JEE Main Part Test - 1 - JEE MCQ

# JEE Main Part Test - 1 - JEE MCQ

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## 75 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Part Test - 1

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JEE Main Part Test - 1 - Question 1

### The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the volume of the sheet to correct significant figures.

Detailed Solution for JEE Main Part Test - 1 - Question 1

Length , l = 4,234 m
Thickness , t = 2.01 x 10-2 m
Volume = l x b x t
⇒ V = 4.234 x 1.005 x 0.0201 = 0.0855289 = 0.0855 m3 (significant figure = 3)

JEE Main Part Test - 1 - Question 2

### Population of a town is reported as 157,900 . Which of the following statements correct?

Detailed Solution for JEE Main Part Test - 1 - Question 2

157900  implies that the population is believed to be within the range of about 157850 to about 157950.

In other words, the population is 157900 ± 50. The “plus-or-minus 50” appended to this number means that we consider the absolute uncertainty of the population measurement to be 50 – (–50) = 100.

We can also say that the relative uncertainty is 100/157900, which we can also express as 1 part in 1579, or 1/1579 = 0.000633, or about 0.06 percent.

JEE Main Part Test - 1 - Question 3

### The product of energy and time is called action. The dimensional formula for action is same as that for

Detailed Solution for JEE Main Part Test - 1 - Question 3

Both are different from each other.
Energy × Time=(M1L2T−2)×(T1)=M1L2T−1
Force × Velocity=(M1L1T−2)×(L1T−1)=M1L2T−3
Impulse × distance=(M1L1T−1)×(L1)=M1L2T−1
Power=M1L2T−3
Angular Energy=M1L1T−2

JEE Main Part Test - 1 - Question 4

The Van der Waal equation for 1 mole of a real gas is

where P is the pressure, V is the volume, T is the absolute temperature, R is the molar gas constant and a, b are Van dar Waal constants. The dimensions of a are the same as those of

Detailed Solution for JEE Main Part Test - 1 - Question 4

Solution,
As we know the Vander Waals equation is {P+(a/V2)} (V−b) =RT
Then,
The dimension of a is,
(a/V2) = (P)
Or, a=PV2

JEE Main Part Test - 1 - Question 5

A particle is projected with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is

Detailed Solution for JEE Main Part Test - 1 - Question 5

JEE Main Part Test - 1 - Question 6

The  range R of projectile is same when its maximum height are h1 and h2. What is the relation between R, h1 and h2?

Detailed Solution for JEE Main Part Test - 1 - Question 6

Range is same for angles of projection θ and 900 - θ

JEE Main Part Test - 1 - Question 7

A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first 2 sec, it rotates through an angle q1; in the next 2 sec it rotates through an additional angle θ2, the ratio θ21 is

Detailed Solution for JEE Main Part Test - 1 - Question 7

JEE Main Part Test - 1 - Question 8

Two cars are moving in same direction with speed of 30 kmph. They are separated by a distance of 5 km. What is the speed of a car moving in opposite direction if it meets the two cars at an interval of 4 min?

Detailed Solution for JEE Main Part Test - 1 - Question 8

Lets say the speed of second car be v. Now with respect to car 1, speed of car 2 is v - 30. Now if the cross each other at just after 4 min then the distance traveled by car 2 relative to car 1 is (v -30) x 4/60 and we know the distance travelled relatively is 5km
Thus we get
(v -30) x 4/60 = 5
V = 300/4 - 30
= 45 km/h

JEE Main Part Test - 1 - Question 9

A body starts from rest, the ratio of distances travelled by the body during 3rd and 4thseconds is :

Detailed Solution for JEE Main Part Test - 1 - Question 9

The velocity after 2 sec = 2a and velocity after 3 sec = 3a
For some constant acceleration a,
Now distance travelled in third second, s3 = 2a + ½ a
= 5/2 a
Similarly distance travelled in fourth second, s4= 3a + ½ a
= 7/2 a
Hence the reqiuired ratio is 5/7

JEE Main Part Test - 1 - Question 10

The dimensional formula for acceleration is

Detailed Solution for JEE Main Part Test - 1 - Question 10

Acceleration (a) = Velocity × Time-1 . . . . . (1)

Since, Velocity (v) = Displacement × Time-1

∴ The dimensional formula of velocity is [M0 L1 T-1] . . . . (2)

On substituting equation (2) in equation (1) we get,

Acceleration = Velocity × Time-1

Or, a = [M0 L1 T-1] × [T]-1 = [M0 L1 T-2]

Therefore, acceleration is dimensionally represented as [M0 L1 T-2].

JEE Main Part Test - 1 - Question 11

An object is said to be in uniform motion in a straight line if its displacement

Detailed Solution for JEE Main Part Test - 1 - Question 11

Explanation:Uniform motion is the kind of motion in which a body covers equal displacement in equal intervals of time. It does not matter how small the time intervals are, as long as the displacements covered are equal.

If a body is involved in rectilinear motion and the motion is uniform, then the acceleration of the body must be zero.

JEE Main Part Test - 1 - Question 12

For motion in 3 dimensions we need

Detailed Solution for JEE Main Part Test - 1 - Question 12

Explanation:Motion is a change in position of an object with time. In order to specify position, we need to use a reference point and a set of axes. It is convenient to choose a rectangular coordinate system consisting of three mutually perpenducular axes, labelled X-, Y-, and Z- axes.

The point of intersection of these three axes is called origin (O) and serves as the reference point. The coordinates (x, y. z) of an object describe the position of the object with respect to this coordinate system.

To measure time, we position a clock in this system. This coordinate system along with a clock constitutes a frame of reference.

*Multiple options can be correct
JEE Main Part Test - 1 - Question 13

A man pulls a block heavier than himself with a light horizontal rope. The coefficient of friction is the same between the man and the ground, and between the block and the ground.

Detailed Solution for JEE Main Part Test - 1 - Question 13

The friction force between the block and ground is more as compared to friction force between man and ground.
such that unless man doesn't move the block will not be moved.
The block of mass say M is heavier than the man of mass say m. The surface is rough with friction coefficient say μ. So when the man applies the force on the block the force cannot exceed the frictional force μmg without moving as μMg>μmg. Now if he starts moving (i.e. the force applied is increased and now the friction between him and the surface is not holding him stationary) there is a possibility that the block may move. Now as there is no other force acting on the system and as the man is lighter than block so he would have greater acceleration than the block when both move.

JEE Main Part Test - 1 - Question 14

Adjoining figure shows a force of 40 N acting at 30° to the horizontal on a body of mass 5 kg resting on a smooth horizontal surface. Assuming that the acceleration of free-fall is 10 ms_2, which of the following statements A, B, C is (are) correct?

[1] The horizontal force acting on the body is 20 N

[2] The weight of the 5 kg mass acts vertically downwards

[3] The net vertical force acting on the body is 30 N

Detailed Solution for JEE Main Part Test - 1 - Question 14

JEE Main Part Test - 1 - Question 15

A man getting down a running bus, falls forward because-

Detailed Solution for JEE Main Part Test - 1 - Question 15

The explanation is the question itself. Inertia is a property of which it resists its change of state of rest or state of motion, so as soon as the person jumps the lower part immediately comes to rest by sharing contact with the ground the upper body due to inertia of motion resists its change in state of motion.

JEE Main Part Test - 1 - Question 16

Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity 2v  in a directionperpendicular to the original direction. The mass A moves after collision in the direction.

Detailed Solution for JEE Main Part Test - 1 - Question 16

There is no external force acting on the spheres. So linear momentum will be conserved.

Before the collision, In direction x, Linear momentum = m2v ....1

In direction y, Linear momentum=0

After the collision, spheres moves as shown in figure. Let velocity of sphere A is v1

In direction x, Linear momentum =m1v1cos(θ) ...2

In direction y, Linear momentum =m2v2−m1v1sin(θ) .......3

Linear momentum will be conserved,

From equation 1 and 2, ⇒ m1v1cos(θ)=m2v ...4

From the equation 2, ⇒ m2v− m1v1sin(θ) = 0 ⇒ m1v1sin(θ) = m2v2 ......5

Dividing equation 5 by 4, ⇒ tan(θ) = 1/2 ⇒ θ = tan−1(1/2)

JEE Main Part Test - 1 - Question 17

A ring of mass 200 gram is attached to one end of a light spring of force constant 100 N/m and natural length 10 cm. The ring is constrained to move on a rough wire in the shape of the quarter ellipse of the major axis 24 cm and the minor axis 16 cm with its centre at the origin. The plane of the ellipse is vertical and wire is fixed at points A and B as shown in the figure. Initially, ring is at A with other end of the spring fixed at the origin. If normal reaction of wire on ring at A is zero and ring is given a horizontal velocity of 10 m/s towards right so that it just reaches point B, then select the correct alternative (s) (g = 10 m/s2)

Detailed Solution for JEE Main Part Test - 1 - Question 17

∴ Spring is compressed by 2 cm at A and will be stretched by 2 cm at B.
By work-energy theorem,
Wmg + Ws + WN + Wf = Kf − Ki
0.16 + 0 + 0 + Wf = 0 - 10
Wf = -10.16

JEE Main Part Test - 1 - Question 18

Two spheres of masses m1 and m2 (m1>m2) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is:

Detailed Solution for JEE Main Part Test - 1 - Question 18

If, r1 and r2 are position vectors of the centres of the positional vector of their centre of mass is given by R=(m1r1+m2r2)/ (m1 + m2)
R= (m1r2+m2r2)/m1+m2)
The acceleration of the centre of mass is given by:
A=d2R/dt
= [m1 d2r/dt+m2 d2r/dt2 ]/(m1+m2)
But, d2r1/dt2 and d2r2/dt2 are the accelerations of masses m1 and m2
Have the same magnitude (m1-m2)/(m1+m2) g
If we take acceleration of [m1(m1-m2)/ (m1+m2)g – m2(m1-m2)/ (m1+m2)/]/ (m1+m2)
On simplifying that we get,
a=[(m1-m2)/ (m1+m2)]2 g

JEE Main Part Test - 1 - Question 19

Which of the following options are correct,
where i, j and k are unit vectors along the x, y and z axis?

