Description

This mock test of JEE Main Question Paper 2018 With Solutions (15th-April Morning) for JEE helps you for every JEE entrance exam.
This contains 90 Multiple Choice Questions for JEE JEE Main Question Paper 2018 With Solutions (15th-April Morning) (mcq) to study with solutions a complete question bank.
The solved questions answers in this JEE Main Question Paper 2018 With Solutions (15th-April Morning) quiz give you a good mix of easy questions and tough questions. JEE
students definitely take this JEE Main Question Paper 2018 With Solutions (15th-April Morning) exercise for a better result in the exam. You can find other JEE Main Question Paper 2018 With Solutions (15th-April Morning) extra questions,
long questions & short questions for JEE on EduRev as well by searching above.

QUESTION: 1

A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius R/2, and the other mass, in a circular orbit of radius 3R/2. The difference between the final and initial total energies is

Solution:

QUESTION: 2

In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:

Solution:

Given,

5 complete rotations of screw = 0.25cm

So 1 rotation of screw = 0.05

Hence, 1 main scale division = 0.05 cm

and 1 circular scale = 0.05/100 division = 5 × 10^{−4} cm.

Now Reading is 4 main scale and 30 circular scale divisions

So , thickness = 4 × 0.05 + 30 × 5 × 10^{−4}

= 0.2150 cm.

QUESTION: 3

The B-H curve for a ferromagnet is shown in the figure. The ferromagnet is placed inside a long solenoid with 1000 turns/ cm. The current that should be passed in the solenoid to demagnetise the ferromagnet completely is:

Solution:

From given diagram

Coercivity = −100 (A/m)

Hence

Reverse magnetic field required to demagnetize the substance

= µ_{o} H

= µ_{o} 100

so µ_{o} 100 = µ_{o}.N.I

I = 1mA

QUESTION: 4

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are λ_{1} and λ_{2}, their de Broglie wavelength in the frame of reference attached to their centre of mass is:

Solution:

QUESTION: 5

Light of wavelength 550 nm falls normally on a slit of width 22.0 x 10^{-5}cm. The angular position of the second minima from the central maximum will be (in radians):

Solution:

b sin θ = nλ

b sin θ = 2λ

sin θ = 1/2

θ = π/6

QUESTION: 6

A uniform rod AB is suspended from a point X, at a variable distance x from A, as shown. To make the rod horizontal, a mass in is suspended from its end A. A set of (m, x) values is recorded. The appropriate variables that give a straight line, when plotted, are:

Solution:

= y = mx + c

QUESTION: 7

A Helmholtz coil has a pair of loops, each with N turns and radius R. They are placed coaxially at distance R and the same current I flows through the loops in the same direction. The magnitude of magnetic field at P, midway between the centres A and C, is given by [Refer to figure given below]:

Solution:

QUESTION: 8

An ideal capacitor of capacitance 0.2 µF is charged to a potential difference of 10 V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self-inductance 0.5 mH. The current at a time when the potential difference across the capacitor is 5 V, is:

Solution:

Q_{o} = 0.2 × 10 µC = 2µC

V_{o} = 10V

E_{i} of capacitor = 1/2 × 0.2 µf × (10v)^{2}

= 10µJ

= E_{f} of capacitor = 1/2 × 0.2µf × (5v)^{2}

= 2.5 µj

⇒ E_{inductor} = 7.5 µJ = 1/2 Li^{2}

⇒ 7.5 × 10^{−6} = 1/2 × 0.5 × 10^{−3} × i^{2}

⇒ 30 × 10^{−3} = i^{2}

⇒ i = √3/10 = 0.17 A

QUESTION: 9

Take the mean distance of the moon and the sun from the earth to be 0.4 x 10^{6} km and 150 x 10^{6} km respectively. Their masses are 8 x 10^{22} kg and 2 x 10^{30 }kg respectively. The radius of the earth is 6400 km. Let ΔF_{1} be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF_{2} be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to ΔF_{1}/ΔF_{2} is:

Solution:

= (375)^{3} (4 × 10^{−8})

= (0.37 × 10^{3})^{3}(4 × 10^{−8})

= 1.64

QUESTION: 10

An automobile, travelling at 40 km/h, can be stopped at a distance of 40m by applying brakes. If the same automobile is travelling at 80 km/h, the minimum stopping distance, in metres, is (assume no skidding):

Solution:

u = 40 Km/hr

μ = 100/9 m/s

V^{2 }− u^{2} = 2a × 40

a = −1.54 m m/s^{2}

Now.

S = 160m

QUESTION: 11

A charge Q is placed at a distance a/2 above the centre of the square surface of edge a as shown in the figure

The electric flux through the square surface is:

Solution:

Through whole cube = Q/∈_{o}

Through one face = Q/6∈_{o}

QUESTION: 12

A force of 40 N acts on a point B at the end of an L-shaped object, as shown in the figure. The angle θ that will produce maximum moment of the force about point A is given by

Solution:

τ_{A }= f cosθ(4) + Fsinθ(2)

τ_{A} = 2F(sinθ + 2cosθ

dτ_{A}/dθ = 2F[cosθ − 2sinθ] = 0

Tan θ = 1/2

QUESTION: 13

The equivalent capacitance between A and B in the circuit given below, is:

Solution:

QUESTION: 14

A monochromatic beam of light has a frequency v = 3/2π × 10^{12} Hz and is propagating along the direction It is polarized along the k̂ direction. The acceptable form for the magnetic field is:

Solution:

Wave propagation vector should be along

So, b is the only option which satisfies the above condition.

QUESTION: 15

A particle is oscillating on the X-axis with an amplitude 2 cm about the point x_{0} = 10 cm, with a frequency ω. A concave mirror of focal length 5 cm is placed at the origin (see figure).

Identify the correct statements.

(A)The image executes periodic motion.

(B)The image executes non-periodic motion.

(C)The turning points of the image are asymmetric w.r.t. the image of the point at x =10 cm.

(D)The distance between the turning points of the oscillation of the image is 100/21 Cm.

Solution:

QUESTION: 16

A tuning fork vibrates with frequency 256 Hz and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (Speed of sound in air is 340 ms^{-1})

Solution:

Third normal mode of frequency in open pipe,

f = 3vs/2l

Where, Vs = 340m/s

Get L = 2m

Or

L = 200 cm

QUESTION: 17

A thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids, whose densities are ρ_{1} and ρ_{2} (ρ_{1}> ρ_{2}), fill half the circle. The angle θ between the radius vector passing through the common interface and the vertical is:

Solution:

h_{1}ρ_{1} − 1 = h_{2}ρ_{2}

(R Cos θ + R sinθ)ρ_{2} = (R cosθ − R sinθ)ρ_{1}

(ρ_{1} + ρ_{2}) Sin θ = (ρ_{1} − ρ_{2} )Cosθ

_{}

QUESTION: 18

In a common emitter configuration with suitable bias, it is given that R_{L} is the load resistance and R_{BE} is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by:

β is current gain, I_{B}, I_{C} and I_{E} are respectively base, collector and emitter currents.

Solution:

QUESTION: 19

A given object takes n times more time to slide down a 45° rough inclined plane as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline Is:

Solution:

So

Sin θ = n^{2} sinθ − µn^{2} cosθ

1 = n^{2} − µn^{2}

1 − n^{2} = −µn^{2}

n^{2} − 1 = −µn^{2}

µ = 1 − 1/n^{2}

QUESTION: 20

One mole of an ideal monatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27°C. The work done on the gas will be:

Solution:

= 300R ln 2

QUESTION: 21

A Carnot's engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is:

Solution:

Given,

Q_{2} = 500 cal.

