JEE Main Practice Test- 10 - JEE MCQ

# JEE Main Practice Test- 10 - JEE MCQ

Test Description

## 75 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Practice Test- 10

JEE Main Practice Test- 10 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Main Practice Test- 10 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Practice Test- 10 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Practice Test- 10 below.
Solutions of JEE Main Practice Test- 10 questions in English are available as part of our Mock Tests for JEE Main and Advanced 2024 for JEE & JEE Main Practice Test- 10 solutions in Hindi for Mock Tests for JEE Main and Advanced 2024 course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt JEE Main Practice Test- 10 | 75 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mock Tests for JEE Main and Advanced 2024 for JEE Exam | Download free PDF with solutions
 1 Crore+ students have signed up on EduRev. Have you?
JEE Main Practice Test- 10 - Question 1

### A solid ball rolls down a parabolic path ABC from a height h as shown in figure. Portion AB of the path is rough while BC is smooth. How high will the ball climb in BC ?

Detailed Solution for JEE Main Practice Test- 10 - Question 1

At B , total kinetic energy = mgh

Here,

m = mass of ball

The ratio of rotational to kinetic energy would be , Kr/Kt = 2/5

where, Kr = 2/7mgh and Kt = 5/7 mgh

In portion BC, friction is absent . Therefore, rotational K.E will remain constant and Translational K.E will convert into potential energy.

Hence, if H be the height to which ball climbs in BC, then

mgH = Kt

mgH = 5/7mgh

H = 5/7h

JEE Main Practice Test- 10 - Question 2

### A heat engine has an efficiency η. Temperatures of source and sink are each decreased by 100 K. Then, the efficiency of the engine.

Detailed Solution for JEE Main Practice Test- 10 - Question 2

where T1 and T2 are the temperatures of a source and sink respectively.
When T1 and T2 both are decreased by 100 K each, (T1 - T2) stays constant. T1 decreases.
∴  η increases.

JEE Main Practice Test- 10 - Question 3

### A particle is moving in a uniform circular motion on a horizontal surface. Particle position and velocity at time t = 0 are shown in the figure in the coordinate system. Which of the indicated variable on the vertical axis is incorrectly matched by the graph shown alongside for particle's motion-

Detailed Solution for JEE Main Practice Test- 10 - Question 3

JEE Main Practice Test- 10 - Question 4

In Coolidge tube experiment, if applied voltage is increased to three times, the short wavelength limit of continuous X - ray spectrum shifts by 20 pm. What is the initial voltage applied to the tube ?

Detailed Solution for JEE Main Practice Test- 10 - Question 4

Cut-off wavelength for continuous x-ray is given as :
∴ hc eV0 = λ  &  hc 3 eV0 = λ - Δλ

⇒   V0 = 2 hc 3 e Δ λ = 41 kV

JEE Main Practice Test- 10 - Question 5

A washer is made of metal having resistivity 10–7 Ωm. The washer has inner radius 1 cm, outer radius 3 cm and thickness 1 mm. A magnetic field, oriented normal to the plane of the washer, has the time dependent magnitude B = (2t) tesla/sec. Find the current (in ampere) around the washer

Detailed Solution for JEE Main Practice Test- 10 - Question 5

Electric field at a general radial distance is E
E = rN/c
J = σE
Current in circular element di = j(dr)t
So net current in washer i =

JEE Main Practice Test- 10 - Question 6

Ends of two wires A and B having resistivity  and  of same cross section area joined in series together to from a single wire. If the resistance of the joined wire does not change with temperature, then find the ratio of their lengths  given that temperature coefficient of resistivity of wire A and B is

αA= 4 ×10–5/ °C and αB = –6 × 10–6/°C. Assume that mechanical dimensions do not change with temperature

