
IIT Mains 2015
Test Name: IIT Mains 2015
Question 1:
Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m s^{1} and 40 m s^{1} respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m s^{2}). (The figures are schematic and not drawn to scale)
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:The displacement of the stones w.r.t to the time t is given by
y=ut12gt2where y is the displacement, u is the initial velocity of the stone g is acceleration due to gravity
Time required by the first stone to reach the ground, t_{1}_{ }isâ€‹
240=10t112×10×t12⇒240=10t15t12⇒5t1210t1240=0⇒t18t1+6=0⇒t1=8, 6 ∴ t1=8 sTime required by the second stone to reach the ground, t_{2}_{ }is
240=40t212×10×t22⇒5t240t2240=0⇒t212t2+8=0⇒t2=12, 8∴t2=12 sFor t = 0 to t = 8 s, both the stones are in air. Thus,
y2y1=30t⇒y2y1∝t∴The graph of y2y1 against t is a straight line for 0≤t≤8After t = 8 s, stone will fall to ground and will remain there. So, y_{1} = 240 and
y2=40t12gt2⇒y2y1=40t12gt2⇒y2=40t12gt2240As stone 2 has acceleration with respect to stone 1. The graph (3) is the correct description.
Hence, the correct answer is option C.
Question 2:
The period of oscillation of a simple pendulum is
T=2πLg.Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g is
Option 1:  2% 
Option 2:  3% 
Option 3:  1% 
Option 4:  5% 
Solutions:
Given, T2=4π2.Lg⇒ g=4π2.LT2Accuracy in determining the value of g can be calculated as:∆gg×100=∆LL×100+2.∆TT×100⇒∆gg×100=0.120×100+2×190×100⇒∆gg×100=2.7 %≃3 %Thus, the accuracy in determination of g is 3%.
Hence, the correct option is (B) [[VIDEO:13819]]
Question 3:
Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is:
Option 1:  100 N 
Option 2:  80 N 
Option 3:  120 N 
Option 4:  150 N 
Solutions:For the given arrangement, the free body diagram of the two blocks will be given as below:
The frictional force f applied by the wall, will be in upward direction. So, for the system to be in equilibrium, we have f = 120 N
Hence, the correct option is (3).
[[VIDEO:13820]]
Question 4:
A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:
Option 1:  44% 
Option 2:  50% 
Option 3:  56% 
Option 4:  62% 
Solutions: As the collision is perfectly inelastic, the momentum will remain conserved. Let v_{x} and v_{y} be the components of final velocity v‘ in x and y directions respectively.
m.2v=m+2mvx⇒vx=23vAlso, applying the conservation of momentum is y direction, we get
2m.v=m+2mvy⇒vy=23vThe resultant velocity after collision will be
v’=vx2+vy2⇒v’=23v2+23v2⇒v’=223vThe kinetic energy, E_{1} before collison is
E1=12m2v2+122m.v2⇒E1=12m.4v2+mv2⇒E1=2mv2+mv2⇒E1=3mv2The kinetic energy, E_{2} after collison is â€‹
E2=123m.22v32⇒E2=32m.8v29=43v2Loss in kinetic energy is given by
E1E2=3v243v2=9v24v23=53v2Percentage loss in Kinetic Energy = Loss in Kinetic EnergyInitial Kinetic Energy ×100 =E1E2E1×100=53v23v2×100⇒Percentage loss=59×100=55.6≈56%Hence, the correct option is (3)
Question 5:
Distance of the centre of mass of a solid uniform cone from its vertex is z_{0}. If the radius of its base is R and its height is h then z_{0} is equal to:
Option 1:  h24R 
Option 2:  3h4 
Option 3:  5h8 
Option 4:  3h28R 
Solutions:
Consider a disc of radius, x of width dy at depth y from the vertex point of the cone. Let M be the mass of the cone. ∴ Mass of the disc will be
dm=M13πR2h×πx2dyNow due to symmetry of the triangle OAB and OCD
xy=RH
⇒x=Ryh∴dm=3MπR2h×πRyh2dy⇒dm=3My2h3dyThe location of centre of mass of the disc from the vertex can be calculated as ycm=1M∫dm.y⇒ycm=1M∫3My2h3dy.y∴Location of centre of mass of the cone from the vertex can be calculated as: z0 =1M∫0h3My2h3dy.y⇒z0 =3h3∫0Hy3dy⇒z0=34hHence, the correct option is (2).
[[VIDEO:13822]]
Question 6:
From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is:
Option 1:  MR2322π 
Option 2:  MR2162π 
Option 3:  4MR293π 
Option 4:  4MR233π 
Solutions:
A cube is cut from a sphere of radius R. Let a be the side of cube. If the cube is of maximum volume then, the diagonal of the cube must be equal to the diameter of the sphere. So, we have
3a=2Ra=2R3The volume of sphere is given by,
V=4π3R3The density of the sphere will be d=MV⇒d=M4π3R3=3M4πR3The density of both the cube and the sphere will remain same. Thus, Mass of the cube, M‘ is given by
M’=da3=3M4πR38R333⇒M’=2M3πThe moment of inertia of the cube is
I=M’a26=2M3π×16×2R32⇒I=4MR293πHence, the correct option is (3).
[[VIDEO:13821]]
Question 7:
From a solid sphere of mass M and radius R, a spherical portion of radius
R2is removed, as shown in the figure. Taking gravitational potential V = 0 at r = ∞, the potential at the centre of the cavity thus formed is: (G = gravitational constant)
Option 1:  GM2R 
Option 2:  GMR 
Option 3:  2GM3R 
Option 4:  2GMR 
Solutions:
Gravitational potential at the centre of cavity is V = V_{1}_{ }– V_{2} where V_{1}_{ }is potential at the centre of cavity due to complete sphere of radius R V_{2}_{ }is potential at the centre of cavity due to sphere of radius R/2.
Potential at an internal point of solid sphere at a distance ‘r‘ is given by
V=GMR32r22R2To calculate V_{1}
r=R2V1=GMR32R28R2=118GMRTo calculate V_{2}
V2=32.GM8R2=38GMR
Net potential, V=V1V2V=118GMR+38GMR=GMRHence, the correct option is (2)
[[VIDEO:13823]]
Question 8:
A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to T_{M}. If the Young’s modulus of the material of the wire is Y then
1Yis equal to: (g = gravitational acceleration)
Option 1:  TMT21AMg 
Option 2:  TMT21MgA 
Option 3:  1TMT2AMg 
Option 4:  1TTM2AMg 
Solutions:Time period of a simple pendulum is given by ,
T=2πlg …..(1)where l is length of the simple pendulumg is the acceleration due to gravityOn increasing the mass of the bob, the length of the pendulum increases. Let Δâ€‹l be the increment in the length of the pendulum Then, new Time period, T_{M }will be
TM=2πl+Δlg …..(2)Dividing (2) by (1), we get
TMT=l+∆ll⇒TMT2=l+∆ll=1+∆ll⇒∆ll=TMT21 …..(3)The young modulus, Yâ€‹ is given by
Y=FA.lΔlY=MgA.lΔl⇒1Y=AMg.ΔllFrom (3), we get
1Y=AMgTMT21Hence, the correct option is (1)
[[VIDEO:13824]]
Question 9:
Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume
u=UV∝T4and pressure
P=13UV.If the shell now undergoes an adiabatic expansion the relation between T and R is:
Option 1:  T∝eR 
Option 2:  T∝e3R 
Option 3:  T∝1R 
Option 4:  T∝1R3 
Solutions:Given
P=13UVAlso, UV∝T4Thus, P=13UV =13kT4where k is some constant
For an ideal gas PV=nR1T⇒ P=nR1TV where R1 is molar gas constantThus, nR1TV∝T4⇒nR1V∝T3⇒1V∝T3or 14π3R3∝T3⇒1R3∝T3T∝1RHence, the correct option is (3)
[[VIDEO:13825]]
Question 10:
A solid of constant heat capacity 1 J/°C is being heated keeping it in contact with reservoirs in two ways: (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat. In both the cases body is brought form initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively is:
Option 1:  ln2, 4ln2 
Option 2:  ln2, ln2 
Option 3:  ln2, 2ln2 
Option 4:  2ln2, 8ln2 
Solutions:Since each reservoir supplies same amount of heat the change in temperature of the body will be equal when kept in contact with both reservoirs. Since it is kept in contact with two reservoirs, the change in temperature will be same and can be calculated as:
Δs=∫T1T2dQT+∫T2T3dQTor⇒Δs=∫T1T2CdTT+∫T2T3CdTT⇒Δs=C∫100150dTT+C∫150200dTT⇒Δs=ClnT100150+ClnT150200⇒Δs=Cln150ln100+Cln200ln150⇒Δs=Cln200ln100=Cln200100=Cln2Similarly, Change in entropy for the system (ii) can be calculated as:
Since it is kept in contact with eight reservoirs such that each reservoir supplies same amount of heat, the change in temperature will be same in each case and can be calculated as
Δs= ∫T1T2dQT+∫T2T3dQT+∫T3T4dQT+∫T4T5dQT+∫T5T6dQT+∫T6T7dQT+∫T7T8dQT+∫T8T9dQT⇒Δs=∫100112.5CdTT+∫112.5125CdTT+∫125137.5CdTT+∫137.5150CdTT+∫150162.5CdTT+∫162.5175CdTT+∫175187.5CdTT+∫187.5200CdTT⇒Δs=ClnT100112.5 +ClnT112.5125 +ClnT125137.50 +ClnT137.5150 +ClnT150162.5 +ClnT162.50175 +ClnT175187.5 +ClnT187.50200 ⇒Δs=Cln112.5ln100+Cln125ln112.5+Cln137.5ln125+Cln150ln137.5+Cln162.5ln150+Cln175ln162.5+Cln187.5ln175+Cln200ln187.5⇒Δs=Cln200ln100=Cln200100=Cln(2)=ln2 ∵C=1 J οC1Hence, the correct option is (2).
