# IIT JEE Advanced 2014 Paper 1 Code 2

### Test Name: IIT JEE Advanced 2014 Paper 1 Code 2

#### Question 1:

Two ideal batteries of emf V1 and V2 and three resistance R1, R2 and R3 are connected, as shown in the figure. The current in resistance R2 would be zero if

 Option 1: V1 = V2 and R1 = R2 = R3 Option 2: V1 = V2 and R1 = 2R2 = R3 Option 3: V1 = 2V2 and 2R1 = 2R2 = R3 Option 4: 2V1 = V2 and 2R1 = R2 = R3

Solutions:

Applying KVL to loop PQRUP, we have:

Given:

${i}_{1}–{i}_{2}=0$

Or,

${i}_{1}={i}_{2}$

…(ii)
Now, equation (i) will become

${i}_{1}=\frac{{V}_{1}}{{R}_{1}}$

…(iii)

Applying KVL to loop STURS, we have:

From equations (iii) and (iv), we get:

$\frac{{V}_{1}}{{R}_{1}}=\frac{{V}_{2}}{{R}_{3}}$

The solutions provided in options A, B and D are possible.

Hence, the correct options are A, B and D.

#### Question 2:

A transparent thin film of uniform thickness and refractive index n1 = 1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index n2= 1.5, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance f1 from the film, while rays of light traversing from glass to air get focused at distance f2 from the film. Then,

 Option 1: |f1| = 3R Option 2: |f1| = 2.8R Option 3: |f2| = 2R Option 4: |f2| = 1.4R

Solutions:Focal length of the thin film of uniform thickness:

Therefore, the film will not show any effect.

Now, from air to glass,

$\frac{{n}_{2}}{v}–\frac{{n}_{\mathrm{air}}}{u}=\frac{{n}_{2}–{n}_{\mathrm{air}}}{R}$

.
Also,
u = ∞

Thus, we have:

$\frac{1.5}{{f}_{1}}–0=\frac{1.5–1}{R}\phantom{\rule{0ex}{0ex}}⇒{f}_{1}=3R$

From glass to air,

$\frac{{n}_{\mathrm{air}}}{v}–\frac{{n}_{2}}{u}=\frac{{n}_{\mathrm{air}}–{n}_{2}}{–R}$

.
Thus, we have:

$\frac{1}{{f}_{2}}=\frac{1–1.5}{–R}\phantom{\rule{0ex}{0ex}}⇒{f}_{2}=2R$

Hence, the correct options are A and C.

#### Question 3:

In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle θ with the horizontal floor. The coefficient of friction between the wall and the ladder is μ1 and that between the floor and the ladder is μ2. The normal reaction of the wall on the ladder is N1 and that of the floor is N2. If the ladder is about slip, then

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

Solutions:
From the figure, we have:

For μ1 = 0:

As the ladder is in static equilibrium, τ = 0.
Or,

Here, l is the assumed length of the ladder.

From equation (i), we can say:
If μ2 = 0, then N1 will not get balance. Thus, to maintain equilibrium, μ2 can never be zero.
Thus, option B is not possible.
For μ1 ≠ 0:
From equations (i) and (ii), we have:

${N}_{2}=\frac{mg}{1+{\mu }_{1}{\mu }_{2}}$

.

Hence, the correct options are C and D.

#### Question 4:

Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K?

 Option 1: 4 if wires are in parallel Option 2: 2 if wires are in series Option 3: 1 if wires are in series Option 4: 0.5 if wires are in parallel

Solutions:Heat (H) is given by

$H=P×t$

$⇒t=\frac{H}{P}=\frac{HR}{{V}^{2}}$

It is clear from the above relation that R

$\propto$

t.

Let R be the resistance of the wire in the heater having length L and diameter d.
Thus, we have:

$R=\rho \frac{L}{A}$

=

$\rho \frac{L}{\mathrm{\pi }{\left(\frac{d}{2}\right)}^{2}}$

Here,
ρ = Resistivity of the wire
A = Area of cross-section of the wire
Time taken by this heater to raise the temperature by 40 K, t = 4 minutes
Let R' be the resistance of the new heater's wire having length L and diameter 2d.
Thus, we have:

When two wires of resistance R' are connected in series, the resultant resistance

$\left({R}_{1}\right)$

is
R1 =

$\frac{R}{4}+\frac{R}{4}=\frac{R}{2}$

Time taken by this combination,

$t\text{'}$

=

$\frac{t}{2}$

= 2 minutes
When two wires of resistance R' are connected in parallel, the resultant resistance

$\left({R}_{2}\right)$

is
R2 =

$\frac{R\text{'}}{2}$

=

$\frac{R}{8}$

Time taken by this combination,

$t\text{'}$

=

$\frac{t}{8}$

= 0.5 minute

Hence, the correct options are B and D.

#### Question 5:

A light source, which emits two wavelengths λ1 = 400 nm and λ2 = 600 nm, is used in a Young's double slit experiment. If recorded fringe widths for λ1 and λ2 are β1 and β2 and the number of fringes for them within a distance y on one side of the central maximum are m1 and m2, respectively m then

 Option 1: β2 > β1 Option 2: m1 > m2 Option 3: From the central maximum, 3rd maximum of λ2 overlaps with 5th minimum of λ1 Option 4: The angular separation of fringes for λ1 is greater than λ2.

Solutions:Fringe width

$\beta$

is given by

$\beta =\frac{\lambda D}{d}$

…(i)
Here,

$\lambda$

= Wavelength of the light used

$D$

= Distance between the sources of light and the screen
d = Distance between two coherent light sources

Given:

${\beta }_{1}$

is the fringe width for wavelength

${\lambda }_{1}$

and

${\beta }_{2}$

is the fringe width for wavelength

${\lambda }_{2}$

.
Also,

${\lambda }_{2}>{\lambda }_{1}$

[From equation (i)]
Number of fringes

$m$

is given by
m =

$\frac{y}{\beta }$

$\because {\beta }_{2}>{\beta }_{1}$

$\therefore$ ${m}_{1}>{m}_{2}$

The 3rd maximum of

${\lambda }_{2}$

will lie at a distance of

${y}_{n}=n{\lambda }_{2}\frac{D}{d}$

${y}_{3}=3\left(600\right)\frac{D}{d}$

=

The 5th minimum of

${\lambda }_{1}$

will lie at a distance of

${y}_{n}\text{'}=\left(2n–1\right){\lambda }_{1}\frac{D}{d}$

${y}_{5}\text{'}=\left(2×5–1\right)×400\frac{D}{2d}$

=

Thus, the 3rd maximum of

${\lambda }_{2}$

overlaps with the 5th minimum of

${\lambda }_{1}$

.
Angular separation is given by

$\frac{\beta }{D}$

=

$\frac{\lambda }{d}$

Angular separation

Therefore, angular separation will be more for

${\lambda }_{2}$

.

Hence, the correct options are A, B and C.

#### Question 6:

At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I(t) = I0 cos (ωt), with I0 = 1 A and ω = 500 rad s–1 starts flowing in it with the initial direction shown in the figure. At

$t=\frac{7\pi }{6\omega }$

, the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20 μF, R = 10 Ω and the battery is ideal with emf of 50 V, identify the correct statement (s).

 Option 1: Magnitude of the maximum charge on the capacitor before $t=\frac{7\pi }{6\omega }$ is 1 × 10–3 C. Option 2: The current in the left part of the circuit just before $t=\frac{7\pi }{6\omega }$ is clockwise. Option 3: Immediately after A is connected to D, the current in R is 10A. Option 4: Q = 2 × 10–3 C.

