# JEE Advanced 2013 Paper-1(Code 0)

### Test Name: JEE Advanced 2013 Paper-1(Code 0)

#### Question 1:

The work done on a particle of mass m by a force, (K being a constant of appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the x-y plane is.

 Option 1: Option 2: Option 3: Option 4: 0

Solutions:

As the force is given as,

Let us take, x2 + y2 = r2

Thus, we get

xdx + ydy = rdr

#### Question 2:

Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure. One of the blocks has thermal conductivity κ and the other 2κ. The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s to transport a certain amount of heat from the hot end to the cold end in the configuration I. The time to transport the same amount of heat in the configuration II is

 Option 1: 2.0 s Option 2: 3.0 s Option 3: 4.5 s Option 4: 6.0 s

Solutions:

For the two blocks in series,

For the two blocks in parallel,

From equation (i) and (ii),

#### Question 3:

The non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is

 Option 1: 1 : 4 Option 2: 1 : 2 Option 3: 6 : 9 Option 4: 8 : 9

Solutions:

Pressure of a gas is given as,

#### Question 4:

A particle of mass m is projected from the ground with an initial speed u0 at an angle α with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed u0. The angle that the composite system makes with the horizontal immediately after the collision is

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

Solutions:

The speed of the first particle at the highest point = u0cosα

#### Question 5:

A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse in 30 mW and the speed of light is 3 ×108 ms−1. The final momentum of the object is

 Option 1: 0.3 × 10−17kg ms−1 Option 2: 1.0 × 10−17kg ms−1 Option 3: 3.0 × 10−17kg ms−1 Option 4: 9.0 × 10−17kg ms−1

Solutions:

Change in momentum is given as;

#### Question 6:

In the Young’s double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is

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

Solutions:

As the intensity is given as

#### Question 7:

The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is times the wavelength in free space. The radius of the curved surface of the lens is

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

Solutions:

The refractive index of the lens,

#### Question 8:

One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is

 Option 1: 0.25 Option 2: 0.5 Option 3: 2 Option 4: 4

Solutions:

The elongation in the wire is given as:

As the force on both the wires is same, thus

#### Question 9:

A ray of light travelling in the direction is incident on a plane mirror. After reflection, it travels along the direction. The angle of incidence is

 Option 1: 30° Option 2: 45° Option 3: 60° Option 4: 75°

Solutions:

The incident ray diagram is as shown in the figure.

Taking the dot product of the two vectors representing the light ray,

#### Question 10:

The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is

 Option 1: 5.112 cm Option 2: 5.124 cm Option 3: 5.136 cm Option 4: 5.148 cm

Solutions:

Least count of the vernier = 1 MSD − 1 VSD = 0.05 cm − 0.049 cm = 0.001 cm

Hence, thickness of the object = main scale reading + vernier scale reading × least count

t = 5.10+24×0.001 = 5.124 cm

#### Question 11:

In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S1 is pressed first to fully charge the capacitor C1 and then released. The switch S2 is then pressed to charge the capacitor C2. After some time, S2 is released and then S3 is pressed. After some time,

 Option 1: the charge on the upper plate of C1 is 2CV0. Option 2: the charge on the upper plate of C1 is CV0. Option 3: the charge on the upper plate of C2 is 0. Option 4: the charge on the upper plate of C2 is −CV0.

Solutions:

When S1 is pressed and released, charge on C1 is 2CV0 (on the upper plate).

When S2 is pressed and released, charge on C1 is CV0 (on upper plate) and charge on C2 is CV0 (on the upper plate).

When S3 is pressed and released, charge on C1 is CV0 (on upper plate) and charge on C2 is −CV0 (on upper plate).

#### Question 12:

A particle of mass M and positive charge Q, moving with a constant velocity, enters a region of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from x = 0 to x = L for all values of y. After passing through this region, the particle emerges on the other side after 10 milliseconds with a velocity. The correct statement(s) is (are)

 Option 1: The direction of the magnetic field is −z direction. Option 2: The direction of the magnetic field is +z direction. Option 3: The magnitude of the magnetic field units. Option 4: The magnitude of the magnetic field isunits.

Solutions:

As the upward force is acting on the particle and the direction of the particle’s motion is in the positive x-direction, thus the direction of the magnetic field is in the negative z-direction.