[[VIDEO:13826]]
Question 11:
Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increase as V^{q}, where V is the volume of the gas. The value of q is :
γ=CpCv
Option 1:  3γ+56 
Option 2:  3γ56 
Option 3:  γ+12 
Option 4:  γ12 
Solutions:
The average time of collision is given by, ς=12πd2NV3RTMThus, ς∝VTAs,TVγ1=K⇒T=KVγ1⇒ς∝VVγ1⇒ς∝VVγ12⇒ς∝V1+γ12⇒ς∝Vγ+12Hence, the correct option is (3).
[[VIDEO:13827]]
Question 12:
For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly? (graph are schematic and not drawn to scale).
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:A simple pendulum performs simple harmonic motion. Potential energy in simple harmonic motion (PE) is given by
PE=12mω2x2Where m = mass of the particle x = displacement from the mean position ω = angular speed of particle
Potential energy is maximum at extreme position, at this position x = A = amplitude of simple harmonic motion
Thus, maximum potential energy will be
PEmax=12mω2A2Thus, potential energy is maximum at extreme position and zero at mean position.
Kinetic energy in simple harmonic motion is given by,
KE=12mω2A2x2Thus, kinetic energy in simple harmonic motion is maximum at mean position and decreases as we go away from the mean position, becomes zero at extreme position.
Only graph (2) satisfies this condition.
Hence, the correct option is (2).
[[VIDEO:13828]]
Question 13:
A train is moving on a straight track with speed 20 ms^{−1}. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 ms^{−1}) close to
Option 1:  6 % 
Option 2:  12 % 
Option 3:  18 % 
Option 4:  24 % 
Solutions:Given: Speed of train, v_{s} = 20 m s^{−1} Speed of sound, v = 320 m s^{−1} Frequency of whistle, f_{0} = 1000 Hz
As the train coming close to the person, frequency (f_{1}) of the whistle heard by person standing near the track is given by
f1=f0vvvs⇒ f1=1000×320(32020)⇒f1=1000×320300=1066.67 HzAs the train goes away from him, the frequency (f_{2}) of the whistle heard by the person is given by
f2=f0vv+vs⇒f2=1000×320320+20⇒f2=1000×320340⇒f2=941.17 Hz Percentage change in frequency, d=f1f2f1×100%d=1066.67941.171066.67×100%d=125.41066.67%=11.76%≈12%Hence, the correct answer is option B.
Question 14:
A long cylindrical shell carries positive surface charge σ in the upper half and negative surface charge −σ in the lower half. The electric field lines around the cylinder will look like figure given in : (figure are schematic and not drawn to scale)
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:Electric field lines always start from the positive charge and terminate at the negative charge. Therefore, options (B) and (C) are not possible . Electric field lines are also be smooth curves. Therefore, option (D) is incorrect.
Option (A) satisfies all the properties of electric field lines.
Hence, the correct answer is option A.
Question 15:
A uniformly charged solid sphere of radius R has potential V_{0} (measured with respect to ∞) on its surface. For this sphere the equipotential surfaces with potentials
3V02, 5V04, 3V04and V04have radius R_{1}, R_{2}, R_{3} and R_{4} respectively. Then
Option 1:  R_{1} = 0 and R_{3} > (R_{4} − R_{3}) 
Option 2:  R_{1} ≠ 0 and (R_{2} − R_{1}) > (R_{4} − R_{3}) 
Option 3:  R_{1} = 0 and R_{2} < (R_{4} − R_{3}) 
Option 4:  2R < R_{4} 
Solutions:Let Q be the charge enclosed by the sphere and R be the radius of the sphere. Potential inside the sphere (V_{in}) is given by
Vin=Q4πε0R32r22R2Where r=distance of the point outside the spherePotential on the surface of sphereVS isVS=Q4πε0R=VO givenPotential outside the sphere isVout=Q4πε0rPotential at the centre of the sphere Vcentre is given byVcentre=k43πR3∫0R4πr2drr⇒V0=32kQR k=14πε0=32V0∴ R1=0If54VO is possible inside the sphere∴54VO=VO32r22R2 ∵ r=R2⇒r=R2=R23V04 is possible outside the sphere∴ 3V04=Q4πε0r⇒34Q4πε0=Q4πε0r⇒r=43R=R3Similarly, V04 is possible outside the sphereV04=Q4πε0r⇒r=4R=R4We can see that R2<R4R3Therefore, the correct option is (3) and (4).
Question 16:
In the given circuit, charge Q_{2} on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF. Q_{2} as a function of C is represented by
(Figures are drawn schematically and are not to scale.)
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:Here, 1 μF and 2 μF are in parallel. So, their equivalent capacitance is given by
C’=1+2=3 μFSo, the given circuit is simplified as given below
Now, the capacitances C and C‘ are in series. So, their equivalent capacitance is given by
1Ceqv=1C+13⇒Ceqv=3CC+3The charge flowing through the equivalent capacitance is given by
Q=Ceqv.EThe charge remains the same in the series connection. Therefore, the charge through C and C‘ individually will be equal to Q. But, the total charge distributes in the ratio of capacitances in parallel connection.
The charge Q_{2} through the capacitance 2 μF is given by
Q2=Q×23=2CEC+3=2E1+3CFor
1≤C≤3, Q_{2} will increase as E is constant. Also,
dQ2dC=C+32E2CEC+32⇒dQdC=6EC+32⇒dQdC>0Obtaining the 2nd derivative, we get
d2QdC2=6E2C+32=12EC+32⇒d2QdC2<0So, from the second derivative test, we can say that the variation is concave downwards.
Hence, the correct answer is option B.
Question 17:
When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10^{–4} m s^{–1}. If the electron density in the wire is 8 × 10^{28} m^{–3}, the resistivity of the material is close to :
Option 1:  1.6 × 10^{–8} Ωm 
Option 2:  1.6 × 10^{–7} Ωm 
Option 3:  1.6 × 10^{–6} Ωm 
Option 4:  1.6 × 10^{–5} Ωm 
Solutions:The drift speed of electrons flowing through the conductor is given by
vd=iAne …..(1)where i is the current through the conductorA is its area of cross sectionn is the electron density through itAlso, the electron density in the wire is given by
j=iA=σE …..(2)where σ is the conductivity of material of conductorE is the Electric field across the length of the conductorFrom (1), we haveiA=nevdFrom (2), we havenevd=σE=σVlwhere V is the applied potential difference across the conductorl is the length of the conductor⇒σ=nevdlVBut, ρ=1σwhere ρ is the resisitivity of the material of the conductor∴ρ=Vnevdl=58×1028×1.6×1019×2.5×104×0.1⇒ρ=1.6×105 ΩmHence, the correct option is (D).
[[VIDEO:13832]]
Question 18:
In the circuit shown, the current in the 1 Ω resistor is :
Option 1:  1.3 A, from P to Q 
Option 2:  0 A 
Option 3:  0.13 A, from Q to P 
Option 4:  0.13 A, from P to Q 
Solutions:
Apply KVL in loop (I), we get
6i1i29=0⇒9=6i1i2 …..(1)Now, applying KVL in loop (II), we get
4i2i16=0⇒6=4i2i1 …..(2)Solving (1) and (2), we get
i1i2=0.13 A∴ The current through 1 Ω resistor is 0.13 A from Q to P.
