IIT JEE Advanced 2014 Paper 2 Code 2

Test Name: IIT JEE Advanced 2014 Paper 2 Code 2

Question 1:

A glass capillary tube is of the shape of a truncated cone with an apex angle α so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height h, where the radius of its cross section is b. If the surface tension of water is S, its density is ρ, and its contact angle with glass is θ, the value of h will be (g is the acceleration due to gravity) 

Option 1: 2Sbρgcosθα
Option 2: 2Sbρgcosθ+α
Option 3: 2Sbρgcosθα/2
Option 4: 2Sbρgcosθ+α/2
Correct Answer: 4

Solutions:Let R be the radius of curvature formed by the water as shown in the figure.

In

ABC:

bR=cosθ+α2

Or

1R=cosθ+α2b

    …(1)

There is a pressure difference between the two sides of the meniscus and that difference is given by

PoPi=2SR

The pressure inside the tube at the point where it is dipped in water, must be same as the atmospheric pressure.

P02SR+hρg=P0h=2SRρg

Using equation (1), we get:

h=2Scosθ+α2bρg

Hence, the correct option is D.

Question 2:

A planet of radius

R=110×

(radius of Earth) has the same mass density as Earth. Scientists dig a well of depth

R5

on it and lower a wire of the same length and of linear mass density 10−3 kgm−1 into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth = 6 × 106m and the acceleration due to gravity on Earth is 10 ms−2)

Option 1: 96 N
Option 2: 108 N
Option 3: 120 N
Option 4: 150 N
Correct Answer: 2

Solutions:Given:
Radius of the planet, Rp = RE/10 = 6 × 105 m
ρp = ρE = ρ
Density of the wire, λ = 10−3 kg m-1
Acceleration due to gravity on earth, g = 10 m/s2
We know that

g=GMR2=GR243πR3ρ=43πGρR

gpgE=RpREor gp=1 m/s2

Now, at a depth x,

gx=1xRgp

The force on a small element of wire at a depth x from the surface of the planet can be written as

dF=(λdx)gx=λ1xRgpdxF=0R5λ1xRgpdx=λ×xx22R0R5×gpF=9λR50gp=108 N

Hence, the correct option is B.

Question 3:

Charges Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/2, R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1, 2 and 3 are E1, E2 and E3 respectively, then

Option 1: E1>E2>E3
Option 2: E3>E1>E2
Option 3: E2>E1>E3
Option 4: E3>E2>E1
Correct Answer: 3

Solutions:Electric field at point P due to sphere 1,

E1=14πε0QR2

Electric field at point P due to sphere 2,

E2=14πε02QR2=214πε0QR2

For sphere 3, we have
Total charge = 4Q
Volume charge density,

ρ=4Q43π(2R)3=3Q8πR3

Electric field at point P inside the sphere at a distance R from centre =

ρR3ε0

Thus, electric field for sphere 3,

E3=133QR8πR3ε0=1214πε0QR2

So, the relation among the electric fields is E2 > E1 > E3.

Hence, the correct option is C.

Question 4:

If λCu is the wavelength of Kα X-ray line of copper (atomic number 29) and λMo is the wavelength of the Kα X-ray line of molybdenum (atomic number 42), then the ratio λCu/λMo is close to

Option 1: 1.99
Option 2: 2.14
Option 3: 0.50
Option 4: 0.48
Correct Answer: 2

Solutions:According to Moseley's law, the frequency of the Kα X-ray for a certain material of atomic number Z is given by

f=C(Z1)2

Here, C is a constant whose value is 4.96

×

107 Hz1/2.

The relation between the wavelength, frequency and velocity of a wave is given by
v = fλ

λ1f

For the wavelength of the Kα X-ray line of copper (Z = 29), we have:

λCu1fCuλCu1C(ZCu1)2λCu1C(291)2λCu1C(28)2

For the wavelength of the Kα X-ray line of molybdenum (Z = 42), we have:

λMo1fMoλMo1C(ZMo1)2λMo1C(421)2λMo1C(41)2

So, the value of λCu/λMo is given by

λCuλMo=(41)2(28)2=2.1442.14

Hence, the correct option is B.

Question 5:

A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is

Option 1: always radially outwards.
Option 2: always radially inwards.
Option 3: radially outwards initially and radially inwards later.
Option 4: radially inwards initially and radially outwards later.
Correct Answer: 4

Solutions:
Let the radius of the quarter be R.
The above figure shows the bead in a position when the line joining the centre and the ring makes an angle θ with the horizontal.
Let the tangential velocity of the bead at the position C be v.
The change in potential energy of the bead from position A to C is given by
mgR

mgRsinθ = mgR(1

sinθ)
Change in kinetic energy of the bead from position A to C = 

12mv2

Applying conservation of mechanical energy:
Change in potential energy = Change in kinetic energy
mgR(1

sinθ) =

12mv2

    …(1)
The forces acting on the bead are: weight mg (downwards) and Normal reaction N (radially outwards)
As the bead is in circular motion, the net radial force on it will act as the centripetal force.

mgsinθN=mv2R

Normal reaction on the bead,

N=mgsinθmv2R

Putting the value of v from equation (1):

N=mgsinθ2mgR(1sinθ)R=mg(3sinθ2)

  
The Normal reaction on the bead will be radially outwards, i.e. N > 0.
∴ 3sinθ > 2
The Normal reaction on the bead will be radially inwards, i.e. N < 0.
∴ 3sinθ < 2
Therefore, it can be concluded that the normal force on the wire due to bead is radially inwards when

θ>sin123

and radially outwards when

θ<sin123

.
When the bead is at point A, θ = 90o (

90o>sin123

) and when the bead is at point B θ = 0o(

0o<sin123

). From the above statement, we can conclude that the normal force on the wire is radially inwards initially and radially outwards later.

Hence, the correct option is D.

Question 6:

A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio u1 : u2 = 2 :1 and hc = 1240 eV nm, the work function of the metal is nearly

Option 1: 3.7 eV
Option 2: 3.2 eV
Option 3: 2.8 eV
Option 4: 2.5 eV
Correct Answer: 1

Solutions:Let the work function of the metal be Φo.
Energy of the photon for the radiation of wavelength 248 nm,

hcλ=1240248=5 ev

Energy of the photon for the radiation of wavelength 310 nm,

hcλ=1240310=4 ev

According to Einstein's equation for photoelectric effect, the maximum kinetic energy of a photoelectron is given by
KEmax =

hcλ

Φo    …(1)
It is given that u1:u2 = 2:1

The ratio of the maximum kinetic energies of the ejected photoelectrons is given by

KE1KE2=u1u22=41

    …(2)
Using equation (1), equation (2) can be written as:

KE1KE2=41=5Φo4Φo164Φo=5ΦoΦo=1133.7 eV

Hence, the correct option is A.

Question 7:

A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with time t most appropriately? The figures are only illustrative and not to the scale.

Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer: 2

Solutions:When the ball is dropped, it is in free fall and its acceleration is equal to g (acceleration due to gravity).
Let us write the equations of motion for the ball.
Taking the direction of acceleration due to gravity to be positive, we have:

i) When the ball is falling down:
v = gt    …(1)

II. When the ball rises after bouncing:
Let u be the speed of the ball when it rebounds. 
Then its velocity at any instant is given by
v = u

gt    …(2)
The graph for the motion of the ball can be drawn as:

The kinetic energy of the ball is given by

K=12mv2

.
From equation (1) and (2), we have:
Kinetic energy of the ball when it is falling,

K=12m(gt)2=12mg2t2

     …(3)
Kinetic energy of the ball when it is rising,

K=12m(ugt)2

    …(4)
If we draw a graph for the kinetic energy versus time, we can see that it is a parabola passing through the origin for equation (3); and a parabola shifted to the right hand side for equation (4).
Out of the given options, option (B) represents the graph for the kinetic energy versus time.

Hence, the correct option is B.

Question 8:

During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90 Ω, as shown in the figure. The least count of the scale used in the metre bridge is 1 mm. The unknown resistance is

Option 1: 60 ± 0.15 Ω
Option 2: 135 ± 0.56 Ω
Option 3: 60 ± 0.25 Ω
Option 4: 135 ± 0.23 Ω
Correct Answer: 3

Solutions:The null point on the metre bridge is calculated by the following formula:

R1R2=l(100l)

Here, R1 = R (unknown resistance)
R2 = 90 Ω

R=90l(100l)

    …(1)
Let the error in the calculation of the resistance R be

R.   
Relative error is given by:

RR=ll+l100lR60=0.140+0.160R=0.52=0.25

Putting l = 40 cm in equation (1), we have:
The value of the unknown resistance is given by

R=90×4060=60 Ω

Hence, the value of the resistance can be written as

R=(60±0.25) Ω

.

