JEE Main Practice Test- 9 - JEE MCQ

# JEE Main Practice Test- 9 - JEE MCQ

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## 75 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Main Practice Test- 9

JEE Main Practice Test- 9 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2025 preparation. The JEE Main Practice Test- 9 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Practice Test- 9 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Practice Test- 9 below.
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JEE Main Practice Test- 9 - Question 1

### A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration 'a' vertically.The tension in the string is equal to :

Detailed Solution for JEE Main Practice Test- 9 - Question 1

(Force diagram in the frame of the car)
Applying Newton's law perpendicular to strin
mg sin θ = ma cos θ
tan  θ = a/g
Applying Newton's law along string

JEE Main Practice Test- 9 - Question 2

### A uniform sphere has a mass M and radius R. Find the pressure p inside the sphere, caused by gravitational compression, as a function of the distance r from its centre. (γ is universal gravitational constant)

Detailed Solution for JEE Main Practice Test- 9 - Question 2

We consider a spherical concentric shell of radius x and thickness dx.
The mass of considered element is dm = (4π x2dx)ρ.
Gravitational field at a point in the shell is

∴

The pressure in the element is

∴
But
By putting the value of ρ, we get,

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JEE Main Practice Test- 9 - Question 3

### A 220 volt input is supplied to a transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is

Detailed Solution for JEE Main Practice Test- 9 - Question 3

Here,
Input voltage, Vp = 220 V
Output voltage, Vs = 440 V
Input current, Ip = Ip
Output current, Is = 2A
Efficiency of the transformer,  η = 80%

Efficiency of the transformer,

JEE Main Practice Test- 9 - Question 4

A stone is dropped from a height of 45 m on a horizontal level ground. There is horizontal wind blowing due to which horizontal acceleration of the stone becomes 10 m/s2.(Take g = 10 m/s2).
The time taken (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- 9 - Question 4

Taking motion in vertical direction,

u = 0, g = 10 m/s, h = 45 m

⇒ h = 0 + (1/2) gt 2
⇒ t =  √(2h/g) = √[(2× 45)/10]
⇒ t = 3 sec.
(b) Taking motion in horizontal direction,

JEE Main Practice Test- 9 - Question 5

The combination of 'NAND' gates shown here (in figure), are equivalent to

Detailed Solution for JEE Main Practice Test- 9 - Question 5

(if a NAND is given only 1 input it s behaves as NOT gate, as it is  (one input is divided in two and then passed through NAND)

i.e. OR gate.
For second circuit C = Not of (A NAND B)
i.e. AND only

JEE Main Practice Test- 9 - Question 6

In a Wheatstone's bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is

Detailed Solution for JEE Main Practice Test- 9 - Question 6

In balanced Wheatstone bridge, the galvanometer arm can be neglected so, equivalent resistance = R.

JEE Main Practice Test- 9 - Question 7

A neutron moving with speed v makes a head-on collision with a hydrogen atom in ground state kept at rest. Find the minimum kinetic energy of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron ≈ mass of hydrogen = 1.67 x 10-27 kg

Detailed Solution for JEE Main Practice Test- 9 - Question 7

Suppose the neutron and the hydrogen atom move at speeds v1 and v2 after the collision. The collision will be inelastic if a part of the kinetic energy is used to excite the atom.  Suppose an energy ΔE is used in this way. Using conservation of linear momentum and energy,
mv = mv1 + mv2 ........(i)
and
From (i),
From (ii),
Thus,
Hence, (v1 - v2)2 = (v1 + v2)2 - 4v1v2 =
As v1 - v2 must be real,

or,
An electron in the ground state has an energy of -13.6 eV. The second energy level is -3.4 eV. Thus it would take E2 − E1 = -3.4 eV − -13.6 eV = 10.2 eV to excite the electron from the ground state to the first excited state.
The minimum energy that can be absorbed by the hydrogen atom in ground state to go in an excited state is 10.2 eV. Thus, the minimum kinetic energy of the neutron needed for an inelastic collision is

JEE Main Practice Test- 9 - Question 8

A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be

Detailed Solution for JEE Main Practice Test- 9 - Question 8

JEE Main Practice Test- 9 - Question 9

During adiabatic change, specific heat is

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

In adiabatic process, ΔH = 0

Hence, zero is correct

JEE Main Practice Test- 9 - Question 10

Compute the bulk modulus of water if its volume changes from 100 litres to 99.5 litre under a pressure of 100 atmosphere.