Detailed Solution for JEE Main Practice Test- 10 - Question 6

As net resistance does not changes on change temp. so

JEE Main Practice Test- 10 - Question 7

A cone of radius = height = r is under a liquid of density d. Its base is parallel to the free surface of the liquid at a depth H from it as shown in the figure. What is the net force due to liquid on its curved surface? (neglect atmospheric pressure)

Detailed Solution for JEE Main Practice Test- 10 - Question 7

Force due to liquid on curved surface = F

F = weight of liquid above the cone

JEE Main Practice Test- 10 - Question 8

A spherical black body has a radius R and steady surface temperature T, heat sources ensure the heat evolution at a constant rate and distributed uniformly over its volume. What would be the new steady surface temperature of the object if the radius is decreased by half? Assume surrounding to be at absolute zero and heat evolution rate through unit volume remain same.

Detailed Solution for JEE Main Practice Test- 10 - Question 8

Net heat getting generated in complete volume of sphere = rate of heat radiated by its surface

JEE Main Practice Test- 10 - Question 9

The smallest length scale known in physics is the Planck length. It is an important ingredient in some current cosmological theories. Which of the following expressions could represent this Planck length? (Symbols has usual meaning)

Detailed Solution for JEE Main Practice Test- 10 - Question 9

Dimensional formula of only option D matches with length.

JEE Main Practice Test- 10 - Question 10

In the figure shown two motors P1 & P2 fixed on a plank which is hanging with light string passing over fixed Pulley P. If the motors start winding the thread with angular velocity ω1 & ω2 then velocity of plank V is (here R1 & R2 are the radii of motor rotor respectively) [Given: ω1 = 2 rad/s, R1 = 2m, ω2 = 2 rad/s, R2 = 3m]

Detailed Solution for JEE Main Practice Test- 10 - Question 10

JEE Main Practice Test- 10 - Question 11

Five identical balls each of mass m and radius r are strung like beads at random and at rest along a smooth, rigid horizontal thin road of length L, mounted between immovable supports. Assume 10r < L and that the collision between balls or between balls and supports are elastic. If one ball is struck horizontally so as to acquire a speed v, the average force felt by the support is

Detailed Solution for JEE Main Practice Test- 10 - Question 11

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 12

A double star is a system of two stars rotating about their centre of mass only under their mutual gravitational attraction. Let the stars have masses m and 2m and let their separation be l. Their time period of rotation about their centre of mass will be proportional to

Detailed Solution for JEE Main Practice Test- 10 - Question 12

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 13

If white light is used in a Young’s double – slit experiment. Point C represents centre of a screen

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 14

Consider a thermodynamic cycle in a PV diagram shown in the figure performed by one mole of a monoatomic gas. The temperature at A is T0 and volume at A and B are related as VB = VC = 2VA. Choose the correct option(s) form the following

Detailed Solution for JEE Main Practice Test- 10 - Question 14

temprature at state 'B' is maximum

Net work done by gas in cyclic process

Heat capacity for process

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 15

If the plank’s constant would be double the present value, in the Bohr’s model for hydrogen atom

Detailed Solution for JEE Main Practice Test- 10 - Question 15

JEE Main Practice Test- 10 - Question 16

In 1906, Robert Millikan devised an experiment that allowed him to determine the charge of an electron. A schematic of Millikan’s set – up is shown below:
Two metal plates are connected by a series of batteries to form a capacitor. There is an electric field between the plates. The metal plates are inside an insulated cylindrical container.
Oil drops are introduced into the container through a small hole in the top. The oil drops acquire a negative charge as they pass through the nozzle of the oil can. Some of the drops fall through a hole in the upper plate. By adjusting the voltage between the plates, certain drops can be suspended between them. The relationship between the electric field between the plates and the voltage across the plates is ∆V = EL
Where E is the electric field and L is the plate separation.
Millikan chose oil because of its relatively low vapour pressure and high charge holding ability. (To answer the following question assume oil drops as to be non-conducting tiny spheres)