Hence, the correct option is C.

Question 9:

Parallel rays of light of intensity I = 912 Wm−2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant σ = 5.7 × 10−8 Wm−2 K−4 and assume that the energy exchange with the surroundings is only through radiation. The final steady-state temperature of the black body is close to

Option 1: 330 K
Option 2: 660 K
Option 3: 990 K
Option 4: 1550 K
Correct Answer: 1

Solutions:Let the temperature of the black body be T.
According to Stefan's Law, the power radiated by a black body is given by

P=σA(T4T04)

.
Here,
σ = Stefan-Boltzmann constant
T = temperature of the black body
T0 = temperature of the surroundings

Rate of radiation energy lost by the sphere is given by,

P=σ4πR2(T43004)

Rate of radiation energy incident on the sphere, P' =

I×πR2

= 912

×

(

πR2

)

Rate of radiation energy lost by the sphere = Rate of radiation energy incident on it.

Therefore,

σ4πR2(T43004)

= 912

×

(

πR2

)

T43004=9124×σT43004=9124×5.7×108T43004=40×108T4=40×108+3004=(40+81)×108T330 K

Hence, the correct option is A.

Question 10:

A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid, as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is

Option 1: 1.21
Option 2: 1.30
Option 3: 1.36
Option 4: 1.42
Correct Answer: 3

Solutions: 
From the figure,

sinic=rr2+h2   ...(i) 

Here,  ic is the critical angle
           r is the radius of the bright circular spot and,
           h is the height of the transparent block

We also know that

sinic=μlμb    ...(ii)   

Here, μl and μb are the refractive indices of the liquid and block respectively.
On comparing equation (i) and (ii), we get:

μlμb=rr2+h2μl=rr2+h2×μbμl=5.775.772+102×2.72μl=1.36    

Hence, the correct option is C.

Question 11:

The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2), all in the plane of the paper. The distance of each wire from the centre of the loop is d. The loop and the wires are carrying the same current I. The current in the loop is in the counterclockwise direction if seen from above.

When da but the wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case,

Option 1: current in wire 1 and wire 2 is the direction PQ and RS, respectively and ha
Option 2: current in wire 1 and wire 2 is the direction PQ and SR, respectively and ha
Option 3: current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ 1.2a
Option 4: current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ 1.2a
Correct Answer: 3

Solutions:Given:
The loop and wires are carrying current = I.
Distance of each wire from the center of the loop = d .

Let the direction of current in wire 1 be from P to Q and in wire 2 be from S to P. From the above figure,

BR

= magnetic field due to the ring

B1

= magnetic field due to wire 1

B2

= magnetic field due to wire 2
Magnitude of

B1

and

B2

are equal.

B1=B2=μ0ia2πa2+h2

∵ Magnetic field due to the circular loop is

μ0Ia22a2+h23/2

.
Resultant of

B1

and

B2

=

2B1cosθ=2×μ0ia2πa2+h2

.

BR=2μ0Iπa24πr2

For zero magnetic field at P,

μ0ia22a2+h23/2=2μ0ia2πa2+h23a22=h2h1.2a

Hence, the correct option is C.

Question 12:

The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop d. The loop and the wires are carrying the same current I. The current in the loop is in the counterclockwise direction if seen from above.

Consider

da

and the loop is rotated about its diameter parallel to the wires by 30° from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)

Option 1: μ0I2a2d
Option 2: μ0I2a22d
Option 3: 3μ0I2a2d
Option 4: 3μ0I2a22d
Correct Answer: 2

Solutions:Magnetic field at the midpoint of the two wires, B =

μ0I2πd+μ0I2πd=μ0Iπd

Magnetic moment of the loop, M =

I(πa2)

Torque on the loop,

τ

=

MBsinθ

τ

=

MBsin150°

τ

=

I(πa2)×μ0Iπd×12 =μ0I2a22d

Hence, the correct option is B.

Question 13:

In the figure, a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are

CV=32R and CP=52R

and those for an ideal diatomic gas are

CV=52R and CP=72R.

Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be

Option 1: 550 K
Option 2: 525 K
Option 3: 513 K
Option 4: 490 K
Correct Answer: 4

Solutions:Condition: The partition is rigidly fixed.
At equilibrium, the heat contents of the two compartments will be equal.
∴ Q2 = Q1
Here, Q1 and Q2 are the heat contents of the monatomic and diatomic gases, respectively.
For the monatomic gas:
Q1 = n1CV1ΔT
For the diatomic gas:
Q2 = n2CP2ΔT

Here,
ΔT  = Change in temperature
n1 and n2 = Numbers of moles of monatomic and diatomic gases
As per the condition,

Q1=Q2n1CV1T=n2CP2T2×3R2700T=2×7R2T400T=490 K

Hence, the correct option is D.

Question 14:

In the figure, a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are

CV=32R and CP=52R

and those for an ideal diatomic gas are

CV=52R and CP=72R.

Now, consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then, total work done by the gases till the time they achieve equilibrium will be

Option 1: 250R
Option 2: 200R
Option 3: 100R
Option 4: −100R
Correct Answer: 4

Solutions:We know that the partition is free to move without friction. Also, pressure in both compartments is the same.
Q2 = Q1
Here, Q1 and Q2 are the heat contents of the monatomic gas and the diatomic gas, respectively.
For the monatomic gas, Q1 = n1Cp1ΔT .
For the diatomic gas, Q2 = n2Cp2ΔT .
Here,
ΔT  = Change in temperature
n1 and n2  = Numbers of moles of monatomic and diatomic gases.
As per the condition,

n1Cp1T=n2Cp2T2×5R2T700=2×7R2400T5T3500=28007T12T=6300T=525 K

Work done:
ΔW = ΔU
       =  n1Cv1ΔT + n2Cv2ΔT
     

=2×3R2×525700+2×5R2×525400=3R175+5R125=525R+625R=100R

ΔW = ΔU = 100R

Hence, the correct option is D.

Question 15:

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle. Radius of the piston and nozzle are 20 mm and 1 mm respectively. The upper end of the container is open to the atmosphere.

If the piston is pushed at a speed of 5 mms−1, the air comes out of the nozzle with a speed of

Option 1: 0.1 ms−1
Option 2: 1 ms1
Option 3: 2 ms−1
Option 4: 8 ms−1
Correct Answer: 3

Solutions:Given:
Radius of the piston = 20 mm
Radius of the nozzle = 1 mm
Piston is pushed with a speed of 5 mms−1.
Using the equation of continuity, we get:
a1v1 = a2v2
Here,
a1 = Area of the cross section of the piston
v1 = Speed at which the piston is pushed
a2 = Area of the cross section of the nozzle
v2 = Speed at which air comes out of the nozzle
Thus, we get:

π×20×1032×5×103=π×1×1032×v22×103×103=v2v2=2 ms1

Hence, the correct option is C.

Question 16:

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle are 20 mm and 1 mm respectively. The upper end of the container is open to the atmosphere.

If the density of air is

ρa

and that of the liquid

ρ

, then for a given piston, speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to

Option 1: ρaρ
Option 2: ρaρ
Option 3: ρρa
Option 4: ρ
Correct Answer: 1

Solutions:
Given:
Density of the air = ρa
Density of the liquid = ρl
From figure, we have:
Pressure at C =

P012ρlvl2

Pressure at B =

P012ρlvl2ρlgh

         (h is the height from C to B.)
Pressure at A =

P012ρava2

We know:
Pressure at A = Pressure at B
Thus, we have:

P012ρava2=P012ρlvl2ρlghvl=ρaρlva22ghvlρaρl

Question 17:

A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In the following, state of the lift's motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in List II and select the correct answer using the code given below the lists.

List I List II
P. Lift is accelerating vertically up. 1. d = 1.2 m
Q. Lift is accelerating vertically down with an acceleration less than the gravitational acceleration. 2. d > 1.2 m
R. Lift is moving vertically up with constant speed. 3. d < 1.2 m
S. Lift is falling freely. 4. No water leaks out of the jar.
Option 1: P-2, Q-3, R-2, S-4
Option 2: P-2, Q-3, R-1, S-4
Option 3: P-1, Q-1, R-1, S-4
Option 4: P-2, Q-3, R-1, S-1
Correct Answer: 3

Solutions:Let H be the depth of the water in the jar and h be the height of the jar from the base of the lift.

