JEE Main Practice Test- 1 - JEE MCQ

JEE Main Practice Test- 1 - JEE MCQ

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75 Questions MCQ Test - JEE Main Practice Test- 1

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JEE Main Practice Test- 1 - Question 1

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

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

JEE Main Practice Test- 1 - Question 3

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.
JEE Main Practice Test- 1 - Question 31

Which of the following is formed when NaCl is heated with conc. H2SO4 and solid K2Cr2O7?

Detailed Solution for JEE Main Practice Test- 1 - Question 31
NaCl is heated with conc. H2SO4 and solid K2Cr2O7 to give chromyl chloride (orange red) vapours.

JEE Main Practice Test- 1 - Question 32

b-particle is emitted in radioactivity by

Detailed Solution for JEE Main Practice Test- 1 - Question 32
Since the nucleus does not contain b-particle, it is produced by the conversion of a neutron to a proton at the moment of emission.
JEE Main Practice Test- 1 - Question 33

Which of the following alkenes will react faster with H2 under catalytic hydrogenation conditions?

Detailed Solution for JEE Main Practice Test- 1 - Question 33
The following order of stability of alkenes has been derived from heat of hydrogenation

R2C = CR2 > R2C = CHR > R2C = CH2 = trans-RCH=CHR > cis-RCH = CHR > RCH = CH2 > CH2 = CH2

The reactivity order is reverse of the stability order.

JEE Main Practice Test- 1 - Question 34

When m-chlorobenzaldehyde is treated with 50% KOH solution, the product(s) obtained is/are

Detailed Solution for JEE Main Practice Test- 1 - Question 34
It is a case of Cannizzaro reaction.

JEE Main Practice Test- 1 - Question 35

Among cellulose, poly(vinylchloride), nylon and natural rubber, the polymer in which the intermolecular force of attraction is the weakest is

Detailed Solution for JEE Main Practice Test- 1 - Question 35
Cellulose and nylon are fibres and their intermolecular forces are the strongest. Poly vinyl chloride is a thermoplastic, the intermolecular force in it is neither strong nor weak.

Natural rubber is an elastomer and the polymers are held by weak van der Waals forces.

Hence, natural rubber has the weakest forces.

JEE Main Practice Test- 1 - Question 36

Name the end product in the following series of reactions

Detailed Solution for JEE Main Practice Test- 1 - Question 36
Here the final product is Benzylamine

JEE Main Practice Test- 1 - Question 37

Which of the following does not illustrate the anomalous properties of lithium?

Detailed Solution for JEE Main Practice Test- 1 - Question 37
The strength of the metallic bond decreases down the group. The hardness of these alkali metals decreases down the group. Hence, 'Li' is harder than other alkali metals.
JEE Main Practice Test- 1 - Question 38

The vapour of Hg absorb some electrons accelerated by a potential difference of 4.5 volts from rest as a result of which light is emitted. If the full energy of a single incident electron is supposed to be converted into light emitted by electron in a single Hg atom, find the wave number (1/λ) of the light

Detailed Solution for JEE Main Practice Test- 1 - Question 38

Potential difference 4.5 V

∴ Potential energy = 4.5 eV energy absorbed and this energy is emitted by electron of Hg atom.

JEE Main Practice Test- 1 - Question 39

Identify the correct statement:

Detailed Solution for JEE Main Practice Test- 1 - Question 39
Corrosion of iron can be minimised by forming a contact with another metal with a lower reduction potential. Corrosion of iron can be minimised by forming an impermeable barrier at its surface.
JEE Main Practice Test- 1 - Question 40

A crystalline hydrated salt on being heated loses 45.6% of its weight. Percentage composition of anhydrous salt is Al - 10.5 %, K - 15.1 %, S - 24.8 %, O - 49.6 %. Find molecular formula of salt (hydrated) (M = 948 g/mol for hydrated salt).

Detailed Solution for JEE Main Practice Test- 1 - Question 40
For anhydrous salt

Empirical formula of anhydrous salt = KAlS2O8

E.F mass of Anhydrous salt = 258 g/mol

Empirical formula of Anhydrous salt KAlS2O8.12H2O

E F Mass = 258 + 216 = 474 g/Empirical formula mass

Molecular Formula = K2Al2S4O16.24H2O

= K2SO4.Al2(SO4)3.24H2O

JEE Main Practice Test- 1 - Question 41

Which one of the following ores is best concentrated by froth flotation method?