Hence, the correct option is (C).
[[VIDEO:13833]]
Question 19:
Two coaxial solenoids of different radii carry current I in the same direction. Let
F→1be the magnetic force on the inner solenoid due to the outer one and
F→2be the magnetic force on the outer solenoid due to the inner one. Then :
Option 1:  F→1=
F→2= 0 
Option 2: 
F→1is radially inwards and F→2is radially outwards 
Option 3:  F→1is radially inwards and
F→2= 0 
Option 4:  F→1is radially outwards and
F→2= 0 
Solutions:For a solenoid having n turns and through which current I is flowing, the magnetic field:
inside the solenoid is given by
B=μ0nIoutside the solenoid is zero
The magnetic field due to the inner coil at external points will be zero. So, the magnetic force on the outer coil
F→2will be zero.
Also, the magnetic field at any current element of the inner coil due to outer coil will be perpendicular to its length. From Fleming’s left hand rule, the force on each current element will be radially outward. So, net force on the inner coil
F→1will be zero.
∴F1→=F2→=0Hence, the correct option is (1).
[[VIDEO:13834]]
Question 20:
Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle ‘θ‘ with the vertical. If wires have mass λ per unit length then the value of I is : (g = gravitational acceleration)
Option 1:  sinθπλgLμ0cosθ 
Option 2:  2sinθπλgLμ0cosθ 
Option 3:  2πgLμ0tanθ 
Option 4:  2πλgLμ0tanθ 
Solutions:
Let l be the length of each long current carrying thin wire. Then, force on each wire due to other will be given by
F=μ02π.I2l2y From figure, y=Lsinθ⇒F=μ0I2l4πLsinθFrom the equilibrium shown in figure, we have
Tsinθ=μ0I2l4πLsinθ andTcosθ =λgDividing both equations, we gettanθ=μ0I2l4πLsinθλlg⇒I2=4πLsin2θλgμ0cosθ⇒I=2sinθπλgLμ0cosθHence, the correct option is (2).
[[VIDEO:13835]]
Question 21:
A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures below:
If there is a uniform magnetic filed of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?
Option 1:  (a) and (b), respectively 
Option 2:  (a) and (c), respectively 
Option 3:  (b) and (d), respectively 
Option 4:  (b) and (c), respectively 
Solutions:The torque on the current carrying loop in an external magnetic field is given by
τ=MBsinθwhere M is the magnetic moment of the loopB is the external magnetic fieldθ is the angle between themAlso, the potential energy of the fieldmagnet system is given by
U=MBcosθFor figure (a),
θ=90°(b),
θ=0°(c),
θ=90°(d),
θ=180°(i) For stable equilibrium, torque on the loop should be zero and the potential energy of the fieldmagnet system should be minimum. This is possible for case (b), where
τ=MBsin0°=0 andU=MBcos0°=MBwhich is the minimum potential energy of the fieldmagnet system. Therefore, for case (b), orientation of the loop will be stable.
(ii) For unstable equilibrium, torque on the loop should be zero and the potential energy of the fieldmagnet system should be maximum. This is possible for case (d), where
τ=MBsin180°=0 andU=MBcos180°=MBwhich is the maximum potential energy of the fieldmagnet system. Therefore, for case (d), orientation of the loop will be unstable.
Hence, the correct answer is option C.
Question 22:
An inductor (L = 0.03 H) and a resistor (R = 0.15 kΩ) are connected in series to a battery of 15 V EMF in a circuit shown below. The key K_{1} bas been kept closed for a long time. Then at t = 0, K_{1} is opened and key K_{2} is closed simultaneously. At t = 1 ms, the current in the circuit will be : (e^{5} â‰Œ 150)
Option 1:  100 mA 
Option 2:  67 mA 
Option 3:  6.7 mA 
Option 4:  0.67 mA 
Solutions:When key K_{1} closed for long time, the current through the inductor will be
I0=ER=150.15×103=0.1 ANow, K_{1} is opened and K_{2} is closed. So, the current through the circuit will be
i=I0et/τFor t=1 ms=1×10–3 s, time constant will beτ=LR=0.030.15×103=2×104tτ=1×10–32×104=5∴ i = 0.1e5⇒i = 0.67 mAHence, the correct option is (4).
Question 23:
A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the LED is
Option 1:  1.73 V/m 
Option 2:  2.45 V/m 
Option 3:  5.48 V/m 
Option 4:  7.75 V/m 
Solutions:Given: Power of the light source, P = 0.1 W Distance of the point from the light source, r = 1 m The intensity of light at a point at distance r from the source of light is given by,
I = P4πr2Let the average energy of the light be U_{av}.
Since, I=Uav×cand Uav=12∈0 E02⇒12∈0 E02c=P4πr2⇒E0=2P∈0 c×4πr2On substituting the values, ⇒E0=2.45 V m1Hence, the correct answer is option B.
Question 24:
Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is µ, then a ray, incident at an angle θ on the face AB, would get transmitted through the face AC of the prism provided
Option 1:  θ>sin1μsinAsin11μ 
Option 2:  θ<sin1μsinAsin11μ 
Option 3:  θ>cos1μ sinA+sin11μ 
Option 4:  θ<cos1μ sinA+sin11μ 
Solutions:
sinθ=μsinr1⇒sinr1=sinθμ⇒r1=sin1sinθμ …..1μsinθc=1×sin 90°⇒sinθc=1μr2<θc⇒sinr2<sinθc⇒ r2<sin11μr1+r2=A⇒r2=Ar1⇒r2=Asin1sinθμ Using 1⇒Asin1sinθμ<sin11μ⇒Asin11μ<sin1sinθμ⇒sinAsin11μ<sinθμ⇒sin1μsinAsin1μ<θHence, the correct answer is option A.
Question 25:
On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygens’ principle leads us to conclude that as it travels, the light beam :
Option 1:  becomes narrower 
Option 2:  goes horizontally without any deflection 
Option 3:  bends downwards 
Option 4:  bends upwards 
Solutions:According to Huygens’ principle, each point on wavefront behaves as a point source of light. It is given here that the refractive index of air is smallest near the ground and increases with height from the ground. The speeds of the different parts of a plane wavefront, moving horizontally are different. The upper parts of the wavefront have lesser speeds than the lower parts. Also, as the refractive index increases with height, the plane wavefront does not remain horizontal as the rays will bend towards the normal. Thus, the beam of light directed horizontally will bend upwards.
Hence, the correct option is (4).
Question 26:
Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is :
Option 1:  1 μm 
Option 2:  30 μm 
Option 3:  100 μm 
Option 4:  300 μm 
Solutions:Given: Radius of human pupil, r = 0.25 cm Wavelength of light, λ = 500 nm = 5 âœ• 10^{−7} m
Diameter of human pupil, d = 2r = 2 âœ• 0.25 = 0.5 cm
Let θ be the resolving angle for a naked eye. Then, sinθ=1.22λd∴ sinθ=1.22 × 5 ×1075 ×103=1.22 ×104The distance of comfortable viewing, D = 25 cm = 0.25 m Let y be the minimum separation between two objects that the human eye can resolve.
∴ sinθ=yD ⇒ x=Dsinθ=0.25 × 1.22 ×104= 3 × 105 m= 30 μmHence, the correct option is (2).
Question 27:
As an electron makes a transition from an excited state to the ground state of a hydrogen − like atom/ion :
Option 1:  its kinetic energy increases but potential energy and total energy decrease 
Option 2:  kinetic energy, potential energy and total energy decrease 
Option 3:  kinetic energy decreases, potential energy increases but total energy remains same 
Option 4:  kinetic energy and total energy decrease but potential energy increases 
Solutions:For an electron in a hydrogenlike atom/ion of atomic number Z.
The centripetal force is provided by the electrostatic force of attraction.
The kinetic energy of the electron is given by, K= 18πε0Ze2rThe potential energy of the electron is given by, P=14πε0Ze2rThe total energy is given by, T=K+P=18πε0Ze2rwhere r is the radius of the stationary state.
As, the electron makes a transition from an excited state to the ground state, the value of r decreases. Thus, the kinetic energy of the electron increases and potential energy and total energy decrease.
Hence, the correct option is (1).