η_{Carnot} = 1 - T_{2}/T_{1}

= 1/6

and (COP)_{HP} = 1/η_{Carnot} = 6

(Cop)_{HP} − (Cop)_{Ref} = 1

So, (cop)_{Ref} = 5 = Q_{2}/w

so w = 100 cal

Or w = 420 J

QUESTION: 22

In the given circuit all resistances are of value R ohm each. The equivalent resistance between A and B is:

Solution:

Due to short circuit current will flow along

B-C-D-E-F-A

So

R eq = R + R

R eq = 2R

QUESTION: 23

A solution containing active cobalt 60/27 Co having activity of 0.8 µCi and decay constant λ is injected in an animal's body. If 1 cm^{3} of blood is drawn from the animal's body after 10 hrs of injection, the activity found was 300 decays per minute. What is the volume of blood that is flowing in the body? (1 Ci = 3.7 x 10^{10 }decays per second and at t = 10 hrs e^{ –λt} = 0.84)

Solution:

(dn/dt) = −λN_{o}

0.8 µ Ci = −λN_{o}

N_{1} = N_{o}(0.84)

v → N_{1}

v cm^{3} → N_{1}/v

v = 3.7/5 × 0.84 × 0.84 × 10^{4} cm^{3}

= 0.5 × 10^{4} _{cm}3 = 5 × 10^{3}_{cm}3 = 5 lit

QUESTION: 24

In a meter bridge, as shown in the figure, it is given that resistance Y =12.5 Ω and that the balance is obtained at a distance 39.5 cm from end A (by Jockey J). After interchanging the resistances X and Y, a new balance point is found at a distance I_{2} from end A. What are the values of X and l_{2}?

Solution:

For balanced wheat stone bridge

x(100 − l_{1}) = y × l_{1}

So x(100 − 39.5) = 12.5(39.5)

x = 8.16 Ω

If x and y inter changed

y(100 − l_{2}) = xl_{2}

12.5(100 − l_{2}) = 8.16 l_{2}

get l_{2} = 60.5 cm

QUESTION: 25

The energy required to remove the electron from a singly ionized Helium atom is 2.2 times the energy required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is:

Solution:

E_{1} = ionization energy of ionized He

E_{2} = 2. 2E_{2}

E_{1} = 13.6 ev

E_{2} = 13.6/2.2 = 6.18 ev

Total = E_{1} + E_{2} = 13.6 ev + 6.18 ev

= 20 ev

QUESTION: 26

The number of amplitude modulated broadcast stations that can be accomodated in a 300 kHz band width for the highest modulating frequency 15 kHz will be:

Solution:

The station will require a band width of 30 khz

So

No. of stations = 300/30

= 10

QUESTION: 27

The velocity-time graphs of a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in 15 s and (ii) the time at which the car will catch up with the scooter are, respectively.

Solution:

s_{1} = 1/2 × 3 × (15)^{2} = 1.5 × 225

= 225 + 112.5

= 337.5 m

s_{2} = v_{2}t

s_{2} = 15 × 30 = 450 m ⟹ s_{2} − s_{1} = 112.5 m

For catching up ⟹ s_{1} = s_{2}

30t = 1/2 × 3 × t^{2}

20 = t

QUESTION: 28

A body of mass M and charge q is connected to a spring of spring constant k. It is oscillating along x direction about its equilibrium position, taken to be at x = 0, with an amplitude A. An electric field E is applied along the x-direction. Which of the following statements is correct?

Solution:

Energy at the extreme

= 1/2 kA^{2} = T. E

After switching on electric field

New mean position ⟹ kx_{o} = qE

x_{o} = qE/k

So entreme position also shifts by qE/k

QUESTION: 29

A planoconvex lens becomes an optical system of 28 cm focal length when its plane surface is silvered and illuminated from left to right as shown in Fig-A.

