JEE Main Part Test - 1 - JEE MCQ

Length , l = 4,234 m
Breadth, b = 1.005 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)

The number 157,900 ends with two zeros, which means it has been rounded to the nearest hundred.
When a number is rounded to the nearest hundred, the possible actual value lies within ±50 of the given number.

This means the actual population could be as low as:

• 157,850 (if rounded up to 157,900)

Or as high as:

• 157,949 (if rounded down to 157,900)

But in standard approximation rules:

• Lower limit = 157,900 – 50 = 157,850

• Upper limit = 157,900 + 50 = 157,950

So, the maximum possible error is ±50.

The range is the same for projection angles θ and 90° − θ:

R = u² sin 2θ / g

Maximum heights for θ and 90° − θ:

Using the identity sin 2θ = 2 sin θ cos θ, we have:

Final Range Expression: 

Thus, the range R for angles θ and 90° - θ is the same, and the relationship between the maximum heights at these two angles leads to this conclusion.

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

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

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].

Given:

• A force of 40 N acts at an angle of 30° to the horizontal.
• The mass of the body is 5 kg.
• The acceleration due to gravity is 10 m/s².
• The surface is smooth, meaning there is no friction.

Analyzing the Statements:
Statement 1: The horizontal force acting on the body is 20 N.
The horizontal component of the force is found using the cosine of 30°: Horizontal force = 40 * cos(30°) cos(30°) = √3 / 2 ≈ 0.866 Horizontal force = 40 * 0.866 ≈ 34.64 N This is not 20 N. So, Statement 1 is incorrect.

Statement 2: The weight of the 5 kg mass acts vertically downwards.
The weight of the body is calculated as: Weight = mass * gravity = 5 * 10 = 50 N This force acts vertically downward. So, Statement 2 is correct.

Statement 3: The net vertical force acting on the body is 30 N.
The vertical component of the force is found using the sine of 30°: Vertical force = 40 * sin(30°) sin(30°) = 1/2 = 0.5 Vertical force = 40 * 0.5 = 20 N

The net vertical force is the difference between the vertical component of the applied force and the weight of the body: Net vertical force = Vertical force - Weight = 20 N - 50 N = -30 N The negative sign indicates the force is acting downward, and the magnitude is 30 N. So, Statement 3 is correct.
Conclusion: The correct statements are Statement 2 and Statement 3.
Correct Answer: C: 2 and 3.

∴ 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.

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

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

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.

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

The possible quantum 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.

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 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.

• 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.

In sp² hybridization, three hybrid orbitals are formed, arranged in a trigonal planar geometry with bond angles of 120°. This is seen in molecules like BF₃ or C₂H₄. Other hybridizations have different angles:
sp³: 109.5° (tetrahedral).
sp: 180° (linear).
90°: Not typical for standard hybridizations.
For sp²: ∠ = 120°.
This explanation clarifies the angles associated with different hybridizations and gives examples of molecules with sp² hybridization.

Given the functional equation:

f(x - y) = f(x)f(y) - f(a - x)f(a + y) and f(0) = 1.

Substitute x = y = 0:

f(0) = f(0)f(0) - f(a)f(a)
⇒ 1 = 1 - [f(a)]²
⇒ [f(a)]² = 0
⇒ f(a) = 0

To find f(2a - x):

f(2a - x) = f(a - (x - a)) = f(a)f(x - a) - f(a - a)f(a + x - a) = 0 - f(0)f(x) = -f(x)

Answer: A.

Step 1: Express z³
We are given:

z = x - iy

Now, let's calculate z³.

z³ = (x - iy)³

Expanding this using the binomial expansion:

z³ = x³ - 3x²(iy) + 3x(iy)² - (iy)³
z³ = x³ - 3x²(iy) + 3x(-y²) - iy³
z³ = x³ - 3xy² - i(3x²y + y³)

So, we have:

z³ = (x³ - 3xy²) - i(3x²y + y³)

Step 2: Take the reciprocal of z³
Next, we find the reciprocal of z³, which is given as 1/z³ = p + iq. To find the reciprocal of a complex number a + bi, we use the formula:

1 / (a + bi) = (a - bi) / (a² + b²)

In our case, a = x³ - 3xy² and b = 3x²y + y³. Therefore:

1 / z³ = ((x³ - 3xy²) + i(3x²y + y³)) / ((x³ - 3xy²)² + (3x²y + y³)²)

Step 3: Match real and imaginary parts
From the equation 1/z³ = p + iq, we compare the real and imaginary parts:

p = (x³ - 3xy²) / ((x³ - 3xy²)² + (3x²y + y³)²)

q = (3x²y + y³) / ((x³ - 3xy²)² + (3x²y + y³)²)

Step 4: Solve the expression
Now, we are asked to find the value of the following expression:

(x + y) / p * 1 / (p² + q²)

We already know that p² + q² is the denominator of 1/z³, which is:

p² + q² = (x³ - 3xy²)² + (3x²y + y³)²

Step 5: Check the possible answers
The possible answers are:

A: -2

B: -1

C: 2

D: 1

After calculating, we find that the value of the expression equals -1, which corresponds to Option B.

Thus, the correct answer is B: -1.

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

|z² - 1|=|z|²+1 ⇒|z² - 1|² = 

⇒  

⇒  

⇒  ⇒ 

⇒ z is purely imaginary

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

= 

Compare with :

a + b = −5 + 9 = 4

The answer (4) is correct.

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)