Detailed Solution for JEE Main Part Test - 1 - Question 19

Dot product of two different unit vectors is 0 and dot product of two same unit vectors is 1. Cross product of two different unit vectors taken according to right hand thumb rule is the other vector. Cross product of two same unit vectors is 0.

JEE Main Part Test - 1 - Question 20

A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in the figure. When the string is cut, the initial angular acceleration of the rod is:
[2013]

Detailed Solution for JEE Main Part Test - 1 - Question 20

Weight of the rod will produce the torque,
τ = mg x L / 2 = I α = mL2 / 3 α (∵ Irod = ML2 / 3)
Hence, angular acceleration, α = 3g / 2L

*Answer can only contain numeric values
JEE Main Part Test - 1 - Question 21

A physical quantity Q is given by

The percentage error in A,B, C, D are 1%, 2%, 4%, 2% respectively. Find the percentage error in Q.

Detailed Solution for JEE Main Part Test - 1 - Question 21

= 22 or 22 %

*Answer can only contain numeric values
JEE Main Part Test - 1 - Question 22

A bullet is fired from a gun at a speed of 5000 m/s. At what height should the gun be aimed above a goal if it has to strike the goal at a distance of 500 m? Take g=10m/s2

Detailed Solution for JEE Main Part Test - 1 - Question 22

Let t be the time taken to cover the horizontal distance of 500 m from the gun to the target ,then
t = 500 / 5000 = .1 s
During this time,bullet will fall down vertically due to acceleration due to gravity.
Vertical distance covered
h = 1/2gt2 = .005 m
= 5cm.
Hence the gun should be aimed at 5cm above the target

*Answer can only contain numeric values
JEE Main Part Test - 1 - Question 23

A bullet of mass 20 g strikes a pendulum of mass 5 kg. The centre of mass of pendulum rises a vertical distance of 10 cm. If the bullet gets embedded into the pendulum, calculate its initial speed.

Detailed Solution for JEE Main Part Test - 1 - Question 23

Mass of the bullet m1 = 20 g = 0.02 kg.

Mass of the pendulum m2 = 5 kg

Centre of mass of pendulum rises to a height = h = 10 cm = 0.1 m

Speed of the bullet = u1

Pendulum is at rest .:. u2 = 0

Common velocity of the bullet and the pendulum after the bullet is embeded into the object = v

From II equation of motion

v = √[2gh] = √[2 x 9.8 x 0.1] = √[1.96] = 1.4 ms-1

Substitute the value of v in equation (1)

1.4 = 0.02 u1/5.02

u= 5.02x1.4 / 0.02
u= 351.4 ms-1

*Answer can only contain numeric values
JEE Main Part Test - 1 - Question 24

A disc has a speed of 1200 rpm and it is made to slow down at a uniform rate of 4 rad/s2.Calculate the number of revolution it makes before coming to rest?

Detailed Solution for JEE Main Part Test - 1 - Question 24

f = 1200 rpm = 20 rotation per sec
Initial angular velocity of the disc (ω0) = 2πf = 2π × 20 = 40π rad/s
Final angular velocity of the disc (ωf) = 0
Angular acceleration ((α) = -4 rad/s2
Now,

Now,

Therefore, Number of revolutions =

= 314

*Answer can only contain numeric values
JEE Main Part Test - 1 - Question 25

In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10–10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Detailed Solution for JEE Main Part Test - 1 - Question 25

Mass of H atom = m
Mass of Cl atom = 35.5m
Let the centre of mass of the system lie at a distance x from the Cl atom.
Distance of the centre of mass from the H atom = (1.27 – x)

Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:
[ m(1.27 – x) + 35.5mx ] / (m + 35.5m)  =  0
m(1.27 – x) + 35.5mx =  0
1.27 - x = -35.5x
∴ x = -1.27 / (35.5 - 1)  =  -0.037 Å
Here, the negative sign indicates that the centre of mass lies at the left of the molecule. Hence, the centre of mass of the HCl molecule lies 0.037Å from the Cl atom.

JEE Main Part Test - 1 - Question 26

Electromagnetic radiations having λ = 310 Å are subjected to a metal sheet having work function = 12.8 eV. What will be the velocity of photoelectrons with maximum Kinetic energy.

Detailed Solution for JEE Main Part Test - 1 - Question 26

V2 = 9.56 × 1012
V = 3.09 × 106 m/sec.

*Multiple options can be correct
JEE Main Part Test - 1 - Question 27

Select the correct statement(s).

Detailed Solution for JEE Main Part Test - 1 - Question 27

Option A: The ideal body, which emits and absorbs all frequencies is called a black body
Option B:  As temperature increases, distribution of frequency (and energy) increases.

Option C: As temperature increases, radiation emitted is shifted to a lower wavelength, i.e. higher frequency.
Option D: All types of radiations have the same speed 3 x 108 ms-1.