In order for an oil drop of mass m, radius r and volume charge density ρ, to be suspended between the plates, the magnitude and direction of the electric field must be:

Detailed Solution for JEE Main Practice Test- 10 - Question 16

JEE Main Practice Test- 10 - Question 17

In 1906, Robert Millikan devised an experiment that allowed him to determine the charge of an electron. A schematic of Millikan’s set – up is shown below:
Two metal plates are connected by a series of batteries to form a capacitor. There is an electric field between the plates. The metal plates are inside an insulated cylindrical container.
Oil drops are introduced into the container through a small hole in the top. The oil drops acquire a negative charge as they pass through the nozzle of the oil can. Some of the drops fall through a hole in the upper plate. By adjusting the voltage between the plates, certain drops can be suspended between them. The relationship between the electric field between the plates and the voltage across the plates is ∆V = EL
Where E is the electric field and L is the plate separation.
Millikan chose oil because of its relatively low vapour pressure and high charge holding ability. (To answer the following question assume oil drops as to be non-conducting tiny spheres)

Suppose the original oil droplet were replaced with a positively charged one that had twice the charge and three times the mass of the original droplet, how would the magnitude of the electric field have to be changed in order for the drop to remain suspended?

JEE Main Practice Test- 10 - Question 18

In a hypothetical atom, a negatively charged particle having a charge of magnitude 3e and mass 3m revolves around a proton. Here, e is the electronic charge and m is the electronic mass. Mass of proton may be assumed to be much larger than that of the negatively charged particle, thus the proton is at rest. This “atom” obeys Bohr’s postulate of quantization of angular momentum, that is  It is given that for the first Bohr orbit of hydrogen atom: radius of orbit is r0 speed of electron is V0, and total energy is –E0. Now answer the following questions.
Speed of the revolving particle is, in the first Bohr orbit.

Detailed Solution for JEE Main Practice Test- 10 - Question 18

If mass of revolving particle is m, and change q. change at nucleus Q

Energy of nth orbit

Radius of first orbit of this atom

JEE Main Practice Test- 10 - Question 19

In a hypothetical atom, a negatively charged particle having a charge of magnitude 3e and mass 3m revolves around a proton. Here, e is the electronic charge and m is the electronic mass. Mass of proton may be assumed to be much larger than that of the negatively charged particle, thus the proton is at rest. This “atom” obeys Bohr’s postulate of quantization of angular momentum, that is  It is given that for the first Bohr orbit of hydrogen atom: radius of orbit is r0 speed of electron is V0, and total energy is –E0. Now answer the following questions.

Detailed Solution for JEE Main Practice Test- 10 - Question 19

If mass of revolving particle is m, and change q. change at nucleus Q

Energy of nth orbit

Radius of first orbit of this atom

JEE Main Practice Test- 10 - Question 20

In a hypothetical atom, a negatively charged particle having a charge of magnitude 3e and mass 3m revolves around a proton. Here, e is the electronic charge and m is the electronic mass. Mass of proton may be assumed to be much larger than that of the negatively charged particle, thus the proton is at rest. This “atom” obeys Bohr’s postulate of quantization of angular momentum, that is  It is given that for the first Bohr orbit of hydrogen atom: radius of orbit is r0 speed of electron is V0, and total energy is –E0. Now answer the following questions.
The momentum of an emitted photon when it makes a transition from the second excited state to ground state, is

Detailed Solution for JEE Main Practice Test- 10 - Question 20

If mass of revolving particle is m, and change q. change at nucleus Q

Energy of nth orbit

Radius of first orbit of this atom

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 21

The given figure shows a plot of the time dependent force x F acting on a particle in motion along the x-axis. What is the total impulse (in kg-m/s) delivered by this force to the particle from time t = 0 to t = 2 second?