If the same lens is instead silvered on the curved surface and illuminated from other side as in Fig. B, it acts like an optical system of focal length 10 cm. The refractive index of the material of lens is:

Solution:

When plane surface is silvered

Focal length f_{1 }=_{ }

and f_{1} = 28 cm

When curved surface is silvered

Focal length f_{2} = R/2μ ... (ii)

µ = 1.55

QUESTION: 30

The relative error in the determination of the surface area of a sphere is α. Then the relative error in the determination of its volume is:

Solution:

Area of sphere (A) = 4πR^{2}

taking log

ln A = In(4π) + 2In(R)

differentiating both sides

now, similarly

QUESTION: 31

The main reduction product of the following compound with NaBH4 in methanol is:

Solution:

Sodium borohydride will reduce ketone to alcohol. It will not reduce amide group and

*C*=*C* double bond.

Hence, option (B) is the correct answer.

QUESTION: 32

A white sodium salt dissolves readily in water to give a solution which is neutral to litmus. When silver nitrate solution is added to the aforementioned solution, a white precipitate is obtained which does not dissolve in dil. nitric acid. The anion is:

Solution:

So, Cl^{- } is answer.

QUESTION: 33

Which of the following statements about colloids is False?

Solution:

QUESTION: 34

A sample of NaCIO_{3} is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl. The mass of AgCI (in g) obtained will be: (Given: Molar mass of AgCl= 143.5 g mol^{ -1})

Solution:

QUESTION: 35

In which of the following reactions, an increase in the volume of the container will favour the formation of -products?

Solution:

Volume increases P decreases reaction proceed in that direction where number of gaseous moles increases.

QUESTION: 36

The reagent(s) required for the following conversion are:

Solution:

QUESTION: 37

An ideal gas undergoes a cyclic process as shown in Figure.

ΔU_{BC} = −5 Kj mol^{−1}, q_{AB} = 2Kj mol^{−1}

W_{AB }= −5Kj mol^{−1}, W_{CA} = 3Kj mol^{−1}

Heat absorbed by the system during process CA is:

Solution:

QUESTION: 38

In graphite and diamond, the percentage of p-characters of the hybrid orbitals in hybridisation are respectively:

Solution:

graphite → sp^{2} → % S → 33 % , % p = 67 %

diamond → Sp^{3} → % S → 25 %, % p = 75 %

QUESTION: 39

The correct combination is:

Solution:

(1) [Ni (Cl)_{4}]^{2} → d^{8} (Ni)^{2+}

Cl^{−} is weak

Field ligand → So due to unpraised is paramagnetic

(2)

Ni(CO)_{4 }→ Tetrachedral

QUESTION: 40

The major product of the following reaction is:

Solution:

The reaction undergoes acylation first followed by substitution Intramolecular.

QUESTION: 41

When an electric current is passed through acidified water, 112 mL of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is:

Solution:

at NTP 1 mol = 22. 4 l

112 ml H_{2} ⇒

⇒ 1/200 mol of H_{2}

H_{2}O → H_{2} + 1/2 O_{2}

2H^{−} + 2e^{− }→ H_{2}

1 mol of H_{2} required 2 mole e̅

1/200 mol of H_{2} require 2/200 = 1/100 mol of e̅

1/100 mol of e̅ = 1/100 × 6.022 × 10^{23}e̅ × 1.6 × 10^{−19}

QUESTION: 42

Which of the following is a Lewis acid?

