JEE Advanced Mix Test 55

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JEE Advanced Mix Test 55

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Weekly Quiz Competition

Question 1
4 / -1

An ideal gas has adiabatic exponent γ. In some process, its molar heat capacity varies as C = α/T , where α is a constant. Work performed by one mole of gas during its heating from T_o to nT_o will be

A
αlog_e n

B
αlog_e n - (n - 1)RT_o

C

D
None of these

Solution
Question 2
4 / -1

Two non-conducting solid spheres of radii R and 2R having uniform volume charge densities ρ₁ and ρ₂ respectively, touch each other. The net electric field at a distance 2R from centre of the smaller sphere is zero. The ratio ρ₁/ρ₂ can be

A
-4/25

B
-32/25

C
32/25

D
-6/25

Solution
Question 3
4 / -1

A thin uniform rod of mass M and length L is hinged at its upper end and released from rest in a horizontal position. Find the tension at a point located at a distance L/3 from the hinge point, when the rod becomes vertical.

A
2 Mg

B
3 Mg

C
4 Mg

D
2 Mg/3

Solution
Question 4
4 / -1

Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of

A
0.42 m from mass of 0.3 kg

B
0.70 m from mass of 0.7 kg

C
0.98 m from mass of 0.3 kg

D
0.98 m from mass of 0.7 kg

Solution
Question 5
4 / -1

Which chloro derivative of benzene among the following would undergo hydrolysis most readily with aqueous sodium hydroxide to furnish the corresponding hydroxyl derivative?

A

B

C

D
C₆H₅Cl

Solution

The polarity of the C-Cl bond in 2, 4, 6-trinitrochlorobenzene is enhanced by three NO₂ groups (EWG) at the o- and p-positions, which activate the molecule towards nucleophilic substitution.

Hence, it undergoes hydrolysis most readily.
Question 6
4 / -1

The radioactive decay ₈₃Bi²¹¹ ⟶ ₈₁Tl²⁰⁷, takes place in 100 L, 27°C. Starting with 2 moles of ₈₃Bi²¹¹(t_1/2=130 sec) sec the pressure development in the vessel after 520 sec will be:

A
1.875 atm

B
0.2155 atm

C
0.4618 atm

D
4.618 atm

Solution

The solution uses the ideal gas law (P = nRT/V), assumes helium is the only gas contributing to pressure, and correctly computes 0.4618 atm.
Question 7
4 / -1

What is the order of reactivity of the following compounds in E1 reaction?

(I) Tertiary-butyl bromide

(II) 1-phenyl-2-bromopropane

(III) Benzyl bromide

(IV) 2-bromopropane

A
III < IV < II < I

B
I < II < III < IV

C
IV < III < I < II

D
IV < III < II < I
Solution
Correct Answer: Option D

IV < III < II < I

Explanation:

The E1 reaction mechanism involves:
1. Formation of a carbocation (slow step).
2. Loss of a β-proton to form an alkene.
So, the stability of the carbocation formed after the halide leaves decides the rate of the E1 reaction.
- (I) Tertiary-butyl bromide → Forms a tertiary carbocation, which is very stable. → Highest reactivity
- (II) 1-Phenyl-2-bromopropane → Forms a benzylic secondary carbocation, stabilized by resonance.
- (III) Benzyl bromide → Forms a primary benzylic carbocation, also resonance stabilized.
- (IV) 2-Bromopropane → Forms a secondary carbocation, less stable than benzylic or tertiary.
Thus, the correct increasing reactivity order is:

IV < III < II < I
Question 8
4 / -1

Let F : R → R be a thrice differentiable function. Suppose that F(1) = 0, F(3) = -4 and F'(x) < 0 for all x ∈ (1/2, 3). Let f(x) = xF(x) for all x ∈ R.

Which of the following statements is incorrect?

A
f'(1) < 0

B
f(2) < 0

C
f '(x) ≠ 0 for any x ∈ (1, 3)

D
f '(x) = 0 for some x ∈ (1, 3)

Solution

Option (1)

f '(x) = F(x) + xF '(x)

⇒ f '(1) = F(1) + F '(1)

⇒ f '(1) = F'(1) < 0 (since F '(x) < 0

⇒ f '(1) < 0

It is correct.

Option (2):

f(x) = x F(x)

⇒ f(2) = 2F(2)

⇒ F(x) is decreasing and F(1) = 0.

Hence, F(2) < 0

⇒ f(2) < 0

It is correct.

Option (3):

f '(x) = F(x) + xF '(x)

F(x) < 0 for all x belongs to (1, 3).

F'(x) < 0 for all x belongs to (1, 3).

⇒ f '(x) < 0 for all x belongs to (1,3)

⇒ f '(x) 0 for any x ∈ (1, 3)

It is correct.

Hence, option (4) is the answer..
Question 9
4 / -1

Football teams T₁ and T₂ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T₁ winning, drawing and losing a game against T₂ are 1/2, 1/6, and 1/3 respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored, respectively, by teams T₁ and T₂, after two games.P(X=Y) is ?

A
1/4

B
13/36

C
1/2

D
7/12

Solution

P(X = Y)

= P(draw)⋅P(draw) + P(T₁ win)⋅P(T₂ win) + P(T₂ win)⋅P(T₁ win)

= (1/6 × 1/6) + (1/2 × 1/3) + (1/3 × 1/2)

= 13/36
Question 10
4 / -1

The value of the function f(x) = 1 + x + ∫^x₁(log²t + 2 log t)dt, where f′(x) vanishes

A

B

C
3e

D
e⁻²

Solution