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

JEE Main Part Test - 1 - Question 28

Consider the ground state of Cr atom (X = 24). The number of electrons with the azimuthal quantum numbers, ℓ = 1 and 2 are, respectively [2004]

Detailed Solution for JEE Main Part Test - 1 - Question 28

Electronic configuration of Cr atom (z = 24) = 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s1
when ℓ = 1, p - subshell,
Numbers of electrons = 12
when ℓ = 2, d - subshell,
Numbers of electrons = 5

JEE Main Part Test - 1 - Question 29

Which of the following sets of quantum numbers is correct for an electron in 4f orbital ? [2004]

Detailed Solution for JEE Main Part Test - 1 - Question 29

The possible quan tum numbers for 4f electron are n = 4, ℓ = 3, m = – 3, –2 –1, 0, 1 , 2 , 3 and s =

Of various possiblities only option (a) is possible.

*Multiple options can be correct
JEE Main Part Test - 1 - Question 30

Radial probability density in the occupied orbital of a hydrogen atom in the ground state (1s) is given below

Detailed Solution for JEE Main Part Test - 1 - Question 30

Radial probability increases as r increase reaches a maximum value when r = a0 (Bohr’s radius) and then falls. When radial probability is very small.
Thus, (a) and (c) are true.

JEE Main Part Test - 1 - Question 31

25.0 g of FeSO4.7H2O was dissolved in water containing dilute H2SO4, and the volume was made up to 1.0 L. 25.0 mL of this solution required 20 mL of an N/10 KMnO4 solution for complete oxidation. The percentage of FeSO4. 7H2O in the acid solution is

Detailed Solution for JEE Main Part Test - 1 - Question 31

JEE Main Part Test - 1 - Question 32

If 6.3 g of NaHCO3 are added to 15.0 g CH3COOH solution, the residue is found to weigh 18.0 g. What will be the mass of CO2 released in the reaction?

Detailed Solution for JEE Main Part Test - 1 - Question 32

The correct answer is Option B.
The chemical reaction will be:
NaHCO3 + CH3COOH →CH3COONa + H2O + CO2
molar mass:
NaHCO3 = 84
CH3COOH=60
CH3COONa=82
CO2 =44
84gNaHCO3 + 60gCH3COOH → 82gCH3COONa + 44gCO2
Moles of NaHCO3 = 6.3/84= 0.075
Moles of CH3COOH = 15/60= 0.25
∴NaHCO3 is the limited reagent.
Moles of CO2 formed = 0.075
Weight of CO2 = 0.075×44 = 3.3g

JEE Main Part Test - 1 - Question 33

Suppose the elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mole of XY2 weighs 10 g and 0.05 mole of X3Y2 weighs 9 g, the atomic weights of X and Y are

Detailed Solution for JEE Main Part Test - 1 - Question 33

Let atomic weight of x = Mx

atomic weight of y = My

we know,

mole = weight /atomic weight

a/c to question,

mole of xy2 = 0.1

so,

0.1 = 10g/( Mx +2My)

Mx + 2My = 100g -------(1)

for x3y2 ; mole of x3y2 = 0.05

0.05 = 9/( 3Mx + 2My )

3Mx + 2My = 9/0.05 = 9× 20 = 180 g ---(2)

solve eqns (1) and (2)

2Mx = 80

Mx = 40g/mol

and My = 30g/mole

JEE Main Part Test - 1 - Question 34

Number of nitrogen atoms present in 1.4 g of N2

Detailed Solution for JEE Main Part Test - 1 - Question 34

Given weight of N₂ gas = 1.4 g

Molar mass of N₂ gas = 28 g

So, mole = given mass/ molar mass

⇒ mole = 1.4/28 = 1/20 mole

Now, number of molecules = mole × avogadro number

⇒ number of molecules = 1/20 × 6.022 × 10²³

⇒ number of molecules = 3.011 × 10²²

Now, we are asked for number of atoms. In N₂, there are 2 atoms, so to obtain number of atoms we will multiply with 2 in number of molecules.

⇒ number of atoms = 2 × 3.011 × 10²²

⇒ number of atoms = 6.022 × 10²²

JEE Main Part Test - 1 - Question 35

The following equations are balanced atomwise and chargewise.

(i) Cr2O72- + 8H+ + 2H2O2 → 2Cr3+ + 7H2O + 2O2

(ii) Cr2O72- + 8H+ + 5H2O2→ 2Cr3+ + 9H2O + 4O2

(iii) Cr2O72- + 8H+ + 7H2O2→ 2Cr+ + 11H2O + 5O2

The precise equationl equations representing the oxidation of H2O2 is/are

Detailed Solution for JEE Main Part Test - 1 - Question 35

The correct answer is option A
Cr2O72- converts into Cr3+ in acidic medium I.e. in H+ medium.
First balance the Cr atom on both sides and then Oxygen atom. H+ is in excess due to acidic medium.
Add H+ as +ve charge to balance the charge on both sides.

JEE Main Part Test - 1 - Question 36

Elements which occupied position in the lother meyer curve, on the peaks, were -

Detailed Solution for JEE Main Part Test - 1 - Question 36

Hydrogen, the first in the list of elements is a special case and can be considered as making up the first period all by itself. The second and third period in Meyer's table included seven elements each, and duplicated Newlands's law of octaves.

