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: $\frac{2S}{b\rho g}\mathrm{cos}\left(\theta –\alpha \right)$ Option 2: $\frac{2S}{b\rho g}\mathrm{cos}\left(\theta +\alpha \right)$ Option 3: $\frac{2S}{b\rho g}\mathrm{cos}\left(\theta –\alpha /2\right)$ Option 4: $\frac{2S}{b\rho g}\mathrm{cos}\left(\theta +\alpha /2\right)$

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

In

$∆$

ABC:

$\frac{b}{R}=\mathrm{cos}\left(\theta +\frac{\alpha }{2}\right)$

Or

$\frac{1}{R}=\frac{\mathrm{cos}\left(\theta +\frac{\alpha }{2}\right)}{b}$

…(1)

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

${P}_{o}–{P}_{\mathrm{i}}=\frac{2S}{R}$

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

${P}_{0}–\frac{2S}{R}+h\mathrm{\rho }g={P}_{0}\phantom{\rule{0ex}{0ex}}⇒h=\frac{2S}{R\mathrm{\rho }g}$

Using equation (1), we get:

$h=\frac{2S\mathrm{cos}\left(\theta +\frac{\alpha }{2}\right)}{b\mathrm{\rho }g}$

Hence, the correct option is D.

Question 2:

$R=\frac{1}{10}×$

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

$\frac{R}{5}$

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

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

$g=\frac{GM}{{R}^{2}}=\frac{G}{{R}^{2}}\left(\frac{4}{3}\pi {R}^{3}\rho \right)=\frac{4}{3}\pi G\rho R$

Now, at a depth x,

${g}_{x}=\left(1–\frac{x}{R}\right){g}_{\mathrm{p}}$

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

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

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

${E}_{1}=\frac{1}{4\pi {\epsilon }_{0}}\frac{Q}{{R}^{2}}$

Electric field at point P due to sphere 2,

${E}_{2}=\frac{1}{4\pi {\epsilon }_{0}}\frac{2Q}{{R}^{2}}=2\left(\frac{1}{4\pi {\epsilon }_{0}}\frac{Q}{{R}^{2}}\right)$

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

$\mathrm{\rho }=\frac{4Q}{\frac{4}{3}\pi \left(2R{\right)}^{3}}=\frac{3Q}{8\pi {R}^{3}}$

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

$\frac{\rho R}{3{\epsilon }_{0}}$

Thus, electric field for sphere 3,

${E}_{3}=\frac{1}{3}\frac{3QR}{8\pi {R}^{3}{\epsilon }_{0}}=\frac{1}{2}\left(\frac{1}{4\pi {\epsilon }_{0}}\frac{Q}{{R}^{2}}\right)$

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: 0.5 Option 4: 0.48

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

$f=C\left(Z–1{\right)}^{2}$

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

$⇒\lambda \propto \frac{1}{f}$

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

${\lambda }_{\mathrm{Cu}}\propto \frac{1}{{f}_{\mathrm{Cu}}}\phantom{\rule{0ex}{0ex}}⇒{\lambda }_{Cu}\propto \frac{1}{C\left({Z}_{\mathrm{Cu}}–1{\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒{\lambda }_{Cu}\propto \frac{1}{C\left(29–1{\right)}^{2}}\phantom{\rule{0ex}{0ex}}\therefore {\lambda }_{\mathrm{Cu}}\propto \frac{1}{C\left(28{\right)}^{2}}$

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

${\lambda }_{\mathrm{Mo}}\propto \frac{1}{{f}_{\mathrm{Mo}}}\phantom{\rule{0ex}{0ex}}⇒{\lambda }_{\mathrm{Mo}}\propto \frac{1}{C\left({Z}_{\mathrm{Mo}}–1{\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒{\lambda }_{\mathrm{Mo}}\propto \frac{1}{C\left(42–1{\right)}^{2}}\phantom{\rule{0ex}{0ex}}\therefore {\lambda }_{\mathrm{Mo}}\propto \frac{1}{C\left(41{\right)}^{2}}$

So, the value of λCu/λMo is given by

$\frac{{\lambda }_{\mathrm{Cu}}}{{\lambda }_{\mathrm{Mo}}}=\frac{\left(41{\right)}^{2}}{\left(28{\right)}^{2}}=2.144\approx 2.14$

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

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 =

$\frac{1}{2}m{v}^{2}$

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

$–$

sinθ) =

$\frac{1}{2}m{v}^{2}$

…(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.

$mg\mathrm{sin}\theta –N=\frac{m{v}^{2}}{R}$

$N=mg\mathrm{sin}\theta –\frac{m{v}^{2}}{R}$

Putting the value of v from equation (1):

$N=mg\mathrm{sin}\theta –\frac{2mgR\left(1–\mathrm{sin}\theta \right)}{R}=mg\left(3\mathrm{sin}\theta –2\right)$

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

$\theta >{\mathrm{sin}}^{–1}\left(\frac{2}{3}\right)$

$\theta <{\mathrm{sin}}^{–1}\left(\frac{2}{3}\right)$

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

${90}^{\mathrm{o}}>{\mathrm{sin}}^{–1}\left(\frac{2}{3}\right)$

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

${0}^{\mathrm{o}}<{\mathrm{sin}}^{–1}\left(\frac{2}{3}\right)$

). 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

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

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

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

$\frac{hc}{\lambda }$

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

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

$\frac{{\mathrm{KE}}_{1}}{{\mathrm{KE}}_{2}}={\left(\frac{{u}_{1}}{{u}_{2}}\right)}^{2}=\frac{4}{1}$

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

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:

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

$K=\frac{1}{2}m{v}^{2}$

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

$K=\frac{1}{2}m\left(gt{\right)}^{2}=\frac{1}{2}m{g}^{2}{t}^{2}$

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

$K=\frac{1}{2}m\left(u–gt{\right)}^{2}$

…(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 â„¦, 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 â„¦ Option 2: 135 ± 0.56 â„¦ Option 3: 60 ± 0.25 â„¦ Option 4: 135 ± 0.23 â„¦

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

$\frac{{R}_{1}}{{R}_{2}}=\frac{l}{\left(100–l\right)}\phantom{\rule{0ex}{0ex}}$

Here, R1 = R (unknown resistance)
R2 = 90 â„¦

$R=\frac{90l}{\left(100–l\right)}\phantom{\rule{0ex}{0ex}}$

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

$∆$

R.
Relative error is given by:

$\frac{∆R}{R}=\frac{∆l}{l}+\frac{∆l}{100–l}\phantom{\rule{0ex}{0ex}}⇒\frac{∆R}{60}=\frac{0.1}{40}+\frac{0.1}{60}\phantom{\rule{0ex}{0ex}}⇒∆R=\frac{0.5}{2}=0.25$

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

Hence, the value of the resistance can be written as

.