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

By definition of bulk modulus,

Now as isothermal elasticity of a gas is equal to its pressure,

i e , bulk modulus of water is very large as compared to air This means that air is about 20,000 times more compressive than water, i.e , the average distance between air molecules is much larger than between water molecules

JEE Main Practice Test- 9 - Question 11

When a train is at a distance of 2km, its engine sounds a whistle. A man near the railway track hears the whistle directly and by placing his ear against the track of the train. If the two sounds are heard at an interval of 5.2 s, find the speed of the sound in iron (material of the rail track). given that velocity of sound in air is 330 ms-1.

Detailed Solution for JEE Main Practice Test- 9 - Question 11

Here, S = 2 km = 2,000 m ; vair = 330 m s-1
Therefore, time taken by sound to travel through air,

As the two sounds (through air and iron rails) are heard at an interval of 5.2 s, time taken by sound to travel through iron rails,
tiron = tair - 5.2
= 6.06 - 5.2 = 0.86 s
Therefore, velocity of sound in iron,

= 2,325.6 m s-1

JEE Main Practice Test- 9 - Question 12

Dispersive powers of materials used in lenses of an achromatic doublet are in the ratio 5:3. If the focal length of concave lens is 15 cm, then the focal length of the other lens will be

Detailed Solution for JEE Main Practice Test- 9 - Question 12

JEE Main Practice Test- 9 - Question 13

As per the diagram a point charge +q is placed at the origin O. Work done in taking another point charge -Q from the point A [coordinates (0, a)] to another point B [coordinates (a, 0)] along the straight path AB is

Detailed Solution for JEE Main Practice Test- 9 - Question 13

Work done is equal to zero because the potential of A and B are the same
= 1 /4πε0  q/a

No work is done if a particle does not change its potential energy.
i.e., initial potential energy = final potential energy.

JEE Main Practice Test- 9 - Question 14

A man crosses the river perpendicular to river flow in time t seconds and travels an equal distance down the stream in T-seconds. The ratio of man's speed in still water to the speed of river water will be :

Detailed Solution for JEE Main Practice Test- 9 - Question 14

Let velocity of man in still water be v and that of water with repect to ground be u. Velocity of man perpendicular to river flow with respect to ground
Velocity of man downstream = v + u

⇒   ⇒
∴

JEE Main Practice Test- 9 - Question 15

The mean lives of a radioactive substance are 1620 years and 405 years for α - emission and β - emission respectively. Find the time during which three-fourth of a sample will decay if it is decaying both by α – emission and β – emission simultaneously.

Detailed Solution for JEE Main Practice Test- 9 - Question 15

The decay constant λ is the reciprocal of the mean life τ
Thus,
and
∴     Total decay constant, λ = λα + λβ
or
We know that
N = N0e–λt
When  part of the sample has disintegrated, N = N0/4
∴
or  eλt = 4
Taking logarithm of both sides, we get

or
= 2×324 × 0.693 = 449 year

JEE Main Practice Test- 9 - Question 16

An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Take π = 3.14

Detailed Solution for JEE Main Practice Test- 9 - Question 16

Here. T = 0.314 s ; m = 1 kg

or
When spring is loaded with a weight 1 kg:
m g = k I
or
or

JEE Main Practice Test- 9 - Question 17

A point object is moving with a speed v in front of an arrangement of two mirrors as shown in figure. If the velocity of image in mirror M1 with respect to image in mirror M2 is n V sinθ, n =

Detailed Solution for JEE Main Practice Test- 9 - Question 17

Angle between Their magnitudes is v.

