
IIT Mains 2014
Test Name: IIT Mains 2014
Question 1:
The pressure that has to be applied at the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is
(For steel, Young’s modulus is 2 × 10^{11} Nm^{–2} and coefficient of thermal expansion is 1.1 × 10^{–5 }K^{–1}.)
Option 1:  2.2 × 10^{7} Pa 
Option 2:  2.2 × 10^{6} Pa 
Option 3:  2.2 × 10^{8} Pa 
Option 4:  2.2 × 10^{9} Pa 
Solutions:Increase in length, ΔL, due to change in temperature = LαΔT Here, L = Original length of the wire = 10 cm = 0.10 m α = Coefficient of thermal expansion = 1.1 × 10^{5} K^{1} ΔT = Change in temperature = 100°C ∴ ΔL = 0.10 × 1.1 × 10^{5 }× 100 Decrease in length, ΔL’, due to the applied pressure at the ends of the steel wire =
FAY×LHere, F = Force applied at both ends A = Area of crosssection of wire Y = Young’s modulus of elasticity = 2 × 10^{11} Nm^{2} ∴ ΔL‘ =
FA2×1011×0.10Given: ΔL‘ = ΔL
⇒FA2×1011×0.10=0.10×1.1×105×100⇒FA=Pressure=1.1×105×100×2×1011 =2.2×108 PaHence, the correct option is (3).
[[VIDEO:13836]]
Question 2:
A conductor lies along the zaxis at –1.5 ≤ z < 1.5 m and carries a fixed current of 10.0 A in
a^zdirection (see figure). For a field
B→=3.0×104e0.2x a^yT, what is the power required to move the conductor at a constant speed to x = 2.0 m and y = 0 m in 5 × 10^{–3} s? Assume parallel motion along the xaxis.
Option 1:  14.85 W 
Option 2:  29.7 W 
Option 3:  1.57 W 
Option 4:  2.97 W 
Solutions:Force acting on the currentcarrying conductor due to magnetic field = ILB Here, I = Current through the conductor L = Length of the conductor B = Magnetic field strength Work done, W, in moving the conductor to x = 2.0 m and y = 0 from x = 0 and y = 0:
W=∫F.dx⇒W= ∫ILB.dx⇒W=∫021033×104e0.2xdx⇒W=9×1031e0.40.2
Power required, P = Work done/Time
⇒P=Wt=9×1031e0.40.2×5×103=2.97 WHence, the correct option is (4).
[[VIDEO:13837]]
Question 3:
A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical. About the point of suspension, the
Option 1:  angular momentum changes in direction but not in magnitude 
Option 2:  angular momentum changes both in direction and magnitude 
Option 3:  angular momentum is conserved 
Option 4:  angular momentum changes in magnitude but not in direction 
Solutions:
Here, the angular momentum about the point of suspension is given as:
L→=mv→×l→From the above equation we can say that,
L→is always perpendicular to the plane containing
v→and
l→. During rotation direction of
v→is changing, therefore direction of
L→will also change. and magnitude of angular momentum remains constant.
Hence, the correct option is (A). [[VIDEO:13838]]
Question 4:
The current voltage relation of a diode is given by I = (e^{1000V}^{/T} – 1) mA, where the applied voltage V is in volts and the temperature T is in Kelvin. If a student makes an error measuring ±0.01 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA?
Option 1:  0.5 mA 
Option 2:  0.05 mA 
Option 3:  0.2 mA 
Option 4:  0.02 mA 
Solutions:Given:
I=e1000V/T1 mA∵ I= 5 mA∴ 5=e1000V/T1⇒e1000V/T=6
Again,I=e1000V/T1⇒dIdV=e1000V/T×1000T⇒dI=6×1000T×dV⇒dI=6×1000300×0.01⇒dI=0.2 mAHence, the correct option is (C).
Question 5:
An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically by additional 46 cm. What will be length of the air column above mercury in the tube now?
(Atmospheric pressure = 76 cm of Hg)
Option 1:  38 cm 
Option 2:  6 cm 
Option 3:  16 cm 
Option 4:  22 cm 
Solutions: Here, P + l = P_{0} = 76 cm of Hg â‡¨ P = 76 − l
P08A=54l76lA⇒l=38 cm and l=92 cm
(l = 92 cm is not possible.)
∴ Length of the air column = 54
38 = 16 cm
Hence, the correct option is (3).
[[VIDEO:13840]]
Question 6:
Match ListI (electromagnetic wave type) with ListII (its association/application) and select the correct option from the choices given below.
ListI  ListII  
(a)  Infrared waves  (i)  To treat muscular strain 
(b)  Radio waves  (ii)  For broadcasting 
(c)  Xrays  (iii)  To detect fracture of bones 
(d)  Ultraviolet rays  (iv)  Absorbed by the ozone layer of the atmosphere 
Option 1: 


Option 2: 


Option 3: 


Option 4: 

Solutions:(a) Infrared rays are used to treat muscular strain. (b) Radio waves are used for broadcasting. (c) Xrays are used to detect bone fractures. (d) Ultraviolet rays are absorbed by the ozone layer of the atmosphere.
Hence, the correct option is (2). [[VIDEO:13841]]
Question 7:
A parallelplate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 10^{4} V/m, the charge density of the positive plate will be close to
Option 1:  3 × 10^{4} C/m^{2} 
Option 2:  6 × 10^{4} C/m^{2} 
Option 3:  6 × 10^{–7} C/m^{2} 
Option 4:  3 × 10^{–7} C/m^{2} 
Solutions:Electric field between the plates of the capacitor:
E=σKε0Here, σ = Charge density of the plate K = Dielectric constant of the dielectric
⇒σ=EKε0⇒σ=3×1042.28.85×1012 =5.841×107 ≈6×107 C/m2Hence, the correct option is (3). [[VIDEO:13842]]
Question 8:
A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?
Option 1:  A screw gauge having 100 divisions in the circular scale and pitch as 1 mm 
Option 2:  A screw gauge having 50 divisions in the circular scale and pitch as 1 mm 
Option 3:  A metre scale 
Option 4:  A vernier calliper where 10 divisions in the vernier scale matches with 9 divisions in the main scale and the main scale has 10 divisions in 1 cm 
Solutions:The value measured is 3.50 cm. So, the least count must be 0.01 cm or 0.1 mm. For the vernier calliper, 1 M.S.D. = 1/10 cm = 1 mm Also, 10 V.S.D. = 9 M.S.D. â‡¨ 1 V.S.D. = 9/10 × M.S.D. = 0.9 mm Least count = 1 M.S.D. − 1 V.S.D. = 1 mm − 0.9 mm = 0.1 mm
Hence, the correct option is (4). [[VIDEO:13843]]
Question 9:
Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is
Option 1:  GMR1+22 
Option 2:  12GMR1+22 
Option 3:  GMR 
Option 4:  22GMR 
Solutions:
The net force on any one particle will be along the xdirection (because components in the ydirection are equal and opposite and will cancel each other), and it is given as:
F=F2+2F1 cos45°⇒F=GM22R2+2GM2R22 cos45° =GM24R2+2GM222R2 =GM2R214+12Because the particles are moving in a circle, this force will provide the necessary centripetal force.
⇒Mu2R=GMR214+12⇒u2 =GMR1+224⇒u=12GMR1+22Hence, the correct option is (2). [[VIDEO:13844]]
Question 10:
In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be
Option 1:  12 A 
Option 2:  14 A 
Option 3:  8 A 
Option 4:  10 A 
Solutions:Power consumption of 15 bulbs of 40 W each = 600 W Power consumption of 5 bulbs of 100 W each = 500 W Power consumption of 5 fans of 80 W each = 400 W Power consumption of 1 heater of 1000 W = 1000 W ∴ Total power consumption, P, by all appliances = 2500 W Voltage of electric mains, V = 220 V Now, Current capacity of the main fuse:
i=PV=2500220=11.36 AThus, the minimum capacity of the fuse must be 12 A.
Hence, the correct option is (1). [[VIDEO:13846]]
Question 11:
A particle moves with simple harmonic motion in a straight line. In the first τ s, after starting from rest, it travels a distance a, and in the next τ s, it travels 2a in the same direction, then :
Option 1:  amplitude of motion is 4a 
Option 2:  time period of oscillations is 6τ 
Option 3:  amplitude of motion is 3a 
Option 4:  time period of oscillations is 8τ 
Solutions:Displacement equation of SHM: x = A(1 − cos ωt) Here, x = Displacement A = Amplitude of motion ω = Angular frequency t = Time
For the first τ s, we have: t = τ and x = a ∴ a = A(1 − cos ωτ)
Or, cos ωτ =
1aA …(i) For the next τ s, we have: t = 2τ and x = (a + 2a) = 3a ∴ 3a = A(1 − cos 2ωτ)
Or, cos 2ωτ =
13aAOr,
2cos2ωτ1=13aAOn substituting the value of cos ωτ from equation (i), we get:
21aA21=13aAOn simplifying, we get:
aAaA12=0Or,
aA=0, aA=12 aA=0is not possible. Thus, the feasible solution is
aA=12.
The amplitude of motion, A = 2a Now, from equation (i), we have:
cos ωτ=112=12=cosπ3Or, ωτ=π3Using
ω=2πT, we get:
2πTτ=π3Or,
T=6τHence, the correct option is (2).
Question 12:
The coercivity of a small magnet where the ferromagnet gets demagnetised is 3 × 10^{3} Am^{–1}. The current required to be passed in a solenoid of length 10 cm and number of turns 100 so that the magnet gets demagnetised when inside the solenoid is
Option 1:  3 A 
Option 2:  6 A 
Option 3:  30 mA 
Option 4:  60 mA 
Solutions:We know:
Coercivity H=Bμ0And,B=μ0nI∴ H=nl …iHere, n is the number of turns per unit length.
n=1000.1 m⇒n=103 m1Given: H = 3 × 10^{3} Am^{1}
On substituting the values of H and n in equation (i), we get: H=nI⇒I=Hn=3×103103=3 AHence, the correct option is (1).