#### Question 13:

A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x,t) = (0.01 m) sin [(62.8 m−1) x] cos [(628 s−1)t].

Assuring π = 3.14, the correct statement(s) is (are)

 Option 1: The number of nodes is 5. Option 2: The length of the string is 0.25 m. Option 3: The maximum displacement of the midpoint of the string, form its equilibrium position is 0.01 m. Option 4: The fundamental frequency is 100 Hz.

Solutions:

As there are 5 complete loops, thus the total number of nodes = 6

As the mid-point is antinode, thus the maximum displacement = 0.01 m

The fundamental frequency is given as:

#### Question 14:

A solid sphere of radius R and density ρ is attached to one end of a mass-less spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3ρ. The complete arrangement is placed in a liquid of density 2ρ and is allowed to reach equilibrium. The correct statement(s) is (are)

 Option 1: the net elongation of the spring is Option 2: the net elongation of the spring is Option 3: the light sphere is partially submerged. Option 4: the light sphere is completely submerged.

Solutions:

At the equilibrium condition, let the net elongation in the spring be x.

For the upper sphere

Since the weight of the system is equal to the total buoyant force, thus the light sphere is completely submerged.

#### Question 15:

Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities ρ1 and ρ2 respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratio can be

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

Solutions:

The position of the two spheres is as shown in the figure.

As at point B, as the resultant electric field is zero

At point A also, the electric field is zero,

#### Question 16:

A bob of mass m, suspended by a string of length l1, is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspended by a string of length l2, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio is

Solutions:

As the speed of the first bob at the highest point,

For the elastic collision between the objects of same mass the velocities are exchanged.

#### Question 17:

A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle. If the initial speed (in ms−1) of the particle is zero, the speed (in ms−1) after 5 s is

Solutions:

Since the power is constant, thus

#### Question 18:

The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is

Solutions:

The maximum kinetic energy of the ejected electrons is given as;

#### Question 19:

A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second. Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is

Solutions:

Since the number of nuclei at any instant is given as: N = N0e-λt

#### Question 20:

A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s−1 about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s−1) of the system is

Solutions:

Initially the moment of inertia of the disc,

When the ring is placed, the moment of inertia of the system

From the conservation of the angular momentum,

#### Question 21:

Consider the following complex ions, P, Q and R.

The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is

 Option 1: R < Q < P Option 2: Q < R < P Option 3: R < P < Q Option 4: Q < P < R

Solutions:

P= [FeF6]3-

Oxidation state of central metal atom = x + (-1) × 6 = -3

or, x = -3 + 6 = +3

Therefore, Fe+3 =

Using spin only formula, μ = [where n = no. of unpaired electrons]

μ =

Now, for Q = [V(H2O)6]2+, oxidation state of vanadium

x + 0 × 6 = +2 (water is a neutral molecule)

x = +2, therefore V2+ =

μ =

Similarly for R = [Fe(H2O)6]2+, oxidation state of Fe is +2,

i.e. Fe2+ =

μ =

Hence correct order of complexes according to the magnetic moment value is Q < R < P

i.e. [V(H2O)6]2+ < [Fe(H2O)6]2+ < [FeF6]3-

#### Question 22:

The arrangement of X ions around A+ ion in solid AX is given in the figure (not drawn to scale). If the radius of X is 250 pm, the radius of A+ is

 Option 1: 104 pm Option 2: 125 pm Option 3: 183 pm Option 4: 57 pm

Solutions:

From the given figure, it can be said that cation A+ occupies the octahedral void formed by the X anions. The radius ratio range is 0.414 – 0.732,

Therefore,

Since minimum value for a cation to accommodate an octahedral void without distortion is 0.414, therefore radius of cation is 103.5 ~ 104 pm.

#### Question 23:

Sulfide ores are common for the metals

 Option 1: Ag, Cu and Pb Option 2: Ag, Cu and Sn Option 3: Ag, Mg and Pb Option 4: Al, Cu and Pb

Solutions:

Sulphide ores are common for silver, copper and lead and is found in the form of Ag2S (silver sulphide), CuFeS2 (copper pyrites) and PbS (lead sulphide) respectively.

