JEE Main 2018 (Code B)

Test Name: JEE Main 2018 (Code B)

Question 1:

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively:

 Option 1: (0, 0) Option 2: (0, 1) Option 3: (.89, .28) Option 4: (.28, .89)

Solutions:Case I: For collision with deuterium Let be the velocity of neutron before collision and let v1 and v2 be the velocity of neutron and deuterium after collision. Following the law of conservation of momentum, we have

mv+0=mv1+2mv2v=v1+2v2   …..(i) Also, for elastic collision, e = 1

⇒v2-v1=v   …..(ii)Solving (i) and (ii), we havev1=-v3   …..(iii)Now, Pd=12mv2-12mv1212mv2⇒Pd=12mv2-12mv2912mv2     …..From (iii)Pd=89=0.89Case II: For collision with carbon nucleus Let be the velocity of neutron before collision and let vand v2 be the velocity of neutron and carbon nucleus after collision. Following the law of conservation of momentum, we have

mv+0=mv1+12mv2v=v1+12v2   …..(i) Also, for elastic collision, e = 1

⇒v2-v1=v   …..(ii)Solving (i) and (ii), we havev1=-11v13   …..(iii)Now, Pc=12mv2-12mv1212mv2⇒Pc=12mv2-12m121v216912mv2     …..From (iii)Pc=48169≈0.28Hence, the correct answer is option C.

Question 2:

The mass of a hydrogen molecule is 3.32 × 10–27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of 45° to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly :

 Option 1: 2.35 × 102 N/m2 Option 2: 4.70 × 102 N/m2 Option 3: 2.35 × 103 N/m2 Option 4: 4.70 × 103 N/m2

Solutions:Given: n = 1023, Mass of hydrogen molecule, m = 3.32

×10-27 kg Let v be the velocity of hydrogen molecule before and after collision (collision with the wall is given elastic). Change in momentum of 1023 hydrogen molecules per second,

∆p =

2×n×m×v×cos 45=2×1023×3.32×10-27×103×12

⇒∆p=0.47 kg-m/sForce exerted by â€‹1023 hydrogen molecules on the wall, F = × 1 s = 0.47 N Pressure exerted by â€‹1023 hydrogen molecules on the wall of area 2 cm2,

P =FA=0.472104=2.35×103 N/m2Hence, the correct answer is option C.

Question 3:

A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere,

drr, is;

 Option 1: mg3Ka Option 2: mgKa Option 3: Kamg Option 4: Ka3mg

Solutions:Increase in pressure on sphere when mass m is place on the surface of piston,

∆P =mgaNow, we know bulk modulus is given as

K =∆P∆VVNow, ∆VV=4πr2dr43πr3=3drr⇒K =∆P3drror, drr=∆P3K=mg3aKHence, the correct answer is option A.

Question 4:

Two batteries with e.m.f. 12 V and 12 V are connected in parallel across a load resistor of 10 Ω. The internal resistances of the two batteries are 1 Ω and 2 Ω respectively. The voltage across the load lies between :

 Option 1: 11.4 V and 11.5 V Option 2: 11.7 V and 11.8 V Option 3: 11.6 V and 11.7 V Option 4: 11.5 V and 11.6 V

Solutions:Let V be the voltage across the load. Thus, using nodal analysis, we have

V-132+V-121=V10⇒V=11.56 VHence, the correct answer is option D.

Question 5:

A particle is moving in a circular path of radius a under the action of an attractive potential

U=-k2r2. Its total energy is :

 Option 1: Zero Option 2: -32 ka2 Option 3: -k4 a2 Option 4: k2 a2

Solutions:Given, P.E., Force acting on the particle, Now,

kr3=

mv2r where, v is the velocity of the particle.

⇒v2=kmr2∴Kinetic energy of the particle, K.E. =12×m×kmr2=k2r2Now, total energy = K.E. + P.E. =

k2r2-k2r2=0Hence, the correct answer is option A.

Question 6:

Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is : Option 1: 43.3 kg Option 2: 10.3 kg Option 3: 18.3 kg Option 4: 27.3 kg

Solutions: For m1 to be at rest,

T=m1g = 5g For (m2 + m) to be at rest, f = T = 5g    …..(i) Now,

f≤μ(m2 + m)gFrom (i), we get5g≤μ(m2 + m)g5μ-m2≤mm≥50.15-10m≥23.33 kg

Hence, the correct answer is option D.

Question 7:

If the series limit frequency of the Lyman series is vL, then the series limit frequency of the P fund series is :

 Option 1: vL / 16 Option 2: vL / 25 Option 3: 25 vL Option 4: 16 vL

Solutions:For series limit of Lyman series, n1 = 1 and n2 . Thus,

νL=RcZ2112-1∞=RcZ2For series limit of Pfund series, n= 5 and n2 = . Thus,

νp=RcZ2152-1∞=RcZ225=νL25Hence, the correct answer is option B.

Question 8:

Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is found to be

I2. Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be

I8. The angle between polarizer A and C is :

 Option 1: 45° Option 2: 60° Option 3: 0° Option 4: 30°

Solutions:When an unpolarized light passes through a polariser for the first time, its intensity reduces by half. Thus, after passing through A, intensity of light becomes I/2. Now, after passing through B,

I2=I2cos2θ⇒θ=0oThus, A and B polarisers are placed parallel to each other. Now, when another identical polariser C is placed between A and B, intensity of light is found to be

I8. Thus,

I8=I2cos4θ⇒cos4θ=14cosθ=12θ=45oHence, the correct answer is option A.

Question 9:

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λn, λg be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let

∧nbe the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)

 Option 1: ∧n2≈A+Bλn2 Option 2: ∧n2≈λ Option 3: ∧n≈A+Bλn2 Option 4: ∧n≈A+Bλn

Solutions:

1∧n=RZ2112-1n2∧n=1RZ2112-1n2-1As n is very large, thus using binomial expansion∧n=1RZ2112+1n2∧n=1RZ2+1RZ21n2∧n = A+Bλn2      …..As, λn=2πn2h24π2mZe21n ∝ nHence, the correct answer is option C.

Question 10:

The reading of the ammeter for a silicon diode in the given circuit is : Option 1: 11.5 mA Option 2: 13.5 mA Option 4: 15 mA

Solutions:The silicon diode is forward biased. The knee/cut-in voltage of silicon diode is 0.7 V. Thus, I=3-0.7200=11.5 mA

Hence, the correct answer is option A.

Question 11:

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii re, rp, rα respectively in a uniform magnetic field B. The relation between re, rp, rα is :

 Option 1: re < rp < rα Option 2: re < rα < rp Option 3: re > rp = rα Option 4: re < rp = rα

Solutions:Force on a charged particle of charge q and mass m moving with velocity v in a magnetic field B = qvB  But,

qvB =mv2r where r is the radius of the circular path traversed by the particle in the magnetic field. Thus,

r =mvqB    …..(i)Now, kinetic energy, K=mv22⇒v=2Km    …..(ii)Putting (ii) in (i), we haver =m2KmqB=2KmqBThus, re =2KmeeBrα =2K×4mp2eBrp =2KmpeBHence, rp= rα>reHence, the correct answer is option D.

Question 12:

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant

K=53is inserted between the plates, the magnitude of the induced charge will be :

 Option 1: 2.4 nC Option 2: 0.9 nC Option 3: 1.2 nC Option 4: 0.3 nC

Solutions:Initial charge on the capacitor, Qi = CV=90×20=1800 pCAfter the insertion of dielectric material, the charge on the capacitor plateQf=KCV=53×1800=3000 pCThus, induced charge on the platesQ=Qf-Qi =3000-1800=1200 pC=1.2 nCHence, the correct answer is option C.

Question 13:

For an RLC circuit driven with voltage of amplitude νm and frequency

ω0=1LCthe current exibits resonance. The quality factor, Q is given by :

 Option 1: Rω0 C Option 2: CRω0 Option 3: ω0 LR Option 4: ω0 RL

Solutions:The quality factor, Q of RLC circuit is Q =

ωoBW Now, BW =

RLQ =

ωoLRHence, the correct answer is option C.

Question 14:

A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz?

 Option 1: 2 × 105 Option 2: 2 × 106 Option 3: 2 × 103 Option 4: 2 × 104

Solutions:Carrier bandwidth utilised for transmission, fc=10100×10×1010=1010 HzNumber of channels transmitted simultaneously,

n=Total bandwidth available for transmissionChannel bandwidth=fc5×103=10105×103=2×105 HzHence, the correct answer is option A.

Question 15:

A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 × 103 kg / m3 and its Young’s modulus is 9.27 × 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations?

 Option 1: 10 kHz Option 2: 7.5 kHz Option 3: 5 kHz Option 4: 2.5 kHz

Solutions: The fundamental frequency of the longitudinal vibrations,

f=12lYρ=12×0.69.27×10102.7×103≈5 kHzHence, the correct answer is option C.

Question 16:

Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is: Option 1: 732MR2 Option 2: 1812MR2 Option 3: 192MR2 Option 4: 552MR2

Solutions:

Io=MR22+6MR22+M(2R)2=MR212+3+24=552MR2Now, O is the centre of mass of the system. Applying parallel axis theorem between O and P,IP=Io+7M(3R)2=552MR2+63MR2=1812MR2Hence, the correct answer is option B.

Question 17:

Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities + σ, – σ and + σ respectively. The potential of shell B is:

 Option 1: σ∈0b2-c2b+a Option 2: σ∈0b2-c2c+a Option 3: σ∈0a2-b2a+c Option 4: σ∈0a2-b2b+c

Solutions: The potential of shell B is

VB=KQAb+KQBb+KQCc     =K(+σ)(4πa2)b+ (-σ)(4πb2)b+(+σ)(4πc2)c     =4πσ4πεoa2b-b2b+c2c=σεoa2-b2b+cHence, the correct answer is option D.

Question 18:

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.

 Option 1: 2 Ω Option 2: 2.5 Ω Option 3: 1 Ω Option 4: 1.5 Ω

Solutions:Let the internal resistance of the cell be r. Here, l1 = 52 cm l2 = 40 cm R = 5 Ω Using the formula,

r = Rl1l2-1we get the internal resistance of the cell as

r =55240-1=1.5 ΩHence, the correct answer is option D.

Question 19:

An EM wave from air enters a medium. The electric fields are

E→1=E01 x^ cos 2πν zc-tin air and

E→2=E02 x^ cos k 2 z-ctin medium, where the wave number k and frequency v refer to their values in air. The medium is non-magnetic. If

∈r1 and ∈r2  refer to relative permittivities of air and medium respectively, which of the following options is correct?

 Option 1: ∈r1∈r2=14 Option 2: ∈r1∈r2=12 Option 3: ∈r1∈r2=4 Option 4: ∈r1∈r2=2

Solutions:From the wave equations, In air: ω=2πν, k=2πνc⇒k=ωcIn medium: ω=kc, k’=2k⇒k’=2ωcLet speed of wave in the medium be c‘. Thus, k’=ωc’

⇒2ωc=ωc’c’=c2⇒1μoμr2εoεr2=12μoμr1εoεr1Since, medium is non-magnetic,⇒μr2=μr1=1Thus, εr1εr2=14Hence, the correct answer is option A.

Question 20:

The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of he slit is 1 µm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.)

 Option 1: 75 µm Option 2: 100 µm Option 3: 25 µm Option 4: 50 µm

Solutions:Semi-angular width = 30o

⇒asin30=λNow, fringe width, β=λDd⇒10-2=10-6×12×0.5dd=52×10-5=25 μmHence, the correct answer is option C.

Question 21:

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number = 6.02 × 1023 gm mole–1)

 Option 1: 2.2 N/m Option 2: 5.5 N/m Option 3: 6.4 N/m Option 4: 7.1 N/m

Solutions:Mass of 1 atom of silver=

108×10-36.02×1023 kg=1.794×10-25 kgTime period, =

2πmk

11012 =

2π1.794×10-25k ⇒= 7.176 N/m

Hence, the correct answer is option D.

Question 22:

From a uniform circular disc of radius R and mass 9 M, a small disc of radius

R3is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is : Option 1: 10 MR2 Option 2: 379MR2 Option 3: 4 MR2 Option 4: 409MR2

Solutions:Moment of inertia of small disc about center of bigger disc =

12MR32+M2R32=MR218+4MR29=MR22Moment of inertia of remaining portion =

9MR22-MR22=4MR2Hence, the correct answer is option C.

Question 23:

In a collinear collision, a particle with an initial speed ν0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :

 Option 1: ν02 Option 2: ν02 Option 3: ν04 Option 4: 2 ν0

Solutions: Following conservation of linear momentum, mv= mvmv2 vvv2       …..(i)

KE KE=

mv122+mv222Given, 1.5 KE= KEf

Question 24:

The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B2. The ratio

B1B2is :

 Option 1: 2 Option 2: 12 Option 3: 3 Option 4: 3

Solutions:Let the radius of loop be R. I R= m For doubling the dipole moment keeping the currect same, radius should become

2R. B1=

μ0I2RB2=

μ0I22R=B12

B1B2=2Hence, the correct asnwer is option A.

Question 25:

The density of  a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is :

 Option 1: 4.5% Option 2: 6% Option 3: 2.5% Option 4: 3.5%

Solutions:Given,

∆mm×100=1.5 %, ∆ll×100=1 %Density,

ρ=mv=ml3

∆ρρ=∆mm+3∆ll∆ρρ×100=∆mm×100+3∆ll×100                    =1.5+3×1=4.5 %Hence, the correct answer is option A.

Question 26:

On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 kΩ. How much was the resistance on the left slot before interchanging the resistances?

 Option 1: 550 Ω Option 2: 910 Ω Option 3: 990 Ω Option 4: 505 Ω

Solutions:Let the resistances on the left and right slots be R1 and R2, respectively, before interchanging the resistances. R1+R2=1000 Ω Initially,

R1R2=xl-xFinally,

R2R1=x-0.1ll-x+0.1l=x-0.1l1.1l-x

x-0.1l1.1l-x=l-xxx2-0.1lx=1.1l2-1.1lx-lx+x22lx=1.1l2x=0.55l

R1R2=0.550.45=119As, R1+R2=1000 ΩR1=550 Ω, R2=450 ΩHence, the correct answer is option A.