Question 28:
Match List − I (Fundamental Experiment) with List − II (its conclusion) and select the correct option from the choices given below the list :
List − I  List − II  
(A)  FranckHertz Experiment.  (i)  Particle nature of light 
(B)  Photoelectric experiment.  (ii)  Discrete energy levels of atom 
(C)  DavisonGermer Experiment  (iii)  Wave nature of electron 
(iv)  Structure of atom 
Option 1:  A − (i), B − (iv), C − (iii) 
Option 2:  A − (ii), B − (iv), C − (iii) 
Option 3:  A − (ii), B − (i), C − (iii) 
Option 4:  A − (iv), B − (iii), C − (ii) 
Solutions:Frank−Hertz experiment demonstrated the existence of excited states in mercury atoms. It confirmed that the electrons occupy only discrete, quantized energy states.
Photo−electric experiment proved the particle nature of light.
The DavissonGermer experiment demonstrated the wave nature of the electron.
Hence, the correct option is (3).
Question 29:
A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are :
Option 1:  2 MHz only 
Option 2:  2005 kHz, and 1995 kHz 
Option 3:  2005 kHz, 2000 kHz and 1995 kHz 
Option 4:  2000 kHz and 1995 kHz 
Solutions:Given frequency of carrier wave, f_{c}_{ }= 2 MHz = 2000 kHz The frequency of modulated wave, f_{M}_{ }= 5 kHz The frequencies of the resultant signal can be f_{c} – f_{M}, f_{M}, f_{c}_{ }+ f_{M} Therefore frequencies of the resultant signal can be 2005 Hz, 2000 Hz, 1995 Hz.
Hence, the correct option is (3).
Question 30:
An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q_{0} and then connected to the L and R as shown below :
If a student plots graphs of the square of maximum charge
QMax2on the capacitor with time (t) for two different values L_{1} and L_{2} (L_{1} > L_{2}) of L then which of the following represents this graph correctly? (plots are schematic and not drawn to scale)
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
(1)
Applying Kirchoff law to the close loop we get
qCiRLdidT=0But, i=dqdt∴ qC+dqdtR+Ld2qdt2=0⇒d2qdt2+RLdqdt+qLC=0
As the generral solution of double differential equation d2xdt2+bmdxdt+kmx=0 isA=A0edt2m
So, the solution of above equation can be written asQmaxt=Q0eRt2L⇒Qmax2=Q02eRtLThus, lesser the value of self induction, more the value of damping force. Therefore, the correct option is (1).
Question 31:
The molecular formula of a commercial resin used for exchanging ions in water softening is C_{8}H_{7}SO_{3}Na (mol. wt. 206). What would be the maximum uptake of Ca^{2}^{+} ions by the resin when expressed in mole per gram of resin?
Option 1:  1103 
Option 2:  1206 
Option 3:  2309 
Option 4:  1412 
Solutions:In synthetic resin method of water softening, Na^{+} ions are exchanged by Ca^{+} ions. This is given as 2 C_{8}H_{7}SO_{3}Na_{(S)} + Ca^{2}^{+}_{(aq) }
→(C_{8}H_{7}SO_{3})_{2}Ca_{(s)} + 2 Na^{+}_{(aq)} ResinMass of 1 mol of resin = 206 g Mass of 2 mol of resin = 412 g
412 g of resin uptakes 1 mol of Ca^{2}^{+} ions 1 g of resin uptakes
1412mol of Ca^{2}^{+} ions Threrfore, the maximum uptake of Ca^{2}^{+} ions by the resin when expressed in mole per gram is
1412. Hence, the correct answer is option D.
Question 32:
Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately :
Option 1:  1.86 Å 
Option 2:  3.22 Å 
Option 3:  5.72 Å 
Option 4:  0.93 Å 
Solutions:Unit cell edge length (a) = 4.29
A∘Radius of sodium atom (r) = ?
For BCC lattice,
r =
34a
=
34 ×4.29
= 1.857
A∘ ~ 1.86
A∘Hence, the correct option is (1).
Question 33:
Which of the following is the energy of a possible excited state of hydrogen?
Option 1:  +13.6 eV 
Option 2:  −6.8 eV 
Option 3:  −3.4 eV 
Option 4:  +6.8 eV 
Solutions:Energy of n^{th} shell of hydrogen atom is given by
E_{n }=
13.6n2eV For n = 2 E_{2}_{ = }
13.622=
3.4eV Therefore,
3.4eV is the energy of a possible excited state of hydrogen.
Hence, the correct option is (3).
Question 34:
The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is :
Option 1:  ion−ion interaction 
Option 2:  ion−dipole interaction 
Option 3:  London force 
Option 4:  hydrogen bond 
Solutions:Hydrogen bond is a special case of dipoledipole interaction in which the intermolecular interaction is inversely proportional to the cube of distance between the molecules.
Hence, the correct option is (4).
Question 35:
The following reaction is performed at 298 K.
2NO(g) + O_{2}(g) â‡Œ 2NO_{2}(g) The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO_{2}(g) at 298 K? (K_{p} = 1.6 âœ• 10^{12})
Option 1:  R(298) â„“n (1.6 âœ• 10^{12}) −86600 
Option 2:  86600 + R(298) â„“n (1.6 âœ• 10^{12}) 
Option 3:  86600 – ℓn (1.6 ×1012)R(298) 
Option 4:  0.5 [2 âœ• 86,600 − R(298) â„“n (1.6 âœ• 10^{12})] 
Solutions:For the reaction, 2NO(g) + O_{2}(g) â‡Œ 2NO_{2}(g)
K_{p} = 1.6
×10^{12}
∆GNO(g)°= 86.6 kJ/mol = 86600 J/mol
∆GReaction∘ = 2∆GNO2(g)°2∆GNO(g)°
∆GNO2(g)°= 12∆GReaction° +2∆GNO(g)° = 0.5RT ln Kp+2×86600 = 0.5R298ln1.6×1012+2×86600Hence, the correct option is (4).
Question 36:
The vapour pressure of acetone at 20°C is 185 torr. When 1.2 g of a non − volatile substance was dissolved in 100 g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol^{−1})
Option 1:  32 
Option 2:  64 
Option 3:  128 
Option 4:  488 
Solutions:
Vapour pressure of pure acetoneP0 = 185 torrVapour pressure of solutionP = 183 torrAccording to Raoult law for nonvolatile solute,Relative lowering in vapour pressure, P0PP0= n1n1+n2where n1 = moles of nonvolatile solute n2 = moles of solvent 185183185=1.2M11.2M1+10058 M1= Molar mass of nonvolatile solute M1= 64 g/molHence, the correct option is (B). [[VIDEO:13875]]
Question 37:
The standard Gibbs energy change at 300 K for the reaction 2A â‡Œ B + C is 2494.2 J. At a given time, the composition of the reaction mixture is [A] =
12, [B] = 2 and [C] =
12. The reaction proceeds in the :
[R = 8.314 J/K/mol, e = 2.718]
Option 1:  forward direction because Q > K_{c} 
Option 2:  reverse direction because Q > K_{c} 
Option 3:  forward direction because Q < K_{c} 
Option 4:  reverse direction because Q < K_{c} 
Solutions:
∆G°=RT 1ne Kc 2494.2=8.314 ×300 1ne Kc Kc=e1 Kc= 1e= 12.718 = 0.36 for reaction below, 2A ⇌ B + C reaction quotient Q=BCA2 Q= 2×12122=4As reaction quotientQ is greater than KC , it means reaction proceed in reverse direction.Hence, the correct option is 2.
Question 38:
Two Faraday of electricity is passed through a solution of CuSO_{4}. The mass of copper deposited at the cathode is : (at. mass of Cu = 63.5 amu)
Option 1:  0 g 
Option 2:  63.5 g 
Option 3:  2 g 
Option 4:  127 g 
Solutions:
According to Faraday law of electrolysis, Mass depositM = EF×QE= Atomic mass nFor Cu2+, n=2 and atomic mass =63.5 amuCharge Q=2×F M=63.52×F×2×F M= 63.5 gHence, the correct option is (2).
Question 39:
Higher order (>3) reactions are rare due to :
Option 1:  low probability of simultaneous collision of all the reacting species 
Option 2:  increase in entropy and activation energy as more molecules are involved 
Option 3:  shifting of equilibrium towards reactants due to elastic collisions 
Option 4:  loss of active species on collision 
Solutions:Probability of simultaneous collision decreases exponentially as order increases. For order greater than 3, probability of simultaneous collision is low for all the reacting species.
Hence, the correct option is (1).
Question 40:
3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is :
Option 1:  18 mg 
Option 2:  36 mg 
Option 3:  42 mg 
Option 4:  54 mg 
Solutions:
Initial no. of moles of acetic acid =50×103 ×0.06 =3×103After adsoption no. of moles of acetic acid =50×103×0.042=2.1×103 Molar mass of acetic acid=60 g/mol Mass of acetic acid adsored=32.1×103×60=54×103 g Thus 3g of charcoal adsorbe 54×103 g of acetic acid , 1g of charcoal adsorbe 54×1033 g of acetic acid. Mass of acetic acid= 18×103g=18 mg.Hence, the correct option is (1).
