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JEE Main 2014 Question Paper With Solutions (6th-April-2014)


90 Questions MCQ Test JEE Main Mock Test Series 2020 & Previous Year Papers | JEE Main 2014 Question Paper With Solutions (6th-April-2014)


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QUESTION: 1

​The current voltage relation of diode is given by I = (e1000V/T – 1) mA, where the applied V is in volts and the temperature T is in degree kelvin. If a student makes an error measuring ± 0.01 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA?

Solution:

 I = (e1000 V/T – 1)mA
When I = 5 mA, e1000 V/T = 6 mA
Also, dI =
 
= 0.2 mA

QUESTION: 2

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:

Solution:

Time taken to reach highest point is t1 = u/g
Speed on reaching ground = 
Now, v = u + at 
                                     

QUESTION: 3

A mass m is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

Solution:

a = Rα
mg  – T = ma
T × R = mR2α
or T = ma
a = g/2

QUESTION: 4

A block of mass m is placed on a surface with a vertical cross-section given by y = x3/6. f the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is
Solution:

tanθ = dy/dx = x2/2
At limiting equilibrium,
μ = tanθ

⇒ x = ±1
Now, y = 1/6

QUESTION: 5

When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx2 where a and b are constants. The work done in stretching the unstretched rubber-band by L is :
Solution:

QUESTION: 6

A bob of mass m attached to an inextensible string of length l is suspended from a vertical support.
The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical. About the point of suspension
Solution:

τ = mg × l sin θ. (Direction parallel to plane of rotation of particle)

as perpendicular to direction of L changes but magnitude remains same.

QUESTION: 7

Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is
Solution:





QUESTION: 8

The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is :
(For steel Young's modulus is 2 × 1011 Nm–2 and coefficient of thermal expansion is 1.1 × 10–5 K–1)
Solution:

As length is constant,
Strain = 
Now pressure = stress = Y × strain
= 2 × 1011 × 1.1 × 10–5 × 100
= 2.2 × 108 Pa

QUESTION: 9

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the tube. Each liquid subtends 90° angle at centre. Radius joining their interface makes an angle α with vertical. Ratio is

  
Solution:

Equating pressure at A

(Rcosα + Rsinα)d2g = (Rcosα – Rsinα)d1g
⇒ 

QUESTION: 10

On heating water, bubbles being formed at the bottom of the vessel detatch and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r << R, and the surface tension of water is T, value of r just before bubbles detatch is (Density of water is ρω)
Solution:

When the bubble gets detached, Buoyant force = force due to surface tension

QUESTION: 11

​Three rods of copper, brass and steel are welded together to form a Y-shaped structure. Area of cross-section of each rod = 4 cm2. End of copper rod is maintained at 100°C whereas ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cm respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is
Solution:


Q = Q1 + Q2

200 – 2T = 2T + T
⇒ T = 40°C
Q =  =  4.8 cal/s

QUESTION: 12

One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement
Solution:


For BC, ΔT = –200 K
ΔU = –500R

QUESTION: 13

​An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be length of the air column above mercury in the tube now?
(Atmospheric pressure = 76 cm of Hg)
Solution:


P + x = P0
P = (76 – x)
8 × A × 76 = (76 – x) × A × (54 – x)
x = 38
Length of air column = 54 – 38 = 16 cm.

QUESTION: 14

A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a, and in next τ s, it travels 2a in same direction then
Solution:

As it starts from rest, we have
x = Acosωt. At t = 0, x = A
when t = , x = A – a
when t = , x = A – 3a

Now, A – a = Acos

QUESTION: 15

A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.
Solution:


∴ Answer is 6.