Solutions:Variation of current in the circuit with time is given as follows:

As shown in the figure, from t = 0 to

$\frac{7\mathrm{\pi }}{6\omega }$

the charge will be maximum at t =

$\frac{\mathrm{\pi }}{2\omega }$

, when current in the circuit is zero.
Charge at this point

$\left(Q\right)$

is given by

At t =

$\frac{7\mathrm{\pi }}{6\omega }$

, the direction of current will be opposite to the initial direction. And, the charge on the upper plate is given by

Applying KVL, when the switch is shifted to D, we get:
50 +

$\frac{{10}^{–3}}{20×{10}^{–6}}$ $–$ $i×10=0$

Final charge on C is given by

$Q\text{'}=CV$

=

Total charge flown from the battery will be 2

.

Hence, the correct options are C and D.

#### Question 7:

A parallel-plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C, while that of the portion with dielectric in between is C1. When the capacitor is charged, the plate area covered by the dielectric gets charge Q1 and the rest of the area gets charge Q2. The electric field in the dielectric is E1 and that in the other portion is E2. Choose the correct option/options, ignoring edge effects.

 Option 1: $\frac{{E}_{1}}{{E}_{2}}=1$ Option 2: $\frac{{E}_{1}}{{E}_{2}}=\frac{1}{K}$ Option 3: $\frac{{Q}_{1}}{{Q}_{2}}=\frac{3}{K}$ Option 4: $\frac{C}{{C}_{1}}=\frac{2+K}{K}$

Solutions:Let C2 be the capacitance of the portion of parallel-plate capacitor without dielectric.
Also, let A be the total area of the plate.
Now, capacitance C1:
C1 =

$\frac{K{\epsilon }_{0}\frac{A}{3}}{d}$

Capacitance C2:
C2 =

$\frac{{\epsilon }_{0}\frac{2A}{3}}{d}$

Total capacitance, C = C1 + C2

$⇒$

C =

$\frac{K{\epsilon }_{0}\frac{A}{3}}{d}$

+

$\frac{{\epsilon }_{0}\frac{2A}{3}}{d}$

$⇒$

C =

$\frac{\left(K+2\right){\epsilon }_{0}A}{3d}$

$⇒$

C =

$\frac{\left(K+2\right){C}_{1}}{K}$

$\therefore \frac{C}{{C}_{1}}=\frac{K+2}{K}$

Electric field: E1 = E2 =

$\frac{V}{d}$

Here,
V = Potential difference between the plates of the capacitor

Hence, the correct options are A and D.

#### Question 8:

One end of a taut string of length 3 m along the x-axis is fixed at x = 0. The speed of the waves in the string is 100 ms–1. The other end of the string is vibrating in the y-direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is (are)

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

Solutions:There should be a node at x = 0 and an antinode at x = 3 m.
That is,
y = 0 at x = 0 and y =

$±$

A at x = 3 m
Also,

We can see that only options A, C and D satisfy this condition.

Hence, the correct options are A, C and D.

#### Question 9:

A student is performing an experiment using a resonance column and a tuning fork of frequency 244 s–1. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is (0.350 ± 0.005) m, the gas in the tube is
(Useful information :

. The molar masses M in grams are given in the options. Take the values of

$\sqrt{\frac{10}{M}}$

for each gas as given there.)

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

Solutions:We know:
Minimum length for the occurrence of resonance, l = λ/4
or λ = 4l

Now, fundamental frequency is given as follows:

$f=\frac{1}{\lambda }\sqrt{\frac{\gamma RT}{M}}=\frac{1}{4l}\sqrt{\frac{\gamma RT}{M}}$

$\therefore \frac{∆f}{f}=\frac{∆l}{l}$

(A) Neon:

(B) Nitrogen:

(C) Oxygen:

(D) Argon:

From the above calculations, the only possible gas is Argon because its frequency of vibration is comparable to the frequency of the tuning fork used.

Hence, the correct option is D.

#### Question 10:

Let E1 (r), E2 (r) and E3 (r) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density λ, and an infinite plane with uniform surface charge density σ. If E1 (r0) = E2 (r0) =  E3 (r0)  at a given distance r0, then

 Option 1: Q = 4 σπr02 Option 2: ${r}_{0}=\frac{\lambda }{2\pi \sigma }$ Option 3: E1 (r0/2) = 2E2 (r0/2) Option 4: E2 (r0/2) = 4E3 (r0/2)

Solutions:Electric field due to different charge distribution at point r0 is given as follows:

${E}_{1}\left(r\right)=\frac{Q}{4{\mathrm{\pi \epsilon }}_{0}{r}^{2}}\phantom{\rule{0ex}{0ex}}⇒{E}_{1}\left({r}_{0}\right)=\frac{Q}{4{\mathrm{\pi \epsilon }}_{0}{{r}_{0}}^{2}}\phantom{\rule{0ex}{0ex}}{E}_{2}\left(r\right)=\frac{\lambda }{2{\mathrm{\pi \epsilon }}_{0}r}\phantom{\rule{0ex}{0ex}}⇒{E}_{2}\left({r}_{0}\right)=\frac{\lambda }{2\mathrm{\pi }{\epsilon }_{0}{r}_{0}}\phantom{\rule{0ex}{0ex}}{E}_{3}\left(r\right)=\frac{\sigma }{2{\epsilon }_{0}}\phantom{\rule{0ex}{0ex}}⇒{E}_{3}\left({r}_{0}\right)=\frac{\sigma }{2{\epsilon }_{0}}$

Given:
E1(r0) = E2(r0) = E3(r0)

Electric field due to different charge distribution at point r0/2 is given as follows:

Hence, the correct option is C.

#### Question 11:

Two parallel wires in the plane of the paper are distance X0 apart. A point charge is moving with speed u between the wires in the same plane at a distance X1 from one of the wires. When the wires carry current of magnitude I in the same direction, the radius of curvature of the path of the point charge is R1. In contrast, if the currents I in two wires have directions opposite to each other, the radius of curvature of the path is R2. If

$\frac{{X}_{0}}{{X}_{1}}=3,$

the value of

$\frac{{R}_{1}}{{R}_{2}}$

is

Solutions:Magnetic field at the location of point charge due to wire at distance X1 from it is given by

${B}_{1}=\frac{{\mu }_{0}I}{2{\mathrm{\pi X}}_{1}}$

Magnetic field at the location of point charge due to other wire at distance X0 X1 from it is given by

${B}_{1}\text{'}=\frac{{\mu }_{0}I}{2\mathrm{\pi }\left({X}_{\mathit{0}}–{X}_{1}\right)}$

When the direction of current is same in both the wires:

The net magnetic field at the given point will be:

When the direction of current is different in both the wires:

The net magnetic field at the given point will be:

We know that the radius of curvature of path of a point charge due to magnetic field is given by

$R=\frac{mv}{qB}\phantom{\rule{0ex}{0ex}}⇒R\propto \frac{1}{B}\phantom{\rule{0ex}{0ex}}\therefore \frac{{R}_{1}}{{R}_{2}}=\frac{{B}_{net2}}{{B}_{net1}}=\frac{{X}_{0}}{{X}_{0}–2{X}_{1}}=\frac{\frac{{X}_{0}}{{X}_{1}}}{\frac{{X}_{0}}{{X}_{1}}–2}=\frac{3}{3–2}=3$

Hence, the correct answer is 3.