JEE Main Practice Test- 9 - Question 18

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 temperature difference between the outer and inner surfaces of the shell in steady state is T.

Detailed Solution for JEE Main Practice Test- 9 - Question 18

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

So Equation (i) becomes,

i.e., thickness of shell (b - a)

JEE Main Practice Test- 9 - Question 19

A body falling freely from a given height ‘H’ hits an inclined plane in its path at a height ‘h’. As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of (h/H) the body will take maximum time to reach the ground?

Detailed Solution for JEE Main Practice Test- 9 - Question 19

For finding the maximum time using the concept of differentiation

JEE Main Practice Test- 9 - Question 20

Two radio antennas radiating waves in phase are located at point A and B, 200 m part (Figure). The radio waves have a frequency of 6 MHz. A radio receiver is moved out from point B along a line perpendicular to the line connecting A and B (line BC shown in figure). At what distances from B will there be destructive interference?

Detailed Solution for JEE Main Practice Test- 9 - Question 20

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JEE Main Practice Test- 9 - Question 21

There is a long solid conductor of resistivity ρ=1x10−6Ωm, surrounded by air. There is a steady electric current along the length of the conductor. At point ‘P’ just outside the cylinder the electric field strength E=10−4V/m is directed at an angle of α to the normal to the surface. Find the current density in the conductor in the vicinity of point ‘P’ (see figure)

Detailed Solution for JEE Main Practice Test- 9 - Question 21

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JEE Main Practice Test- 9 - Question 22

A parallel beam of light falls normally on the first face of a prism of small angle A. At the second face it is partly transmitted and partly reflected. Then the reflected beam strike the first face again and emerges from first surface in a direction making an angle 6030′ with the normal at the first surface. The refracted beam is found to have undergone a deviation of 1
Calculate the angle of prism in degree.

Detailed Solution for JEE Main Practice Test- 9 - Question 22

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JEE Main Practice Test- 9 - Question 23

The speed of sound in a mixture of n1=2 moles of He, n2=2 moles of H2 at temperature  is  Find  (Take )

Detailed Solution for JEE Main Practice Test- 9 - Question 23

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JEE Main Practice Test- 9 - Question 24

Six identical parallel metallic large plates are located in air at equal distances d to neighbouring plates. The area of each plate is A. Some of the plates are connected by conducting wires to each other. The capacitance of the system of plates between two points P and Q in pF is:
(Take A = 0.05 m2d = 17.7 mm, ε0=8.85x10−12F/mε0=8.85x10−12F/m).

Detailed Solution for JEE Main Practice Test- 9 - Question 24

The equivalent capacitance between points PQ is capacitance between two neighboring plates by wheat stone bridge.

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JEE Main Practice Test- 9 - Question 25

A long wire PQR is made by joining two wires PQ and QR of equal radii. PQ has a length 4.8 m and mass 0.06 kg. QR has length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80 N. A sinusoidal wave pulse of amplitude 3.5 cm is sent along the wire PQ from the end P. No power is dissipated during the propagation of wave pulse. Find amplitude (in mm) of reflected pulse from junction Q.

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JEE Main Practice Test- 9 - Question 26

If helium and methane are allowed to diffuse out of the container under the similar conditions of temperature and pressure, then the ratio of rate of diffusion of helium to methane is .

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JEE Main Practice Test- 9 - Question 27

For a monatomic gas kinetic energy = E, the relation with rms velocity is

Detailed Solution for JEE Main Practice Test- 9 - Question 27

For a monatomic gas with only Translational degrees of freedom
E⁡a = 3/2 kbT⁡ (Avg K E per molecule)

JEE Main Practice Test- 9 - Question 28

In this balanced equartion x, y, z are

Detailed Solution for JEE Main Practice Test- 9 - Question 28

eqn (2) multiply with 3 and adding with eqn (1) we get

So x = 1
y = 3
z = 6

JEE Main Practice Test- 9 - Question 29

Propyne and propene can be distinguished by

Detailed Solution for JEE Main Practice Test- 9 - Question 29

The terminal hydrogen is acidic in
H3C - C ≡ CH
(propyne) and it reacts with ammoniacal AgNO3. In propene, CH3CH = CH2 there is no acidic hydrogen.

