JEE Main 2024 1st February Evening Shift Question Paper Question Paper with Solutions | JEE Main Mock Test | JeeMate

Q 1PhysicsSection A • MCQelectrostatics

+4 / -1

C_1

and

C_2

are two hollow concentric cubes enclosing charges

2 Q

and

3 Q

respectively as shown in figure. The ratio of electric flux passing through

C_1

and

C_2

is :

qImage-electric-flux-and-gauss-law-electrostatics-electricity-Jeemate

A

3: 2

B

5: 2

C

2: 5

D

2: 3

Q 2PhysicsSection B • Numericalatoms and nuclei

+4 / -1

A particular hydrogen-like ion emits the radiation of frequency

3 \times 10^{15} \mathrm{~Hz}

when it makes transition from

n=2

to

n=1

. The frequency of radiation emitted in transition from

n=3

to

n=1

is

\frac{x}{9} \times 10^{15} \mathrm{~Hz}

, when

x=

________ .

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:32

Step-by-Step Solution:

The emission frequency of radiation from a hydrogen-like ion during electron transitions can be understood using the formula derived from Rydberg's equation for hydrogen-like atoms, which is given as:

$E = \frac{E_0}{h} \left( \frac{1}{n^2_1} - \frac{1}{n^2_2} \right)$

where:

$E$ is the energy of the emitted photon,
$E_0$ is the Rydberg constant for hydrogen,
$h$ is Planck's constant,
$n_1$ and $n_2$ are the principal quantum numbers for the initial and final energy levels, respectively, with $n_2 < n_1$ .

Given a transition from $n=2$ to $n=1$ emits radiation with a frequency of $3 \times 10^{15} \mathrm{~Hz}$ , we can write:

$\nu_1 = 3 \times 10^{15} \mathrm{~Hz}$

For the transition from $n=3$ to $n=1$ , we can use the same principle to find the frequency of the emitted radiation, $\nu_2$ , which will depend on the difference in energy levels involved in the transition. The frequency is directly proportional to this energy difference, so we can compare the two transitions using their respective frequencies:

$\frac{\nu_2}{\nu_1} = \frac{\left( \frac{1}{n_{1f}^2} - \frac{1}{n_{1i}^2} \right)}{\left( \frac{1}{n_{2f}^2} - \frac{1}{n_{2i}^2} \right)} = \frac{\left( \frac{1}{1^2} - \frac{1}{3^2} \right)}{\left( \frac{1}{1^2} - \frac{1}{2^2} \right)} = \frac{\left(1 - \frac{1}{9}\right)}{\left(1 - \frac{1}{4}\right)}$

After simplification:

$\frac{\nu_2}{3 \times 10^{15}} = \frac{\left(\frac{8}{9}\right)}{\left(\frac{3}{4}\right)} = \frac{32}{27}$

Thus, to find the frequency $ u_2 $for the transition from$ n=3 $to$ n=1 $:<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>ν</mi><mn>2</mn></msub><mo>=</mo><mfrac><mn>32</mn><mn>9</mn></mfrac><mo>×</mo><msup><mn>10</mn><mn>15</mn></msup><mrow><mi mathvariant="normal">H</mi><mi mathvariant="normal">z</mi></mrow></mrow><annotation encoding="application/x-tex">\nu_2 = \frac{32}{9} \times 10^{15} \mathrm{Hz}</annotation></semantics></math>ν2=932×1015HzHence, in the given formula$ \frac{x}{9} \times 10^{15} \mathrm{~Hz} $for the frequency,$ x $is equal to$ 32$.

Q 3PhysicsSection A • MCQheat and thermodynamics

+4 / -1

If the root mean square velocity of hydrogen molecule at a given temperature and pressure is

2 \mathrm{~km} / \mathrm{s}

, the root mean square velocity of oxygen at the same condition in

\mathrm{km} / \mathrm{s}

is :

A

1.0

B

1.5

C

2.0

D

0.5

Official Correct Answer:Option (D)

Step-by-Step Solution:

Here is your text with LaTeX notation and paragraph tags converted as requested:

To calculate the root mean square (rms) velocity of gas molecules, we can use the formula:

$v_{\text{rms}} = \sqrt{\frac{3kT}{m}}$

where:

$v_{\text{rms}}$ is the root mean square velocity of the gas molecules,
$k$ is the Boltzmann constant,
$T$ is the absolute temperature in Kelvin, and
$m$ is the mass of one molecule of the gas.

Since the temperature and pressure are the same for hydrogen and oxygen, we can ignore the constant and temperature parts of the equation because they will cancel out in the comparison between the two gases.

Now, we need to compare the mass of one molecule of hydrogen to that of one molecule of oxygen. The molecular mass of hydrogen (H₂) is approximately 2 g/mol, while the molecular mass of oxygen (O₂) is approximately 32 g/mol.

We know the rms speed of hydrogen is 2 km/s, so let's find the mass ratio and then determine the speed of oxygen molecules. Using the rms velocity formula and considering the ratio of masses:

$\frac{v_{\text{rms, H₂}}}{v_{\text{rms, O₂}}} = \sqrt{\frac{m_{\text{O₂}}}{m_{\text{H₂}}}}$

We know the mass m is proportional to the molecular weight for each gas, so we can substitute:

$\frac{v_{\text{rms, H₂}}}{v_{\text{rms, O₂}}} = \sqrt{\frac{M_{\text{O₂}}}{M_{\text{H₂}}}}$

Now we plug in the values:

$\frac{2 \text{ km/s}}{v_{\text{rms, O₂}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$

Solving for $v_{\text{rms, O₂}}$ :

$v_{\text{rms, O₂}} = \frac{2 \text{ km/s}}{4} = 0.5 \text{ km/s}$

Therefore, the rms velocity of oxygen molecules at the same temperature and pressure conditions as that of hydrogen with a rms velocity of 2 km/s is 0.5 km/s.

The correct answer is Option D : 0.5.

Q 4PhysicsSection A • MCQdual nature of radiation

+4 / -1

Conductivity of a photodiode starts changing only if the wavelength of incident light is less than

660 \mathrm{~nm}

. The band gap of photodiode is found to be

\left(\frac{\mathrm{X}}{8}\right) \mathrm{eV}

. The value of

\mathrm{X}

is :

(Given,

\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}

)

A

11

B

13

C

15

D

21

Official Correct Answer:Option (C)

Step-by-Step Solution:

To find the value of $X$ in the band gap $\left(\frac{\mathrm{X}}{8}\right) \mathrm{eV}$ , we need to understand the relationship between the wavelength of light that can result in changes in the conductivity of a photodiode and the photodiode's band gap energy.

The band gap energy ( $E_{g}$ ) of a material is the minimum energy required for an electron to transition from the valence band to the conduction band, thus creating a hole-electron pair and allowing conductivity to occur. When light with a certain wavelength ( $\lambda$ ) is incident upon the photodiode, if the energy of the photons is greater than or equal to the band gap energy of the photodiode, then the photons can excite electrons and change the conductivity of the photodiode.

The energy of a photon ( $E_{photon}$ ) is given by the equation:

$E_{photon} = \frac{hc}{\lambda}$

where:

$h$ is Planck's constant, which is given as $6.6 \times 10^{-34}$ J·s.
$c$ is the speed of light in a vacuum ( $\approx 3 \times 10^{8}$ m/s).
$\lambda$ is the wavelength of the incident light.

Then the band gap energy in terms of electron volts is found by converting the energy from joules to electron volts (eV) using the charge of an electron $e$ ( $1.6 \times 10^{-19}$ C):

$E_{g} = \frac{hc}{\lambda e}$

Given that the photodiode starts conducting when the wavelength of light is less than $660 \mathrm{~nm}$ , we should use this wavelength as the threshold wavelength $\lambda_{threshold}$ :

$E_{g} = \frac{hc}{\lambda_{threshold}\times e}$

$E_{g} = \frac{6.6 \times 10^{-34} \text{ J·s} \cdot 3 \times 10^{8} \text{ m/s}}{660 \times 10^{-9} \text{ m} \times 1.6 \times 10^{-19} \text{ C}}$

$E_{g} = \frac{6.6 \times 3 \times 10^{-26}}{660 \times 1.6} \mathrm{eV}$

$E_{g} = \frac{19.8}{660 \times 1.6} \mathrm{eV}$

$E_{g} = \frac{19.8}{1056} \mathrm{eV}$

$E_{g} = \frac{18.75}{1000} \mathrm{eV}$

$E_{g} = 1.875 \mathrm{eV}$

Now, we have to compare this value to $\left(\frac{\mathrm{X}}{8}\right) \mathrm{eV}$ to find the value of $X$ :

$1.875 = \frac{X}{8}$

$X = 1.875 \times 8$

$X = 15$

Therefore, the value of $X$ is 15. The correct answer is Option C.

Q 5PhysicsSection A • MCQelectronic devices

+4 / -1

To measure the temperature coefficient of resistivity

\alpha

of a semiconductor, an electrical arrangement shown in the figure is prepared. The arm BC is made up of the semiconductor. The experiment is being conducted at

25^{\circ} \mathrm{C}

and resistance of the semiconductor arm is

3 \mathrm{~m} \Omega

. Arm

\mathrm{BC}

is cooled at a constant rate of

2^{\circ} \mathrm{C} / \mathrm{s}

. If the galvanometer

\mathrm{G}

shows no deflection after

10 \mathrm{~s}

, then

\alpha

is :

qImage-semiconductor-and-p-n-junction-diode-electronic-devices-modern-physics-Jeemate

A

-1 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}

B

-2 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}

C

-2.5 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}

D

-1.5 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}

Q 6PhysicsSection B • Numericalproperties of matter

+4 / -1

One end of a metal wire is fixed to a ceiling and a load of

2 \mathrm{~kg}

hangs from the other end. A similar wire is attached to the bottom of the load and another load of

1 \mathrm{~kg}

hangs from this lower wire. Then the ratio of longitudinal strain of upper wire to that of the lower wire will be ________.

[Area of cross section of wire

=0.005 \mathrm{~cm}^2, \mathrm{Y}=2 \times 10^{11} \mathrm{Nm}^{-2}

and

\mathrm{g}=10 \mathrm{~ms}^{-2}

]

Answer Format:Enter numerical value rounded to the nearest integer

Q 7PhysicsSection A • MCQrotational motion

+4 / -1

A disc of radius

\mathrm{R}

and mass

\mathrm{M}

is rolling horizontally without slipping with speed

v

. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is :

qImage-combined-translational-and-rotational-motion-rotational-motion-mechanics-Jeemate

qImage-combined-translational-and-rotational-motion-rotational-motion-mechanics-Jeemate

A

\frac{3}{4} \frac{v^2}{\mathrm{~g}}

B

\frac{v^2}{g}

C

\frac{2}{3} \frac{v^2}{\mathrm{~g}}

D

\frac{1}{2} \frac{v^2}{\mathrm{~g}}

Q 8PhysicsSection B • Numericalsimple harmonic motion

+4 / -1

A mass

m

is suspended from a spring of negligible mass and the system oscillates with a frequency

f_1

. The frequency of oscillations if a mass

9 \mathrm{~m}

is suspended from the same spring is

f_2

. The value of

\frac{f_1}{f_2} \mathrm{i}

________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 9PhysicsSection A • MCQdual nature of radiation

+4 / -1

Monochromatic light of frequency

6 \times 10^{14} \mathrm{~Hz}

is produced by a laser. The power emitted is

2 \times 10^{-3} \mathrm{~W}

.

How many photons per second on an average, are emitted by the source ?

