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# JEE Main 2015 Question Paper With Solutions (4th-April-2015)

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JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 1

### Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s2) (The figures are schematic and not drawn to scale)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 1

Till both are in air (From t = 0 to t = 8 sec)

When second stone hits ground and first stone is in air Δx decreases.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 2

### The period of oscillation of a simple pendulum isMeasured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 2

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 3

### Given in the figure are two blocks A and B of weight 20 N and 100 N,  respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 3

Clearly fs = 120 N (for vertical equilibrium of the system)

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 4

A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction wth speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 4

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 5

Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius of its base is R and its height is h then z0 is equal to

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 5

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 6

From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 6

d = 2R= a√3

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 7

From a solid sphere of mass M and radius R, a spherical portion of radius R/2 is removed, as shown in the figure. Taking gravitational potential V = 0 at r = ∞, the potential at the centre of the cavity thus formed is (G = gravitational constant)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 7

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 8

A pendulum made of a uniform wire of crosssectional area A has time period .  When an additional mass M is added to its bob, the time period changes to TM. If the Young's modulus of the material of the wire is Y then 1/Y is equal to (g = gravitational acceleration)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 8

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 9

Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume and pressure.If the shell now undergoes an adiabatic expansion the relation between T and R is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 9

PV =μRT ------ (ii)

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 10

A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways :
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat.
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat.
In both the cases body is brought from initial temperature 100°C to final temperature 200°C.
Entropy change of the body in the two cases respectively is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 10

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 11

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq, where V is the volume of the gas. The value of q is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 11

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 12

For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly?
(Graphs are schematic and not drawn to scale)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 12

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 13

A train is moving on a straight track with speed 20 ms–1. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 ms–1) close to

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 13

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 14

A long cylindrical shell carries positive surface charge σ in the upper half and negative surface charge -σ in the lower half. The electric field lines around the cylinder will look like figure given in
(figures are schematic and not drawn to scale)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 14

The field line should resemble that of a dipole.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 15

A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ¥) on its surface. For this sphere the equipotential surfaces with potentials  have radius R1, R2, R3 and R4 respectively. Then

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 15

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 16

In the given circuit, charge Q2 on the 2μF capacitor changes as C is varied from 1 μF to 3 μF. Q2 as a function of C is given properly by : (Figures are drawn schematically and are not to scale)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 16

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 17

When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10–4 ms–1. If the electron density in the wire is 8 × 1028 m–3, the resistivity of the material is close to

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 17

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 18

In the circuit shown, the current in the 1 W resistor is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 18

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 19

Two coaxial solenoids of different radii carry current I in the same direction. Letbe the magnetic force on the inner solenoid due to the outer one andbe the magnetic force on the outer solenoid due to the inner one. Then

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 19

Net force on each of them would be zero.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 20

Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle θ with the vertical. If wires have mass λ per unit length then the value of I is (g = gravitational acceleration)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 20

Tcosθ = λgl ...........(1)
Tsinθ = ...........(2)

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 21

A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures below:
(a)  (b)
(c)    (d)

If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 21

Stable equilibrium

Unstable equilibrium

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 22

An inductor (L = 0.03 H) and a resistor (R = 0.15 kΩ) are connected in series to a battery of 15V EMF in a circuit shown below. The key K1 has been kept closed for a long time. Then at t = 0, K1 is opened and key K2 is closed simultaneously. At t = 1 ms, the current in the circuit will be (e ≌150)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 22

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 23

A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 23

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 24

Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is μ, a ray, incident at an angle θ, on the face AB would get transmitted through the face AC of the prism provided.

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 24

sin θ = μ sin r1
sin r1

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 25

On a hot summer night, the refractive index of air is smallest near the ground and increases with height form the ground. When a light beam is directed horizontally, the Huygen's principle leads us to conclude that as it travels, the light beam

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 25

Consider a plane wavefront travelling horizontally.
As it moves, its different parts move with different speeds. So, its shape will change as shown Þ Light bends upward

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 26

Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 26

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 27

As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 27

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 28

Match List-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the choices given below the list:

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 28

1. Franck-Hertz exp.– Discrete energy level.
2. Photo-electric effect– Particle nature of light
3. Davison-Germer exp.– Diffraction of electron beam.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 29

A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 29

Frequencies of resultant signal are

(2000 + 5) kHz, 2000 kHz, (2000 – 5) kHz, 2005 kHz, 2000 kHz, 1995 kHz

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 30

An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0 and then connected to the L and R as shown below :

If a student plots graphs of the square of maximum charge (Q2Max) on the capacitor with time (t) for two different values L1 and L2 (L1 > L2) of L then which of the following represents this graph correctly? (Plots are schematic and not drawn to scale)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 30

For a damped pendulum, A =

(Since L plays the same role as m)

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 31

The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (mol. wt. 206). What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin?