Detailed Solution for JEE Main Practice Test- 1 - Question 41

Froth flotation method is used for concentration of sulphide ores.

Galena → PbSGalena → PbS

Malachite is Cu2(CO3)2(OH)2

Magnetite is Fe3O4

Siderite is FeCO3

JEE Main Practice Test- 1 - Question 42

Which of the following is a biradical?

Detailed Solution for JEE Main Practice Test- 1 - Question 42

JEE Main Practice Test- 1 - Question 43

At 500 K, the half life period of a gaseous reaction at an initial pressure of 80 kPa is 350 s350 s. When the pressure is 40 kPa, the half life period is 175 s; the order of the reaction is :

Detailed Solution for JEE Main Practice Test- 1 - Question 43

JEE Main Practice Test- 1 - Question 44

Consider the following three halides :

(1) CH3 − CH2 − ClCH3 - CH2 - Cl

(2) CH2 = CH − ClCH2 = CH - Cl

(3) C6H5 − ClC6H5 - Cl

Arrange C − Cl bond length of these compounds in decreasing order -

Detailed Solution for JEE Main Practice Test- 1 - Question 44
(1) CH3 − CH2 − Cl

Resonating structure = 0

(2) CH2 = CH − Cl

Resonating structure = 2

(3) C6H5 − Cl

Resonating structure = 4

JEE Main Practice Test- 1 - Question 45

Ratio of ΔTb/Kb of 10 g AB2 and 14 g A2B per 100 g of solvent in their respective, solution (AB2 and A2B both are non-electroytes) is 1 mol/kg in both cases. Hence, atomic weight of A and B are respectively :

Detailed Solution for JEE Main Practice Test- 1 - Question 45

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 46

Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is

Detailed Solution for JEE Main Practice Test- 1 - Question 46

In the redox reaction, Cr2O72- oxidises Mohr's salt, FeSO4.(NH4)2SO4.6H2O.

6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O

Mohr's salt [FeSO4.(NH4)2SO4.6H2O] and dichromate reacts in 6 : 1 molar ratio.

One mole of Cr2O72- will require six moles of Fe2+ ions.

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 47

The pH of a 0.01 M solution of a weak acid having degree of dissociation 12.5% is ____. (Answer up to 1 decimal place)

Detailed Solution for JEE Main Practice Test- 1 - Question 47

[H+] = Molarity × Degree of dissociation

We know,

pH = -log [H+]

= -log (1.25 × 10-3)

= -log 1.25 + 3

= 3 - 0.0970

= 2.9030

= 2.9

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 48

For the reaction A → C + D, the initial concentration of A was 0.01 M. After 100 s, the concentration of A becomes 0.001 M. The rate constant of the reaction has the numerical value 9.0. The order of reaction is

Detailed Solution for JEE Main Practice Test- 1 - Question 48

For a second order reaction,

= 1000 - 100 = 9 × 100

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 49

The number of stereo isomers possible for a compound of the molecular formula is CH3−CH=CH−CH(NH2)−CH3 is

Detailed Solution for JEE Main Practice Test- 1 - Question 49

As about the double bond, two geometrical isomers are possible and the compound is having one chiral carbon so there are two optical isomers. It means it has total four stereo isomers.

CH3−CH=CH−*CH(NH2)−CH3

Total G.I = 2n = 21 = 2

Total O.I = 2n = 21 = 2

Here n = number of double bonds or number of chiral carbon atoms.

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 50

Total number of optically inactive stereoisomers of tartaric acid is/are

Detailed Solution for JEE Main Practice Test- 1 - Question 50

Only Meso form of succinic acid is optically inactive due to presence of plane of symmetry. Enantiomers are optically active isomers of tartaric acid. Mixture of enantiomers in equal amount is known as racemic mixture. Racemic mixture is optically inactive.

JEE Main Practice Test- 1 - Question 51

Let A and B be two sets in a universal set U. Then,

Detailed Solution for JEE Main Practice Test- 1 - Question 51

Clearly, option (1) is not correct. Also, from (i), option (3) is not correct.