(Given

\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}

)

A

5 \times 10^{15}

B

7 \times 10^{16}

C

6 \times 10^{15}

D

9 \times 10^{18}

Q 10PhysicsSection A • MCQheat and thermodynamics

+4 / -1

A diatomic gas

(\gamma=1.4)

does

200 \mathrm{~J}

of work when it is expanded isobarically. The heat given to the gas in the process is :

A

800 \mathrm{~J}

B

600 \mathrm{~J}

C

700 \mathrm{~J}

D

850 \mathrm{~J}

Q 11PhysicsSection A • MCQelectromagnetic waves

+4 / -1

If frequency of electromagnetic wave is

60 \mathrm{~MHz}

and it travels in air along

z

direction then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other and the wavelength of the wave (in

\mathrm{m}

) is :

A

2.5

B

5

C

10

D

2

Q 12PhysicsSection A • MCQwave optics

+4 / -1

A microwave of wavelength

2.0 \mathrm{~cm}

falls normally on a slit of width

4.0 \mathrm{~cm}

. The angular spread of the central maxima of the diffraction pattern obtained on a screen

1.5 \mathrm{~m}

away from the slit, will be :

A

60^{\circ}

B

45^{\circ}

C

15^{\circ}

D

30^{\circ}

Q 13PhysicsSection A • MCQmotion

+4 / -1

Train A is moving along two parallel rail tracks towards north with speed $72 \mathrm{~km} / \mathrm{h}$ and train B is moving towards south with speed $108 \mathrm{~km} / \mathrm{h}$ . Velocity of train B with respect to A and velocity of ground with respect to B are (in $\mathrm{ms}^{-1}$ ):

A

-50 and -30

B

-50 and 30

C

-30 and 50

D

50 and -30

Official Correct Answer:Option (B)

Step-by-Step Solution:

To find the velocity of Train B with respect to Train A, we have to subtract the velocity of Train A from the velocity of Train B, keeping in mind that they are moving in opposite directions. Since they are moving in opposite directions, the relative velocity is calculated by adding their magnitudes when converting into the same unit, which in this case is meters per second.

First, let's convert the speeds from km/h to m/s by multiplying by the conversion factor $\frac{1000 \text{ m/km}}{3600 \text{ s/h}} = \frac{5}{18} \text{ m/s}$ .

For Train A:

v_A = 72 \frac{\text{km}}{\text{h}} \times \frac{5}{18} \frac{\text{m/s}}{(\text{km/h})} = 20 \frac{\text{m}}{\text{s}}

For Train B:

v_B = 108 \frac{\text{km}}{\text{h}} \times \frac{5}{18} \frac{\text{m/s}}{(\text{km/h})} = 30 \frac{\text{m}}{\text{s}}

To find the velocity of B relative to A ( $v_{B/A}$ ), we consider the direction: Train B is moving towards the south and Train A is moving towards the north. Therefore, relative to Train A, Train B is moving even faster towards the south, we calculate:

v_{B/A} = v_B + v_A = 30 \frac{\text{m}}{\text{s}} + 20 \frac{\text{m}}{\text{s}} = 50 \frac{\text{m}}{\text{s}}

Since Train B is moving towards the south and Train A towards the north, we take the southward direction as negative in our coordinate system for this calculation. That means the velocity of B with respect to A is:

v_{B/A} = -50 \frac{\text{m}}{\text{s}}

Next, we calculate the velocity of the ground with respect to Train B ( $v_{ground/B}$ ). The ground is stationary, thus it has a velocity of 0 m/s in any direction. The velocity of an object with respect to another object moving is just the opposite of the second object's velocity. Thus:

v_{ground/B} = -v_B = -30 \frac{\text{m}}{\text{s}}

However, because we are considering the southward direction as negative, the negative of a southward velocity is a northward velocity. Hence we get:

v_{ground/B} = 30 \frac{\text{m}}{\text{s}}

So the velocity of Train B with respect to Train A is -50 m/s, and the velocity of the ground with respect to Train B is 30 m/s. Therefore, the correct answer is:

Option A: -50 m/s and -30 m/s. (Incorrect, because the velocity of ground with respect to B is positive in our chosen coordinate system)

Option B: -50 m/s and 30 m/s. (Correct)

Option C: -30 m/s and 50 m/s. (Incorrect)

Option D: 50 m/s and -30 m/s. (Incorrect)

Thus, the correct answer is Option B: -50 m/s and 30 m/s.

Q 14PhysicsSection A • MCQcurrent electricity

+4 / -1

In an ammeter,

5 \%

of the main current passes through the galvanometer. If resistance of the galvanometer is

\mathrm{G}

, the resistance of ammeter will be :

A

199 \mathrm{~G}

B

200 \mathrm{~G}

C

\frac{G}{20}

D

\frac{\mathrm{G}}{199}

Official Correct Answer:Option (C)

Step-by-Step Solution:

When an ammeter is designed, a shunt resistor ( $R_s$ ) is placed in parallel with the galvanometer to ensure that only a small fraction of the total current passes through the galvanometer itself. This is done because the galvanometer is usually a sensitive instrument designed to measure small currents, and passing a large current through it could damage it.

In this scenario, we are told that $5\%$ of the main current passes through the galvanometer. This means that the remaining $95\%$ of the current must pass through the shunt. Let's denote the total current as $I$ , the current through the galvanometer as $I_g$ , and the current through the shunt as $I_s$ . Therefore, we have:

$I_g = \frac{5}{100} I$

$I_s = I - I_g = I - \frac{5}{100} I = \frac{95}{100} I$

Since the galvanometer and shunt are in parallel, the voltage across each must be the same:

$V_g = V_s$

According to Ohm's law, $V = IR$ , where $V$ is the voltage, $I$ is the current, and $R$ is the resistance. Hence, for the galvanometer and the shunt:

$I_g G = I_s R_s$

By substituting $I_g$ and $I_s$ from the above proportionality, we get:

$\left(\frac{5}{100} I\right) G = \left(\frac{95}{100} I\right) R_s$

Let's solve for $R_s$ :

$R_s = \frac{5}{95} G$

Further simplifying this:

$R_s = \frac{G}{19}$

The total resistance of the ammeter $R_a$ can be found using the parallel resistance formula:

$\frac{1}{R_a} = \frac{1}{G} + \frac{1}{R_s}$

Substitute $R_s$ with $\frac{G}{19}$ :

$\frac{1}{R_a} = \frac{1}{G} + \frac{19}{G}$

$\frac{1}{R_a} = \frac{20}{G}$

Thus the resistance of the ammeter $R_a$ is:

$R_a = \frac{G}{20}$

Hence, the correct answer is Option C: $\frac{G}{20}$ .

Q 15PhysicsSection B • Numericalmagnetics

+4 / -1

A moving coil galvanometer has 100 turns and each turn has an area of

2.0 \mathrm{~cm}^2

. The magnetic field produced by the magnet is

0.01 \mathrm{~T}

and the deflection in the coil is 0.05 radian when a current of

10 \mathrm{~mA}

is passed through it. The torsional constant of the suspension wire is

x \times 10^{-5} \mathrm{~N}-\mathrm{m} / \mathrm{rad}

. The value of

x

is _______ .

Answer Format:Enter numerical value rounded to the nearest integer

Q 16PhysicsSection B • Numericalmotion

+4 / -1

A particle initially at rest starts moving from reference point

x=0

along

x

-axis, with velocity

v

that varies as

v=4 \sqrt{x} \mathrm{~m} / \mathrm{s}

. The acceleration of the particle is __________

\mathrm{ms}^{-2}

.

Answer Format:Enter numerical value rounded to the nearest integer

Q 17PhysicsSection A • MCQunits and measurements

+4 / -1

Match List - I with List - II.

List I (Number)	List II (Significant figure)
(A) 1001	(I) 3
(B) 010.1	(II) 4
(C) 100.100	(III) 5
(D) 0.0010010	(IV) 6

Choose the correct answer from the options given below :

A

(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

B

(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

C

(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

D

(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Q 18PhysicsSection A • MCQlaws of motion

+4 / -1

A body of mass

4 \mathrm{~kg}

experiences two forces

\vec{F}_1=5 \hat{i}+8 \hat{j}+7 \hat{k}

and

\overrightarrow{\mathrm{F}}_2=3 \hat{i}-4 \hat{j}-3 \hat{k}

. The acceleration acting on the body is :

A

2 \hat{i}+\hat{j}+\hat{k}

B

4 \hat{i}+2 \hat{j}+2 \hat{k}

C

-2 \hat{i}-\hat{j}-\hat{k}

D

2 \hat{i}+3 \hat{j}+3 \hat{k}

Q 19PhysicsSection A • MCQatoms and nuclei

+4 / -1

From the statements given below :

(A) The angular momentum of an electron in

n^{\text {th }}

orbit is an integral multiple of

\hbar

.

(B) Nuclear forces do not obey inverse square law.

(C) Nuclear forces are spin dependent.

(D) Nuclear forces are central and charge independent.

(E) Stability of nucleus is inversely proportional to the value of packing fraction.

Choose the correct answer from the options given below :

A

(B), (C), (D), (E) only

B

(A), (C), (D), (E) only

C

(A), (B), (C), (E) only

D

(A), (B), (C), (D) only

Official Correct Answer:Option (C)

Step-by-Step Solution:

Let's analyze each of the given statements to determine which ones are correct:

(A) The angular momentum of an electron in $n^{\text{th}}$ orbit is an integral multiple of $\hbar$ .

This statement reflects Bohr's quantization rule for angular momentum in the Bohr model of the hydrogen atom. According to this rule, the angular momentum of an electron in a stationary orbit is quantized and given by:

L = n\hbar

where $n$ is a principal quantum number (which can be any positive integer), and $\hbar$ is the reduced Planck's constant. Therefore, this statement (A) is correct.

(B) Nuclear forces do not obey inverse square law.

Nuclear forces, specifically strong nuclear forces, do act over short ranges within the nucleus but do not obey the inverse square law, which is characteristic of the electromagnetic and gravitational forces. Thus, statement (B) is correct.

(C) Nuclear forces are spin dependent.

The strength of the nuclear force can depend on the spin alignment of the nucleons. This is why some isotopes are more stable than others depending on spin-related factors. Thus, statement (C) is correct.

(D) Nuclear forces are central and charge independent.

The strong nuclear force is indeed charge independent, meaning it is the same regardless of the types of nucleons involved (neutrons or protons). However, nuclear forces are not always central; they can be tensor forces too, which involve more complex interactions that are not purely central. Therefore, the entirety of statement (D) is not correct; the statement should specify that nuclear forces are charge independent but may not always be central.

(E) Stability of nucleus is inversely proportional to the value of packing fraction.

This statement (E) is correct.

Based on the above explanations, the correct statements are (A), (B), (C) and (E). Hence, the correct answer is:

\text{Option C : (A), (B), (C), (D) only}

Q 20PhysicsSection A • MCQcapacitor

+4 / -1

A galvanometer

(G)

of

2 \Omega

resistance is connected in the given circuit. The ratio of charge stored in

C_1

and

C_2

is :

qImage-combination-of-capacitors-capacitor-electricity-Jeemate

A

1

B

\frac{2}{3}

C

\frac{3}{2}

D

\frac{1}{2}

Q 21PhysicsSection B • Numericalcapacitor

+4 / -1

In an electrical circuit drawn below the amount of charge stored in the capacitor is _______

\mu

C.

qImage-capacitance-capacitor-electricity-Jeemate

Answer Format:Enter numerical value rounded to the nearest integer

Q 22PhysicsSection A • MCQlaws of motion

+4 / -1

A cricket player catches a ball of mass

120 \mathrm{~g}

moving with

25 \mathrm{~m} / \mathrm{s}

speed. If the catching process is completed in

0.1 \mathrm{~s}

then the magnitude of force exerted by the ball on the hand of player will be (in SI unit) :

A

30

B

24

C

12

D

25

Official Correct Answer:Option (A)

Step-by-Step Solution:

The first step in solving this problem is to calculate the change in momentum of the ball when it is caught. The change in momentum, or impulse, is the product of the mass of the ball and the change in velocity (as momentum is mass times velocity).

The ball is initially moving with a velocity of $v_i = 25 \mathrm{~m/s}$ before the catch and finally comes to rest with a velocity of $v_f = 0 \mathrm{~m/s}$ after the catch. Since the ball is caught, the final velocity is zero. The change in velocity $\Delta v = v_f - v_i = 0 - 25 = -25 \mathrm{~m/s}.$ Remember that the direction of the force exerted by the ball on the hand will be opposite to the direction of the ball's initial motion.

The mass of the ball $m$ is given as $120 \mathrm{~g}$ which needs to be converted into kilograms to maintain SI units: $m = 120 \mathrm{~g} = 120 \times 10^{-3} \mathrm{~kg} = 0.12 \mathrm{~kg}.$

Now we can calculate the change in momentum (impulse): $\Delta p = m \Delta v = 0.12 \mathrm{~kg} \times (-25 \mathrm{~m/s}).$

Substituting the values we get: $\Delta p = 0.12 \times -25 = -3 \mathrm{~kg \cdot m/s}.$

The negative sign indicates that the change in momentum is in the opposite direction of the ball's initial motion, which makes sense because the ball's velocity is reduced to zero.

The magnitude of the impulse is independent of the sign and is $3 \mathrm{~kg \cdot m/s}$ .