QUESTION: 16

Assume that an electric field exists in space. Then the potential difference VA – VO, where VO is the potential at the origin and VA the potential at x = 2 m is
Solution:

QUESTION: 17

A parallel plate capacitor is made of two circular plates separated by a distance 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close to
Solution:


= 2.2 × 8.85 ×10–12 × 3 ×104 ≈ 6 × 10–7  C/m2

QUESTION: 18

​In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be:
Solution:

15 × 40 + 5 × 100 + 5 × 80 + 1000
= V × I
600 + 500 + 400 + 1000 = 220 I

I = 12 A.

QUESTION: 19

A conductor lies along the z-axis at –1.5 ≤ z < 1.5 m and carries a fixed current of 10.0 A indirection (see figure). For a fieldT, find the power required to move the conductor at constant speed to x = 2.0 m, y = 0 m in 5 × 10–3 s. Assume parallel motion along the x-axis
Solution:

Average Power = work/time


= 2.97 × 10–3 J

QUESTION: 20

The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 × 103 A m–1. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is
Solution:


I = 3 A.

QUESTION: 21

In the circuit shown here, the point 'C' is kept connected to point 'A' till the current flowing through the circuit becomes constant. Afterward, suddenly, point 'C' is disconnected from point 'A' and connected to point 'B' at time t = 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to
Solution:

Applying Kirchhoff's law in closed loop, –VR – VC = 0
⇒ VR/VC = –1

QUESTION: 22

During the propagation of electromagnetic waves in a medium
Solution:

Energy is equally divided between electric and magnetic field

QUESTION: 23

​A thin convex lens made from crown glass has focal length f. When it is measured in two different liquids having refractive indices 4/3 and 5/3, it has the focal lengths f1 and f2 respectively. The correct relation between the focal lengths is
Solution:

By Lens maker's formula

⇒ f1 = 4f & f2 = –5f

QUESTION: 24

A green light is incident from the water to the air-water interface at the critical angle(θ). Select the correct statement
Solution:

sin θc = 1/μ

For greater  wavelength (i.e. lesser frequency) μ is less
So, θc would be more. So, they will not suffer reflection and come out at angles less then 90°.

QUESTION: 25

Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams are IA and IB respectively, then equals
Solution:

By law of Malus, I = I0cos2θ
Now, IA′ = IAcos230
IB′ = IBcos260
As IA′ = IB′

IA/IB = 1/3

QUESTION: 26

​The radiation corresponding to 3→2 transitions of hydrogen atoms falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10–4 T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to
Solution:



= 0.8 V
For transition between 3 to 2,

QUESTION: 27

​Hydrogen (1H1), Deuterium (1H2), singly ionised Helium (2He4)+ and doubly ionised lithium (3Li6)++ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are λ1, λ2, λ3 and λ4 respectively then approximately which one of the following is correct?
Solution:


QUESTION: 28

The forward biased diode connection is
Solution:


For forward Bias, p-side must be at higher potential than n-side.

QUESTION: 29

Match List-I (Electromagnetic wave type) with List-II (Its association/application) and select the correct option from the choices given below the lists :
Solution:

(a) Infrared rays are used to treat muscular strain
(b) Radiowaves are used for broadcasting
(c) X-rays are used to detect fracture of bones
(d) Ultraviolet rays are absorbed by ozone

QUESTION: 30

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?
Solution:

As measured value is 3.50 cm, the least count must be 0.01 cm = 0.1 mm
For vernier scale with 1 MSD = 1 mm and 9 MSD = 10 VSD,
Least count = 1 MSD – 1 VSD = 0.1 mm

QUESTION: 31

The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is
Solution:

37 →1s22s22p63s23p63d104s24p65s1
So last electron enters 5s orbital
Hence n = 5, l = 0, m1 = 0, ms = 

QUESTION: 32

If Z is a compressibility factor, van der Waals equation at low pressure can be written as
Solution:

Compressibility factor 
(For one mole of real gas)
van der Waal equation

QUESTION: 33

CsCl crystallises in body centred cubic lattice. If ‘a’ is its edge length then which of the following expressions is correct?
Solution:


QUESTION: 34

​For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20 mL of M/10 sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is
Solution:

As per question


QUESTION: 35

​Resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution is 1.4 S m–1. The resistance of 0.5 M solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in S m2 mol–1 is
Solution:

For 0.2 M solution
R = 50 Ω
σ = 1.4 S m–1 = 1.4 × 10–2 S cm–1

Now, 

For 0.5 M solution
R = 280 Ω
σ = ?

