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JEE Main Practice Test- 1 - JEE MCQ


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30 Questions MCQ Test - JEE Main Practice Test- 1

JEE Main Practice Test- 1 for JEE 2024 is part of JEE preparation. The JEE Main Practice Test- 1 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Practice Test- 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Practice Test- 1 below.
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JEE Main Practice Test- 1 - Question 1
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The self inductance of an electric fan motor is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of

Detailed Solution for JEE Main Practice Test- 1 - Question 1
For maximum power,

Or, C = 1 μ F

JEE Main Practice Test- 1 - Question 2
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A car of mass 800 kg moves on a circular track of radius 40 m. If the coefficient of friction is 0.5, then the maximum velocity with which the car can move is

Detailed Solution for JEE Main Practice Test- 1 - Question 2
When a car goes round a curved road, it requires some centripetal force. While rounding the curve, the wheels of the car have a tendency to leave the curved path and regain the straight line path. Force of friction between the wheels and the road opposes this tendency of the wheels. This force (of friction) therefore acts, towards the centre of the circular track and provides the necessary centripetal force.

Hence, the maximum velocity with which a vehicle can go round a level curve, without skidding is Here, r = 40 m

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JEE Main Practice Test- 1 - Question 3
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Directions: This question has two statements, Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement 1: Davisson-Germer experiment proved the wave nature of electrons.

Statement 2: If electrons have wave nature, then they can interfere and show diffraction.

Detailed Solution for JEE Main Practice Test- 1 - Question 3
Davisson-Germer experiment showed that electron beams can undergo diffraction, when passed through atomic crystals. This shows the wave nature of electrons as waves can exhibit interference and diffraction.
JEE Main Practice Test- 1 - Question 4
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A semi circular wire of radius 'r' is charged uniformly with a charge 'Q' as shown in the figure. Then, the force experienced by a charge 'q' kept at its centre is
Detailed Solution for JEE Main Practice Test- 1 - Question 4

Consider a small charge element dQ at an angle θ subtending a small angle element dθ with centre.

Force between the charges q and dQ is

Resolving the components of the force dF cosθ along x-axis and dF sinθ along y-axis, the components dF sinθ will cancel each other, dF cosθ will add up.

Therefore, resultant force on the charge q,

The direction of the force is along x-axis.

JEE Main Practice Test- 1 - Question 5
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A shell is fired from a cannon with a speed of 100 ms-1 at an angle 30° with the vertical (y-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio 1 : 2. The lighter fragment moves vertically upwards with an initial speed of 200 ms-1. What is the speed of the heavier fragment at the time of explosion?
Detailed Solution for JEE Main Practice Test- 1 - Question 5

JEE Main Practice Test- 1 - Question 6
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Consider a solid sphere of mass M and radius R. The energy released in forming this solid sphere is
Detailed Solution for JEE Main Practice Test- 1 - Question 6

Self potential energy of the sphere having mass m and radius r,

Hence, energy released in formation of the sphere is

JEE Main Practice Test- 1 - Question 7
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A block of mass m is placed on a surface with a vertical cross-section given by y = x3/6. If the coefficient of friction is 0.5, then the maximum height above the ground at which the block can be placed without slipping is
Detailed Solution for JEE Main Practice Test- 1 - Question 7
Block is under limiting friction, so

μ = tan θ ... (i)

Equation of the surface, y = x3/6

From equations (i) and (ii), we get

JEE Main Practice Test- 1 - Question 8
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The logic gate circuit shown in the figure realises which of the following functions?
Detailed Solution for JEE Main Practice Test- 1 - Question 8

This represents the XOR function.

JEE Main Practice Test- 1 - Question 9
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A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is-
Detailed Solution for JEE Main Practice Test- 1 - Question 9

Frictional force balances the weight of the body

Frictional force f = μN = mg

f= 0.2 X 10 = 2N

Therefore, the weight of the block mg= 2N

JEE Main Practice Test- 1 - Question 10
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The magnetism of a magnet is due to which of the following?
Detailed Solution for JEE Main Practice Test- 1 - Question 10

The magnetism of the magnet is due to the spin motion of electrons. Each electron in an atom is revolving in an orbit around the nucleus. The revolving electron is equivalent to a tiny loop of current. Therefore, it possesses some orbital magnetic dipole moment = current × area of the loop.

In addition to the orbital motion, every electron is assumed to have a spin motion around its axis. Therefore, another dipole magnetic moment called spin magnet is also associated with electron. The vector sum of provides the net magnetic dipole moment , to the atom. is much greater than , hence magnetism of magnet is due to spin motion of electrons.