Impulse is also equal to the average force exerted on the ball times the time interval during which the force is exerted. We can use the formula: $\Delta p = F_{avg} \Delta t$

Where $F_{avg}$ is the average force and $\Delta t$ is the time interval of $0.1 \mathrm{~s}$ . Re-arranging the formula to solve for $F_{avg}$ gives us: $F_{avg} = \frac{\Delta p}{\Delta t}.$

Substituting the known values we have: $F_{avg} = \frac{3}{0.1} = 30 \mathrm{~N}.$

The magnitude of the average force exerted by the hand of the player to catch the ball is $30 \mathrm{~N}$ .

Q 23PhysicsSection B • Numericalrotational motion

+4 / -1

A uniform rod

A B

of mass

2 \mathrm{~kg}

and length

30 \mathrm{~cm}

at rest on a smooth horizontal surface. An impulse of force

0.2 \mathrm{~Ns}

is applied to end B. The time taken by the rod to turn through at right angles will be

\frac{\pi}{x} \mathrm{~s}

, where

x=

_______ .

Answer Format:Enter numerical value rounded to the nearest integer

Q 24PhysicsSection A • MCQgravitation

+4 / -1

A light planet is revolving around a massive star in a circular orbit of radius

\mathrm{R}

with a period of revolution T. If the force of attraction between planet and star is proportional to

\mathrm{R}^{-3 / 2}

then choose the correct option :

A

\mathrm{T}^2 \propto \mathrm{R}^{7 / 2}

B

\mathrm{T}^2 \propto \mathrm{R}^3

C

\mathrm{T}^2 \propto \mathrm{R}^{5 / 2}

D

\mathrm{T}^2 \propto \mathrm{R}^{3 / 2}

Q 25PhysicsSection B • Numericalelectromagnetic induction

+4 / -1

A coil of 200 turns and area

0.20 \mathrm{~m}^2

is rotated at half a revolution per second and is placed in uniform magnetic field of

0.01 \mathrm{~T}

perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is

\frac{2 \pi}{\beta}

volt. The value of

\beta

is _______.

Answer Format:Enter numerical value rounded to the nearest integer

Q 26PhysicsSection A • MCQcurrent electricity

+4 / -1

In a metre-bridge when a resistance in the left gap is

2 \Omega

and unknown resistance in the right gap, the balance length is found to be

40 \mathrm{~cm}

. On shunting the unknown resistance with

2 \Omega

, the balance length changes by :

A

62.5

B

22.5 \mathrm{~cm}

C

20 \mathrm{~cm}

D

65 \mathrm{~cm}

Official Correct Answer:Option (B)

Step-by-Step Solution:

To solve this problem, let's first understand that a meter bridge setup is based on the principle of a Wheatstone bridge, in which two unknown resistances are in such a configuration that if the bridge is balanced, the ratio of the resistances on one side is equal to the ratio of resistances on the other side. In a balanced condition, no current flows through the galvanometer that is connected diagonally across the bridge.

We can express the condition of the balance as follows:

$\frac{R_1}{R_2} = \frac{L_1}{L_2}$

where $R_1$ is the known resistance $2 Ω$ , $R_2$ is the unknown resistance, $L_1$ is the balance length $40 cm$ and $L_2$ is the length of the remaining wire on the meter bridge (100 cm - 40 cm = 60 cm).

Let's calculate the initial unknown resistance ( $R_2$ ) using the balance condition:

$\frac{2}{R_2} = \frac{40}{60}$

$\Rightarrow R_2 = \frac{2 \times 60}{40} = 3 \Omega$

Now, when the unknown resistance $R_2$ is shunted with a 2 Ω resistor, the new combined resistance ( $R'_2$ ) can be calculated using the parallel resistance formula:

$\frac{1}{R'_2} = \frac{1}{R_2} + \frac{1}{2}$

$\Rightarrow \frac{1}{R'_2} = \frac{1}{3} + \frac{1}{2} = \frac{2 + 3}{6} = \frac{5}{6}$

Therefore, the new combined resistance ( $R'_2$ ) is:

$R'_2 = \frac{6}{5} \Omega$

If $L'_1$ is the new balance length and $L'_2$ is the remaining length, we now have:

$\frac{2}{\frac{6}{5}} = \frac{L'_1}{100 - L'_1}$

$\Rightarrow \frac{L'_1}{100 - L'_1} = \frac{5}{3}$

Now, solve for (L'_1):

$3L'_1 = 5(100 - L'_1)$

$\Rightarrow 3L'_1 = 500 - 5L'_1$

$\Rightarrow 8L'_1 = 500$

$\Rightarrow L'_1 = 62.5 \mathrm{~cm}$

The balance length has changed from 40 cm to 62.5 cm, so the change by $L'_1 - L_1$ is:

$62.5 - 40 = 22.5 \mathrm{~cm}$

Therefore, the balance length changes by 22.5 cm, which corresponds to Option B.

Q 27PhysicsSection A • MCQelectromagnetic induction

+4 / -1

A transformer has an efficiency of

80 \%

and works at

10 \mathrm{~V}

and

4 \mathrm{~kW}

. If the secondary voltage is

240 \mathrm{~V}

, then the current in the secondary coil is :

A

1.33 \mathrm{~A}

B

13.33 \mathrm{~A}

C

1.59 \mathrm{~A}

D

15.1 \mathrm{~A}

Q 28PhysicsSection B • Numericalwave optics

+4 / -1

In Young's double slit experiment, monochromatic light of wavelength 5000 Å is used. The slits are

1.0 \mathrm{~mm}

apart and screen is placed at

1.0 \mathrm{~m}

away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is _________

\times 10^{-6}

\mathrm{m}

.

Answer Format:Enter numerical value rounded to the nearest integer

Q 29PhysicsSection A • MCQproperties of matter

+4 / -1

A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become :

A

\frac{1}{100}

th

B

\frac{1}{10}

th

C

100 times

D

10 times

Official Correct Answer:Option (B)

Step-by-Step Solution:

To answer this question, we need to understand the relationship between the surface area of the droplets and the surface energy involved.

Surface energy is directly proportional to the surface area of the liquid. The surface energy, $E$ , for a droplet is given by:

$E = \gamma \times A$

Where:

$E$ is the surface energy,
$\gamma$ is the surface tension of the liquid, and
$A$ is the surface area of the droplet.

When multiple droplets coalesce, they form a larger droplet with a certain volume. Since the volume is conserved, the volume of the large droplet will be equal to the sum of the volumes of the small droplets.

Let's denote:

$r$ as the radius of a small droplet,
$R$ as the radius of the large droplet,
$V_{\text{small}}$ as the volume of a small droplet, and
$V_{\text{large}}$ as the volume of the large droplet.

The volume of one small droplet is:

$V_{\text{small}} = \frac{4}{3}\pi r^3$

The total volume of 1000 small droplets is:

$1000 \times V_{\text{small}} = 1000 \times \frac{4}{3}\pi r^3$

Since the volume is conserved, the volume of the large droplet formed by the coalescence of 1000 small droplets is:

$V_{\text{large}} = 1000 \times \frac{4}{3}\pi r^3$

Now, if $R$ is the radius of the large droplet, then:

$V_{\text{large}} = \frac{4}{3}\pi R^3$

Equating the volumes, we have:

$\frac{4}{3}\pi R^3 = 1000 \times \frac{4}{3}\pi r^3$

$R^3 = 1000 \times r^3$

$R = 10r$

Now, let's look at the surface area. The surface area for a small droplet is $A_{\text{small}} = 4\pi r^2$ and for a large droplet is $A_{\text{large}} = 4\pi R^2$ . Substitute $R = 10r$ :

$A_{\text{large}} = 4\pi (10r)^2$

$A_{\text{large}} = 4\pi \times 100r^2$

$A_{\text{large}} = 100 \times 4\pi r^2$

$A_{\text{large}} = 100 \times A_{\text{small}}$

So, the large droplet has 100 times the surface area of one small droplet.

The surface energy of 1000 small droplets would be $1000 \times E_{\text{small}}$ because each small droplet has an energy $E_{\text{small}} = \gamma \times A_{\text{small}}$ .

The surface energy of the big droplet is $E_{\text{large}} = \gamma \times A_{\text{large}}$ . But we have just shown that $A_{\text{large}} = 100 \times A_{\text{small}}$ , so:

$E_{\text{large}} = \gamma \times 100 \times A_{\text{small}}$

This means the surface energy of the big droplet is 100 times the surface energy of one small droplet. Since there were 1000 small droplets originally, the surface energy of the big droplet is:

$\frac{E_{\text{large}}}{1000 \times E_{\text{small}}} = \frac{\gamma \times 100 \times A_{\text{small}}}{1000 \times \gamma \times A_{\text{small}}} = \frac{1}{10}$

Therefore, the correct answer is:

Option B: The surface energy will become $\frac{1}{10}$ th of the original.

Q 30PhysicsSection B • Numericalelectrostatics

+4 / -1

Suppose a uniformly charged wall provides a uniform electric field of

2 \times 10^4 \mathrm{~N} / \mathrm{C}

normally. A charged particle of mass

2 \mathrm{~g}

being suspended through a silk thread of length

20 \mathrm{~cm}

and remain stayed at a distance of

10 \mathrm{~cm}

from the wall.

Then the charge on the particle will be

\frac{1}{\sqrt{x}} \mu \mathrm{C}

where

x=

___________ . [use

\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2

]

Answer Format:Enter numerical value rounded to the nearest integer

Q 31ChemistrySection B • Numericalbiomolecules

+4 / -1

The number of tripeptides formed by three different amino acids using each amino acid once is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Q 32ChemistrySection A • MCQchemistry in everyday life

+4 / -1

Match List - I with List - II.

List I (Compound)	List II (Use)
(A) Carbon tetrachloride	(I) Paint remover
(B) Methylene chloride	(II) Refrigerators and air conditioners
(C) DDT	(III) Fire extinguisher
(D) Freons	(IV) Non Biodegradable insecticide

Choose the correct answer from the options given below :

A

(A)-(II), (B)-(III), (C)-(I), (D)-(IV)

B

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

C

(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

D

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Q 33ChemistrySection A • MCQchemical bonding and molecular structure

+4 / -1

Given below are two statements :

Statement (I) : A

\pi

bonding MO has lower electron density above and below the inter-nuclear axis.

Statement (II) : The

\pi^*

antibonding MO has a node between the nuclei.

In the light of the above statements, choose the most appropriate answer from the options given below :

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

D

Statement I is false but Statement II is true

Q 34ChemistrySection A • MCQalcohols phenols and ethers

+4 / -1

Match List - I with List - II.

List I (Reactants)	List II (Product)
(A) Phenol, Zn/Δ	(I) Salicylaldehyde
(B) Phenol, CHCl₃, NaOH, HCl	(II) Salicylic acid
(C) Phenol, CO₂, NaOH, HCl	(III) Benzene
(D) Phenol, Conc. HNO₃	(IV) Picric acid

Choose the correct answer from the options given below :

A

(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

B

(A)-(III), (B)-(I), (C)-(II), (D)-(IV)

C

(A)-(IV), (B)-(II), (C)-(I), (D)-(III)

D

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Q 35ChemistrySection B • Numericalelectrochemistry

+4 / -1

The amount of electricity in Coulomb required for the oxidation of

1 \mathrm{~mol}

of

\mathrm{H}_2 \mathrm{O}

to

\mathrm{O}_2

is __________

\times 10^5 \mathrm{C}

.

Answer Format:Enter numerical value rounded to the nearest integer

Q 36ChemistrySection A • MCQp block elements

+4 / -1

Given below are two statements :

Statement (I) :

\mathrm{SiO}_2

and

\mathrm{GeO}_2

are acidic while

\mathrm{SnO}

and

\mathrm{PbO}

are amphoteric in nature.

Statement (II) : Allotropic forms of carbon are due to property of catenation and

\mathrm{p} \pi-\mathrm{d} \pi

bond formation.

In the light of the above statements, choose the most appropriate answer from the options given below :

A

Statement I is false but Statement II is true

B

Both Statement I and Statement II are false

C

Both Statement I and Statement II are true

D

Statement I is true but Statement II is false

Q 37ChemistrySection A • MCQcoordination compounds

+4 / -1

Which of the following compounds show colour due to d-d transition?