Detailed Solution for JEE Main Practice Test- 10 - Question 21

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 22

In the figure shown a small block B of mass m is released from the top of a smooth movable wedge A of the same mass m. The height of wedge A shown in figure is h = 16 cm. B ascends another movable smooth wedge C of the same mass. Neglecting friction anywhere find the maximum height (in cm) attained by block B on wedge C.

Detailed Solution for JEE Main Practice Test- 10 - Question 22

Let u and v be the speed of wedge A and block B at just after the Block B get off the wedge A. Applying conservation of momentum
in horizontal direction, we get
mu = mv   ...(1)
Applying conservation of energy between initial and final state as shown in figure (1), we get

At the instant block B reaches amximum height h' on the wedge C (figure 2) the speed of block B and wedge C are v'
Applying conservation of momentum in horizontal direction, we get
mv=(m+m)v'  ...(4)

Applying conservation of energy between initial and final state

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 23

Mass 2m is kept on a smooth circular track (R = 9 meters) of mass m which is kept on a smooth horizontal surface. The circular track is given a horizontal velocity  towards left and released. Find the maximum height reached by 2m in meters.

Detailed Solution for JEE Main Practice Test- 10 - Question 23

Let v be the final speed of block when it is at maximum height. At that instant the speed of circular track shall also be v

From conservation of momentum

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 24

Two blocks of masses m1 and m2 are connected by spring of constant K such that m2/m1=9. The spring is initially compressed and the system is released from rest at t = 0 second. The work done by spring on the blocks m1 and m2 be ω1 and ω2 respectively by time t. The speeds of both the blocks at time ‘t’ are non-zero. Then find the value of ω12.

Detailed Solution for JEE Main Practice Test- 10 - Question 24

Here in this question when we released the system this will start moving

So ω1 =  work done by spring on the block 1 = change in kinetic energy of block 1

ω2 = work done by spring on the block 2 = change in kinetic energy of block 2

So only spring will do the work

So

and

so

And we know there is no external force on this system so momentum will be conserved

So           m1v1 = m2v2

so     v1/v2 = m1/m2

So  now put this in the ratio

We get  ω12  = m2/m1 = 9

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 25

Two simple pendulums of length 5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed .... oscillations.

Detailed Solution for JEE Main Practice Test- 10 - Question 25

If t is the time taken by pendulums to come in same phase again first time after t=0.

and NS= Number of oscillations made by shorter length pendulum with time period TS.

NL= Number of oscillations made by longer length pendulum with time period TL

Then t=NSTS=NLTL

⇒ NS=2NL i.e. if NL=1
⇒ NS=2

JEE Main Practice Test- 10 - Question 26

Copper reduces  into NO and NO2 depending upon concentration of HNO3 in solution. Assuming  [Cu2+] = 0.1M, and PNO = PNO2 = 10–3 bar. At which concentration of HNO3, thermodynamic tendency for reduction of  into NO and NO2 by copper is same?

Detailed Solution for JEE Main Practice Test- 10 - Question 26

JEE Main Practice Test- 10 - Question 27

A radioactive material (t1/2 = 30 days) gets spilled over the floor of a room. If initial activity is ten times the permissible value, after how many days will it be safe to enter the room

Detailed Solution for JEE Main Practice Test- 10 - Question 27

It will be safe to enter the room after the nuclei decreases to 1/10 of its initial amount.

→ Half life = 30 days

→ Using integration law,

∴ After 100 days, it will be safe to enter the room.

JEE Main Practice Test- 10 - Question 28

Detailed Solution for JEE Main Practice Test- 10 - Question 28

Trinitro benzene diagenium ionis strong electrophile type and show coupling even with mesitylene

JEE Main Practice Test- 10 - Question 29

A mixture of all possible stereoisomers from the above structure is subjected to fractional distillation, which of the following statements is correct

JEE Main Practice Test- 10 - Question 30

Which of the following about SF4, SOF4 and COF2 molecules is correct?