Question 27:

In an a.c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t

i=20 sin 30 t-π4In one cycle of a.c., the average power consumed by the circuit and the watt-less current are, respectively:

 Option 1: 502, 0 Option 2: 50, 0 Option 3: 50, 10 Option 4: 10002, 10

Solutions:From e and i, power factor = cos

π4=

12Power consumed = erms irms cos

π4                         =

1002×202×12=10002Watt-less current= irms

×sin

π4 =

202×12=10 AHence, the correct answer is option D.

Question 28:

All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

 Option 1: Option 2: Option 3: Option 4: Solutions:Graphs 1, 2, 3 can represent the motion of a ball thrown vertically upwards. Initially speed decreases with time, becomes 0 at an instant and then increases with time. Graph 4 does not explain this motion.

Hence, the correct answer is option D.

Question 29:

Two moles of an ideal mono-atomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

 Option 1: (a) 189 K (b) – 2.7 kJ Option 2: (a) 195 K (b) 2.7 kJ Option 3: (a) 189 K (b) 2.7 kJ Option 4: (a) 195 K (b) – 2.7 kJ

Υ-1 = constant For mono-atomic gas, =

53300

×V53-1 = T2

×(2V)53-1T=

1223 ×300 = 189 K

∆U=nCv∆T=2×32×8.314×(189-300)=-2.7 kJHence, the correct answer is option A.

Question 30:

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then :

 Option 1: T ∝ R(n + 1)/2 Option 2: T ∝ Rn/2 Option 3: T ∝ R3/2 for any n. Option 4: T∝Rn2+1

Solutions:

F∝1Rn

mv2R∝1Rnv2∝1Rn-1v ∝ R1-n2T =

2πRv

T∝RvT∝RR1-n2T∝ R1-1-n2T∝R1+n2Hence, the correct answer is option A.

Question 31:

If the tangent at (1, 7) to the curve x2 = y – 6 touches the circle x2 + y2 + 16x + 12y + c = 0 then the value of c is :

 Option 1: 85 Option 2: 95 Option 3: 195 Option 4: 185

Solutions: Equation of the curve:

x2=y-6Differentiating the given equation we get

2x=dydxThe tangent to the curve is at point (1, 7). So,

dydx1,7=2×1=2Equation of the tangent is thus,

y-7=2x-1⇒2x-y+5=0The equation of the circle touching the tangent is

x2+y2+16x+12y+c=0.

⇒x2+16x+y2+12y+c=0⇒x2+16x+64+y2+12y+36+c=0+64+36⇒x+82+y+62=100-c⇒x–82+y–62=100-cThus, centre of the circle is

100-c. Distance between the centre of the circle

-8,-6 and the tangent 2x − y + 5 = 0 is the radius

100-c.

-8×2+-1×-6+522+-12=100-c⇒-16+6+55=100-c⇒-55=100-c⇒5=100-cSquaring both sides we get5=100-c⇒c=95Hence, the correct answer is option B.

Question 32:

If L1 is the line of intersection of the planes 2x – 2y + 3z – 2 = 0, xy + z + 1 = 0 and L2 is the line of intersection of the planes x + 2yz – 3 = 0, 3xy + 2z – 1 = 0, then the distance of the origin from the plane, containing the lines L1 and L2, is :

 Option 1: 122 Option 2: 12 Option 3: 142 Option 4: 132

Solutions:Given: line Lpasses through the intersection of the planes

2x-2y+3z-2=0 and

x-y+z+1=0So,

2+λx-2+λy+3+λz-2+λ=0                          …..(i) Also, line Lpasses through the intersection of the planes

x+2y-z-3=0 and 3x-y+2z-1=0So,

1+3μx-2-μy+2μ-1z-3-μ=0                        …..(ii) From (i) and (ii) we have

2+λ1+3μ=-2+λ2-μ⇒2+λ2-μ=-2+λ1+3μ⇒μ-2=1+3μ⇒-3=2μ⇒μ=-32Putting the value of

μ=-32 in (i) we get

1-92x+2+32y+-3-1z-3+32=0⇒7x-7y+8z+3=0                                      …..iiiDistance of the origin from the plane containing the lines Land L2 will be the distance between equation (iii) and the origin. So,

d=Ax1+By1+Cz1+DA2+B2+C2=7×0+-7×0+8×0+372+-72+82=3162=132Hence, the correct answer is option D.

Question 33:

If α, β ∈ C are the distinct roots, of the equation x2x + 1 = 0, then α101 + β107 is equal to :

 Option 1: 1 Option 2: 2 Option 3: –1

Solutions:Given:

α,

β are the distinct roots of the equation

x2-x+1=0

ω is a non-real cube root of unity. So,

α=-ω and β=-ω2

α101+β107=-ω101+-ω2107=-ω99×ω2+-ω214=-ω333×ω2+-ω213×ω=-ω333×ω2+-ω371×ω=-1×ω2+-1×ω                               Since ω3=1=-ω+-ω2=-ω+ω2                                               Since ω2+ω+1=0=–1=1Hence, the correct answer is option A.

Question 34:

Tangents are drawn to the hyperbola 4x2y2 = 36 at the points P and Q. If these tangents intersect at the point T(0, 3) then the area (in sq. units) of ΔPTQ is :

 Option 1: 603 Option 2: 365 Option 3: 455 Option 4: 543

Solutions: The given hyperbola is

4×2-y2=36                    …..(1)

⇒x29-y236=1Tangents are drawn at the points P and Q. Line PQ is the chord of contact and we know that the chord of contact can be written as T = 0. So,

4×2-y2=36⇒4x×x-y×y=36⇒4xx1-yy1=36⇒4×0-y3=36                          Since x1=0, y1=3⇒0-3y=36⇒y=-12 Putting the value of y = −12 in (1) we get

4×2–122=36⇒4×2-144=36⇒4×2=180⇒x=±35So, the points where the tangents meet the hyperbola are P

-35,-12 and Q

35,-12. Distance between the point T(0, 3) and the line y = −12 will be 3 + 12 = 15 units PQ =

2×35=65 units

Area of △TPQ=12×PQ×TS=12×65×15=455 square units. Hence, the correct answer is option C.

Question 35:

If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other at right angles, then the value of b is :

 Option 1: 4 Option 2: 92 Option 3: 6 Option 4: 72

Solutions:The given curves are

y2=6x                    …..(1) and

9×2+by2=16                                      …..(2) Slope of curve (1) will be

2ydydx=6⇒dydx=3y=m1Differentiating (2) we get

18x+2bydydx=0⇒9x+bydydx=0⇒dydx=-9xby=m2The two curves intersect each other at right angles so,

m1m2=-1⇒3y×-9xby=-1⇒27xby2=1⇒27x=by2⇒b=92 Hence, the correct answer is option B.