A

\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7

B

\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}

C

\mathrm{KMnO}_4

D

\mathrm{K}_2 \mathrm{CrO}_4

Official Correct Answer:Option (B)

Step-by-Step Solution:

Among the given options, the compounds that show color due to d-d transitions are those that have partially filled d-orbitals when in the form of complex ions or compounds.

Let's examine each option:

Option A: $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$

Chromium in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ exists in the +6 oxidation state as the dichromate ion ( $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ ). In this oxidation state, chromium does not have any electrons in the d-orbitals (it has a $\mathrm{[Ar]}\;3\mathrm{d}^0$ configuration) so there is no possibility for d-d transitions. Instead, the color is due to charge transfer transitions.

Option B: $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$

Copper(II) sulfate pentahydrate, $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ , contains copper in a +2 oxidation state. In this state, copper has a $\mathrm{[Ar]}\;3\mathrm{d}^9$ electron configuration, which means there is one unpaired electron in the d-orbitals. Due to the presence of this unpaired electron, d-d transitions can occur, and this is why $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is blue in color.

Option C: $\mathrm{KMnO}_4$

In potassium permanganate ( $\mathrm{KMnO}_4$ ), manganese is in the +7 oxidation state, and it has a $\mathrm{[Ar]}\;3\mathrm{d}^0$ electron configuration. Like chromium in the +6 state, manganese in the +7 state does not have any d-electrons available for d-d transitions. The deep purple color of $\mathrm{KMnO}_4$ is also due to charge transfer transitions.

Option D: $\mathrm{K}_2 \mathrm{CrO}_4$

Potassium chromate ( $\mathrm{K}_2 \mathrm{CrO}_4$ ) contains chromium in the +6 oxidation state as the chromate ion ( $\mathrm{CrO}_4^{2-}$ ). Like dichromate, chromium does not have d-electrons and therefore cannot undergo d-d transitions in this compound. The color is due to charge transfer transitions.

So the only compound from the options listed that shows color due to d-d transitions is Option B: $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ .

Q 38ChemistrySection B • Numericalsome basic concepts of chemistry

+4 / -1

10 \mathrm{~mL}

of gaseous hydrocarbon on combustion gives

40 \mathrm{~mL}

of

\mathrm{CO}_2(\mathrm{~g})

and

50 \mathrm{~mL}

of water vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is _________ .

Answer Format:Enter numerical value rounded to the nearest integer

Q 39ChemistrySection A • MCQperiodic classification of elements

+4 / -1

Given below are two statements :

Statement (I) : Both metals and non-metals exist in p and d-block elements.

Statement (II) : Non-metals have higher ionisation enthalpy and higher electronegativity than the metals.

In the light of the above statements, choose the most appropriate answer from the options given below :

A

Both Statement I and Statement II are false

B

Both Statement I and Statement II are true

C

Statement I is false but Statement II is true

D

Statement I is true but Statement II is false

Official Correct Answer:Option (C)

Step-by-Step Solution:

To answer this question, let's investigate both statements in detail.

Statement (I): Both metals and non-metals exist in p and d-block elements.

The periodic table is divided into blocks based on the electron configuration of the elements. These blocks are labeled s, p, d, and f. The p-block contains a mix of metals, non-metals, and metalloids. For example, the p-block contains nonmetals such as oxygen (O) and nitrogen (N), as well as metals like aluminum (Al) and lead (Pb).

The d-block, also known as the transition metals block, primarily consists of metals. However, the d-block does not typically contain elements that are traditionally classified as non-metals. The elements in the d-block have a wide range of properties but are generally characterized by their metallic traits such as conductivity and malleability.

Therefore, Statement I is not entirely true because while both metals and nonmetals exist in the p-block of the periodic table, the d-block is principally composed of metals.

Statement (II): Non-metals have higher ionisation enthalpy and higher electronegativity than the metals.

Ionization enthalpy (or ionization energy) is the energy required to remove an electron from a gaseous atom or ion. Non-metals generally have higher ionization energies compared to metals because non-metals have more tightly bound electrons to their nucleus. This is partly due to the fact that non-metals tend to have higher electronegativities and smaller atomic radii, meaning that their outer electrons are closer to the nucleus and more strongly attracted to it.

Electronegativity is a measure of an atom's ability to attract and bond with electrons. Non-metals have higher electronegativities typically because they are more eager to gain electrons to achieve a full valence shell, reflecting their position on the right side of the periodic table. Metals, on the other hand, are more inclined to lose electrons and form positive ions, indicating their lower electronegativities.

Hence, Statement II is true as non-metals indeed have higher ionisation enthalpy and higher electronegativity than metals.

Given the analysis:

Statement I is not true.
Statement II is true.

Thus, the most appropriate answer is:

Option C - Statement I is false but Statement II is true.

Q 40ChemistrySection B • Numericalpractical organic chemistry

+4 / -1

Following Kjeldahl's method,

1 \mathrm{~g}

of organic compound released ammonia, that neutralised

10 \mathrm{~mL}

of

2 \mathrm{M} ~\mathrm{H}_2 \mathrm{SO}_4

. The percentage of nitrogen in the compound is ___________

\%

.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:56

Step-by-Step Solution:

To determine the percentage of nitrogen using the Kjeldahl's method, we first need to find out how much ammonia (NH₃) was produced and then calculate the equivalent amount of nitrogen in the sample.

The ammonia released reacts with sulfuric acid (H₂SO₄) in the following stoichiometry :

2 \mathrm{NH}_3 + \mathrm{H}_2 \mathrm{SO}_4 \rightarrow (\mathrm{NH}_4)_2 \mathrm{SO}_4

Each mole of H₂SO₄ reacts with 2 moles of NH₃. Since 10 mL of 2 M sulfuric acid is neutralized by the ammonia, we can calculate the amount (in moles) of ammonia :

n(\mathrm{NH}_3) = 2 \times n(\mathrm{H}_2 \mathrm{SO}_4)

Now calculate the moles of H₂SO₄ used :

$n(\mathrm{H}_2 \mathrm{SO}_4) = M \times V$

$n(\mathrm{H}_2 \mathrm{SO}_4) = 2 \, \mathrm{M} \times 10 \, \mathrm{mL} \times \frac{1 \, \mathrm{L}}{1000\,\mathrm{mL}}$

$n(\mathrm{H}_2 \mathrm{SO}_4) = 0.02 \, \mathrm{mol}$

Therefore, the number of moles of NH₃ released will be twice that of the moles of H₂SO₄ neutralized :

$n(\mathrm{NH}_3) = 2 \times 0.02 \, \mathrm{mol}$

$n(\mathrm{NH}_3) = 0.04 \, \mathrm{mol}$

Next, we use the molar mass of nitrogen (14 g/mol) to find the mass of nitrogen :

$m(N) = n(N) \times M(N)$

$m(N) = 0.04 \, \mathrm{mol} \times 14 \, \mathrm{g/mol}$

$m(N) = 0.56 \, \mathrm{g}$

To find the percentage of nitrogen in the compound, we take the mass of nitrogen divided by the mass of the original sample and multiply by 100% :

$\mathrm{Percent \, nitrogen} = \left( \frac{m(N)}{m(\mathrm{sample})} \right) \times 100\%$

$\mathrm{Percent \, nitrogen} = \left( \frac{0.56\, \mathrm{g}}{1\, \mathrm{g}} \right) \times 100\%$

$\mathrm{Percent \, nitrogen} = 56\%$

The percentage of nitrogen in the compound is 56%.

Q 41ChemistrySection A • MCQcoordination compounds

+4 / -1

Given below are two statements :

Statement (I) : Dimethyl glyoxime forms a six-membered covalent chelate when treated with

\mathrm{NiCl}_2

solution in presence of

\mathrm{NH}_4 \mathrm{OH}

.

Statement (II) : Prussian blue precipitate contains iron both in

(+2)

and

(+3)

oxidation states.

In the light of the above statements, choose the most appropriate answer from the options given below :

A

Statement I is false but Statement II is true

B

Both Statement I and Statement II are true

C

Statement I is true but Statement II is false

D

Both Statement I and Statement II are false

Q 42ChemistrySection A • MCQbasics of organic chemistry

+4 / -1

The set of meta directing functional groups from the following sets is :

A

-\mathrm{CN},-\mathrm{NH}_2,-\mathrm{NHR},-\mathrm{OCH}_3

B

-\mathrm{CN},-\mathrm{CHO},-\mathrm{NHCOCH}_3,-\mathrm{COOR}

C

-\mathrm{NO}_2,-\mathrm{NH}_2,-\mathrm{COOH},-\mathrm{COOR}

D

-\mathrm{NO}_2,-\mathrm{CHO},-\mathrm{SO}_3 \mathrm{H},-\mathrm{COR}

Official Correct Answer:Option (D)

Step-by-Step Solution:

The term meta-directing refers to substituents on a benzene ring that direct incoming electrophiles to the meta position (relative to themselves) during an electrophilic aromatic substitution reaction.

Meta directors are typically deactivating groups with respect to the aromatic ring and they often posses an atom with a partial positive charge or a full formal positive charge directly attached to the ring, which tends to withdraw electron density from the ring through the inductive effect or resonance. This decreased electron density in the ring makes the ortho and para positions less reactive, thereby directing new substituents to the meta position. Furthermore, the presence of these withdrawing groups can stabilize the intermediate cation (whichever is formed) better when the substituent is at the meta position relative to the ortho or para positions.

Let's examine the options:

Option A (Incorrect):

$-\mathrm{CN}$ (cyanide) is a strong meta director due to its strong electron-withdrawing nature.
$-\mathrm{NH}_2$ (amino), $-\mathrm{NHR}$ (alkylamino), and $-\mathrm{OCH}_3$ (methoxy) are strong ortho/para directors because they donate electron density to the ring through resonance.

Option B (Possibly correct):

$-\mathrm{CN}$ (cyanide) is a strong meta director as stated before.
$-\mathrm{CHO}$ (formyl) is a weak meta director because the carbonyl group withdraws electron density via resonance and induction.
$-\mathrm{NHCOCH}_3$ (acetamido) is a bit tricky. While $-\mathrm{NH}_2$ usually directs to the ortho/para position, when it is part of the amide group, the resonance with the carbonyl can make it withdraw electron density, thus Acetamido become a meta-director.
$-\mathrm{COOR}$ (ester) is a meta director due to the electron-withdrawing nature of the carbonyl group.

Option C (Incorrect):

$-\mathrm{NO}_2$ (nitro) is a very strong meta director due to its strong electron-withdrawing nature both through induction and resonance.
$-\mathrm{NH}_2$ (amino) and $-\mathrm{COOH}$ (carboxylic acid) are strong ortho/para directors. Carboxylic acid groups can be deactivating due to their overall electron-withdrawing nature, but they still direct to the ortho/para positions because the electronegative oxygen can donate electron density by resonance.
$-\mathrm{COOR}$ (ester) is a meta director.

Option D (Correct):

$-\mathrm{NO}_2$ (nitro) is a strong meta director as before.
$-\mathrm{CHO}$ (formyl) is a meta director.
$-\mathrm{SO}_3 \mathrm{H}$ (sulphonic acid) is a strong meta director because the sulphonic acid group is very electron-withdrawing due to the presence of several electronegative oxygen atoms.
$-\mathrm{COR}$ (ketone) is a meta director because of the electron-withdrawing nature of the carbonyl group.

Based on the description above, the set of functional groups that are meta directing from the given options is Option D: $-\mathrm{NO}_2$ , $-\mathrm{CHO}$ , $-\mathrm{SO}_3 \mathrm{H}$ , $-\mathrm{COR}$ . These groups are all electron-withdrawing and would direct incoming substituents to the meta position during electrophilic aromatic substitution.

Q 43ChemistrySection B • Numericalcompounds containing nitrogen

+4 / -1

Number of compounds which give reaction with Hinsberg's reagent is _________.

qImage-aliphatic-amines-compounds-containing-nitrogen-organic-chemistry-Jeemate

qImage-aliphatic-amines-compounds-containing-nitrogen-organic-chemistry-Jeemate

Answer Format:Enter numerical value rounded to the nearest integer

Q 44ChemistrySection A • MCQcoordination compounds

+4 / -1

\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}

and

\left[\mathrm{CoF}_6\right]^{3-}

are respectively known as :

A

Inner orbital Complex, Spin paired Complex

B

Spin paired Complex, Spin free Complex

C

Spin free Complex, Spin paired Complex

D

Outer orbital Complex, Inner orbital Complex

Q 45ChemistrySection A • MCQchemical bonding and molecular structure

+4 / -1

Select the compound from the following that will show intramolecular hydrogen bonding.