JEE Main Practice Test- 1 - Question 11
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The resistance of the rod is 1 Ω. It is bent in form of a square. What is the resistance across adjacent corners?
Detailed Solution for JEE Main Practice Test- 1 - Question 11
When the rod is bent in the form of a square, then each side has a resistance of 1/4 Ω. As shown R1, R2 and R3 are connected in series, so, their equivalent resistance,

Now, R' and R4 are connected in parallel, so, the equivalent resistance of the circuit is

JEE Main Practice Test- 1 - Question 12
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Figure shows the vertical cross-section of a vessel filled with a liquid of density ρ. The normal thrust per unit area on the walls of the vessel at point P, as shown will be
Detailed Solution for JEE Main Practice Test- 1 - Question 12
Depth of point P from the free surface of water in the vessel = (H − h). Since, the liquid exerts equal pressure in all direction at one level, hence, the pressure at P = (H − h)ρg
JEE Main Practice Test- 1 - Question 13
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A conducting ring of mass 2 kg and radius 0.5 m is placed on a smooth horizontal plane. The ring carries a current i = 4 A. A horizontal magnetic field B = 10 T is switched on at time t = 0 as shown in the figure. The initial angular acceleration of the ring will be
Detailed Solution for JEE Main Practice Test- 1 - Question 13

τ = Iα

iAB = Iα

(∵ Ring is rotating about its diameter)

JEE Main Practice Test- 1 - Question 14
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A stone is dropped from a height of 45 m on a horizontal level ground. There is the horizontal wind blowing due to which horizontal acceleration of the stone becomes 10 m s−2. (Take g = 10 m s−2). The time taken (t)t by stone to reach the ground and the net horizontal displacement (x) of the stone from the time it is dropped and till it reaches the ground are respectively
Detailed Solution for JEE Main Practice Test- 1 - Question 14
Taking motion in vertical direction,

Taking motion in the horizontal direction,

JEE Main Practice Test- 1 - Question 15
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The current (I) in the inductance is varying with time according to the plot shown in the figure.

The correct variation of voltage with time in the coil is
Detailed Solution for JEE Main Practice Test- 1 - Question 15
From the graph given in the question, we can see that the current is rising up to the time T/2 from 0, and decaying up to the time T from T/2.

Now, voltage induced in an inductor is given as;

Now, voltage induced in an inductor is given as;

If we see the slope for up to t = T/2 i.e; dIdtdIdt is positive and constant (L is already constant term) which shows that the value of voltage is not changing from time t = 0 to t = T/2.

And again, if we see the slope for up to t = T i.e; dI/dt is negative but constant which shows that the value of voltage is not changing from time t= T/2 to t = T.

Hence, the graph between voltage and time is as shown below;

JEE Main Practice Test- 1 - Question 16
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Two charges q1 and q2 are placed 30 cm apart, as shown in the figure. A third charge q3 is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is where k is
Detailed Solution for JEE Main Practice Test- 1 - Question 16
The potential energy when q, is at point C

The potential energy when q3 is at point D

Thus change in potential energy is

JEE Main Practice Test- 1 - Question 17
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A point source of heat of power P is placed at the centre of a spherical shell of mean radius R. The material of the shell has thermal conductivity K. Calculate the thickness of the shell if the temperature difference between the outer and inner surfaces of the shell in steady-state is T.
Detailed Solution for JEE Main Practice Test- 1 - Question 17
Consider a concentric spherical shell of radius r and thickness dr as shown in diagram. The radial rate of flow of heat through this shell in steady state will be,

[Negative sign is used as with increase in r, θ decreases.]

Now as for spherical shell A = 4πr2

which on integration and simplification gives

Now in steady state as no heat is absorbed, rate of loss of heat by conduction is equal to that of supply, i.e., H = P, and here

θ1 - ​θ2 = T and a ≃ b = Ra ≃ b = R

So Equation (i) becomes,

i.e., thickness of shell

JEE Main Practice Test- 1 - Question 18
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Two condensers, one of capacity C and other of capacity C/2 are connected to a V - volt battery, as shown in the figure. The work done in charging fully both the condensers is
Detailed Solution for JEE Main Practice Test- 1 - Question 18
As the condensers are connected in parallel, therefore potential difference across both the condensers remains the same.

Work done in charging fully both the condensers is given by

JEE Main Practice Test- 1 - Question 19
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Two radio antennas radiating waves in phase are located at points A and B, 200 m apart (Figure). The radio waves have a frequency of 6 MHz. A radio receiver is moved out from point BB along a line perpendicular to the line connecting A and B (line BC as shown in the figure). At what distances from B will there be destructive interference?
Detailed Solution for JEE Main Practice Test- 1 - Question 19

path difference at P = AP - BP

path difference at

For destructive interference at P

path difference

From (1) and (2)

For first minima n = 1,

For second minima n = 2,

For third minima n = 3,

JEE Main Practice Test- 1 - Question 20
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With the assumption of no slipping, determine the mass mm of the block which must be placed on the top of a 6 kg cart in order that the system period is 0.75 s. What is the minimum coefficient of static friction μS for which the block will not slip relative to the cart if the cart is displaced m50 mm from the equilibrium positions and released? Take (g = 9.8 m s-2).
Detailed Solution for JEE Main Practice Test- 1 - Question 20

= 2.55 kg

Maximum acceleration of SHM is,

amax = ω2A (A = amplitude)

i.e., maximum force on mass 'm' is m ω2 A which is being provided by the force of friction between the mass and the cart. Therefore,

μsmg ≥ mω2 A

Thus, the minimum value of μs should be 0.358.