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

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

$P=\sigma A\left({T}^{4}–{{T}_{0}}^{4}\right)$

.
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,

$P=\sigma 4\pi {R}^{2}\left({T}^{4}–{300}^{4}\right)$

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

$I×\pi {R}^{2}$

= 912

$×$

(

$\pi {R}^{2}$

)

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

Therefore,

$\sigma 4\pi {R}^{2}\left({T}^{4}–{300}^{4}\right)$

= 912

$×$

(

$\pi {R}^{2}$

)

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: 1.3 Option 3: 1.36 Option 4: 1.42

Solutions:
From the figure,

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

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

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 h ≈ a Option 2: current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ a 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

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,

$\stackrel{\to }{{B}_{R}}$

= magnetic field due to the ring

$\stackrel{\to }{{B}_{1}}$

= magnetic field due to wire 1

$\stackrel{\to }{{B}_{2}}\phantom{\rule{0ex}{0ex}}$

= magnetic field due to wire 2
Magnitude of

$\stackrel{\to }{{B}_{1}}$

and

$\stackrel{\to }{{B}_{2}}$

are equal.

$\stackrel{\to }{{B}_{1}}=\stackrel{\to }{{B}_{2}}=\frac{{\mu }_{0}ia}{2\mathrm{\pi }\left({a}^{2}+{h}^{2}\right)}$

âˆµ Magnetic field due to the circular loop is

$\frac{{\mu }_{0}I{a}^{2}}{2{\left({a}^{2}+{h}^{2}\right)}^{3/2}}$

.
Resultant of

$\stackrel{\to }{{B}_{1}}$

and

$\stackrel{\to }{{B}_{2}}$

=

$2\stackrel{\to }{{B}_{1}}\mathrm{cos}\theta =\frac{2×{\mu }_{0}ia}{2\mathrm{\pi }\left({a}^{2}+{h}^{2}\right)}$

.

$\stackrel{\to }{{B}_{R}}=\frac{2{\mu }_{0}I{\mathrm{\pi a}}^{2}}{4{\mathrm{\pi r}}^{2}}$

For zero magnetic field at P,

$\frac{{\mu }_{0}i{a}^{2}}{2{\left({a}^{2}+{h}^{2}\right)}^{3/2}}=\frac{2{\mu }_{0}ia}{2\mathrm{\pi }\left({a}^{2}+{h}^{2}\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{3{a}^{2}}{2}={h}^{2}\phantom{\rule{0ex}{0ex}}\therefore h\approx 1.2a$

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

$d\gg a$

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: $\frac{{\mu }_{0}{I}^{2}{a}^{2}}{d}$ Option 2: $\frac{{\mu }_{0}{I}^{2}{a}^{2}}{2d}$ Option 3: $\frac{\sqrt{3}{\mu }_{0}{I}^{2}{a}^{2}}{d}$ Option 4: $\frac{\sqrt{3}{\mu }_{0}{I}^{2}{a}^{2}}{2d}$

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

$\frac{{\mu }_{0}I}{2\mathrm{\pi }d}+\frac{{\mu }_{0}I}{2\mathrm{\pi }d}=\frac{{\mu }_{0}I}{\mathrm{\pi }d}$

Magnetic moment of the loop, M =

$I\left({\mathrm{\pi a}}^{2}\right)$

Torque on the loop,

$\tau$

=

$MB\mathrm{sin}\theta$

$\tau$

=

$MB\mathrm{sin}150°$

$\tau$

=

$I\left(\pi {a}^{2}\right)×\frac{{\mu }_{0}I}{\mathrm{\pi }d}×\frac{1}{2}$ $=\frac{{\mathrm{\mu }}_{0}{I}^{2}{a}^{2}}{2d}$

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

and those for an ideal diatomic gas are

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

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,

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

and those for an ideal diatomic gas are

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

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,

Work done:
ΔW = ΔU
=  n1Cv1ΔT + n2Cv2ΔT

$=2×\frac{3R}{2}×\left(525–700\right)+2×\frac{5R}{2}×\left(525–400\right)\phantom{\rule{0ex}{0ex}}=–3R\left(175\right)+5R\left(125\right)\phantom{\rule{0ex}{0ex}}=–525R+625R\phantom{\rule{0ex}{0ex}}=–100R$

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 ms−1 Option 3: 2 ms−1 Option 4: 8 ms−1

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:

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

${\rho }_{a}$

and that of the liquid

${\rho }_{\mathrm{\ell }}$

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

 Option 1: $\sqrt{\frac{{\rho }_{a}}{{\rho }_{\mathrm{\ell }}}}$ Option 2: $\sqrt{{\rho }_{a}{\rho }_{\mathrm{\ell }}}$ Option 3: $\sqrt{\frac{{\rho }_{\mathrm{\ell }}}{{\rho }_{a}}}$ Option 4: ${\rho }_{\mathrm{\ell }}$