#### Question 12:

A thermodynamic system is taken from an initial state i with internal energy Ui = 100 J to the final state f along two different paths iaf and ibf, as schematically shown in the figure. The work done by the system along the paths af, ib and bf are Waf = 200 J, Wib = 50 J and Wbf = 100 J respectively. The heat supplied to the system along the path iaf, ib and bf are Qiaf, Qib and Qbf respectively. If the internal energy of the system in the state b is Ub = 200 J and Qiaf = 500 J, the ratio Qbf /Qib is

Solutions:Since ia is an isochoric process, we have:
Wia = 0
Given:
Waf = 200 J, Ui = 100 J , Ub = 200 J
Wib = 50 J, Wbf = 100 J
Qiaf = 500 J,
Wiaf = WiaWaf  = 200 J

So, we have:
UfUi = QiafWiaf
= 500 − 200 = 300 J
∴  Uf = 300 + Ui
= 300 + 100 = 400 J
Also, we have:
Qib = Uib + Wib = (UbUi) + Wib
Qib = (200 − 100) + 50 = 150 J

Qbf = Ubf + Wbf = (UfUb) + Wbf
Qbf = (400 − 200) + 100 = 300 J

$\therefore \frac{{Q}_{bf}}{{Q}_{ib}}=\frac{300}{150}=2$

Hence, the correct answer is 2.

#### Question 13:

A rocket is moving in a gravity-free space with a constant acceleration of 2 ms−2 along +x-direction (see figure). The length of a chamber inside the rocket is 4 m. A ball is thrown from the left end of the chamber in +x-direction with a speed of 0.3 ms−1 relative to the rocket. At the same time, another ball is thrown in −x-direction with a speed of 0.2 ms−1 from its right end relative to the rocket. The time in seconds when the two balls hit each other is

Solutions:The net acceleration of the balls will be due to the acceleration of the rocket in positive x-direction.
Let after time t, the balls collide.
Then, the distance travelled by the ball thrown from the left end of the chamber is given by

${S}_{1}=0.3t+\frac{1}{2}×2×{t}^{2}$

And, the distance travelled by the ball thrown from the right end of the chamber is given by

${S}_{2}=0.2t–\frac{1}{2}×2×{t}^{2}$

For balls to collide,

Hence, the correct answer is 8.

#### Question 14:

Consider an elliptically shaped rail PQ in the vertical plane with OP = 3 m and OQ = 4 m. A block of mass 1 kg is pulled along the rail from P to Q with a force of 18 N, which is always parallel to line PQ (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches Q is (n × 10) Joules. The value of n is (take acceleration due to gravity = 10 ms−2)

Solutions:

Change in kinetic energy = Total work done from P to Q
= Work done against gravity + Work done by force
Work done against gravity = −mgh = −1 × 10 × 4 = −40 J
Work done by force to cause displacement PQ = F × d = 18 × 5 = 90 J
Change in kinetic energy = −40 + 90 = 50 J
We know that the initial kinetic energy was zero at P.
∴ Change in kinetic energy = Kinetic energy at Q = 50 J = 5 × 10 J

Hence, the correct answer is 5.

#### Question 15:

During Searle's experiment, zero of the Vernier scale lies between 3.20 × 10−2 m and 3.25 × 10−2 m of the main scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 × 10−2 m and 3.25 × 10−2 m of the main scale but now the 45th division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 × 10−7 m2. The least count of the Varnier scale is 1.0 × 10−5 m. The maximum percentage error in the Young's modulus of the wire is

Solutions:Let the initial weight be W and extension be Δl. Then,

Let the weight be W' now and extension be Δl'. Then,

Now,

$W\text{'}–W=\frac{YA}{l}\left(∆l\text{'}–∆l\right)\phantom{\rule{0ex}{0ex}}⇒Y=\frac{\left(W\text{'}–W\right)l}{A\left(∆l\text{'}–∆l\right)}$

W'W = 2 kg, l = 2 m and A = 8×10-7 m2 are constant.

$\therefore Y=\frac{\mathrm{constant}}{\left(∆l\text{'}–∆l\right)}$

The error in measurement of Y will be due to error in measurement of extension produced in the wire. The maximum percentage error is thus given by

$\frac{∆Y}{Y}×100=\left(\frac{∆l+∆l\text{'}}{∆l\text{'}–∆l}\right)×100=\frac{2×{10}^{–5}}{25×{10}^{–5}}×100=8%$

Hence, the correct answer is 8.

#### Question 16:

Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30° and 60°, respectively, with respect to the horizontal, as shown in the figure. The speed of A is

At time t = 0 s, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t = t0, A just escapes being hit by B, t0 in seconds is

Solutions:The observer sees B moving with a constant velocity perpendicular to the line of motion. This means

Given, relative distance between the airplanes = 500 m.
Relative velocity = vBsin30

$°$

Hence, the correct answer is 5.

#### Question 17:

A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg are attached to the platform at a distance 0.25 m from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have horizontal speed of 9 ms−1 with respect to the ground. The rotational speed of the platform in rad s−1 after the balls leave the platform is

Solutions:Both the balls when fired will produce a torque in clockwise direction.
From,

$L=I\omega \phantom{\rule{0ex}{0ex}}$

Using law of conservation of angular momentum.

Hence, the correct answer is 4.

#### Question 18:

A uniform circular disc of mass 1.5 kg and radius 0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude F = 0.5 N are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in rad s−1 is

Solutions:
From the figure, it is clear that only the sine components of all the forces will produce a torque in the disc in the anticlockwise direction.

The angular velocity after 1 s can be computed as follows:

Hence, the correct answer is 2.

#### Question 19:

A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 â„¦ resistance, it can be converted into a voltmeter of range 0−30 V. If connected to a

$\frac{2n}{249}\mathrm{\Omega }$

resistance, it becomes an ammeter of range 0−1.5 A. The value of n is

Solutions:Since the galvanometer gives full scale deflection with 0.006 A, we have:
Ig = 0.006 A
Let the internal resistance of the galvanometer be Rg.
We convert galvanometer into a voltmeter of voltage V = 30 V by connecting a resistance R = 4990 Ω in series as shown in the following figure:

Now, we convert galvanometer into a ammeter of current I = 15 A by connecting a resistance R' =

$\frac{2n}{249}\mathrm{\Omega }$

parallelly as shown in the following figure.

${I}_{g}×{R}_{g}=\left(I–{I}_{g}\right)×R\text{'}\phantom{\rule{0ex}{0ex}}⇒0.006×10=\left(1.5–0.006\right)\frac{2n}{249}\phantom{\rule{0ex}{0ex}}⇒\frac{0.06}{1.494}=\frac{2n}{249}\phantom{\rule{0ex}{0ex}}⇒2n=\frac{0.06}{1.494}×249\phantom{\rule{0ex}{0ex}}⇒2n=10\phantom{\rule{0ex}{0ex}}⇒n=5$

Hence, the correct answer is 5.

#### Question 20:

To find the distance d over which a signal can be seen clearly in foggy conditions, a railways engineer uses dimensional analysis and assumes that the distance depends on the mass density ρ of the fog, intensity (power/area) S of the light from the signal and its frequency f. The engineer finds that d is proportional to S1/n. The value of n is

Solutions:Since the distance (d) depends on the mass density (ρ) of the fog, the intensity (S) of the light from the signal and its frequency (f), we have
d = ρaSbfc    …(1)
And, we know
[d] = M0L1T0
[ρ] = ML−3
[S] = MT−3
[f] = T−1
Using dimensional inequality in equation (1), we get:
M0L1T0 = MaL−3a × MbT−3b × Tc
⇒ M0L1T0 = Ma + bL−3aT–3b c    …(2)
Equating the powers of M and L on both sides of equation (2), we get:
a + b = 0,  −3a = 1

$a=–\frac{1}{3}$

And,

$b=\frac{1}{3}$

According to the question:

$n=\frac{1}{b}$

or, n = 3

Hence, the correct answer is 3.