A

B

\mathrm{H}_2 \mathrm{O}

C

\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}

D

\mathrm{NH}_3

Q 46ChemistrySection A • MCQaldehydes ketones and carboxylic acids

+4 / -1

Which among the followng has highest boiling point?

A

\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{OH}

B

\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3

C

\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO}

D

\mathrm{H}_5 \mathrm{C}_2-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5

Official Correct Answer:Option (A)

Step-by-Step Solution:

To predict the boiling point of these compounds, we need to consider the types of intermolecular forces present in each molecule. These intermolecular forces include van der Waals forces, dipole-dipole interactions, and hydrogen bonding.

The compounds given are:

Option A : Butanol ( $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{OH}$ ) - a primary alcohol

Option B : Butane ( $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ ) - an alkane

Option C : Butyraldehyde ( $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO}$ ) - an aldehyde

Option D : Diethyl ether ( $\mathrm{H}_5 \mathrm{C}_2-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5$ ) - an ether

Now let's analyze each one:

Option A: Butanol is capable of hydrogen bonding because it has an $\mathrm{-OH}$ group. Hydrogen bonding is the strongest intermolecular force among the forces affecting these molecules. Therefore, butanol will have the highest boiling point among the compounds without considering the others.

Option B: Butane only has van der Waals forces because it is a non-polar molecule. Van der Waals forces are the weakest intermolecular forces, resulting in a lower boiling point compared to molecules that can hydrogen bond or have permanent dipoles.

Option C: Butyraldehyde can form dipole-dipole interactions because of its polar carbonyl ( $\mathrm{C=O}$ ) group, but it cannot form hydrogen bonds with itself since there is no $\mathrm{OH}$ group. Dipole-dipole interactions are stronger than van der Waals forces but weaker than hydrogen bonds. As a result, butyraldehyde will have a higher boiling point than butane but a lower boiling point than butanol.

Option D: Diethyl ether has an oxygen atom, which makes it polar and enables dipole-dipole interactions. However, ethers cannot form hydrogen bonds with themselves, so diethyl ether will have a higher boiling point than butane but lower than butanol.

Considering the intermolecular forces, the compound with the highest boiling point is the one that can form hydrogen bonds, which in this list is the primary alcohol, butanol.

Therefore, the compound with the highest boiling point is:

Option A : Butanol ( $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{OH}$ ).

Q 47ChemistrySection A • MCQatomic structure

+4 / -1

The number of radial node/s for

3 p

orbital is :

A

3

B

2

C

1

D

4

Q 48ChemistrySection A • MCQd and f block elements

+4 / -1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : In aqueous solutions

\mathrm{Cr}^{2+}

is reducing while

\mathrm{Mn}^{3+}

is oxidising in nature.

Reason (R) : Extra stability to half filled electronic configuration is observed than incompletely filled electronic configuration.

In the light of the above statements, choose the most appropriate answer from the options given below :

A

(A) is true but (R) is false

B

Both (A) and (R) are true and (R) is the correct explanation of (A)

C

Both (A) and (R) are true but (R) is not the correct explanation of (A)

D

(A) is false but (R) is true

Official Correct Answer:Option (B)

Step-by-Step Solution:

Let us analyze both the Assertion (A) and the Reason (R) to determine the correct answer.

Assertion (A): In aqueous solutions, $\mathrm{Cr}^{2+}$ is reducing while $\mathrm{Mn}^{3+}$ is oxidising in nature. This assertion is true. The $\mathrm{Cr}^{2+}$ ion has a tendency to be oxidized to $\mathrm{Cr}^{3+}$ because $\mathrm{Cr}^{3+}$ has a more stable electronic configuration. Therefore, $\mathrm{Cr}^{2+}$ acts as a reducing agent. On the other hand, $\mathrm{Mn}^{3+}$ tends to stabilize by being reduced to $\mathrm{Mn}^{2+}$ , which is a more stable electronic configuration due to having a half-filled d5 subshell ( $\mathrm{Mn}^{2+}$ : [Ar] 3d⁵). Thus, $\mathrm{Mn}^{3+}$ acts as an oxidizing agent.

Reason (R): Extra stability to half filled electronic configuration is observed than incompletely filled electronic configuration. This reason is also true. Half-filled and fully filled electron configurations, like those found in $\mathrm{Mn}^{2+}$ and $\mathrm{Cr}^{0}$ respectively, are particularly stable due to symmetry and exchange energy considerations. This is a reason why certain ions like $\mathrm{Mn}^{3+}$ want to be reduced to $\mathrm{Mn}^{2+}$ , and why $\mathrm{Cr}^{0}$ (with a [Ar] 3d⁵ 4s¹ configuration) is not as common as $\mathrm{Cr}^{3+}$ (which has a [Ar] 3d³ configuration, and while not half-filled, is favored due to crystal field stabilization energy considerations in octahedral complexes).

If we evaluate these statements together, we see that Assertion (A) is directly related to the electronic configurations and their stability, as asserted by Reason (R). The stability of half-filled and full orbitals contributes to the observed redox behavior of $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ in aqueous solutions.

Therefore, the correct answer is Option B: Both (A) and (R) are true and (R) is the correct explanation of (A).

Q 49ChemistrySection A • MCQbasics of organic chemistry

+4 / -1

The functional group that shows negative resonance effect is :

A

-\mathrm{OH}

B

-\mathrm{OR}

C

-\mathrm{COOH}

D

-\mathrm{NH}_2

Q 50ChemistrySection A • MCQionic equilibrium

+4 / -1

Solubility of calcium phosphate (molecular mass, M) in water is

\mathrm{W_{g}}

per

100 \mathrm{~mL}

at

25^{\circ} \mathrm{C}

. Its solubility product at

25^{\circ} \mathrm{C}

will be approximately.

A

10^7\left(\frac{W}{M}\right)^3

B

10^3\left(\frac{\mathrm{W}}{\mathrm{M}}\right)^5

C

10^7\left(\frac{W}{M}\right)^5

D

10^5\left(\frac{\mathrm{W}}{\mathrm{M}}\right)^5

Official Correct Answer:Option (C)

Step-by-Step Solution:

To determine the solubility product (K_sp) of calcium phosphate, we need to consider its chemical formula and how it dissociates in water. The formula for calcium phosphate is $\text{Ca}_3(\text{PO}_4)_2$ . When dissolved in water, it dissociates according to the following equation:

$\text{Ca}_3(\text{PO}_4)_2 (s) \rightleftharpoons 3 \text{Ca}^{2+} (aq) + 2 \text{PO}_4^{3-} (aq)$

Let $S$ be the molar solubility of calcium phosphate in $\text{mol}/L$ . According to the stoichiometry of the dissociation, if $S$ moles per liter of calcium phosphate dissolve, $3S$ moles per liter of calcium ions and $2S$ moles per liter of phosphate ions are produced.

The expression for the solubility product constant (Ksp) for calcium phosphate in water is the product of the concentrations of the ions raised to the power of their respective coefficients in the balanced equation:

$K_{sp} = [\text{Ca}^{2+}]^3 [\text{PO}_4^{3-}]^2$

Substituting the concentrations with $3S$ for $\text{Ca}^{2+}$ and $2S$ for $\text{PO}_4^{3-}$ , we get:

$K_{sp} = (3S)^3 (2S)^2 = 27S^3 \cdot 4S^2 = 108S^5$

If the solubility of calcium phosphate is $\text{W}_{g}$ per $100 \text{ mL}$ of water at $25^\circ\text{C}$ , we need to convert grams to moles to find $S$ . Solubility in moles per liter (which is the same as moles per $1000 \text{ mL}$ ) is:

$S = \frac{\text{W}}{\text{M}} \times \frac{1000 \text{ mL}}{100 \text{ mL}} = 10\frac{\text{W}}{\text{M}}$

Now, substituting $S$ with $10\frac{\text{W}}{\text{M}}$ into the expression for $K_{sp}$ :

$K_{sp} = 108 \left(10\frac{\text{W}}{\text{M}}\right)^5$

This simplifies to:

$K_{sp} = 108 \times 10^5 \left(\frac{\text{W}}{\text{M}}\right)^5$

Since $108$ is on the order of $10^2$ , we can approximate this to:

$K_{sp} \approx 10^7 \left(\frac{\text{W}}{\text{M}}\right)^5$

Therefore, the correct answer based on the options provided is:

Option C: $10^7 \left(\frac{\text{W}}{\text{M}}\right)^5$

Q 51ChemistrySection B • Numericalelectrochemistry

+4 / -1

Consider the following redox reaction :

\mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2

The standard reduction potentials are given as below

\left(\mathrm{E}_{\text {red }}^0\right)

:

\begin{aligned} & \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\circ}=+1.51 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{CO}_2 / \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4}^{\circ}=-0.49 \mathrm{~V} \end{aligned}

If the equilibrium constant of the above reaction is given as

\mathrm{K}_{\mathrm{eq}}=10^x

, then the value of

x=

__________ (nearest integer)

Answer Format:Enter numerical value rounded to the nearest integer

Q 52ChemistrySection A • MCQp block elements

+4 / -1

The strongest reducing agent among the following is :

A

\mathrm{SbH}_3

B

\mathrm{NH}_3

C

\mathrm{BiH}_3

D

\mathrm{PH}_3

Q 53ChemistrySection A • MCQhydrocarbons

+4 / -1

qImage-reaction-of-alkenes-hydrocarbons-organic-chemistry-Jeemate

Acid D formed in above reaction is :

A

Malonic acid

B

Oxalic acid

C

Succinic acid

D

Gluconic acid

Q 54ChemistrySection A • MCQbasics of organic chemistry

+4 / -1

Lassaigne's test is used for detection of :

A

Phosphorous and halogens only

B

Nitrogen, Sulphur and Phosphorous only

C

Nitrogen, Sulphur, phosphorous and halogens

D

Nitrogen and Sulphur only

Q 55ChemistrySection B • Numericalbasics of organic chemistry

+4 / -1

Total number of isomeric compounds (including stereoisomers) formed by monochlorination of 2-methylbutane is _______ .

Answer Format:Enter numerical value rounded to the nearest integer

Q 56ChemistrySection A • MCQhydrocarbons

+4 / -1

In the given reactions identify

A

and

B

qImage-reaction-of-alkynes-hydrocarbons-organic-chemistry-Jeemate

A

A : n-Pentane B : Cis - 2 - butene

B

A : 2-Pentyne B : Cis-2-butene

C

A : n-Pentane B : trans-2-butene

D

A : 2-Pentyne B : trans-2-butene

Q 57ChemistrySection B • Numericalsolutions

+4 / -1

Mass of ethylene glycol (antifreeze) to be added to

18.6 \mathrm{~kg}

of water to protect the freezing point at

-24^{\circ} \mathrm{C}

is ________

\mathrm{kg}

(Molar mass in

\mathrm{g} ~\mathrm{mol}^{-1}

for ethylene glycol

62, \mathrm{~K}_f

of water

=1.86 \mathrm{~K}

\mathrm{kg} ~\mathrm{mol}^{-1}

)

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:15

Step-by-Step Solution:

To determine the mass of ethylene glycol (antifreeze) needed to lower the freezing point of water to $-24^{\circ} \mathrm{C}$ , we can use the freezing point depression equation, which is a colligative property given by:

\Delta T_f = i \cdot K_f \cdot m

where:

$\Delta T_f$ is the depression in the freezing point.
$i$ is the van't Hoff factor, which is the number of particles the solute splits into in solution. For ethylene glycol, which does not dissociate in solution, $i = 1$ .
$K_f$ is the cryoscopic constant (freezing point depression constant) of water, which is given as $1.86 \mathrm{~K} ~\mathrm{kg} ~\mathrm{mol}^{-1}$ .
$m$ is the molality of the solution.

However, molality is defined as the number of moles of solute per kilogram of solvent, so we first need to find the molality that corresponds to the desired freezing point depression:

\Delta T_f = -24^{\circ} \mathrm{C} - (0^{\circ} \mathrm{C}) = -24^{\circ} \mathrm{C}

Now, we can solve for $m$ :

m = \frac{\Delta T_f}{i \cdot K_f}

Substituting the given values:

m = \frac{-24^{\circ} \mathrm{C}}{1 \cdot 1.86 \mathrm{~K} ~\mathrm{kg} ~\mathrm{mol}^{-1}}

Note that $\mathrm{C}$ and $\mathrm{K}$ are interchangeable when calculating changes in temperature. Now, we can compute the molality:

m = \frac{-24}{1.86}

Next, we calculate the molality of the ethylene glycol:

m \approx -12.903 \mathrm{~mol} ~\mathrm{kg}^{-1}

Keep in mind that molality is always positive, but the negative sign indicates the direction of the temperature change. Since we're calculating the amount needed, we can consider it as 12.903 moles per kilogram of water.

Now we calculate the moles of ethylene glycol required for $18.6 \mathrm{~kg}$ of water:

\text{moles of ethylene glycol} = m \cdot \text{mass of water (in kg)}

\text{moles of ethylene glycol} = 12.903 \mathrm{~mol} ~\mathrm{kg}^{-1} \cdot 18.6 \mathrm{~kg}

\text{moles of ethylene glycol} \approx 240.0158 \mathrm{~mol}

Next, we find the mass of ethylene glycol needed using its molar mass:

\text{mass of ethylene glycol} = \text{moles of ethylene glycol} \cdot \text{molar mass of ethylene glycol}

Given the molar mass of ethylene glycol is $62 \mathrm{~g} ~\mathrm{mol}^{-1}$ , we then convert it to $\mathrm{kg} ~\mathrm{mol}^{-1}$ by dividing by 1000:

\text{molar mass of ethylene glycol} = 0.062 \mathrm{~kg} ~\mathrm{mol}^{-1}

Now we can calculate the mass:

\text{mass of ethylene glycol} = 240.0158 \mathrm{~mol} \cdot 0.062 \mathrm{~kg} ~\mathrm{mol}^{-1}

\text{mass of ethylene glycol} \approx 14.88098 \mathrm{~kg}

Therefore, approximately $14.881 \mathrm{~kg}$ of ethylene glycol needs to be added to $18.6 \mathrm{~kg}$ of water to lower the freezing point to $-24^{\circ} \mathrm{C}$ .

Q 58ChemistrySection B • Numericalchemical kinetics and nuclear chemistry

+4 / -1

The following data were obtained during the first order thermal decomposition of a gas A at constant volume :

\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})

S.No.	Time /s	Total pressure /(atm)
1.	0	0.1
2.	115	0.28

The rate constant of the reaction is ________

\times 10^{-2} \mathrm{~s}^{-1}

(nearest integer)

Answer Format:Enter numerical value rounded to the nearest integer

Q 59ChemistrySection B • Numericalthermodynamics

+4 / -1

For a certain reaction at

300 \mathrm{~K}, \mathrm{~K}=10

, then

\Delta \mathrm{G}^{\circ}

for the same reaction is - ____________

\times 10^{-1} \mathrm{~kJ} \mathrm{~mol}^{-1}

.

(Given

\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}

)

Answer Format:Enter numerical value rounded to the nearest integer

Q 60ChemistrySection A • MCQperiodic classification of elements

+4 / -1

The transition metal having highest

3^{\text {rd }}

ionisation enthalpy is :

A

\mathrm{Mn}

B

\mathrm{Fe}

C

\mathrm{Cr}

D

V

Official Correct Answer:Option (A)

Step-by-Step Solution:

The third ionization enthalpy refers to the energy required to remove the third electron from a di-positive ion ( $M^{2+}$ ) to form a tri-positive ion ( $M^{3+}$ ). In the case of transition metals, the energies involved in removing the third electron are typically higher than for the first and second electrons due to an increasing effective nuclear charge and the electrons being removed from closer energy levels to the nucleus or from a half-filled or full-filled d-subshell which is relatively more stable.

Let us look at the electronic configurations of the ions of each metal listed to determine which would require the highest third ionisation enthalpy. Since we are talking about $3^{\text{rd}}$ ionization, we would be considering the removal of three electrons, ending up with a $+3$ oxidation state for each metal.

Manganese ( $\mathrm{Mn}$ ): [Ar] $3d^5 4s^2$

Iron ( $\mathrm{Fe}$ ): [Ar] $3d^6 4s^2$

Chromium ( $\mathrm{Cr}$ ): [Ar] $3d^5 4s^1$

Vanadium ( $V$ ): [Ar] $3d^3 4s^2$

After removing two electrons (assuming they come from the 4s orbital and the next available d orbital), the configurations would be:

$\mathrm{Mn^{2+}}$ : [Ar] $3d^5$

$\mathrm{Fe^{2+}}$ : [Ar] $3d^6$

$\mathrm{Cr^{2+}}$ : [Ar] $3d^4$

$V^{2+}$ : [Ar] $3d^3$

Now we look at the energy required to remove the third electron from the $M^{2+}$ ions:

For $\mathrm{Mn^{2+}}$ , removing the third electron would break the half-filled $3d^5$ configuration and require significant energy.
For $\mathrm{Fe^{2+}}$ , removing the third electron would form half-filled $3d^5$ configuration, which forms stable configuration.
For $\mathrm{Cr^{2+}}$ , the third electron removal will disrupt the $3d^4$ configuration, which is not particularly stable.
For $V^{2+}$ , the third electron removal will disrupt the $3d^3$ configuration.

In terms of stability, the half-filled $d^5$ subshell in manganese ( $\mathrm{Mn^{2+}}$ ) provide extra stability. So, the $3^{\text{rd}}$ ionisation enthalpy of $\mathrm{Mn^{2+}}$ is the highest because it involves breaking a half-filled configuration which is more stable.

Thus, out of the given options, manganese ( $\mathrm{Mn}$ ) would have the highest third ionization enthalpy. Therefore, the correct answer is:

Option A : $\mathrm{Mn}$

Q 61MathematicsSection A • MCQ3d geometry

+4 / -1

Consider a

\triangle A B C

where

A(1,3,2), B(-2,8,0)

and

C(3,6,7)

. If the angle bisector of

\angle B A C

meets the line

B C

at

D

, then the length of the projection of the vector

\overrightarrow{A D}

on the vector

\overrightarrow{A C}

is :

A

\frac{37}{2 \sqrt{38}}

B

\sqrt{19}

C

\frac{39}{2 \sqrt{38}}

D

\frac{\sqrt{38}}{2}

Q 62MathematicsSection A • MCQmatrices and determinants

+4 / -1

Let the system of equations

x+2 y+3 z=5,2 x+3 y+z=9,4 x+3 y+\lambda z=\mu

have infinite number of solutions. Then

\lambda+2 \mu

is equal to :

A

22

B

17

C

15

D

28

Q 63MathematicsSection A • MCQsets and relations

+4 / -1

Consider the relations

R_1

and

R_2

defined as

a R_1 b \Leftrightarrow a^2+b^2=1

for all

a, b \in \mathbf{R}

and

(a, b) R_2(c, d) \Leftrightarrow

a+d=b+c

for all

(a, b),(c, d) \in \mathbf{N} \times \mathbf{N}

. Then :

A

R_1

and

R_2

both are equivalence relations

B

Only

R_1

is an equivalence relation

C

Only

R_2

is an equivalence relation

D

Neither

R_1

nor

R_2

is an equivalence relation

Official Correct Answer:Option (C)

Step-by-Step Solution:

To determine if the given relations $R_1$ and $R_2$ are equivalence relations, we need to check whether each of them satisfies the three defining properties of an equivalence relation: reflexivity, symmetry, and transitivity.

Let's start by analysing $R_1$ :

Reflexivity: A relation $R$ on a set $S$ is reflexive if every element is related to itself, that is, for every $a \in S$ , the pair $(a,a)$ is in $R$ . In the case of $R_1$ , for an arbitrary $a \in \mathbf{R}$ , we need to check if $a R_1 a$ holds, which translates to checking if $a^2 + a^2 = 1$ . This would mean $2a^2 = 1$ or $a^2 = \frac{1}{2}$ . Since this is not true for every real number $a$ , $R_1$ is not reflexive.

Symmetry: A relation $R$ is symmetric if whenever $a R b$ , then $b R a$ . For $R_1$ , if $a^2 + b^2 = 1$ , then it is also true that $b^2 + a^2 = 1$ . Thus, $R_1$ is symmetric.

Transitivity: A relation $R$ is transitive if whenever $a R b$ and $b R c$ , then $a R c$ . For $R_1$ , suppose $a R_1 b$ and $b R_1 c$ , this means $a^2 + b^2 = 1$ and $b^2 + c^2 = 1$ . However, if we add these two, we get $a^2 + 2b^2 + c^2 = 2$ . There is no guarantee that $a^2 + c^2 = 1$ . Therefore, $R_1$ is not transitive.

Conclusion: Relation $R_1$ is not reflexive or transitive, and hence it is not an equivalence relation.

Now let's consider $R_2$ :

Reflexivity: For any $(a, b) \in \mathbf{N} \times \mathbf{N}$ , it's clear that $a + b = b + a$ , which is just a reiteration of the commutative property of addition. Therefore, $(a, b) R_2 (a, b)$ , and $R_2$ is reflexive.

Symmetry: If $(a, b) R_2 (c, d)$ , meaning $a + d = b + c$ , then by reordering the terms we can similarly have $c + b = d + a$ , which means $(c, d) R_2 (a, b)$ , so $R_2$ is symmetric.

Transitivity: Suppose $(a, b) R_2 (c, d)$ and $(c, d) R_2 (e, f)$ , i.e., $a + d = b + c$ and $c + f = d + e$ , we want to show that $(a, b) R_2 (e, f)$ . Adding the two equations, we get $a + d + c + f = b + c + d + e$ . By rearranging and simplifying, we get $a + f = b + e$ , thus, $(a, b) R_2 (e, f)$ .

So $R_2$ is reflexive, symmetric, and transitive, and therefore it is an equivalence relation.

Conclusion: According to the above examination, only $R_2$ is an equivalence relation. Thus, the correct answer is:

Option C: Only $R_2$ is an equivalence relation.

Q 64MathematicsSection A • MCQdefinite integration

+4 / -1

If

\int\limits_0^{\frac{\pi}{3}} \cos ^4 x \mathrm{~d} x=\mathrm{a} \pi+\mathrm{b} \sqrt{3}

, where

\mathrm{a}

and

\mathrm{b}

are rational numbers, then

9 \mathrm{a}+8 \mathrm{b}

is equal to :

A

2

B

1

C

3

D

\frac{3}{2}

Official Correct Answer:Option (A)

Step-by-Step Solution:

To solve the given integral $\int\limits_0^{\frac{\pi}{3}} \cos ^4 x \mathrm{~d} x$ , we'll apply a known power-reduction formula that allows us to express even powers of sine and cosine functions in terms of cosine of multiple angles. Specifically for $\cos^4 x$ , we can write it in terms of double angles as:

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

We can then expand and simplify the integral using this formula. Let's proceed with this:

$\int\limits_0^{\frac{\pi}{3}} \cos^4 x \mathrm{~d}x = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1 + \cos(2x)}{2}\right)^2 \mathrm{~d}x$

Now, let's expand the integrand and then integrate term by term:

$= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \cos^2(2x)\right)\mathrm{~d}x$

For the $\cos^2(2x)$ term, we again use the power reduction formula:

$\cos^2(2x) = \frac{1 + \cos(4x)}{2}$

Let's substitute this into the integral and continue:

$= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \left(\frac{1 + \cos(4x)}{2}\right)\right)\mathrm{~d}x$

Simplify and integrate:

$= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{8} + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x$

$= \int\limits_0^{\frac{\pi}{3}} \left(\frac{3}{8} + \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x$

$= \left. \left(\frac{3}{8} x + \frac{1}{4} \sin(2x) + \frac{1}{32} \sin(4x)\right) \right|_0^{\frac{\pi}{3}}$

Evaluating this from $0$ to $\frac{\pi}{3}$ :

$= \left(\frac{3}{8} \cdot \frac{\pi}{3} + \frac{1}{4} \sin\left(2 \cdot \frac{\pi}{3}\right) + \frac{1}{32} \sin\left(4 \cdot \frac{\pi}{3}\right)\right) - \left(\frac{3}{8} \cdot 0 + \frac{1}{4} \sin(0) + \frac{1}{32} \sin(0)\right)$

$= \frac{\pi}{8} + \frac{1}{4} \sin\left(\frac{2\pi}{3}\right) + \frac{1}{32} \sin\left(\frac{4\pi}{3}\right)$

$\sin\left(\frac{2\pi}{3}\right)$ is $\frac{\sqrt{3}}{2}$ and $\sin\left(\frac{4\pi}{3}\right)$ is $-\frac{\sqrt{3}}{2}$ :

$= \frac{\pi}{8} + \frac{1}{4} \cdot \frac{\sqrt{3}}{2} - \frac{1}{32} \cdot \frac{\sqrt{3}}{2}$

$= \frac{\pi}{8} + \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{64}$

Now, combining terms we get the final result:

$= \frac{\pi}{8} + \frac{8\sqrt{3} - \sqrt{3}}{64}$

$= \frac{\pi}{8} + \frac{7\sqrt{3}}{64}$

Now, let's match this result to the form $\mathrm{a}\pi + \mathrm{b}\sqrt{3}$ and find $a$ and $b$ :

$a = \frac{1}{8}, \quad b = \frac{7}{64}$

Now we find $9a + 8b$ :

$9a + 8b = 9 \cdot \frac{1}{8} + 8 \cdot \frac{7}{64}$

$= \frac{9}{8} + \frac{7}{8}$

$= \frac{16}{8}$

$= 2$

Therefore, the value of $9a + 8b$ is 2, which correspond to Option A.

