JEE Advanced Mix Test 50

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JEE Advanced Mix Test 50

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Weekly Quiz Competition

Question 1
4 / -1

A sphere of mass m and radius r is placed inside a rotating horizontal cylinder of radius R. As the cylinder's angular acceleration, β, gradually increases, find the maximum value of β that enables the sphere to reach from A to the horizontal point B.

A
2g/R

B
5g/R

C
(2g) / (5R)

D
(5g) / (2R)
Solution
Concept Used:

A small sphere of mass m and radius r is placed inside a rotating horizontal cylinder of radius R . The sphere undergoes pure rolling motion inside the cylinder due to friction.

The motion of the sphere is governed by Newton’s Laws and the rolling condition :
- The cylinder’s angular acceleration (β) causes an effective acceleration in the rotating frame.
- The sphere will be at rest with respect to the ground when the maximum acceleration (α) is reached.
- The downward acceleration in the rotating frame is given by:
Calculation:

Applying the rolling condition:

⇒ βr = (g sinθ + βr) / (1 + 2/5)

⇒ (7/5) βr - βr = g sinθ

⇒ θ = sin⁻¹ (2βr / 5g)

The maximum angular acceleration (β) required for the sphere to reach point B is:

⇒ β = (5g) / (2R)

Correct Option: Option 4
Question 2
4 / -1

Imagine you have a cup of Starbucks coffee in a moving car. If the car suddenly brakes

A
The coffee in the middle of the cup will move forward relative to the cup.

B
The coffee will move backward relative to the cup.

C
The coffee will remain at the same horizontal position in the cup.

D
Depending on the car's speed, the coffee might move forward or backward.

Solution

Explanation:

When a car is moving forward and suddenly brakes , the objects inside the car tend to maintain their motion due to inertia (Newton’s First Law of Motion).

The liquid inside the cup is not rigidly attached to the cup and will continue moving forward relative to the ground. However, the cup experiences the braking force applied by the car.

The cup slows down along with the car because it is in contact with it.

The liquid inside does not experience the same immediate braking force.

As a result, the liquid appears to move backward relative to the cup .

Correct Option: Option 2
Question 3
4 / -1

A square frame is constructed from four uniform rods. The rods, AB, BC, CD, and DA, have masses 2m, 3m, 4m, and 5m respectively, and each has a length of 2a. Where is the center of mass of this frame located in the x-y plane, specifically, which quadrant or region defined by x and y coordinates.

A
For a = 1m, the region of center of mass are 1< y<2 and 0< x <1

B
For a = 2m, the region of center of mass are 1< y<2 and 2< x <3

C
For a = 1m, the region of center of mass are 2< y<3 and 1< x <2

D
For a = 2m, the region of center of mass are 1< y<2 and 1< x <2

Solution

Concept Used:

The center of mass (COM) of a system of rods is determined using the weighted average formula:

X_COM = (Σm_ix_i) / Σm_i

Y_COM = (Σm_iy_i) / Σm_i

where:

m_i = mass of each rod

(x_i, y_i) = center of each rod

Calculation:

Given masses and lengths :

Rod AB : Mass = 2m , Center at ( a, 0 )

Rod BC : Mass = 3m , Center at ( 2a, a )

Rod CD : Mass = 4m , Center at ( a, 2a )

Rod DA : Mass = 5m , Center at ( 0, a )

Summing the mass values:

⇒ Total mass, M = 2m + 3m + 4m + 5m = 14m

Finding X_COM:

⇒ X_COM = [(2m × a) + (3m × 2a) + (4m × a) + (5m × 0)] / 14m

⇒ X_COM = (2a + 6a + 4a) / 14

⇒ X_COM = (12a) / 14 = 6a/7

Finding Y_COM:

⇒ Y_COM = [(2m × 0) + (3m × a) + (4m × 2a) + (5m × a)] / 14m

⇒ Y_COM = (0 + 3a + 8a + 5a) / 14

⇒ Y_COM = (16a) / 14 = 8a/7

The center of mass is located at (6a/7, 8a/7).

Thus the value of a =1m gives option 1.
Question 4
4 / -1

A ball of 8 kg, moving at 15 m/s, enters a 22 kg spring gun at rest. If the surfaces are frictionless and the spring is massless, determine the gun's speed after the ball stops moving relative to the gun.

A
2 m/s

B
4 m/s

C
6 m/s

D
8 m/s

Solution

Concept Used:

This problem involves the principle of conservation of linear momentum . Since the surfaces are frictionless and the spring is massless, there are no external forces acting on the system.

The law of conservation of momentum states:

Initial Momentum = Final Momentum

Before the collision:

Mass of the ball, m₁ = 8 kg

Initial velocity of the ball, v₁ = 15 m/s

Mass of the gun, m₂ = 22 kg

Initial velocity of the gun, v₂ = 0

Calculation:

Using the conservation of momentum equation :

⇒ (m₁ v₁) + (m₂ v₂) = (m₁ + m₂) v

⇒ (8 × 15) + (22 × 0) = (8 + 22) v

⇒ 120 = 30v

⇒ v = 4 m/s

The gun's speed after the ball stops moving relative to the gun is 4 m/s .