#### Question 21:

In a galvanic cell, the salt bridge

 Option 1: does not participate chemically in the cell reaction. Option 2: stops the diffusion of ions from one electrode to another. Option 3: is necessary for the occurrence of the cell reaction. Option 4: ensures mixing of the two electrolytic solution.

Solutions:Salt bridge  does not participate  chemically in the cell reaction, however it helps in keeping the solution of two electrode separate, so that ions do not mix freely but diffusion of ions can't be stopped by it.

Hence, the correct option is A.

#### Question 22:

The pair(s) of reagents that yield paramagnetic species is/are

 Option 1: Na and excess of NH3 Option 2: K and excess of O2 Option 3: Cu and dilute HNO3 Option 4: O2 and 2-ethylanthraquinol

Solutions:Na, on reaction with excess of NH3, gives a solution of alkali metals containing free ammoniated electrons that are paramagnetic in nature.

K, on reaction with excess of O2, gives superoxide, which is paramagnetic in nature due to the presence of a single unpaired electron.

Cu and dilute HNO3 give NO, which is paramagnetic in nature.

O2 and 2-ethylanthraquinol will give 2-ethylanthraquinone and hydrogen peroxide. Hydrogen peroxide is diamagnetic in nature.

Hence, the correct options are (A), (B) and (C).

#### Question 23:

For the reaction :

${\mathrm{I}}^{–}+{{\mathrm{ClO}}_{3}}^{–}+{\mathrm{H}}_{2}{\mathrm{SO}}_{4}\to {\mathrm{Cl}}^{–}+{{\mathrm{HSO}}_{4}}^{–}+{\mathrm{I}}_{2}$

The correct statement(s) in the balanced equation is/are :

 Option 1: Stoichiometric coefficient of HSO4− is 6. Option 2: Iodide is oxidised. Option 3: Sulphur is reduced. Option 4: H2O is one of the products.

Solutions:6I + ClO3 + 6H2SO4 → Cl + 6HSO4 + 3I2 + 3H2O
From the balanced chemical equation, it is clear that the stoichiometric coefficient of HSO4 is 6; iodide (I) is oxidised to I2 and water (H2O) is one of the products.

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

#### Question 24:

Upon heating with Cu2S, the reagent(s) that give(s) copper metal is/are

 Option 1: CuFeS2 Option 2: CuO Option 3: Cu2O Option 4: CuSO4

Solutions:When Cu2S is heated with Cu2O, it gives Cu and releases SO2.

Both CuSO4 and CuO, on heating, produce Cu2O. On further reaction with Cu2S, they give Cu.

Cu is not obtained when CuFeS2 is heated with Cu2S.

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

#### Question 25:

The correct statement(s) for orthoboric acid is/are

 Option 1: It behaves as a weak acid in water due to self ionisation. Option 2: Acidity of its aqueous solution increases upon addition of ethylene glycol. Option 3: It has a three dimensional structure due to hydrogen bonding. Option 4: It is a weak electrolyte in water.

Solutions:Orthoboric acid is H3BO3 or B(OH)3.
It partially reacts with water and forms H3O+ and [B(OH)4]. It is soluble in water and behaves as a weak monobasic acid. Orthoboric acid is a type of acid which does not donate proton; rather, it accepts hydroxyl ion and behaves as a Lewis acid.
B(OH)+ 2 H2O â‡Œ H3O+ + [B(OH)4]

Thus, orthoboric acid does not behave as a weak acid due to self ionisation.

In aqueous solution of orthoboric acid, if a certain organic polyhydroxy compound such as glycerol, mannitol or sugar is added, then it behaves as strong monobasic acid.

BO33− units have a planar structure. In solid state, BO33− units are hydrogen bonded, giving rise to a two-dimensional sheet with almost hexagonal symmetry.

Orthoboric acid is less soluble in cold water and furnishes BO33− ions, so it is a weak electrolyte in water.

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

#### Question 26:

The correct combination of names for isomeric alcohols with molecular formula C4H10O is/are

 Option 1: tert-butanol and 2-methylpropan-2-ol Option 2: tert-butanol and 1, 1-dimethylethan-1-ol Option 3: n-butanol and butan-1-ol Option 4: isobutyl alcohol and 2-methylpropan-1-ol

Solutions:In option B, the common name and IUPAC name of the isomeric alcohol with molecular formula C4H10O is incorrect.

Hence, the correct options are (A), (C) and (D).

#### Question 27:

The reactivity of compound with different halogens under appropriate conditions is given below:

The observed pattern of electrophilic substitution can be explained by

 Option 1: the steric effect of the halogen Option 2: the steric effect of the tert-butyl group Option 3: the electronic effect of the phenolic group Option 4: the electronic effect of the tert-butyl group

Solutions:In this organic compound, due to electronic effect of phenolic group, the most probable positions where substitution reactions can occur are I, II and III.

Position I is sterically hindered by the tert-butyl group and hydroxy group while the position II is sterically hindered by tert-butyl group. Thus, the heavy group like iodide will only get substituted at III position to avoid the steric hinderance while bromide group will be substituted at II and III and the light group like chloride will be substituted at I, II and III positions.

Hence, the correct options are (A), (B) and (C).

#### Question 28:

An ideal gas in a thermally insulated vessel at internal pressure = P1, volume = V1 and absolute temperature = T1 expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are P2, V2 and T2, respectively. For this expansion.

 Option 1: q = 0 Option 2: T2 = T1 Option 3: P2V2 = P1V1 Option 4:

Solutions:

Since,

w = −Pexâˆ†V,

The work done will also be zero when the external pressure is zero.
As the system is thermally insulated, q is zero.

Now,

ΔU = 0     [âˆµ ΔU = q + w]
Also,

ΔT = 0, so T1 = T2

Since T is constant and mass of the gas is fixed, according to Boyle's law, we have:
P2V2 = P1V1

Hence, the correct options are (A), (B) and (C).

#### Question 29:

Hydrogen bonding plays a central role in the following phenomena:

 Option 1: Ice floats in water. Option 2: Higher Lewis basicity of primary amines than tertiary amines in aqueous solutions. Option 3: Formic acid is more acidic than acetic acid. Option 4: Dimerisation of acetic acid in benzene.

Solutions:When water freezes, hydrogen bonding in water molecules gives rise to a cage-like structure as a result the density of ice is lower than that of water. Thus, ice floats on water.
Basicity of amines depends on the extent of hydrogen bonding among the amines. Primary amines form H-bonds more effectively than tertiary amines, thus they are more basic than tertiary amines in aqueous solutions. However, basicity of amines in aqueous solutions also depends on solvation of cations formed in the solution.
Dimerisation of acetic acid in benzene is also the result of hydrogen bonding taking place between the acetic acid molecules.

Hence, the correct options are A, B and D.

#### Question 30:

In the reaction shown below, the major product(s) formed is/are

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

Solutions:

Hence, the correct option is (A).

#### Question 31:

MX2 dissociates into M2+ and X ions in an aqueous solution, with a degree of dissociation (α) of 0.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is

Solutions:

The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is known as van't Hoff factor.

Dissociation of MX2 in the solution would take place as follows:

MX2 â‡Œ M2+ + 2 X

Degree of dissociation (α) = 0.5

For an ionisable molecule, the van't Hoff factor (i) can be calculated as

$\mathrm{i}=\frac{1+\left(n–1\right)\mathrm{\alpha }}{1}$

Here, i = van't Hoff factor and n = number of ions produced in the solution

So, the van't Hoff factor for MX2 is 2.