Now, consider the case when the lift is accelerating upwards.


Velocity of the jet in this case is given as

v=2H(a+g)

Initially, the vertical component of velocity was zero.
Let t be the time taken by the jet to hit the floor.
Using the equation of motion, we get:

h=uyt+12(a+g)t2h=0+12(a+g)t2t=2h(a+g)

Horizontal range of the jet is given as

d=v×t

d=2H(a+g)×2h(a+g)d=2Hh

This shows that the distance at which the water jet hits the floor of the lift is independent of the acceleration of the lift. Similar is the case when the lift is accelerating downwards or moving at a constant speed.

When the lift is falling freely, no water leaks out of the jar because the acceleration of both the jar and water is the same; hence, water does not exert any pressure on the jar.

Hence, the correct option is C.

Question 18:

Four charges Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x-axis at x = −2a, −a, +a and +2a, respectively. A positive charge q is placed on the positive y-axis at a distance b > 0. Four options of the signs of these charges are given in List I. The direction of the forces on the charge q is given in List II. Match List I with List II and select the correct answer using the code given below the lists.
 

List I List II
P. Q1, Q2, Q3, Q4 all positive 1. +x
Q. Q1, Q2 positive; Q3, Q4 negative 2. x
R. Q1, Q4 positive; Q2, Q3 negative 3. +y
S. Q1, Q3 positive; Q2, Q4 negative 4. y

Option 1: P-3, Q-1, R-4, S-2
Option 2: P-4, Q-2, R-3, S-1
Option 3: P-3, Q-1, R-2, S-4
Option 4: P-4, Q-2, R-1, S-3
Correct Answer: 1

Solutions: First case (P): When all four changes are positive:

     

We can see that the net force on charge q is along the positive y-axis.

Second case (Q): When Q1 and Q2 are positive and Q3 and Q4 are negative:

     

We can see that the net force on charge q is along the positive x-axis.

Third case (R): When Q1 and Q4 are positive Q2 and Q3 are negative:

     

We can see that the net force on charge q is along the negative y-axis.

Fourth case (S): When Q1 and Q3 are positive and Q2 and Q4 are negative:

    

We can see that the net force on charge q is along the negative x-axis.

Hence, the correct option is A.

Question 19:

Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal lengtha in List II and select the correct answer using the code given below the lists.

List I List II
P. 1. 2r
Q. 2. r/2
R. 3. −r
S. 4. r
Option 1: P-1, Q-2, R-3, S-4
Option 2: P-2, Q-4, R-3, S-1
Option 3: P-4, Q-1, R-2, S-3
Option 4: P-2, Q-1, R-3, S-4
Correct Answer: 2

Solutions:Given:
The radius of the curvature of all curved surfaces is r.
The refractive index of all lenses (

μ

r) is 1.5.

For P:

    

Focal length of each lens:

1f=(μr1)1r11r21f=(1.51)1r+1r              (r1=r and r2=r)1f=(0.5)2rf=r

Focal length of the combination:

1fR=1f+1f1fR=1r+1rfR=r2

∴ P

2

For Q:

     

Focal length of each lens:

1f=(μr1)1r11r21f=(1.51)1+1r          (r1= and r2=r)1f=(0.5)1rf=2r

Focal length of the combination:

1fR=1f+1f1fR=12r+12rfR=r

∴ Q

4

For R:

    

Focal length of each lens:

1f=(μr1)1r11r21f=(1.51)11r           (r1= and r2=r)1f=(0.5)1rf=2r

Focal length of the combination:

1fR=1f+1f1fR=12r12rfR=r

∴ R

3

For S:

    

Focal length of the first lens:

1f1=(μr1)1r11r21f1=(1.51)1r+1r        (r1=r and r2=r)1f1=(0.5)1rf1=r

Focal length of the second lens:

1f2=(μr1)1r11r21f2=(1.51)11r           (r1= and r2=r)1f2=(0.5)1rf2=2r

Focal length of the combination:

1fR=1f1+1f21fR=1r12r=12rfR=2r

∴ S

1

Hence, the correct option is B.

Question 20:

A block of mass m1 =1 kg another mass m2 = 2 kg, are placed together (see figure) on an inclined plane with angle of inclination θ. Various values of θ are given in List I. The coefficient of friction between the block m1 and the plane is always zero. The coefficient of static and dynamic friction between the block m2 and the place are equal to μ = 0.3. In List II, expressions for the friction on block m2 are given. Match the correct expression of the friction in List II with the angles given in List I and choose the correct option. The acceleration due to gravity is denoted by g.

[Useful information: tan (5.5°) ≈ 0.1; tan (11.5°) ≈ 0.2; tan (16.5°) ≈ 0.3]


 

  List I   List II
P. θ = 5° 1. m2gsin θ
Q. θ = 10° 2. (m1 + m2)gsin θ
R. θ = 15° 3. μm2gcos θ
S. θ = 20° 4. μ(m1 + m2)gcos θ
Option 1:
P-1, Q-1, R-1, S-3
Option 2: P-2, Q-2, R-2, S-3
Option 3: P-2, Q-2, R-2, S-4
Option 4: P-2, Q-2, R-3, S-3
Correct Answer: 4

Solutions:Given:
Mass of the first block, m1 = 1 kg
Mass of the second block, m2 = 2 kg

The coefficient of static and dynamic friction between block m2 and the plane,

μ

= 0.3

        

The coefficient of friction between block m1 and the plane is zero.

From the figure, we can say that the block will not move when

μNm1gsinθ+m2gsinθμ(m2gcosθ)m1gsinθ+m2gsinθtanθμm2m1+m2=0.2

We know:

tan(11.5°)0.2

Therefore, block m2 will start sliding when

θ>11.5°

.

So, when

θ=5° and θ=10°

, blocks are at rest. Thus, the frictional force between the block of mass m2 and the plane is

(m1+m2)gsinθ

.
When

θ=15° and θ=20°

, the block starts sliding. Thus, the frictional force between the block of mass m2 and the plane is

μm2gcosθ

.

Hence, the correct option is D. 

Question 21:

Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is

Option 1: Be2
Option 2: B2
Option 3: C2
Option 4: N2
Correct Answer: 3

Solutions:The electronic configurations of the given molecules, assuming that 2s-2p mixing is not operative and considering the Z-axis as the internuclear axis, are as follows:

1. The total number of electrons in a Be2 molecule is eight.
Electronic configuration of Be2:

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2

As all the electrons are paired, Be2 is diamagnetic in nature.

2. The total number of electrons in a B2 molecule is ten.
Electronic configuration of B2:

(σ1s)2)(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2

As all the electrons of B2 are paired, it is also diamagnetic in nature.

3. The total number of electrons in a C2 molecule is twelve.
The electronic configuration of C2 can be written in two ways.
a) Involving 2s-2p mixing:

σ1s2σ*1s2σ2s2σ*2s2π2px2 = π2py2

b) Not involving mixing of 2s and 2p orbitals:

σ1s2σ*1s2σ2s2σ*2s2σ2pz2π2px1 = π2py1

As the C2 molecule in case B contains two unpaired electrons, it is paramagnetic in nature.

4. The total number of electrons in N2 molecule is fourteen.

Electronic configuration of N2 molecule involving 2s-2p mixing:

σ1s2σ*1s2σ2s2σ*2s2π2px2 = π2py2σ2pz2

Electronic configuration of N2 assuming that 2s-2p mixing is not operative:

σ1s2σ*1s2σ2s2σ*2s2σ2pz2π2px2 = π2py2

In both the cases, all the electrons of N2 are paired. Therefore, it is also diamagnetic in nature.

Hence, the correct option is (C).

Question 22:

For the process

H2O (l) H2O (g)

at T = 100 °C and 1 atmosphere pressure, the correct choice is

Option 1:  Ssystem >0 and Ssurroundings >0
Option 2:  

Ssystem >0 and Ssurroundings <0
Option 3: Ssystem<0 and Ssurroundings>0
Option 4: Ssystem <0 and Ssurroundings <0
Correct Answer: 2

Solutions:The total entropy change for a system and the surrounding for any reaction is given by

Stotal = Ssystem + Ssurrounding 

                …(1)

At 100 oC and 1 atm pressure, H2O changes from liquid to gaseous state. This temperature is termed as the boiling point of water. At this temperature, the liquid water and  the gaseous water vapour exist in equilibrium.