Li Be B C N O F

Na Mg Al Si P S Cl

However, the next wave had more than seven elements. The third wave had about 17 to 18 elements. This clearly showed where Newlands’ law had failed. One could not force the law of octaves to hold strictly throughout the table of elements, with seven elements in each row. After the first two periods, the length of the period had to be longer.so all the correct answer.

JEE Main Part Test - 1 - Question 37

In the general electronic configuration -

(n - 2)f1-14 (n - 1)d0-1 ns2, if value of n = 7 the configuration will be -

Detailed Solution for JEE Main Part Test - 1 - Question 37

The correct answer is Option B.

General electronic configuration is given:
(n − 2)f1-14(n − 1)d01ns2 , where(n=7)
5f1-146d01ns2
The seventh period (n=7) is similar to the sixth period with the successive filling up of the 7s, 5f, 6d and 7p orbitals and includes most of the man-made radioactive elements. This period will end at the element with atomic number 118 which would belong to the noble gas family. Filling up of the 5f orbitals after actinium (Z=89) gives the 5f-inner transition series known as the Actinide Series.

JEE Main Part Test - 1 - Question 38

In which group of the modern periodic table are halogens placed?

Detailed Solution for JEE Main Part Test - 1 - Question 38

Periods:

• Elements are arranged in increasing the atomic number of elements in a period.
• One extra electron gets added to the outermost shell as we move along the periods from left to right.
• The electron gets added to the same shell or orbit and thus the electrons present for bonding increase by one unit.
• Thus, the shell number remains the same but the number of electrons present for bonding increases along a period.

Groups:

• Elements having the same number of outer electrons are put in the same group of the periodic table.
• When we move down a group, one extra shell gets added to the elements.
• The outermost shell has electrons present for bonding.
• Though the number of shells increases as we go down in a group, the number of electrons in the outermost shell remains the same.
• For example, the Halogens F, Cl, Br, I, At all belong to group 17 and have 7 electrons in the outermost shell.
• Similarly, Group 16 elements have 6 electrons in the outermost shell, group 15 has 5 electrons in the outermost shell, and so on.

Hence, group 17 is called halogens.

JEE Main Part Test - 1 - Question 39

The symbol for element with atomic number 111 and name Unununium  is

Detailed Solution for JEE Main Part Test - 1 - Question 39
• The element was to be called unununium (with the corresponding symbol of Uuu),a systematic element name as a placeholder, until the element was discovered (and the discovery then confirmed) and a permanent name was decided on.
• Although widely used in the chemical community on all levels, from chemistry classrooms to advanced textbooks, the recommendations were mostly ignored among scientists in the field, who called it element 111, with the symbol of E111, (111) or even simply 111.
JEE Main Part Test - 1 - Question 40

Azide ion [N3-] exhibits an (N−N) bond order of 2 and may be represented by the resonance structures I,II and III given below:

The most stable resonance structure is

Detailed Solution for JEE Main Part Test - 1 - Question 40

JEE Main Part Test - 1 - Question 41

C—Cl bond in (vinyl chloride) is stabilised in the same way as in

Detailed Solution for JEE Main Part Test - 1 - Question 41

Due to delocalisation of π-electrons, (C—Cl) bond is stable and it does not show SN reactions; Cl directly attached (C=C) bond, i.e. vinyl group.

JEE Main Part Test - 1 - Question 42

Hydrogen bonds are formed in many compounds e.g., H2O, HF, NH3. The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points of above compounds is :

Detailed Solution for JEE Main Part Test - 1 - Question 42

Strength of H-bonding depends on the electronegativity of the atom which follows the order: F > O > N .
Strength of H-bond is in the order:
H……. F > H…….. O > H…….. N
But each H2O molecule is linked to 4 other H2O molecules through H-bonds whereas each HF molecule is linked only to two other HF molecules.
Hence, correct decreasing order of the boiling points is HzO > HF > NH3.

JEE Main Part Test - 1 - Question 43

Which of the following angle corresponds to sp2 hybridisation?

Detailed Solution for JEE Main Part Test - 1 - Question 43

sp2 hybridisation gives three sp2 hybrid orbitals which are planar triangular forming an angle of 120° with each other.
The electronic configurations of three elements A, B and C are given below.
Answer the questions from 14 to 17 on the basis of these configurations.
A ls22s22p6
B ls22s22p63s23p3
C ls22s22p63s23p

JEE Main Part Test - 1 - Question 44

bond in between (C2— C3) vinyl acetylene is formed by ...... overlapping

Detailed Solution for JEE Main Part Test - 1 - Question 44

Vinyl acetylene is

bond are at equal position, numbering of C-chains based on IUPAC nomenclature is done from side.

JEE Main Part Test - 1 - Question 45

Direction (Q. Nos. 1-11) This section contains 11 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

Q.  Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is

Detailed Solution for JEE Main Part Test - 1 - Question 45

*Answer can only contain numeric values
JEE Main Part Test - 1 - Question 46

In white phosphorous (P4) if
x is total number of triangle
y is total number of plane of symmetry z is total number of P–P bond
then calculate value of (y + z)/x.