Hence, the correct answer is 2.

#### Question 32:

In an atom, the total number of electrons having quantum numbers

is

Solutions:

Here,

n = 4 and

This means that the orbital is 4p. Since the p-orbital has three subshells (px, py and pz), the number of electrons with ms= −1/2 would be three.

Hence, the correct answer is 3.

#### Question 33:

The total number of distinct naturally occurring amino acids obtained by complete acidic hydrolysis of the peptide shown below is

Solutions:By complete acidic hydrolysis of the given compound, only one distinct naturally occurring amino acid is obtained.

These are the three types of products obtained by complete acidic hydrolysis.

Glycine is a naturally occurring amino acid obtained by hydrolysis of the given compound.

Hence, the correct answer is 1.

#### Question 34:

If the value of Avogadro number is 6.023 × 1023 mol−1 and the value of Boltzmann constant is 1.380 × 10−23 J K−1, then the number of significant digits in the calculated value of the universal gas constant is

Solutions:Given:
Avogadro number (NA) = 6.023 × 1023 mol1
Boltzmann constant (KB) = 1.380 × 1023 JK1

The relation between the universal gas constant (R), Avogadro number (NA) and Boltzmann constant (KB) is

${\mathrm{K}}_{\mathrm{B}}=\frac{\mathrm{R}}{{\mathrm{N}}_{\mathrm{A}}}$

∴ The value of the universal gas constant has four significant figures.

Hence, the correct answer is 4.

#### Question 35:

A compound H2X with molar weight of 80 g is dissolved in a solvent having density of 0.4 g ml−1. Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is

Solutions:Molality,

w2×1000M2×w1
Here,
w2 = mass of the solute
w1 = mass of the solvent
M2 = molar mass of the solute

Density of the solvent = 0.4 g/mL
Mass of 1 mL of solvent = 0.4 g
Mass of 1000 mL of solvent = 0.4 × 1000 = 400 g
Molarity of the solution = 3.2 M
Mass of the solute, w2 = 3.2 × 80 = 256 g
m= 256×100

Hence, the correct answer is 8.

#### Question 36:

The total number(s) of stable conformers with non-zero dipole moment for the following compound is (are)

Solutions:The total number of stable conformers of the given compound with non-zero dipole moment is three.

These are the staggered conformations of the given compound.

Hence, the correct answer is 3.

#### Question 37:

A list of species having the formula XZ4 is given below.

.
Defining shape on the basis of the location of X and Z atoms, the total number of species having a square planar shape is

Solutions:Among the given list of species having the formula XZ4, XeF4, BrF4 , [Cu(NH3)4]2+ and [PtCl4]2− have square planar (dsp2) shape.

 Molecule Shape Geometry Hybridisation XeF4 Square planar sp3d2 SF4 See-saw sp3d SiF4 Tetrahedral sp3 BF4− Tetrahedral sp3 BrF4− Square planar sp3d2 [Cu(NH3)4]2+ Square planar dsp2 [FeCl4]2− Tetrahedral sp3 [CoCl4]2− Tetrahedral sp3 [PtCl4]2− Square planar dsp2

Hence, the correct answer is (4).

#### Question 38:

Consider the following list of reagents :
Acidified K2Cr2O7, alkaline KMnO4, CuSO4, H2O2, Cl2, O3, FeCl3, HNO3 and Na2S2O3.
The total number of reagents that can oxidise aqueous iodide to iodine is

Solutions:The reagents that can oxidise I to I2 are acidified KMnO4, CuSO4, H2O2, Cl2, O3, FeCl3 and HNO3.
The reactions are as follows:
K2Cr2O4 + KI + H2SO4 → K2SO4 + Cr2(SO4)3 + I2 + H2O
KMnO4 + NaOH + KI → No reaction
2CuSO4 + 4KI → 2CuI↓ + I2 + 2K2SO4
H2O2 + 2KI → 2KOH + I2
Cl2 + 2KI → 2KCl + I2
O3 + H2O + KI → 2KOH + O2 +I2
FeCl3 + 2KI → 2KCl + FeCl2 + I2
HNO3 + KI → KNO3 + I2 + NO↑
Na2S2O3 + KI → No reaction

Hence, the correct answer is 7.

#### Question 39:

Consider all possible isomeric ketones, including stereoisomers of MW = 100. All these isomers are independently reacted with NaBH4 (NOTE: stereoisomers are also reacted separately). The total number of ketones that give a racemic products(s) is/are

Solutions:The isomeric ketones with molar mass 100 are:
(I)

(II)

(III)

(IV)

(V)

(VI)

Ketone having a chiral centre, on reaction with NaBH4, gives diastereoisomers. Thus, the number of ketones that give racemic mixture on reaction with NaBH4 is five.

#### Question 40:

Among PbS, CuS, HgS, MnS, Ag2S, NiS, CoS, Bi2S3 and SnS2, the total number of BLACK coloured sulphides is

Solutions:Out of the given sulphides, CuS, PbS, CoS, NiS, HgS, Ag2S and Bi2S3 are black in colour.

Hence, the correct answer is 7.

#### Question 41:

Let M be a 2 × 2 symmetric matrix with integer entires. Then M is invertible if

 Option 1: the first column of M is the transpose of the second row of M Option 2: the second row of M is the transpose of the first column of M Option 3: M is a diagonal matrix with non-zero entries in the main diagonal Option 4: the product of entries in the main diagonal of M is not the square of an integer

Solutions:Let M =

$\left[\begin{array}{cc}a& b\\ b& c\end{array}\right]$

, where a, b, c

$\in$

I.

For invertible matrix,
det(M)

$\ne$

0
ac

$–$

b2

$\ne$

0
⇒ ac

$=$

b2

Now,
Option (A):

a

$=$

b

$=$

c
⇒ det(M)

$=$

0
M is not invertible.

Option (B):

$\left[\begin{array}{cc}b& c\end{array}\right]={\left[\begin{array}{c}a\\ b\end{array}\right]}^{T}=\left[\begin{array}{cc}a& b\end{array}\right]$

b

$=$

a

$=$

c
⇒ det(M)

$=$

0
M is not invertible.

Option (C):
det(M)

$=$

$\left|\begin{array}{cc}a& 0\\ 0& c\end{array}\right|=ac–0=ac\ne 0$

M is invertible.
Thus, M is invertible if it is a diagonal matrix with non-zero entries in the main diagonal.

Option (D):
det(M)

$=$

ac

$–$

b2
If M is invertible, then ac

$–$

b2

$\ne$

0 ⇒ ac

$\ne$

b2 , where b is an integer.
Thus, M is invertible if the product of entries in the main diagonal of M is not the square of an integer.

Hence, the correct options are C and D.

#### Question 42:

A circle S passes through the point (0, 1) and is orthogonal to the circles (x − 1)2 + y2 = 16 and x2 + y2 = 1. Then

 Option 1: radius of S is 8 Option 2: radius of S is 7 Option 3: centre of S is (−7, 1) Option 4: centre of S is (−8, 1)

Solutions:

Let the equation of the circle S be x2 + y2 + 2gx + 2fy + C = 0.
S is orthogonal with (x − 1)2 + y2 = 16. Then,

2{g(−1) + f(0)} = C − 15         (âˆµ If two circles are orthogonal, then 2gg' + 2ff' = C + C')
⇒ −2g = C −15         …….(i)

Also, S is orthogonal to x2 + y2 = 1.Then,

2{g(0) + f(0)} = C − 1
C = 1

Substituting the value of C in equation (i), we get:

−2g = 1 − 15
⇒ −2g = −14
g = 7

Also, the circle passes through (0, 1).
∴ 1 + 2f + 1 = 0
f = − 1

Hence, the equation of the circle S is x2 + y2 + 14x − 2y + 1 = 0.
⇒ (x + 7)2 + (y − 1)2 = 49 = 72

∴ Centre of S = (−7, 1) and radius of S = 7.
Hence, the correct options are B and C.