For an equilibrium reaction,

ΔStotal = 0

On substituting the value of

Stotal 

in equation (1), we get:

ΔSsystem + ΔSsurrounding   = 0 ΔSsystem  =   ΔSsurrounding

 
As the water is undergoing a change in phase (i.e., from liquid to gas), the degree of randomness increases. Hence, for this process,

ΔSsystem > 0

As the change in entropy of the system is positive, the change in entropy of the surrounding will be negative (

ΔSsurrounding < 0

).

Hence, the correct option is (B). 

Question 23:

For the elementary reaction M → N, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is

Option 1: 4
Option 2: 3
Option 3: 2
Option 4: 1
Correct Answer: 2

Solutions:

Consider the elementary reaction M → N, where M is the reactant and N is the product.

The rate law for the above reaction is given by
R  = k [M]a               …(i)

Here, R is the rate of reaction; [M] is the concentration of M, k is rate constant and a is the order of reaction with respect to M.

Let R1 be the new rate of reaction when the concentration of M is doubled.

i.e., R1 = k[2M]a        …(ii)

As the rate of reaction increases by a factor of 8 on doubling the concentration M, we have:
 R1 = 8R

On substituting value of R1 in equation (ii), we get:
8R = k[2M]a        …(iii)

Dividing equation (iii) by (i), we get:

8R R = 2MMa8 = 2a23 = 2a  

On comparing, a = 3

Thus, the order of the reaction with respect to M is three.

Hence, the correct option is (B).

Question 24:

For the identification of β-naphthol using dye test, it is necessary to use

Option 1: dichloromethane solution of β-naphthol.
Option 2: acidic solution of β-naphthol.
Option 3: neutral solution of β-naphthol.
Option 4: alkaline solution of β-naphthol.
Correct Answer: 4

Solutions:

During the dye test for the identification of β-naphthol, weakly alkaline solution is required. In weakly alkaline solution, β-naphthol ionises to form sodium 2-naphthoxide. This activates the naphthalene ring and makes it more susceptible for electrophilic substitution reaction, thereby increasing the yield and the rate of reaction.

Hence, the correct option is (D).

Question 25:

Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure.

Option 1: I >II >III
Option 2: III > II >I
Option 3: II > III > I
Option 4: III > I > II
Correct Answer: 2

Solutions:Among the given isomers of hexane, III will have the highest boiling point and I will have the lowest boiling point.
Reason: As the branching increases, the intermolecular forces between the molecules of hexane decreases. As a result, the boiling point decreases.

Hence, the correct option is (B).

Question 26:

The acidic hydrolysis of ether X shown below is fastest when

Option 1: one phenyl group is replaced by a methyl group
Option 2: one phenyl group is replaced by a para-methoxyphenyl group
Option 3: two phenyl groups are replaced by two para-methoxyphenyl groups
Option 4: no structural change is made to X
Correct Answer: 3

Solutions:The acid-catalysed hydrolysis of the given ether follows the given mechanism:

The greater the stability of the carbocation formed, the faster is the rate of reaction. If two phenyl groups in the given compound are replaced by two para-methoxyphenyl groups (+R effect), then the electron-releasing effect of these groups will increase the stability of the carbocation; as a result, the reaction will be fast.

Hence, the correct option is C.

Question 27:

Hydrogen peroxide, in its reaction with KlO4 and NH2OH, respectively, is acting as a

Option 1: reducing agent, oxidising agent
Option 2: reducing agent, reducing agent
Option 3: oxidising agent, oxidising agent
Option 4: oxidising agent, reducing agent
Correct Answer: 1

Solutions:Reaction of hydrogen peroxide (H2O2) with KIO4 and NH2OH can be written as:

KIO4 + H2O2 → KIO3 + H2O + O2

2NH2OH + H2O2 → N2 + 4H2O

From the given reactions, it is evident that hydrogen peroxide acts as a reducing agent with KIO4 and as an oxidsing agent with NH2OH.

Hence, the correct option is (A).

Question 28:

The major product in the following reaction is

Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer: 4

Solutions:

The reaction mechanism for the given reaction can be represented in the following manner:

First, the nucleophilic attack of the Grignard reagent (CH3MgBr) at the carbonyl carbon occurs to form an adduct, which then undergoes hydrolysis to form an alcohol.

The alcohol, so formed, then undergoes dehydrohalogenation reaction to form a cyclic ether.

Thus, the complete reaction is as follows:

Hence, the correct option is (D).

Question 29:

Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is

Option 2: 1
Option 3: 2
Option 4: 3
Correct Answer: 3

Solutions:Complete hydrolysis of xenon hexafluoride results in the formation of xenon trioxide and hydrogen fluoride.
XeF6 + 3H2O → XeO3 + 6HF

So, the compound P is XeO3.
XeO3, in the presence of an aqueous alkali, forms the hydrogen xenate ion (

HXeO4

).

XeO3 + OH HXeO4

So, the compound Q is

HXeO4

.

The hydrogen xenate ion (

HXeO4

) then undergoes disproportionation to form xenon and the perxenate ion (

XeO64

).

2HXeO4 + 2OH XeO64 + Xeg + 2H2O + O2g

Thus, the two gases Xe and O2 are released as products in the final step of the given reaction mechanism.

Hence, the correct option is (C).

Question 30:

The product formed in the reaction of SOCl2 with white phosphorous is

Option 1: PCl3
Option 2: SO2Cl2
Option 3: SCl2
Option 4: POCl3
Correct Answer: 1

Solutions:White phosphorus (P4), on treatment with thionyl chloride (SOCl2), undergoes the following reaction to form phosphorus trichloride.

P4 + 8SOCl2 → 4PCl3 + 4SO2 + 2S2Cl2

Hence, the correct option is (A).

Question 31:

 X and Y are two volatile liquids with molar weights 10 g mol−1 and 40 g mol−1 , respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product, which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.

The value of d in cm (shown in the figure), as estimated from Graham's Law, is

Option 1: 8
Option 2: 12
Option 3: 16
Option 4: 20
Correct Answer: 3

Solutions:According to Graham's Law of Diffusion, "under similar conditions of temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density."

r  1density

In other words, it can also be written as

r  1M

, where M is the molecular mass of the gas.

In the given reaction, vapours of gas X and gas Y diffuse in the tube to form a product at distance d from the plug soaked in X.

Thus, according to Graham's Law of Diffusion,

rXrY = MYMXTime taken for diffusion by X = Time taken for diffusion by Y drX = 24  drYrXrY = d24 d = MXMYd24  d = 4010 = 2d = 48  2dd = 16 cm

Hence, the correct option is (C).

Question 32:

X and Y are two volatile liquids with molar weights 10 g mol−1 and 40 g mol1 , respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product, which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.

The experimental value of d is found to be smaller than the estimate obtained using Graham's Law. This is due to

Option 1: larger mean free path for X compared to that of Y
Option 2: larger mean free path for Y compared to that of X
Option 3: increased collision frequency of Y with the inert gas as compared to that of X with the inert gas
Option 4: increased collision frequency of X with the inert gas as compared to that of Y with the inert gas
Correct Answer: 4

Solutions:Mean free path (λ) of a gas is given by the following expression:

λ = RT2πd2NAP

, where

d : molecular diameter of the gas
P : pressure the gas experiences
T : temperature of the gas

At constant temperature, λ is dependent upon the molecular diameter of the gas and the pressure exerted on the gas.

Since X and Y have equal molecular diameters and pressure of the inert gas inside the tube is constant, the mean free path of gas X is same as that of gas Y.

Now, greater the collision frequency of a gas, the more hindrance it experiences in its flow. The increased collision frequency of gas X makes it suffer more collisions per second with the inert gas inside the tube, compared to gas Y. As a result, gas X covers less distance inside the tube than that predicted by Graham's Law. Thus, the experimental value of d is found to be smaller than the estimate obtained using Graham's Law.

Hence, the correct option is (D).

Question 33:

Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both the schemes.

The product X is

Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer: 1

Solutions:The complete sequence of reactions in scheme-1 is given below:

Hence, the correct option is A.

Question 34:

Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both the schemes.

The correct statement with respect to product Y is

Option 1: It gives a positive Tollens' test and is a functional isomer of X.
Option 2: It gives a positive Tollens' test and is a geometrical isomer of X.
Option 3: It gives a positive iodoform test and is a functional isomer of X.
Option 4: It gives a positive iodoform test and is a geometrical isomer of X.
Correct Answer: 3

Solutions:The complete sequence of reactions in scheme-2 is given below:

Compound Y contains the -COCH3 group; therefore, it will give positive iodoform test. Also, from the structure of compound Y, it is evident that it is a functional isomer of compound X.

Hence, the correct option is C.