#### Question 24:

The standard enthalpies of formation of CO2­(g), H2O(l) and glucose(s) at 25°C are −400 kJ/mol, −300 kJ/mol and −1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25°C is

 Option 1: +2900 kJ Option 2: −2900 kJ Option 3: −16.11 kJ Option 4: +16.11 kJ

Solutions:

The required chemical equation for the combustion of glucose is

#### Question 25:

Upon treatment with ammoniacal H2S, the metal ion that precipitates as a sulfide is

 Option 1: Fe(III) Option 2: Al(III) Option 3: Mg(II) Option 4: Zn(II)

Solutions:

Ammonical H2S is group IV reagent and therefore will cause the precipitation of group IV elements. Among given metal ions, Zn(II) is group IV element and will precipitate as ZnS (white ppt.). Fe3+ and Al3+ will also formed their respective hydroxide but not sulphides.

#### Question 26:

Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 25°C. For this process, the correct statement is

 Option 1: The adsorption requires activation at 25°C. Option 2: The adsorption is accompanied by a decrease in enthalpy. Option 3: The adsorption increases with increase of temperature. Option 4: The adsorption is irreversible.

Solutions:

The adsorption of a methylene blue on an activated charcoal is physical adsorption and physical adsorption is accompanied by a decrease in enthalpy.

#### Question 27:

KI in acetone, undergoes SN2 reaction with each of P, Q, R and S. The rates of the reaction vary as

 Option 1: P > Q > R > S Option 2: S > P > R > Q Option 3: P > R > Q > S Option 4: R > P > S > Q

Solutions:

The relative rate for the SN2 reaction is: 10 > 20 > 30 halides. The presence of hetero- atom and carbonyl group at α-carbon increases the rate of SN2 reaction. Therefore correct order is:

#### Question 28:

In the reaction,

P + Q R + S

the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is

 Option 1: 2 Option 2: 3 Option 3: 0 Option 4: 1

Solutions:

For P, let reaction follows the first order kinetics.

Then,

From the above expression it can be said that t75%= 2 t­50% according to first order reaction, hence order with respect to P is 1.

From the graph, concentration of Q decreases with time, and therefore order with respect Q is zero.

Hence overall order of the reaction is =1 + 0 = 1.

#### Question 29:

Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of

 Option 1: NO Option 2: NO2 Option 3: N2O Option 4: N2O4

Solutions:

Pure nitric acid is a colourless liquid, but on exposure to light it turns into yellow-brown because of slight decomposition into NO2 and O2

4 HNO3 → 4 NO2 + O2 + 2 H2O

#### Question 30:

The compound that does NOT liberate CO2, on treatment with aqueous sodium bicarbonate solution, is

 Option 1: Benzoic acid Option 2: Benzenesulphonic acid Option 3: Salicylic acid Option 4: Carbolic acid (Phenol)

Solutions:

Carbolic acid (Phenol) is a weaker acid than carboxylic acid and its derivative like benzoic acid and salicylic acid. Phenol is also weaker acid than benzene sulphonic acid and therefore it will not liberate CO2 gas upon treatment with sodium bicarbonate.

#### Question 31:

The initial rate of hydrolysis of methyl acetate (1 M) by a weak acid (HA,1M) is 1/100th of that of a strong acid (HX, 1M), at 25°C. The Kaof HA is

 Option 1: 1 × 10−4 Option 2: 1 × 10−5 Option 3: 1 × 10−6 Option 4: 1 × 10−3

Solutions:

Rate with respect to weak acid (HA, 1M)

R1 = K[H+]HA [ester] …(1)

Rate with respect to strong acid (HX, 1M)

R2 = K[H+]HX [ester] …(1)

Here, ester hydrolysis follows first order kinetics with respect to [H+]

Therefore, The Kaof HA is 10-4

#### Question 32:

The hyper-conjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to

 Option 1: σ → p(empty) and σ → π* electron delocalisations. Option 2: σ → σ* and σ → π electron delocalisations. Option 3: σ → p(filled) and σ → π electron delocalisations. Option 4: p(filled) → σ* and σ → π* electron delocalisations.

Solutions:

Hyper conjugation stability of tertiary butyl cation

Here carbocation has one vacant p-orbital so it shows σ − p (empty) electron delocalization.

Hyper conjugation stability of 2-butene

Here σ − π* electron delocalization takes place.

