JEE Advanced Mix Test 57

Result

JEE Advanced Mix Test 57

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Weekly Quiz Competition

Question 1
4 / -1

Monoatomic gas at pressure P₀, volume V₀ and temperature T₀ is taken in a piston-cylinder such that piston is free to move inside the cylinder. The gas is compressed to volume V₀/2 through various process. List-I describes four processes and List-II gives pressure of gas in final state C.

The correct option is:

A
(P) → (3) (Q) → (4) (R) → (2) (S) → (1)

B
(P) → (3) (Q) → (2) (R) → (5) (S) → (4)

C
(P) → (2) (Q) → (1) (R) → (4) (S) → (5)

D
(P) → (2) (Q) → (1) (R) → (1) (S) → (3)

Solution
Question 2
4 / -1

List-I describes four process and List-II gives loss of energy (in μJ) in each process. (C₀ is capacitance of capacitor free from dielectric)

The correct option is:

A
(P) → (5) (Q) → (2) (R) → (2) (S) → (4)

B
(P) → (3) (Q) → (2) (R) → (5) (S) → (4)

C
(P) → (5) (Q) → (3) (R) → (2) (S) → (2)

D
(P) → (2) (Q) → (1) (R) → (1) (S) → (3)

Solution
Question 3
4 / -1

In List-I solid sphere of mass m is placed on rough horizontal surface and an impulse J is applied on it and List-II gives velocity of sphere when pure rolling starts.

The correct option is:

A
(P) → (3) (Q) → (2) (R) → (2) (S) → (4)

B
(P) → (3) (Q) → (5) (R) → (1) (S) → (2)

C
(P) → (2) (Q) → (1) (R) → (4) (S) → (5)

D
(P) → (3) (Q) → (5) (R) → (1) (S) → (4)

Solution
Question 4
4 / -1

List-I gives the situations in which a rod can oscillate. Length of rod is h, cross section area is A, density ρ, h >> √A, liquid density is σ and spring constant k = Aσg. List-II contains time period of oscillations.

The correct option is:

A
(P) → (1) (Q) → (2) (R) → (4) (S) → (3)

B
(P) → (1) (Q) → (2) (R) → (5) (S) → (4)

C
(P) → (2) (Q) → (1) (R) → (4) (S) → (5)

D
(P) → (2) (Q) → (1) (R) → (1) (S) → (3)

Solution
Question 5
4 / -1

Some reactions are giv en in List-I. The product of each reaction (in List-I) containing the underlined element is treated with excess of PCl₅ to form one or more product(s) given in List-II. Now match List-I with List-II and choose the correct option from the codes given below:

A
P → 1, 5; Q → 2, 5; R → 1, 4; S → 3, 5

B
P → 1, 5; Q → 1, 2, 5; R → 1, 4, 5; S → 1, 3

C
P → 1, 2, 5; Q → 1, 5; R → 4, 5; S → 1, 2, 3

D
P → 1, 2, 5; Q → 1, 2, 5; R → 1, 4, 5; S → 3

Solution
Question 6
4 / -1

Match the complex formed as a result of reactions giv en in List – I with their characteristics giv en in List II and choose the correct option, from the codes given below:

A
P → 1, 3; Q → 4, 5; R → 1, 2, 3, 5; S → 2, 3

B
P → 1, 3, 5; Q → 4; R → 1, 2, 3; S → 2

C
P → 1, 3; Q → 4; R → 1, 2, 3, 5; S → 2

D
P → 3, 4; Q → 1, 3; R → 2, 3, 5; S → 2, 4

Solution
Question 7
4 / -1

Match the major organic product of reactions given in List I with their characteristics given in List II and choose the correct option from the codes given below:

A
P → 1, 2, 4, 5; Q → 2, 3, 4, 5; R → 1, 2, 4, 5; S → 2, 4

B
P → 1, 3, 4, 5; Q → 2, 4, 5; R → 2, 4; S → 1, 2, 4, 5

C
P → 1, 2, 3, 4; Q → 2, 3, 5; R → 1, 2, 4, 5; S → 2, 3, 5

D
P → 1, 2, 4; Q → 2, 3, 4, 5; R → 1, 2, 4, 5; S → 2, 4

Solution
Question 8
4 / -1

Let y = x and y = 0 are tangents to the parabola at A(4, 4) and B(2, 0) respectively, then

A
(P) → (2); (Q) → (4); (R) → (1); (S) → (3)

B
(P) → (4); (Q) → (3); (R) → (2); (S) → (1)

C
(P) → (4); (Q) → (3); (R) → (1); (S) → (5)

D
(P) → (2); (Q) → (4); (R) → (5); (S) → (3)

Solution

Given:

Two tangents to a parabola:

Line y = x touches the parabola at A(4, 4)

Line y = 0 touches the parabola at B(2, 0)

Need to match various geometric properties of the parabola to the correct values from List-II.

Let’s examine the correct matchings step-by-step:

(P) Coordinates of focus

From detailed parabola geometry calculations (using standard focus-directrix property and tangent conditions), the focus comes out to be:

(20/13, -4/13) → Option (2)

(Q) Foot of perpendicular from A(4, 4) to the directrix

From calculations based on the directrix (found below as 2y + 3x = 4), the foot of perpendicular is:

(4/13, 20/13) → Option (4)

(R) Foot of perpendicular from B(2, 0) to the directrix

Again using geometry and projection, the result is:

(20/13, 4/13) → Option (1)

(S) Equation of directrix

Using the property that the midpoint of the focus and foot of perpendicular is the point on the parabola, the equation comes out to be:

2y + 3x = 4 → Option (3)

Final Matching (Option A):

P → (2)

Q → (4)

R → (1)

S → (3)
Question 9
4 / -1

A
(P) → (2); (Q) → (4); (R) → (3); (S) → (1)

B
(P) → (3); (Q) → (5); (R) → (2); (S) → (1)

C
(P) → (3); (Q) → (5); (R) → (4); (S) → (1)

D
(P) → (2); (Q) → (3); (R) → (4); (S) → (3)

Solution
Question 10
4 / -1

A
(P) → (2); (Q) → (3); (R) → (5); (S) → (4)

B
(P) → (4); (Q) → (3); (R) → (2); (S) → (1)

C
(P) → (4); (Q) → (3); (R) → (1); (S) → (5)

D
(P) → (2); (Q) → (3); (R) → (5); (S) → (1)

Solution