JEE Main Practice Test- 9 - Question 30

A piston is cleverly designed so that it extracts the maximum amount of work out of a chemical reaction, by matching Pexternal to the Pinternal at all times. This 8cm diameter piston initially holds back 1 mol of gas occupying 1 L, and comes to rest after being pushed out a further 2 L at 25oC .After exactly half of the work has been done, the piston has travelled out a total of

Detailed Solution for JEE Main Practice Test- 9 - Question 30

Piston change in volume

V = of cylinder = πr2h

JEE Main Practice Test- 9 - Question 31

A monatomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and equal to 1. What is the molar heat capacity of the gas?

Detailed Solution for JEE Main Practice Test- 9 - Question 31

C is molar heat capacity for process (given)
Integrate,

JEE Main Practice Test- 9 - Question 32

Which of the following carboxylic acids undergoes decarboxylation easily ?

Detailed Solution for JEE Main Practice Test- 9 - Question 32

JEE Main Practice Test- 9 - Question 33

The most stable resonating structure of following compound is

Detailed Solution for JEE Main Practice Test- 9 - Question 33

Complete octet & Extended conjugation

JEE Main Practice Test- 9 - Question 34

Heating mixture of Cu2O and Cu2S will give

Detailed Solution for JEE Main Practice Test- 9 - Question 34

Cu2S + 2Cu2O ⟶ 6Cu + SO2
This is an example of auto - reduction.

JEE Main Practice Test- 9 - Question 35

A solution which is 10–3 M each in Mn2 +, Fe2 +, Zn2 + and Hg2 + is treated with 10-16 M sulphide ion. If Ksp of MnS, FeS, ZnS and HgS are 10-13, 10–18, 10–24 and 10–53 respectively, which one will precipitate first ?

Detailed Solution for JEE Main Practice Test- 9 - Question 35

[S2-] need for precipitation of

Thus minimum [S2-] for precipitation is for HgS it will be precipitated first

JEE Main Practice Test- 9 - Question 36

In which of the following arrangements, the sequence is not strictly according to the property written against it?

Detailed Solution for JEE Main Practice Test- 9 - Question 36

Drago's Rule :-
In case of NH3, PH3, - - -  bond angle decrease ,NH3  (≈107) & PH3 (≈90) So, %S charactor of lp in NH3 is less & %s charactor of lp top to bottom increase so basic charactor decrease.

JEE Main Practice Test- 9 - Question 37

A crystal made up of particles X, Y, and Z. X forms fcc packing. Y occupies all octahedral voids of X and Z occupies all tetrahedral voids of X. It all particles along one body diagonal are removed, then the formula of the crystal would be

Detailed Solution for JEE Main Practice Test- 9 - Question 37

For fcc, number of X atoms = 4/unit cell
Number of Tetrahedral Voids = Z = 8
Number of Octahedral Voids = Y = 4
Number of atoms removed along one body diagonal = 2X (corner) and 2Z (TVs) and 1 Y (OV at body centre)
∴    Number of X atoms left = 4 - [2 × (1/8) ]= (15/4)
Number of Y atom left = 4 - (1 × 1) = 3
Number of Z atom left = 8 - (2 × 1) = 6
The simplest formula
⇒ X5 Y4Z8

JEE Main Practice Test- 9 - Question 38

In  the formal charge on each oxygen atom and the P – O bond order are respectively

Detailed Solution for JEE Main Practice Test- 9 - Question 38

There are Four Resonating Structures B.O = 1.25

= -0.75
π Bond of P = 0 is delocalised due to resonance on four O-atom
bond order

JEE Main Practice Test- 9 - Question 39

Based on following information determine the value of x and y

atomic masses of Al,Cl and Ag are respectively 27,35.5,108.