Detailed Solution for JEE Main Practice Test- 10 - Question 30

Equatorial FSF bond angle is less in SFthan in SOF4 since lone pair repulsion is more than two electron pairs in double bond
In both SF4 and SOF4 the hybridisation state of S is same sp3d
The OCF bond angle on COF2 is more than 120° since two electron pairs in double bond repel more than one electron pair in C-F bonds.
Due to repulsion by lone pair the axial FSF bond angle is less than 180°

JEE Main Practice Test- 10 - Question 31

B2O3 substitutes nonmetal oxides from several metal salts because

Detailed Solution for JEE Main Practice Test- 10 - Question 31

Less volatile with more melting point (B2O3) acidic oxide can substitute more volatile acidic oxides.

JEE Main Practice Test- 10 - Question 32

Which of these is most stable?

Detailed Solution for JEE Main Practice Test- 10 - Question 32

NCl3, NBr3 and NI3 are explosive.

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 33

The correct statement regarding various types of molecular speeds are

Detailed Solution for JEE Main Practice Test- 10 - Question 33

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 34

The compounds that should be used to prepare glycine and β – alanine by Gabriel phthalimide synthesis are

Detailed Solution for JEE Main Practice Test- 10 - Question 34

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 35

In which of these compounds, Nitrogen can be estimated by Duma’s method?

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 36

A black mineral (A) in solid state is fused with KOH and KNO3 and the mixture extracted with water to get a green coloured solution (B). On passing CO2 gas through the solution the colour changes to pink with a black residue (C). Which of the following is/are correct

Detailed Solution for JEE Main Practice Test- 10 - Question 36

JEE Main Practice Test- 10 - Question 37

A is

Detailed Solution for JEE Main Practice Test- 10 - Question 37

JEE Main Practice Test- 10 - Question 38

Formation of “B” through the attack of first two reagents involve respectively

Detailed Solution for JEE Main Practice Test- 10 - Question 38

JEE Main Practice Test- 10 - Question 39

Product “C” is

Detailed Solution for JEE Main Practice Test- 10 - Question 39

JEE Main Practice Test- 10 - Question 40

A white substance (A) reacts with dilute H2SO4 to produce a colourless gas (B) and a colourless solution (C). The reaction between (B) and acidified K2Cr2O7 solution produces a green solution and a slightly coloured precipitate (D). The substance (D) burns in air to produce a gas (E), which reacts with (B) to yield (D) and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colourless liquid. Addition of aqueous NH3 or NaOH to (C) produces first a precipitate which dissolves in the excess of the respective reagent to produce a clear solution in each case
(B) and (D) are respectively

Detailed Solution for JEE Main Practice Test- 10 - Question 40

JEE Main Practice Test- 10 - Question 41

A white substance (A) reacts with dilute H2SO4 to produce a colourless gas (B) and a colourless solution (C). The reaction between (B) and acidified K2Cr2O7 solution produces a green solution and a slightly coloured precipitate (D). The substance (D) burns in air to produce a gas (E), which reacts with (B) to yield (D) and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colourless liquid. Addition of aqueous NH3 or NaOH to (C) produces first a precipitate which dissolves in the excess of the respective reagent to produce a clear solution in each case
The precipitate obtained by addition of aqueous NH3 or NaOH to (C) initially is _____ which dissolves in excess reagent to produce ________

Detailed Solution for JEE Main Practice Test- 10 - Question 41

JEE Main Practice Test- 10 - Question 42

Which of the following is correct regarding solutions of sodium metal in liquid ammonia.

Detailed Solution for JEE Main Practice Test- 10 - Question 42

All the statements are true regarding solutions of alkali metals in liquid NH3.