Question 41:
The ionic radii (in Å) of N^{3−}, O^{2−} and F^{−} are respectively :
Option 1:  1.36, 1.40 and 1.71 
Option 2:  1.36, 1.71 and 1.40 
Option 3:  1.71, 1.40 and 1.36 
Option 4:  1.71, 1.36 and 1.40 
Solutions:
As charge on anion increases, its ionic radii also increases. N3 has more charge than O2 and F, so it have larger ionic radii order areN3 > O2 > FHence, the correct option is (3).
Question 42:
In the context of the Hall−Heroult process for the extraction of Aâ„“, which of the following statements is false?
Option 1:  CO and CO_{2} are produced in this process 
Option 2:  Aâ„“_{2}O_{3} is mixed with CaF_{2} which lowers the melting point of the mixture and brings conductivity. 
Option 3:  Aâ„“^{3}^{+} is reduced at the cathode to form Aâ„“. 
Option 4:  Na_{3}Aâ„“F_{6} serves as the electrolyte. 
Solutions:In the the Hall−Heroult process for the extraction of aluminium, Na_{3}AlF_{6} added to reduced the melting point alumina (Al_{2}O_{3}).
Hence, the correct option is (4).
Question 43:
From the following statements regarding H_{2}O_{2}, choose the incorrect statement :
Option 1:  It can act only as an oxidizing agent 
Option 2:  It decomposes on exposure to light 
Option 3:  It has to be stored in plastic or wax lined glass bottles in dark 
Option 4:  It has to be kept away from dust 
Solutions:H_{2}O_{2} can acts as both oxidizing as well as reducing agent.
Hence, the correct option is (1).
Question 44:
Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy?
Option 1:  CaSO_{4} 
Option 2:  BeSO_{4} 
Option 3:  BaSO_{4} 
Option 4:  SrSO_{4} 
Solutions:For solubility, Hydration enthalpy > Lattice enthalpy BeSO_{4} is most soluble among the all alkali earth metal sulphate due to its small size. So, BeSO_{4} has greater hydration enthalpy than its lattice enthalpy.
Hence, the correct option is (2).
Question 45:
Which among the following is the most reactive?
Option 1:  Cl_{2} 
Option 2:  HF 
Option 3:  I_{2} 
Option 4:  ICI 
Solutions:Among the given compounds, ICl is more reactive. Because intra halogen bond is stronger than inter halogen bond. Therefore, interhalogen compound (ICl) is more reactive than intrahalogen compounds (except F_{2}). In HF, intermolecular hydrogen bonding is present. So, it is less reactive.
Hence, the correct answer is option D.
Question 46:
Match the catalysts to the correct processes:
Catalyst  Process  
(A)  TiCl_{3}  (i)  Wacker process 
(B)  PdCl_{2}  (ii)  Ziegler − Natta polymerization 
(C)  CuCl_{2}  (iii)  Contact process 
(D)  V_{2}O_{5}  (iv)  Deacon’s process 
Option 1:  (A) − (iii), (B) − (ii), (C) − (iv), (D) − (i) 
Option 2:  (A) − (ii), (B) − (i), (C) − (iv), (D) − (iii) 
Option 3:  (A) − (ii), (B) − (iii), (C) − (iv), (D) − (i) 
Option 4:  (A) − (iii), (B) − (i), (C) − (ii), (D) − (iv) 
Solutions:
Catalyst  Process  
(A)  TiCl_{3}  (ii)  Ziegler − Natta polymerization 
(B)  PdCl_{2}  (i)  Wacker process 
(C)  CuCl_{2}  (iv)  Deacon’s process 
(D)  V_{2}O_{5}  (iii)  Contact process 
Hence, the correct option is (2).
Question 47:
Which one has the highest boiling point?
Option 1:  He 
Option 2:  Ne 
Option 3:  Kr 
Option 4:  Xe 
Solutions:Noble gas molecules have van der Waals interactions between their atoms. Greater the molecular mass, stronger will be the van der Waals forces between the atoms and higher will be the boiling point. Thus, Xe has the highest boiling point.
Hence, the correct option is (4). [[VIDEO:13876]]
Question 48:
The number of geometric isomers that can exist for square planar [Pt (Cl) (py) (NH_{3}) (NH_{2}OH)]^{+} is (py = pyridine) :
Option 1:  2 
Option 2:  3 
Option 3:  4 
Option 4:  6 
Solutions:[Pt(Cl) (py) (NH_{3}) (NH_{2}OH)]^{+} exists in three geometrical isomers.
Hence, the correct option is (2). [[VIDEO:13877]]
Question 49:
The color of KMnO_{4} is due to:
Option 1:  M → L charge transfer transition 
Option 2:  d → d transition 
Option 3:  L → M charge transfer transition 
Option 4:  σ − σ* transition 
Solutions:In KMnO_{4}, Mn has +7 oxidation state, that means it has no delectrons. Thus the colour of KMnO_{4} is not due to dd transition. The colour of KMnO_{4} is due to ligand to metal charge transfer transition.
Hence, the correct option is (3).
Question 50:
Assertion : Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen. Reason : The reaction between nitrogen and oxygen requires high temperature.
Option 1:  Both assertion and reason are correct, and the reason is the correct explanation for the assertion 
Option 2:  Both assertion and reason are correct, but the reason is not the correct explanation for the assertion 
Option 3:  The assertion is incorrect, but the reason in correct 
Option 4:  Both the assertion and reason are incorrect 
Solutions:Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen as the reaction requires high temperature. Thus both the assertion and reason are correct.
Hence, the correct option is (1). [[VIDEO:13879]]
Question 51:
In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is : (at. mass Ag = 108; Br = 80)
Option 1:  24 
Option 2:  36 
Option 3:  48 
Option 4:  60 
Solutions:Mass of organic compound = 250 mg = 0.25 g Mass of AgBr = 141 mg = 0.141 g Molar mass of AgBr = 108 + 80 = 188 g/mol 188 g AgBr contains 80 g bromine 0.141 g AgBr contains
80188×0.141 gbromine.
Percentage of bromine=80×0.141×100188×0.25=24Hence, the correct option is (1).
Question 52:
Which of the following compounds will exhibit geometrical isomerism?
Option 1:  1−Phenyl−2−butene 
Option 2:  3−Phenyl−1−butene 
Option 3:  2−Phenyl−1−butene 
Option 4:  1,1−Diphenyl−1−propane 
Solutions:Of the given organic compounds, 1−Phenyl−2−butene shows the geometrical isomerism
Hence, the correct option is (1). [[VIDEO:13880]]
Question 53:
Which compound would give 5−keto−2−methyl hexanal upon ozonolysis?
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
Hence, the correct option is (2). [[VIDEO:13881]]
Question 54:
The synthesis of alkyl fluorides is best accomplished by :
Option 1:  Free radical fluorination 
Option 2:  Sandmeyer’s reaction 
Option 3:  Finkelstein reaction 
Option 4:  Swarts reaction 
Solutions:The synthesis of alkyl fluorides is best accomplished by Swarts reaction.
Hence, the correct option is (4).
Question 55:
In the following sequence of reactions :
Toluene →KMnO4 A →SOCl2 B →BaSO4H2/Pd C, the product C is :
Option 1:  C_{6}H_{5}COOH 
Option 2:  C_{6}H_{5}CH_{3} 
Option 3:  C_{6}H_{5}CH_{2}OH 
Option 4:  C_{6}H_{5}CHO 
Solutions:
Hence, the correct option is (4). [[VIDEO:13882]]
Question 56:
In the reaction
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
Hence, the correct option is (3).
Question 57:
Which polymer is used in the manufacture of paints and lacquers?
Option 1:  Bakelite 
Option 2:  Glyptal 
Option 3:  Polypropene 
Option 4:  Poly vinyl chloride 
Solutions:Glyptal is used in the manufacture of paints and lacquers.
Hence, the correct option is (2).
Question 58:
Which of the vitamins given below is water soluble?
Option 1:  Vitamin C 
Option 2:  Vitamin D 
Option 3:  Vitamin E 
Option 4:  Vitamin K 
Solutions:Out of given vitamins, only vitamin C is water soluble.
Hence, the correct option is (1).
Question 59:
Which of the following compounds is not an antacid?