Q 65MathematicsSection A • MCQprobability

+4 / -1

Let Ajay will not appear in JEE exam with probability

\mathrm{p}=\frac{2}{7}

, while both Ajay and Vijay will appear in the exam with probability

\mathrm{q}=\frac{1}{5}

. Then the probability, that Ajay will appear in the exam and Vijay will not appear is :

A

\frac{9}{35}

B

\frac{3}{35}

C

\frac{24}{35}

D

\frac{18}{35}

Official Correct Answer:Option (D)

Step-by-Step Solution:

We are given that the probability of Ajay not appearing in the JEE exam is $\mathrm{p}=\frac{2}{7}$ , and the probability that both Ajay and Vijay will appear in the exam is $\mathrm{q}=\frac{1}{5}$ .

We are asked to find the probability that Ajay will appear in the exam and Vijay will not. Let's denote this probability as $\mathrm{r}$ .

To find $\mathrm{r}$ , we need to use the concept of complementary events. The probability that Ajay will appear in the exam is the complement of the probability that he will not appear. So,

$P(\text{Ajay appears}) = 1 - P(\text{Ajay does not appear}) = 1 - \mathrm{p} = 1 - \frac{2}{7} = \frac{5}{7}.$

The event that both Ajay and Vijay appear in the exam is independent of the event that only Ajay appears (and Vijay does not). Therefore, we can express the probability that only Ajay will appear (and Vijay will not) as the difference of Ajay appearing minus both Ajay and Vijay appearing, because the probability of both appearing ( $\mathrm{q}$ ) is included in the probability of Ajay appearing:

$\mathrm{r} = P(\text{Ajay appears}) - P(\text{Both Ajay and Vijay appear}) = \frac{5}{7} - \frac{1}{5}.$

To subtract these two fractions, we need a common denominator, which would be $35$ in this case. So,

$\mathrm{r} = \frac{5}{7} \cdot \frac{5}{5} - \frac{1}{5} \cdot \frac{7}{7} = \frac{25}{35} - \frac{7}{35} = \frac{25 - 7}{35} = \frac{18}{35}.$

Therefore, the probability that Ajay will appear in the exam and Vijay will not appear is $\mathrm{r} = \frac{18}{35}$ , which corresponds to Option D.

Q 66MathematicsSection A • MCQcomplex numbers

+4 / -1

If

z

is a complex number such that

|z| \leqslant 1

, then the minimum value of

\left|z+\frac{1}{2}(3+4 i)\right|

is :

A

2

B

\frac{5}{2}

C

\frac{3}{2}

D

3

Official Correct Answer:Option (C)

Step-by-Step Solution:

To find the minimum value of $\left|z+\frac{1}{2}(3+4i)\right|$ , where $z$ is a complex number with $|z| \leqslant 1$ , we can think of this geometrically as the distance from any point inside or on the boundary of the unit circle in the complex plane to the fixed point $\frac{1}{2}(3+4i)$ .

To make this more clear, first write the expression as follows: $\left|z+\frac{1}{2}(3+4i)\right| = \left|z+\frac{3}{2}+\frac{4}{2}i\right| = \left|z+1.5+2i\right|$

This means we are looking for the distance between any point on the complex plane represented by $z$ (where $z$ has a magnitude of up to 1) and the point $1.5+2i$ .

Since $z$ is within the unit circle, $|z| \leqslant 1$ and $|-1.5 - 2i| = \sqrt{1.5^2 + 2^2} = \sqrt{2.25 + 4} = \sqrt{6.25} = 2.5$ .

So now we have: $\left|z + 1.5 + 2i\right| \geqslant \left| 1 - 2.5 \right|$

Which simplifies to: $\left|z + 1.5 + 2i\right| \geqslant 1.5$

Therefore, the minimum value of $\left|z+\frac{1}{2}(3+4i)\right|$ given that $|z| \leqslant 1$ is $\frac{3}{2}$ .

So the correct answer is:

Option C: $\frac{3}{2}$

Q 67MathematicsSection A • MCQfunctions

+4 / -1

If the domain of the function

f(x)=\frac{\sqrt{x^2-25}}{\left(4-x^2\right)}+\log _{10}\left(x^2+2 x-15\right)

is

(-\infty, \alpha) \cup[\beta, \infty)

, then

\alpha^2+\beta^3

is equal to :

A

140

B

175

C

125

D

150

Official Correct Answer:Option (D)

Step-by-Step Solution:

To find the domain of the function $f(x) = \frac{\sqrt{x^2-25}}{(4-x^2)}+\log_{10}(x^2+2x-15),$ we need to consider the domain conditions for both the square root function and the logarithmic function.

The square root function $\sqrt{x^2-25}$ requires that the argument of the square root be non-negative, so $x^2 - 25 \geq 0.$ This inequality is satisfied when $x \leq -5 \quad \text{or} \quad x \geq 5.$

The denominator of the rational part of $f(x)$ , $(4-x^2)$ , cannot be zero, otherwise, the function will become undefined due to division by zero. Thus, we must have $4 - x^2 \neq 0.$ This inequality is violated when $x = \pm2.$

Combining these conditions gives us the domain for the rational part of the function: $x \in (-\infty, -5] \cup (5, \infty) \quad \text{and} \quad x \neq 2,-2.$

Moving on to the logarithmic function, $\log_{10}(x^2+2x-15)$ , the argument must be positive: $x^2 + 2x - 15 > 0.$ This is a quadratic inequality, which we can factor to find the solution: $(x+5)(x-3) > 0.$ From this, we see that the inequality is satisfied for $x < -5 \quad \text{or} \quad x > 3.$

The overall domain of $f(x)$ is the intersection of the domains for each piece. Taking the intersection of the two sets gives us: $x \in (-\infty, -5) \cup (5, \infty),$

Since the question states that the domain is of the form $(-\infty, \alpha) \cup [\beta, \infty)$ , we can infer that $\alpha = -5 \quad \text{and} \quad \beta = 5.$

We calculate $\alpha^2 + \beta^3$ as follows: $\alpha^2 + \beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150.$

So the correct answer, representing the sum of $\alpha^2$ and $\beta^3$ , is: Option D $150$ .

Q 68MathematicsSection A • MCQquadratic equation and inequalities

+4 / -1

Let

\alpha

and

\beta

be the roots of the equation

p x^2+q x-r=0

, where

p \neq 0

. If

p, q

and

r

be the consecutive terms of a non constant G.P. and

\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}

, then the value of

(\alpha-\beta)^2

is :

A

8

B

9

C

\frac{20}{3}

D

\frac{80}{9}

Q 69MathematicsSection A • MCQsequences and series

+4 / -1

Let

S_n

denote the sum of the first

n

terms of an arithmetic progression. If

S_{10}=390

and the ratio of the tenth and the fifth terms is

15: 7

, then

\mathrm{S}_{15}-\mathrm{S}_5

is equal to :

A

800

B

890

C

790

D

690

Official Correct Answer:Option (C)

Step-by-Step Solution:

To solve this problem, we will start by using the properties of an arithmetic progression (AP).

The sum of the first $n$ terms of an AP can be calculated using the formula: $S_n = \frac{n}{2} (2a + (n-1)d)$ where $S_n$ is the sum of the first $n$ terms, $a$ is the first term, and $d$ is the common difference between the terms.

Given the information: $S_{10} = 390$

We can plug $n=10$ into the sum formula to get:

$S_{10} = \frac{10}{2} (2a + (10-1)d)$

$390 = 5(2a + 9d)$

$390 = 10a + 45d$

$78 = 2a + 9d \quad .........\text{(1)}$

Next, we're given the ratio of the tenth term ( $T_{10}$ ) to the fifth term ( $T_5$ ): $\frac{T_{10}}{T_5} = \frac{15}{7}$

The $n$ th term of an AP is given by:

$T_n = a + (n-1)d$

So, for the tenth term: $T_{10} = a + (10-1)d = a + 9d$

And for the fifth term:

$T_5 = a + (5-1)d = a + 4d$

Now we can write the ratio as:

$\frac{a + 9d}{a + 4d} = \frac{15}{7}$

$7(a + 9d) = 15(a + 4d)$

$7a + 63d = 15a + 60d$

$63d - 60d = 15a - 7a$

$3d = 8a \quad .........\text{(2)}$

Now we have two equations (1) and (2):

$78 = 2a + 9d \quad \text{(1)}$

$3d = 8a \quad \text{(2)}$

We can solve these equations simultaneously.

From equation (2):

$d = \frac{8}{3}a$

Plugging this back into (1):

$78 = 2a + 9\left(\frac{8}{3}a\right)$

$78 = 2a + 24a$

$78 = 26a$

$a = 3$

Now we can find $d$ :

$d = \frac{8}{3}a$

$d = \frac{8}{3} \times 3$

$d = 8$

Now we can find $S_{15}$ and $S_5$ using the formula for the sum of an AP.

For $S_{15}$ :

$S_{15} = \frac{15}{2} (2 \cdot 3 + (15-1) \cdot 8)$

$S_{15} = \frac{15}{2} (6 + 14 \cdot 8)$

$S_{15} = \frac{15}{2} (6 + 112)$

$S_{15} = \frac{15}{2} \cdot 118$

$S_{15} = 15 \cdot 59$

$S_{15} = 885$

For $S_5$ :

$S_5 = \frac{5}{2} (2 \cdot 3 + (5-1) \cdot 8)$

$S_5 = \frac{5}{2} (6 + 4 \cdot 8)$

$S_5 = \frac{5}{2} (6 + 32)$

$S_5 = \frac{5}{2} \cdot 38$

$S_5 = 5 \cdot 19$

$S_5 = 95$

The difference $S_{15} - S_{5}$ is:

$S_{15} - S_{5} = 885 - 95$

$S_{15} - S_{5} = 790$

Therefore, the correct answer is Option C, which is 790.

Q 70MathematicsSection A • MCQlimits continuity and differentiability

+4 / -1

Let

f(x)=\left\{\begin{array}{l}x-1, x \text { is even, } \\ 2 x, \quad x \text { is odd, }\end{array} x \in \mathbf{N}\right.

.

