
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 p_{d} ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is p_{c}. The values of p_{d} and p_{c} 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 v be the velocity of neutron before collision and let v_{1}_{ }and v_{2} 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
⇒v2v1=v …..(ii)Solving (i) and (ii), we havev1=v3 …..(iii)Now, Pd=12mv212mv1212mv2⇒Pd=12mv212mv2912mv2 …..From (iii)Pd=89=0.89Case II: For collision with carbon nucleus Let v be the velocity of neutron before collision and let v_{1 }and v_{2} 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
⇒v2v1=v …..(ii)Solving (i) and (ii), we havev1=11v13 …..(iii)Now, Pc=12mv212mv1212mv2⇒Pc=12mv212m121v216912mv2 …..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 10^{23} hydrogen molecules strike, per second, a fixed wall of area 2 cm^{2} at an angle of 45° to the normal, and rebound elastically with a speed of 10^{3} m/s, then the pressure on the wall is nearly :
Option 1:  2.35 × 10^{2 }N/m^{2} 
Option 2:  4.70 × 10^{2 }N/m^{2} 
Option 3:  2.35 × 10^{3 }N/m^{2} 
Option 4:  4.70 × 10^{3 }N/m^{2} 
Solutions:Given: n = 10^{23}, 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 10^{23} hydrogen molecules per second,
∆p =
2×n×m×v×cos 45=2×1023×3.32×1027×103×12
⇒∆p=0.47 kgm/sForce exerted by â€‹10^{23} hydrogen molecules on the wall, F =
× 1 s = 0.47 N Pressure exerted by â€‹10^{23} hydrogen molecules on the wall of area 2 cm^{2},
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
V132+V121=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. =
k2r2k2r2=0Hence, the correct answer is option A.
Question 6:
Two masses m_{1} = 5 kg and m_{2} = 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 m_{2} 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 m_{1} to be at rest,
T=m1g = 5g For (m_{2} + m) to be at rest, f = T = 5g …..(i) Now,
f≤μ(m2 + m)gFrom (i), we get5g≤μ(m2 + m)g5μm2≤mm≥50.1510m≥23.33 kg
Hence, the correct answer is option D.
Question 7:
If the series limit frequency of the Lyman series is v_{L}, then the series limit frequency of the P fund series is :
Option 1:  v_{L} / 16 
Option 2:  v_{L} / 25 
Option 3:  25 v_{L} 
Option 4:  16 v_{L} 
Solutions:For series limit of Lyman series, n_{1 }= 1 and n_{2} = . Thus,
νL=RcZ21121∞=RcZ2For series limit of Pfund series, n_{1 }= 5 and n_{2} = . Thus,
νp=RcZ21521∞=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 λ_{n}, λ_{g} be the de Broglie wavelength of the electron in the n^{th} state and the ground state respectively. Let
∧nbe the wavelength of the emitted photon in the transition from the n^{th} 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=RZ21121n2∧n=1RZ21121n21As 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/cutin voltage of silicon diode is 0.7 V. Thus,
I=30.7200=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 r_{e}, r_{p}, r_{α }respectively in a uniform magnetic field B. The relation between r_{e}, r_{p}, r_{α} is :
Option 1:  r_{e} < r_{p} < r_{α} 
Option 2:  r_{e} < r_{α} < r_{p} 
Option 3:  r_{e} > r_{p} = r_{α} 
Option 4:  r_{e} < r_{p} = r_{α} 
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=QfQi =30001800=1200 pC=1.2 nCHence, the correct answer is option C.
Question 13:
For an RLC circuit driven with voltage of amplitude ν_{m} 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 × 10^{5} 
Option 2:  2 × 10^{6} 
Option 3:  2 × 10^{3} 
Option 4:  2 × 10^{4} 
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 × 10^{3} kg / m^{3} and its Young’s modulus is 9.27 × 10^{10} 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∈0b2c2b+a 
Option 2:  σ∈0b2c2c+a 
Option 3:  σ∈0a2b2a+c 
Option 4:  σ∈0a2b2b+c 
Solutions: The potential of shell B is
VB=KQAb+KQBb+KQCc =K(+σ)(4πa2)b+ (σ)(4πb2)b+(+σ)(4πc2)c =4πσ4πεoa2bb2b+c2c=σεoa2b2b+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, l_{1} = 52 cm l_{2} = 40 cm R = 5 Ω Using the formula,
r = Rl1l21we get the internal resistance of the cell as
r =552401=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 zctin air and
E→2=E02 x^ cos k 2 zctin medium, where the wave number k and frequency v refer to their values in air. The medium is nonmagnetic. 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 nonmagnetic,⇒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:Semiangular width = 30^{o}
⇒asin30=λNow, fringe width, β=λDd⇒102=106×12×0.5dd=52×105=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 10^{12}/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 × 10^{23} 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=
108×1036.02×1023 kg=1.794×1025 kgTime period, T =
2πmk
11012 =
2π1.794×1025k ⇒k = 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 MR^{2} 
Option 2:  379MR^{2} 
Option 3:  4 MR^{2} 
Option 4:  409MR^{2} 
Solutions:Moment of inertia of small disc about center of bigger disc =
12MR32+M2R32=MR218+4MR29=MR22Moment of inertia of remaining portion =
9MR22MR22=4MR2Hence, the correct answer is option C.
Question 23:
In a collinear collision, a particle with an initial speed ν_{0} 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_{0 }= mv_{1 }+ mv_{2} v_{0 }= v_{1 }+ v_{2 }…..(i)
KE_{i }=
KE_{f }=
mv122+mv222Given, 1.5 KE_{i }= KE_{f}
1.5×mv022=mv122+mv22232v02=v12+v2232v02=v12+(v0v1)21.5v02=v12+v02+v122v0v12v122v0v10.5v02=0Solving quadratic equationv1=1+2v02v2=12v02v1v2=2v0Hence, the correct answer is option D.
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 B_{1}. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B_{2}. The ratio
B1B2is :
Option 1:  2 
Option 2:  12 
Option 3:  3 
Option 4:  3 
Solutions:Let the radius of loop be R. IR^{2 }= m For doubling the dipole moment keeping the currect same, radius should become
2R. B_{1}=
μ0I2RB_{2}=
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 R_{1 }and_{ }R_{2}, respectively, before interchanging the resistances. R_{1}+R_{2}=1000 Ω Initially,
R1R2=xlxFinally,
R2R1=x0.1llx+0.1l=x0.1l1.1lx
x0.1l1.1lx=lxxx20.1lx=1.1l21.1lxlx+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 tπ4In one cycle of a.c., the average power consumed by the circuit and the wattless 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 = e_{rms}i_{rms}cos
π4 =
1002×202×12=10002Wattless current= i_{rms}
×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 monoatomic 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 
Solutions:For adiabatic process, TV
Υ1 = constant For monoatomic gas, =
53300
×V531 = T_{2}
×(2V)531T_{2 }=
1223 ×300 = 189 K
∆U=nCv∆T=2×32×8.314×(189300)=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 n^{th} power of R. If the period of rotation of the particle is T, then :
Option 1:  T ∝ R^{(n + 1)/2} 
Option 2:  T ∝ R^{n/2} 
Option 3:  T ∝ R^{3/2 }for any n. 
Option 4:  T∝Rn2+1 
Solutions:
F∝1Rn
mv2R∝1Rnv2∝1Rn1v ∝ R1n2T =
2πRv
T∝RvT∝RR1n2T∝ R11n2T∝R1+n2Hence, the correct answer is option A.
Question 31:
If the tangent at (1, 7) to the curve x^{2} = y – 6 touches the circle x^{2} + y^{2} + 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=y6Differentiating 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,
y7=2x1⇒2xy+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=100c⇒x–82+y–62=100cThus, centre of the circle is
8,6 and radius is
100c. Distance between the centre of the circle
8,6 and the tangent 2x − y + 5 = 0 is the radius
100c.
8×2+1×6+522+12=100c⇒16+6+55=100c⇒55=100c⇒5=100cSquaring both sides we get5=100c⇒c=95Hence, the correct answer is option B.
Question 32:
If L_{1} is the line of intersection of the planes 2x – 2y + 3z – 2 = 0, x – y + z + 1 = 0 and L_{2} is the line of intersection of the planes x + 2y – z – 3 = 0, 3x – y + 2z – 1 = 0, then the distance of the origin from the plane, containing the lines L_{1} and L_{2}, is :
Option 1:  122 
Option 2:  12 
Option 3:  142 
Option 4:  132 
Solutions:Given: line L_{1 }passes through the intersection of the planes
2x2y+3z2=0 and
xy+z+1=0So,
2+λx2+λy+3+λz2+λ=0 …..(i) Also, line L_{2 }passes through the intersection of the planes
x+2yz3=0 and 3xy+2z1=0So,
1+3μx2μy+2μ1z3μ=0 …..(ii) From (i) and (ii) we have
2+λ1+3μ=2+λ2μ⇒2+λ2μ=2+λ1+3μ⇒μ2=1+3μ⇒3=2μ⇒μ=32Putting the value of
μ=32 in (i) we get
192x+2+32y+31z3+32=0⇒7x7y+8z+3=0 …..iiiDistance of the origin from the plane containing the lines L_{1 }and L_{2} 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 x^{2} – x + 1 = 0, then α^{101} + β^{107} is equal to :
Option 1:  1 
Option 2:  2 
Option 3:  –1 
Solutions:Given:
α,
β are the distinct roots of the equation
x2x+1=0
ω is a nonreal 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 4x^{2} – y^{2} = 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
4×2y2=36 …..(1)
⇒x29y236=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,
4×2y2=36⇒4x×xy×y=36⇒4xx1yy1=36⇒4×0y3=36 Since x1=0, y1=3⇒03y=36⇒y=12 Putting the value of y = −12 in (1) we get
4×2–122=36⇒4×2144=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 y^{2} = 6x, 9x^{2} + by^{2} = 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 nonzero 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 nonzero solution
1k33k2243=0⇒13k+8k9+4+3122k=0⇒3k+8+5k+366k=0⇒444k=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 ⇒ −2y − z = 0 ⇒ z = −2y We need to find the value of
xzy2
⇒xzy2=5y2yy2=10y2y2=10Hence, the correct answer is option D.
Question 37:
Let
S=x∈R : x ≥0and
2x3+xx6+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:
x3≥0⇒x≥3⇒x≥9
2x3+xx6+6=0⇒2x3+x6x+6=0⇒2x6+x6x+6=0⇒x4x=0⇒x=4x⇒x216x=0⇒xx16=0⇒x=0,16But
x≥9 so, x = 16 Case II:
x3<0⇒x<3
2x3+xx6+6=0⇒23x+x6x+6=0⇒62x+x6x+6=0⇒128x+x=0⇒x6x2=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· cosπ6x12=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·cosπ6x12=1
⇒8cosx·cosπ6+x·cosπ6x12=1⇒8cosx·cosπ62+cos2x212=1⇒8cosx·122+cos2x212=1⇒8cosx·14+cos2x212=1
⇒8cosx·cos2x214=1⇒2cosx2cos2x1=1⇒2cosx22cos2x11=1⇒2cosx4cos2x3=1⇒24cos3x3cosx=1⇒2cos3x=1⇒cos3x=12
⇒3x=2nπ±π3⇒x=2nπ3±π9x∈0,π so, x=π9,2π3+π9,2π3π9Sum =
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=x1x, 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=x1x,
hx=fxgx=x2+1x2x1x=x1×2+2x1xLet
x1x=p
⇒ht=p2+2p=p+2ph’t=12p2
h’t=0⇒12p2=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 = {(a, b) ∈ R × R : a – 5 < 1 and b – 5 < 1}; a – 5 < 1 Let
a5=p⇒p<1 ⇒p<1 …..(1) and b – 5 < 1 Let
b5=q⇒q<1 ⇒q<1 …..(2)
Also, B = {(a, b) ∈ 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
p129+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 y^{2} = 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 =
16016–16=1632=12Equation of the tangent will be
y16=12x16⇒2y32=x16⇒x2y+16=0Slope of normal =
2Equation of the normal will be
y16=2x16⇒y16=2x+32⇒2x+y48=0Coordinate of point B will be satisfying the equation of normal so,
2x+048=0⇒2x=48⇒x=24So, B(24, 0) is the point where the normal intersects the xaxis. 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 =
160164=1612=43 Slope of PB,
mPB =
1601624=168=2
tanθ=mPCmPB1+mPCmPB=43–21+43×2=2=2Hence, the correct answer is option D.
Question 44:
If
x42x2x2xx42x2x2xx4=A+Bx xA2,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:
x42x2x2xx42x2x2xx4=A+Bx xA2, C1→C1+C2+C3=
5x42x2x5x4x42x5x42xx4
=5x412x2x1x42x12xx4R3→R3R1,R2→R2R1=5x412x2x0x4000x4=5x44x2=A+BxxA2So,
A=4,B=5Hence, the correct answer is option A.
Question 45:
The sum of the coefficients of all odd degree terms in the expansion of
x+x315+xx315, x>1is :
Option 1:  1 
Option 2:  2 
Option 3:  –1 
Solutions:Given:
x+x315+xx315, x>1
Let x31=p⇒x31=p2
x+p5+xp5=C05x5+C05x4p+…+C05p5+C05x5C05x4p+…C05p5=2C05x5+C25x3p2+C45xp4=2×5+10x3x31+5xx312 From i=2×5+10×610×3+5×7+5x10x4We take only the odd degree terms2x5,20×3,+10×7,10xSum of the odd degree terms will be=220+10+10=2Hence, the correct answer is option B.
Question 46:
Let a_{1}, a_{2}, a_{3}, …., a_{49}, 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+1314d=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+r1d2=140m⇒∑r=1178+r1.12=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=∫π2π2sin2x1+2xdx ∵∫abfxdx=∫abfa+bxdx⇒I=∫π2π22x.sin2x1+2xdx …iiAdding i and ii,2I=∫π2π2sin2xdx2I=2∫0π2sin2xdx2I=2×214sin2x0π22I=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 18x^{2} – 9πx + π^{2} = 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:  1232 
Option 2:  1221 
Option 3:  1231 
Option 4:  123+1 
Solutions:
18×29πx+π2=0⇒6xπ3xπ=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 =1231 sq unitsHence, the correct answer is option C.
Question 50:
For each t ∈R, 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+x1x1x+2x2x+…+15x15x ∵for any GIF x=xx=limx→0+x1+2+…+15xlimx→0+x1x+2x+…+15x=1200 ∵0≤x<1=120Hence, the correct answer is option A.
Question 51:
If
∑i=19xi5=9and
∑i=19xi52=45,then the standard deviation of the 9 items x_{1}, x_{2}, …., x_{9} is :
Option 1:  2 
Option 2:  3 
Option 3:  9 
Option 4:  4 
Solutions:We have,
∑i=19xi5=9⇒∑i=19xi9×5=9⇒∑i=19xi=54 …i
∑i=19xi52=45⇒∑i=19xi210∑i=19xi+25=45⇒∑i=19xi210×54+9×25=45⇒⇒∑i=19xi2=360 …ii
Var(X)=1n∑i=1nxi21n∑i=1nxi2 =36095492 =4036 =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 cos^{6}x.
∫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=xπ·ex1 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=xπ·ex1 sin xFor this function we need to check differentiablty at
x=π and x=0. At x = π
LHD at x = π=limx→πfxf0xπ =limh→0πhπeπh1sinπh0πhπ =0RHD at x = π=limx→π+fxf0xπ =limh→0π+hπeπ+h1sinπ+h0π+hπ =0âˆµ LHD = RHD so function is differentiable at x = π. At x = 0
LHD at x = 0=limx→0fxf0x0 =limh→00hπe0h1sin0h00h0 =0RHD at x = 0=limx→0+fxf0x0 =limh→00+hπe0+h1sin0+h00+h0 =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=2×2π22sinx
yπ6=2π62π22sinπ6=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⇒xyz231011=0⇒2xy+z=0 …iIf
u→is perpendicular to
a→. Then,
u→·a→=0⇒2x+3yz=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
54, 1+1, 43 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 =2123=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 midpoint 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 1^{2} + 2·2^{2} + 3^{2} + 2·4^{2} + 5^{2} + 2·6^{2} + …. 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 = 1^{2} + 2·2^{2} + 3^{2} + 2·4^{2} + 5^{2} + 2·6^{2} + …. + 19^{2} + 2·20^{2} = 1^{2} + (2^{2} + 2^{2}) + 3^{2} + (4^{2} + 4^{2}) + 5^{2} + (6^{2} + 6^{2}) + …. + 19^{2} + (20^{2} + 20^{2}) = [1^{2} + 2^{2} + 3^{2} + 4^{2} + … + 20^{2}] + [2^{2} + 4^{2} + 6^{2} + … + 20^{2}] = [1^{2} + 2^{2} + 3^{2} + 4^{2} + … + 20^{2}] + 2^{2} × [1^{2} + 2^{2} + 3^{2} + … + 10^{2}] =2021416+22×1011216=2870+1540=4410Similarly,
B = [1^{2} + 2^{2} + 3^{2} + 4^{2} + … + 40^{2}] + [2^{2} + 4^{2} + 6^{2} + … + 40^{2}] = [1^{2} + 2^{2} + 3^{2} + 4^{2} + … + 40^{2}] + 2^{2} × [1^{2} + 2^{2} + 3^{2} + … + 20^{2}] =4041816+22×2021416=22140+2870=33620Now,
B2A=100λ⇒3362024410=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=332+532=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
Radius=3102=352. Hence, the correct answer is option A.
Question 61:
Total number of Ione pair of electrons in
I3ion is :
Option 1:  9 
Option 2:  12 
Option 3:  3 
Option 4:  6 
Solutions:
Total number of lone pairs in I^{3} are:
Hybridisation: sp^{3}d 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:  FeCl_{3} 
Option 2:  Pb(CH_{3}COO)_{2} 
Option 3:  Al(CN)_{3} 
Option 4:  CH_{3}COOK 
Question 63:
Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with Br_{2} to form product B. A and B are respectively :
Option 1:  
Option 2:  
Option 3:  
Option 4: 
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: 


Option 2: 


Option 3: 


Option 4: 

Question 66:
The transalkenes are formed by the reduction of alkynes with :
Option 1:  Na / liq. NH_{3} 
Option 2:  Sn – HCl 
Option 3:  H_{2} – Pd / C, BaSO_{4} 
Option 4:  NaBH_{4} 
Question 67:
The ratio of mass percent of C and H of an organic compound (C_{X}H_{Y}O_{Z}) is 6 : 1. If one molecule of the above compound (C_{X}H_{Y}O_{Z}) contains half as much oxygen as required to burn one molecule of compound C_{X}H_{Y} completely to CO_{2} and H_{2}O. The empirical formula of compound C_{X}H_{Y}O_{Z} is :
Option 1:  C_{3}H_{4}O_{2} 
Option 2:  C_{2}H_{4}O_{3} 
Option 3:  C_{3}H_{6}O_{3} 
Option 4:  C_{2}H_{4}O 
Solutions:O_{2} 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:  H_{2}O and (H_{2}O + O_{2}) 
Option 2:  H_{2}O and (H_{2}O + OH^{–}) 
Option 3:  (H_{2}O + O_{2}) and H_{2}O 
Option 4:  (H_{2}O + O_{2}) and (H_{2}O + OH^{–}) 
Solutions:
During reduction H2O12(OA) +[Fe(CN)6 ]4→H+ [Fe(CN)6 ]3 +H2O2During 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: 
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 +O2nfactor 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∆SRTlnK= ∆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:  6iodohexanal 
Option 3:  nHexane 
Option 4:  1Hexene 
Solutions:
Glucose + HI →∆ nHexane 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 cisisomer. (II) Two isomers are produced if the reactant complex ion is a transisomer. (III) Only one isomer is produced if the reactant complex ion is a transisomer. (IV) Only one isomer is produced if the reactant complex ion is a cisisomer.
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 CO_{2} in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with (CH_{3}CO)_{2}O in the presence of catalytic amount of H_{2}SO_{4} produces :
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:The first reaction is Kolbe’s reaction. Salicylic acid undergoes acylation to produce aspirin (nonnarcotic analgesic).
Hence, the correct answer is option C.
Question 77:
An aqueous solution contains an unknown concentration of Ba^{2}^{+}. When 50 mL of a 1 M solution of Na_{2}SO_{4} is added, BaSO_{4} just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO_{4} is 1 × 10^{–10}. What is the original concentration of Ba^{2}^{+}?
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 Na_{2}SO_{4} is mixed with an unknown concentration of BaSO_{4}, the volume of resultant solution is 500 mL. Therefore, volume of BaSO_{4} solution is (50050)=450 mL. Now, concentration of SO_{4}^{2} in Ba^{2+} solution can be written as: M_{1}V_{1}=M_{2}V_{2}
⇒1×50=M2×500⇒M2=50500=101
BaSO4→Ba2+s+SO42sFor just precipitation, ionic product=K_{sp} [Ba^{2+}][SO_{4}^{2}]=K_{sp}
[Ba2+]×101=1010⇒[Ba2+]=109 M in 500 mL solutionIn order to calculate the concentration of original solution of BaSO_{4} (volume=450 mL),
M1×450=109×500M1=109×500450⇒M1=1.11×109 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+3H2OAl_{2}O_{3} 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 H_{2}S and 0.20 M HCl. If the equilibrium constants for the formation of HS^{–} from H_{2}S is 1.0 × 10^{–7} and that of S^{2–} from HS^{–} ions is 1.2 × 10^{–13} then the concentration of S^{2–} 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:H_{2}S H^{+} + HS^{–} 0.1 0 0 0.1(1x) 0.1x+0.2+0.1xy 0.1x(1y)
HS^{–} H^{+} + S^{2} 0.1x(1y) 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
Ka_{1}=10^{7}=[0.20.1x(1y)][0.1(1x)] Since,x and y are very small. x=510^{7}
Ka_{2}=(0.20.1xy)0.1x y=6*10^{13}
[S^{2}]=0.1xy=310^{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 [3Ca_{3}(PO_{4})_{2}.Ca(OH)_{2}] to :
Option 1:  [3Ca_{3}(PO_{4})_{2}.CaF_{2}] 
Option 2:  [3{Ca(OH)_{2}}.CaF_{2}] 
Option 3:  [CaF_{2}] 
Option 4:  [3(CaF_{2}).Ca(OH)_{2}] 
Solutions:
[3Ca_{3}(PO )_{4}.Ca(OH)_{2}] when reacts with two moles of flouride ions form [3Ca_{3}(PO_{4})_{2}.CaF_{2}].
Hence, the correct answer is option A.
Question 82:
The compound that does not produce nitrogen gas by the thermal decomposition is
Option 1:  NH_{4}NO_{2} 
Option 2:  (NH_{4})_{2}SO_{4} 
Option 3:  Ba(N_{3})_{2} 
Option 4:  (NH_{4})_{2}Cr_{2}O_{7} 
Solutions: Ammonium sulphate on thermal decomposition gives ammonia (NH_{4})_{2}SO_{4}
→∆ 2NH_{3}(g) + H_{2}SO_{4} Hence, the correct answer is option B.
Question 83:
The predominant form of histamine present in human blood is (pK_{a}, Histidine = 6.0)
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Question 84:
The oxidation states of Cr in [Cr(H_{2}O)_{6}] Cl_{3}, [Cr(C_{6}H_{6})_{2}], and K_{2}[Cr(CN)_{2}(O)_{2}(O_{2}) (NH_{3})] 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}]Cl_{3} , x + 0 × 6 + 3 ×(–1) = 0 Therefore, x=+3
In [Cr(C_{6}H_{6})_{2}], y + 2 × 0 = 0 Therefore, y = 0
In K_{2}[Cr(CN)_{2}(O)_{2}(O_{2})(NH_{3})]
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 
Question 86:
The combustion of benzene (l) gives CO_{2}(g) and H_{2}O(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:  PH_{3} and SiCI_{4} 
Option 2:  BCI_{3} and AICI_{3} 
Option 3:  PH_{3} and BCI_{3} 
Option 4:  AICI_{3} and SiCI_{4} 
Question 88:
Which of the following compounds contain(s) no covalent bond(s) ? KCl, PH_{3}, O_{2}, B_{2}H_{6}, H_{2}SO_{4}
Option 1:  KCl 
Option 2:  KCl, B_{2}H_{6} 
Option 3:  KCl, B_{2}H_{6}, PH_{3} 
Option 4:  KCl, H_{2}SO_{4} 
Question 89:
For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?
Option 1:  [Co(H_{2}O)_{4}Cl_{2}]Cl.2H_{2}O 
Option 2:  [Co(H_{2}O)_{3}Cl_{3}].3H_{2}O 
Option 3:  [Co(H_{2}O)_{6}]Cl_{3} 
Option 4:  [Co(H_{2}O)_{5}Cl]Cl_{2}.H_{2}O 
Solutions:[Co(H_{2}O)_{4}Cl_{2}]Cl. 2H_{2}O
→[Co(H_{2}O)_{4}Cl_{2}]^{+ }+Cl^{– }(i = 2) [Co(H_{2}O)_{3}Cl_{3}]3H_{2}O
→No dissociation (i = 1) [Co(H_{2}O)_{6}]Cl_{3}
→[Co(H_{2}O)_{6}]^{3+} +3Cl^{– } (i = 4) [Co(H_{2}O)_{5}Cl]Cl_{2}.H_{2}O
→[Co(H_{2}O)_{5}Cl]^{2+}+2Cl^{–} (i = 3)
∆Tf α i∆Tf =TfTf’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+ 
Solutions:H_{2}^{2 }
→σ1s2 σ*1s2So, Bond order =
222= 0Therefore, H_{2}^{2 } doesnot exist. Hence, the correct answer is option B.