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JEE Main Practice Test- 9 - Question 40

Aqueous ammonia is used as a precipitating reagent for Al3+ ions as Al(OH)3 rather than aqueous NaOH, because

Detailed Solution for JEE Main Practice Test- 9 - Question 40

Al(OH)3 formed with NaOH dissolves in excess of NaOH to form aluminate ion.
which is stable.

JEE Main Practice Test- 9 - Question 41

Bakelite is made from phenol and formaldehyde. The initial reaction between them is the example of :

Detailed Solution for JEE Main Practice Test- 9 - Question 41

The reaction of phenol and formaldehyde, is called Lederer Mannase reaction. It follows electrophilic substitution and rearrangement reaction.

JEE Main Practice Test- 9 - Question 42

For an exothermic reaction, following two steps are involved.
Step 1. A + B → I   (slow)
Step 2. I  → AB   (fast)

Q. Which of the following graphs correctly represent this reaction ?

Detailed Solution for JEE Main Practice Test- 9 - Question 42

In any reaction mechanism the no. of activated complexes showing maxima of potential energies is the no. of steps in that mechanism and in between these maxima is a valley where lies a more stable intermediate. There is one reaction intermediate lying at valley of two maxima representing two steps. Ea for first step should be higher than second step as first is slow step. These requirements fulfilled by choice (ii) & not choice (i) the choice (iv) is not possible because that is for endothermic reaction (Δ H + ve) & (ii) is single step equation so not possible.

JEE Main Practice Test- 9 - Question 43

An electron moving with velocity 'v' is found to have a certain value of de-Broglie wavelength. The velocity to be possessed by the neutron to have the same de-Broglie wavelength is

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JEE Main Practice Test- 9 - Question 44

Chlorobenzene on heating with NaOH at 300oC and high pressure followed by reaction with dil. HCl gives

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JEE Main Practice Test- 9 - Question 45

A sample of an ideal gas is expanded from original volume of 1m3 to twice its volume in a reversible process for which p = αV2  (α = 5 atm m- 6)
Calculate (i)  Work done on P - V diagram
(ii)  Calculate ΔSm ifC Vm = 20J K -1 m -1 , R = 8JK–1 m–1 and ln2 = 0.7
The values of work done & ΔSm may be

Detailed Solution for JEE Main Practice Test- 9 - Question 45

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JEE Main Practice Test- 9 - Question 46

E has how many halogen.

Detailed Solution for JEE Main Practice Test- 9 - Question 46

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JEE Main Practice Test- 9 - Question 47

Calculate EMF of the cell in millivolt.

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JEE Main Practice Test- 9 - Question 48

8.21dm3 of NH3 at 500K and 1 atm expands adiabatically upto 0.5 atm against a constant external pressure of 0.5 atm. Calculate magnitude of change in internal energy (in calories). Assume that NH3 behaves as an ideal gas and vibration degrees of freedom are inactive.
(R=2 calmol−1K−1=0.0821atm−litK−1mol−1).

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JEE Main Practice Test- 9 - Question 49

The total number of sigma and pi bonds in tricyclometaphosphoric acid, (HPO3)3 is:

Detailed Solution for JEE Main Practice Test- 9 - Question 49

It has 15 sigma bonds, and 3 Pi bonds.

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JEE Main Practice Test- 9 - Question 50

If the dipole moment of AB molecule is given by 1.2 D and A–B the bond length is 4.8 D then % covalent character of the bond is:

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JEE Main Practice Test- 9 - Question 51

A vector of magnitude 9 and perpendicular to both the vectors

Detailed Solution for JEE Main Practice Test- 9 - Question 51

Hence a unit vector perpendicular to

Hence the required vector is

JEE Main Practice Test- 9 - Question 52

If x = 9 is the chord of contact of the hyperbola x2 - y2 = 9, then the equation of the corresponding pair of tangents is