#### Question 43:

Let

be three vectors each of magnitude

$\sqrt{2}$

and the angle between each pair of them is

$\frac{\mathrm{\pi }}{3}$

. If

$\stackrel{\to }{a}$

is a nonzero vector per perpendicular to

is a nonzero vector perpendicular to

$\stackrel{\to }{y}\mathrm{and}\stackrel{\to }{z}×\stackrel{\to }{x}$

, then

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

Solutions:Here, magnitude of vectors x, y, and z are equal.

So,

Let the angle between each pair be

$\theta$

.
i.e.,

$\theta =\frac{\pi }{3}$

then,

Also,

$\stackrel{\to }{a}=k\left(\stackrel{\to }{x}×\left(\stackrel{\to }{y}×\stackrel{\to }{z}\right)\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{b}=l\left(\stackrel{\to }{y}×\left(\stackrel{\to }{z}×\stackrel{\to }{x}\right)\right)$

Using triple product of vectors,

Option (A):

$\stackrel{\to }{b}.\stackrel{\to }{z}\left(\stackrel{\to }{z}–\stackrel{\to }{x}\right)=l\left(\stackrel{\to }{z}–\stackrel{\to }{x}\right)=\stackrel{\to }{b}$

Option A is correct

Option (B):

$\stackrel{\to }{a}.\stackrel{\to }{y}\left(\stackrel{\to }{y}–\stackrel{\to }{z}\right)=k\left(\stackrel{\to }{y}–\stackrel{\to }{z}\right)=\stackrel{\to }{a}$

Option B is correct

Option (C):

Option C is correct.

Option (D):

$\stackrel{\to }{a}.\stackrel{\to }{y}\left(\stackrel{\to }{y}–\stackrel{\to }{z}\right)=k\left(\stackrel{\to }{y}–\stackrel{\to }{z}\right)=\stackrel{\to }{a}$

$\stackrel{\to }{a}.\stackrel{\to }{y}\left(\stackrel{\to }{z}–\stackrel{\to }{y}\right)=–\stackrel{\to }{a}$

Hence, the correct options are A, B and C.

#### Question 44:

From a point P(λ, λ, λ), perpendiculars PQ and PR are drawn, respectively, on the lines y = x, z = 1 and y = −x, z = −1. If P is such that ∠QPR is a right angle, then the possible value(s) of λ is (are)

 Option 1: $\sqrt{2}$ Option 2: 1 Option 3: −1 Option 4: $–\sqrt{2}$

Solutions:The equation of the line passing through point Q(y = x, z = 1) is

$\frac{x}{1}=\frac{y}{1}=\frac{z–1}{0}$

= k              …(1)
Any point on the line passing through Q is (k, k, 1).

Similarly, equation of the line passing through point R(y = −x, z = −1) is

$\frac{x}{–1}=\frac{y}{1}=\frac{z+1}{0}$

= m            ….(2)
Any point on the line passing through R is (

$–\mathrm{m}$

, m,

$–1$

).

Since PQ is perpendicular to line (1),

$⇒$

Q

Similarly, PR is perpendicular to line (2),

$⇒$

R

Also, PQ

$\perp$

PR.

$\lambda$

= 1 will be rejected. Otherwise, P will coincide with Q.
Thus, λ = −1

Hence, the correct option is C.

#### Question 45:

For every pair of continuous functions f, g: [0, 1] → â„ such that max{f(x) : x ∈ [0, 1]} = max {g(x) : x ∈ [0, 1]}, the correct statement (s) is(are)

 Option 1: (f(c))2 + 3f(c) = (g(c))2 + 3g(c) for some c ∈ [0, 1] Option 2: (f(c))2 + f(c) = (g(c))2 + 3g(c) for some c ∈ [0, 1] Option 3: (f(c))2 + 3f(c) = (g(c))2 + g(c) for some c ∈ [0, 1] Option 4: (f(c))2 = (g(c))2 for some c ∈ [0, 1]

Solutions:Let

and

${c}_{2}$

be the points of maxima of f(x) and g(x), respectively, such that

.
Let h(x) be a function, such that
h(x) = f(x) − g(x)
h(

) = f(

) − g(

)
Now, f(

) > g(

)
So, h(

) >0      …(1)
h(

${c}_{2}$

) = f(

${c}_{2}$

) − g(

${c}_{2}$

)
Now, f(

${c}_{2}$

) < g(

${c}_{2}$

)
So, h(

${c}_{2}$

) < 0       …(2)
So, a point c lies between

and

${c}_{2}$

, such that h(c) = 0.
Therefore, f(c) = g(c).
Option A:
(f(c))2 + 3f(c) = (g(c))2 + 3g(c)
(f(c))2 + 3f(c) = (f(c))2 + 3f(c)
So, option A is correct.
Option B:
(f(c))2 + f(c) = (g(c))2 + 3g(c)
(f(c))2 + f(c) ≠ (f(c))2 + 3f(c)
So, option B is not correct.
Option C:
(f(c))2 + 3f(c) = (g(c))2 + g(c)
(f(c))2 + 3f(c) ≠ (f(c))2 + f(c)
So, option C is not correct.
Option D
(f(c))2 − (g(c))2= 0
So, option D is correct
Hence, the correct options are A and D.

#### Question 46:

Let M and N be two 3 × 3 matrices, such that MN = NM. Further, if M ≠ N2 and M2 = N4, then

 Option 1: determinant of (M2 + MN2) is 0 Option 2: there is a 3 × 3 non-zero matrix ∪, such that (M2 + MN2) ∪ is the zero matrix Option 3: determinant of (M2 + MN2) ≥ 1 Option 4: for a 3 × 3 matrix ∪, if (M2 + MN2) ∪ equals the zero matrix, then ∪ is the zero matrix

Solutions:It is given that MN2 and M2 = N4.
Now, MN2 ⇒ M − N2≠ 0
and M2 = N4M2N4 = 0
⇒ (MN2)(M + N2) = 0 (âˆµ MN = NM)

Three cases arise:
(i) (MN2) = 0 and (M + N2) ≠ 0, which is not possible (âˆµM − N2≠ 0) .
(ii) (M + N2) = 0 and M − N2≠ 0
(iii) âˆ£MN2âˆ£ = 0 and âˆ£M + N2âˆ£ = 0

Now,
Option (A):
âˆ£M2 + MN2âˆ£ = âˆ£Mâˆ£âˆ£M + N2âˆ£ = 0 {Using case (iii)}

Option (B):
âˆ£M + N2âˆ£ = 0 ⇒ âˆ£M2 + MN2âˆ£ = 0
If âˆ£M2 + MN2âˆ£ = 0, then (M2 + MN2)U = 0 will have infinitely many solutions (U is a non-zero matrix).

Option (C):
âˆ£M2 + MN2âˆ£ = 0

$\ngeqq$

1

Option (D):
If âˆ£M2 + MN2âˆ£ = 0 and (M2 + MN2)U = 0, then U is a non-zero matrix.

Hence, the correct options are A and B.