Detailed Solution for JEE Main Part Test - 1 - Question 46

x = 4
y = 6
z = 6
Ans. (3)

*Answer can only contain numeric values
JEE Main Part Test - 1 - Question 47

How many chiral centers are in the following compound?

Detailed Solution for JEE Main Part Test - 1 - Question 47

*Answer can only contain numeric values
JEE Main Part Test - 1 - Question 48

The energy of electron in the second and third Bohr orbit of the hydrogen atom is –5.42 x  10–12 erg and –2.41 x  10–12erg, respectively. Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.

Detailed Solution for JEE Main Part Test - 1 - Question 48

E3 – E2 = hv = hc/λ – 2.41 ´ 10–12 – (– 5.42 ´ 10–12)
= 6.626 x 10-27 x 3 x 1010/ λ
∴ λ = 6.626 x 10-27 x 3 x 1010 / 3.01 x 10-12
= 6.604 x 10–5 cm = 6.604 x 10–5 x 108
= 6604 Å

*Answer can only contain numeric values
JEE Main Part Test - 1 - Question 49

29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. The molecular weight of HCl is 36.5 g mol-1. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is

Detailed Solution for JEE Main Part Test - 1 - Question 49

Let mass of the stock solution = 100g

Mass of HCl in 100 g of 29.2 % (w/w) HCl stock solution = 29.2 g

Volume of the stock solution = 100g/1.25 g mL= 80 g

Number of moles of HCl in stock solution = 29.2/36.5 = 0.8

Molarity of the stock solution = (0.82/80)×1000 = 10 M

Using,

M1V1 = M2V2

10× V1 =0.4×200

or

V1 = 8mL

Hence, the volume required is 8 mL.

JEE Main Part Test - 1 - Question 50

How many spectral lines are emitted by atomic hydrogen excited to  nth energy level?

Detailed Solution for JEE Main Part Test - 1 - Question 50

JEE Main Part Test - 1 - Question 51

The domain of the function √(log1/3 log4 ([x]2 - 5)) is (where [x] denotes greatest integer function)

Detailed Solution for JEE Main Part Test - 1 - Question 51

JEE Main Part Test - 1 - Question 52

Domain of definition of the function

Detailed Solution for JEE Main Part Test - 1 - Question 52

JEE Main Part Test - 1 - Question 53

A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to

Detailed Solution for JEE Main Part Test - 1 - Question 53

JEE Main Part Test - 1 - Question 54

Let R be a relation defined on the set of N natural numbers as R = {(x, y): y is a factor of x, x, y∈ N} then,

Detailed Solution for JEE Main Part Test - 1 - Question 54

R = {(x, y): y is a factor of x, x, y∈ N}
As we know that 2 is a factor of 4
So, according to the options (4, 2) ϵ R

JEE Main Part Test - 1 - Question 55

If= {x is a multiple of 3} and= {x is a multiple of 5}, then A - B is

Detailed Solution for JEE Main Part Test - 1 - Question 55

JEE Main Part Test - 1 - Question 56

If z = x- iy and = p+ iq, then  is equal to [2004]

Detailed Solution for JEE Main Part Test - 1 - Question 56

= p+ iq ⇒ z = p3 + (iq)3 + 3 p (iq)( p+ iq)

⇒ x - iy = p3 - 3pq2 + i(3p2q-q3)

∴ x = p3 - 3pq2 ⇒

y = q3 - 3p2q ⇒

JEE Main Part Test - 1 - Question 57

If z ≠ 1 and is real, then the point represented by the complex number z lies : [2012]

Detailed Solution for JEE Main Part Test - 1 - Question 57

Now   is real ⇒  Im  = 0

⇒

⇒​  Im [(x2 – y2 + 2ixy) (x – 1) – iy)] = 0

⇒  2xy (x – 1) – y (x2 – y2) = 0

⇒ y(x2 + y2 – 2x) = 0 ⇒  y = 0;  x2 + y2 – 2x = 0

∴ z lies either on real axis or on a circle through origin.

JEE Main Part Test - 1 - Question 58

The value of is                  [2006]

Detailed Solution for JEE Main Part Test - 1 - Question 58

= i × 0 – i                [∵ e-2πi = 1] = – i

JEE Main Part Test - 1 - Question 59

If | z2 - 1 |=|z |2+1, then z lies on              [2004]

Detailed Solution for JEE Main Part Test - 1 - Question 59

|z2 - 1|=|z|2+1 ⇒|z2 - 1|2

⇒

⇒

⇒  ⇒

⇒ z is purely imaginary

JEE Main Part Test - 1 - Question 60

If b ≠ 1 be any nth root of unity then 1 + 3β + 5β2 + ........n terms equals

Detailed Solution for JEE Main Part Test - 1 - Question 60

β = (1)1/n ⇒ βn = 1 ........(1)
If S be the sum of the given series which is arithmetic geometric series, then

JEE Main Part Test - 1 - Question 61

Find the number of integral values of x for which

(5x - 1) < (x + 1)2 < (7x – 3)

Detailed Solution for JEE Main Part Test - 1 - Question 61

Taking

⇒ (x + 1)2  > (5x - 1)

⇒ x2 – 3x + 2 > 0

⇒ (x – 1 )( x – 2 ) > 0

X < 1 or x > 2        ……(i)

Taking

⇒ (x + 1)2 < (7x – 3)

⇒ x2 – 5x + 4 > 0

⇒ (x – 1 )( x – 4 ) > 0

⇒ 1 < x < 4         ……(ii)

Combining (1) and (2) we get 2 < x < 4

Hence, x will only take one integer i.e. 3

JEE Main Part Test - 1 - Question 62

If (5x + 2) < (8x -1) and (7x - 2) > 3(x + 6). What will be the solution set of x ( x ∈ N?