Solution:

Lewis acid → which has vacant orbital,

So B(CH_{3})_{3}

QUESTION: 43

The correct match between List-I and List-II is:

Solution:

O – P ⇒ difference in Boiling point ⇒ Steam distillation

Coloured impurity → Chromatography

QUESTION: 44

Xenon hexafluoride on partial hydrolysis produces compounds 'X' and 'Y'. Compounds 'X' and 'Y' and the oxidation state of Xe are respectively:

Solution:

Xe F_{6} + H_{2}O → XeO F_{4} + 2HF

XeF_{6} + 2H_{2}O → XeO_{2}F_{2} + 4HF

QUESTION: 45

N_{2}O_{5} decomposes to NO_{2} and O_{2} and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mmHg to 87.5 mmHg. The pressure of the gaseous mixture after 100 minute at constant temperature will be:

Solution:

2N_{2}O_{5} → 4NO_{2} + O_{2}

p − 2x - 4x - x

pt = p − 2x + 4x + x

pt = p + 3x

at t = 0, pt = p = 50 mmHg

at t = 50 mm, pt = 87.5 mmHg

p + 3x = 87.5

p = 87.5 − 3x

50 = 87.5 − 3x

12.5 = x

p − 2x = 50 − 2(12.5) = 25

Since K will remain same

50 = 50 × 4 − 8y

50 = 200 − 8y

8y = 150

y = 18.75

pt = p + 3y

= 50 + 3 (18.73) = 106.25 mmHg

QUESTION: 46

For Na^{+}, mg^{2+}, F^{-} and O2^{-}; the correct order of increasing ionic radii is:

Solution:

When negative charge increase, the radius of ion increases.

QUESTION: 47

Which of the following will not exist in zwitter ionic form at pH =7?

Solution:

The N atom of amide is not basic

QUESTION: 48

The IUPAC name of the following compound is:

Solution:

Basic Nomenclature.

QUESTION: 49

The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution (K_{sp} of PbCl_{2} = 3.2 x 10^{-8}; atomic mass of Pb= 207 u) is:

Solution:

Ksp of PbCl_{2} is 3.2 × 10^{-8}

PbCl_{2} is 3.2 × 10^{−8}

Pbcl_{2}(s) ⇌ Pb^{2+}(aq) + 2Cl^{−}(aq)

t = 0 1 0 0.

At equilibrium 1 – S S 2S.

K_{sp} = [S] [2S]^{2}

3.2 × 10^{−8} = 4s^{3}

S^{3 }= 0.8 × 10^{−8}

S^{3} = 8 × 10^{−9}

S = 2 × 10^{−3}

Solubility = W/V

∴ Solubility of PbCl_{2} in gL^{−1} = 2 × 10^{−3} × 278

= 556 × 10^{−3}gL^{−1}

0.556 gL^{−1}

0.556/0.1 = 1/x

x = 0.1/0.556

= 0.18L

QUESTION: 50

Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is:

Solution:

= 4.375 eV

QUESTION: 51

Which of the following arrangements shows the schematic alignment of magnetic moments of antiferromagnetic substance?

Solution:

Basic knowledge of Antiferromagnetic

QUESTION: 52

In hydrogen azide (above) the bond orders of bonds (I) and (II) are:

Solution:

Hydrogen azide: HN_{3}

Both works correct

QUESTION: 53

The copolymer formed by addition polymerization of styrene and acrylonitrile in the presence of peroxide is:

Solution:

(C_{6}H_{8})_{n} − (C_{3}H_{3}N)_{m}

QUESTION: 54

Which of the following will most readily give the dehydrohalogenation product?

Solution:

Most probable is C_{6}H_{5} Br

QUESTION: 55

In the molecular orbital diagram for the molecular ion, N_{2}^{+}, the number of electrons in the σ2_{p} molecular orbital is:

Solution:

QUESTION: 56

Which of the following is the correct structure of Adenosine?

Solution:

QUESTION: 57

Identify the pair in which the geometry of the species is T-shape and square-pyramidal, respectively:

Solution:

QUESTION: 58

The decreasing order of bond angles in BF_{3}, NH_{3}, PF_{3} and I_{3}^{-} is:

Solution:

I_{3}^{-} :180

QUESTION: 59

For which of the following reactions, ΔH is equal to ΔU?

Solution:

ΔH = ΔU + Δn_{g}RT.