Question 35:

An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral complexes with these reagents. Aqueous solution of M2, on reaction with reagent S, gives white precipitate, which dissolves in excess of S. The reactions are summarised in the scheme given below:

M1, Q and R, respectively, are

Option 1: Zn2+, KCN and HCl
Option 2: Ni2+ HCl and KCN
Option 3: Cd2+, KCN and HCl
Option 4: Co2+, HCl and KCN
Correct Answer: 2

Solutions:Metal M1 forms tetrahedral complex when it reacts with reagent Q and forms square planar complex when it reacts with reagent R.
In tetrahedral complexes, the hybridisation is sp3, whereas in the case of square planar complexes, the hybridisation is dsp2.
The electronic configuration of Ni2+ is d8.
In the presence of weak-field ligand, it undergoes sp3 hybridisation. And in the case of strong-field ligand, pairing of electrons occurs, which vacates one d orbital to form a square planar complex.
Therefore, the metal M1 should be Ni2+. Q and R should be HCl and KCN, respectively, as Cl is a weak-field ligand and CN is a strong-field ligand.

The scheme for the reaction is as follows:

Hence, the correct option is (B).

Question 36:

An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral complexes with these reagents. Aqueous solution of M2, on reaction with reagent S, gives white precipitate, which dissolves in excess of S. The reactions are summarised in the scheme given below:

Reagent S is

Option 1: K4FeCN6
Option 2: Na2HPO4
Option 3: K2CrO4
Option 4: KOH
Correct Answer: 4

Solutions:The metal ion M 1, when treated with Q and R in excess, forms tetrahedral complex only.
This clearly indicates that the geometry of the metal complex remains unaffected by the nature of the ligand. Hence, in both the cases, the hybridisation is sp3.
In order to justify all the conditions, metals should have either d9 or d10 configuration. Therefore, the metal ion could be Cu or Zn.
Zn, on reaction with KOH, forms a white precipitate of zinc hydroxide (white precipitate), whereas Cu, on reaction with KOH, forms copper hydroxide, which is green in colour. Hence, the metal M2 should be Zn.

Moreover, when Zn(OH)2 is further treated with excess KOH, it dissolves to form zincate ions.

Zn      KOH    ZnOH2white precipitate        KOH(excess)           ZnOH42

The reactions are summarised in the following scheme:

Hence, the correct option is (D).

Question 37:

Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select the correct answer using the code given below the lists.
{en = H2NCH2CH2NH2; atomic numbers: Ti = 22; Cr = 24; Co = 27; Pt = 78}
 

  List-I   List-II
P. [Cr(NH3)4Cl2]Cl 1. Paramagnetic and exhibits ionisation isomerism
Q. [Ti(H2O)5Cl](NO3)2 2. Diamagnetic and exhibits cis-trans isomerism
R. [Pt(en)(NH3)Cl]NO3 3. Paramagnetic and exhibits cis-trans isomerism
S. [Co(NH3)4(NO3)2]NO3 4. Diamagnetic and exhibits ionisation isomerism
Option 1:
  P Q R S
(A) 4 2 3 1
Option 2:
  P Q R S
(B) 3 1 4 2
Option 3:
  P Q R S
(C) 2 1 3 4
Option 4:
  P Q R S
(D) 1 3 4 2
Correct Answer: 2

Solutions:

[Cr(NH3)4Cl2]Cl has chromium ion in +3 oxidation state with valence d3 electronic configuration. All the three electrons are unpaired and hence, the complex exhibits paramagnetism. The complex also shows cis-trans isomerism.

[Ti(H2O)5Cl](NO3)2 has titanium ion in +3 oxidation state with valence d1 electronic configuration. The unpaired electron gives rise to paramagnetic character. The complex shows ionisation isomerism. The two ionisation isomers are

[Ti(H2O)5Cl](NO3)2 and [Ti(H2O)5(NO3)](NO3)Cl .

[Pt(en)(NH3)Cl]NO3 has platinum ion in +2 oxidation state with valence d8 electronic configuration. Now, the complex exhibits square planar geometry due to the presence of strong-field ligands. As a result, the electrons in the metal d-orbitals get paired up against Hund's rule of maximum multiplicity. Thus, the complex is diamagnetic. It also exhibits ionisation isomerism. The two ionisation isomers are

[Pt(en)(NH3)Cl]NO3 and [Pt(en)(NH3)(NO)3]Cl .

[Co(NH3)4(NO3)2]NO3 has cobalt ion in +3 oxidation state with valence d6 electronic configuration. Again, due to the presence of strong-field ligands, the electrons in the metal d-orbitals get paired up, giving rise to diamagnetic character in the complex. The complex exhibits cis-trans isomerisation.

Hence, the correct option is (B).

Question 38:

Match the orbital overlap figures shown in List-I with the description given in List-II and select the correct answer using the code given below the lists.

Option 1:
  P Q R S
(A) 2 1 3 4
Option 2:
  P Q R S
(B) 4 3 1 2
Option 3:
  P Q R S
(C) 2 3 1 4
Option 4:
  P Q R S
(D) 4 1 3 2
Correct Answer: 3

Solutions:

Assuming the shaded portion of the figure represents the positive phase of the orbital and the unshaded portion represents the negative phase of the orbital, the orbital overlap figure shown in P,

depicts a positive overlap of two d-orbitals along the inter-nuclear axis. This results in dd

σ

bonding.

The orbital overlap figure shown in Q,

depicts a positive lateral overlap of one p and one d-orbital. This results in p

π

bonding.

The orbital overlap figure shown in R,

depicts a negative lateral overlap of one p and one d-orbital. This results in p

π

anti-bonding.

The orbital overlap figure shown in S,

depicts a negative overlap of two d-orbitals along the internuclear axis. This results in dd

σ

anti-bonding.

Hence, the correct option is (C).

Question 39:

Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway from List I with an appropriate structure from List II and select the correct answer using the code given below the lists.

Option 1:
  P Q R S
(A) 1 3 4 2
Option 2:
  P Q R S
(B) 2 4 3 1
Option 3:
  P Q R S
(C) 4 1 2 3
Option 4:
  P Q R S
(D) 3 2 1 4
Correct Answer: 1

Solutions:Pyrolysis (thermal decomposition) of the given peroxyesters can be shown as follows:

1.

2.

3.

4.

It is evident that the respective pathways best suitable for the peroxyesters are as follows:

1 → Pathway P

2 → Pathway S

3 → Pathway Q

4 → Pathway R

Hence, the correct option is (A).

Question 40:

Match the four starting materials (P, Q, R, S) given in List I with the corresponding reaction schemes (I, II, III, IV) provided in List II and select the correct answer using the code given below the lists.

Option 1:
  P Q R S
(A) 1 4 2 3
Option 2:
  P Q R S
(B) 3 1 4 2
Option 3:
  P Q R S
(C) 4 1 3 2
Option 4:
  P Q R S
(D) 1 4 2 3
Correct Answer: 3

Solutions:The complete reaction sequences for all the four schemes are given below:

1. Scheme I:

2. Scheme II:

3. Scheme III:

4. Scheme IV:

 Hence, the correct option is (C).

Question 41:

The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation p(p(x)) = 0 has

Option 1: only purely imaginary roots
Option 2: all real roots
Option 3: two real and two purely imaginary roots
Option 4: neither real nor purely imaginary roots
Correct Answer: 4

Solutions:It is given that p (x) has purely imaginary roots.

Let 

p(x)=x2+k=0

, where k 

R, such that k > 0.

Now,

ppx=x2+k2+k=0

             â€‹= 

x4+2kx2+k2+k=0

The above equation is quadratic in 

x2

.

 x2= 2k±2k24k2+k2

        

=2k±4k24k24k2

       â€‹  

=2k±2k2

         â€‹

=k ± i k x=±k ± ik

  
So, the roots are neither real nor purely imaginary.

Hence, the correct option is D.

Question 42:

Three boys and two girls stand in a queue. The probability that the number of boys ahead of every girl is at least one more than the number of girls ahead of her is

Option 1: 12
Option 2: 13
Option 3: 23
Option 4: 34
Correct Answer: 1

Solutions:Let the three boys and two girls be denoted by B1, B2, B3, G1, G2.

B1B2B3

As per the given condition, a girl cannot occupy the 4th position.

B1B2B3

Either the girls can occupy two of the 1st, 2nd or 3rd positions, or they can both be at the 1st or 2nd position.

∴ Total number of ways in which the girls can be placed

C23×2!×3! + C12×2!×3!

= 36 + 24 = 60

Also, the total number of ways in which the three boys and two girls can be seated = 5! = 120.