Question 36:

If the system of linear equations x + ky + 3z = 0 3x + ky – 2z = 0 2x + 4y – 3z = 0 has a non-zero solution (x, y, z) then

xzy2is equal to :

 Option 1: –30 Option 2: 30 Option 3: –10 Option 4: 10

Solutions:The system of linear equations given are x + ky + 3z = 0 3x + ky – 2z = 0 2x + 4y – 3z = 0 For these equations to have a non-zero solution

1k33k-224-3=0⇒1-3k+8-k-9+4+312-2k=0⇒-3k+8+5k+36-6k=0⇒44-4k=0⇒k=11The equations will become x + 11y + 3z = 0                  …..(1) 3x + 11y – 2z = 0                …..(2) 2x + 4y – 3z = 0                  …..(3) Adding (1) and (3) we get 3x + 15y = 0 ⇒ x + 5y = 0 ⇒ x = −5y                           …..(4) Putting the value of x in (3) −10y + 4y −3z = 0 ⇒ −6y − 3z = 0 ⇒ −2yz = 0 ⇒ z = −2y We need to find the value of

xzy2

⇒xzy2=-5y-2yy2=10y2y2=10Hence, the correct answer is option D.

Question 37:

Let

S=x∈R : x ≥0and

2x-3+xx-6+6=0.Then S :

 Option 1: contains exactly two elements. Option 2: contains exactly four elements. Option 3: is an empty set. Option 4: contains exactly one element.

Solutions:Case I:

x-3≥0⇒x≥3⇒x≥9

2x-3+xx-6+6=0⇒2x-3+x-6x+6=0⇒2x-6+x-6x+6=0⇒x-4x=0⇒x=4x⇒x2-16x=0⇒xx-16=0⇒x=0,16But

x≥9 so, x = 16 Case II:

x-3<0⇒x<3

2x-3+xx-6+6=0⇒23-x+x-6x+6=0⇒6-2x+x-6x+6=0⇒12-8x+x=0⇒x-6x-2=0⇒x=6,2But

x<3 so,

x=2⇒x=4Thus, x = 16, 4. So, these are exactly two elements in S. Hence, the correct answer is option A.

Question 38:

If sum of all the solutions of the equation

8 cos x·cosπ6+x· cosπ6-x-12=1in [0, π] is kπ, then k is equal to :

 Option 1: 89 Option 2: 209 Option 3: 23 Option 4: 139

Solutions:Given:

8cosx·cosπ6+x·cosπ6-x-12=1

⇒8cosx·cosπ6+x·cosπ6-x-12=1⇒8cosx·cosπ62+cos2x2-12=1⇒8cosx·122+cos2x2-12=1⇒8cosx·14+cos2x2-12=1

⇒8cosx·cos2x2-14=1⇒2cosx2cos2x-1=1⇒2cosx22cos2x-1-1=1⇒2cosx4cos2x-3=1⇒24cos3x-3cosx=1⇒2cos3x=1⇒cos3x=12

⇒3x=2nπ±π3⇒x=2nπ3±π9x∈0,π so, x=π9,2π3+π9,2π3-π9Sum =

13π9

kπ=13π9⇒k=139 kπ=13π9⇒k=139Hence, the correct answer is option D.

Question 39:

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :

 Option 1: 15 Option 2: 34 Option 3: 310 Option 4: 25

Solutions:Number of red balls = 4 Number of black balls = 6 Total balls = 4 + 6 = 10 Probability of selecting a red ball =

410=25Probability of selecting a black ball =

610=35Case 1: when a red ball is selected When a red ball is selected 2 additional red balls are put back So, total red balls = 4 + 2 = 6 Total black balls = 6 Total balls = 12 Probability of selecting a red ball =

612=12Case 2: when a black ball is selected When a black ball is selected, 2 additional black balls are put back Total red balls = 4 Total black balls = 6 + 2 = 8 Total balls = 12 Probability of selecting the red ball =

412=13Required probability =

25×12+35×13=15+15=25Hence, the correct answer is option D.

Question 40:

Let

fx=x2+1x2and

gx=x-1x, x∈R–1, 0, 1.If

hx=fxgx,then the local minimum value of h(x) is :

 Option 1: -22 Option 2: 22 Option 3: 3 Option 4: –3

Solutions:

fx=x2+1×2,

gx=x-1x,

hx=fxgx=x2+1x2x-1x=x-1×2+2x-1xLet

x-1x=p

⇒ht=p2+2p=p+2ph’t=1-2p2

h’t=0⇒1-2p2=0⇒1=2p2⇒p2=2⇒p=±2So, the critical points are

±2Local minimum value occurs at

p=2and the local minimum value of h(x) is

h2=2+22=2+2=22Hence, the correct answer is option B.

Question 41:

Two sets A and B are as under :

A = {(a, b) ∈ R × R : |a – 5| < 1 and |b – 5| < 1};

B = {(a, b) ∈ R × R : 4(a – 6)2 + 9 (b – 5)2 ≤ 36}. Then :

 Option 1: A ∩ B = Ï• (an empty set) Option 2: neither A ⊂ B nor B ⊂ A Option 3: B ⊂ A Option 4: A ⊂ B

Solutions:A = {(ab) ∈ R × R : |a – 5| < 1 and |b – 5| < 1}; |a – 5| < 1 Let

a-5=p⇒p<1 ⇒p<1                  …..(1) and |b – 5| < 1 Let

b-5=q⇒q<1 ⇒q<1                  …..(2)

Also, B = {(ab) ∈ R × R : 4(a – 6)2 + 9 (b – 5)2 ≤ 36} 4(a – 6)2 + 9 (b – 5)2 ≤ 36

(a – 6)2 9+(b – 5)24≤1

p-129+q24≤1              From 1 and 2This equation represents an ellipse. And since we have

p<1 and q<1 which represents a rectangle so they lie inside the ellipse. Hence, the correct answer is option D.

Question 42:

The Boolean expression ~ (p ∨ q) ∨ (~ p ∧ q) is equivalent to :

 Option 1: q Option 2: ~q Option 3: ~p Option 4: p

Solutions:Given: ~ (p ∨ q) ∨ (~ p ∧ q)

 p q p∨q ~p∨q ~p∧q ~ (p ∨ q) ∨ (~ p ∧ q) ~p T T T F F F F T F T F F F F F T T F T T T F F F T F T T

Hence, the correct answer is option C.

Question 43:

Tangent and normal are drawn at P(16, 16) on the parabola y2 = 16x, which intersect the axis of the parabola at A and B, respectively. If C is the centre of the circle through points P, A and B and ∠CPB = θ, then a value of tan θ is :

 Option 1: 3 Option 2: 43 Option 3: 12 Option 4: 2

Solutions: Equation of the parabola:

y2=16xThe tangent passes through the point P(16, 16) and meets the axis of the parabola at point A. The coordinate of the point A will be (−16, 0). Slope of AP =

16-016–16=1632=12Equation of the tangent will be

y-16=12x-16⇒2y-32=x-16⇒x-2y+16=0Slope of normal =

-2Equation of the normal will be

y-16=-2x-16⇒y-16=-2x+32⇒2x+y-48=0Coordinate of point B will be satisfying the equation of normal so,

2x+0-48=0⇒2x=48⇒x=24So, B(24, 0) is the point where the normal intersects the x-axis. Given that C is the centre of the circle through the point A, B and P so,

Ca,b=-16+242,0⇒Ca,b=4,0Slope of PC,

mPC =

16-016-4=1612=43 Slope of PB,

mPB =

16-016-24=16-8=-2

tanθ=mPC-mPB1+mPCmPB=43–21+43×-2=-2=2Hence, the correct answer is option D.

Question 44:

If

x-42x2x2xx-42x2x2xx-4=A+Bx x-A2,then the ordered pair (A, B) is equal to :

 Option 1: (–4, 5) Option 2: (4, 5) Option 3: (–4, –5) Option 4: (–4, 3)

Solutions:Given:

x-42x2x2xx-42x2x2xx-4=A+Bx x-A2, C1→C1+C2+C3=

5x-42x2x5x-4x-42x5x-42xx-4

=5x-412x2x1x-42x12xx-4R3→R3-R1,R2→R2-R1=5x-412x2x0-x-4000-x-4=5x-4-4-x2=A+Bxx-A2So,

A=-4,B=5Hence, the correct answer is option A.

Question 45:

The sum of the co-efficients of all odd degree terms in the expansion of

x+x3-15+x-x3-15, x>1is :

 Option 1: 1 Option 2: 2 Option 3: –1

Solutions:Given:

x+x3-15+x-x3-15, x>1

Let x3-1=p⇒x3-1=p2

x+p5+x-p5=C05x5+C05x4p+…+C05p5+C05x5-C05x4p+…-C05p5=2C05x5+C25x3p2+C45xp4=2×5+10x3x3-1+5xx3-12                   From i=2×5+10×6-10×3+5×7+5x-10x4We take only the odd degree terms2x5,-20×3,+10×7,10xSum of the odd degree terms will be=2-20+10+10=2Hence, the correct answer is option B.

Question 46:

Let a1, a2, a3, …., a49, be in A.P. such that

∑k=012a4k+1=416 and a9+a43=66.If

a12+a22+….+a172=140 m,then m is equal to :

 Option 1: 34 Option 2: 33 Option 3: 66 Option 4: 68

Solutions:We have,

∑k=012a4k+1=416⇒a1+a5+a9+…+a49=416⇒a+a+4d+a+8d+…+a+48d=416⇒1322a+13-14d=416    ⇒a+24d=32        …..iAnd,

a9+a43=66⇒a+8d+a+42d=66⇒a+25d=33        …..iiNow solving (i) and (ii) we get,

a=8 and d=1

a12+a22+…+a172=140m⇒∑ar2r=117=140m⇒∑r=117a+r-1d2=140m⇒∑r=1178+r-1.12=140m⇒∑r=117r+72=140m⇒82+92+…+242=140m⇒4760=140m⇒m=34Hence, the correct answer is option A.

Question 47:

A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is :

 Option 1: 3x + 2y = xy Option 2: 3x + 2y = 6xy Option 3: 3x + 2y = 6 Option 4: 2x + 3y = xy

Solutions:Let P(a, 0), Q(0, b) and R(a, b). Equation of PQ is

xa+yb=1 Now, if PQ pass through (2, 3), then

2a+3b=1⇒3a+2b=abSo locus of R is 3x+2y=xy Hence, the correct answer is option A.

Question 48:

The value of

∫-π2π2sin2x1+2x dxis :

 Option 1: 4π Option 2: π4 Option 3: π8 Option 4: π2

Solutions:

Let I=∫-π2π2sin2x1+2xdx     …i⇒I=∫-π2π2sin2-x1+2-xdx        ∵∫abfxdx=∫abfa+b-xdx⇒I=∫-π2π22x.sin2x1+2xdx       …iiAdding i and ii,2I=∫-π2π2sin2xdx2I=2∫0π2sin2xdx2I=2×2-14sin2x0π22I=π2I=π4Hence, the correct answer is option B.

Question 49:

Let

gx=cos x2, fx=x,and α, β (α < β) be the roots of the quadratic equation 18x2 – 9πx + π2 = 0. Then the area (in sq. units) bounded by the curve y = (gof) (x) and the lines x = α, x = β and y = 0, is :

 Option 1: 123-2 Option 2: 122-1 Option 3: 123-1 Option 4: 123+1

Solutions:

18×2-9πx+π2=0⇒6x-π3x-π=0⇒x=π6,π3So, α=π6 and β=π3We also have,

fx=x and gx=cosx2∴gofx=cosxNow, the area bounded by the curve y = cosx and line

x=π6,

x=π3and y = 0.

Required area=∫π6π3cosxdx                          =sinxπ6π3                          =123-1 sq unitsHence, the correct answer is option C.

Question 50:

For each tR, let [t] be the greatest integer less than or equal to t. Then

limx→0+x1x+2x+….+15x

 Option 1: is equal to 120. Option 2: does not exist (in R). Option 3: is equal to 0. Option 4: is equal to 15.

Solutions:

limx→0+x1x+2x+…+15x=limx→0+x1x-1x+2x-2x+…+15x-15x    ∵for any GIF x=x-x=limx→0+x1+2+…+15x-limx→0+x1x+2x+…+15x=120-0     ∵0≤x<1=120Hence, the correct answer is option A.

Question 51:

If

∑i=19xi-5=9and

∑i=19xi-52=45,then the standard deviation of the 9 items x1, x2, …., x9 is :

 Option 1: 2 Option 2: 3 Option 3: 9 Option 4: 4

Solutions:We have,

∑i=19xi-5=9⇒∑i=19xi-9×5=9⇒∑i=19xi=54      …i

∑i=19xi-52=45⇒∑i=19xi2-10∑i=19xi+25=45⇒∑i=19xi2-10×54+9×25=45⇒⇒∑i=19xi2=360      …ii

Var(X)=1n∑i=1nxi2-1n∑i=1nxi2            =3609-5492            =40-36            =4∴Standard Deviation=Var=4=2Hence, the correct answer is option A.

Question 52:

The integral

∫sin2x cos2xsin5x+cos3x sin2x+sin3x cos2x+cos5x2dxis equal to :

(where C is a constant of integration)

 Option 1: 11+cot3x+C Option 2: -11+cot3x+C Option 3: 131+tan3x+C Option 4: -131+tan3x+C

Solutions:We have,

∫sin2x cos2xsin5x+cos3x sin2x+sin3x cos2x+cos5x2dx

=∫sin2x cos2xsin2xsin3x+cos3x+cos2xsin3x +cos3x2dx=∫sin2x cos2xsin3x+cos3xsin2x +cos2x2dx=∫sin2x cos2xsin3x+cos3x2dxDividing the numerator and denominator by cos6x.

∫tan2x sec2x1+tan3x2dxLet 1+tan3x=t3tan2x.sec2xdx=dt=13∫1t2dt=-13t+C∴I=-131+tan3x+CHence, the correct answer is option D.

Question 53:

Let

S=t∈R : fx=x-π·ex-1 sin x is not differentiable at t. Then the set S is equal to :

 Option 1: {π} Option 2: {0, π} Option 3: Ï• (an empty set) Option 4: {0}

Solutions:We have,

fx=x-π·ex-1 sin xFor this function we need to check differentiablty at

x=π and x=0. At x = π

LHD at x = π=limx→π-fx-f0x-π                           =limh→0π-h-πeπ-h-1sinπ-h-0π-h-π                           =0RHD at x = π=limx→π+fx-f0x-π                           =limh→0π+h-πeπ+h-1sinπ+h-0π+h-π                           =0âˆµ LHD = RHD so function is differentiable at x = π. At x = 0

LHD at x = 0=limx→0-fx-f0x-0                           =limh→00-h-πe0-h-1sin0-h-00-h-0                           =0RHD at x = 0=limx→0+fx-f0x-0                           =limh→00+h-πe0+h-1sin0+h-00+h-0                           =0âˆµ LHD = RHD so function is differentiable at x = 0. So, set S is an empty set. Hence, the correct answer is option C.

Question 54:

Let y = y (x) be the solution of the differential equation

sin xdydx+y cos x=4x, x∈0, π. If

yπ2=0,then

yπ6is equal to :

 Option 1: -89π2 Option 2: -49π2 Option 3: 493π2 Option 4: -893π2

Solutions:

dydx+y cotx=4xcosecx,Clearly, it is a linear differential equation of the form

dydx+Py=Q, where P=cotx and Q= 4xcosecxNow,

I.F.=e∫cotxdx=elogsinx=sinxUsing:

yI.F.=∫QI.F.dx+C

ysinx=∫4xdx+C⇒ysinx=2×2+CFor

yπ2=0,we get

0=π22+C⇒C=-π22Now,

y=2×2-π22sinx

yπ6=2π62-π22sinπ6=-89π2Hence, the correct answer is option A.

Question 55:

Let

u→be a vector coplanar with the vectors

a→=2i^+3j^−k^and

b→=j^+k^. If

u→is perpendicular to

a→and

u→·b→=24, then

u→2is equal to :

 Option 1: 256 Option 2: 84 Option 3: 336 Option 4: 315

Solutions:Let

u→=xi^+yj^+zk^. If

u→be a vector coplanar with the vectors

a→and

b→. Then,

u→ a→ b→=0⇒xyz23-1011=0⇒2x-y+z=0        …iIf

u→is perpendicular to

a→. Then,

u→·a→=0⇒2x+3y-z=0       …iiAnd,

u→·b→=24

⇒y+z=24      …iiiFrom (i), (ii) and (iii) we get, x = −4, y = 8 and z = 16

u→=-4i^+8j^+16k^u→2=42+82+162=336Hence, the correct answer is option C.

Question 56:

The length of the projection of the line segment joining the points (5, –1, 4) and (4, –1, 3) on the plane, x + y + z = 7 is :

 Option 1: 13 Option 2: 23 Option 3: 23 Option 4: 23

Solutions: Direction ratio of

AB→are

5-4, -1+1, 4-3 i.e.1, 0, 1. Let θ be the angle between

AB→and normal to the plane. Vector normal to the plane is

i^+j^+k^. ∴ Direction ratio of the normal to plane are (1, 1, 1). Now,

cosθ=1×1+0×1+1×112+02+1212+12+12=223=23CD is the length of projection of

AB→on the plane.

Projection of AB→ on the plane =AB→sinθ                                                       =21-23=23Hence, the correct answer is option B.

Question 57:

PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45°, 30° and 30°, then the height of the tower (in m) is :

 Option 1: 1003 Option 2: 502 Option 3: 100 Option 4: 50

Solutions:Let the height of the tower, MN be h. In rt. âˆ†QMN,

tan30°=13=hx⇒x=3h              …..1In rt. âˆ†PMN,

tan45°=MNPM⇒PM=MN=h              …..2Now, In rt. âˆ†PMR,

PM2+MR2=PR2⇒h2+x2=2002⇒h2+3h2=40000                       Using 1⇒h2=10000⇒h=100 mHence, the correct answer is option C.

Question 58:

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangement is :

 Option 1: at least 500 but less than 750 Option 2: at least 750 but less than 1000 Option 3: at least 1000 Option 4: less than 500

Solutions:Number of ways selecting 4 novels out of 6 novels =

C46=15Number of ways selecting 1 dictionary out of 3 dictionaries =

C13=3. Now, we need to arrange these novels and dictionary in a row on a shelf so that the dictionary is always in the middle. So, Total number of ways = 15 × 3 × 4! = 1080. Hence, the correct answer is option C.

Question 59:

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 12 + 2·22 + 32 + 2·42 + 52 + 2·62 + …. If B – 2A = 100 λ, then λ is equal to :

 Option 1: 464 Option 2: 496 Option 3: 232 Option 4: 248

Solutions:Given: A = 12 + 2·22 + 32 + 2·42 + 52 + 2·62 + …. + 192 + 2·202 = 12 + (22 + 22) + 32 + (42 + 42) + 52 + (62 + 62) + …. + 192 + (202 + 202) = [12 + 22 + 32 + 42 + … + 202] + [22 + 42 + 62 + … + 202] = [12 + 22 + 32 + 42 + … + 202] + 22 × [12 + 22 + 32 + … + 102] =2021416+22×1011216=2870+1540=4410Similarly,

B = [12 + 22 + 32 + 42 + … + 402] + [22 + 42 + 62 + … + 402] = [12 + 22 + 32 + 42 + … + 402] + 22 × [12 + 22 + 32 + … + 202] =4041816+22×2021416=22140+2870=33620Now,

B-2A=100λ⇒33620-24410=100λ⇒24800=100λ⇒λ=248Hence, the correct answer is option D.

Question 60:

Let the orthocentre and centroid of a triangle be A(–3, 5) and B(3, 3)  respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is :

 Option 1: 352 Option 2: 352 Option 3: 10 Option 4: 210

Solutions:If A(–3, 5) and B(3, 3) are orthocentre and centroid of a triangle and C is the circumcentre of this triangle.

AB=-3-32+5-32=210Now, we know that, The orthocentre, centroid and circumcentre of any triangle are collinear. The centroid divides the distance from the orthocentre to the circumcentre in the ratio 2 : 1. ⇒BCAB=12⇒BCAB+1=12+1⇒ACAB=32⇒AC=32AB⇒AC=32×210=310If line segment AC is the diameter of that circle, then

Question 61:

Total number of Ione pair of electrons in

I3-ion is :

 Option 1: 9 Option 2: 12 Option 3: 3 Option 4: 6

Solutions:

Total number of lone pairs in I3- are:

Hybridisation: sp3d Shape: Linear the total number of lone pairs are 9. Hence, the correct answer is option A.

Question 62:

Which of the following salts is the most basic in aqueous solution?

 Option 1: FeCl3 Option 2: Pb(CH3COO)2 Option 3: Al(CN)3 Option 4: CH3COOK

Solutions:Salts that are from strong bases and weak acids hydrolyze, and gives a pH greater than 7. The anion in the salt is derived from a weak acid and will accept the proton from the water in the reaction. This will make the water act as an acid that will, in this case, leaving a hydroxide ion (OH). CH3COOK is the most basic salt. Hence, the correct answer is option D.