Detailed Solution for JEE Main Practice Test- 9 - Question 52

Let (h, k) be point whose chord of contact with respect to hyperbola x2 - y2 = 9 is x = 9.
W e know that, chord of contact of (h, k) with respect to hyperbola x2 - y2 = 9 is T = 0

But it is the equation of the line x = 9.
This is possible when h = 1, k = 0 (by comparing both equations).
Againe quation of pair of tangents is T2 = SS1

JEE Main Practice Test- 9 - Question 53

Detailed Solution for JEE Main Practice Test- 9 - Question 53

Roots of the corresponding equation are
∴ by sign - scheme :

JEE Main Practice Test- 9 - Question 54

If α, β and γ are the angles which a directed line makes with the positive directions of the co-ordinates axes, then find the value of  sin2α + sin2β + sin2γ.

Detailed Solution for JEE Main Practice Test- 9 - Question 54

The direction cosines of the line are

JEE Main Practice Test- 9 - Question 55

In an acute - angled triangle ABC, points D, E and F are the feet of the perpendiculars from A, B and C onto BC, AC and AB, respectively. H is the orthocentre of ΔABC. If sin A = 3/5 and BC = 39, then the length of AH is

Detailed Solution for JEE Main Practice Test- 9 - Question 55

JEE Main Practice Test- 9 - Question 56

then f(x) is equal to

Detailed Solution for JEE Main Practice Test- 9 - Question 56

JEE Main Practice Test- 9 - Question 57

If x is rational and ,⁡  then value of x is equal to

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JEE Main Practice Test- 9 - Question 58

The line x = c cuts the triangle with corners (0, 0), (1, 1) and (9, 1) into two regions. For the area of the two regions to be the same c must be equal to :

Detailed Solution for JEE Main Practice Test- 9 - Question 58

The area of the triagle formed by three given vertices is
Now the line x = c bisect the area of triangle in two equal parts so the area of the both the parts of the triangle must be four

(or c = 15 which is not possible)

JEE Main Practice Test- 9 - Question 59

The number of solutions of the equation a f(x) + g(x) = 0, where a > 0, g (x) ≠ 0  and has least value 1/2 is

Detailed Solution for JEE Main Practice Test- 9 - Question 59

Which is impossible.
Hence, no solutions.
∴ Number of solutions = zero.

JEE Main Practice Test- 9 - Question 60

The area of the curvilinear triangle bounded by the y-axis & the curve y = tan x & y = (2/3) cos x is

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JEE Main Practice Test- 9 - Question 61

The equations of the tangents of the parabola y2 = 12x, which passes through the point (2,5).

Detailed Solution for JEE Main Practice Test- 9 - Question 61

y= 12 x
4a = 12  (∵ y= 4ax)
⇒  a = 3
Equation of tangent in slope form is

It passes through (2,5)
⇒ 5=2m+3/m
⇒ 5m=2m2+3
⇒ 2m2-5m+3=0
⇒ (m-1)(2m-3)=0
⇒ m = 1,3/2

So equation of tangents are

JEE Main Practice Test- 9 - Question 62

The focal chord to y2 = 16xis tangent to (x - 6)2 + y= 2, then the possible values of the slope of this chord, are

Detailed Solution for JEE Main Practice Test- 9 - Question 62

The focal chord of y2 = 16 x is tangent to circle (x - 6)2 + y2 = 2 .
Focus of parabola is (a; 0) i.e., (4, 0) is the Focus
Now, tangents are drawn from (4, 0) to (x - 6)2 + y2 = 2 Since, PA is tangent to circle.
Since, tanθ = slope of tangent

∴ Slope of focal chord as tangent to circle
= ±1

JEE Main Practice Test- 9 - Question 63

If P(1) = 0 and  then value of x for which is P(x) > 0 satisfy interval

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JEE Main Practice Test- 9 - Question 64

If 5a + 4b + 20c = t, then the value of t for which the line ax + by + c - 1 = 0 always passes through a fixed point is