Option 1:  Aluminium hydroxide 
Option 2:  Cimetidine 
Option 3:  Phenelzine 
Option 4:  Ranitidine 
Solutions:Phenelzine is an antidepressant drug. It is not an antacid.
Hence, the correct option is (3). [[VIDEO:13884]]
Question 60:
Which of the following compounds is not colored yellow?
Option 1:  Zn_{2}[Fe(CN)_{6}] 
Option 2:  K_{3}[Co(NO_{2})_{6}] 
Option 3:  (NH_{4})_{3} [As (Mo_{3} O_{10})_{4}] 
Option 4:  BaCrO_{4} 
Solutions:Zn_{2} [Fe(CN)_{6}] is bluish white ppt.
Hence, the correct option is (1). [[VIDEO:13885]]
Question 61:
Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A âœ• B, each having at least three elements is :
Option 1:  219 
Option 2:  256 
Option 3:  275 
Option 4:  510 
Solutions:Number of elements in set A = 4 Number of elements in set B = 2 ∴ Number of elements in set A × B = 4 × 2 = 8 ⇒ Total number of subsets of A × B = 2^{8} = 256 Now, Number of subsets of A × B having 0 elements =
C08=1Number of subsets of A × B having 1 element =
C18=8Number of subsets of A × B having 2 elements
=C28=8!2! 6!=8×72=28∴ Number of subsets having at least 3 elements = Total number of subsets of A × B − Number of subsets of A × B having 0 elements − Number of subsets of A × B having 1 element − Number of subsets of A × B having 2 elements = 256 − 1 − 8 − 28 = 256 − 37 = 219
Hence, the correct option is (1).
Question 62:
A complex number z is said to be unimodular if
z=1. Suppose z_{1} and z_{2} are complex numbers such that
z12z22z1z2is unimodular and z_{2} is not unimodular. Then the point z_{1} lies on a :
Option 1:  straight line parallel to xaxis 
Option 2:  straight line parallel to yaxis 
Option 3:  circle of radius 2 
Option 4:  circle of radius
2 
Solutions:Since,
z12z22z1z2is unimodular.
∴z12z22z1z2=1⇒z12z22=2z1z22⇒z12z2 z12z2=2z1z2 2z1z2 ∵zz=z2⇒z12z2 z12z2=2z1z2 2z1z2
⇒z1z12z1z22z2z1+4z2z2=42z1z22z1z2+z1z1z2z2⇒z1z1+4z2z2=4+z1z1z2z2⇒4+z12 z224 z22z12=0 ∵zz=z2⇒z221z224=0But, z2≠1 ∵z2 is not unimodular∴z2=2This represents a circle with centre (0, 0) and radius 2 units.
Hence, the correct option is (3).
Question 63:
Let α and β be the roots of equation x^{2} − 6x − 2 = 0. If a_{n} = α^{n} − β^{n}, for n ≥ 1, then the value of
a102a82a9is equal to :
Option 1:  6 
Option 2:  −6 
Option 3:  3 
Option 4:  −3 
Solutions:The given equation is
x26x2=0.
Multiplying the given equation by
x8, we get
x106×92×8=0Since α and β are the roots of the given equation, so
α106α92α8=0 …..1β106β92β8=0 …..2Subtracting (2) from (1), we get
α10β106α9β92α8β8=0⇒a106a92a8=0 an=αnβn⇒a102a82a9=3Hence, the correct option is (3).
Question 64:
If
A=122212a2bis a matrix satisfying the equation AA^{T} = 9I, where I is 3 âœ• 3 identity matrix, then the ordered pair (a, b) is equal to:
Option 1:  (2, −1) 
Option 2:  (−2, 1) 
Option 3:  (2, 1) 
Option 4:  (−2, −1) 
Solutions:We have,
AAT=9I⇒122212a2b 12a21222b=9I⇒90a+4+2b092a+22ba+4+2b2a+22ba2+4+b2=900090009So,
a+4+2b=0⇒a+2b=4 …..12a+22b=0⇒2a2b=2 …..2a2+4+b2=0⇒a2+b2=5 …..3Solving 1, 2 and 3, we geta=2, b=1Hence, the correct option is (4).
Question 65:
The set of all values of λ for which the system of linear equations : 2x_{1} − 2x_{2} + x_{3} = λx_{1} 2x_{1} − 3x_{2} + 2x_{3} = λx_{2} −x_{1} + 2x_{2} = λx_{3} has a nontrivial solution,
Option 1:  is an empty set 
Option 2:  is a singleton 
Option 3:  contains two elements 
Option 4:  contains more than two elements 
Solutions:The given system of equations can be rewritten as
(2 − λ)x_{1} − 2x_{2} + x_{3} = 0 2x_{1} − (3 + λ)x_{2} + 2x_{3} = 0 −x_{1} + 2x_{2} − λx_{3} = 0
Now, this system of equations has a nontrivial solution if
∆=2λ2123λ212λ=0⇒ (2 − λ)(3λ + λ^{2} − 4) + 2(−2λ + 2) + (4 − 3 − λ) = 0 (Expanding along R_{1}) ⇒ 6λ + 2λ^{2} − 8 − 3λ^{2} − λ^{3} + 4λ − 4λ + 4 + 1 − λ = 0 ⇒ λ^{3} + λ^{2} − 5λ + 3 = 0 ⇒ λ^{3} − λ^{2} + 2λ^{2} − 2λ − 3λ + 3 = 0 ⇒ λ^{2} (λ − 1) + 2λ(λ − 1) − 3(λ − 1) = 0 ⇒ (λ − 1) (λ^{2} + 2λ − 3) = 0 ⇒ (λ − 1) (λ + 3) (λ − 1) = 0 ⇒ λ = 1, 1, −3
Thus, the set of all values of λ for which the system of linear equations has a nontrival solution contains two elements.
Hence, the correct option is (3).
Question 66:
The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is:
Option 1:  216 
Option 2:  192 
Option 3:  120 
Option 4:  72 
Solutions:Case I: We know All 5digit numbers are greater than 6,000.
∴ Total numbers greater than 6,000 formed using 5 digits = 5! = 120
Case II: Using 4 digits
6,7 or 84 ways3 ways2 waysi.e. 3 ways∴ Total numbers greater than 6,000 formed using 4 digits = 3 × 4 × 3 × 2 = 72
Total number greater than 6,000 formed using the given digits = 120 + 72 = 192
Hence, the correct option is (2). [[VIDEO:13752]]
Question 67:
The sum of coefficient of integral power of x in the binomial expansion of
12x50is :
Option 1:  12350+1 
Option 2:  12350 
Option 3:  123501 
Option 4:  12250+1 
Solutions:
12×50=50C050C12x1+50C22x250C32x3+50C42x4 …50C492x49 +50C502x50Sum of coefficients of integral powers of x in the above expansion,
S=50C0+50C2·22+50C4·24+…+50C50.250Now, 1+x50=50C0+50C1 x+50C2 x2+50C3 x3+50C4 x4+…+50C50 x50 …..1Putting x = 2 and x = −2 in (1), we get
350=50C0+50C1·2+50C2·22+50C3·23+50C4·24+…+50C50·220 …..21=50C050C1·2+50C2·2250C3·23+50C4·24…+50C50·250 …..3Adding (2) and (3), we get
350+1=250C0+50C2.22+50C4.24+…+50C50·250⇒50C0+50C2·22+50C4·24+…+50C50·250=350+12Thus, the sum of the coefficients of integral powers of x in the given binomial expansion is
12350+1.
Hence, the correct option is (1). [[VIDEO:13753]]
Question 68:
If m is the A.M. of two distinct real numbers â„“ and n (â„“, n > 1) and G1, G2 and G3 are three geometric means between â„“ and n, then
G14+2G24+G34equals,
Option 1:  4â„“^{2} mn 
Option 2:  4â„“m^{2}n 
Option 3:  4â„“mn^{2} 
Option 4:  4â„“^{2}m^{2}n^{2} 
Solutions: It is given that m is the A.M. of â„“ and n.
⇒ 2m = â„“ + n …..(1)
Also, â„“, G_{1}, G_{2}, G_{3}, n are in G.P.
Let r be the common ratio of G.P.