Question 13:
The forwardbiased diode connection is
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:For the p–n junction to be forward bias, the p side should be at higher potential than the n side.
Hence, the correct option is (3).
Question 14:
During the propagation of electromagnetic waves in a medium,
Option 1:  electric energy density is equal to the magnetic energy density 
Option 2:  both electric and magnetic energy densities are zero 
Option 3:  electric energy density is double of the magnetic energy density 
Option 4:  electric energy density is half of the magnetic energy density 
Solutions:The average electric energy density of an electromagnetic wave propagating in a medium is given by
uE=12εE2=14εE02 ( using E=E02). The average magnetic energy density of an electromagnetic wave propagating in a medium is given by
uB=12B2μ=14B02 μ ( using B=B02). Using the relation
E0=B0με, it can be easily shown that
uE=uB.
Hence, the correct option is (1).
Question 15:
In the circuit shown here, point C is kept connected to point A till the current flowing through the circuit becomes constant. Afterwards, suddenly, point C is disconnected from point A and connected to point B at time t = 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to
Option 1:  –1 
Option 2:  1ee 
Option 3:  e1e 
Option 4:  1 
Solutions:When point C is connected to point B, the resistor and inductor become parallel to each other. On applying Kirchhoff’s law, we get:
VRVL=0⇒VRVL=1Hence, the correct option is (1).
Question 16:
A mass ‘m‘ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?
Option 1:  5g6 
Option 2:  g 
Option 3:  2g3 
Option 4:  g2 
Solutions:
If the string does not slip on the cylinder, it’s a case of pure rolling, i.e.
a=Rα Let the the mass ‘m’ moves downward with acceleration a. Then,
ma=mgT…..(i) For the cylinder: Torque acting due to tension in the string is
TRand torque can be written as
Iα, where I is the moment of inertia. For a hollow cylinder:
I=mR2Therefore,
TR=mR2α
⇒T=mRα
⇒T=ma………(ii) Solving equations (i) and (ii),we get:
a=g2Hence, the correct option is (4).
Question 17:
One mole of a diatomic ideal gas undergoes a cyclic process ABC, as shown in the figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K, respectively. Choose the correct statement.
Option 1:  The change in internal energy in process AB is −350 R. 
Option 2:  The change in internal energy in process BC is −500 R. 
Option 3:  The change in internal energy in the whole cyclic process is 250 R. 
Option 4:  The change in internal energy in process CA is 700 R. 
Solutions:Process AB is an isochoric process, as shown. The change in internal energy in process AB is
∆UAB=nCVTBTA =1×5R2800400=1000 RProcess CA is isobaric. The change in internal energy in process CA is
∆UCA=nCVTATC =1×5R2400600=500 RSince, process ABC is cyclic,
∆UABC=0Process BC is an adiabatic process. So, the change in internal energy in process BC:
∆UAB+∆UBC+∆UCA=0⇒1000R+∆UBC500R=0⇒∆UBC=500R
Hence, the correct option is (2).
Question 18:
From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path.
The relation between H, u and n is
Option 1:  2gH = nu^{2}(n − 2) 
Option 2:  gH = (n − 2)u^{2} 
Option 3:  2gH = n^{2}u^{2} 
Option 4:  gH = (n − 2)^{2}u^{2} 
Solutions:At maximum height, velocity v will be zero. Therefore, time taken ‘t’ by the particle to reach the maximum height can be calculated using the first equation of motion as:
v=ugtv=0⇒t=ugTime taken ‘t’ by the particle to hit the ground can be calculated using the second equation of motion as:
H=ut’12gt’2Given, t’ = nt
⇒H=unt12gn2t2Substituting the value of t from above, we get:
H=unug12gn2u2g2⇒2gH=nu2n2Hence, the correct option is (1).
Question 19:
A thin convex lens made from crown glass
μ=32has focal length f. When it is measured in two different liquids of refractive indices
43and
53, it has focal lengths f_{1} and f_{2}, respectively. The correct relation between the focal lengths is
Option 1:  f_{2} > f and f_{1} becomes negative 
Option 2:  f_{1} and f_{2} become negative 
Option 3:  f_{1} = f_{2} < f 
Option 4:  f_{1} > f and f_{2} becomes negative 
Solutions:The relation between the focal length of a lens and the refractive index of material of glass is expressed as:
1f=μ11R11R2 …(i) When placed in a medium other than air, the focal length is expressed as:
1fm=μμm11R11R2 …(ii) Here, μ_{m} is the refractive index of the medium. Dividing (i) by (ii), we get:
fmf=μ1μμm1For μ_{m} = 4/3 and f_{m} = f_{1}:
⇒f1f=3213/24/31=4⇒f1=4fFor μ_{m} = 5/3 and f_{m }= f_{2}:
⇒f2f=3213/25/31=5⇒f2<0Hence, the correct option is (4).
Question 20:
Three rods of copper, brass and steel are welded together to form a Y – shaped structure. Area of cross – section of each rod is 4 cm^{2}. The end of the copper rod is maintained at 100°C, whereas the ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cm, respectively. The rods are thermally insulated from the surroundings, except at the ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units, respectively. Rate of heat flow through the copper rod is
Option 1:  4.8 cal/s 
Option 2:  6.0 cal/s 
Option 3:  1.2 cal/s 
Option 4:  2.4 cal/s 
Solutions:Rate of heat flow through a rod =
dQdt=KA∆TlAs the rods form a Yshape, Rate of heat flow through the copper rod_{(1)} = Rate of heat flow through the brass rod_{(2)} + Rate of heat flow through the steel rod_{(3)}
dQ1dt=dQ2dt+dQ3dt …iLet the temperature of the junction be
T °C. Now, substituting the values of K, A, l and âˆ†T in (i), we get:
0.92100T46=0.26T013+0.12T012⇒T=40 °CSo, the rate of heat flow through the copper rod =
dQ1dt=0.92×41004046=4.8 cal/sHence, the correct option is (1).
Question 21:
A pipe of length 85 cm is closed at one end. Find the number of possible natural oscillations of the air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.
Option 1:  6 
Option 2:  4 
Option 3:  12 
Option 4:  8 
Solutions:We know that in fundamental mode,
λ4=lHere,
lis the length of the pipe.
λ=4×0.85The fundamental frequency of any wave:
f0=vλHere, v is the velocity of wave. Substituting the values of v and λ:
f0=vλ=3404×0.85=100 HzSo, frequency of the nth normal mode of vibration is given as
fn=(2n1)f0Thus, the possible frequencies are, 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz and 1100 Hz, as all these frequencies lie below 1250 Hz.
Hence, the correct option is (1). [[VIDEO:13847]]
Question 22:
There is a circular tube in a vertical plane. Two liquids that do not mix, and of densities d_{1} and d_{2,} are filled in the tube. Each liquid subtends an angle of 90° at the centre. The radius joining their interface makes an angle α with vertical. The ratio
d1d2is
Option 1:  1+tan α1tan α 
Option 2:  1+sin α1cos α 
Option 3:  1+sin α1sin α 
Option 4:  1+cos α1cos α 
Solutions:Consider the following figure given below: In
∆OCD, OC=Rsinαand in
∆OBA, OB=Rcosα.
Equating pressure at P: Using
P=ρgh
Here, ρ=density, g=accerlation due to grravity and h is the height. d1gR(cosαsinα)=d2gR(cosα+sinα)d1d2=cosα+sinαcosαsinα=1+tanα1tanαHence, the correct option is A.
Question 23:
A green light is incident from water on the air – water interface at critical angle (θ). Select the correct statement.
Option 1:  The spectrum of visible light whose frequency is more than that of green light will come out to the air medium. 
Option 2:  The entire spectrum of visible light will come out of water at various angles to the normal. 
Option 3:  The entire spectrum of visible light will come out of water at an angle of 90° to the normal. 
Option 4:  The spectrum of visible light whose frequency is less than that of green light will come out to the air medium. 
Solutions:The relation between refractive index and wavelength is expressed as:
μ∝1λ …(i)For light of frequency (f‘) greater than green light i.e.
f’>fgreen, its wavelength (λ‘) will be smaller than that of the green light, i.e.
λ'<λgreen. (Using relation,
f=cλ) Now, from equation (i),
μ’>μgreenUsing Snell’s Law,
sinθ=1μ As frequency of visible light increases, refractive index increases. With the increase of refractive index, the critical angle decreases. So, the light with frequency greater than green light will undergo total internal reflection and the light with frequency less than green light will pass into the air.
Hence, the correct option is (4). [[VIDEO:13848]]
Question 24:
Each of hydrogen (_{1}H^{1}), deuterium (_{1}H^{2}), singlyionised helium (_{2}He^{4})^{+} and doublyionised lithium (_{3}Li^{6})^{++} have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiations are λ_{1}, λ_{2}, λ_{3} and λ_{4}, respectively, then approximately which one of the following is correct?
Option 1:  λ_{1} = λ_{2} = 4λ_{3} = 9λ_{4} 
Option 2:  λ_{1} = 2λ_{2} = 3λ_{3} = 4λ_{4} 
Option 3:  4λ_{1} = 2λ_{2} = 2λ_{3} = λ_{4} 
Option 4:  λ_{1} = 2λ_{2} = 2λ_{3} = λ_{4} 
Solutions:We know that
1λ=RZ21n121n22Given, n1=1 and n2=21λ=RZ2112122∴λ=43RZ2For hydrogen and deutrium, Z = 1,
⇒λ1=43R⇒λ2=43RFor helium, Z = 2,
⇒λ3=412RFor Lithium, Z = 3,
⇒λ4=427R⇒λ1=λ2=4λ3=9λ4Hence, the correct option is (1). [[VIDEO:13849]]
Question 25:
The radiation corresponding to 3 → 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10^{−4} T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to
Option 1:  0.8 eV 
Option 2:  1.6 eV 
Option 3:  1.8 eV 
Option 4:  1.1 eV 
Solutions:The radius of the circular path followed by a particle is
r=mvqB…(i) And the maximum kinetic energy attained by the particle is
K(max)=p22m=(mv)22mFrom equation (i),
K(max)=(qrB)22mSubstituting q = 1.6 × 10^{19} C, r = 10 mm = 10^{2} m, B = 3 × 10^{4} T and m = 9.1 × 10^{31} kg, we get:
K(max)=(1.6×1019×102×3×104)22×9.1×1031×1.6×1019=0.8 eVFor transition between 3 to 2:
E=hν=13.61419⇒hν=13.6536
∴Work function =hνK(max) =13.65360.8=1.1 eVHence, the correct option is (4). [[VIDEO:13850]]
Question 26:
A block of mass m is placed on a surface with a vertical crosssection given by
y=x36.If the coefficient of friction is 0.5, the maximum height above the ground, at which the block can be placed without slipping, is
Option 1:  13 m 
Option 2:  12 m 
Option 3:  16 m 
Option 4:  23 m 
Solutions:Slope:
tanθ=dydx⇒tanθ=dx36dx⇒tanθ=x22
At limiting equilibrium,
μ=tanθ⇒0.5=x22⇒x=±1 ∴y=16 mHence, the correct option is (3). [[VIDEO:13851]]
Question 27:
When a rubberband is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx^{2} , where a and b are constants. The work done in stretching the unstretched rubberband by L is
Option 1:  aL22 +bL33 
Option 2:  12aL22 +bL33 
Option 3:  aL^{2} + bL^{3} 
Option 4:  12 aL^{2} +bL^{3} 
Solutions:Given,
F=ax+bx2We know thatdW=F.dlOn integrating both sides of the above equation from 0 to L:
W=∫0L(ax+bx2)dx⇒W=aL22+bL33Hence, the correct option is (1). [[VIDEO:13852]]
Question 28:
On heating water, bubbles being formed at the bottom of the vessel detach and rise. Consider the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r << R, and the surface tension of water is T, value of r just before the bubbles detach is (density of water is ρ_{w})
Option 1:  R2ρw gT 
Option 2:  R23ρw gT 
Option 3:  R2ρw g3T 
Option 4:  R2ρw g6T 
Solutions:We know that on heating water, the water bubbles get detached from the surface of the vessel and rise.
So, the buoyant force = force due to surface tension
∫T×dlsinθ+∆P(πr2)=43πR3ρwgHere, dl is the small length element of the bubble touching the surface of the vessel. T is the tension of the water.
∆Pis the excess pressure inside the bubble.
ρwis the density of the water From
∆ABC, sinθ=rR
⇒T×2πr×rR+2TR(πr2)=43πR3ρwg⇒4πTr2R=43πR3ρwg⇒r2=R4ρwg3T∴r=R2ρwg3THence, the correct option is (3).
Question 29:
Two beams, A and B, of planepolarised light with mutually perpendicular planes of polarisation are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of the polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams are I_{A} and I_{B} respectively, then
IAIBequals
Option 1:  1 
Option 2:  13 
Option 3:  3 
Option 4:  32 
Solutions:Using Malus’ law,
I=I0cos2θ: For plane polarised beams of light A and B, after rotation of the polaroid, the intensities are
IA’=IAcos230°IB’=IBcos260°Given,
IA’=IB’
⇒IA×34=IB×14⇒IAIB=13Hence, the correct option is (2). [[VIDEO:13853]]
Question 30:
Assume that an electric field
E→=30 x2 i^exists in space. Then, the potential difference V_{A} − V_{O} (where V_{O} is the potential at the origin and V_{A }the potential at x = 2 m) is
Option 1:  −80 J 
Option 2:  80 J 
Option 3:  120 J 
Option 4:  −120 J 
Solutions:We know that
dV=E.→dx→On integrating both sides of the above equation, according to the limits given in the question, we get:
∫V0VAdV=∫0230x2dx⇒VAV0=30×3302=80 VNone of the options given in the question correctly matches with the answer, as the unit of potential difference is volt or joule per coulomb and the options are given in joules.
Question 31:
The image of the line
x – 13 = y – 31 = z – 45in the plane
2x − y + z + 3 = 0 is the line :
Option 1:  x + 33 = y – 51 = z – 2 5 
Option 2:  x + 3 3 = y – 5 – 1 = z + 25 
Option 3:  x – 33 = y + 5 1 = z – 2 5 
Option 4:  x – 3 3 = y + 5 1 = z – 25 
Solutions:The equations of the given line and the plane are:
x13=y31=z45 … (1)
2x
–y + z + 3 = 0 … (2)
The given line passes through the point (1, 3, 4) and is parallel to the vector, whose direction ratios are 3, 1 and
5.
∵ al+bm+cn=2×3+1×1+1×5=615=0Thus, the given line and the plane are parallel to each other.
So, the direction ratios of the image of the given line are 3, 1 and
5.
Let (x_{1}, y_{1}, z_{1}) be the image of the point (1, 3, 4) in the given plane.
∴x112=y131=z141=2ax2+by2+cz2+da2+b2+c2Here,
a=2, b=1, c=1, d=3, x2, y2, z2=1, 3, 4 ∴x112=y131=z141=22×13+4+322+12+12=2 ⇒x1=3, y1=5, z1=2Thus, the equation of the image of the given line is
x+33=y51=z25.
Hence, the correct option is (1).
Question 32:
If the coefficients of x^{3} and x^{4} in the expansion of (1 + ax + bx^{2}) (1 − 2x)^{18} in powers of x are both zero, then (a, b) is equal to :
Option 1:  16, 2513 
Option 2:  14, 2513 
Option 3:  14, 2723 
Option 4:  16, 2723 
Solutions:Let us expand the given expression.
1+ax+bx212×18=1+ax+bx2C018+C1182×1+C2182×2+C3182×3+C4182×4+…Now, coefficient of x^{3} =
C31823+a×C21822+b×C11821=0 ⇒4×17×163×2+a×172×2b=0 ⇒51a3b544=0 … (1)
Coefficient of x^{4} =
C41824+a×C31823+b×C21822=0 ⇒4×16×154×3×22a×163×2+b×12=0 ⇒32a3b240=0 … (2)
Solving (1) and (2), we get:
a=16, b=2723Hence, the correct option is (4).
Question 33:
If a Ïµ R and the equation
−3(x − [x])^{2 }+ 2 (x − [x]) + a^{2} = 0
(where [x] denotes the greatest integer ≤ x) has no integral solution, then all possible values of a lie in the interval :
Option 1:  (−1,0) ∪ (0, 1) 
Option 2:  (1, 2) 
Option 3:  (−2, −1) 
Option 4:  (∞, 2) ∪( 2, ∞) 
Solutions:The given equation is as follows:
3x[x]2+2x[x]+a2=0 , a∈R … (1)
We know that,
{x}=x[x].
∴3×2+2x+a2=0⇒x223xa23=0 ⇒x132=a23+19 ∵x<1 ⇒ x13<113⇒x132<49and
a23+19>0 ∴0<a23+19<49 ⇒13<a2<1⇒0≤a2<1 ∵a2≥0 … (2)
Let a = 0.
From equation (1), we have:
3x[x]2+2x[x]+0=0 ⇒x[x]3x+3[x]+2=0⇒x[x]=0 or 3x+3[x]+2=0Take
x[x]=0⇒x=[x], which is possible only when x is an integer.
But it is given that, the quadratic equation has no integral solution.
∴a≠0Thus, from equation (2), we have
a∈1, 0∪0, 1.
Hence, the correct option is (1).
Question 34:
If a→×b→ b→×c→ c→×a→ = λa→ b→ c→2 then λis equal to :
Option 1:  2 
Option 2:  3 
Option 4:  1 
Solutions:
a→×b→ b→×c→ c→×a→=a→×b→. b→×c→×c→×a→=a→×b→.b→×c→ .a→c→b→×c→.c→a→=a→×b→.b→×c→.a→c→0=a→×b→.c→b→×c→.a→=a→ b→ c→b→ c→ a→=a→ b→ c→a→ b→ c→=a→ b→ c→2Comparing this with the given expression, we get:
λ=1Hence, the correct option is (4).
Question 35:
The variance of first 50 even natural numbers is:
Option 1:  8334 
Option 2:  833 
Option 3:  437 
Option 4:  4374 
Solutions:The first 50 even numbers are 2, 4, 6 … 100.
We know: Variance,
σ2=∑xi2Nx2∴ Variance of the first 50 even natural numbers
=22+42+…+1002502+4+…+100502=412+22+…+5025021+2+…+50502=4×505110150×62×50512×502=34342601=833Hence, the correct option is (2). [[VIDEO:13771]]
Question 36:
A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then the speed (in m/s) of the bird is :
Option 1:  40 (2 – 1) 
Option 2:  40(3 – 2) 
Option 3:  202 
Option 4:  20(3 1) 
Solutions:
Let AB be the height of the pole.
∴ AB = 20 m
Suppose the bird is at point C after 1 second.
∴ CD = 20 m
Let the speed of the bird be v m/s.
Distance covered by the bird in 1 sec = BD = v m/s × 1 s = v m
In âˆ†OAB:
tan45°=ABOB⇒1= 20 OB⇒OB=20 mIn âˆ†OCD:
tan30°=CDOD⇒13=20 20+v⇒20+v=203⇒v=2031Thus, speed of the bird is
2031m/s.