Detailed Solution for JEE Main Practice Test- 9 - Question 64

Equation of given line is ax + by + c - 1 = 0

From given relation, substuting value of c

Clearly for t = 20, the given line will pass through the point  for all values of a & b

JEE Main Practice Test- 9 - Question 65

If f(x) is a polynomial satisfying f (x) f (1/x) = f (x) + f (1/x) and f (2) > 1,

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Then by comparing the coefficients of like powers, we get

JEE Main Practice Test- 9 - Question 66

The equation of the pair of straight lines parallel to the y - axis and which are tangents to the circle x2 + y2 - 6x - 4y - 12 = 0 is :

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If a line parallel to y-axis, x = λ touches the given circle, then

JEE Main Practice Test- 9 - Question 67

The number of common tangents that can be drawn to the circles x2 + y2 - 4x - 6y - 3 = 0 and x2 + y2 + 2x + 2y + 1 = 0 is :

Detailed Solution for JEE Main Practice Test- 9 - Question 67

Centre and radius of S1 are C1 (2, 3) and r1 = 4 and centre and radius of S2 are C2 (-1, -1) and r2 = 1.

Hence the given circles are externally touching each other Hence, number of common tangents = 3

JEE Main Practice Test- 9 - Question 68

The distance between the circumcentre and orthocentre of the triangle whose vertices are (0, 0), (6, 8) and (-4, 3) is

Detailed Solution for JEE Main Practice Test- 9 - Question 68

ΔOBC is right angled at O
Circum centre = mid point of hypotenuse BC
Orthocentre = vertex O (0, 0)

JEE Main Practice Test- 9 - Question 69

Set of values of x for which sin-1x > cos-1x is

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JEE Main Practice Test- 9 - Question 70

If n is a positive integer and Ck = nCk, then

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JEE Main Practice Test- 9 - Question 71

If A is a square matrix of order n such that |adj adj(adjA)|=[A]27, then find the possible value of n.

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JEE Main Practice Test- 9 - Question 72

Normals are drawn from the point P with slopes m1,m2,m3 to the parabola y2=4x, if locus of P with m1m2=α  is a part of the parabola itself then find the value of a.

Detailed Solution for JEE Main Practice Test- 9 - Question 72

The equation of the normal to the parabola y2 = 4ax. is y = mx – 2am – am3.
Hence, the equation of the normal to the parabola y2 = 4x is  y = mx -2m – m3
Now, if it passes through (h, k), then we have k = mh - 2m - m3
This can be written as m3 + m(2-h) + k = 0 .... (1)
Here, m1 +  m2 + m3 = 0
m1m2 + m2m3+ m3m = 2 - h
Also, m1m2m3 = -k,
where m1m2 = α
Now, this gives m3 = -k/α and this must satisfy equation (1).

Hence, we have (-k/α)3 - (k/α)(2-h) + k = 0
Solving this, we get k2 = α2h - 2α2 + α3
And so we have y2 = α2x - 2α2 - α3
On comparing this equation  with y2 = 4x we get
α2 = 4 and -2a2 + a3 = 0
This give  α = 2

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JEE Main Practice Test- 9 - Question 73

Line segments drawn from the vertex opposite to the hypotenuse of a right angles triangle to the point trisecting the hypotenuse have length sin x and cos x where 0 < x < π/2.  If the length of hypotenuse is  where p and are coprime, then find 2p+q.

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JEE Main Practice Test- 9 - Question 74

If the equation of the curve on reflection of the ellipse   about the line x−y−2=0 is

16x2+6y2+k1x−36y+k2=0, then find the value of |2k1+k2||2k1+k2|

Detailed Solution for JEE Main Practice Test- 9 - Question 74

Let p(4(1+cos θ)), 3(1+sin θ) is any point on the curve its reflection about the line x - y - 2= 0

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JEE Main Practice Test- 9 - Question 75

If f(x) is a polynomial of degree 4 with leading coefficient 1 satisfies the relation f(1)=1,f(2)=2, f(3)=3  and f(12)+f(−8) is equal to λ. Then find the digit at unit place of λ.

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