Detailed Solution for JEE Main Part Test - 1 - Question 62

Natural number (N): Natural numbers include all the whole numbers excluding the number 0.

Calculation:

We have, (5x + 2) < (8x -1)

⇒ 8x - 5x >2 + 1  ⇒ 3x > 3

⇒ x > 1            ------(1)

Also, (7x - 2) > 3(x + 6)

⇒ 7x - 2 > 3x + 18 ⇒ 7x - 3x > 18 + 2

⇒ 4x > 20 ⇒ x > 5         ------(2)

Taking intersection (resultant) of both, we get

x > 5 i.e.  {5, 6, 7, 8, ......}

Hence, option 5 is correct.

JEE Main Part Test - 1 - Question 63

Calculate the least whole number, which when subtracted from both the terms of the ratio 5 : 6 gives a ratio less than 17 : 22.

Detailed Solution for JEE Main Part Test - 1 - Question 63

Given:

Initial ratio = 5 ∶ 6

Final ratio should be less than 17 ∶ 22

Calculation:

Let the least whole number that is needed to be subtracted be a.

According to the question,

(5 - a)/(6 - a) < 17/22

⇒ 5 × 22 - 22a < 17 × 6 - 17a

⇒ 110 - 22a < 102 - 17a

⇒ 110 - 102 < - 17a + 22a

⇒ 8 < 5a

⇒ 8/5 = 1.6 < a

∴ The least whole number must be 2.

JEE Main Part Test - 1 - Question 64

The equations ax + 9y = 1 and 9y - x - 1 = 0 represent the same line if a =

Detailed Solution for JEE Main Part Test - 1 - Question 64

Given:

Equation1 = ax + 9y = 1

Equation2 = 9y - x - 1 = 0

Concept used:

If linear equations are a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. Here, the equations have an infinite number of solutions, if

a1/a2 = b1/b2 = c1/c2

Calculation:

We have equations,

ax + 9y = 1

⇒ ax + 9y - 1 = 0

and, 9y - x - 1 = 0

⇒ x - 9y + 1 = 0

Here, a1 = a, b1 = 9, c1 = -1

and, a2 = 1, b2 = -9, c2 = 1

As we know that

a1/a2 = b1/b2 = c1/c2

⇒ a/1 = 9/-9 = -1/1

⇒ a = -1 = -1

⇒ a = -1

∴ The value of a is -1.

JEE Main Part Test - 1 - Question 65

Given, 6x + 2(6 - x) > 2x - 2 < 5x/2 - 3x/4, then x can take which of the following values?

Detailed Solution for JEE Main Part Test - 1 - Question 65

⇒ 6x + 2(6 - x) > 2x - 2

⇒ 6x + 12 - 2x > 2x - 2

⇒ 2x > - 14

⇒ x > - 7     ----(1)

⇒ 2x - 2 < 5x/2 - 3x/4

⇒ 2x - 2 < 7x/4

⇒ 8x - 7x < 8

⇒ x < 8     ----(2)

From (1) and (2),

- 7 < x < 8

∴ x = 5 satisfies the given conditions from the above options.

JEE Main Part Test - 1 - Question 66

If 2nC3 : nC2 = 12 : 1, then the value of n is ?

Detailed Solution for JEE Main Part Test - 1 - Question 66

Given:

If 2nC3 : nC2 = 12 : 1

Formula used :

nCr = n!/r!(n - r)!

Calculation:

If 2nC3 : nC2 = 12 : 1      ----(1)

Using the formula

2nC2 = {2n!/3! (2n - 3)!}

⇒ 2n(2n - 1)(2n - 2)(2n - 3)!/(2n - 3)! × 3 × 2

⇒ n(2n - 1)2(n - 1)/3      ----(2)

nC2 = {n!/2! (n - 2)!

⇒ n(n - 1)(n - 2)!/(n - 2)! × 2

⇒ n(n - 1)/2      ----(3)

Putting equation 2 and 3 in equation 1

⇒ {n(2n - 1)2(n - 1)/3}/{n(n - 1)/2} = 12/1

⇒ 2n - 1 = 9

⇒ 2n = 10

⇒ n = 5

∴ The value of n is 5.

JEE Main Part Test - 1 - Question 67

How many possible two-digit numbers can be formed by using the digits 3, 5 and 7 (repetition of digits is allowed)?