∴ G_{1} = â„“r, G_{2} = â„“r^{2}, G_{3} = â„“r^{3}, n = â„“r^{4}
n=ℓr4 ⇒r4=nℓ ∴ G14+2G24+G34=ℓ4r4 1+2r4+r8 =ℓ4.nℓ 1+2nℓ+nℓ2 =nℓ31+nℓ2 =nℓ3n+ℓ2ℓ2 =nℓ2m2 =4ℓm2nHence, the correct option is (2). [[VIDEO:13754]]
Question 69:
The sum of first 9 terms of the series
131+13+231+3+13+23+331+3+5+…is
Option 1:  71 
Option 2:  96 
Option 3:  142 
Option 4:  192 
Solutions:n^{th} term of the given series,
an=13+23+…+n31+3+5+…+2n1
=nn+122n21+2n1 Sn=n2a+l=n2n+124n2=n+124 ∴∑n=19 an=14∑n=19 n+12=1422+32+…+102=1412+22+…+1021=141010+12×10+161 12+22+32+…+n2=nn+12n+16=96Thus, the sum of first nine terms of the series is 96.
Hence, the correct option is (2). [[VIDEO:13755]]
Question 70:
limx→0 1cos2x 3+cosxxtan4xis equal to :
Option 1:  4 
Option 2:  3 
Option 3:  2 
Option 4:  12 
Solutions:
limx→0 1cos2x3+cosxxtan 4x=limx→0 2 sin2x3+cosxxtan4x=limx→02sin2xx23+cosx4xtan4x4x2=2limx→0sinxx23+limx→0cosx4limx→0tan4x4x=2×12×3+14×1=2Hence, the correct option is (3).
Question 71:
If the function,
gx=kx+1 , 0 ≤ x ≤ 3mx+2 , 3 < x ≤ 5is differentiable, then the value of k + m is:
Option 1:  2 
Option 2:  165 
Option 3:  103 
Option 4:  4 
Solutions:We have,
gx=kx+1, 0≤x≤3mx+2, 3<x≤5It is given that g(x) is differentiable function.
∴ g(x) is continuous function.
⇒limx→3gx=limx→3+gx=g3⇒limx→3kx+1=limx→3+mx+2=g3⇒2k=3m+2=2k …..1Also,
g’x=k2x+1, 0≤x≤3 m, 3<x≤5 L.H.D at x=3=limx→3g’x=k4R.H.D at x=3=limx→3+g’x=mSince g(x) is differentiable function.
∴ L.H.D=R.H.D.⇒k4=m …..2From 1 and 2, we getm=25, k=85∴ k+m=2Hence, the correct option is (1).
Question 72:
The normal to the curve, x^{2} + 2xy − 3y^{2} = 0, at (1, 1) :
Option 1:  does not meet the curve again 
Option 2:  meets the curve again in the second quadrant 
Option 3:  meets the curve again in the third quadrant 
Option 4:  meets the curve again in the fourth quadrant 
Solutions:Given:
x^{2} + 2xy − 3y^{2} = 0 …..(1)
Differentiating on both sides with respect to x, we get
2x+2xdydx+y6ydydx=0⇒x+xdydx+y3ydydx=0⇒x3ydydx=x+y⇒dydx=x+yx3y⇒dydx1,1 =1∴ Slope of the normal at (1, 1) =
1dydx1, 1−1
Now, the equation of normal to the curve at (1, 1) is
y − 1 = −1(x − 1)
⇒ x + y = 2 …..(2)
Solving (1) and (2), we get
x^{2} + 2x (2 − x) −3 (2 − x)^{2} = 0 ⇒ x^{2} + 4x − 2x^{2} − 3 (4 + x^{2} − 4x) = 0 ⇒ x^{2} + 4x − 2x^{2} − 12 − 3x^{2} + 12x = 0 ⇒ −4x^{2} + 16x − 12 = 0 ⇒ x^{2} − 4x + 3 = 0 ⇒ (x − 1)(x − 3) = 0 ⇒ x = 1, 3
From (2), we have
When x = 1, y = 1 x = 3, y = −1
So, the point of intersection of normal and the curve are (1, 1) and (3, −1).
Thus, the normal meets the curve again at (3, −1). This point lies in the fourth quadrant.
Hence, the correct option is (4).
Question 73:
Let f(x) be a polynomial of degree four having extreme values at x = 1 and x = 2. If
limx→01+fxx2=3,then f(2) is equal to:
Option 1:  − 8 
Option 2:  − 4 
Option 4:  4 
Solutions:We have
limx→01+fxx2=3⇒limx→0fxx2=2 ∴ fxx2=ax2+bx+2, where a, b∈R and a≠0⇒fx=ax4+bx3+2×2⇒f’x=4ax3+3bx2+4xNow, f’1=4a+3b+4=0 …..1and f’2=32a+12b+8=0⇒8a+3b+2=0 …..2Solving 1 and 2, we geta=12, b=2∴ fx=x422×3+2x2So, f2=816+8=0Hence, the correct option is (3). [[VIDEO:13756]]
Question 74:
The integral
∫dxx2x4+13/4equals :
Option 1:  x4+1×414+c 
Option 2:  x4+114+c 
Option 3:  x4+114+c 
Option 4:  x4+1×414+c 
Solutions:
I=∫dxx2x4+134=∫dxx2·x31+1×434=∫dxx51+1×434 Let 1+1×4=t⇒4x5dx=dt ∴ I=14∫dtt34=14t34+134+1+C=14t1414+C=1+1×414+C=x4+1×414+CHence, the correct option is (4).
Question 75:
The integral
∫24log x2log x2+log3612x+x2dx is equal to :
Option 1:  2 
Option 2:  4 
Option 3:  1 
Option 4:  6 
Solutions:
I=∫24log x2log x2+log3612x+x2dx⇒I=∫24log x2log x2+log6x2dx⇒I=∫24log xlog x+log6xdx …..1 logab=bloga ⇒I=∫24log6xlog6x+log xdx …..2 ∫abfxdx=∫abfa+bxdxAdding (1) and (2), we get
2I=∫24log x+log 6xlog6x+log xdx⇒2I=∫241dx⇒2I=x24⇒2I=42=2⇒I=1Hence, the correct option is (3).
Question 76:
The area (in sq. units) of the region described by {(x, y) : y^{2} ≤ 2x and y ≥ 4x − 1} is :
Option 1:  732 
Option 2:  564 
Option 3:  1564 
Option 4:  932 
Solutions:We have
y^{2} ≤ 2x and y ≥ 4x − 1
The region enclosed by the given curves is the shaded region.
Solving y^{2} = 2x and y = 4x − 1, we get
y = 1,
12
∴ Required area=∫121y+14y22dy=y28+y4y36121=18+141613218+148=932 square unitsHence, the correct option is (4). [[VIDEO:13757]]
Question 77:
Let y(x) be the solution of the differential equation
xlogxdydx+y=2xlogx, x≥1.Then y(e) is equal to :
Option 1:  e 
Option 3:  2 
Option 4:  2e 
Solutions:Given:
xlogxdydx+y=2xlogx, x≥1 …..1 ⇒dydx+yxlogx=2This is a linear differential equation.
Its integrating factor is given by
I.F.=e∫1xlogxdx=eloglogx=logxHence, the solution of the given differential equation is
y·log x=∫2log x dx+C⇒y·logx=2xlog xx +C⇒y·logx=2xlogx1 +CNow, when x = 1, y = 0 [Using (1)] ∴0=2×1× 01+C⇒C=2 ∴ y·lnx=2xlogx1+2 …..2Putting x = e in (2), we get
yloge=2eloge1+2⇒y×1=2e11+2⇒y=2Hence, the correct option is (3).
Question 78:
The number of points, having both coordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0), is:
Option 1:  901 
Option 2:  861 
Option 3:  820 
Option 4:  780 
Solutions:The equation of line AB is given by y − 0 =
410041(x − 41) ⇒ y − 0 = (−1)(x − 41) ⇒ x + y = 41
For y = 1, The abscissa of the required points are 1, 2, 3, …, 39. So, the number of required points for y = 1 is 39.
Similarly, Number of required points for y = 2 is 38.
. . .
And, Number of required points for y = 39 is 1.
∴ Total number of required points
= 39+38+37+ … +1 =3939+12=780Hence, the correct option is (4). [[VIDEO:13758]]
Question 79:
Locus of the image of the point (2, 3) in the line (2x − 3y + 4) + k(x − 2y + 3) = 0, k
∈R, is a :
Option 1:  straight line parallel to xaxis. 
Option 2:  straight line parallel to yaxis. 
Option 3:  circle of radius
2. 
Option 4:  circle of radius
3. 
Solutions:The equation of the given line is (2x − 3y + 4) + k(x − 2y + 3) = 0.
This represents a family of straight lines passing through the point (1, 2).
Let (α, β) be the image of the point (2, 3) in the given line.
Now, âˆ†BAD
≅âˆ†BCD (By SAS congruence criterion)
∴ BC = AB (CPCT)
⇒α12+β22=212+322=2⇒α12+β22=22Thus, the locus of image of the point (2, 3) in the given line is
x12+y22=22This represents a circle of radius
2.