#### Question 47:

Let a ∈ â„ and let f : â„ → â„ be given by f(x) = x5 − 5x + a. Then

 Option 1: f(x) has three real roots if a > 4 Option 2: f(x) has only one real root if a > 4 Option 3: f(x) has three real roots if a < −4 Option 4: f(x) has three real roots if −4 < a < 4

Solutions:Given:

$f\left(x\right)={x}^{5}–5x+a$

$f\text{'}\left(x\right)=5{x}^{4}–5$

For critical point,

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

.

When a > 4,

.
So, the graph of the function is as shown:

Since the graph crosses the axis one time, it has one real root.
When

$a<–4$

,

.
So, the graph of the function is as shown:

Since the graph crosses the axis one time, it has one real root.
When

$–4

,

.
So, the graph of the function is as shown:

Since, the graph crosses the axis three times, it has three real roots.

Hence, the correct options are B and D.

#### Question 48:

Let f : (0, ∞) → â„ be given by

. Then

 Option 1: f(x) is monotonically increasing on [1, ∞) Option 2: f(x) is monotonically decreasing on (0, 1) Option 3: Option 4: f(2x) is an odd function of x on â„

Solutions:f(x) =

${\int }_{1}{x}}^{x}\frac{{e}^{–\left(t+\frac{1}{t}\right)}}{t}dt$

Using Leibniz Integral rule, we get:

f '(x) =

${e}^{–\left(x+\frac{1}{x}\right)}\left(\frac{1}{x}\right)–{e}^{–\left(x+\frac{1}{x}\right)}\left(x\right)\left(–\frac{1}{{x}^{2}}\right)$

=

$\frac{2}{x}{e}^{–\left(x+\frac{1}{x}\right)}$

Option (A):
f '(x) =

$\frac{2}{x}{e}^{–\left(x+\frac{1}{x}\right)}$

> 0 ∀ x > 0
Thus, f(x) is monotonically increasing on [1,

$\infty$

).

Option (B):
f '(x) =

$\frac{2}{x}{e}^{–\left(x+\frac{1}{x}\right)}$

> 0 ∀ x > 0
Thus, f(x) is not monotonically decreasing on (0, 1).

Option (C):

Option (D):
Since

$f\left(x\right)+f\left(\frac{1}{x}\right)=0$

,
f(2x) + f(2x) = 0
Thus, f(2x) is an odd function.

Hence, the correct options are A, C and D.

#### Question 49:

Let

be given by f(x) = (log (sec x + tan x))3. Then

 Option 1: f(x) is and odd function Option 2: f(x) is one-one function Option 3: f(x) is an onto function Option 4: f(x) is an even function

Solutions:Option(A):

It is given that,

$f\left(x\right)={\left(\mathrm{log}\left(\mathrm{sec}x+\mathrm{tan}x\right)\right)}^{3}$

Replacing x by −x, we get

Therefore, f(x) is an odd function.

Option(B):

Differentiating

$f\left(x\right)={\left(\mathrm{log}\left(\mathrm{sec}x+\mathrm{tan}x\right)\right)}^{3}$

, we get

Therefore, f(x) is a one-one function.

Option(C):

The function f(x) is an onto function if the range of f(x) = R.

So, the range of f(x) is R.

Thus, f(x) is an onto function.

Option(D):

Here, f(x) is an odd function, so option D is false.

Hence, the correct options are A, B and C.

#### Question 50:

Let f : [a, b] → [1, ∞) be a continuous function and let g : â„ → â„ be defined as

Then

 Option 1: g(x) is continuous but not differentiable at a Option 2: g(x) is differentiable on â„ Option 3: g(x) is continuous but not differentiable at b Option 4: g(x) is continuous and differentiable at either a or b but not both

Solutions:First, we check the continuity at x = a and x = b.

We have,
LHL at x = a
=

$\underset{x\to {a}^{–}}{\mathrm{lim}}g\left(x\right)=0$

RHL at x = a
=

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

g(a) =

${\int }_{a}^{a}f\left(t\right)dt=0$

So, g(x) is continuous at x = a.

Now,
LHL at x = b
=

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

RHL at x = b
=

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

g(b) =

${\int }_{a}^{b}f\left(t\right)dt$

So, g(x) is continuous at x = b.

Now, we check the differentiability at x = a and x = b.

We have,
LHD at x = a
=

$\underset{x\to {a}^{–}}{\mathrm{lim}}\frac{g\left(x\right)–g\left(a\right)}{x–a}=0\phantom{\rule{0ex}{0ex}}$

RHD at x = a
=

$\underset{x\to a+}{\mathrm{lim}}\frac{g\left(x\right)–g\left(a\right)}{x–a}=\underset{h\to 0}{\mathrm{lim}}\frac{{\int }_{a}^{a+h}f\left(t\right)dt–0}{h}=\underset{h\to 0}{\mathrm{lim}}f\left(a+h\right)=f\left(a\right)\phantom{\rule{0ex}{0ex}}$

So, g(x) is not differentiable at x = a.
Now,
LHD at x = b

RHD at x = b
=

$\underset{x\to {b}^{+}}{\mathrm{lim}}\frac{g\left(x\right)–g\left(b\right)}{x–b}=\underset{h\to 0}{\mathrm{lim}}\frac{{\int }_{a}^{b}f\left(t\right)dt–{\int }_{a}^{b}f\left(t\right)dt}{h}=0\phantom{\rule{0ex}{0ex}}$

So, g(x) is not differentiable at x = b.

Hence, the correct options are A and C.

#### Question 51:

Let

be three non-coplanar unit vectors such that the angle between every pair of them is

where p,q are r are scalars, then the value of

$\frac{{p}^{2}+2{q}^{2}+{r}^{2}}{{q}^{2}}$

is

Solutions:If

and

$\stackrel{\to }{c}$

be three non-coplanar unit vectors and the angle between each pair is

$\frac{\mathrm{\pi }}{3}$

, then

and

$\stackrel{\to }{a}.\stackrel{\to }{b}=\stackrel{\to }{b}.\stackrel{\to }{c}=\stackrel{\to }{c}.\stackrel{\to }{a}=\frac{1}{2}$

Now,

=

$\frac{1}{2}$

Now,

$\stackrel{\to }{a}×\stackrel{\to }{b}+\stackrel{\to }{b}×\stackrel{\to }{c}=p\stackrel{\to }{a}+q\stackrel{\to }{b}+r\stackrel{\to }{c}$

Taking dot product with

and

$\stackrel{\to }{c}$

, we get:

Solving equations (i), (ii) and (iii), we get:

Then,

$\frac{{p}^{2}+2{q}^{2}+{r}^{2}}{{q}^{2}}=4$

.

#### Question 52:

Let

be defined by f(x) = cos−1 (cos x). The number of points x

$\left[0,4\pi \right]$

satisfying the equation

$f\left(x\right)=\frac{10–x}{10}$

is

Solutions:The graph of

$f\left(x\right)={\mathrm{cos}}^{–1}\left(\mathrm{cos}x\right)$

and

$f\left(x\right)=\frac{10–x}{10}$

intersects at three points.

Thus, the number of points x

$\in$

[0, 4π] satisfying the equation

$f\left(x\right)=\frac{10–x}{10}$

is 3.