∴ Required probability = 

60120=12

Hence, the correct option is A.

Question 43:

Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is

Option 1: 264
Option 2: 265
Option 3: 53
Option 4: 67
Correct Answer: 3

Solutions:Number of ways of dearrangement of 6 cards

=6!111!+12!13!+14!15!+16!=720111+1216+1241120+1720=360120+306+1=265

In these dearrangement of cards, there are 5 ways in which card numbered 1 is going into wrong envelops.

Thus, the number of ways when card number 1 is placed into envelop numbered 2 is

2655=53

ways.

Hence, the correct option is C.

Question 44:

In a triangle, the sum of two sides is x and the product of the same two sides is y. If x2c2 = y, where c is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is

Option 1: 3y2xx+c
Option 2: 3y2cx+c
Option 3: 3y4xx+c
Option 4: 3y4cx+c
Correct Answer: 2

Solutions:Let ab and c be the sides of the given triangle.

Given :

a + b = x;  ab = y

Now,

 

x2c2=y a+b2c2=ab a2+b2+2abc2=ab a2+b2c2=ab a2+b2c22ab=12 cos C = 12 C=120°

Now, the circum-radius, R, is given by

R=c2 sin C=c2 32 R=c3

                                 

Also, the in-radius, r, is given by

r=sc tan C2    =a+b+c2c tan 60°  =x+c2c·3  =xc2·3 r=32xc

                        

Thus, the ratio of in-radius to circum-radius is given by

rR=32xcc3

    

=32cxc

    

=3xcx+c2cx+c=3x2c22cx+c rR=3y2cx+c         x2c2=y

Hence, the correct option is B.

Question 45:

The common tangents to the circle x2 + y2 = 2 and the parabola y2 = 8x touch the circle at the points P, Q and the parabola at the points R, S. Then, the area of the quadrilateral PQRS is

Option 1: 3
Option 2: 6
Option 3: 9
Option 4: 15
Correct Answer: 4

Solutions:The equation of the given circle is

x2+y2=2

.

Also, the equation of the given parabola is

y2=8x

.

We know :

Equation of a tangent to a circle, 

x2+y2 = r2

 is

y = mx±r1+m2

Thus, equation of a tangent to a circle 

x2+y2=2

is

y = mx±2 1+m2

      …(1)

Also, equation of a tangent to a parabola, 

y2  = 4ax

 is

y = mx+am

Thus, equation of a tangent to a parabola, 

y2=8x

is

y = mx+2m

      …(2)

Since (1) and (2) represent the same lines,

± 21+m2 = 2m

Squaring both sides, we get

21+m2= 4m2m4+m22  =0m2+2m21 = 0m = ±1

So, the equations of the common tangent are

y = x+2 and y = x2

Now, solving these equations with the equation of circle, we get:

x2+x+22= 2x2+ 2x + 1 = 0x+12= 0x= 1,1 y = 1,  1So, P(1, 1) and Q(1, 1).

Solving the equations of common tangents with the equation of parabola, we get:

x+22 = 8xx24x+4 = 0x22 = 0x = 2, 2 y = 4, 4So,  R(2, 4)  and S(2,4) .

Using distance formula, we get:

PQ = 1+12+1+12 = 2RS = 222+442 = 8                      

Height of the trapezium PQRS = AB = 3

 Area of trapezium PQRS = 12×AB×PQ+RS =12×3×2+8 = 15                       

Hence, the correct option is D.

Question 46:

The function y = f(x) is the solution of the differential equation

dydx+xyx21=x2+4x1x2

in (−1, 1) satisfying f(0) = 0. Then

3333 fx dx

is

Option 1: π332
Option 2: π334
Option 3: π634
Option 4: π632
Correct Answer: 2

Solutions:The given differential equation is

 

dydx+xyx21 = x4+2x1x2dydxyx1x2 = x4+2x1x2This is of the form dydx+Py = Q.Its solution is given by y ×I.F. = Q×I.F. dx, where I.F. = ePdx.I.F. = exx21dx=1x2

The solution of the given differential equation is

y1x2 = x4+2x1x2×1x2 dx     y1x2 =x4+2x dxy1x2 = x55+x2+C Given: f0= 0C= 0 y = fx = x55+x21x2 Now, I = 3232 fx dxI = 3232 x55+x21x2 dxI = 3232 x551x2 dx + 3232x21x2dxI = 0 + 2032x21x2dxLet x = sinα dx = cosα dαWhen x0, α 0x32, απ3 I = 20π3 sin2α cosαcosα dαI = 0π31cos2αdαI = α12sin2α0π3I = π312×sin2π30I =π334

Hence, the correct option is B.

Question 47:

Let f : [0, 2] → ℝ be a function which is continuous on [0, 2] and is differentiable on (0, 2) with f(0) = 1. Let

Fx=0x2 ft dt

for x ∈ [0, 2]. If F'(x) = f'(x) for all x ∈ (0, 2), then F(2) equals

Option 1: e2 −1
Option 2: e4 − 1
Option 3: e − 1
Option 4: e4
Correct Answer: 2

Solutions:Given:

F(x) = 0x2f(t) dt                  ...1

Differentiating both sides, we get:

F'(x) = f(x2)×2xf0×0 

                 (Using Leibnitz's rule)

F'(x) = 2x f(x)

It is also given that,

F'x=f'x f'x=2x fx f'(x)f(x) = 2x

Integrating both sides, we get:

ln fx  = x2+c

When x = 0, f (0) = 1

c=0 lnfx=x2 fx=ex2               ...2

From (1) and (2), we have

Fx  =0x2ex dx Fx = ex21

For x = 2,

F(2)  =e41

Hence, the correct option is B.

Question 48:

Coefficient of x11 in the expansion of (1 + x2)4 (1 + x3)7 (1 + x4)12 is

Option 1: 1051
Option 2: 1106
Option 3: 1113
Option 4: 1120
Correct Answer: 3

Solutions:The given expansion is

1+x241+x371+x412

.

Coefficient of x11  in

1+x241+x371+x412

= Coefficient of x11  in

C04+C14x2+C24x4+C34x6+C44x8C07+C17x3+C27x6+C37x9+...C012+C112x4+C212x8+...

=

C14×C37×C012+C24×C17×C112+C44×C17×C012+C04×C17×C212

=  (4 × 35 × 1) + (6 × 7 × 12) + (1 × 7 × 1) + (1 ×7 × 66)

= 140 + 504 + 7 + 462

= 1113

Hence, the correct option is C.
[[VIDEO:13816]]

Question 49:

For x ∈ (0, π), the equation sin x + 2 sin 2x − sin 3x = 3 has

Option 1: infinitely many solutions
Option 2: three solutions
Option 3: one solution
Option 4: no solution
Correct Answer: 4

Solutions:Given,

sinx+2sin2xsin3x=3

i.e. 

sinxsin3x+2sin2x=3 2cos2xsinx+4sinxcosx=3 2sinx 2cosxcos2x=3 2sinx 2cosx2cos2x1=3 2sinx 12cos2x+2cosx=3 2sinx 12cos2xcosx=3 2sinx 322cosx122=3

This is possible only when 

sinx=1 and cosx122=0

.

Now, there does not exist any 

x0,π

 for which 

sinx=1 and cosx=12

.

Thus, the given equation has no solution.

Hence, the correct option is D.
[[VIDEO:13817]]

Question 50:

The following integral

π4π2 2 cosec x17 dx

is equal to

Option 1: 0log 1+2 2eu+eu16 du
Option 2: 0log 1+2 2eu+eu17 du
Option 3: 0log 1+2 2eueu17 du
Option 4: 0log 1+2 2eueu16 du
Correct Answer: 1

Solutions:Let I =

π4π22cosecx17dx

        =

π4π22sinx162cosecxdx

Now,

 

2sinx=21+cosxsinx1+cosx          =1+sin2x+cos2x+2cosxsinx1+cosx          =1+cosx+cosx+cos2x+sin2xsinx1+cosx         =1sinx+cosxsinx+sinx1+cosx        =cosecx+cotx+1cosecx+cotx

I

π4π2cosecx+cotx+1cosecx+cotx162cosecxdx

Put cosecx + cotx = eu.

cosecxcotxcosec2x=eududxcosecxcotx+cosecx=eududxcosecxdx=du

When

xπ4, eu2+1uln2+1 xπ2, eu1u0

I

2ln2+10eu+eu16du=0ln2+12eu+eu16du

Hence, the correct option is A.