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

${P}_{0}–\frac{1}{2}{\rho }_{l}{{v}_{l}}^{2}$

Pressure at B =

${P}_{0}–\frac{1}{2}{\rho }_{l}{{v}_{l}}^{2}–{\rho }_{l}gh$

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

${P}_{0}–\frac{1}{2}{\rho }_{a}{{v}_{a}}^{2}$

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

${P}_{0}–\frac{1}{2}{\rho }_{a}{{v}_{a}}^{2}={P}_{0}–\frac{1}{2}{\rho }_{l}{{v}_{l}}^{2}–{\rho }_{l}gh\phantom{\rule{0ex}{0ex}}⇒{v}_{l}=\sqrt{\frac{{\rho }_{a}}{{\rho }_{l}}{{v}_{a}}^{2}–2gh}\phantom{\rule{0ex}{0ex}}⇒{v}_{l}\propto \sqrt{\frac{{\rho }_{a}}{{\rho }_{l}}}$

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

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

$v=\sqrt{2H\left(a+g\right)}$

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:

$h={u}_{y}t+\frac{1}{2}\left(a+g\right){t}^{2}\phantom{\rule{0ex}{0ex}}h=0+\frac{1}{2}\left(a+g\right){t}^{2}\phantom{\rule{0ex}{0ex}}⇒t=\sqrt{\frac{2h}{\left(a+g\right)}}$

Horizontal range of the jet is given as

$d=v×t$

$⇒d=\sqrt{2H\left(a+g\right)}×\sqrt{\frac{2h}{\left(a+g\right)}}\phantom{\rule{0ex}{0ex}}⇒d=2\sqrt{Hh}$

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

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

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

$\mu$

r) is 1.5.

For P:

Focal length of each lens:

Focal length of the combination:

$\frac{1}{{f}_{\mathrm{R}}}=\frac{1}{f}+\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{f}_{\mathrm{R}}}=\frac{1}{r}+\frac{1}{r}\phantom{\rule{0ex}{0ex}}⇒{f}_{\mathrm{R}}=\frac{r}{2}$

∴ P

$\to$

2

For Q:

Focal length of each lens:

Focal length of the combination:

$\frac{1}{{f}_{\mathrm{R}}}=\frac{1}{f}+\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{f}_{\mathrm{R}}}=\frac{1}{2r}+\frac{1}{2r}\phantom{\rule{0ex}{0ex}}⇒{f}_{\mathrm{R}}=r$

∴ Q

$\to$

4

For R:

Focal length of each lens:

Focal length of the combination:

$\frac{1}{{f}_{\mathrm{R}}}=\frac{1}{f}+\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{f}_{\mathrm{R}}}=–\frac{1}{2r}–\frac{1}{2r}\phantom{\rule{0ex}{0ex}}⇒{f}_{\mathrm{R}}=–r$

∴ R

$\to$

3

For S:

Focal length of the first lens:

Focal length of the second lens:

Focal length of the combination:

$\frac{1}{{f}_{\mathrm{R}}}=\frac{1}{{f}_{1}}+\frac{1}{{f}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{f}_{\mathrm{R}}}=\frac{1}{r}–\frac{1}{2r}=\frac{1}{2r}\phantom{\rule{0ex}{0ex}}⇒{f}_{\mathrm{R}}=2r$

∴ S

$\to$

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

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,

$\mu$

= 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

$\mu N\ge {m}_{1}g\mathrm{sin}\theta +{m}_{2}g\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}\mu \left({m}_{2}g\mathrm{cos}\theta \right)\ge {m}_{1}g\mathrm{sin}\theta +{m}_{2}g\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\theta \ge \frac{\mu {m}_{2}}{{m}_{1}+{m}_{2}}=0.2$

We know:

$\mathrm{tan}\left(11.5°\right)\approx 0.2$

Therefore, block m2 will start sliding when

$\theta >11.5°$

.

So, when

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

$\left({m}_{1}+{m}_{2}\right)g\mathrm{sin}\theta$

.
When

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

$\mu {m}_{2}g\mathrm{cos}\theta$

.

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

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:

$\left(\sigma 1s{\right)}^{2}\left(\sigma *1s{\right)}^{2}\left(\sigma 2s{\right)}^{2}\left(\sigma *2s{\right)}^{2}$

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:

$\left(\sigma 1s{\right)}^{2}\right)\left(\sigma *1s{\right)}^{2}\left(\sigma 2s{\right)}^{2}\left(\sigma *2s{\right)}^{2}\left(\sigma 2{p}_{z}{\right)}^{2}$

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:

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

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:

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

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

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

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

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

…(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,

On substituting the value of

in equation (1), we get:

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

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

).

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

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:

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.

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

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

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

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:

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

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 (

${\mathrm{HXeO}}_{4}^{–}$

).

So, the compound Q is

${\mathrm{HXeO}}_{4}^{–}$

.

The hydrogen xenate ion (

${\mathrm{HXeO}}_{4}^{–}$

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

${\mathrm{XeO}}_{6}^{4–}$

).

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

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

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."

In other words, it can also be written as

, 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,

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

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

, 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:

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.

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

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: ${\mathrm{K}}_{4}\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_{6}\right]$ Option 2: ${\mathrm{Na}}_{2}{\mathrm{HPO}}_{4}$ Option 3: ${\mathrm{K}}_{2}{\mathrm{CrO}}_{4}$ Option 4: $\mathrm{KOH}$

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.

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

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

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

$\sigma$

bonding.

The orbital overlap figure shown in Q,

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

$\pi$

bonding.

The orbital overlap figure shown in R,

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

$\pi$

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

$\sigma$

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

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

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

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

Let

$p\left(x\right)={x}^{2}+k=0\phantom{\rule{0ex}{0ex}}$

, where k

$\in$

R, such that k > 0.

Now,

$p\left(p\left(x\right)\right)={\left({x}^{2}+k\right)}^{2}+k=0$

â€‹=

${x}^{4}+2k{x}^{2}+{k}^{2}+k=0$

The above equation is quadratic in

${x}^{2}$

.