#### Question 53:

For a point P in the plane, let d1(P) and d2(P) be the distances of the point P from the lines xy = 0 and x + y = 0, respectively. The area of the region R, consisting of all points P lying in the first quadrant of the plane and satisfying

$2\le {d}_{1}\left(P\right)+{d}_{2}\left(P\right)\le 4,$

is

Solutions:

Let

$P\left(\alpha ,\beta \right)$

be a point in the first quadrant.

and

$⇒$

$2⩽\left|\frac{\alpha –\beta }{\sqrt{2}}\right|+\left|\frac{\alpha +\beta }{\sqrt{2}}\right|⩽4$

Case 1: Let

$\alpha >\beta$ $⇒$ $⇒$

Case 2: Let

$\beta >\alpha$ $⇒$ $⇒$

So, the required region is as shown below:

$=\left\{{\left(2\sqrt{2}\right)}^{2}–{\left(\sqrt{2}\right)}^{2}\right\}$

sq. units
= (8 − 2) sq. units
= 6 sq. units

Hence, the required area is 6 sq. units.

#### Question 54:

The largest value of the non-negative integer a for which

$\underset{x\to 1}{\mathrm{lim}}{\left\{\frac{–ax+\mathrm{sin}\left(x–1\right)+a}{x+\mathrm{sin}\left(x–1\right)–1}\right\}}^{\frac{1–x}{1–\sqrt{x}}}=\frac{1}{4}$

Solutions:Given:

$\underset{x\to 1}{\mathrm{lim}}{\left\{\frac{–ax+\mathrm{sin}\left(x–1\right)+a}{x+\mathrm{sin}\left(x–1\right)–1}\right\}}^{\frac{1–x}{1–\sqrt{x}}}=\frac{1}{4}$

….(1)

Now,

$\underset{x\to 1}{\mathrm{lim}}{\left\{\frac{–ax+\mathrm{sin}\left(x–1\right)+a}{x+\mathrm{sin}\left(x–1\right)–1}\right\}}^{\frac{1–x}{1–\sqrt{x}}}$

=

$\underset{x\to 1}{\mathrm{lim}}{\left\{\frac{–ax+\mathrm{sin}\left(x–1\right)+a}{x+\mathrm{sin}\left(x–1\right)–1}\right\}}^{\frac{\left(1–\sqrt{x}\right)\left(1+\sqrt{x}\right)}{1–\sqrt{x}}}$

=

$\underset{x\to 1}{\mathrm{lim}}{\left\{\frac{–a\left(x–1\right)+\mathrm{sin}\left(x–1\right)}{x–1+\mathrm{sin}\left(x–1\right)}\right\}}^{\left(1+\sqrt{x}\right)}$

Putting x = 1 + h

=

$\underset{h\to 0}{\mathrm{lim}}{\left\{\frac{–ah+\mathrm{sin}h}{h+\mathrm{sin}h}\right\}}^{\left(1+\sqrt{1+h}\right)}$

=

$\underset{h\to 0}{\mathrm{lim}}{\left\{\frac{h\left(–a+\frac{\mathrm{sin}h}{h}\right)}{h\left(1+\frac{\mathrm{sin}h}{h}\right)}\right\}}^{\left(1+\sqrt{1+h}\right)}$

=

${\left\{\frac{\left(–a+1\right)}{2}\right\}}^{2}$

….(2)

Putting the value of equation (2) in equation (1), we get:

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

Hence, the value of a is 2.

#### Question 55:

Let

$n\ge 2$

be an integer. Take n distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then the value of n is

Solutions:

It is given that n ≥ 2

Thus, total number of lines = nC2.

Also, number of adjacent lines (blue line segments) = n.

Thus, number of non-adjacent lines (red line segments) = nC2n .

It is given that blue line segments = red line segments.

nC2n = n

nC2 = 2n

Thus, n = 5 (âˆµ n ≥ 2) .

Thus, the value of n is 5.

#### Question 56:

Let a, b and c be positive integers, such that

$\frac{b}{a}$

is an integer. If a, b and c are in geometric progression and the arithmetic mean of a, b and c is b + 2, then the value of

$\frac{{a}^{2}+a–14}{a+1}$

is

Solutions:It is given that a, b and c are in geometric progression and a, b, c

$\in$

Z.

Let a = a, b = ar and c = ar2.

Since

Thus, r is an integer.

Now,

A.M. of a, b and c = b + 2.

$⇒\frac{a+b+c}{3}=b+2\phantom{\rule{0ex}{0ex}}⇒\frac{a+ar+a{r}^{2}}{3}=ar+2\phantom{\rule{0ex}{0ex}}⇒a+ar+a{r}^{2}=3ar+6\phantom{\rule{0ex}{0ex}}⇒a–2ar+a{r}^{2}=6\phantom{\rule{0ex}{0ex}}⇒a{\left(1–r\right)}^{2}=6\phantom{\rule{0ex}{0ex}}⇒a=\frac{6}{{\left(1–r\right)}^{2}}=\mathrm{integer}$

Since a and r are integers, the values of a and r must be 6 and 2, respectively.

Now,

$\frac{{a}^{2}+a–14}{a+1}=\frac{{\left(6\right)}^{2}+6–14}{6+1}=\frac{36+6–14}{7}=\frac{28}{7}=4$

Hence, the value of

$\frac{{a}^{2}+a–14}{a+1}$

is 4.

#### Question 57:

Let n1 < n2 < n3 < n4 < n5 be positive integers, such that n1 + n2 + n3 + n4 + n5 = 20. Then the number of such distinct arrangements (n1, n2, n3, n4, n5) is

Solutions:Since

and

${n}_{1}+{n}_{2}+{n}_{3}+{n}_{4}+{n}_{5}=20$

, the possible arrangements are as shown below:

 n5 n1 n2 n3 n4 10 1 2 3 4 9 1 2 3 5 8 1 2 3 6 8 1 2 4 5 7 1 2 4 6 7 1 3 4 5 6 2 3 4 5

Hence, the total number of distinct arrangements is 7.

#### Question 58:

Let

be respectively given by f(x) = |x| + 1 and g(x) = x2 + 1. Define

$h:\mathrm{ℝ}\to \mathrm{ℝ}$

by

The number of points at which h(x) is not differentiable is

Solutions:Given:
f(x) =

and g(x) = x2 + 1

The graph of f(x) =

$\left|x\right|+1$

is as shown:

The graph of g(x) = x2 + 1 is as shown:

Combined graph of f(x) and g(x) is as shown:

h(x) =

It can be observed that there are three non-differentiable points.
Hence, the number of points at which h(x) is not differentiable is 3.

#### Question 59:

The slope of the tangent to the curve (yx5)2 = x (1 + x2)2 at the point (1, 3) is

Solutions:Given: (y − x5)2 = x(1 + x2)2

Differentiating both sides, we get:

$2\left(y–{x}^{5}\right)\left(\frac{dy}{dx}–5{x}^{4}\right)={\left(1+{x}^{2}\right)}^{2}+2x\left(1+{x}^{2}\right).2x$

At point (1, 3),

$2\left(3–1\right)\left(\frac{dy}{dx}–5\left(1\right)\right)={\left(1+1\right)}^{2}+2\left(1\right)\left(1+1\right).2\left(1\right)\phantom{\rule{0ex}{0ex}}⇒4\left(\frac{dy}{dx}–5\right)=4+8\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}–5=3\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=8$

Hence, the slope of the tangent to the curve (y − x5)2 = x(1 + x2)2 at point (1, 3) is 8.

#### Question 60:

The value of

is

${\int }_{0}^{1}4{x}^{3}\left\{\frac{{\mathrm{d}}^{2}}{d{x}^{2}}{\left(1–{x}^{2}\right)}^{5}\right\}dx$
${\int }_{0}^{1}4{x}^{3}\left\{\frac{{\mathrm{d}}^{2}}{d{x}^{2}}{\left(1–{x}^{2}\right)}^{5}\right\}dx$