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 31

The maximum uptake =

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 32

Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 32

Edge length of BCC is 4.29 Å.
In BCC,
edge length =

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 33

Which of the following is the energy of a possible excited state of hydrogen?

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 33

Energy of excited state is negative and correspond to n>1.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 34

The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 34

H-bond is one of the dipole-dipole interaction and dependent on inverse cube of distance between the molecules.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 35

The following reaction is performed at 298 K.

The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2(g) at 298 K ? (Kp = 1.6 × 1012)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 35

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 36

The vapour pressure of acetone at 20°C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol-1) of the substance is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 36

Vapour pressure of pure acetone P°A = 185 torr
Vapour pressure of solution, PS = 183 torr
Molar mass of solvent, MA = 58 g/mole as we know

= 63.68 g/mole

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 37

The standard Gibbs energy change at 300 K for the reaction  At a given time, the composition of the reaction mixture is . The reaction proceeds in the : [R = 8.314 J/K/mol, e = 2.718]

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 37

Now
as QC > KC, hence reaction will shift in backward direction.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 38

Two faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is (at. mass of Cu = 63.5 amu)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 38

So, 2 F charge deposite 1 mol of Cu. Mass deposited = 63.5 g.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 39

Higher order (>3) reactions are rare due to

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 39

Higher order greater than 3 for reaction is rare because there is low probability of simultaneous collision of all the reacting species.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 40

3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 40

Number of moles of acetic acid adsorbed

∴ Weight of acetic acid adsorbed
= 0.9 × 60 mg = 54 mg
Hence, the amount of acetic acid adsorbed per g of charcoal =
Hence, option (1) is correct.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 41

The ionic radii (in Å) of N3- , O2- and F- are respectively

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 41

Radius of N3–, O2– and F follow order N3– >  O2– > F
As per inequality only option (3) is correct that is 1.71 Å, 1.40 Å and 1.36 Å

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 42

In the context of the Hall-Heroult process for the extraction of Al, which of the following statement is false?

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 42

In Hall-Heroult process Al2O3 (molten) is electrolyte.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 43

From the following statement regarding H2O2, choose the incorrect statement

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 43

H2O2  can be reduced or oxidised. Hence, it can act as reducing as well as oxidising agent.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 44

Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy?

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 44

BeSO4 has hydration energy greater than its lattice energy.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 45

Which among the following is the most reactive?

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 45

Because of polarity and weak bond interhalogen compounds are more reactive.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 46

Match the catalysts to the correct processes :
Catalyst                 Process
a. TiCl3          (i) Wacker process
b. PdCl       (ii) Ziegler-Natta polymerization
c. CuCl2        (iii) Contact process
d. V2O5         (iv) Deacon's process

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 46

TiCl3 - Ziegler-Natta polymerisation
V2O5 - Contact process
PdCl2 - Wacker process
CuCl2 - Deacon's process

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 47

​Which one has the highest boiling point?

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 47

Down the group strength of van der Waal's force of attraction increases hence Xe have highest boiling point.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 48

The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]+ is (py = pyridine)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 48

as per question a = Cl, b = py, c = NH3 and d = NH2OH are assumed.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 49

The color of KMnO4 is due to

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 49

Charge transfer spectra from ligand (L) to metal (M) is responsible for color of KMnO4.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 50

Assertion : Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form  oxides of nitrogen.
Reason : The reaction between nitrogen and oxygen requires high temperature.

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 50

N2 + O2 --> 2NO
Required temperature for above reaction is around 3000°C which is a quite high temperature. This reaction is observed during thunderstorm.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 51

In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is (At. mass Ag = 108; Br = 80)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 51

Percentage of Br

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 52

Which of the following compounds will exhibit geometrical isomerism?

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 52

For geometrical isomerism doubly bonded carbon must be bonded to two different groups which is only satisfied by 1 - Phenyl - 2 - butene.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 53

Which compound would give 5-keto-2-methyl hexanal upon ozonolysis?

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 53

5-keto-2-methylhexanal is

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 54

The synthesis of alkyl fluorides is best accomplished by

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 54

Swart's reaction

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 55

In the following sequence of reactions :
the product C is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 55

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 56

In the reaction  the product E is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 56

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 57

Which polymer is used in the manufacture of paints and lacquers?