[If Δn_{g} = 0 then ΔH =ΔU]

QUESTION: 60

The increasing order of nitration of the following compounds is:

Solution:

Nitration is electrophilic aromatic substitution reaction. Methoxy and amino groups are strongly activating groups. Methyl group is weakly activating group.

Since among methyl and methoxy group, methoxy group is more reactive than methyl group, (c) is more reactive than (d).

Even-though amino group is strongly activating group, it gets protonated in presence of acid to form anilinium ion which is strongly deactivating. Hence, (a) is less reactive than (c) and (d).

Note:

The activating groups increases the electron density on benzene ring and increases the rate of electrophilic aromatic substitution reaction. The deactivating groups decreases the electron density on benzene ring and decreases the rate of electrophilic aromatic substitution reaction.

QUESTION: 61

The set of all α ∈ R, for which w =is a purely imaginary number , for all z ∈ C Satifying |z| = 1 and Re z ≠ 1, is:

Solution:

As ω is purely imaginary

If Re(z)≠1

then, α = 0

QUESTION: 62

Solution:

QUESTION: 63

An aeroplane flying at a constant speed, parallel to the horizontal ground, √3 km above it, is observed at an elevation of 60° from a point on the ground. If, after five seconds, its elevation from the same point, is 30°, then the speed (in km/hr) of the aeroplane, is:

Solution:

M 5 sec = 2km

Speed = 2/5 km/sec

QUESTION: 64

The value of the integral

Solution:

QUESTION: 65

A variable plane passes through a fixed point (3, 2, 1) and meets x, y and z axes at A, B and C respectively. A plane is drawn parallel to yz-plane through A, a second plane is drawn parallel zx-plane through B and a third plane is drawn parallel to xy-plane through C. Then the locus of the point of intersection of these three planes, is:

Solution:

E. q. , of variable plane

a(x − 3) + b(y − 2) + c(2 − 1) = 0

Plane paroled to y2 plane passing through !

Intersection of the x three

QUESTION: 66

An angle between the plane, x + y + z = 5 and the line of intersection of the planes, 3x + 4y + z – 1 = 0 and 5x + 8y + 2z + 14 = 0, is:

Solution:

Perpendicular vector to plane

Î +̂J + k̂

Parallel vector to line

= I(8 − 8) − Ĵ (6 − 5) + k̂ (24 − 20)

= −Ĵ + 4k̂

QUESTION: 67

If n is the degree of the polynomial, m is the coefficient of x^{n }in it, then the ordered pair (n, m)is equal to ∶

Solution:

Rationalize and get

Degree = 12

Coff = 2 [ ^{8}C_{0} + ^{8}C_{r} + − ^{8}C_{8}]5^{4}

^{8}C_{0} + ^{8}C_{2} + ^{8}C_{4} + − ^{8}C_{4} = 2^{4−1}

Coff = 2 (2^{8−1}) 5^{4}

Coff = 2 2^{7}5^{4}

= 2 × 2^{3} × 2^{4} × 5^{4}

= 16 (10)^{4}

= (20)^{4}

QUESTION: 68

If β is one of the angles between the normals to the ellipse, x^{2} + 3y^{2} = 9 at the points (3 cosθ, √3 sinθ)and(−3 sin θ√3 cosθ) ′; θϵ (0, π/2) ; then 2 cot β /sin 2θ is equal to:

Solution:

2x dx + 6y dy = 0

2x dx = −cy dy

QUESTION: 69

In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, x + y = 5 and x = 4. Then area of ∆ABC (in sq. units) is :

Solution:

QUESTION: 70

The mean of a set of 30 observations is 75. If each observation is multiplied by a non-zero number λ and then each of them is decreased by 25, their mean remains the same. Then λ is equal to :

Solution:

QUESTION: 71

A circle passes through the points (2, 3) and (4, 5). If its centre lies on the line, y — 4x + 3 = 0, then its radius is equal to :

Solution:

Let A(2, 3) & B(4, 5) be the points through which circle is passing & radius lies on the line y-4x + 3 = 0.