$=\frac{–2k±\sqrt{4{k}^{2}–4{k}^{2}–4k}}{2}$

â€‹

$=\frac{–2k±2\sqrt{–k}}{2}$

â€‹

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: $\frac{1}{2}$ Option 2: $\frac{1}{3}$ Option 3: $\frac{2}{3}$ Option 4: $\frac{3}{4}$

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

$\square {\mathrm{B}}_{1}\square {\mathrm{B}}_{2}\square {\mathrm{B}}_{3}\square$

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

$\square {\mathrm{B}}_{1}\square {\mathrm{B}}_{2}\square {\mathrm{B}}_{3}\overline{)\square }$

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

= 36 + 24 = 60

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

∴ Required probability =

$\frac{60}{120}=\frac{1}{2}$

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

Solutions:Number of ways of dearrangement of 6 cards

$=6!\left(1–\frac{1}{1!}+\frac{1}{2!}–\frac{1}{3!}+\frac{1}{4!}–\frac{1}{5!}+\frac{1}{6!}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=720\left(1–\frac{1}{1}+\frac{1}{2}–\frac{1}{6}+\frac{1}{24}–\frac{1}{120}+\frac{1}{720}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=360–120+30–6+1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=265\phantom{\rule{0ex}{0ex}}$

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

$\frac{265}{5}=53$

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: $\frac{3y}{2x\left(x+c\right)}$ Option 2: $\frac{3y}{2c\left(x+c\right)}$ Option 3: $\frac{3y}{4x\left(x+c\right)}$ Option 4: $\frac{3y}{4c\left(x+c\right)}$

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

Given :

a + b = x;  ab = y

Now,

${x}^{2}–{c}^{2}=y$ $⇒{\left(a+b\right)}^{2}–{c}^{2}=ab$ $⇒{a}^{2}+{b}^{2}+2ab–{c}^{2}=ab$ $⇒{a}^{2}+{b}^{2}–{c}^{2}=–ab$ $⇒\frac{{a}^{2}+{b}^{2}–{c}^{2}}{2ab}=–\frac{1}{2}$ $⇒\angle C=120°$

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

$\therefore R=\frac{c}{\sqrt{3}}$

Also, the in-radius, r, is given by

$\therefore r=\frac{\sqrt{3}}{2}\left(x–c\right)$

$\frac{r}{R}=\frac{\frac{\sqrt{3}}{2}\left(x–c\right)}{\frac{c}{\sqrt{3}}}$

$=\frac{3}{2c}\left(x–c\right)$

$=\frac{3\left(x–c\right)\left(x+c\right)}{2c\left(x+c\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{3\left({x}^{2}–{c}^{2}\right)}{2c\left(x+c\right)}$

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

Solutions:The equation of the given circle is

${x}^{2}+{y}^{2}=2$

.

Also, the equation of the given parabola is

${y}^{2}=8x$

.

We know :

Equation of a tangent to a circle,

is

Thus, equation of a tangent to a circle

${x}^{2}+{y}^{2}=2$

is

…(1)

Also, equation of a tangent to a parabola,

is

Thus, equation of a tangent to a parabola,

${y}^{2}=8x$

is

…(2)

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

Squaring both sides, we get

So, the equations of the common tangent are

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

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

Using distance formula, we get:

Height of the trapezium PQRS = AB = 3

Hence, the correct option is D.

Question 46:

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

$\frac{dy}{dx}+\frac{xy}{{x}^{2}–1}=\frac{{x}^{2}+4x}{\sqrt{1–{x}^{2}}}$

in (−1, 1) satisfying f(0) = 0. Then

is

 Option 1: $\frac{\mathrm{\pi }}{3}–\frac{\sqrt{3}}{2}$ Option 2: $\frac{\mathrm{\pi }}{3}–\frac{\sqrt{3}}{4}$ Option 3: $\frac{\mathrm{\pi }}{6}–\frac{\sqrt{3}}{4}$ Option 4: $\frac{\mathrm{\pi }}{6}–\frac{\sqrt{3}}{2}$

Solutions:The given differential equation is

The solution of the given differential equation is

Hence, the correct option is B.

Question 47:

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

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

Solutions:Given:

Differentiating both sides, we get:

(Using Leibnitz's rule)

It is also given that,

$F\text{'}\left(x\right)=f\text{'}\left(x\right)$

Integrating both sides, we get:

When x = 0, f (0) = 1

$⇒c=0$

From (1) and (2), we have

For x = 2,

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

Solutions:The given expansion is

${\left(1+{x}^{2}\right)}^{4}{\left(1+{x}^{3}\right)}^{7}{\left(1+{x}^{4}\right)}^{12}$

.

Coefficient of x11  in

${\left(1+{x}^{2}\right)}^{4}{\left(1+{x}^{3}\right)}^{7}{\left(1+{x}^{4}\right)}^{12}$

= Coefficient of x11  in

$\left({}^{4}C_{0}+{}^{4}C_{1}{x}^{2}+{}^{4}C_{2}{x}^{4}+{}^{4}C_{3}{x}^{6}+{}^{4}C_{4}{x}^{8}\right)\left({}^{7}C_{0}+{}^{7}C_{1}{x}^{3}+{}^{7}C_{2}{x}^{6}+{}^{7}C_{3}{x}^{9}+...\right)\left({}^{12}C_{0}+{}^{12}C_{1}{x}^{4}+{}^{12}C_{2}{x}^{8}+...\right)$

=

$\left({}^{4}C_{1}×{}^{7}C_{3}×{}^{12}C_{0}\right)+\left({}^{4}C_{2}×{}^{7}C_{1}×{}^{12}C_{1}\right)+\left({}^{4}C_{4}×{}^{7}C_{1}×{}^{12}C_{0}\right)+\left({}^{4}C_{0}×{}^{7}C_{1}×{}^{12}C_{2}\right)$

=  (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

Solutions:Given,

$\mathrm{sin}x+2\mathrm{sin}2x–\mathrm{sin}3x=3$

i.e.

$\left(\mathrm{sin}x–\mathrm{sin}3x\right)+2\mathrm{sin}2x=3$ $⇒2\mathrm{cos}2x\left(–\mathrm{sin}x\right)+4\mathrm{sin}x\mathrm{cos}x=3$

This is possible only when

.

Now, there does not exist any

$x\in \left(0,\mathrm{\pi }\right)$

for which

.

Thus, the given equation has no solution.

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

Question 50:

The following integral

is equal to

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

Solutions:Let I =

${\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}{\left(2\mathrm{cosec}x\right)}^{17}dx$

=

${\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}{\left(\frac{2}{\mathrm{sin}x}\right)}^{16}2\mathrm{cosec}xdx$

Now,

I

${\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}{\left[\left(\mathrm{cosec}x+\mathrm{cot}x\right)+\frac{1}{\left(\mathrm{cosec}x+\mathrm{cot}x\right)}\right]}^{16}2\mathrm{cosec}xdx$

Put cosecx + cotx = eu.

$\therefore –\mathrm{cosec}xcotx–{\mathrm{cosec}}^{2}x={e}^{u}\frac{du}{dx}\phantom{\rule{0ex}{0ex}}⇒–\mathrm{cosec}x\left(\mathrm{cot}x+\mathrm{cosec}x\right)={e}^{u}\frac{du}{dx}\phantom{\rule{0ex}{0ex}}⇒–\mathrm{cosec}xdx=du$

When

I

$–2{\int }_{\mathrm{ln}\left(\sqrt{2}+1\right)}^{0}{\left({e}^{u}+{e}^{–u}\right)}^{16}du={\int }_{0}^{\mathrm{ln}\left(\sqrt{2}+1\right)}2{\left({e}^{u}+{e}^{–u}\right)}^{16}du$

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: $\frac{29}{105}$ Option 2: $\frac{53}{105}$ Option 3: $\frac{57}{105}$ Option 4: $\frac{1}{2}$

Solutions:

Now,

${x}_{1}+{x}_{2}+{x}_{3}$

is odd if

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

Case I

If

${x}_{1}$

is odd,

are even.

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

Case II

If

${x}_{2}$

is odd,

are even.

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

Case III

If

${x}_{3}$

is odd,

are even.

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

Case IV

If

${x}_{1}$

,

are all odd.

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

$\therefore$

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

Also,

Total outcomes = 3 × 5 × 7 = 105

∴ Required probability =

$\frac{53}{105}$

Hence, the correct option is B.

Question 52:

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

 Option 1: $\frac{9}{105}$ Option 2: $\frac{10}{105}$ Option 3: $\frac{11}{105}$ Option 4: $\frac{7}{105}$

Solutions:

If

are in arithmetic progression, then

${x}_{2}–{x}_{1}={x}_{3}–{x}_{2}$ $⇒{x}_{1}+{x}_{3}$

must always be even.

$⇒$

both must be odd or both must be even.

Case I:

If

are both odd, then

${x}_{1}\in \left\{1,3\right\}$

and

${x}_{3}\in \left\{1,3,5,7\right\}$

.

$\therefore$

Possible outcomes =

.

Case II:

If

are both even, then

${x}_{1}\in \left\{2\right\}$

and

${x}_{3}\in \left\{2,4,6\right\}$

.

$\therefore$

Possible outcomes =

.

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

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

$\therefore$

â€‹Required probability =

$\frac{11}{105}$

.

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: $–\frac{1}{t}$ Option 2: Option 3: $\frac{1}{\mathrm{t}}$ Option 4:

Solutions:Given:

Let the coordinates of Q be

.

Since PQ is the focal chord,

Thus, the coordinates of Q are

$\left(\frac{a}{{t}^{2}},–\frac{2a}{t}\right)$

â€‹.

Now,

It is given that, QR is parallel to PK.

$\therefore$

Slope of PK = Slope of QR

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: Option 2: Option 3: Option 4:

Solutions:

Given: st = 1

$\therefore$

Coordinates of S

$\equiv \left(\frac{a}{{t}^{2}},\frac{2a}{t}\right)$

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

${y}^{2}=4ax$

is

So, the equation of the tangent at P

$\left(a{t}^{2},2at\right)$

is

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

${y}^{2}=4ax$

isâ€‹

$y–{y}_{1}=–\frac{{y}_{1}}{2a}\left(x–{x}_{1}\right)$

So, the equation of the normal at S

$\left(\frac{a}{{t}^{2}},\frac{2a}{t}\right)$

is

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

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

Question 55:

Given that for each a ∈ (0, 1),

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

is

 Option 1: π Option 2: 2π Option 3: $\frac{\mathrm{\pi }}{2}$ Option 4: $\frac{\mathrm{\pi }}{4}$

Solutions:Given:
g(a) =

Hence, the correct option is A.

Question 56:

Given that for each a ∈ (0, 1),

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

is

 Option 1: $\frac{\mathrm{\pi }}{2}$ Option 2: π Option 3: $–\frac{\mathrm{\pi }}{2}$

Solutions:Given:

$g\left(a\right)=\underset{h\to {0}^{+}}{\mathrm{lim}}{\int }_{h}^{1–h}{t}^{–a}{\left(1–t\right)}^{a–1}dt$

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

$g\text{'}\left(a\right)=\underset{h\to {0}^{+}}{\mathrm{lim}}{\int }_{h}^{1–h}\frac{d}{da}\left[{t}^{–a}{\left(1–t\right)}^{a–1}\right]dt$

$=\underset{h\to {0}^{+}}{\mathrm{lim}}{\int }_{h}^{1–h}\left[–{t}^{–a}\mathrm{ln}\left(t\right){\left(1–t\right)}^{a–1}+{t}^{–a}{\left(1–t\right)}^{a–1}\mathrm{ln}\left(1–t\right)\right]dt$

$=\underset{h\to {0}^{+}}{\mathrm{lim}}{\int }_{h}^{1–h}{t}^{–a}{\left(1–t\right)}^{a–1}\left[–\mathrm{ln}\left(t\right)+\mathrm{ln}\left(1–t\right)\right]dt$

$=\underset{h\to {0}^{+}}{\mathrm{lim}}{\int }_{h}^{1–h}{t}^{–a}{\left(1–t\right)}^{a–1}\left[\mathrm{ln}\left(\frac{1–t}{t}\right)\right]dt$

Now,

$g\text{'}\left(\frac{1}{2}\right)=\underset{h\to {0}^{+}}{\mathrm{lim}}{\int }_{h}^{1–h}{t}^{–\frac{1}{2}}{\left(1–t\right)}^{–\frac{1}{2}}\left[\mathrm{ln}\left(\frac{1–t}{t}\right)\right]dt$

…(1)

Using

${\int }_{a}^{b}f\left(x\right)dx={\int }_{a}^{b}f\left(a+b–x\right)dx$

, we get:

$g\text{'}\left(\frac{1}{2}\right)=\underset{h\to {0}^{+}}{\mathrm{lim}}{\int }_{h}^{1–h}{t}^{–\frac{1}{2}}{\left(1–t\right)}^{–\frac{1}{2}}\left[\mathrm{ln}\left(\frac{t}{1–t}\right)\right]dt$

…(2)

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

$2g\text{'}\left(\frac{1}{2}\right)=\underset{h\to {0}^{+}}{\mathrm{lim}}{\int }_{h}^{1–h}{t}^{–\frac{1}{2}}{\left(1–t\right)}^{–\frac{1}{2}}\left[\mathrm{ln}\left(\frac{t}{1–t}\right)+\mathrm{ln}\left(\frac{1–t}{t}\right)\right]dt$ $⇒2g\text{'}\left(\frac{1}{2}\right)=\underset{h\to {0}^{+}}{\mathrm{lim}}{\int }_{h}^{1–h}{t}^{–\frac{1}{2}}{\left(1–t\right)}^{–\frac{1}{2}}\mathrm{ln}\left(1\right)dt$

$⇒2g\text{'}\left(\frac{1}{2}\right)=\underset{h\to {0}^{+}}{\mathrm{lim}}{\int }_{h}^{1–h}{t}^{–\frac{1}{2}}{\left(1–t\right)}^{–\frac{1}{2}}\left(0\right)dt$ $⇒2g\text{'}\left(\frac{1}{2}\right)=0$

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 is 1. 8 Q. The number of points in the interval at which f(x) = sin(x2) + cos(x2) attains its maximum value is 2. 2 R. equals 3. 4 S. 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

Solutions:P.

Let

$f\left(x\right)={a}_{0}{x}^{2}+{a}_{1}x+{a}_{2}$

, where

.

Given,  f(0) = 0

${a}_{2}=0$

…(1)

Also,

${\int }_{0}^{1}f\left(x\right)dx=1$ $⇒{\left(\frac{{a}_{0}{x}^{3}}{3}+\frac{{a}_{1}{x}^{2}}{2}+{a}_{2}x\right)}_{0}^{1}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{a}_{0}}{3}+\frac{{a}_{1}}{2}+{a}_{2}=1$

From (1), we have:

$\frac{{a}_{0}}{3}+\frac{{a}_{1}}{2}=1$ $⇒2{a}_{0}+3{a}_{1}=6$

For non – negative integral solutions,

${a}_{0}=0$

and

Or

and

${a}_{1}=0$

$f\left(x\right)=2x$

or

$f\left(x\right)=3{x}^{2}$

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

Q.

Let

$f\left(x\right)=\mathrm{sin}\left({x}^{2}\right)+\mathrm{cos}\left({x}^{2}\right)$

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

$f\text{'}\left(x\right)=2x\left(\mathrm{cos}{x}^{2}–\mathrm{sin}{x}^{2}\right)$

…(1)

For maxima or minima,

$f\text{'}\left(x\right)=0$

.

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

$f\text{'}\text{'}\left(x\right)=2\left[–2{x}^{2}\left(\mathrm{sin}{x}^{2}+\mathrm{cos}{x}^{2}\right)+c\mathrm{os}{x}^{2}–\mathrm{sin}{x}^{2}\right]$

Now,

$f\text{'}\text{'}\left(0\right)=2>0$ $f\text{'}\text{'}\left(±\sqrt{\frac{\mathrm{\pi }}{4}}\right)=–4\sqrt{2}\left(\frac{\mathrm{\pi }}{4}\right)<0$ $f\text{'}\text{'}\left(±\sqrt{\frac{5\mathrm{\pi }}{4}}\right)=4\sqrt{2}\left(\frac{5\mathrm{\pi }}{4}\right)>0$ $f\text{'}\text{'}\left(±\sqrt{\frac{9\mathrm{\pi }}{4}}\right)=–4\sqrt{2}\left(\frac{9\mathrm{\pi }}{4}\right)<0\phantom{\rule{0ex}{0ex}}$ $f\text{'}\text{'}\left(±\sqrt{\frac{13\mathrm{\pi }}{4}}\right)=4\sqrt{2}\left(\frac{13\mathrm{\pi }}{4}\right)>0$

f is maximum at

$x=±\sqrt{\frac{\mathrm{\pi }}{4}}$

and

$x=±\sqrt{\frac{9\mathrm{\pi }}{4}}$

.

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

R.

Let I =

${\int }_{–2}^{2}\frac{3{x}^{2}}{1+{e}^{x}}dx$

…(1)

We know that

I =

${\int }_{–2}^{2}\frac{3{x}^{2}}{1+{e}^{–x}}dx$

…(2)

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

I = 8

S.

Suppose I =

${\int }_{–\frac{1}{2}}^{\frac{1}{2}}\mathrm{cos}2x\mathrm{log}\left(\frac{1+x}{1–x}\right)dx$

Let

$f\left(x\right)=c\mathrm{os}2x\mathrm{log}\left(\frac{1+x}{1–x}\right)$

Then,

$f\left(–x\right)=c\mathrm{os}2\left(–x\right)\mathrm{log}\left(\frac{1–x}{1+x}\right)=c\mathrm{os}2x\mathrm{log}{\left(\frac{1+x}{1–x}\right)}^{–1}=–c\mathrm{os}2x\mathrm{log}\left(\frac{1+x}{1–x}\right)=f\left(x\right)$

Thus,

$f\left(x\right)=c\mathrm{os}2x\mathrm{log}\left(\frac{1+x}{1–x}\right)$

is an odd function.

⇒

${\int }_{–\frac{1}{2}}^{\frac{1}{2}}\mathrm{cos}2x\mathrm{log}\left(\frac{1+x}{1–x}\right)dx=0$

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. 1. 1 Q. 2. 2 R. If the normal from the point P (h, 1) on the ellipse is perpendicular to the line x + y = 8, the the value of h is 3. 8 S. Number of positive solutions satisfying the equation 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

Solutions:P.

The given function is

… (1)

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

$\frac{dy}{dx}=–\mathrm{sin}\left(3{\mathrm{cos}}^{–1}x\right)×\frac{–3}{\sqrt{1–{x}^{2}}}$ $⇒\sqrt{1–{x}^{2}}\frac{dy}{dx}=3\mathrm{sin}\left(3{\mathrm{cos}}^{–1}x\right)$

Now, squaring both sides, we get:

$\left(1–{x}^{2}\right){\left(\frac{dy}{dx}\right)}^{2}=9{\mathrm{sin}}^{2}\left(3{\mathrm{cos}}^{–1}x\right)$ $⇒\left(1–{x}^{2}\right){\left(\frac{dy}{dx}\right)}^{2}=9\left(1–{\mathrm{cos}}^{2}\left(3{\mathrm{cos}}^{–1}x\right)\right)$

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

$\left(1–{x}^{2}\right)×2\frac{dy}{dx}×\frac{{d}^{2}y}{d{x}^{2}}–2x{\left(\frac{dy}{dx}\right)}^{2}=–18y\frac{dy}{dx}$ $\therefore \frac{1}{y}\left\{\left({x}^{2}–1\right)\frac{{d}^{2}y}{d{x}^{2}}+x\left(\frac{dy}{dx}\right)\right\}=\frac{9y}{y}=9$

To justify

$\frac{dy}{dx}\ne 0$

, let us suppose

$\frac{dy}{dx}=0$

. Integrating

$\frac{dy}{dx}=0$

with respect to x, we get:

But it is given that

.

$\therefore \frac{dy}{dx}\ne 0$

Q.

It is given that

$\left|\sum _{k=1}^{n–1}\left(\overline{){a}_{k}}×\overline{){a}_{k+1}}\right)\right|=\left|\sum _{k=1}^{n–1}\left(\overline{){a}_{k}}·\overline{){a}_{k+1}}\right)\right|$

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

$\theta =\frac{2\mathrm{\pi }}{n}$

.

Here,

$\left|\overline{){a}_{1}}\right|=\left|\overline{){a}_{2}}\right|=\left|\overline{){a}_{3}}\right|...=\left|\overline{){a}_{n}}\right|$

Let

$\left|\overline{){a}_{1}}\right|=\left|\overline{){a}_{2}}\right|=\left|\overline{){a}_{3}}\right|...=\left|\overline{){a}_{n}}\right|$

= R

$\therefore \left|\sum _{k=1}^{n–1}\left(\overline{){a}_{k}}×\overline{){a}_{k+1}}\right)\right|=\left|\sum _{k=1}^{n–1}\left(\overline{){a}_{k}}·\overline{){a}_{k+1}}\right)\right|$ $⇒\left|\sum _{k=1}^{n–1}\left(\left|\overline{){a}_{k}}\right|\left|\overline{){a}_{k+1}}\right|\mathrm{sin\theta }\stackrel{^}{n}\right)\right|=\left|\sum _{k=1}^{n–1}\left(\left|\overline{){a}_{k}}\right|\left|\overline{){a}_{k+1}}\right|\mathrm{cos\theta }\right)\right|$

,  where

$\stackrel{^}{n}$

is a unit vector perpendicular to the vectors

.

$⇒{R}^{2}\mathrm{sin\theta }\left|\sum _{k=1}^{n–1}\left(\stackrel{^}{n}\right)\right|={R}^{2}\mathrm{cos\theta }\left|\sum _{k=1}^{n–1}1\right|$ $⇒\left(n–1\right){R}^{2}\mathrm{sin\theta }=\left(n–1\right){R}^{2}\mathrm{cos\theta }$ $⇒\mathrm{sin}\frac{2\mathrm{\pi }}{n}=\mathrm{cos}\frac{2\mathrm{\pi }}{n}⇒\mathrm{tan}\frac{2\mathrm{\pi }}{n}=1\phantom{\rule{0ex}{0ex}}$

For minimum value of n, we must have

$\frac{2\mathrm{\pi }}{n}=\frac{\mathrm{\pi }}{4}⇒n=8$

R.

The equation of the ellipse is

$\frac{{x}^{2}}{6}+\frac{{y}^{2}}{3}=1$

.

Here,

.

We know that the equation of the normal to the ellipse

$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$

is

$\frac{ax}{\mathrm{cos}\theta }–\frac{by}{\mathrm{sin}\theta }={a}^{2}–{b}^{2}$

.

So, the equation of the normal to the ellipse

$\frac{{x}^{2}}{6}+\frac{{y}^{2}}{3}=1$

is

$\frac{\sqrt{6}x}{\mathrm{cos\theta }}–\frac{\sqrt{3}y}{\mathrm{sin\theta }}=3$

It is given that

$\frac{\sqrt{6}x}{\mathrm{cos\theta }}–\frac{\sqrt{3}y}{\mathrm{sin\theta }}=3$

is perpendicular to  x + y = 8.

$\therefore –1×\sqrt{2}\mathrm{tan\theta }=–1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan\theta }=\frac{1}{\sqrt{2}}$

It is also given that

$\frac{\sqrt{6}x}{\mathrm{cos\theta }}–\frac{\sqrt{3}y}{\mathrm{sin\theta }}=3$

passes through P(h, 1).

$\therefore \frac{\sqrt{6}h}{\frac{\sqrt{2}}{\sqrt{3}}}–\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}=3$ $\therefore \frac{3\sqrt{2}h}{\sqrt{2}}–3=3⇒h=2$

S.

The given equation is

${\mathrm{tan}}^{–1}\left(\frac{1}{2x+1}\right)+{\mathrm{tan}}^{–1}\left(\frac{1}{4x+1}\right)={\mathrm{tan}}^{–1}\left(\frac{2}{{x}^{2}}\right)$

.