JEE Main Practice Test- 10 - Question 43

For the given reaction the correct statement is

Choose the correct option

Detailed Solution for JEE Main Practice Test- 10 - Question 43

Free radical obtained from x is stabilised due to resonance with one phenyl group and free radical from y is stabilised by resonance with two phelnyl rings. Also free radical from z is stabilised by resonance with one phenyl group and hyperconjugation with CH,Igroup But free radical from w is not stabilised by any effect. So ease of abstraction or order of reactivity of different H-atoms is y > z > x > w. Abstraction from x, y, z and w give 1, 2 (enantiomers), 2 (enantiomers) and 1 product respectively so total no. of products are 6. But both enantiomers appear in one fraction so total no. of fractions is 4.

JEE Main Practice Test- 10 - Question 44

For the cell (at 1 bar H2 pressure) Pt/H2(g) H X (m1), NaX(m2), NaCl(m3)/AgCl/Ag/Pt it is found that the value of E

approaches 0.2490 in the limit of zero concentration. Calculate  for the acid HX at 27°C. (R = 8.3 Jmole–1K–1, F = 96500C)

Detailed Solution for JEE Main Practice Test- 10 - Question 44

JEE Main Practice Test- 10 - Question 45

During the titration of 100 ml of a weak monobasic acid solution using 0.1 M NaOH, the solution became neutral at 40 mL addition of NaOH and equivalence point was obtained at 50 mL NaOH addition. The Ka of the acid is (log 2 = 0.3)

Detailed Solution for JEE Main Practice Test- 10 - Question 45

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 46

The sum of no of cyclic transition states and intermediates in the above reaction during the formation of product is/are....

Detailed Solution for JEE Main Practice Test- 10 - Question 46

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 47

How many carbon atoms (In A, B and (c) changed their hybridization till the formation of D? (Consider each reaction and do not consider stereoisomerism)

Detailed Solution for JEE Main Practice Test- 10 - Question 47

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 48

No of stereoisomers possible for X is

Detailed Solution for JEE Main Practice Test- 10 - Question 48

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 49

Index of hydrogen of the major product is….

Detailed Solution for JEE Main Practice Test- 10 - Question 49

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 50

No of 1° carbons in product is …..

Detailed Solution for JEE Main Practice Test- 10 - Question 50

JEE Main Practice Test- 10 - Question 51

Let a, b and c be positive constants. The value of ‘a’ in terms of ‘c’ if the value of integral   is independent of ‘b’ equals

Detailed Solution for JEE Main Practice Test- 10 - Question 51

JEE Main Practice Test- 10 - Question 52

Let  are three vectors along the adjacent edges of a tetrahedron,  if  and  then volume of tetrahedron is

Detailed Solution for JEE Main Practice Test- 10 - Question 52

JEE Main Practice Test- 10 - Question 53

Let ω be a complex cube root of unity with ω ≠ 1. A fair die is thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that

Detailed Solution for JEE Main Practice Test- 10 - Question 53

A dice is thrown thrice n(s)=6 x 6 x 6

Favorable events, ωr1 + ωr2 + ωr3

(r1​,r2​,r3​) are ordered 3 triples which can take values.

(1,2,3)(1,5,3)(4,2,3)(4,5,3)

(1,2,6)(1,5,6)(4,2,6)(4,5,6)

i.e, 8 ordered pairs and each can be arranged in 3! ways.

3!=6

JEE Main Practice Test- 10 - Question 54

A ray of light travels along a line y = 4 and strikes the surface of a curves y2 = 4 (x + y), then equations of the line along which reflected ray travel is

Detailed Solution for JEE Main Practice Test- 10 - Question 54

Focus is (0,2)
Point os interflection property of parabola, reflected ray passes through the focus.
X = 0 is required line.