Let centre be (n, k), this point must satisfy the line y – 4x - + 3 = 0. Hence

k − 4n + 3 = 0

k = 4n − 3.

So, Centre coordinates be 0(n, 4n -3)

Now, OA = OB (both one radius)

A(2, 3) B(4, 5)

⇒ OA = OB

⇒ OA^{2} = OB^{2}

(n − 2)^{2} + (4n − 3 − 3)^{2} = (n − 4)^{2} + (4n − 3 − 5)^{2}

n^{2} + 4 − 4n + 16n^{2} + 36 − 48n = n^{2} + 16 − 8n + 16n2 + 64 − 64n

40 − 52n = 80 − 72n

72n − 52n = 80 − 40

20n = 40

n = 2

Hence k = 4n − 3

k = 8 − 3

k = 5

Centre coordinates O(2, 5)& !(2, 3)

radius = OA = (2 − 2)^{2} + (5 − 3)^{2 }= 2

Hence = radius = 2

QUESTION: 72

Let S be the set of all real values of k for which the system of linear equations

x+ y + z = 2

2x + y - z = 3

3x + 2y + kz = 4

has a unique solution.

Then S is :

Solution:

Therefore,set S=equal to R-{0}

QUESTION: 73

If tanA and tanB are the roots of the quadratic equation, 3x^{2} —10x — 25 = 0, then the value of 3 sin^{2}(A +B) —10 sin(A +B).cos(A+B) —25 cos^{2}(A + B)

Solution:

QUESTION: 74

If x^{2} + y^{2} + sin y = 4, then the value of d^{2}y/dx^{2} at the point ( -2,0) is:

Solution:

x^{2} + y^{2} + sin y = 4

Diff above eqn with respect to x.

QUESTION: 75

If x_{1} , x_{2} … . , x_{n} and 1/h_{1}, 1/h_{2 }......1/h_{n }are two A. P. s such that x_{3} = h_{2} = 8 and x_{8} = h_{7} = 20 , then x_{5} . h_{10} equals:

Solution:

QUESTION: 76

If are unit vectors such that then is equal to:

Solution:

QUESTION: 77

If the tangents drawn to the hyperbola 4y^{2} = x^{2} + 1 intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid-point of AB is:

Solution:

4y^{2} = x^{2} + 1

Point 4yy_{1} = xx_{1} + 1 with 4y_{1}^{2} = x_{1}^{2} + 1

x axis

y axis

Mid point h = k =

QUESTION: 78

The area (in sq. units) of the region {x ϵ R ∶ x ≥ 0, y ≥ x − 2 and y ≥ √x} is ∶

Solution:

QUESTION: 79

Let A be a matrix such that A ∙ isa scalar matrix and |3A| = 108. Then A^{2 }equals:

Solution:

a = 2c + 3d

c = 0

2a + 3b = 0

ad − bc = 12

|3A| = 108

|A| = 12

A^{2} =

c = 0

d = 2

a = 6

b = -4

A^{2} =

QUESTION: 80

If f = 2x + 1, (x ϵ R − {1, −2}), then ∫ f(x)dx is equal to:

(where C is a constant of integration)

Solution:

= 2x + 1

Let t =

tx + 2t = x − 4

2t + 4 = x(1 − t)

x =

QUESTION: 81

If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm^{2}) of this cone is:

Solution:

(h − R)^{2} + r^{2} = R^{2}

r^{2} = R^{2} − (h − R)^{2}

R = 3

r^{2} = q − (h − 3)^{2}

h(−3h + 12) = 0

3h = 12, h = 0

h = 4. r = 2√2.