Question 51:

Box 1 contains three cards bearing numbers 1, 2 and 3; box 2 contains five cards bearing numbers 1, 2, 3, 4 and 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6 and 7. A card is drawn from each of the boxes. Let xi be the number on the card drawn from the ith box, i = 1, 2, 3.
The probability that x1 + x2 + x3 is odd is

Option 1: 29105
Option 2: 53105
Option 3: 57105
Option 4: 12
Correct Answer: 2

Solutions:

Now, 

x1+x2+x3

is odd if

  • two of xi are even and the third is odd
  • all of xi are odd

Case I

If

x1

is odd, 

x2 and x3

 are even.

Then, favourable outcomes = 2 × 2 × 3 = 12

Case II

If

x2

is odd, 

x1 and x3

 are even.

Then, favourable outcomes = 1 × 3 × 3 = 9

Case III

If

x3

is odd,

x1 and x2

 are even.

Then, favourable outcomes = 1 × 2 × 4 = 8

Case IV

If

x1

,

x2, x3

  are all odd.

Then, favourable outcomes = 2 × 3 × 4 = 24

 Total favourable outcomes = 24 + 8 + 9 + 12 = 53

Also,

Total outcomes = 3 × 5 × 7 = 105

∴ Required probability =

53105

Hence, the correct option is B.

Question 52:

The probability that x1, x2 and x3 are in arithmetic progression is

Option 1: 9105
Option 2: 10105
Option 3: 11105
Option 4: 7105
Correct Answer: 3

Solutions:

If

x1, x2, x3

are in arithmetic progression, then

x2x1=x3x2  2x2=x3+x1                  ...1 x1+x3

 must always be even.

x1 and x2

 both must be odd or both must be even.

Case I:

If

x1 and x3

 are both odd, then

x11,3

and

x31,3,5,7

.

 Possible outcomes =

1,1, 1,3, 1,5, 1,7, 3,1, 3,3, 3,5, 3,7

.

Case II:

If 

x1 and x3

 are both even, then

x12

and

x32,4,6

.

 Possible outcomes =

2,2, 2,4, 2,6

.

Thus, total favourable outcomes = 8 + 3 = 11.

Also, total outcomes = 3 × 5 × 7 = 105.

 â€‹Required probability =

11105

.

Hence, the correct option is C.

Question 53:

Let a, r, s and t be non-zero real numbers. Let P(at2, 2at), Q, R(ar2, 2ar) and S(as2, 2as) be distinct points on the parabola y2 = 4ax. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (2a, 0).
The value of r is

Option 1: 1t
Option 2: t2 + 1t
Option 3: 1t
Option 4: t2  1t
Correct Answer: 4

Solutions:Given:

P(at2, 2at), R(ar2, 2ar), S (as2, 2as), F(a, 0), K(2a, 0)

Let the coordinates of Q be 

 (at12, 2at1)

.

Since PQ is the focal chord,

 t·t1=1t1=1t 

Thus, the coordinates of Q are 

at2,2at

 â€‹.

Now, 

Slope of PK = 02at2aat2 = 2tt22       Slope of QR = 2ar+2atar2at2 = 2r1t 

It is given that, QR is parallel to PK.

 Slope of PK = Slope of QR

2tt22 = 2r1t2tr1t = 2t22tr1 = t22rt =  t21r = t21t

Hence, the correct option is D.

Question 54:

If st = 1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is

Option 1: (t2 + 1)22t3
Option 2: a(t2 + 1)22t3
Option 3: a(t2 + 1)2t3
Option 4: a(t2 + 2)2t3
Correct Answer: 2

Solutions:

Given: st = 1

s = 1t

 Coordinates of S

at2,2at

Now, we know that the equation of a tangent at a point (x1, y1) on the parabola 

y2=4ax

 is

yy1 = 2a(x+x1)

So, the equation of the tangent at P 

(at2,2at)

 is

y(2at) =2a(x+at2)ty =(x+at2)         x=tyat2                            ...(1)

Also, the equation of a normal at a point (x1y1) on the parabola 

y2=4ax

 is​

yy1=y12axx1

So, the equation of the normal at S 

at2,2at

 is

y2at = 2a2atxat2ty2a = at2xx = ty+2a+at2                     ...(2)

From (1) and (2), we get:

tyat2 = ty+2a+at22ty = 2a+at2+at22ty = at+1t2y =  at+1t22ty = a(t2+1)22t3

Hence, the correct option is B.
[[VIDEO:13818]]

Question 55:

Given that for each a ∈ (0, 1),

limh0+ h1hta1ta1dt

exists. Let this limit be g(a). In addition, it is given that the function g(a) is differentiable on (0, 1).

The value of

g 12

is

Option 1: π
Option 2: 2π
Option 3: π2
Option 4: π4
Correct Answer: 1

Solutions:Given:
g(a) =

limh0+ h1hta 1ta1dt  g12 = limh0+h1ht121t12dt

    

           =limh0+h1h 1t1  tdt

     

          =limh0+h1h 1t2  tdt            

   

           =limh0+h1h 1t2  t + 14 14dt

   

          = limh0+h1h 114 t122dt

  

           = limh0+sin1 t  1212h1h

  

           = limh0+sin12t  1h1h

  

           = limh0+sin121  h  1  sin12h  1

  

           = limh0+sin11 2h  sin12h  1

             

= π2 π2= π2+ π2=π

       
Hence, the correct option is A.

Question 56:

Given that for each a ∈ (0, 1),

limh0+ h1hta1ta1dt

exists. Let this limit be g(a). In addition, it is given that the function g(a) is differentiable on (0, 1).

The value of

g' 12

is

Option 1: π2
Option 2: π
Option 3: π2
Correct Answer: 4

Solutions:Given:

ga=limh0+h1hta1ta1dt

Differentiating on both sides with respect to a, we get:

g'a=limh0+h1hddata1ta1dt

        

=limh0+h1htalnt1ta1+ta1ta1ln1tdt

        

=limh0+h1hta1ta1lnt+ln1tdt

        

=limh0+h1hta1ta1ln1ttdt

Now,

g'12=limh0+h1ht121t12ln1ttdt

     …(1)

Using

abfxdx=abfa+bxdx

, we get:

g'12=limh0+h1ht121t12lnt1tdt

    …(2)

Adding (1) and (2), we get:

2g'12=limh0+h1ht121t12lnt1t+ln1ttdt 2g'12=limh0+h1ht121t12ln1dt

2g'12=limh0+h1ht121t120dt 2g'12=0  g'12=0

 Hence, the correct option is D.

Question 57:

List-I List-II
P. The number of polynomials f(x) with non-negative integer coefficients of degree ≤ 2, satisfying f(0) = 0 and

01 fx dx=1,

is

1. 8
Q. The number of points in the interval

13, 13

at
which f(x) = sin(x2) + cos(x2) attains its maximum value is

2. 2
R. 22 3x21+ex dx

equals

3. 4
S. 1212cos 2x log 1+x1xdx012cos 2x log 1+x1xdx 

equals

4. 0
Option 1:
  P Q R S
(A) 3 2 4 1
Option 2:
  P Q R S
(B) 2 3 4 1
Option 3:
  P Q R S
(C) 3 2 1 4
Option 4:
  P Q R S
(D) 2 3 1 4
Correct Answer: 4

Solutions:P.

Let

fx=a0x2+a1x+a2

, where

a0, a1, a2>0

.

Given,  f(0) = 0

a2=0

              …(1)

Also,

01fxdx=1 a0x33+a1x22+a2x01=1a03+a12+a2=1

From (1), we have:

a03+a12=1 2a0+3a1=6

For non – negative integral solutions,

a0=0

and

a1= 2

Or

a0= 3

and

a1=0

fx=2x

or

fx=3x2

Thus, the number of polynomials satisfying the given conditions is 2.

Q.

Let

fx=sinx2+cosx2

On differentiating both sides with respect to x, we get:

f'x=2xcosx2sinx2

              …(1)

For maxima or minima,

f'x=0

.

x=0, tanx2=1 x=0, x2=π4, 5π4, 9π4, 13π4

Now, differentiating (1) with respect to x, we get:

f''x=22x2sinx2+cosx2+cosx2sinx2

Now,

f''0=2>0 f''±π4=42π4<0 f''±5π4=425π4>0 f''±9π4=429π4<0 f''±13π4=4213π4>0

f is maximum at 

x=±π4

and 

x=±9π4

.

Thus, there are 4 points at which f attains maximum value.

R.