Now, 

= 5 S cm2 mol–1
= 5 × 10–4 S m2 mol–1

QUESTION: 36

For complete combustion of ethanol,

the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol–1 at 25°C. Assuming ideality the enthalpy of combustion, ΔcH, for the reaction will be (R = 8.314 kJ mol–1)
Solution:


Bomb calorimeter gives ΔU of the reaction
So, as per question

QUESTION: 37

​The equivalent conductance of NaCl at concentration C and at infinite dilution are λC and λ∞, respectively. The correct relationship between λC and λ∞ is given as
(Where the constant B is positive)
Solution:

According to Debye Huckle onsager equation,

QUESTION: 38

​Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and 0.125 M Na3PO4(aq) at 25°C. Which statement is true about these solutions, assuming all salts to be strong electrolytes?
Solution:

π= iCRT

QUESTION: 39

For the reaction If KP = KC(RT)x where the symbols have usual meaning then the value of x is (assuming ideality)
Solution:


x = Δng = no. of gaseous moles in product – no. of gaseous moles in reactant
= 

QUESTION: 40

For the non-stoichiometre reaction 2A + B → C + D, the following kinetic data were obtained in three separate experiments, all at 298 K.

The rate law for the formation of C is
Solution:

2A + B → C+ D
Rate of Reaction = 

Let rate of Reaction = k[A]x[B]y

Now from table,
1.2 × 10–3 = k [0.1]x[0.1]y .....(i)
1.2 × 10–3 = k [0.1]x[0.2]y ...(ii)
2.4 × 10–3 = k [0.2]x[0.1]y ...(iii)
Dividing equation (i) by (ii)


Now Dividing equation (i) by (iii)

QUESTION: 41

Among the following oxoacids, the correct decreasing order of acid strength is
Solution:


Resonance produced conjugate base.



(iv) ClO– is not resonance stabilized.
As per resonance stability order of conjugate base is

QUESTION: 42

The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is
Solution:

On electrolysis only in case of Ca2+ salt aqueous solution H2 gas discharge at Cathode.
Case of Cr
At cathode : Cr3+ + 2e– → Cr
So, Cr is deposited.
Case of Ag
At cathode : Ag+ + e– → Ag So, Ag is deposited.
Case of Cu
At cathode : Cu2+ + 2e– → Cu
Case of Ca2+
At cathode : 

QUESTION: 43

The octahedral complex of a metal ion M3+ with four monodentate ligands L1, L2, L3 and L4 absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is
Solution:


The energy of red light is less than that of violet light.
So energy order is
Red < Yellow < Green < Blue
The complex absorbs lower energy light lower will be its strength. So order of ligand strength is
L1 < L3 < L2 < L4

QUESTION: 44

Which one of the following properties is not shown by NO?
Solution:

Nitric oxide is paramagnetic in the gaseous state as it has one unpaired electron in its outermost shell.
The electronic configuration of NO is

However, it dimerises at low temperature to become diamagnetic.

Its bond order is 2.5 and it combines with O2 to give nitrogen dioxide.

QUESTION: 45

In which of the following reactions H2O2 acts as a reducing agent?
(a) H2O2 + 2H+ + 2e– → 2H2O
(b) H2O2 – 2e– → O2 + 2H+
(c) H2O2 + 2e– → 2OH–
(d) H2O2 + 2OH– – 2e– → O2 + 2H2O
Solution:

The reducing agent oxidises itself.



Note : Powers of 'O' are oxidation number of 'O' in the compound.

QUESTION: 46

The correct statement for the molecule, CsI3, is
Solution:

It contains Cs+ and I3– ions.

QUESTION: 47

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1:4. The ratio of number of their molecule is
Solution:

Let the mass of O2 = x
Mass of N2 = 4x
Number of moles of O2 = x/32
Number of moles of N2 = 4x/28 = x/7

QUESTION: 48

Given below are the half-cell reactions


The E° for 3 Mn2+ → Mn + 2Mn3+ will b
Solution:


(1) – 2 × (2)
3Mn2+ → Mn + 2Mn3+ ;

= [2.36 – 2(–1.51)] F
= (2.36 + 3.02) F
= 5.38 F
But 
⇒ 5.38F = –2FE°
⇒ E° = –2.69 V
As E° value is negative reaction is non spontaneous.

QUESTION: 49

Which series of reactions correctly represents chemical reactions related to iron and its compound?
Solution:


This reaction is corresponding to the combustion of Fe.

These reactions correspond to the production of Fe by reduction of Fe3O4 in blast furnace.

QUESTION: 50

The equation which is balanced and represents the correct product(s) is
Solution:

The complex
[CoCl(NH3)5]+ decomposes under acidic medium, so

QUESTION: 51

In SN2 reactions, the correct order of reactivity for the following compounds
CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is
Solution:

Rate of SN2 reaction depends on steric crowding of alkyl halide. So order is
CH3Cl > (CH3)CH2 – Cl > (CH3)2CH – Cl > (CH3)3CCl

QUESTION: 52

On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is
Solution:

QUESTION: 53

The most suitable reagent for the conversion of R – CH2 – OH → R – CHO is
Solution:

PCC is mild oxidising agent, it will convert

QUESTION: 54

​The major organic compound formed by the reaction of 1, 1, 1-trichloroethane with silver powder is
Solution:


1, 1, 1-trichloroethane

QUESTION: 55

Sodium phenoxide when heated with CO2 under pressure at 125°C yields a product which on acetylation produces C.

The major product C would be
Solution:

QUESTION: 56

Considering the basic strength of amines in aqueous solution, which one has the smallest pKb value?​
Solution:

Among C6H5NH2, CH3NH2, (CH3)2NH,
(CH3)3N. C6H5NH2 is least basic due to resonance.

Out of (CH3)3N, CH3NH2, (CH3)2NH. (CH3)2NH is most basic due to +I effect and hydrogen bonding in H2O.

QUESTION: 57

For which of the following molecule significant μ ≠ 0?
Solution:


QUESTION: 58

Which one is classified as a condensation polymer?
Solution:

Dacron is polyester formed by condensation polymerisation of terephthalic acid and ethylene glycol.


Acrylonitrile, Neoprene and Teflon are addition polymers of acrylonitrile, isoprene and tetrafluoroethylene respectively.

QUESTION: 59

Which one of the following bases is not present in DNA?
Solution:

DNA contains ATGC bases
A – Adenine
T – Thymine
G – Guanine
C – Cytocine
So quinoline is not present.

QUESTION: 60

​In the reaction,the product C is
Solution:

Ethylene

QUESTION: 61

If X = {4n – 3n – 1 : n ∈ N} and Y = {9(n – 1) : n ∈ N}, where N is the set of natural numbers, then X ∪ Y is equal to​
Solution:

QUESTION: 62

If z is a complex number such that |z| ≥ 2, then the minimum value of 
Solution:



So, 

QUESTION: 63

If a ∈ R and the equation
–3(x – [x])2 + 2 (x – [x]) + a2 = 0
(where [x] denotes the greatest integer ≤ x) has no integral solution, then all possible values of a lie in the interval
Solution:

–3(x – [x])2 + 2[x – [x]) + a2 = 0
3 {x}2 – 2{x} – a2 = 0




For non-integral solution
0 < a2< 1 and a ∈ (–1, 0) ∪ (0, 1)
Alternative
–3{x}2 + 2{x} + a2 = 0
Now, –3{x}2 + 2{x}

to have no integral roots 0 < a2 < 1
∴ a∈(–1, 0) ∪ (0, 1)