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 57

Glyptal is used in manufacture of paints and lacquires.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 58

Which of the vitamins given below is water soluble?

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 58

Vitamin C is water soluble vitamin.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 59

Which of the following compounds is not an antacid?

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 59

Phenelzine is not antacid, it is anti-depressant.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 60

Which of the following compounds is not colored yellow?

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 60

(NH4)3[As (Mo3O10)4], BaCrO4 and K3[Co(NO2)6] are yellow colored compounds but Zn2[Fe(CN)6] is not yellow colored compound.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 61

Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A × B, each having at least three elements is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 61

n(A) = 4, n(B) = 2
n(A × B) = 8
Required numbers = 8C3 + 8C4 + ...... + 8C8
= 28 – (8C0 + 8C1 + 8C2)
= 256 – 37
= 219

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 62

A complex number z is said to be unimodular if |z| = 1. Suppose z1 and z2 are complex numbers such thatis unimodular and z2 is not unimodular. Then the point z1 lies on a

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 62

|z| = 2 i.e. z lies on circle of radius 2.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 63

Let a and b be the roots of equation x2 - 6x - 2 = 0. If an = αn - βn, for n ≥ 1, then the value of  is equal to

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 63

From equation,
α + β = 6
αβ = - 2
The value of

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 64

If A =  is a matrix satisfying the equation AAT = 9I, where I is 3 × 3 identity matrix, then the ordered pair (a, b) is equal to

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 64

a + 4+2b = 0
2a +2- 2b = 0
a + 1-b =0
2a -2b= -2
a + 2b= -4
-----------------
3a =-6
a =-2
-2+1 - b = 0
b = -1
a = -2
(-2, -1)

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 65

The set of all values of l for which the system of linear equations
2x1 - 2x2 + x3 = λx1
2x1 - 3x2 + 2x3 = λx2
-x1 + 2x2  = λx3
has a non-trivial solution

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 65

x1(2 - λ) - 2x2 + x3 = 0
2x1 +x2(-λ - 3) + 2x3 = 0

(2 - λ)(λ2 + 3λ- 4)+ 2(-2λ + 2)+(4 - λ- 3) = 0
+ 6λ - 8 - λ3 - 3λ2 + 4λ - 4λ + 4 -λ + 1 =0

=> λ = 1, 1, -3
Two elements.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 66

The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 66

4 digit numbers

5 digit numbers

5 × 4 × 3 × 2 × 1 = 120
Total number of integers
= 72 + 120 = 192

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 67

The sum of coefficients of integral powers of x in the binomial expansion of is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 67

Sum of coefficient of integral power of x

We know that

Then,

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 68

If m is the A.M. of two distinct real numbers l and n (l, n > 1) and G1, G2 and G3 are three geometric means between l and n, then G14 + 2G24+ G34 equals.

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 68

Now

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 69

The sum of first 9 terms of the series

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 69

= 96

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 70

is equal to

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 70

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 71

If the function.
is differentiable, the value of k + m is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 71

R.H.D.

and 3m – 2k + 2 = 0
L.H.D.

From above,
k/4 = m and 3m – 2k + 2 = 0
m = 2/5 and k = 8/5
k + m = 8/5 + 2/5 + 10/5 = 2
Alternative Answer

g is constant at x = 3
k√4 =3m+ 2
2k = 3m + 2 .......(i)
Also
k/4 = m
k = 4 m...........(ii)
8 m = 3 m + 2

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 72

The normal to the curve, x2 +2xy – 3y2 = 0 at (1,1)

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 72

Curve is x2 + 2xy – 3y2 = 0
Differentiate wr.t. x,

So equation of normal at (1, 1) is
y – 1 = – 1 (x – 1) Þ y = 2 – x Solving it with the curve, we get
x2 + 2x(2 – x) – 3(2 – x)2 = 0
–4x2 + 16x – 12 = 0
x2 – 4x + 3 = 0
x = 1, 3
So points of intersections are (1, 1) & (3, –1) i.e. normal cuts the curve again in fourth quadrant.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 73

Let f(x) be a polynomial of degree four having extreme values at x = 1 and x = 2. If = 3, then f(2) is equal to

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 73

Let f(x) = a0 + a1x + a2x2 + a3x3 + a4x4
Using

So, a0 = 0, a1 = 0, a2 = 2
i.e., f(x) = 2x2 + a3x3 + a4x4
Now, f'(x)= 4x + 3a3x2 + 4a4x3
= x[4 + 3a3x + 4a4x2]
Given, f'(1) = 0 and f'(2) = 0
=> 3a3 + 4a4 + 4 = 0 …(i)
and 6a3 + 16a4 + 4 = 0 …(ii)
Solving, a4 = 1/2, a3 = –2
i.e.,
i.e., f (2) = 0

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 74

The integral equals

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 74

Let
So,

So, option (4).