Question 63:

Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with Br2 to form product B. A and B are respectively :

 Option 1: Option 2: Option 3: Option 4: Solutions: (B)
Hence, the correct answer is option A.

Question 64:

The increasing order of basicity of the following compounds is : (a) (b) (c) (d) Option 1: (b) < (a) < (d) < (c) Option 2: (d) < (b) < (a) < (c) Option 3: (a) < (b) < (c) < (d) Option 4: (b) < (a) < (c) < (d)

Solutions:The basicity of any compound depends upon the ease with which it accepts a proton to form the protonated product. Amidines are stronger organic bases. In the case of the amidine, we can draw two important (the fact that they are equivalent resonance structures is why they both are important) resonance structures to describe the protonated product. The protonated amidine where the positive charge resides on the less electronegative nitrogen. Since we can draw more significant resonance structures for the amidine case then for the amines and hence, protonated amidine is most stable among the all. Amongst the amines, 2

° amines are more stable than 1

° amines whereas, in the case of the imine, resonance is possible which would indicate a more stable product so it would be more basic as compared to the amine which lacks resonance stabilization in its protonated form. (b)< (a)< (d)< (c) Hence, the correct answer is option A.

Question 65:

An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination?

Option 1:
 Base Acid End point Weak Strong Yellow to pinkish red
Option 2:
 Base Acid End point Strong Strong Pink to colourless
Option 3:
 Base Acid End point Weak Strong Colourless to pink
Option 4:
 Base Acid End point Strong Strong Pinkish red to yellow

Solutions:Methyl orange is used for titration of strong acid and weak base because in acidic solution methyl orange gives pinkish-red colour but in basic solution it gives yellow colour. Hence, the correct answer is option A.

Question 66:

The trans-alkenes are formed by the reduction of alkynes with :

 Option 1: Na / liq. NH3 Option 2: Sn – HCl Option 3: H2 – Pd / C, BaSO4 Option 4: NaBH4

Solutions:In this reaction, electrons from Na metal sequentially add to the alkyne, resulting in an anion that is protonated by the NHsolvent. The stereochemistry is due to electronic repulsion the geometry of the resulting alkene is trans. Hence, the correct answer is option A.

Question 67:

The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of compound CXHYOZ is :

 Option 1: C3H4O2 Option 2: C2H4O3 Option 3: C3H6O3 Option 4: C2H4O

Solutions:O2 has z oxygen atom  â€‹

CxHy + x+y4 O2→ x CO2 +y2H2 OO atoms required for combustion=  2x+y4                                              z=12[2 x+y4]                                             z=x +y4Hence, the correct answer is option B.

Question 68:

Hydrogen peroxide oxidises [Fe(CN)6]4– to [Fe(CN)6]3– in acidic medium but reduces [Fe(CN)6]3– to [Fe(CN)6]4– in alkaline medium. The other products formed are, respectively :

 Option 1: H2O and (H2O + O2) Option 2: H2O and (H2O + OH–) Option 3: (H2O + O2) and H2O Option 4: (H2O + O2) and (H2O + OH–)

Solutions:

During reduction H2O-12(OA) +[Fe(CN)6 ]4-→H+ [Fe(CN)6 ]3- +H2O-2During oxidation H2O2(RA) + [Fe(CN)6 ]3-→[Fe(CN)6 ]4- +O2 Hence, the correct answer is option A.

Question 69:

The major product formed in the following reaction is : Option 1: Option 2: Option 3: Option 4: Solutions: Hence, the correct answer is option B.

Question 70:

How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B = 10.8 u)

 Option 1: 3.2 hours Option 2: 1.6 hours Option 3: 6.4 hours Option 4: 0.8 hours

Solutions:

B2H6 + 3O→ B2O3 + 3H2O

B2H6=27.6627.66=1nO2 required =32H2O →2H2 +O2n-factor for O2 =4 Number of equivalent = 3 ×4 =12F= 12×96500Ci×t= 12×96500t=12×96500100s =12×96500100×3600h =3.2 hr

Hence, the correct answer is option A.

Question 71:

Which of the following lines correctly show the temperature dependence of equilibrium constant, K, for an exothermic reaction? Option 1: C and D Option 2: A and D Option 3: A and B Option 4: B and C

Solutions:

∆G= ∆H -T∆S-RTlnK= ∆H -T∆SlnK=-∆HRT +∆SRslope is -∆HRSince, ∆H is -veslope is positive.

Hence, the correct answer is option C.

Question 72:

At 518°C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s–1 when 5% had reacted and 0.5 Torr s–1 when 33% had reacted. The order of the reaction is:

 Option 1: 1 Option 3: 2 Option 4: 3

Solutions: Rate of reaction= k[A]n

R1 = 1 torr/sec =  k365×95 100n R2 =0.5 torr/sec =  k365×67 100n 10.5=9567n⇒ n=2Hence, the correct answer is option C.

Question 73:

Glucose on prolonged heating with HI gives :

 Option 1: Hexanoic acid Option 2: 6-iodohexanal Option 3: n-Hexane Option 4: 1-Hexene

Solutions:

Glucose + HI →∆ n-Hexane Hence, the correct answer is option C.

Question 74:

Consider the following reaction and statements :

CoNH34Br2++Br-→CoNH33Br3+NH3(I) Two isomers are produced if the reactant complex ion is a cis-isomer. (II) Two isomers are produced if the reactant complex ion is a trans-isomer. (III) Only one isomer is produced if the reactant complex ion is a trans-isomer. (IV) Only one isomer is produced if the reactant complex ion is a cis-isomer.

The correct statements are :

 Option 1: (III) and (IV) Option 2: (II) and (IV) Option 3: (I) and (II) Option 4: (I) and (III)

Solutions:Two isomers (fac and mer) are produced if reactant complex ion is a cis isomer. One isomer (fac) is formed if reactant complex ion is a trans isomer.â€‹ Hence, the the correct answer is option D.

Question 75:

The major product of the following reaction is : Option 1: Option 2: Option 3: Option 4: Solutions: Hence, the correct answer is option D.

Question 76:

Phenol on treatment with CO2 in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with (CH3CO)2O in the presence of catalytic amount of H2SO4 produces :

 Option 1: Option 2: Option 3: Option 4: Solutions:The first reaction is Kolbe’s reaction. Salicylic acid undergoes acylation to produce aspirin (non-narcotic analgesic). Hence, the correct answer is option C.

Question 77:

An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1 × 10–10. What is the original concentration of Ba2+?

 Option 1: 1.1 × 10–9 M Option 2: 1.0 × 10–10 M Option 3: 5 × 10–9 M Option 4: 2 × 10–9 M

Solutions:When 50 mL of 1 M solution of Na2SO4 is mixed with an unknown concentration of BaSO4, the volume of resultant solution is 500 mL. Therefore, volume of BaSO4 solution is (500-50)=450 mL. Now, concentration of SO42- in Ba2+ solution can be written as: M1V1=M2V2

⇒1×50=M2×500⇒M2=50500=10-1

BaSO4→Ba2+s+SO42-sFor just precipitation, ionic product=Ksp [Ba2+][SO42-]=Ksp

[Ba2+]×10-1=10-10⇒[Ba2+]=10-9 M in 500 mL solutionIn order to calculate the concentration of original solution of BaSO4 (volume=450 mL),

M1×450=10-9×500M1=10-9×500450⇒M1=1.11×10-9 M, where M1=molarity of original BaSO4 solutionHence, the correct answer is option A.