Hence, the correct option is (4). [[VIDEO:13772]]
Question 37:
The integral
∫0π1 + 4 sin2x2 – 4 sinx 2 dxequals :
Option 1:  π − 4 
Option 2:  2π3 – 4 43 
Option 3:  43 – 4 
Option 4:  43 – 4 – π3 
Solutions:Let
I = ∫0π1+4sin2x24sinx2 dx I = ∫0π1+4sin2x24sinx2 dx
=∫0π12sinx22 dx
=∫0π12sinx2 dx
∴∫0π12sinx2 dx
= ∫0π31 – 2sinx2 dx + ∫π3π 2sinx2 – 1 dx
= x+ 4cosx20π3 + 4cosx2 xπ3π
= π3 + 4cosπ6 4cos0+ 4cosπ2 – π +4cosπ6+ π3
= π3 + 4·32 – 4 π + 4·32 + π3
= 43 – 4 π3
Hence, the correct option is (4).
[[VIDEO:13773]]
Question 38:
The statement
~(p↔ ~q)is :
Option 1:  equivalent to
p ↔ q 
Option 2:  equivalent to
~p ↔ q 
Option 3:  a tautology 
Option 4:  a fallacy 
Solutions:
p  q  ∼q  p↔~q  ~p↔~q  p↔q 
T  T  F  F  T  T 
T  F  T  T  F  F 
F  T  F  T  F  F 
F  F  T  F  T  T 
Since columns V and VI are same, the statement
~p↔~qis equivalent to
p↔q.
Hence, the correct option is (1). [[VIDEO:13774]]
Question 39:
If A is an 3 × 3 non − singular matrix such that AA’ = A’A and B = A^{−1} A’, the BB’ equals :
Option 1:  I + B 
Option 2:  I 
Option 3:  B^{−1} 
Option 4:  (B^{−1})’ 
Solutions:Given:
B = A^{−1} A’
AA’ = A’A
On premultiplying A with both the sides of the equation (B = A^{−1} A’), we get:
AB = A( A^{−1} A’) = (AA^{−1})A’ = IA’ = A’
∴ ABB’ = A’B’ = (BA)’ = (A^{−1} A’A)’ = (IA’)’ = A
⇒^{ }BB’ = I
Hence, the correct option is (2). [[VIDEO:13775]]
Question 40:
The integral
∫ 1 + x – 1x e x+1x dxis equal to :
Option 1:  (x – 1) e x+1x + c 
Option 2:  x e x+1x +c 
Option 3:  (x + 1) e x+1x + c 
Option 4:  x e x+1x + c 
Solutions:
∫1+x1xex+1xdx=∫ex+1x + x1xex+1xdx=∫ex+1x+x11x2ex+1xdxLet f(x)=ex+1x⇒f'(x)=ex+1x(11×2)We know: ∫f(x)+xf'(x)dx=xf(x)+c∴ ∫ex+1x+x11x2ex+1xdx=xex+1x+cHence, the correct option is (2).
Question 41:
If z is a complex number such that z≥2, then the minimum value of
z + 12:
Option 1:  is equal to
52 
Option 2:  lies in the interval (1, 2) 
Option 3:  is strictly greater than
52 
Option 4:  is strictly greater than
32but less then 52 
Solutions:It is given that
z≥2.
Now, we need to evaluate the minimum value of
z+12.
We know:
x+y≥xy ∴z +12⩾z12 ⇒z +12⩾212⇒z +12⩾32Thus, the minimum value of
z+12is
32, which lies in the interval (1, 2).
Hence, the correct option is (2).
Question 42:
If g is the inverse of a function f and f’ (x) =
11 + x5, then g’ (x) is equal to :
Option 1:  1 + x^{5} 
Option 2:  5x^{4} 
Option 3:  11 + {g(x)}5 
Option 4:  1 + {g(x)}^{5} 
Solutions: Given:
f’x=11+x5 …1Also, g is the inverse of the function f.
⇒fogx=x⇒fgx=xDifferentiating both the sides with respect to x, we get:
f’gxg’x=1⇒g’x=1f’gx …2 f’gx=11+gx5 Using 1⇒g’x=111+gx5 =1+gx5
⇒g’x = 1 + gx5Hence, the correct option is (4).
Question 43:
If α, β≠0, and f(n) = αn + βnand
31 + f(1)1 + f(2)1 + f(1)1 + f(2)1 + f(3)1 + f(2)1 + f(3)1 + f(4)= K
(1 – α)2 (1 – β)2 (α – β)2,then K is equal to :
Option 1:  αβ 
Option 2:  1αβ 
Option 3:  1 
Option 4:  −1 
Solutions:Given:
fn=αn+βn, α, β≠0 … (1)
Putting n = 1, 2, 3 and 4 in equation (1), we get:
f1=α+β, f2=α2+β2, f3=α3+β3, f4=α4+β4∴
31+f11+f21+f11+f21+f31+f21+f31+f4=K1α21β2αβ2
⇒31+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4=K1α21β2αβ2 ⇒1111αβ1α2β21111αβ1α2β2=K1α21β2αβ2 ⇒1111αβ1α2β22=K1α21β2αβ2 ⇒1α21β2αβ2=K1α21β2αβ2⇒K=1Hence, the correct option is (C). [[VIDEO:13778]]
Question 44:
Let
fk(x) = 1k (sink x + cosk x)where
x ε R and k≥ 1.Then f_{4}(x) − f_{6}(x) equals :
Option 1:  16 
Option 2:  13 
Option 3:  14 
Option 4:  112 
Solutions:
fkx=1ksinkx+coskx, x∈R, k≥1 ∴f4xf6x=14sin4x+cos4x16sin6x+cos6x ⇒f4xf6x=1123sin4x+3cos4x2sin6x+cos6xOn using the identity
x3+y3=x+yx2xy+y2, we get:
⇒f4xf6x=1123sin4x+3cos4x2sin2x+cos2xsin4x+cos4xsin2xcos2x ⇒f4xf6x=1123sin4x+3cos4x2sin4x2cos4x+2sin2xcos2x ⇒f4xf6x=112sin2x+cos2x+2sin2xcos2x ⇒f4xf6x=112sin2x+cos2x2 ⇒f4xf6x=112Hence, the correct option is (4).
Question 45:
Let
α and βbe the roots of equation px^{2} + qx + r = 0,
p≠0. If p, q, r are in A.P. and
1α+ 1β = 4, then the value of
α – βis :
Option 1:  619 
Option 2:  2179 
Option 3:  349 
Option 4:  2139 
Solutions:The given quadratic equation is
px2+qx+r=0.
αand
βare the roots of the given quadratic equation.
∴ α+β=qp, αβ=rp
1α+1β=4⇒α+βαβ=4⇒qprp=4
⇒ q=4r …(1)
It is also given that p, q and r are in A.P.
∴ 2q=p+r …(2)
From (1) and (2), we get:
p=2qr=8rr=9r …(3)
Now,
αβ=α+β24αβ=q2p24rp ∴ αβ=16r281r2+4r9r=2139 [Using (1) and (3)]
Hence, the correct option is (4).
Question 46:
Let A and B be two events such that
PA∪B=16, PA∩B=14 and PA=14,where
Astands for the complement of the event A. Then the events A and B are
Option 1:  mutually exclusive and independent. 
Option 2:  equally likely but not independent. 
Option 3:  independent but not equally likely. 
Option 4:  independent and equally likely. 
Solutions:Given:
PA∪B=16, PA∩B=14and
PA=14 PA∪B=16⇒PA∪B=116=56 PA=14⇒PA=114=34Now, we have:
PA∪B=PA+PBPA∩B⇒PB=PA∪BPA+PA∩B⇒PB=5634+14⇒PB=13Now,
PA≠PBHence, the events A and B are not equally likely.
Also,
PAPB=34×13=14=PA∩BSo, A and B are independent events.
Thus, the events A and B are independent but not equally likely.
Hence, the correct option is (3). [[VIDEO:13779]]
Question 47:
If f and g are differentiable functions is [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6, then for some c∈[0, 1]:
Option 1:  2f‘ (c) = g‘(c) 
Option 2:  2f‘ (c) = 3g‘ (c) 
Option 3:  f‘(c) = g‘(c) 
Option 4:  f‘(c) = 2g‘ (c) 
Solutions:It is given that the functions f and g are differentiable functions in [0, 1]. So, f and g are continuos in [0, 1].
Using mean value theorem, we have:
f’c=f1f010=62=4 …1Also,
g’c=g1g010=20=2 …2From (1) and (2), we have:
f’(c) = 2 g’(c)
Hence, the correct option is (4). [[VIDEO:13780]]
Question 48:
Let the population of rabbits surviving at a time t be governed by the differential equation
dptdt=12pt200. If p(0) = 100, then p(t) equals:
Option 1:  400 − 300 e^{t/2} 
Option 2:  300 − 200 e^{−t}^{/2} 
Option 3:  600 − 500 e^{t/2} 
Option 4:  400 − 300 e^{−t}^{/2} 
Solutions:It is given that the population of rabbits surviving at a time t be governed by the following differential equation:
dptdt= 12pt – 200
⇒ dpdt= p – 4002
⇒ dpp – 400= 12dt
Integrating both sides, we get:
log p – 400 = 12t + c
Also, p(0) = 100
⇒ log 300 = c
∴
logp – 400 = t2+ log 300
logp – 400300= t2
⇒p – 400 = 300et2 ⇒p400=300et2 or 400p=300et2
Now, since p(0) = 100, we neglect
p400=300et2
⇒400 – p = 300et2
⇒p = 400 – 300 et2
Hence, the correct option is (1).
[[VIDEO:13781]]
Question 49:
Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centred at (0, y), passing through origin and touching the circle C externally, then the radius of T is equal to:
Option 1:  32 
Option 2:  32 
Option 3:  12 
Option 4:  14 
Solutions: The given information can be represented diagrammatically as follows:
In the given figure, the smaller circle with centre (0, y) represents the circle T.
From figure, we have:
AD^{2} = BD^{2} + AB^{2} …(1)
The radius of the smaller circle is y.
∴ AD = 1+ y
From equation (1), we get:
1+y2=
1+1y2 ⇒1+y2+2y = 1 + 1 + y22y⇒2y = 1 2y⇒y = 14
Therefore, the radius of circle T is
14units.
Hence, the correct option is (4). [[VIDEO:13782]]
Question 50:
The area of the region described by A = {(x, y) : x^{2} + y^{2} ≤ 1 and y^{2} ≤ 1 − x} is :
Option 1:  π2+43 
Option 2:  π243 
Option 3:  π223 
Option 4:  π2+23 
Solutions:
A =
x,y: x2+y2≤1, y2≤1xOn putting
y2=1xin the equation
x2+y2=1, we get:
x2x=0⇒x(x1) = 0⇒x=0,1For x = 0, we get: y = 1, −1
For x = 1, we get: y = 0
Thus, the points of intersection are (0, 1), (1, 0), (0, −1).
∴ Required area = A =
2∫101x2dx+2∫011xdx =
2×21x2 – 12sin1x10 43 1x3201 =
sin11431=π2+43 sq unitsHence, the correct option is (1). [[VIDEO:13783]]
Question 51:
Let a, b, c and d be nonzero numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and 5bx + 2by + d = 0 lies in the fourth quadrant and is equidistant from the two axes then :
Option 1:  2bc − 3ad = 0 
Option 2:  2bc + 3ad = 0 
Option 3:  3bc − 2ad = 0 
Option 4:  3bc + 2ad = 0 
Solutions:The equations of the given lines are
4ax + 2ay + c = 0and
5bx + 2by + d =0.
The point of intersection of the lines lies in the fourth quadrant and is equidistant from the two axes. Let us assume the point of intersection to be (h, −h). Therefore, we have:
4ah – 2ah + c = 0 …15bh – 2bh + d =0 …2On solving the two equations, we have:
h2ad2bc=h5bc4ad=18ab10ab ⇒2ad2bc = 4ad 5bc ⇒3bc2ad =0Hence, the correct option is (3). [[VIDEO:13784]]
Question 52:
Let PS be the median of the triangle with vertices P(2, 2), Q(6, −1) and R(7, 3). The equation of the line passing through (1, −1) and parallel to PS is
Option 1:  4x − 7y − 11 = 0 
Option 2:  2x + 9y + 7 = 0 
Option 3:  4x + 7y + 3 = 0 
Option 4:  2x − 9y − 11 = 0 
Solutions:Since PS is the median of the triangle, S is the midpoint of QR.
∴ Coordinates of S:
6+72,1+32=132,1The line is parallel to the median PS, therefore both the slopes will be equal.
Slope of PS=y2y1x2x1=212132=29Since the line passes through
1, 1and its slope is
29, we can use the point slope form.
Equation of the line:
y+1=29x1 ⇒9y+9=2x+2⇒2x+9y+7=0Hence, the correct option is (2).
Question 53:
limx→0sin π cos2 xx2is equal to:
Option 1:  π2 
Option 2:  1 
Option 3:  −π 
Option 4:  π 
Solutions:
limx→0sinπcos2xx2=limx→0sinπ1sin2xx2=limx→0sinππsin2xx2We know:
sinπx=sin x ∴limx→0sinππsin2xx2=limx→0sinπsin2xx2Multiplying the numerator and the denominator by
πsin2x, we get:
limx→0sinπsin2xx2 =limx→0sinπsin2xπsin2x×limx→0πsin2xx2 =1 ×πlimx→0sinxx2=π ∵ limx→0sinxx=1Hence, the correct option is (4). [[VIDEO:13785]]
Question 54:
If X = {4^{n} − 3n − 1 : n ∈ N} and Y = {9(n − 1) : n ∈ N}, where N is the set of natural numbers, then X ∪ Y is equal to:
Option 1:  N 
Option 2:  Y − X 
Option 3:  X 
Option 4:  Y 
Solutions:Given: X = {4^{n} − 3n − 1 : n ∈ N}= {(1+3)^{n} − 3n − 1 : n ∈ N}
Expanding binomially, we get:
4n3n1=C0n+C1n3+C2n32……………..+Cnn3n3n1 =1+3n+C2n32……………..+3n3n1 =C2n32+C3n33…………….+3nTaking 9 as common, we get:9C2n+C3n3………………..+3n2The set X has natural numbers that are multiples of 9.
Also, Y = {9(n − 1) : n ∈ N}
⇒ Y has all the multiples of 9
i.e.,
X⊆Y∴
X∪Y=YHence the correct option is (4). [[VIDEO:13786]]
Question 55:
The locus of the foot of perpendicular drawn from the centre of the ellipse x^{2} + 3y^{2} = 6 on any tangent to it is:
Option 1:  (x^{2} − y^{2})^{2} = 6x^{2} + 2y^{2} 
Option 2:  (x^{2} − y^{2})^{2} = 6x^{2} − 2y^{2} 
Option 3:  (x^{2} + y^{2})^{2} = 6x^{2} + 2y^{2} 
Option 4:  (x^{2} + y^{2})^{2} = 6x^{2} − 2y^{2} 
Solutions:x^{2} + 3y^{2} = 6
⇒x26+y22=1 … (1)
The equation of the tangent to the ellipse x^{2} + 3y^{2} = 6 is as follows:
y=mx±a2m2+b2 ∵a^{2} = 6, b^{2} = 2
∴y=mx±6m2+2 … (2)
Let P(h, k) be the foot of the perpendicular to the tangent.
The equation of the line that passes through the centre of the ellipse x^{2} + 3y^{2} = 6, i.e. (0, 0), and is perpendicular to
y=mx±6m2+2is as follows:
y=1mx … (3)
Eliminating m from equations (2) and (3), we get the required locus as follows:
y=xy×x±6×x2y2+2⇒x2+y22=6×2+2y2Hence, the correct option is (3).
Question 56:
Three positive numbers from an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is :
Option 1:  2+3 
Option 2:  3+2 
Option 3:  23 
Option 4:  2+3 
Solutions:Let the numbers in the G.P. be a, ar, ar^{2}.
According to the question:
a, 2ar, ar^{2} are in A.P.
∴ 2(2ar) = a + ar^{2}
⇒ 4r = 1 + r^{2}
⇒ r^{2}− 4r + 1 = 0
⇒
r=2±3Since three positive numbers form an increasing G.P.,
∴
r=2 +3Hence, the correct option is (4).
Question 57:
If (10)^{9} + 2(11)^{1} (10)^{8} + 3(11)^{2 }(10)^{7} + …..10 (11)^{9} = k(10)^{9}, then k is equal to :
Option 1:  12110 
Option 2:  441100 
Option 3:  100 
Option 4:  110 
Solutions:Let
S = 109 + 2· 111·108 + 3· 112· 107+ …… + 10·119 … (1)
1110·S = 111·108 + 2.112·107 + ……..+ 9·119 + 1110 … (2)
Subtracting (2) from (1), we get:
⇒ 110S = 109 + 111·108 +112·107+ ……. + 119 – 1110 ⇒ 110S=1091+1110+11102+…+111091110
⇒ 110S = 109 111010 – 11110 1 – 1110
⇒ 110S = 1110 – 1010 – 1110
S = 1011S = 100.109
Comparing with
k109, we get:
k = 100
Hence, the correct option is (C).
[[VIDEO:13787]]
Question 58:
The angle between the lines whose direction cosines satisfy the equations l + m + n = 0 and l^{2} = m^{2} + n^{2} is :
Option 1:  π3 
Option 2:  π4 
Option 3:  π6 
Option 4:  π2 
Solutions:The direction cosines of the two lines satisfy the relation
l+m+n=0 …1 l2=m2+n2 …2From (1) and (2), we have:
mn2=m2+n2⇒m2+n2+2mn=m2+n2⇒mn=0⇒m=0 or n=0Now, we know:
l2+m2+n2=1 …3
Here, m = 0, l = − n
Using (3), we have
n2+0+n2=1⇒2n2=1⇒n2=12⇒n=±12∴
l1, m1, n1=12, 0, 12 or 12, 0, 12Here, n = 0, l = − m
Using (3), we have:
m2+m2+0=1⇒2m2=1⇒m2=12⇒m=±12∴
l2, m2, n2=12, 12, 0 or 12, 12,0Let
θbe the angle between the two lines.
∴ cosθ=l1l2+m1m2+n1n2=12×12+0×12+12×0i.e. cosθ=12⇒θ=π3Hence, the correct option is (1). [[VIDEO:13788]]
Question 59:
The slope of the line touching both the parabolas y^{2} = 4x and x^{2} = −32y is :
Option 1:  12 
Option 2:  32 
Option 3:  18 
Option 4:  23 
Solutions:Let A (t^{2}, 2t) be a point on the parabola y^{2} = 4x.
The equation of tangent at A, yt = x + t^{2}, is tangent to the parabola x^{2} + 32y = 0 at B.
∴
x2+32xt+t=0 ⇒x2+32xt+32t=0Now, the discriminant of the above quadratic equation should be zero.
∴ 32t2432t=0 ⇒3232t24t=0 ⇒32t2=4t⇒t3=8⇒t=2∴ Slope of tangent =
1t=12Hence, the correct option is (1). [[VIDEO:13789]]
Question 60:
If x = −1 and x = 2 are extreme points of f(x) = α log x = βx^{2} + x then :
Option 1:  α=6, β=12 
Option 2:  α=6, β=12 
Option 3:  α=2, β=12 
Option 4:  α=2, β=12 
Solutions:
fx=αlogx+βx2+x ⇒f’x=αx+2βx+1=2βx2+α+xxNow, x = −1 and x = 2 are extreme points of f (x).
∴ f’1=0 and f’2=0 ⇒2β12+α+1=0 and 2β22+α+2=0 ⇒2β+α1=0 and 8β+α+2=0On solving the two equations, we get:
α=2 and β=12Hence, the correct option is (3).
Question 61:
Which one of the following properties is not shown by NO?
Option 1:  It combines with oxygen to form nitrogen dioxide. 
Option 2:  It’s bond order is 2.5. 
Option 3:  It is diamagnetic in gaseous state. 
Option 4:  It is a neutral oxide. 
Solutions:NO is a neutral oxide. It combines with oxygen to form NO_{2}.
2NOg + O2g → 2NO2g
Number of electrons in an NO molecule = 7 + 8 = 15
The molecular electronic configuration of NO is
σ1s2 , σ*1s2, σ2s2, σ*2s2, σ2pz2 , π2px2=π2py2, π*2px1 =π*2py0Bond order of NO =
NbNa2=10 – 52= 2.5It has one unpaired electron in the
πantibonding molecular orbital. As a result, it exhibits paramagnetic character in its gaseous state.
Hence, the correct option is (3).
Question 62:
If Z is a compressibility factor, van der Waals’ equation at low pressure can be written as
Option 1:  Z=1PbRT 
Option 2:  Z=1+PbRT 
Option 3:  Z=1+RTPb 
Option 4:  Z=1aVRT 
Solutions:Vander Waals’ equation for one mole of a gas is
P + aV2Vb = RTAt low pressure, the volume of a gas is extremely large. As a result, van der Waals’ constant b can be neglected in comparison to V. (V – b) = V
P + aV2V = RT⇒PV = RT – aV⇒PVRT = 1 – aRTV∵PVRT=Z∴Z=1aRTVHere, Z is compressibility factor.
Hence, the correct option is (4). [[VIDEO:13886]]
Question 63:
The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is
Option 1:  Cu 
Option 2:  Cr 
Option 3:  Ag 
Option 4:  Ca 
Solutions:Among the given metals, only Ca is an sblock element. It is known that during the electrolysis of sblock element, H_{2} gas is discharged at the cathode.
2H2O + 2e→H2g + 2OHaq. Eo = 0.83 VCa2+ + 2e →Cas Eo = 2.87 V
Since, the reaction for the higher value of E^{o} is preferred hence, during electrolysis at cathode the following reaction will occur.
2H2O + 2e → H2 (g) + 2OH (aq)Therefore, Ca cannot be obtained by the electrolysis of an aqueous solution of its salts.
Hence, the correct option is (4). [[VIDEO:13887]]
Question 64:
Resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution is 1.4 S m^{−1}. The resistance of 0.5 M solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in S m^{2} mol^{−1} is
Option 1:  5 × 10^{3} 
Option 2:  5 × 10^{2} 
Option 3:  5 × 10^{−4} 
Option 4:  5 × 10^{−3} 
Solutions:For 0.2 M solution of the electrolyte: Resistance, R = 50 ohm Specific conductance, σ = 1.4 Sm^{−1}= 1.4
×10^{−2} S cm^{−1}
Specific conductance can be expressed as: σ=la×1R la = R×σla= 50×1.4×102 cm1(where l/a is called cell constant)
Now, for 0.5 M solution of the electrolyte: Resistance, R = 280 ohm Molar conductivity, λ is expressed as:
λ=σ×1000M σ=la×1R =50×1.4×102280 = 2.5×103 S cm1 λ = σ×1000M = 2.5×103×10000.5 =5 S cm2 mol1 = 5×104 S m2 mol1Hence, the correct option is (3).
Question 65:
CsCI crystallises in bodycentred cubic lattice. If ‘a‘ is its edge length, then which of the following expressions is correct?
Option 1:  rCs++rCl=32a 
Option 2:  rCs++rCl=3a 
Option 3:  rCs++rCl=3a 
Option 4:  rCs++rCl=3a2 
Solutions:CsCl crystallises in a bcc lattice, in which Cs^{+} ion is present at the centre and the Cl^{−} ions are present at the edges of the unit cell. For a bcc lattice, the relation between the nearest neighbour distance (d) and the edge length (a) is expressed as:
d = 32a
Now, the nearest neighbour distance is equal to the sum of the radii of the cation and the anion.
d = rCs+ + rCl⇒rCs+ + rCl = 32aHence, the correct option is (1).
Question 66:
Consider separate solutions of 0.500 M C_{2}H_{5}OH(aq), 0.100 M Mg_{3}(PO_{4})_{2}(aq), 0.250 M KBr(aq) and 0.125 M Na_{3}PO_{4}(aq) at 25° C. Which statement is true about these solutions, assuming all the salts to be strong electrolytes?
Option 1:  0.125 M Na_{3}PO_{4}(aq) has the highest osmotic pressure. 
Option 2:  0.500 M C_{2}H_{5}OH(aq) has the highest osmotic pressure. 
Option 3:  They all have the same osmotic pressure. 
Option 4:  0.100 M Mg_{3}(PO_{4})_{2}(aq) has the highest osmotic pressure. 
Solutions:
Osmotic pressure is given by the formula mentioned below π = iCRT â€‹Where
π is osmotic pressure, i is van’t Hoff factor, C is concentration, R is universal gas constant and T is the temperature. We know,
α = i1n1Where
α is the dissociation constant and n is the number of particles released on dissociation. As
α= 1, Assuming all salts as strong electrolyte, therefore the relation between n and i can be written as: n = i
Electrolyte  n  i 

C_{2}H_{5}OH  1  1 
Mg_{3}(PO_{4})_{2}  5  5 
KBr  2  2 
Na_{3}PO_{4}  4  4 
(i) For 0.5 M C_{2}H_{5}OH:
π=1×0.5×R×T= 0.5 RT(ii) For 0.1 M Mg_{3}(PO_{4})_{2}:
π=5×0.1×R×T= 0.5 RT (iii) For 0.25 M KBr:
π=2×0.25×R×T= 0.5 RT(iv) For 0.125 M Na_{3}PO_{4}:
π=4×0.125×R×T= 0.5 RTAll the given solutions have the same osmotic pressure.
Hence, the correct option is (3).
Question 67:
In which of the following reactions, H_{2}O_{2} acts as a reducing agent? (a) H_{2}O_{2} + 2H^{+} + 2e^{−} → 2H_{2}O (b) H_{2}O_{2} − 2e^{−} → O_{2} + 2H^{+} (c) H_{2}O_{2} + 2e^{−} → 2OH^{−} (d) H_{2}O_{2} + 2OH^{−} − 2e^{−} → O_{2} + 2H_{2}O
Option 1:  (a), (c) 
Option 2:  (b), (d) 
Option 3:  (a), (b) 
Option 4:  (c), (d) 
Solutions:
The substance which itself undergoes oxidation acts as a reducing agent. (a) (b) (c) (d)
In equations (b) and (d), H_{2}O_{2} undergoes oxidation thus in both reactions it will acts as a reducing agent.
Hence, the correct option is (2).
Question 68:
In S_{N}2 reactions, the correct order of reactivity for the compounds provided in brackets is
CH3Cl, CH3CH2Cl,CH32CHCl and CH33CCl
Option 1:  CH3CH2Cl>CH3Cl>CH32CHCl>CH33CCl 
Option 2:  CH32CHCl>CH3CH2Cl>CH3Cl>CH33CCl 
Option 3:  CH3Cl>CH32CHCl>CH3CH2Cl>CH33CCl 
Option 4:  CH3Cl>CH3CH2Cl>CH32CHCl>CH33CCl 
Solutions:The rate of bimolecular nucleophilic substitution reaction (S_{N}2) depends on the stericcrowding in the alkyl halide. More the number of alkyl groups present around the carbon atom bearing the chloride group, lesser is the possibility of the attack of the nucleophile. So, the correct order of reactivity is CH_{3}Cl > CH_{3}CH_{2}Cl > (CH_{3})_{2}CHCl > (CH_{3})_{3}CCl
Hence, the correct option is (4).
Question 69:
The octahedral complex of a metal iron M^{3}^{+} with four monodentate ligands L_{1}, L_{2}, L_{3} and L_{4} absorbs wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is
Option 1:  L_{3} < L_{2} < L_{4} < L_{1} 
Option 2:  L_{1} < L_{2} < L_{4} < L_{3} 
Option 3:  L_{4} < L_{3} < L_{2} < L_{1} 
Option 4:  L_{1} < L_{3} < L_{2} < L_{4} 
Solutions:
The energy order of light in the visible region is Violet > Indigo > Blue > Green > Yellow > Orange > Red.
Strongfield ligands absorb higher energy radiations and weakfield ligands absorb lower energy radiations. Therefore, the ligand that absorbs blue colour will be the strongfield ligand and the one that absorbs red colour will be the weakfield ligand. So, the increasing order of strength of ligands is L_{1}<l_{3<l2<l4.</l</l</l}
Hence, the correct option is (4).
Question 70:
For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of
M10sulphuric acid. The unreacted acid required 20 mL of
M10sodium hydroxide for complete neutralisation. The percentage of nitrogen in the compound is
Option 1:  3% 
Option 2:  5% 
Option 3:  6% 
Option 4:  10% 
Solutions:The percentage of nitrogen can be found by the following formula:
% N = 1.4 × Molarity of H2SO4 × 2(Volume of of H2SO4 required for the neutralization of ammonia)Mass of compound taken for estimation …(i)
Mass of compound taken = 1.4 g Molarity of H_{2}SO_{4} = M/10 Molarity of NaOH = M/10 Volume of NaOH used up for complete neutralisation = 20 mL Total volume of H_{2}SO_{4} = 60 mL Volume of H_{2}SO_{4} used for complete neutralisation with ammonia can be determined by the following steps: We know that n_{a} M_{a} V_{a} = n_{b} M_{b} V_{b} …. (ii) Where n_{a} and n_{b} are the basicity and acidity of acid and base, respectively. M_{a }and M_{b} are the molarities and Va and V_{b} are the volumes of acid and base, respectively. M_{a} = 1/ 10 M M_{b} = 1/10 M n_{a} = 2, since H_{2}SO_{4} is a dibasic acid n_{b} = 1 as NaOH is monoacidic base V_{b} = 20 mL
On substituting the above values in formula (ii), we get:
2
× (1/10 )M
× V_{a} = 1
× (1/10 )M
× 20 mL
V_{a} = 10 mL
Hence, the volume of H_{2}SO_{4} left unused is 10mL. Therefore, the volume of H_{2}SO_{4} used up for neutralisation = (60 – 10) mL Volume of H_{2}SO_{4} used for complete neutralisation of NH_{3} is 50 mL. â€‹ Now,
% N = 1.4 × (1/10) × 2(50)1.4 = 10This implies that the percentage of nitrogen present in 1.4 g of compound is 10.
Hence, the correct option is (4).
Question 71:
The equivalent conductance of NaCl at concentration C and at infinite dilution are λ_{C} and λ_{∞}, respectively. The correct relationship between λ_{C} and λ_{∞} is (where the constant B is positive)
Option 1:  λC=λ∞BC 
Option 2:  λC=λ∞+BC 
Option 3:  λC=λ∞+BC 
Option 4:  λC=λ∞BC 
Solutions:
The molar conductivity of a strong electrolyte varies with the concentration, according to the DebyeHuckelOnsager equation mentioned below.
λc = λ∞ (B) CHere,
λc is the molar conductivity at any concentration,
λ∞ is the molar conductivity at infinite dilution, B is constant (whose nature depends upon electrolyte, temperature and nature of solvent) and C is the concentration of electrolyte.
Hence, the correct option is (1).
Question 72:
For the reaction
SO2g+12O2g⇌SO3g, if KP=KCRTx, where the symbols have usual meaning, then the value of x is (Assuming ideality)
Option 1:  12 
Option 2:  1 
Option 3:  −1 
Option 4:  12 
Solutions:
Consider the following reaction: SO_{2}(g) +
12O_{2}(g) â‡Œ SO_{3} (g) For the above reaction, the relation between K_{p} and K_{c} is expressed as K_{p} = K_{c} (RT)^{x} Here, K_{p} is the equilibrium constant in terms of pressure. K_{c} is the equilibrium constant in term of concentration. R is the gas constant T is the temperature. x is the change in the number of moles of gaseous substance.
x = Number of moles of gaseous product − Number of moles of gaseous reactant …………….(i) Number of moles of gaseous product = 1 Number of moles of gaseous reactant = ( 1 + 1/2) = 3/2 On substituting the above values in equation (i), we get: x = 1 − 3/2 x = −1/2
Hence, the correct option is (4).
Question 73:
In the reaction,
CH3COOH→ LiAIH4 A→ PCI5 B→ Alc.KOH C,the product C is
Option 1:  ethylene 
Option 2:  acetyl chloride 
Option 3:  acetaldehyde 
Option 4:  acetylene 
Solutions:LiAlH_{4} ( Lithium Aluminium hydride ) is a strong reducing agent and generally employed for the reduction of organic compounds. It reduces carboxylic acid, aldehyde and ketones to their respective alcohols.
CH3COOH →LiAlH4 CH3CH2OHPCl_{5}, on reacting with alcohols, cleaves the carbonoxygen bond to produce alkyl halide.
CH3CH2OH →PCl5 CH3CH2ClAlcoholic KOH is a wellknown regent for the conversion of alkyl halide to alkenes.
CH3CH2Cl →Alc. KOH CH2=CH2When ethanoic acid is treated with LiAlH_{4}, it yields ethanol (A). Ethanol, on treatment with PCl_{5}_{,} results in 1Chloroethane (B). When 1Chloroethane is treated with alcoholic KOH, it undergoes elimination reaction to produce ethene (C). The equation for the above reaction is as follows:
CH3COOH → LiAlH4 CH3CH2OH → PCl5 CH3CH2Cl → Alc. KOH CH2=CH2Ethanoic acid Ethanol Ethyl chloride EthyleneThe common name of ethene is ethylene.
Hence, the correct option is (1).
Question 74:
Sodium phenoxide, when heated with CO_{2} under pressure at 125°C, yields a product that on acetylation produces C.
The major product C would be
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:Sodium phenoxide ion, when treated with carbon dioxide under high pressure (5 atm) and high temperature (125 ^{o}C), yields sodium salicylate (B). When sodium salicylate is treated with acetic anhydride, it results in the acetylation of OH group. The equation of the reaction is given below. The mechanism for the above reaction is as follows:
Hence, the correct option is (3). [[VIDEO:13889]]
Question 75:
On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is
Option 1:  an alkyl cyanide 
Option 2:  an alkyl isocyanide 
Option 3:  an alkanol 
Option 4:  an alkanediol 
Solutions:On heating an aliphatic primary amine with chloroform and ethanolic KOH, alkyl isocyanide is produced. This reaction is known as Carbylamine reaction or Isocyanide test. The alkyl isocyanide, thus formed, has very unpleasant or foul odour. This reaction is exhibited by only aliphatic and aromatic primary amines. Therefore, this test is used to distinguish primary amines from secondary and tertiary amines. The general equation for the reaction is as follows:
RNH2 + CHCl3 + 3KOHethanolic → ∆ RNCalkyl isocyanide + 3KCl + 3H2OHence, the correct option is (2). [[VIDEO:13890]]
Question 76:
Which is the correct statement for the molecule CsI_{3}?
Option 1:  It contains Cs^{3}^{+} and I^{−} ions. 
Option 2:  It contains Cs^{+}, I^{−} and lattice I_{2} molecule. 
Option 3:  It is a covalent molecule. 
Option 4:  It contains Cs^{+} and
I3ions. 
Solutions:Cesium triiodide is an ionic polyhalide compound that on dissociation produces Cs^{+} and
I3ions. Therefore, it contains Cs^{+} and
I3ions only. Moreover, Cs belongs to alkali metals. Therefore, it exhibits only +1 oxidation state.
Hence, the correct option is (4). [[VIDEO:13891]]
Question 77:
The equation which is balanced and represents the correct product(s) is
Option 1:  [Mg (H_{2}O)_{6}]^{2+} + (EDTA)^{4−}
→ excess NaOH MgEDTA2++6H2O 
Option 2:  CuSO_{4} + 4KCN → K_{2}[Cu(CN)_{4}] + K_{2}SO_{4} 
Option 3:  Li_{2}O + 2KCl → 2LiCl + K_{2}O 
Option 4:  [CoCl(NH_{3})_{5}]^{+} + 5H^{+} → Co^{2}^{+} + 5NH_{4}^{+} + Cl^{−} 
Solutions:Among the given reactions, equation (4) is balanced and represents the correct product because [CoCl(NH_{3})_{5}]^{+ }decomposes under acidic medium. [CoCl(NH_{3})_{5}]^{+} + 5H^{+} → Co^{2}^{+} +
5NH4++ Cl^{−}
Hence, the correct option is (4). [[VIDEO:13892]]
Question 78:
For which of the following molecule. significant μ≠0 ? (a) (b) (c) (d)
Option 1:  Only (c) 
Option 2:  (c) and (d) 
Option 3:  Only (a) 
Option 4:  (a) and (b) 
Solutions:1, 4−benzenedithiol and 1, 4−dihydroxybenzene have non zero dipole moments. Because the two hydroxyl and two thiol groups are not symmetrical about C−O−H and C−S−H respectively, these do not lie in the same plane as the benzene ring. Hence, 1, 4−benzenedithiol and 1, 4−dihydroxybenzene show some value of dipole moment.
1, 4−dichlorobenzene and 1, 4−dicynobenzene are symmetrical and chlorine and cyanide groups lie in the same plane of benzene ring. So, the direction of the bond moment vectors cancel each other and show zero dipole moment.
Hence, the correct option is (2). [[VIDEO:13893]]
Question 79:
For the non−stoichiometric reaction 2A + B → C + D, the following kinetic data were obtained in three separate experiments, all at 298 K.
Initial Concentration (A)  Initial Concentration (B)  Initial rate of formation of C (mol L^{−1}S^{−1} ) 
0.1 M 0.1 M 0.2 M  0.1 M 0.2 M 0.1 M  1.2 × 10^{−3} 1.2 × 10^{−3} 2.4 × 10^{−3} 
The rate law for the formation of C is
Option 1:  dcdt= k[A][B]^{2} 
Option 2:  dcdt= k[A] 
Option 3:  dcdt= k[A][B] 
Option 4:  dcdt= k[A]^{2}[B] 
Solutions:The nonstoichiometric reaction is 2A + B → C + D. Now,
Rate of reaction =12dAdt=dBdt=dCdt=dDdtLet, the rate law for the formation of C be
dCdt= k [A]^{a }[B]^{b} where, a and b are order of reaction with respect to reactant A and B respectively. Data in the given table can be written as: k [0.1]^{a }[0.1]^{b} =
1.2×103 ……..(1) k [0.1]^{a }[0.2]^{b} =
1.2×103 ……..(2) k [0.2]^{a }[0.1]^{b} =
2.4×103 ………(3)
Dividing equation (1) by (2), we get:
k0.1a0.1bk0.1a0.2b=1.2×1031.2×10312b=1So,
12b=120b =0Dividing equation (1) by (3) we get,
k0.1a0.1bk0.2a0.1b=1.2×1032.4×10312a=121a=1Therefore,
dCdt=kA1B0dCdt=kAHence, the correct option is (2).
Question 80:
Which series of reactions correctly represents the chemical relations related to iron and its compounds ?
Option 1:  Fe→ Cl2, heat FeCl3→ heat, air FeCl2→ Zn Fe 
Option 2:  Fe→ O2, heat Fe3O4→ CO, 600°C FeO→ CO, 700°C Fe 
Option 3:  Fe→ dil H2SO4 FeSO4→ H2SO4, O2 Fe2SO43→ heat Fe 
Option 4:  Fe→ O2, heat FeO→ dil H2SO4 FeSO4→ heat Fe 
Solutions:When dry chlorine gas is passed over heated iron fillings, anhydrous ferric chloride is formed. 2Fe + 3Cl_{2} → 2FeCl_{3} On heating FeCl_{3} in the presence of air, FeCl_{2} and Cl_{2} are produced.
2FeCl3 →∆2FeCl2 + Cl2On reduction of FeCl_{2} with Zn, Fe metal is obtained. FeCl_{2} + Zn → Fe + ZnCl_{2}
In option (4), on heating Fe in the presence of oxygen and heat, Fe_{3}O_{4} is obtained. So, this sequence is wrong. In option (3), on heating Fe_{2}(SO_{4})_{3}, oxide is formed. Thus, the given sequence of reactions is also wrong. In option (2), according to Ellingham diagram, the reduction of Fe_{3}O_{4} with CO occurs in the range of 600700 ^{o}C and reduction of FeO to Fe occurs at a higher temperature range. So, the given sequence is not correct as well.
Hence, the correct option is (1). [[VIDEO:13894]]
Question 81:
Considering the basic strength of amines in an aqueous solution, which one has the smallest pK_{b} value ?
Option 1:  (CH_{3})_{3}N 
Option 2:  C_{6}H_{5}NH_{2} 
Option 3:  (CH_{3})_{2}NH 
Option 4:  CH_{3}NH_{2} 
Solutions:The greater the basic strength of a compound, the smaller is its pK_{b} value. In the case of aromatic amines, the lone pair of electrons on the N atom is involved in delocalisation with the aromatic
πelectron cloud of the benzene ring. Thus, aromatic amines are less basic than aliphatic amines. Among the alkyl amines, the secondary amines are most basic due to +I effect and formation of hydrogen bonding in aqueous solution. Thus, among the given compounds, (CH_{3})_{2}NH is most basic and has the smallest pK_{b} value.
Hence, the correct option is (3). [[VIDEO:13895]]
Question 82:
Which one of the following bases is not present in DNA?
Option 1:  Cytosine 
Option 2:  Thymine 
Option 3:  Quinoline 
Option 4:  Adenine 
Solutions:The following bases are present in DNA. A – Adenine G – Guanine T – Thymine C – Cytosine The compound, quinoline, is not present in DNA.
â€‹Hence, the correct option is (3). [[VIDEO:13896]]
Question 83:
The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is
Option 1:  5, 1, 1, +
12 
Option 2:  5, 0, 1, +
12 
Option 3:  5, 0, 0, +
12 
Option 4:  5, 1, 0, +
12 
Solutions:The electronic configuration of Rb is 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}. â€‹So, the last electron enters the 5s atomic orbital. For this orbital, the values of quantum numbers are: n = 5 l = 0 m_{l} = 0 m_{s} =
+12Hence, the correct option is (3). [[VIDEO:13897]]
Question 84:
The major organic compound formed by the reaction of 1, 1, 1 − trichloroethane with silver powder is
Option 1:  2 − Butyne 
Option 2:  2 − Butene 
Option 3:  Acetylene 
Option 4:  Ethene 
Solutions:On reaction of 1, 1, 1− trichloroethane with silver powder, 2−butyne is obtained as the product.
Hence, the correct option is (1).\ [[VIDEO:13898]]
Question 85:
Given below are half – cell reactions. Mn^{2}^{+} + 2e^{−} → Mn ; E^{o} = − 1.18 V 2(Mn^{3}^{+} + e^{−} → Mn^{2}^{+}) ; E^{o} = + 1.51 V The E^{o} for 3Mn^{2}^{+} → Mn + 2Mn^{3}^{+} will be
Option 1:  −0.33 V ; the reaction will not occur 
Option 2:  −0.33 V ; the reaction will occur 
Option 3:  −2.69 V ; the reaction will not occur 
Option 4:  −2.69 V ; the reaction will occur 
Solutions:
Mn2++ 2e→Mn; Eo=1.18V ………………….. 1Mn3++ e→Mn2+; Eo=+1.51 V …………………….2For 3Mn2+ → Mn + 2Mn3+, we need to multiply equation 2 with 2 and reverse the resultant equation.2Mn2+ →2Mn3+ + 2e ; Eo=1.51 V …………………..3On adding equations 1 and 3, we get:3Mn2+ →Mn + 2Mn3+ ; Eo=2.69 V∵ ∆G = nFEo.Value of Eo is negative, the value of ∆G will be positive, which means that the reaction will be nonspontaneous.Hence, the correct option is (3). [[VIDEO:13899]]
Question 86:
The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecules is
Option 1:  1 : 8 
Option 2:  3 : 16 
Option 3:  1 : 4 
Option 4:  7 : 32 
Solutions:Given, mass of oxygen : mass of nitrogen = 1 : 4
Let, the mass of oxygen be x g. ∴ Mass of nitrogen = 4x g Number of moles =
Given massMolar massNumber of moles of
O2 = x32∴ Number of molecules of
O2 = x32×NAThe number of moles of
N2 = 4×28∴ Number of molecules of
N2 = 4×28×NASo, the ratio of number of molecules of oxygen to number of molecules of nitrogen is
=x32×NA : 4×28×NA=x32 : x7=7 : 32Hence, the correct option is (4).
Question 87:
Which one is classified as a condensation polymer?
Option 1:  Teflon 
Option 2:  Acrylonitrile 
Option 3:  Dacron 
Option 4:  Neoprene 
Solutions:Dacron is a polymer that is formed by the condensation of terephthalic acid and ethylene glycol. Neoprene and teflon are the addition polymers of isoprene and tetrafluoroethylene, respectively and acrylonitrile is a monomer of the polymer polyacrylonitrile.
Hence, the correct option is (3).
Question 88:
Among the following oxoacids, the correct decreasing order of acid strength is
Option 1:  HClO_{4} > HClO_{3} > HClO_{2} > HOCl 
Option 2:  HClO_{2} > HClO_{4} > HClO_{3} > HOCl 
Option 3:  HOCl > HClO_{2} > HClO_{3} > HClO_{4} 
Option 4:  HClO_{4} > HOCl > HClO_{2} > HClO_{3} 
Solutions:Acidic strength is defined by the resonance stability of the conjugated base obtained by the loss of hydrogen atom from the given acid.
HClO4 → ClO4 + H+ ……. iHClO3 → ClO3 + H+ ……. iiHClO2 → ClO2 + H+ ……. iiiHOCl → ClO + H+ ……..ivFactors that affect the acidic strength:
1. More the resonating structures of the conjugated base more will be the acidic strength
2. More the number of surrounding oxygen atoms, more will be the oxidation number of the central Cl atom and hence, more will be the acidic strength
1. More the resonating structures of the conjugated base more will be the acidic strength
Conjugated base  Number of resonating structures  Resonance structures 
4  
3  
2  
0  Not possible 
On observing the resonance structures of the conjugated bases, the order of the stability of the conjugated bases obtained can be written as:
ClO4 > ClO3 > ClO2 >ClO2. More the number of surrounding oxygen atoms, more will be the oxidation number of the central Cl atom and hence, more will be the acidic strength
Conjugated base  Number of surrounding oxygen atoms  Oxidation number of central Cl atom 
ClO4 
4  +7 
ClO3 
3  +5 
ClO2 
2  +3 
ClO 
1  +1 
So, on the basis of above two factors the order of the acidic strength is given as HClO_{4} > HClO_{3}_{ }> HClO_{2} > HClO
Hence, the correct option is (1). [[VIDEO:13900]]
Question 89:
For complete combustion of ethanol, C_{2}H_{5}OH(l) + 3O_{2}(g) → 2CO_{2}(g) + 3H_{2}O(l), the amount of heat produced, as measured in a bomb calorimeter, is 1364.47 kJ mol^{−1} at 25°C. Assuming ideality of the enthalpy of combustion, âˆ†_{c}H, for the reaction will be (R = 8.314 kJ mol^{−1})
Option 1:  −1460.50 kJ mol^{−1} 
Option 2:  −1350.50 kJ mol^{−1} 
Option 3:  −1366.95 kJ mol^{−1} 
Option 4:  −1361.95 kJ mol^{−1} 
Solutions:The complete combustion of ethanol is represented as:
C2H5OHl + 3O2g → 2CO2 g + 3H2OlGiven, ∆U = 1364.47 kJ mol1∆ng = 2 – 3 = 1R = 8.314 J K1 mol1 = 8.314 × 103 kJ K1 mol1T= 298 K∆H = ∆U + ∆ngRT = 1364.47 + (1×8.314×2981000) = – 1366.947 kJ mol1Hence, the correct option is (3). [[VIDEO:13901]]
Disclaimer: Given value of R is wrong, standard value of R = 8.314 J K^{–1} mol^{–1}
Question 90:
The most suitable reagent for the conversion of R−CH_{2}−OH → R−CHO is
Option 1:  CrO_{3} 
Option 2:  PCC (Pyridinium Chlorochromate) 
Option 3:  KMnO_{4} 
Option 4:  K_{2}Cr_{2}O_{7} 
Solutions:The most suitable reagent for the oxidation of primary alcohols to aldehydes in good amount is PCC. It is a mild oxidising agent.
RCH2OH →PCC RCHO
Reagents like CrO_{3}, KMnO_{4} and K_{2}Cr_{2}O_{7} oxidise the alcohol directly to carboxylic acid.
Hence, the correct option is (2). [[VIDEO:13902]]