#### Question 33:

The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is (are)

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

Solutions:

are Ma2b4 and Ma2b2 type, so they show geometrical isomerism.

show ionization isomerism.

The other pair(s) of coordination complexes/ions do not exhibiting the same kind of isomerism.

#### Question 34:

Among P, Q, R and S, the aromatic compound(s) is/are

 Option 1: P Option 2: Q Option 3: R Option 4: S

Solutions:

A compound is said to be aromatic, if

• It is planar

• There is complete delocalization of π-electrons in the ring

• It obeys Huckel Rule (4n+2)

On the basis of these facts, the products of each of the following reactions are aromatic.

#### Question 35:

Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is (are)

 Option 1: ΔG positive Option 2: ΔSsystem is positive Option 3: ΔSsurrounding = 0 Option 4: ΔH = 0

Solutions:

For ideal solution

ΔH = 0

ΔSsystem > 0

ΔSsurrounding = 0

ΔG = −ve

#### Question 36:

The atomic masses of He and Ne are 4 and 20 a.m.u., respectively. The value of the de Broglie wavelength of He gas at −73°C is “M” times that of the de Broglie wavelength of Ne at 727°C. M is

Solutions:

Dividing (1) and (2)

In term of temperature

Since, the value of the de Broglie wavelength of He gas at −73°C is “M” times that of the de Broglie wavelength of Ne at 727°C. So, M is 5.

#### Question 37:

EDTA4− is ethylenediaminetetraacetate ion. The total number of N−Co−O bond angles in [Co(EDTA)]1− complex ion is

Solutions:

The structure of [Co(EDTA)]1−

From the above structure, the bond angles N − Co − O are

So, there are total 8 N − Co − O bond angles.

#### Question 38:

The total number of carboxylic acid groups in the product P is

Solutions:

Using the given reagents, the sequence of reactions to get product P is

So the total number of carboxylic groups in the product P is 2.

#### Question 39:

A tetra peptide has −COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this tetra peptide, the number of possible sequences (primary structures) with −NH2 group attached to a chiral center is

Solutions:

As per the question, the C-terminal of the tetra-peptide must have alanine and the N-terminal should have optically active amino-acid residue. Since glycine is optically inactive, it must be not present on N-terminal.

So, possible combinations of tetra-peptide may be

Val − Gly − Phe − Ala

Val − Phe − Gly − Ala

Phe− Gly − Val − Ala

Phe − Val − Gly − Ala

Therefore, the number of possible sequences (primary structures) with −NH2 group attached to a chiral center is 4.

#### Question 40:

The total number of lone-pairs of electrons in melamine is

Solutions:

The structure of melamine is

So, total number of lone pairs of electrons is 6.

#### Question 41:

For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines ax + by + c = 0 and bx + ay + c = 0 is less than. Then

 Option 1: a + b − c > 0 Option 2: a − b + c < 0 Option 3: a − b + c > 0 Option 4: a + b − c < 0

Solutions:

We have,

ax + by + c = 0      …(1)

bx + ay + c = 0      …(2)

On solving (1) and (2), we get x = y =

∴ Point of intersection lies on the line y = x and the point of intersection is .

It is given that, distance between (1, 1) and the point of intersection of (1) and (2) is less than .

I.e., distance between (1, 1) and <

<

<

< 2

a + b + c < 2a + 2b

a + bc > 0

Hence, the correct option is A.

#### Question 42:

The area enclosed by the curves y = sin x + cos x and y = |cos x − sin x| over the interval is

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

Solutions:

The given curves over the intervalare

… (1)

… (2)

With the help of shifting of origin the graphs ofandis

Now,

∴ Area enclosed by the given curves, A is

Hence, the correct option is B.

#### Question 43:

The number of points in (−∞,∞), for which x2x sin x − cos x = 0, is

 Option 1: 6 Option 2: 4 Option 3: 2 Option 4: 0

Solutions:

x2x sin x − cos x = 0

x2= x sin x + cos x

Let g(x) = x2 and h(x) = x sin x + cos x

Now, h′(x) = x cos x + sin x sin x

h′(x) = x cos x

Now, the graph of g(x) and h(x) can be plotted as follows:

It can be seen from the above graph that g(x) and h(x) intersects at two points.

∴ There are only two solutions.

#### Question 44:

The value of cot is

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

Solutions:

Consider,

Hence, the correct option is B.

#### Question 45:

A curve passes through the point . Let the slope of the curve at each point. Then the equation of the curve is

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

Solutions:

It is given that slope at (x, y) is .

Consider, .

Put y = vx

Thus, we have

Integrating both sides

It passes through .

c =

Thus,

Hence, the correct option is A.

#### Question 46:

Let (the set of all real numbers) be a positive, non-constant and differentiable function such that . Then the value of lies in the interval

 Option 1: (2e − 1, 2e) Option 2: (e − 1, 2e − 1) Option 3: Option 4:

Solutions:

It is given that f ′(x) < 2 f(x).

Since x, we have e−1 > ye−2x

Hence, the correct option is D.

#### Question 47:

Let determine diagonals of a parallelogram PQRS and be another vector. Then the volume of the parallelepiped determined by the vectors is

 Option 1: 5 Option 2: 20 Option 3: 10 Option 4: 30

Solutions:

It is given that:

and

Take a point T shown in the following figure

From the figure, we have

… (1)

… (2)

Adding (1) and (2), we get

Now, subtracting (2) from (1), we get

Therefore, volume V of the parallelepiped determined byandis

Hence, the correct option is C.

#### Question 48:

Perpendiculars are drawn from points on the line to the plane x + y + z = 3. The feet of perpendiculars lie on the line

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

Solutions:

Let .

∴ Any point on the given line is of the form (2μ − 2, −μ − 1, 3μ).

The point (2μ − 2, −μ − 1, 3μ) also lie on the plane x + y + z = 3.

⇒ (2μ − 2) + (−μ − 1) + (3μ) = 3

⇒ μ =

So, the point of intersection is .

Let P denotes the point .

Let M ≡ (−2, −1, 0) be the point on the given line.

Let N (α, β, γ) be the foot of the perpendicular from M (−2, −1, 0) on the plane.

∴

Since the point N (α, β, γ) lies on the plane x + y + z = 3, so

(r − 2) + (r − 1) + r = 3

r = 2

Thus, N ≡ (0, 1, 2)

So, equation of NP is

#### Question 49:

Four persons independently solve a certain problem correctly with probabilities. Then the probability that the problem is solved correctly by at least one of them is

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

Solutions:

Let us define the following events:

A: Problem is solved correctly by 1st person

B: Problem is solved correctly by 2nd person

C: Problem is solved correctly by 3rd person

D: Problem is solved correctly by 4th person

It is given that,

P (A) = , P (B) = , P (C) = and P (D) =

P (problem is solved correctly by atleast one of the 4 persons) = P (ABC D)

= 1 −

= 1 −

= 1 −

=

$1–\left(1–P\left(A\right)\right)\left(1–P\left(B\right)\right)\left(1–P\left(C\right)\right)\left(1–P\left(D\right)\right)$

=

$1–\frac{1}{2}×\frac{1}{4}×\frac{3}{4}×\frac{7}{8}$

= 1 −

=

Hence, the correct option is A

#### Question 50:

Let complex numbers α and lie on circles , respectively.If satisfies the equation

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

Solutions:

Equation of the two circles is:

(xxo)2 + (yyo)2 = r2      …(1)

(xxo)2 + (yyo)2 = 4r2    …(2)

Equation of circles (1) and (2) can be written in complex form using zo (= xo + iyo)

as and .

It is given that, α and lies on the circles and respectively.

∴ We have and

Consider.

…(3)

Now, consider .

On subtracting (3) from (4), we get

$⇒\left({\left|\alpha \right|}^{2}–1\right)=2\left(4{\left|\alpha \right|}^{2}–1\right)\phantom{\rule{0ex}{0ex}}⇒{\left|\alpha \right|}^{2}–1=8{\left|\alpha \right|}^{2}–2\phantom{\rule{0ex}{0ex}}⇒{\left|\alpha \right|}^{2}=\frac{1}{7}\phantom{\rule{0ex}{0ex}}⇒\left|\alpha \right|=\frac{1}{\sqrt{7}}$

Hence, the correct option is C.