CSA = πrl

= π × 2√2 × √24

= π 2 × √48 = 8π√3

QUESTION: 82

If b is the first term of an infinite G.P. whose sum is five, then b lies in the interval:

Solution:

=5

b = 5 (1 − r)

b ε (0,10) [−1 < r < 1]

QUESTION: 83

Consider the following two binary relations on the set A = {a, b, c}: R_{1} = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)} and R_{2} = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}.Then:

Solution:

R_{1} = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)}

b, c ϵR_{1} c, a ∉ R_{1} R_{1} is not symmetric (b, c), (c, a)ϵR_{1}(b, a) ∉ R_{1} , R_{1} is not transitive

R _{2} = {(a, b), (b, a), (c, c), (c, a), (c, a), (a, a), (b, b), (a, c)}

∀(a, b)ϵR_{2}(b, a) × R_{2}

Therefore it is symmetric

(c, a), (a, b)ϵR_{2}(c, b) ∉ R _{2}

Therefore R _{2} is not transitive

QUESTION: 84

Let S = {λ, µ)ϵ R × R ∶ = f(t) = (|λ|e^{|t|} − µ). sin (2(2|t|), t ϵ R, is a differentiable function}. Then S is a subset of ∶

Solution:

f(t) = (|λ|e^{t }− µ) sin 2t t > 0

−(|λ|e̅^{t} − µ) sin 2t t < 0

f ′ (t) = 2 cos 2t (|λ|e^{t} − µ) + |λ|e^{t }sin 2t t > 0

−2 cos 2t (|λ|e^{t} − µ) + |λ|e̅^{t} sin 2t t > 0

f ′(t → o^{t}) = 2(|λ| − µ)

f ′(t → o̅ ^{t}) = −2(|λ| − µ)

for differentiability LHD = RHD

2(|λ| − µ) = −2(|λ| − µ)

|λ| = µ

⟹ λ ϵ R µ ϵ R t

(λ, µ) C R × [0, ∞)

QUESTION: 85

A box 'A' contains 2 white, 3 red and 2 black balls. Another box 'B' contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box 'B' is:

Solution:

QUESTION: 86

If (p ∧ ~q) ∧ (p ∧ r) → ~ p ∨ q is false, then the truth values of p, q and r are, respectively:

Solution:

QUESTION: 87

If λ ϵ R is such that the sum of the cubes of the roots of the equation, x^{2 }+ (2 - λ)x + (10 - λ) = 0 is minimum, then the magnitude of the difference of the roots of this equation is:

Solution:

α^{3} + β^{3} = (α + β)(α^{2} + β^{2} − αβ)

= −(2 − λ)((λ + β)^{2} − 3αβ)

= (λ − 2)((λ − 2)^{2} + 3(λ − 10))

= (λ − 2)(b^{2} − 4λ + 4 + 3λ − 30)

QUESTION: 88

If f(x) =

Solution:

= cos x − tan x [x^{2} − 2x^{2}]

= x^{2} tan x − x^{2} cos x

f ′(x) = 2x tan x + x^{2} sec^{2} x − 2x cos x + x^{2} sin x

= 2 tan x + x sec^{2} x − 2 cos x + x sin x

= 0 + 0 − 2 + 0 = −2

QUESTION: 89

Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is :

Solution:

4y = −4x − 3

4(x + y) + 3 = 0

QUESTION: 90

n-digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed, is:

Solution:

For each place we have 3 choices

(i)for n − digits 3 × 3 … n times = 3^{n} > 900

n = 7

### NCERT Solutions - Good Morning

Doc | 2 Pages

- JEE Main Question Paper 2018 With Solutions (15th-April Morning)
Test | 90 questions | 180 min

- JEE Main Question Paper 2018 With Solutions (16th-April Morning)
Test | 90 questions | 180 min

- JEE Main Question Paper 2020 With Solutions (9th January - Morning)
Test | 75 questions | 180 min

- JEE Main Question Paper 2019 With Solutions (11th January - Morning)
Test | 90 questions | 180 min

- JEE Main Question Paper 2020 With Solutions (8th January - Morning)
Test | 74 questions | 180 min