Correct Option: Option 2
Question 5
4 / -1

A
only S1

B
All of these

C
S2 and S3 only

D
S1, S2 and S3
Solution
CONCEPT:

Magnetism and Complex Isomerism
- A coordination complex's magnetic properties can be determined by the number of unpaired electrons in the metal's d-orbitals.
- Complexes can show different types of isomerism such as linkage isomerism and geometrical isomerism.
EXPLANATION:
- S1: Both Coox₃³⁻ and CoF₆³⁻ are paramagnetic
  
  CoF₆³⁻ is paramagnetic with 4 unpaired electrons because Co(III) in a high-spin complex has electrons in the 3d orbitals.
  
  Coox₃³⁻ is diamagnetic with no unpaired electrons as it forms a low-spin complex.
- S2: CoCl₃(NH₃)₃ complex is non-conducting
  
  This complex exists as a neutral molecule [Co(NH₃)₃Cl₃], so it does not conduct electricity.
- S3: The number of possible isomers for the complex PtNO₂pyOHNH₃ is six
  
  This complex can show linkage and geometrical isomerism resulting in a total of 6 possible isomers.
- S4: The oxidation state of iron in the brown ring complex [Fe(H₂O)₅NO]SO₄ is +II where NO is NO⁺
  
  In the brown ring complex, NO has a +1 charge and the overall charge of the complex is +1, thus iron has an oxidation state of +2.
Therefore, the correct answer is S2 and S3 only
Question 6
4 / -1

Gold number of haemoglobin is 0.03. Hence, 100mL of gold sol will require how much haemoglobin so that gold is not coagulated by 10mL of 10% NaCl solution?

A
0.03mg

B
30mg

C
0.30mg

D
3mg
Solution
CONCEPT:

Gold Number and Coagulation
- The Gold number is a measure of the protective power of colloids. It is defined as the amount of colloidal substance (in milligrams) required to prevent the coagulation of 10 mL of a gold sol by 1 mL of a 10% NaCl solution.
- In this problem, the Gold number for haemoglobin is given as 0.03, which means 0.03 mg of haemoglobin is required to prevent the coagulation of 10 mL of gold sol by 1 mL of 10% NaCl solution.
- The given amount of gold sol is 100 mL, and we are asked to calculate how much haemoglobin is required to prevent coagulation by 10 mL of 10% NaCl solution.
EXPLANATION:

Gold number = 0.03

Gold sol volume = 100 ml

Let the amount of hemoglobin required is x mg

Since the gold number is the amount in milligrams required to prevent the coagulation of 10 ml of gold sol by 1 ml of 10% NaCl.
Question 7
4 / -1

The EAN of metal atoms in [Fe(CO)₂(NO⁺)₂] and Co₂(CO)₈ respectively are :

A
34,35

B
34,36

C
36,36

D
36,35
Solution
CONCEPT:

Effective Atomic Number (EAN)

EAN = Z - X + Y
- The Effective Atomic Number (EAN) is a concept used to determine the stability of metal complexes. It is calculated using the formula:
  
  Z = Atomic number of the metal atom
  
  X = Number of electrons lost by the metal during bonding (oxidation state of the metal)
  
  Y = Number of electrons donated by the ligands
- The ligands in a complex contribute electrons based on their donor atoms. For example, CO (carbon monoxide) donates 2 electrons, and NO+ (nitrosonium ion) donates 2 electrons as well.
- The concept of EAN is especially useful for organometallic compounds, where transition metals form complexes with ligands like CO and NO+.
EXPLANATION:

Thus, both complexes have an EAN of 36, making option C the correct answer.
- For the complex [Fe(CO)₂(NO)₂]:
  
  Fe has an atomic number (Z) of 26.
  
  Fe is in the 0 oxidation state, so it does not lose any electrons (X = 0).
  
  Each CO ligand donates 2 electrons, and each NO+ ligand donates 2 electrons. Therefore, Y = (2 × 2) + (2 × 2) = 8 electrons donated by ligands.
  
  Hence, EAN = 26 + 0 + 8 = 36.
- For the complex Co₂(CO)₈:
  
  Co has an atomic number (Z) of 27.
  
  Co is in the 0 oxidation state, so it does not lose any electrons (X = 0).
  
  Each CO ligand donates 2 electrons, and there are 8 CO ligands in total. Therefore, Y = 8 × 2 = 16 electrons donated by ligands.
  
  Hence, EAN = 27 + 0 + 16 = 36.
Thus, both complexes have an EAN of 36, making option C the correct answer.
Question 8
4 / -1

A

B

C

D

Solution
Question 9
4 / -1

Given that out of these three vectors two are equal in magnitude and the magnitude of the third vector times as that of either of the two having equal magnitude. Then the angles between vectors are given by :-

A
30^∘,60^∘,90^∘

B
45^∘,45^∘,90^∘

C
45^∘,60^∘,90^∘

D
90^∘,135^∘,135^∘

Solution
Question 10
4 / -1

A box contains the following three coins.

I. A fair coin with head on one face and tail on the other face.

II. A coin with heads on both the faces.

III. A coin with tails on both the faces.

A coin is picked randomly from the box and tossed. Out of the two remaining coins in the box, one coin is then picked randomly and tossed. If the first toss results in a head, the probability of getting a head in the second toss is

A
2/5

B
1/3

C
2/3

D
1/2

Solution

Application:

Let event A is defined as:

A = Getting head in the first toss

Event B is defined as:

B = Getting head in the second toss

According to the question, we need to find the probability of getting a head in the second toss when

Already a head has occurred in the first toss, i.e.So, the probability of getting head in the first toss will be: