate hyperbola, so

The equation of the lines e’x + ey - ee’ = 0
It is tangent to the circle x+ y= r2  

(R) The equation of common tangents is x  2x + 4 = 0
(a + b) = 5
(S) The equation of chord to ellipse whose midpoint is (0, 3) is (T = S1)

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