Hence, the correct option is (3).
Question 80:
The number of common tangents to the circles x^{2} + y^{2} − 4x − 6y − 12 = 0 and x^{2} + y^{2} + 6x + 18y + 26 = 0, is:
Option 1:  1 
Option 2:  2 
Option 3:  3 
Option 4:  4 
Solutions:We have
x2+y24x6y12=0⇒x22+y32=25∴ C1=2, 3 and r1=5And,x2+y2+6x+18y+26=0⇒x+32+y+92=64∴ C2=3, 9 and r2=8So, dC1, C2=52+122=13Also, r1+r2=8+5=13∵ dC1, C2=r1+r2So, the circles touch each other externally. Therefore, the number of common tangents to the given circles is 3.
Hence, the correct option is (3). [[VIDEO:13759]]
Question 81:
The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse
x29+y25=1,is :
Option 1:  274 
Option 2:  18 
Option 3:  272 
Option 4:  27 
Solutions:
The equation of the ellipse is
x29+y25=1.
Here,
a=3 and b=5Now,
b2=a21e2⇒5=91e2⇒e2=159=49⇒e=23Foci of the ellipse =
±ae, 0=±3×23, 0=±2, 0Also,
b2a=53So, one end point of the latus rectum =
2, 53∴ Equation of the tangent to the ellipse at
2, 53is
2×9+53y5=1 xx1a2+yy1b2=1⇒x92+y3=1This intersects the xaxis and yaxis at A
92, 0and B(0, 3), respectively.
The quadrilateral formed by the tangents at the end points of the latera racta of the given ellipse is a rhombus.
∴ Area of the rhombus formed by tangents
= 4 × Area of âˆ†AOB
=4×12×OA×OB=4×12×92×3= 27 square units Thus, the area of the quadrilateral formed by the tangents at the end points of the latera recta to the given ellipse is 27 square units.
Hence, the correct option is (4). [[VIDEO:13760]]
Question 82:
Let O be the vertex and Q be any point on the parabola, x^{2} = 8y. If the point P divides the line segment OQ internally in the ratio 1 : 3, then the locus of P is:
Option 1:  x^{2} = y 
Option 2:  y^{2} = x 
Option 3:  y^{2} = 2x 
Option 4:  x^{2} = 2y 
Solutions:Let any point on the parabola x^{2} = 8y be Q(4t, 2t^{2}). Let the coordinates of the point P be (x, y).
Since, P divides OQ internally in the ratio 1 : 3,
∴ x,y=14t+301+3,12t2+301+3 ⇒x,y=t,t22∴ x=t, y=t22⇒2y=x2Hence, the correct option is (4). [[VIDEO:13761]]
Question 83:
The distance of the point (1, 0, 2) from the point of intersection of the line
x23=y+14=z212and the plane x – y + z = 16, is:
Option 1:  214 
Option 2:  8 
Option 3:  321 
Option 4:  13 
Solutions:
Let
x23=y+14=z212=λ∴ x = 3
λ+ 2, y = 4
λ– 1 and z = 12
λ+ 2
So, the coordinates of any point on the line is (3
λ+ 2, 4
λ– 1, 12
λ+ 2). Also, this point lies on the plane x – y + z = 16.
∴ (3
λ+ 2) – (4
λ– 1) + (12
λ+ 2) = 16
⇒ 3
λ+ 2 – 4
λ+ 1 + 12
λ+ 2 = 16
⇒
λ= 1
∴ (x, y, z) = (5, 3, 14)
Required distance between the points (1, 0, 2) and (5, 3, 14)
=512+302+1422=42+32+122=13 unitsHence, the correct option is (4). [[VIDEO:13762]]
Question 84:
The equation of the plane containing the line 2x – 5y + z = 3; x + y + 4z = 5, and parallel to the plane, x + 3y + 6z = 1, is :
Option 1:  2x + 6y + 12z = 13 
Option 2:  x + 3y + 6z = –7 
Option 3:  x + 3y + 6z = 7 
Option 4:  2x + 6y + 12z = –13 
Solutions:Putting z = 0 in first two planes, we get
∴ 2x – 5y = 3 and x + y = 5
⇒ x = 4, y = 1, when z = 0
⇒ (4, 1, 0) lies on the given plane.
Let x + 3y + 6z = k be a plane parallel to given plane x + 3y + 6z =1.
∴ 4 + 3 + 0 = k
⇒ k = 7
∴ x + 3y + 6z = 7, which is the equation of the required plane
Hence, the correct option is (3). [[VIDEO:13763]]
Question 85:
Let
a→,
b→and
c→be three nonzero vectors such that no two of them are collinear and
a→×b→×c→=13b→c→a→.If θ is the angle between vectors
b→and
c→then a value of sin θ is:
Option 1:  223 
Option 2:  23 
Option 3:  23 
Option 4:  233 
Solutions:
a→×b→×c→=13b→c→a→⇒c→×a→×b→=13b→c→a→⇒c→·b→a→c→·a→b→=13b→c→a→⇒c→·a→b→c→·b→a→=13b→c→a→Equating the components of
a→, we get
c→·b→=13b→c→⇒c→b→cosθ=13b→c→⇒cosθ=13∴ sinθ=1cos2θ=1–132=223Hence, the correct option is (1). [[VIDEO:13764]]
Question 86:
If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is :
Option 1:  553 2311 
Option 2:  552310 
Option 3:  2201312 
Option 4:  221311 
Solutions:Disclaimer: There seems to be ambiguity in the given question. Instead of identical balls, 12 different balls should be used to find the required probability.
Total number of ways in which 12 balls can be placed in 3 identical boxes = 3^{12}
Now, Number of ways of selecting 3 balls out of 12 =
C312=220
So, number of ways of placing these 3 selected balls in one box = 220
Number of ways of ways in which 9 remaining balls can be placed in 2 identical boxes = 2^{9}
∴ Favourable number of ways = 220 × 2^{9}
Probability that one of the boxes contains exactly 3 balls
=Favourable number of waysTotal number of ways=220×29312=2204×3×211311=5532311Hence, the correct option is (1).
Question 87:
The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is :
Option 1:  16.8 
Option 2:  16.0 
Option 3:  15.8 
Option 4:  14.0 
Solutions:It is given that the mean of 16 observations is 16.
⇒∑xi16=16⇒∑xi=256Now, from the given 16 observations, one observation is deleted and three other are added. So, the total number of observations become 16
1 + 3 i.e. 18.
Mean of the resultant data
=∑xi16+3+4+518=25616+3+4+518=25218=14Hence, the correct option is (4).
Question 88:
If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the foot of the tower, are 30°, 45° and 60° respectively, then the ratio AB : BC is :
Option 1:  3:1 
Option 2:  3:2 
Option 3:  1 :3 
Option 4:  2 : 3 
Solutions:Let P be the top of the tower.
Suppose the height of the tower PQ is h units.
In
∆PAQ ,
PQAQ=tan30° ⇒hAQ=13⇒AQ=3hIn
∆PBQ,
PQBQ=tan45°=1 BQ=PQ=hAlso, in
∆PCQ,
PQCQ=tan60°⇒hCQ=3⇒CQ=h3 ∴ ABBC=AQBQBQCQ=3hhhh3=31⇒AB : BC=3 : 1Hence, the correct option is (1). [[VIDEO:13768]]
Question 89:
Let
tan1y=tan1x+tan12×1x2,where
x<13. Then a value of y is:
Option 1:  3xx313×2 
Option 2:  3x+x313×2 
Option 3:  3xx31+3×2 
Option 4:  3x+x31+3×2 
Solutions:
tan1y=tan1x+tan12×1x2
⇒tan1y=tan1x+2×1x21x×2×1x2⇒tan1y=tan13xx313×2⇒y=3xx313x2Hence, the correct option is (1). [[VIDEO:13769]]
Question 90:
The negation of ~s â‹ (~r â‹€ s) is equivalent to :
Option 1:  s â‹€ ~r 
Option 2:  s â‹€ (r â‹€ ~s) 
Option 3:  s â‹ (r â‹ ~s) 
Option 4:  s â‹€ r 
Solutions:~ [~s ∨ (~r ∧ s)] = ~ (~s) ∧ ~ (~r ∧ s) (DeMorgan’s Law) = s ∧ ( r ∨ ~s) = (s ∧ r) ∨ (s ∧ ~s) = (s ∧ r) ∨ F = s ∧ r
Hence, the correct option is (4). [[VIDEO:13770]]