Hence, option (4) is correct.

JEE Main Practice Test- 1 - Question 52

Detailed Solution for JEE Main Practice Test- 1 - Question 52

= 4 + 5ω334 + 3ω365

= 4 + 5(ω3)111. ω + 3(ω3)1212

= 4 + 5. ω + 3ω2 (∴ ω3 = 1)

= 1 + 2ω + 3(1 + ω + ω2) = 1 + 2ω + 3(0)

= 1 - 1 + ω√3i

= √3i

JEE Main Practice Test- 1 - Question 53

If the normal at the point (bt12, 2bt1) on a parabola meets the parabola again at the point (bt22, 2bt2), then

Detailed Solution for JEE Main Practice Test- 1 - Question 53

The normal at (bt12, 2bt1) on the parabola y2 = 4bx meets the parabola again at (bt22, 2bt2).

∴ t1x + y = 2bt1 + bt13 passes through (bt22, 2bt2).

⇒ t1bt22 + 2bt2 = 2bt1 + bt13

⇒ t1(t22 - t12) = 2(t1 - t2)

⇒ t1(t2 + t1) = -2

JEE Main Practice Test- 1 - Question 54

Detailed Solution for JEE Main Practice Test- 1 - Question 54

JEE Main Practice Test- 1 - Question 55

The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitrary constants, is of

Detailed Solution for JEE Main Practice Test- 1 - Question 55
Given: Ax2 + By2 = 1

As the solution has two constants, the order of the differential equation is 2,

So, our choices (2) and (3) are discarded from the list, only choices (1) and (4) are possible.

Again, Ax2 + By2 = 1 ... (*)

Differentiating (*) w.r.t. x,

Again, differentiating (i),

From (i) and (ii), we get

⇒ Order = 2 and degree = 1

JEE Main Practice Test- 1 - Question 56

The volume of a parallelepiped whose sides are given by = 2i - 3j, = i + j - k, = 3i - k is

Detailed Solution for JEE Main Practice Test- 1 - Question 56

= 2(-1) + 3(-1 + 3) + 0(0 - 3) = -2 + 6 = 4

JEE Main Practice Test- 1 - Question 57

In an experiment with 15 observations on x, the following results were available.

∑x2 = 2830, ∑x = 170

One observation, which was 20, was found to be wrong and was replaced by the correct value 30. Then, the corrected variance was

Detailed Solution for JEE Main Practice Test- 1 - Question 57
∑x = 170 and ∑x2 = 2830

Increase in ∑x = 10 ∑x' = 170 + 10 = 180

Increases in ∑x2 = 900 - 400 = 500

Then,

∑(x')2 = 2830 + 500 = 3330

JEE Main Practice Test- 1 - Question 58

The general solution of sin x - 3 sin 2x + sin 3x = cos x - 3 cos 2x + cos 3x is

Detailed Solution for JEE Main Practice Test- 1 - Question 58

(sin x + sin 3x) - 3 sin 2x = (cos x + cos 3x) - 3 cos 2x

∴ sin 2x (2 cos x - 3) = cos 2x (2 cos x - 3)

⇒ cos x = 3/2 or tan 2x = 1

cos x = 3/2 is impossible.

∴ tan 2x = 1

JEE Main Practice Test- 1 - Question 59

If the pth term of an A.P. be q and qth term is p, then its rth term will be

Detailed Solution for JEE Main Practice Test- 1 - Question 59

Given that, Tp = a + (p − 1)d = q …….(i) and

Tq = a + (q − 1)d = p ……. (ii)

From (i) and (ii), we get d = [−(p − q)] / [(p − q)] = −1

Putting value of d in equation (i), then a = p + q − 1

Now, rth term is given by A.P. Tr = a + (r − 1)d

= (p + q − 1) + (r − 1) (−1)

= p + q − r

JEE Main Practice Test- 1 - Question 60

The function f(x) = x1/x has a maximum value at

Detailed Solution for JEE Main Practice Test- 1 - Question 60

⇒ x = e ; which is a point of maxima as f(e - h) > 0 and f(e + h) < 0="" for="" small="" and="" positive="" />

JEE Main Practice Test- 1 - Question 61

For a random variable X, if E(X) = 5 and Var(X) = 6, then E(X2) is equal to

Detailed Solution for JEE Main Practice Test- 1 - Question 61
We know that, variance of a random variate X is equal to difference of expectation of the variate X2 and square of the expectation of the variate X.