JEE Main Practice Test- 10 - Question 55

A triangle ABC having vertices A (5, 1), B (–1, –7) and C (1, 4) respectively. If L1 be the line mirror passing through C and parallel to AB, a light ray emanating from point A and goes along the direction of internal bisector of the angle A, which meets the mirror at E and BC at D. Then sum of the area of ΔACE and ΔABC is

Detailed Solution for JEE Main Practice Test- 10 - Question 55

JEE Main Practice Test- 10 - Question 56

The number of different ways in which the persons A, B, C having 6 one rupee coins, 7 one rupee coins, 8 one rupee coins respectively donate 10 one rupee coins collectively is equal

Detailed Solution for JEE Main Practice Test- 10 - Question 56

JEE Main Practice Test- 10 - Question 57

Let a = cos–1 (cos 20), b = cos–1 (cos 30) and c = sin–1 sin (a + b) then maximum value of sin (2 (a+b+c) x) + cos2 ((a+b+c) x) is

Detailed Solution for JEE Main Practice Test- 10 - Question 57

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 58

If ‘A1’ is the area bounded by |x – aiai| + |y| = bibi, i ∈ N, where  and  then

Detailed Solution for JEE Main Practice Test- 10 - Question 58

JEE Main Practice Test- 10 - Question 59

A rectangle ABCD of dimensions r and 2r is folded along diagonal BD such that planes ABD and CBD are perpendicular to each other. Let the position of the vertex A remains unchanged and C1 is the new position of C.
The distance of C1 from A is equal to

Detailed Solution for JEE Main Practice Test- 10 - Question 59

Let the rectangle ABCD initially lies in xy plane with B King at origin BC' along x - axis and BA
Along y-axis.
Equation of BD in xy-plane is y = 2x
So. the coordinates of foot N of C on BD are
Clearly. CN = C1N
Hence, the coordinates of various points in 3-D are
A(0, r, 0), C(r, 0 , 0) , D (r, 2r, 0),

JEE Main Practice Test- 10 - Question 60

A rectangle ABCD of dimensions r and 2r is folded along diagonal BD such that planes ABD and CBD are perpendicular to each other. Let the position of the vertex A remains unchanged and C1 is the new position of C.
If ∠∠ABC1 = θ, then cos θ is equal to

Detailed Solution for JEE Main Practice Test- 10 - Question 60

Let the rectangle ABCD initially lies in xy plane with B King at origin BC' along x - axis and BA
Along y-axis.
Equation of BD in xy-plane is y = 2x
So. the coordinates of foot N of C on BD are
Clearly. CN = C1N
Hence, the coordinates of various points in 3-D are
A(0, r, 0), C(r, 0 , 0) , D (r, 2r, 0),

JEE Main Practice Test- 10 - Question 61

A square ABCD of diagonal 2a is folded along the diagonal AC so that the planes DAC and BAC are at right angle. The shortest distance between DC and AB is

Detailed Solution for JEE Main Practice Test- 10 - Question 61

When folded coordinates will be D(0,0,a);C(a,0,0);A(−a,0,0);B(0,−a,0)

Equation of DC is,

Equation of ABAB is,

∴ Shortest distance = 2a/√3

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 62

Family of curves which makes an angle of π/4 with the family of hyperbola xy = a, is (a > 0, and a is a parameter)

Detailed Solution for JEE Main Practice Test- 10 - Question 62

xy=a

Take log on both sides

log(xy)=loga

⇒logx+logy=loga

Differentiate with respect to x

which is the differential equation governing the given family of rectangular hyperbola Replacing y' by

In the Ordinary differential equation(ODE), we obtain the ODE  govern in the required isogonal trajectories as

which simplifies as

Differentiate with respect to x

As derivative of denominator is present in numerator in L.H.S., hencewe can directly integrate to get

ln(v2+2v−1) =−2lnx + lnA'

⇒ ln(v2+2v−1) + 2lnx = lnA'

⇒ ln(v2+2v−1) + lnx2 = lnA'

⇒ ln[x2(v2+2v−1)] = lnA'

⇒ x2(v2+2v−1)=A'

Now y=vx

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 63

Let position vector of the orthocenter of ΔABC be  Then, which of the following statement(s) is/are correct (Given position vector of points A, B, C are