#### Question 51:

A line l passing through the origin is perpendicular to the lines

Then, the coordinate(s) of the point(s) on l2, at a distance of from the point of intersection of l and l1 is (are)

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

Solutions:

The given lines are:

Let equation of linebe

… (1)

It is given that,is perpendicular to the linesand, therefore,

Thus, equation of the lineis

For the point of intersection of lineand, let. Thus, any point A on the line is . Point A satisfies line, therefore,

Comparing the coefficients of unit vectors, we get

Solving the above equations, we get r1 = − 1

Thus, the point of intersection of lineand is (2, − 3, 2).

Let be any point on.

Therefore,

Thus, the points onare

(3 – 4, 3 – 4, 2 – 2) and

I.e., (– 1, – 1, 0) and .

Hence, the correct options are (B) and (D).

#### Question 52:

Let f(x) = x sin π x, x > 0. Then for all natural numbers n, f '(x) vanishes at

 Option 1: a unique point in the interval Option 2: a unique point in the interval Option 3: a unique point in the interval (n, n + 1) Option 4: two points in the interval (n, n + 1)

Solutions:

f (x) = x sin πx, x > 0

On differentiating, we get

f (x) = πx cos πx + sin πx

Put f (x) = 0

⇒ πx cos πx + sin πx = 0

⇒ πx cos πx = − sin πx

⇒ πx = − tan πx

The graph of y = πx and y = − tan πx can be plotted as follows:

It can be seen that there is a unique point of intersection in the interval.

Also, there is a unique point of intersection in the interval (n, n + 1).

Hence, the correct options are B and C.

#### Question 53:

Let . Then Sn can take value(s)

 Option 1: 1056 Option 2: 1088 Option 3: 1120 Option 4: 1332

Solutions:

= − 12− 22 + 32+ 42− 52− 62 + 72 … − (4n − 3)2− (4n − 2)2 + (4n − 1)2 + (4n)2

= (32− 12) + (42− 22) + (72− 52) + … + (4n − 1)2− (4n − 3)2 + (4n)2− (4n − 2)2

= 2 [4 + 6 + 12 + 14 + 20 + 22+ …+ 4n]              [The bracket contains 2n terms]

= 2 [4 + 12 + 20 + …] + 2 [6 + 14 + 22 + …]              [Each of the bracket contains n terms]

= 2[A + B], where A = 4 + 12 + 20 + … and B = 6 + 14 + 22 + …

The series given by A forms an A.P. with a = 4 and d = 8, having n number of terms.

Thus, A =

The series given by B forms an A.P. with a = 6 and d = 8.

Thus, B =

Sn = 2 [4n2 + n (4n + 2)] = 4n (4n + 1)

Consider option A.

Let 4n (4n + 1) = 1056

⇒ 4n2 + n − 264 = 0

n = 8

Consider option B.

Let 4n (4n + 1) = 1088

Now 1088 = 32 × 34

On comparing we get 4n (4n + 1) = 32 × 34, but such an integral value of n does not exist.

Consider option C.

Let 4n (4n + 1) = 1120

Now 1088 = 32 × 35

On comparing we get 4n (4n + 1) = 32 × 35, but such an integral value of n does not exist.

Consider option D.

Let 4n (4n + 1) = 1332

⇒ 4n2 + n − 333 = 0

n = 9

Hence, the correct options are A and D.

#### Question 54:

For 3 × 3 matrices M and N, which of the following statement(s) is (are) NOT correct

 Option 1: NTMN is symmetric or skew symmetric, according as M is symmetric or skew symmetric Option 2: MN − NM is skew symmetric for all symmetric matrices M and N Option 3: MN is symmetric for all symmetric matrices M and N Option 4: (adj M) (adj N) = adj (MN) for all invertible matrices M and N

Solutions:

(A) Consider (NTM N)T = NT MT (NT)T = NT MT N

So, (NT M N)T = NT M N, if M is symmetric

If M is skew symmetric, then (NT M N)T = NT MT (NT)T = NT(−M) N = −NTM N

Thus, we have the following result:

(NT M N) is symmetric if M is symmetric and (NT M N) is skew symmetric if M is skew symmetric.

(B) (MNNM)T = (MN)T − (NM)T = NTMTMT NT = NMMN = − (MNNM)

Thus, MNNM is skew symmetric.

(C) Let A = MN

Consider AT = (MN)T = NT MT = NM

AT A

Thus, A is not symmetric and hence MN is not symmetric.