Hence the result.

JEE Main Practice Test- 1 - Question 62

If f(x) = cos(ln x), then

Detailed Solution for JEE Main Practice Test- 1 - Question 62

JEE Main Practice Test- 1 - Question 63

The area bounded by y = xe|x| and the lines |x| = 1, y = 0 is

Detailed Solution for JEE Main Practice Test- 1 - Question 63
Since, |x| = 1

∴ x = ±1

JEE Main Practice Test- 1 - Question 64

The integral is equal to (where [.]. denotes the greatest integer function)

Detailed Solution for JEE Main Practice Test- 1 - Question 64

JEE Main Practice Test- 1 - Question 65

The domain of definition of the function

Detailed Solution for JEE Main Practice Test- 1 - Question 65
x2 − 1 ≥ 0

⇒ x2 ≥ 1

⇒ x ∈(−∞​, −1] ∪ [1, ∞)…….(1)

and x − 1 > 0

⇒ x > 1

⇒ x ∈ (1, ∞) ……(2)

From (1) and (2)

x∈(1, ∞)

JEE Main Practice Test- 1 - Question 66

A plane passes through (1, 2, 2) and is perpendicular to two planes 2x – 2y + z = 0 and x – y + 2z = 4. Square of the distance of the plane from the point (52, 53, 57) is

Detailed Solution for JEE Main Practice Test- 1 - Question 66

Let the plane be a(x – 1) + b(y – 2) + c(z – 2) = 0

Since the plane is perpendicular to the given planes, we have

2a – 2b + c = 0, a – b + 2c = 0

So the equation of the plane is x + y – 3 = 0

d = distance from the point (52, 53, 57)

JEE Main Practice Test- 1 - Question 67

Detailed Solution for JEE Main Practice Test- 1 - Question 67

As sinθ is increasing in [0, π/2]

JEE Main Practice Test- 1 - Question 68

Detailed Solution for JEE Main Practice Test- 1 - Question 68

Applying componendo and dividendo, we get,

Differentiating, we get,

JEE Main Practice Test- 1 - Question 69

is equal to (where C is an arbitrary constant)

Detailed Solution for JEE Main Practice Test- 1 - Question 69

JEE Main Practice Test- 1 - Question 70

If A is the set of even natural numbers less than 8 and B is the set of prime numbers less than 7, then the number of relations from A to B are

Detailed Solution for JEE Main Practice Test- 1 - Question 70
A = {2,4,6 ; B = {2,3,5}

∴ A × B contains 3 × 3 = 9 elements.

Hence, the number of relations from A to B = 29.

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 71

Let If is the inverse of matrix A, then α is

Detailed Solution for JEE Main Practice Test- 1 - Question 71

Given, A-1 = B ⇒ 10A-1 = 10B

Equate element a21 on both sides of (1).

⇒ (-5)(1) + 0(2) + α(1) = 0

⇒ -5 + α = 0

⇒ α = 5

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 72

Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is p/q, then p is equal to

Detailed Solution for JEE Main Practice Test- 1 - Question 72

Number of horses = 5

∴ Probability that 'A' cannot win the race

Probability that 'A' must win the race = 1 - Probability that 'A' cannot win the race

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 73

20C4 + 2.20C3 + 20C18 - 22C18 is equal to

Detailed Solution for JEE Main Practice Test- 1 - Question 73

20C4 + 2.20C3 + 20C18 - 22C18

20C4 + 2.20C3 + 20C2 - 22C18

= 21C4 + 20C3 + 20C2 - 22C18

= 21C4 + 21C3 - 22C18

= 22C4 - 22C18

= 22C18 - 22C18

= 0

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 74

If coefficient of variation of a distribution is 60% and its standard deviation is 18, then its arithmetic mean is

Detailed Solution for JEE Main Practice Test- 1 - Question 74

Given

(CV)A = 60

σA = 18

μA = ?

Now,

*Answer can only contain numeric values
JEE Main Practice Test- 1 - Question 75

Detailed Solution for JEE Main Practice Test- 1 - Question 75

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