QUESTION: 64

Let α and β be the roots of equation px2 + qx + r = 0, p ≠ 0. If p, q, r are in A.P. and  then the value of |α – β| is
Solution:

∵ p, q, r  are in AP
2q = p + r ...(i)
Also 

................. (ii)
From (i)
2(–4r) = p + r
p = – 9r
q = – 4r
r = r
Now 

QUESTION: 65

If α, β ≠ 0, and f(n) = αn + βn and = K(1 – α)2  (1 – β)2 (α – β)2, then K is equal to
Solution:



= [(1 – α)(1 – β)(1 – β)]2
So, k = 1

QUESTION: 66

If A is an 3 × 3 non-singular matrix such that AA′ = A′A and B =A−1 A′ , then BB′ equals
Solution:

BB ' =( A−1.A ')( A( A−1) ')
= A–1.A.A'.(A–1)1 {as AA' = A'A}
= I(A–1A)'
= I.I = I2 = I

QUESTION: 67

If the coefficients of x3 and x4 in the expansion of (1 + ax + bx2) (1 – 2x)18 in powers of x are both zero, then (a, b) is equal to
Solution:

(1 + ax + bx2) (1 – 2x)18
(1 + ax + bx2)[18C0 – 18C1(2x) + 18C2(2x)2 – 18C3(2x)3 + 18C4(2x)4 – .......]
Coeff. of x3 = –18C3.8 + a × 4.18C2 – 2b × 18 = 0

= –51 × 16 × 8 + a × 36 × 17 – 36b = 0
= –34 × 16 + 51a – 3b = 0
= 51a – 3b = 34 × 16 = 544
= 51a – 3b = 544 ... (i)
Only option number (2) satisfies the equation number (i).

QUESTION: 68

If (10)9 + 2(11)1 (10)8 + 3(11)2 (10)7+ ... + 10(11)9 = k(10)9, then k is equal to
Solution:

109 + 2⋅(11)(10)8 + 3(11)2(10)7 +... + 10(11)9 = k(10)9
x = 109 + 2⋅(11)(10)8 + 3(11)2(10)7+ ... +10(11)9
=  11⋅108 + 2⋅(11)2⋅(10)7 +... + 9(11)9 + 1110

= 109 + 11(10)8 + 112×(10)7 +... +119 – 1110

x/10 = (1110 − 1010 ) − 1110 = − 1010
 x = 1011 = k⋅109
k = 100

QUESTION: 69

Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is
Solution:

a, ar, ar2 → G.P.
a, 2ar, ar2 → A.P.
2 × 2ar = a + ar2
4r = 1 + r2
r2 – 4r + 1 = 0

r = 2+√3
r = 2−√3 is rejected
∵ (r > 1)
G.P. is increasing.

QUESTION: 70

is equal to
Solution:


QUESTION: 71

If g is the inverse of a function f and f'(x) =  then g′(x) is equal to
Solution:


QUESTION: 72

If f and g are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6, then for some c ∈]0, 1[
Solution:

Using, mean value theorem

so, f'(c) = 2g'(c)

QUESTION: 73

​If x = –1 and x = 2 are extreme points of f (x) =α log|x|+ βx2 + x then
Solution:

f(x) =α log|x|+ βx2 + x
f ′(x) = α/x + 2βx + 1= 0 at x = –1, 2
−α − 2β +1= 0 ⇒ α+ 2β =1 ...(i)
α/2 + 4β +1= 0
⇒ α+ 8β =−2 ...(ii)

6β =−3 ⇒ β =− 1/2
∴ α= 2

QUESTION: 74

The integral dx is equal to
Solution:


QUESTION: 75

The integral equals
Solution:






= 4√3 − 4− π/3

QUESTION: 76

The area of the region described by A = {(x, y) : x2 + y2 ≤ 1 and y2 ≤ 1 – x} is
Solution:


Shaded area

QUESTION: 77

​Let the population of rabbits surviving at a time t be governed by the differential equation . If p(0) = 100, then p(t) equals
Solution:



Using given condition p(t) = 400 – 300 et/2

QUESTION: 78

​Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R (7, 3). The equation of the line passing through (1, –1) and parallel to PS is
Solution:


S is mid-point of QR
So S

Slope of PS =
 
Equation of line ⇒ y – (–1) = – 2/9 ( x – 1)
9y + 9 = – 2x + 2 ⇒ 2x + 9y + 7 = 0

QUESTION: 79

​Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and 5bx + 2by + d = 0 lies in the fourth quadrant and is equidistant from the two axes then
Solution:

Let (α, -α) be the point of intersection
∴ 4aα – 2aα + c = 0 ⇒ α = -c/2a
and 5bα – 2bα + d = 0
⇒ α= 
⇒ 3bc = 2ad
⇒ 3bc – 2ad = 0
Alternative method :
The point of intersection will be

∵ Point of intersection is in fourth quadrant so x is positive and y is negative.
Also distance from axes is same
So x = – y (∵ distance from x-axis is – y as y is negative)


2ad – 2bc = – 5bc + 4ad
⇒ 3bc – 2ad = 0 ...(i)

QUESTION: 80

The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it is
Solution:

Here ellipse is  where a2 = 6, b2 = 2
Now, equation of any variable tangent is
...... (i)
where m is slope of the tangent
So, equation of perpendicular line drawn from centre to tangent is
............... (ii)
Eliminating m, we get

QUESTION: 81

Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centred at (0, y), passing through origin and touching the circle C externally, then the radius of T is equal to
Solution:



Radius of T = |y|
T touches C externally


QUESTION: 82

The slope of the line touching both the parabolas y2 = 4x and x2 = –32y is
Solution:

y2 = 4x …(1)
x2 = –32y …(2)
m be slope of common tangent Equation of tangent (1)
y = mx + 1/m …(i)
Equation of tangent (2)
y = mx + 8m2 …(ii)
(i) and (ii) are identical
1/m  = 8m2
⇒ m3 = 1/8
m = 1/2
Alternative method :
Let tangent to y2 = 4x be y =mx+ 1/m
as this is also tangent to x2 =−32 y
Solving x2 + 32mx + 32/m = 0
Since roots are equal 
∴ D = 0

QUESTION: 83

The image of the line in the plane 2x – y + z + 3 = 0 is the line
Solution:




3λ + 6 = 0 ⇒ λ = – 2
a = – 3, b = 5, c = 2
So the equation of the required line is

QUESTION: 84

The angle between the lines whose direction consines satisfy the equations l + m + n = 0 and l2 = m2 + n2 is
Solution:

l + m + n = 0
l2 = m2 + n2
Now, (–m – n)2 = m2 + n2
⇒ mn = 0
m = 0 or n = 0


∴ cos θ= 1/2
θ= π/3

QUESTION: 85

​If then λ is equal to
Solution:


QUESTION: 86

​Let A and B be two events such that wherestands for the complement of the event A. Then the events A and B are
Solution:

QUESTION: 87

The variance of first 50 even natural numbers is
Solution:

Variance = 


= 3434 – 2601
⇒σ2 = 833

QUESTION: 88

Let where x ∈R and k ≥ 1. Then f4(x) – f6(x) equals
Solution:


QUESTION: 89

A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then the speed (in m/s) of the bird is
Solution:


t = 1 s
From figure tan 45°= 20/x
and tan30°= 20/x+y
so, y = 20(√3− 1)
i.e., speed = 20(√3 − 1) m/s.

QUESTION: 90

The statement ~(p ↔ ~q) is
Solution:


Clearly equivalent to p ↔ q

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