Detailed Solution for JEE Main Practice Test- 10 - Question 63

JEE Main Practice Test- 10 - Question 64

Let f (x) = x3 + 2x2 – x + 1, then which of the following statement(s) is/are correct

Detailed Solution for JEE Main Practice Test- 10 - Question 64

f (–2) = 3 and f (–3) is negative
So, equation has one real root in between (–3, –2)

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 65

where b ∈ R. The value of ‘I’ can be

Detailed Solution for JEE Main Practice Test- 10 - Question 65

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 66

The equations of AB, BC, AC, the three sides of ΔABC are –x + y –1 = 0, x + y –1 = 0 and x = –4. If (α, 0) and (0, β) lie inside the triangle where α, β ∈ Z, then

Detailed Solution for JEE Main Practice Test- 10 - Question 66

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 67

a, b ∈ I satisfies equation a(b – l) = 3 + b – b2, then a + b is equal to

Detailed Solution for JEE Main Practice Test- 10 - Question 67

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 68

Let  be a sequence of sets defined by an = (n2 - 1)/n. Then,

Detailed Solution for JEE Main Practice Test- 10 - Question 68

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 69

A right angle triangle ABC, right angle at A is inscribed in hyperbola xy = c2 (c > 0) such that slope of BC is 2. If distance of point A from centre of xy = c2 is √10, then which of the following is/are correct for xy = c2

Detailed Solution for JEE Main Practice Test- 10 - Question 69

Let the coordinates of point A are (ct. c/t)
So. the slope of normal at A will be t2.
And normal will be parallel to BC

JEE Main Practice Test- 10 - Question 70

Let P(a, b) be a variable point satisfying  Let R be the complete region represented in x-y plane in which P can lie.
Area of region R is

Detailed Solution for JEE Main Practice Test- 10 - Question 70

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 71

Let z1, z2, z3 be complex numbers such that |z1| = |z2| = |z3| = 3&(z1 ≠ z3). Then find the value of

Detailed Solution for JEE Main Practice Test- 10 - Question 71

[min{|az2,+ (1 - a)z- z1|} = height, from z1 to the line, joining z2 , z3

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 72

Let z represents a variable point in complex plane such that z − z1 is real, where z1 is a fixed point in the same plane? Let “m” be the number of “z” values such that |z| = λ, where λ >|Im(z1)| & let “n” be the number of “z” values such that |z| = λ, where λ = |Im(z1)| then value of m + n =__________

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 73

The least possible degree of a polynomial equation, with real coefficients having 2ω2, 3 + 4ω, 3 + 4ω2, 5 − ω − ω2 as roots is ________
(ω, ω2 are non-real cube roots of 1).

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 74

A curve is defined as  Two spiders, one male and other female were moving together along the curve. The female spider suddenly realizes that the male spider is a rogue spider and immediately tries to get away as far as possible from it. Hence it moved onto the another point on the curve. The maximum distance between two final points when both spiders try out all possibilities, is k. Then the value of  __________[.] Greatest integer function

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 75

If α,β,γ are the roots of x3 − 3x2 + 3x + 7 = 0 and w is a non-real cube root of 1, and the value of  then number of ordered pairs (p, q) such that p + q = 15 is _______

## Mock Tests for JEE Main and Advanced 2024

357 docs|148 tests
Information about JEE Main Practice Test- 10 Page
In this test you can find the Exam questions for JEE Main Practice Test- 10 solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Main Practice Test- 10, EduRev gives you an ample number of Online tests for practice

### Up next

 Test | 54 ques
 Test | 54 ques
 Test | 54 ques
 Test | 54 ques
 Test | 54 ques

## Mock Tests for JEE Main and Advanced 2024

357 docs|148 tests

### Up next

 Test | 54 ques
 Test | 54 ques
 Test | 54 ques
 Test | 54 ques
 Test | 54 ques