If for some

\mathrm{a} \in \mathbf{N}, f(f(f(\mathrm{a})))=21

, then

\lim\limits_{x \rightarrow \mathrm{a}^{-}}\left\{\frac{|x|^3}{\mathrm{a}}-\left[\frac{x}{\mathrm{a}}\right]\right\}

, where

[t]

denotes the greatest integer less than or equal to

t

, is equal to :

A

169

B

121

C

225

D

144

Q 71MathematicsSection A • MCQcircle

+4 / -1

Let the locus of the midpoints of the chords of the circle

x^2+(y-1)^2=1

drawn from the origin intersect the line

x+y=1

at

\mathrm{P}

and

\mathrm{Q}

. Then, the length of

\mathrm{PQ}

is :

A

\frac{1}{2}

B

1

C

\frac{1}{\sqrt{2}}

D

\sqrt{2}

Q 72MathematicsSection A • MCQ3d geometry

+4 / -1

If the mirror image of the point

P(3,4,9)

in the line

\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}

is

(\alpha, \beta, \gamma)

, then 14

(\alpha+\beta+\gamma)

is :

A

102

B

138

C

132

D

108

Q 73MathematicsSection A • MCQbinomial theorem

+4 / -1

Let

m

and

n

be the coefficients of seventh and thirteenth terms respectively

in the expansion of

\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}

. Then

\left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}}

is :

A

\frac{1}{9}

B

\frac{1}{4}

C

\frac{4}{9}

D

\frac{9}{4}

Q 74MathematicsSection B • Numericalvector algebra

+4 / -1

Let

\overrightarrow{\mathrm{a}}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=-\hat{i}-8 \hat{j}+2 \hat{k}

and

\overrightarrow{\mathrm{c}}=4 \hat{i}+\mathrm{c}_2 \hat{j}+\mathrm{c}_3 \hat{k}

be three vectors such that

\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}

. If the angle between the vector

\overrightarrow{\mathrm{c}}

and the vector

3 \hat{i}+4 \hat{j}+\hat{k}

is

\theta

, then the greatest integer less than or equal to

\tan ^2 \theta

is _______________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 75MathematicsSection B • Numericaldifferential equations

+4 / -1

If

\frac{\mathrm{d} x}{\mathrm{~d} y}=\frac{1+x-y^2}{y}, x(1)=1

, then

5 x(2)

is equal to __________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 76MathematicsSection A • MCQdifferential equations

+4 / -1

Let

\alpha

be a non-zero real number. Suppose

f: \mathbf{R} \rightarrow \mathbf{R}

is a differentiable function such that

f(0)=2

and

\lim\limits_{x \rightarrow-\infty} f(x)=1

. If

f^{\prime}(x)=\alpha f(x)+3

, for all

x \in \mathbf{R}

, then

f\left(-\log _{\mathrm{e}} 2\right)

is equal to :

A

7

B

9

C

3

D

5

Q 77MathematicsSection B • Numericalarea under the curves

+4 / -1

The sum of squares of all possible values of

k

, for which area of the region bounded by the parabolas

2 y^2=\mathrm{k} x

and

\mathrm{ky}^2=2(y-x)

is maximum, is equal to :

Answer Format:Enter numerical value rounded to the nearest integer

Q 78MathematicsSection A • MCQellipse

+4 / -1

Let

\mathrm{P}

be a point on the ellipse

\frac{x^2}{9}+\frac{y^2}{4}=1

. Let the line passing through

\mathrm{P}

and parallel to

y

-axis meet the circle

x^2+y^2=9

at point

\mathrm{Q}

such that

\mathrm{P}

and

\mathrm{Q}

are on the same side of the

x

-axis. Then, the eccentricity of the locus of the point

R

on

P Q

such that

P R: R Q=4: 3

as

P

moves on the ellipse, is :

A

\frac{13}{21}

B

\frac{\sqrt{139}}{23}

C

\frac{\sqrt{13}}{7}

D

\frac{11}{19}

Q 79MathematicsSection B • Numericalstraight lines and pair of straight lines

+4 / -1

Let

A B C

be an isosceles triangle in which

A

is at

(-1,0), \angle A=\frac{2 \pi}{3}, A B=A C

and

B

is on the positve

x

-axis. If

\mathrm{BC}=4 \sqrt{3}

and the line

\mathrm{BC}

intersects the line

y=x+3

at

(\alpha, \beta)

, then

\frac{\beta^4}{\alpha^2}

is __________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 80MathematicsSection B • Numericalmatrices and determinants

+4 / -1

Let

A=I_2-2 M M^T

, where

M

is a real matrix of order

2 \times 1

such that the relation

M^T M=I_1

holds. If

\lambda

is a real number such that the relation

A X=\lambda X

holds for some non-zero real matrix

X

of order

2 \times 1

, then the sum of squares of all possible values of

\lambda

is equal to __________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 81MathematicsSection A • MCQdefinite integration

+4 / -1

The value of

\int\limits_0^1\left(2 x^3-3 x^2-x+1\right)^{\frac{1}{3}} \mathrm{~d} x

is equal to :

A

-1

B

2

C

0

D

1

Q 82MathematicsSection A • MCQtrigonometric functions and equations

+4 / -1

The number of solutions of the equation

4 \sin ^2 x-4 \cos ^3 x+9-4 \cos x=0 ; x \in[-2 \pi, 2 \pi]

is :

A

0

B

3

C

1

D

2

Q 83MathematicsSection A • MCQlimits continuity and differentiability

+4 / -1

Let

f(x)=\left|2 x^2+5\right| x|-3|, x \in \mathbf{R}

. If

\mathrm{m}

and

\mathrm{n}

denote the number of points where

f

is not continuous and not differentiable respectively, then

\mathrm{m}+\mathrm{n}

is equal to :

A

5

B

3

C

2

D

0

Q 84MathematicsSection A • MCQ3d geometry

+4 / -1

Let

\mathrm{P}

and

\mathrm{Q}

be the points on the line

\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}

which are at a distance of 6 units from the point

\mathrm{R}(1,2,3)

. If the centroid of the triangle PQR is

(\alpha, \beta, \gamma)

, then

\alpha^2+\beta^2+\gamma^2

is :

A

18

B

24

C

26

D

36

Q 85MathematicsSection A • MCQstatistics

+4 / -1

Consider 10 observations

x_1, x_2, \ldots, x_{10}

such that

\sum\limits_{i=1}^{10}\left(x_i-\alpha\right)=2

and

\sum\limits_{i=1}^{10}\left(x_i-\beta\right)^2=40

, where

\alpha, \beta

are positive integers. Let the mean and the variance of the observations be

\frac{6}{5}

and

\frac{84}{25}

respectively. Then

\frac{\beta}{\alpha}

is equal to :

A

2

B

1

C

\frac{5}{2}

D

\frac{3}{2}

Official Correct Answer:Option (A)

Step-by-Step Solution:

We have given

\bar{x}(

mean

)=\frac{6}{5}

\begin{aligned} & \text { Variance }=\frac{84}{25} \\ & \sum_{i=1}^{10}\left(x_i-\alpha\right)=2 \\ & \Rightarrow x_1+x_2+\ldots+x_{10}-10 \alpha=2 \\ & \Rightarrow \frac{x_1+x_2+\ldots+x_{10}}{10}-\alpha=\frac{2}{10} \\ & \Rightarrow \frac{6}{5}-\alpha=\frac{2}{10} \\ & \Rightarrow \alpha=1 \end{aligned}

\begin{aligned} & \text { and } \sum_{i=1}^{10}\left(x_i-\beta\right)^2=40 \\ & \left(x_1-\beta\right)^2+\left(x_2-\beta\right)^2+\ldots+\left(x_{10}-\beta\right)^2=40 \\ & x_1^2+x_2^2+\ldots+x_{10}^2+10 \beta^2-2 \beta\left(x_1+x_2+\ldots+x_{10}\right)=40 \\ & \Rightarrow \frac{x_1^2+x_2^2+\ldots+x_{10}^2}{10}+\beta^2-\frac{2 \beta\left(x_1+x_2+\ldots+x_{10}\right)}{10}=4 \\ & \Rightarrow \frac{x_1^2+x_2^2+\ldots+x_{10}^2}{10}-\frac{36}{25}+\frac{36}{25}+\beta^2-2 \beta \times \frac{6}{5}=4\end{aligned}

\left[\right.

Variance

\left.=\frac{\sum_{i=1}^n x_i^2}{n}-(\bar{x})^2\right]

\begin{aligned} & \Rightarrow \frac{84}{25}+\frac{36}{25}+\beta^2-\frac{12 \beta}{5}-4=0 \\ & \Rightarrow \frac{120}{25}+\beta^2-\frac{12 \beta}{5}-4=0 \\ & \Rightarrow 25 \beta^2-60 \beta+20=0 \\ & \Rightarrow 5 \beta^2-12 \beta+4=0 \\ & \Rightarrow \beta=2, \frac{2}{5}\end{aligned}

Take

\beta=2

\frac{\beta}{\alpha}=\frac{2}{1}=2

Q 86MathematicsSection B • Numericalarea under the curves

+4 / -1

Three points

\mathrm{O}(0,0), \mathrm{P}\left(\mathrm{a}, \mathrm{a}^2\right), \mathrm{Q}\left(-\mathrm{b}, \mathrm{b}^2\right), \mathrm{a}>0, \mathrm{~b}>0

, are on the parabola

y=x^2

. Let

\mathrm{S}_1

be the area of the region bounded by the line

\mathrm{PQ}

and the parabola, and

\mathrm{S}_2

be the area of the triangle

\mathrm{OPQ}

. If the minimum value of

\frac{\mathrm{S}_1}{\mathrm{~S}_2}

is

\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1

, then

\mathrm{m}+\mathrm{n}

is equal to __________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 87MathematicsSection B • Numericaldefinite integration

+4 / -1

Let

f:(0, \infty) \rightarrow \mathbf{R}

and

\mathrm{F}(x)=\int\limits_0^x \mathrm{t} f(\mathrm{t}) \mathrm{dt}

. If

\mathrm{F}\left(x^2\right)=x^4+x^5

, then

\sum\limits_{\mathrm{r}=1}^{12} f\left(\mathrm{r}^2\right)

is equal to ____________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 88MathematicsSection B • Numericalsequences and series

+4 / -1

If three successive terms of a G.P. with common ratio

\mathrm{r}(\mathrm{r}>1)

are the lengths of the sides of a triangle and

[r]

denotes the greatest integer less than or equal to

r

, then

3[r]+[-r]

is equal to _____________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:1

Step-by-Step Solution:

To solve this problem, let's first denote the three successive terms of a geometric progression (G.P.) with common ratio $r$ as $a$ , $ar$ , and $ar^2$ , where $a$ is the first term and $r > 1$ . These three terms represent the lengths of the sides of a triangle.

According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Therefore, for the three terms to form a triangle, the following inequalities must hold:

$1) \ a + ar > ar^2 \\$

$2) \ a + ar^2 > ar$

$3) \ ar + ar^2 > a$

Given that $r > 1$ , inequalities 2 and 3 will always hold because:

$ar < ar^2 \ \text{and} \ a < ar,$

indicating that both $a + ar^2$ and $ar + ar^2$ will be greater than $ar$ and $a$ respectively. Therefore, we only need to check the first inequality to ensure that the three terms can form a triangle:

$a + ar > ar^2$

Simplifying this, we get:

$a(1 + r) > a r^2$

Since $a$ is positive (as it represents the length of a side of a triangle), we can divide both sides of the inequality by $a$ without changing the direction of the inequality:

$1 + r > r^2$

We can then subtract $r$ from both sides:

$1 > r^2 - r$

Simplifying the right side by factoring $r$ :

$1 > r(r - 1)$

Given that $r > 1$ , the quantity $(r - 1)$ is positive; hence, $r(r - 1)$ is also positive. This means the actual value for $r$ to satisfy the inequality is within the interval $(1, \sqrt{2})$ because $r(r - 1)$ increases with increasing $r$ , and it would be 1 when $r = \sqrt{2}$ . It should be greater than 1, and less than $\sqrt{2}$ such that $r^2 - r$ stays below 1.

Now let’s consider the expressions $[r]$ and $[-r]$ . The symbol $[x]$ denotes the greatest integer less than or equal to $x$ (also known as the floor function).

Since $1 < r < \sqrt{2}$ , $[r] = 1$ , because 1 is the greatest integer less than $r$ within that interval.

For $[-r]$ , we need the greatest integer less than or equal to $-r$ . Since $-r$ is negative and less than $-1$ (because $r > 1$ ), $[-r] = -2$ , as this is the greatest integer that does not exceed the negative value of $r$ (which lies between $-\sqrt{2}$ and $-1$ ).

Now we can substitute these values into the expression:

$3[r] + [-r] = 3 \cdot 1 + (-2) = 3 - 2 = 1$

Therefore $3[r] + [-r]$ is equal to 1.

Q 89MathematicsSection B • Numericalstraight lines and pair of straight lines

+4 / -1

The lines

\mathrm{L}_1, \mathrm{~L}_2, \ldots, \mathrm{L}_{20}

are distinct. For

\mathrm{n}=1,2,3, \ldots, 10

all the lines

\mathrm{L}_{2 \mathrm{n}-1}

are parallel to each other and all the lines

L_{2 n}

pass through a given point

P

. The maximum number of points of intersection of pairs of lines from the set

\left\{\mathrm{L}_1, \mathrm{~L}_2, \ldots, \mathrm{L}_{20}\right\}

is equal to ___________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 90MathematicsSection B • Numericaldifferentiation

+4 / -1

If

y=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x

, then

96 y^{\prime}\left(\frac{\pi}{6}\right)

is equal to :

Answer Format:Enter numerical value rounded to the nearest integer

JEE Main 2024 1st February Evening Shift Question Paper with Solutions