Using the formula,

${\mathrm{tan}}^{–1}a+{\mathrm{tan}}^{–1}b={\mathrm{tan}}^{–1}\left(\frac{a+b}{1–ab}\right)$

, we get:

${\mathrm{tan}}^{–1}\left(\frac{\frac{1}{2x+1}+\frac{1}{4x+1}}{1–\frac{1}{2x+1}·\frac{1}{4x+1}}\right)={\mathrm{tan}}^{–1}\left(\frac{2}{{x}^{2}}\right)$ $⇒{\mathrm{tan}}^{–1}\left(\frac{6x+2}{8{x}^{2}+6x}\right)={\mathrm{tan}}^{–1}\left(\frac{2}{{x}^{2}}\right)$ $⇒\frac{6x+2}{8{x}^{2}+6x}=\frac{2}{{x}^{2}}$

The positive solution of

${\mathrm{tan}}^{–1}\left(\frac{1}{2x+1}\right)+{\mathrm{tan}}^{–1}\left(\frac{1}{4x+1}\right)={\mathrm{tan}}^{–1}\left(\frac{2}{{x}^{2}}\right)$

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

 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: $\begin{array}{ccccc}& \mathrm{P}& \mathrm{Q}& \mathrm{R}& \mathrm{S}\\ \left(\mathrm{A}\right)& 3& 1& 4& 2\end{array}$ Option 2: $\begin{array}{ccccc}& \mathrm{P}& \mathrm{Q}& \mathrm{R}& \mathrm{S}\\ \left(\mathrm{B}\right)& 1& 3& 4& 2\end{array}$ Option 3: $\begin{array}{ccccc}& \mathrm{P}& \mathrm{Q}& \mathrm{R}& \mathrm{S}\\ \left(\mathrm{C}\right)& 3& 1& 2& 4\end{array}$ Option 4: $\begin{array}{ccccc}& \mathrm{P}& \mathrm{Q}& \mathrm{R}& \mathrm{S}\\ \left(\mathrm{D}\right)& 1& 3& 2& 4\end{array}$

Solutions:The given functions are

${f}_{1}:\mathrm{ℝ}\to \mathrm{ℝ}$

,

${f}_{2}\left(x\right)={x}^{2}$ ${f}_{3}:\mathrm{ℝ}\to \mathrm{ℝ}$

,

,

P.

The function f4 can be redefined as

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

Q.

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

$\because {f}_{3}\text{'}\left({0}^{–}\right)={f}_{3}\text{'}\left({0}^{+}\right)=1$

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

R.

The composition function

The graph of f2of1(x) is

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

S.

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

.

 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. equals 3. 1 S. equals 4. 2
 Option 1: $\begin{array}{ccccc}& \mathrm{P}& \mathrm{Q}& \mathrm{R}& \mathrm{S}\\ \left(\mathrm{A}\right)& 1& 2& 4& 3\end{array}$ Option 2: $\begin{array}{ccccc}& \mathrm{P}& \mathrm{Q}& \mathrm{R}& \mathrm{S}\\ \left(\mathrm{B}\right)& 2& 1& 3& 4\end{array}$ Option 3: $\begin{array}{ccccc}& \mathrm{P}& \mathrm{Q}& \mathrm{R}& \mathrm{S}\\ \left(\mathrm{C}\right)& 1& 2& 3& 4\end{array}$ Option 4: $\begin{array}{ccccc}& \mathrm{P}& \mathrm{Q}& \mathrm{R}& \mathrm{S}\\ \left(\mathrm{D}\right)& 2& 1& 4& 3\end{array}$

Solutions:Given:

P.

${z}_{k}=\mathrm{cos}\left(\frac{2k\mathrm{\pi }}{10}\right)+i\mathrm{sin}\left(\frac{2k\mathrm{\pi }}{10}\right)$ $⇒{z}_{k}={e}^{\frac{i2k\mathrm{\pi }}{10}}$

So, there exists a

${z}_{j}$

, such that

.

Thus, the given statement is true.

Q.

${z}_{1}z={z}_{k}\phantom{\rule{0ex}{0ex}}⇒z=\frac{{z}_{k}}{{z}_{1}}={e}^{\frac{i2\left(k–1\right)\mathrm{\pi }}{10}}$

Thus, the equation

has a solution in the set of complex numbers.

Thus, the given statement is false.

R.

$⇒\frac{{z}^{10}–1}{z–1}=1+z+{z}^{2}+{z}^{3}+...+{z}^{9}$ $⇒\left(z–{z}_{1}\right)\left(z–{z}_{2}\right)\left(z–{z}_{3}\right)...\left(z–{z}_{9}\right)=1+z+{z}^{2}+{z}^{3}+...+{z}^{9}$

Substituting z = 1, we get:

$\left(1–{z}_{1}\right)\left(1–{z}_{2}\right)\left(1–{z}_{3}\right)...\left(1–{z}_{9}\right)=1+1+1+1+...+1=10$ $⇒\frac{\left|1–{z}_{1}\right|\left|1–{z}_{2}\right|\left|1–{z}_{3}\right|...\left|1–{z}_{9}\right|}{10}=1$

S.

$1–\sum _{k=1}^{9}\mathrm{cos}\left(\frac{2k\mathrm{\pi }}{10}\right)=1–\left(\mathrm{cos}\frac{2\mathrm{\pi }}{10}+\mathrm{cos}\frac{4\mathrm{\pi }}{10}+\mathrm{cos}\frac{6\mathrm{\pi }}{10}+...+\mathrm{cos}\frac{18\mathrm{\pi }}{10}\right)$

We know that

$\mathrm{cos}a+\mathrm{cos}\left(a+d\right)+\mathrm{cos}\left(a+2d\right)+...+\mathrm{cos}\left(a+\left(n–1\right)d\right)=\frac{\mathrm{sin}\frac{nd}{2}}{\mathrm{sin}\frac{d}{2}}\mathrm{cos}\left(\frac{2a+\left(n–1\right)d}{2}\right)$

Thus, the correct match of the columns is

 P Q R S 1 2 3 4

Hence, the correct option is C.