Let I =

223x21+exdx

        …(1)

We know that

abfxdx =abfa+bxdx

I =

223x21+exdx

         …(2)

Adding (1) and (2), we get:

 

2I=223x211+ex+11+exdx   =223x21+ex+1+ex1+ex1+exdx   =223x2dx   =x322=16

I = 8

S.

Suppose I =

1212cos2xlog1+x1xdx

       

Let

fx=cos2xlog1+x1x

Then,

fx=cos2xlog1x1+x=cos2xlog1+x1x1=cos2xlog1+x1x=fx

Thus,

fx=cos2xlog1+x1x

is an odd function.

⇒ 

1212cos2xlog1+x1xdx=0   1212cos2xlog1+x1xdx012cos2xlog1+x1xdx= 0

Thus, the correct match of the columns is
 

P Q R S
2 3 1 4

Hence, the correct option is D.

Question 58:

  List I   List II
P. Let y(x) = cos(3 cos1 x) , x [1, 1], x ±32. Then 1y(x) (x2  1 d2y(x)dx2+ x dy(x)dx equals 1. 1
Q. Let A1, A2, ..., An (n > 2) be the vertices of a regular polygon of n sides with its centre at the origin. Let ak bethe position vector of the point Ak, k = 1, 2, ..., n. Ifk=1n1(ak×ak+1) = k=1n1(ak .ak+1), the the minimum value of n is 2. 2
R. If the normal from the point P (h, 1) on the ellipse

x26+ y23 = 1

is perpendicular to the line x + y = 8, the the value of h is

3. 8
S. Number of positive solutions satisfying the equation

tan112x+1+ tan114x+1 = tan12x2
4. 9
Option 1:
  P Q R S
(A) 4 3 2 1
Option 2:
  P Q R S
(B) 2 4 3 1
Option 3:
  P Q R S
(C) 4 3 1 2
Option 4:
  P Q R S
(D) 2 4 1 3
Correct Answer: 1

Solutions:P.

The given function is

yx=cos3cos1x  ,   x1,1

        … (1)

Differentiating (1) with respect to x, we get:

dydx=sin3cos1x×31x2 1x2dydx=3sin3cos1x

Now, squaring both sides, we get:

1x2dydx2=9sin23cos1x 1x2dydx2=91cos23cos1x 1x2dydx2=91y2                   y=cos3cos1x

Again differentiating both sides with respect to x, we get:

1x2×2dydx×d2ydx22xdydx2=18ydydx x21d2ydx2+xdydx=9y                      dydx0 1yx21d2ydx2+xdydx=9yy=9

To justify

dydx0

, let us suppose

dydx=0

. Integrating

dydx=0

with respect to x, we get:

y=k constant

But it is given that

yx=cos3cos1x  ,   x1,1

.

dydx0

Q.

It is given that

k=1n1ak×ak+1=k=1n1ak·ak+1

Let the angle between any two consecutive position vectors of an n-sided polygon be

θ=2πn

.

Here,

a1=a2=a3...=an

Let

a1=a2=a3...=an

= R

k=1n1ak×ak+1=k=1n1ak·ak+1 k=1n1akak+1sinθn^=k=1n1akak+1cosθ

,  where

n^

is a unit vector perpendicular to the vectors

a1, a2, a3, ..., an

.

R2sinθk=1n1n^=R2cosθk=1n11 n1R2sinθ=n1R2cosθ sin2πn=cos2πntan2πn=1

For minimum value of n, we must have

2πn=π4n=8

R.

The equation of the ellipse is

x26+y23=1

.

Here,

a=6 and b=3

.

We know that the equation of the normal to the ellipse

x2a2+y2b2=1

is

axcosθbysinθ=a2b2

.

So, the equation of the normal to the ellipse

x26+y23=1

is

6xcosθ3ysinθ=3

It is given that 

6xcosθ3ysinθ=3

is perpendicular to  x + y = 8.

1×2tanθ=1tanθ=12

It is also given that 

6xcosθ3ysinθ=3

passes through P(h, 1).

6h23313=3 32h23=3h=2

S.

The given equation is

tan112x+1+tan114x+1=tan12x2

.

Using the formula,

tan1a+tan1b=tan1a+b1ab

, we get:

tan112x+1+14x+1112x+1·14x+1=tan12x2 tan16x+28x2+6x=tan12x2 6x+28x2+6x=2x2 6x+28x+6=2x     x0 3x27x6=03x+2x3=0x=3, 23

The positive solution of 

tan112x+1+tan114x+1=tan12x2

is x = 3.

Thus, the correct match of the columns is
 

P Q R S
4 3 2 1

Hence, the correct option is A.

Question 59:

Let f1 : ℝ → ℝ, f2 : [0, ∞) → ℝ, f3 : â„ → ℝ and f4 : ℝ → [0, ∞) be defined by

f1x=xif x<0,exif x0;f2x=x2;f3x=sin xif x<0,xif x0andf4x=f2f1xif x<0,f2f1x1if x0.
  List I   List II
P. f4 is 1. onto but not one-one
Q. f3 is 2. neither continuous nor one-one
R. f20f1 is 3. differentiable but not one-one
S. f2 is 4. continuous and one-one
Option 1: PQRSA3142
Option 2: PQRSB1342
Option 3: PQRSC3124
Option 4: PQRSD1324
Correct Answer: 4

Solutions:The given functions are

f1:

f1x=x       ,if  x<0ex       ,if  x0   f2:[0, )

,

f2x=x2 f3:

,

f3x=sinx       ,if  x<0x            ,if  x0   f4:[0, )

,

f4x=f2f1x            ,if  x<0f2f1x1      ,if  x0  

P.

f4:[0, )

The function f4 can be redefined as

f4x=x2             ,if  x<0e2x1     ,if  x0  

Clearly, the function is onto but not one-one. It is continuous but not differentiable at x = 0 as

f4'x=2x             ,if  x<02e2x         ,if  x0   f4'0=0 and f4'0+=2

Q.

f3 :  f3x=sinx      ,if  x<0x           ,if  x0  

Differentiating f3(x) with respect to x, we get:

f3'x=cosx       ,if  x<01             ,if  x0   f3'0=f3'0+=1

Therefore, the function f3(x) is differentiable everywhere but it is not one-one and onto.

R.

The composition function

f2of1x=f2x      , if x<0f2ex     ,if x0  =x2       ,if  x<0e2x     ,if  x0  

The graph of f2of1(x) is

Clearly, f2of1(x) is neither continuous nor one-one.

S.

f2: [0, ),  f2x=x2

Clearly, the function  f2(x) is continuous and one-one.

Thus, the correct match of the columns is
 

P Q R S
1 3 2 4

Hence, the correct option is D.

Question 60:

Let

zk=cos 2kπ10+i sin 2kπ10; k=1, 2,...,9

.
 

  List I   List II
P. For each zk there exists a zj , such that zk.zj = 1. 1. True
Q. There exists a k ∈ {1, 2,….,9), such that z1.z = zk has no solution z in the set of complex numbers. 2. False
R. 1z1 1z2....1z910

equals

3. 1
S. 1k=19  cos 2kπ10

equals

4. 2
Option 1: PQRSA1243
Option 2: PQRSB2134
Option 3: PQRSC1234
Option 4: PQRSD2143
Correct Answer: 3

Solutions:Given:

zk=cos2kπ10+isin2kπ10; k=1, 2, 3,...,9

P.

zk=cos2kπ10+isin2kπ10 zk=ei2kπ10 zk zj=ei2k+jπ10=1 cos2π j+k10+i sin2π j+k10=1 cos2πj+k10=1 and sin2πj+k10=0 j+k=10n, j+k=5m

So, there exists a

zj

, such that

zk zj=1

.

Thus, the given statement is true.

Q.

z1z=zkz=zkz1=ei2k1π10

Thus, the equation

z1z=zk, k=1, 2, 3,...,9

has a solution in the set of complex numbers.

Thus, the given statement is false.

R.

zk=cos2kπ10+isin2kπ10; k=1,2,3,...,10  z10=1z101=0 z101z1=1+z+z2+z3+...+z9 zz1zz2zz3...zz9=1+z+z2+z3+...+z9

Substituting z = 1, we get:

1z11z21z3...1z9=1+1+1+1+...+1=10 1z11z21z3...1z910=1

S.

1k=19cos2kπ10=1cos2π10+cos4π10+cos6π10+...+cos18π10

We know that

cosa+cosa+d+cosa+2d+...+cosa+n1d=sinnd2sind2cos2a+n1d2  1k=19cos2kπ10=1cos4π10+8×2π102=11=2

Thus, the correct match of the columns is
 

P Q R S
1 2 3 4

Hence, the correct option is C.

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