Question 78:

Which of the following compounds will be suitable for Kjeldahl’s method for nitrogen estimation?

 Option 1: Option 2: Option 3: Option 4: Solutions:Nitrogen in aniline is estimated by Kjeldahl’s method. This method is used for quantitative analysis of nitrogen (N) in organic compounds.

Hence, the correct answer is option D.

Question 79:

When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal ‘M’ is :

 Option 1: Al Option 2: Fe Option 3: Zn Option 4: Ca

Solutions:

Al+3H2O→NaOHAl(OH)3(X)↓+32H2(g)￼￼Al(OH)3 is soluble in excess of NaOH and forms Na[Al(OH)4].

2Al(OH)3→∆Al2O3+3H2OAl2O3 is used as adsorbent in chromatography. Thus, the metal is Al.

Hence, the correct answer is A.

Question 80:

An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the formation of HS from H2S is 1.0 × 10–7 and that of S2– from HS ions is 1.2 × 10–13 then the concentration of S2– ions in aqueous solution is :

 Option 1: 6 × 10–21 Option 2: 5 × 10–19 Option 3: 5 × 10–8 Option 4: 3 × 10–20

Solutions:H2S H+             +                   HS 0.1                         0                                     0 0.1(1-x)            0.1x+0.2+0.1xy                    0.1x(1-y)

HS H+              +                   S2- 0.1x(1-y)         0.1x+0.2+0.1xy                  0.1xy

HCl       →         H+                +                 Cl 0.1x+0.2+0.1xy                   0.2

[H+]= 0.1x + 0.2 + 0.1xy =0.2

Ka1=10-7=[0.2 0.1x(1-y)] [0.1(1-x)] Since,x and y are very small. x=5 10-7

Ka2=(0.2 0.1xy) 0.1x y=6*10-13

[S2-]=0.1xy=3 10-20 Hence, the correct answer is option D.

Question 81:

The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is required to make teeth enamel harder by converting [3Ca3(PO4)2.Ca(OH)2] to :

 Option 1: [3Ca3(PO4)2.CaF2] Option 2: [3{Ca(OH)2}.CaF2] Option 3: [CaF2] Option 4: [3(CaF2).Ca(OH)2]

Solutions:

[3Ca3(PO )4.Ca(OH)2] when reacts with two moles of flouride ions form [3Ca3(PO4)2.CaF2].

Hence, the correct answer is option A.

Question 82:

The compound that does not produce nitrogen gas by the thermal decomposition is

 Option 1: NH4NO2 Option 2: (NH4)2SO4 Option 3: Ba(N3)2 Option 4: (NH4)2Cr2O7

Solutions: Ammonium sulphate on thermal decomposition gives ammonia (NH4)2SO4

→∆ 2NH3(g) + H2SO4 Hence, the correct answer is option B.

Question 83:

The predominant form of histamine present in human blood is (pKa, Histidine = 6.0)

 Option 1: Option 2: Option 3: Option 4: Solutions: Since, blood is basic in nature (pH=7.35). Therefore, N will get protonated. Terminal NH2 is most basic and will get protonated to NH3+. Hence, the correct answer is option C

Question 84:

The oxidation states of Cr in [Cr(H2O)6] Cl3, [Cr(C6H6)2], and K2[Cr(CN)2(O)2(O2) (NH3)] respectively are :

 Option 1: + 3, 0, and + 6 Option 2: + 3, 0, and + 4 Option 3: + 3, + 4, and + 6 Option 4: + 3, + 2, and + 4

Solutions: In [Cr(H2O)6]Cl3 , x + 0 × 6 + 3 ×(–1) = 0 Therefore, x=+3

In [Cr(C6H6)2], y + 2 × 0 = 0 Therefore, y = 0

In K2[Cr(CN)2(O)2(O2)(NH3)]

z + 2 (–1)+ 2(–2) + (–2) + 0 = -2 z = + 6

Hence, the correct answer is option B

Question 85:

Which type of ‘defect’ has the presence of cations in the interstitial sites?

 Option 1: Frenkel defect Option 2: Metal deficiency defect Option 3: Schottky defect Option 4: Vacancy defect

Solutions: Frenkel defect  has the presence of cations in the interstitial sites. Hence, the correct answer is option A.

Question 86:

The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 25°C; heat of combustion (in kJ mol–1) of benzene at constant pressure will be : (R = 8.314 JK–1 mol–1)

 Option 1: 3260 Option 2: –3267.6 Option 3: 4152.6 Option 4: –452.46

Solutions:

C6H6(l)+ 7.5 O2(g)  → 6CO2(g) + 3H2O(l)∆ng = 6- 7.5 = -1.5∆H= ∆U + ∆ngRT = -3263.9 kJ – 1.5 ×8.314 ×2981000kJ                                         = -3267.6 kJ Hence, the correct answer is option B.

Question 87:

Which of the following are Lewis acids?

 Option 1: PH3 and SiCI4 Option 2: BCI3 and AICI3 Option 3: PH3 and BCI3 Option 4: AICI3 and SiCI4

Solutions:As there is presence of vacant p-orbitals and incomplete octet in BCl3 and AlCl3, so they both act as Lewis acids. Hence, the correct answer is option B.

Question 88:

Which of the following compounds contain(s) no covalent bond(s) ? KCl, PH3, O2, B2H6, H2SO4

 Option 1: KCl Option 2: KCl, B2H6 Option 3: KCl, B2H6, PH3 Option 4: KCl, H2SO4

Solutions:KCl is an ionic compound. So, it doesn’t contain any covalent bond. Rest of them are covalent compounds. Hence, the correct answer is option A.

Question 89:

For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

 Option 1: [Co(H2O)4Cl2]Cl.2H2O Option 2: [Co(H2O)3Cl3].3H2O Option 3: [Co(H2O)6]Cl3 Option 4: [Co(H2O)5Cl]Cl2.H2O

Solutions:[Co(H2O)4Cl2]Cl. 2H2O

→[Co(H2O)4Cl2]+Cl– (i = 2) [Co(H2O)3Cl3]3H2O

→No dissociation  (i = 1) [Co(H2O)6]Cl3

→[Co(H2O)6]3+ +3Cl–   (i = 4) [Co(H2O)5Cl]Cl2.H2O

→[Co(H2O)5Cl]2++2Cl   (i = 3)

∆Tf α i∆Tf =Tf-Tf’So, i increases, freezing point of solution (Tf’) decreases.Hence, the correct answer is option B.

Question 90:

According to molecular orbital theory, which of the following will not be a viable molecule ?

 Option 1: H2- Option 2: H22- Option 3: He22+ Option 4: He2+