Detailed Solution for JEE Main Part Test - 1 - Question 67

⇒ Number of possible two-digit numbers which can be formed by using the digits 3, 5 and 7 = 3 × 3.

∴ 9 possible two-digit numbers can be formed.

The 9 possible two-digit numbers are:

33, 35, 37, 53, 55, 57, 73, 75, 77

JEE Main Part Test - 1 - Question 68

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that atleast 3 men are there on the committee. In how many ways can it be done ?

Detailed Solution for JEE Main Part Test - 1 - Question 68

Given:

(7 men + 6 women) 5 persons are to be chosen for a committee.

Formula used: nCr = n!/(n - r)! r!

Calculation:

Ways in which at least 3 men are selected;

⇒ 3 men + 2 women

⇒ 4 men + 1 woman

⇒ 5 men + 0 woman

Number of ways = 7C3 × 6C2 + 7C4 × 6C1 + 7C5 × 6C0

⇒ 7!/(3! × 4!) × 6!/(2! × 4!) + 7!/(4! × 3!) × 6!/(1! × 5!) + 7!/(5! × 2!) × 6!/(6!× 0!)

⇒ 35 × 15 + 35 × 6 + 21

⇒ 735 + 21 = 756

∴ The required no of ways = 756.

JEE Main Part Test - 1 - Question 69

In how many different ways can the letters of the word 'GEOGRAPHY' be arranged such that the vowels must always come together?

Detailed Solution for JEE Main Part Test - 1 - Question 69

Given:

The given number is 'GEOGRAPHY'

Calculation:

The word 'GEOGRAPHY' has 9 letters. It has the vowels E, O, A in it, and these 3 vowels must always come together. Hence these 3 vowels can be grouped and considered as a single letter. That is, GGRPHY(EOA).

Let 7 letters in this word but in these 7 letters, 'G' occurs 2 times, but the rest of the letters are different.

Now,

The number of ways to arrange these letters = 7!/2!

⇒ 7 × 6 × 5 × 4 × 3 = 2520

In the 3 vowels(EOA), all vowels are different

The number of ways to arrange these vowels = 3!

⇒ 3 × 2 × 1 = 6

Now,

The required number of ways = 2520 × 6

⇒ 15120

∴ The required number of ways is 15120.

JEE Main Part Test - 1 - Question 70

The number of ways of arrangements of 10 persons in four chairs is -

Detailed Solution for JEE Main Part Test - 1 - Question 70

Given:

The number of ways of arrangements of 10 persons in four chairs

Formula used:

nPr = n!/(n – r)!

Where, n = Number of persons

r = Number of chairs

Calculation:

According to the question

nPr = n!/(n – r)!

⇒ 10!/(10 – 4)!

⇒ 10!/6!

⇒ (10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(6 × 5 × 4 × 3 × 2 × 1)

⇒ (10 × 9 × 8 × 7)

⇒ 5040

∴ The required value is 5040

*Answer can only contain numeric values
JEE Main Part Test - 1 - Question 71

If the number of distinct positive rational numbers p/q smaller than 1, where  p, q ∈ {1, 2, 3 ....., 6} is k then k is :

Detailed Solution for JEE Main Part Test - 1 - Question 71

Out of numbers   will result only 3 distinct rational numbers.
⇒ Total numbers = 6C2 – 7 + 3 = 11

*Answer can only contain numeric values
JEE Main Part Test - 1 - Question 72

If the function ƒ(x) = 2 + x2 – e–x and g(x) = ƒ–1(x), then the value of  equals

Detailed Solution for JEE Main Part Test - 1 - Question 72

g(ƒ(x)) = x ⇒ g'(ƒ(x)). ƒ'(x) = 1
ƒ(x) = 1 ⇒ x = 0
g'(1). ƒ'(0) = 1
ƒ'(x) = 2x + e–x
ƒ'(0) = 1
g'(1) = 1

*Answer can only contain numeric values
JEE Main Part Test - 1 - Question 73

Let   then the value of a + b is:

Detailed Solution for JEE Main Part Test - 1 - Question 73

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JEE Main Part Test - 1 - Question 74

Let y = g(x) be the inverse of a bijective mapping f : R → R, f(x) = 3x3 + 2x. The area bounded by graph of g(x), the x-axis and the ordinate at x = 5 is -

Detailed Solution for JEE Main Part Test - 1 - Question 74

*Answer can only contain numeric values
JEE Main Part Test - 1 - Question 75

If function ƒ(x) = x3 + ax2 + bx + c is monotonically increasing ∀ x ∈ R, where a & b are prime numbers less than 10, then number of possible ordered pairs (a,b) is

Detailed Solution for JEE Main Part Test - 1 - Question 75

a, b ∈ {2, 3, 5, 7}
ƒ '(x) = 3x2 + 2ax + b ≥ 0    ∀ x ∈ R
D ≤ 0
4a2 – 12b ≤ 0
4(a2 – 3b) ≤ 0
a = 2    b = 2,3,5,7
a = 3    b = 3,5,7
a = 5    No solution
a = 7    No solution
Ordered pairs (a,b) are (2,2), (2,3), (2,5), (2,7), (3, 3), (3,5) & (3,7)

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