Hence, the correct options are C and D.

#### Question 55:

A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100 the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are

 Option 1: 24 Option 2: 32 Option 3: 45 Option 4: 60

Solutions:

Let each side of the removed squares be m unit.

Area of the removed 4 squares = 100 unit2

⇒ 4m2 = 100 unit2

m = 5 unit

Since the sides of the rectangular sheet are in the ratio 8: 15, let breadth and length of the rectangular sheet be 8x and 15x respectively.

∴ Volume of the box, V = m (15xm) (8xm) = 4m3 − 46m2x + 120mx2

Box has the maximum volume for m = 5.

Thus, x = 3 is the point of maxima and x = is the point of minima.

So, for maximum volume, x = 3.

∴ Length of the rectangular sheet = 15x = 45 unit

Breadth of the rectangular sheet = 8x = 24 unit

The correct options are (A) and (C).

#### Question 56:

Consider the set of eight vectors . Three non-coplanar vectors can be chosen from V in 2pways. Then p is

Solutions:

Given,

Three vectors are coplanar if one of them can be written as a linear combination of the remaining two vectors.

All vectors are listed below:

Consider the vectors (v1, v5), (v2, v6), (v3, v7) and (v4, v8) in pair. If we take any one of the four pairs consisting two vectors and a third vector out of remaining vectors then theses three vectors will be coplanar. Thus for each pair we have 6 options.

Therefore, number of coplanar vectors = 6 + 6 + 6 + 6 = 24

Out of 8 vectors 3 vectors can be chosen indifferent ways. Theses 56 groups consisting three vectors are the collection of coplanar and no-coplanar vectors.

Therefore,

Number of number of non-coplanar vectors = 56 − 24 = 32

Thus,

#### Question 57:

Of the three independent events E1, E2 and E3, the probability that only E1 occurs is α, only E2 occurs is β and only E3 occurs is γ. Let the probability p that none of events E1, E2 or E3 occurs satisfy the equations.

All the given probabilities are assumed to lie in the interval (0, 1).

Solutions:

It is given that, E1, E2 and E3 are three independent events.

Therefore,

Let x, y and z be the probabilities of E1, E2 and E3 respectively. Then,

x (1 − y) (1 − z) = α      …(1)

(1 − x) y (1 − z) = β      …(2)

(1 − x) (1 − y) z = γ     …(3)

Also,

= p.

⇒ (1 − x) (1 − y) (1 − z) = p

It is given that,

(α − 2β) p = αβ        …(4)

(β− 3γ) p = 2βγ        …(5)

On substituting the values of α, β and γ from (1), (2) and (3) in (4) and (5), we get x = 2y and y = 3z respectively.

Thus, x = 2 (3z) = 6z

Then,

#### Question 58:

The coefficients of three consecutive terms of (1 + x)n + 5 are in the ratio 5 : 10 : 14. Then n =

Solutions:

Let be the coefficients of three consecutive terms of .

It is given that,

Consider .

Consider .

On equating (1) and (2), we get r = 4.

Substituting r = 4 in (2), we get

$n=4+7–5=6$

Hence, the value of n is 6.

#### Question 59:

A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224.

If the smaller of the numbers on the removed cards is k, then k − 20 =

Solutions:

There are 1 to n cards.

Let the two cards removed from n cards be having numbers k and k + 1.

Sum of the numbers from 1 to n = .

Sum of the numbers after removing two cards = 1224

= 1224 + k + (k + 1)

n2 + n − 4k = 2450

n2 + n − 2450 = 4k

⇒ (n + 50) (n − 49) = 4k

n > 49

For, n = 50, = 1275 ( > 1224)

Also, for n = 50, k =

Thus, k − 20 = 5

#### Question 60:

A vertical line passing through the point (h, 0) intersects the ellipse at the point P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If Δ (h) = area of the triangle PQR, and , then

Solutions:

Let the coordinate of P and Q be (h, k) and (h, −k). Then, the equation of the tangent at P(h, k) is and the equation of the tangent at Q(h, −k) is . Thus, the point of intersection is .

Thus, we get

Since the point P(h, k) lies on the ellipse , we get

Thus, we get

This is a decreasing function in the interval.

Thus, we get

And,

Thus, we get