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 75

The integral  is equal to

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 75

2I = 2
I = 1

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 76

The area (in sq. units) of the region described by  is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 76

After solving y = 4x – 1 and y2 = 2x

2y2 – y – 1 = 0

y = 1, -1/2

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 77

Let y(x) be the solution of the differential equation  Then y(e) is equal to

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 77

It is best option. Theoretically question is wrong, because initial condition is not given.
x log x dy/dx + y = 2x logx
If x = 1 then y = 0

Solution is y.log x =
y log x = 2(x logx- x) + c
x = 1, y = 0
Then, c = 2, y(e) = 2

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 78

The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0), is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 78

Total number of integral coordinates as required
= 39 + 38 + 37 + ....... + 1
= = 780

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 79

Locus of the image of the point (2, 3) in the line (2x – 3y + 4) + k(x – 2y + 3) = 0, k ∈ R, is a

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 79

After solving equation (i) & (ii)
2x  –3y +  4= 0 .....(i)
2x  –4y +  6= 0 .....(ii)
x = 1 and y = 2
Slope of AB × Slope of MN = – 1

(y – 3)(y – 1) = –(x – 2)x
y2 – 4y + 3 = –x2 + 2x
x2 + y2 – 2x – 4y + 3 = 0
Circle of radius = √2

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 80

The number of common tangents to the circles x2 + y2 – 4x – 6y – 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0, is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 80

x2 + y2 – 4x – 6y – 12 = 0
C1(center) = (2,3), r =
x2+ y2 + 6x + 18y + 26 = 0
C2(center) (– 3, –9),

C1C2 = 13, C1C2 = r1 + r2
Number of common tangent is 3.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 81

The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse  , is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 81

Ellipse is
i.e., a2 = 9, b2 = 5
So, e = 2/3
As, required area

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 82

Let O be the vertex and Q be any point on the parabola, x2 = 8y. If the point P divides the line segment OQ internally in the ratio 1 : 3, then the locus of P is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 82

x2 = 8y
Let Q be (4t, 2t2)

Let P be (h, k)
∴ h = t,
∴ 2 k =h2
∴ Locus of (h, k) is x2 = 2y.

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 83

The distance of the point (1, 0, 2) from the point of intersection of the line and the
plane x – y + z = 16, is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 83

P(3λ+2 , 4λ -1, 12λ+ 2)
Lies on plane x – y + z = 16
Then,
3λ+2 - 4λ +1+12λ+ 2 = 16
11λ+ 5 = 16
λ = 1  P(5, 3, 14)
Distance =

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 84

The equation of the plane containing the line 2x – 5y + z = 3; x + y + 4z = 5, and parallel to the plane, x + 3y + 6z = 1, is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 84

Required plane is
(2x – 5y + z – 3) + λ(x + y + 4z - 5) = 0
It is parallel to x + 3y + 6z = 1

Solving λ =
∴ Required plane is
(2x – 5y + z – 3) – 11/2 (x + y + 4z – 5) = 0
∴ x + 3y + 6z – 7 = 0

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 85

Letbe three non-zero vectors such that no two of them are collinear and If θ is the angle between vectors and then a value of sinθ is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 85

∴ cos θ = - 1/3
∴ sin θ =

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 86

If 12 identical balls are to be placed in 3 identical boxes, then the probability that one the boxes contains exactly 3 balls is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 86

Question is wrong but the best suitable option is (1).
Required probability =

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 87

The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 87

Mean = 16
Sum = 16 × 16 = 256
New sum = 256 – 16 + 3 + 4 + 5 = 252
Mean == 14

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 88

If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the foot of the tower, are 30º, 45º and 60º respectively, then the ratio, AB : BC, is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 88

AO = h cot30º
= h√3
BO = h

CO = h/√3

= √3

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 89

Let  where |x| < 1/√3 x . Then a value of y is

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 89

JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 90

The negation of is equivalent to

Detailed Solution for JEE Main 2015 Question Paper With Solutions (4th-April-2015) - Question 90

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