JEE Main Part Test - 1 - JEE MCQ

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.

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

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.

∴ Spring is compressed by 2 cm at A and will be stretched by 2 cm at B.
By work-energy theorem,
W_mg + W_s + W_N + W_f = K_f − K_i
0.16 + 0 + 0 + W_f = 0 - 10
W_f = -10.16 J

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.

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.

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.

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.

Let atomic weight of x = Mx

atomic weight of y = M_y

we know,

mole = weight /atomic weight

a/c to question,

mole of xy₂ = 0.1

so,

0.1 = 10g/( M_x +2M_y)

M_x + 2M_y = 100g -------(1)

for x₃y₂ ; mole of x₃y₂ = 0.05

0.05 = 9/( 3M_x + 2M_y )

3M_x + 2M_y = 9/0.05 = 9× 20 = 180 g ---(2)

solve eqns (1) and (2)

2M_x = 80

M_x = 40g/mol

and M_y = 30g/mole

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

The correct answer is option A
Cr₂O₇^2- 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.

The correct answer is Option B.

General electronic configuration is given:
(n − 2)f^1-14(n − 1)d⁰¹ns² , where(n=7)
5f^1-146d⁰¹ns²
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.
 

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.

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

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 H_zO > HF > NH₃.

Vinyl acetylene is

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

• The compound shows several carbon atoms attached to various groups.
• To be a chiral center, a carbon must have four different groups attached to it.

Upon re-inspection of the structure, there are 3 chiral centers in this compound.

So, the correct number of chiral centers is 3.

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

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

⇒ x - iy = p³ - 3pq² + i(3p²q-q³)

∴ x = p³ - 3pq² ⇒ 

y = q³ - 3p²q ⇒ 

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.

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

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

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