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 21
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In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric main is 220 V. The minimum capacity of the main fuse of the building will be (Key in the integral value, in Amperes)

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

Power of 15 bulbs of 40 W = 15 × 40 = 600 W

Power of 5 bulbs of 100 W = 5 × 100 = 500 W

Power of 5 fan of 80 W = 5 × 80 = 400 W

Power of 1 heater of 1 kW = 1,000 W

∴ Total power, P = 600 + 500 + 400 + 1,000 = 2,500 W

When this combination of bulbs, fans and heater is connected to 220 V mains, current in the main fuse of building,

So, the minimum capacity of the main fuse of the building will be 12 A.

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 22
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The resistance of a wire is 5 ohm at 50°C and 6 ohm at 100°C. The resistance of the wire at 0°C will be (in ohm)

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

Given, R50 = 5Ω and R100 = 6Ω

Rt = R0(1 + α t), where Rt = Resistance of a wire at t°C, R0 = Resistance of a wire at 0°C and α = Temperature coefficient of resistance.

∴ R50 = R0[1 + α 50]

and

R100 = R0[1 + α100]

or, R50 - R0 = R0 α(50) ... (i)

R100 - R0 = R0 α(100) ... (ii)

Divide (i) by (ii), we get

or, 10 - 2R0 = 6 - R0

or, R0 = 4Ω

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 23
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A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T (in °C) is given by:

(Given: room temperature = 30°C, specific heat of copper = 0.1 cal/gm°C)

Detailed Solution for JEE Main Practice Test- 1 - Question 23

Heat lost by copper ball = Heat gained by calorimeter + Heat gained by water

(100)(0.1)(T - 75) = (100)(0.1)(45) + (170)(1) (45)

10(T - 75) = 450 + 7650 = 8100

T - 75 = 810

T = 885°C

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 24
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The circuit contains two diodes each with a forward resistance of 50 Ω and with infinite reverse resistance. If the battery voltage is,6 V, the current through the 120 Ω resistance is mA.

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

D2 is reverse bias so current does not flow through D,.

D, is forward bias.

I = 6/300 ⇒ 0.02 A

= 20 mA

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 25
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The impulse imparted to a ball of mass m, when it strikes a rigid wall with a speed u and rebounds with the same speed is −αmu, what is the value of α?

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

The situation is as shown in the figure.

First, calculate the initial linear momentum and final linear momentum,

Pinitial = mu, Pfinal = −mu

Now we know that change at the moment is equal to impulse.

It means,

Impulse imparted to the ball = change in momentum

I = Pfinal − Pinitial = −mu − mu = −2mu

JEE Main Practice Test- 1 - Question 26
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If 5.08 g of a cupric salt is dissolved in water and excess of KI is added to the solution, and the liberated I2 requires 25 ml of M/5 Na2S2O3, where it gets converted into S4O62-, then find the weight percentage of copper in the salt.
Detailed Solution for JEE Main Practice Test- 1 - Question 26
Using,

G. Equivalents of Cu2+ = G .Equivalents of Na2S2O3

∴ Weight of Cu2+ = 0.15875 g

Percentage of Cu+2 in the sample

JEE Main Practice Test- 1 - Question 27
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The compound having the highest dipole moment is
Detailed Solution for JEE Main Practice Test- 1 - Question 27
Out of the given compounds, formaldehyde has the highest dipole moment due to the presence of highly electronegative oxygen atom.

JEE Main Practice Test- 1 - Question 28
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Which of the following is not a buffer solution?
Detailed Solution for JEE Main Practice Test- 1 - Question 28
A buffer solution is that which can resist a change in pH on addition of small amount of acid or base.

It is of two types:

(i) An acidic buffer (weak acid in excess + salt of its conjugate base)

(ii) A basic buffer (weak base in excess + salt of its conjugate acid)

KClO4 + HClO4 is not a buffer solution because it a mixture of a strong acid (HClO4) and the salt of its conjugate base (KClO4).

Note: All the other mixtures in the options are perfect buffers.

JEE Main Practice Test- 1 - Question 29
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A cell utilises the reaction Zn (s) + 2H+ (aq) Zn2+ (aq) + H2 (g). Addition of H2SO4 to the cathode compartment will
Detailed Solution for JEE Main Practice Test- 1 - Question 29

and

If H2SO4 is added to cathodic compartment, (towards reactant side), then Q decreases (due to increase in H+). Hence equilibrium is displaced towards right and Ecell increases.

JEE Main Practice Test- 1 - Question 30
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Which of the following series of reactions correctly represents the chemical relations related to iron and its compound?
Detailed Solution for JEE Main Practice Test- 1 - Question 30
Formation of Fe3O4 from Fe corresponds to the combustion of Fe and rest of the reactions correspond to the production of Fe by reduction of Fe3O4 in the blast furnace.
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