JEE Advanced Mix Test 42

Result

JEE Advanced Mix Test 42

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

A
1.(i), 2.(ii), 3.(iii)

B
1.(ii), 2.(iii), 3.(i)

C
1.(i), 2.(iii), 3.(ii)

D
None of these

Solution
Question 2
4 / -1

A body is thrown with a velocity of 10 m s⁻¹ at an angle of 45° to the horizontal. The radius of curvature of its trajectory in t = 1/√2 s after the body began to move is,

A
0 m

B
2.5 m

C
5 m

D
none of these.

Solution
Question 3
4 / -1

The figure below shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance 2 Ω/m. The position of the conducting rod at t = 0 is shown. A time-dependent magnetic field B = 4t T is switched on at t = 0.

The current in the loop at t = 0 due to induced emf is

A
0.21 A, clockwise

B
0.42 A, clockwise

C
0.21 A, anticlockwise

D
zero

Solution

Using Lenz's law, the upper end of the rod is negative which makes the current flow clockwise.
Question 4
4 / -1

Directions: The following question has four choices, out of which ONLY ONE is correct.

Consider the cell:

Cd(s) | Cd²⁺ (1.0 M) || Cu²⁺ (1.0, M) | Cu(s)

If we want to make a cell with a more positive voltage using the same substances, we should

A
increase both Cd²⁺ and Cu²⁺ to 2.00 M

B
increase only Cd²⁺ to 2.00 M

C
decrease both Cd²⁺ and Cu²⁺ to 0.10 M

D
decrease only Cd²⁺ to 0.100 M

Solution

According to the equation, decrease in the concentration of Cd⁺² or increase in the concentration of Cu⁺² would increase the voltage.

Hence, option (d) is correct.
Question 5
4 / -1

Given below are two cleavage reactions:

(i) (CH₃)₃COCH₃ → CH₃I + (CH₃)₃COH

(ii) (CH₃)₃COCH₃ → CH₃OH + (CH₃)₃CI

A
The reagent used in reaction (i) is anhydrous HI in ether and that in reaction (ii) is concentrated HI.

B
The reagent used in reaction (i) is concentrated HI and that in reaction (ii) is anhydrous HI in ether.

C
The reagent used in both reactions (i) and (ii) is concentrated HI.

D
The reagent used in both reactions (i) and (ii) is anhydrous HI in ether.

Solution
Question 6
4 / -1

In which of the following molecules / ions resonance structures are equivalent:

A
HCOO^⊖

B
CH₂=CH−CH=CH₂

C

D

Solution

Only format ion has equivalent resonance structures. Charge is delocalised on the similar type of elements in the equivalent resonating structures. Resonating structures are not equivalent in other compounds. Equivalent resonance is more dominant than normal resonance. Equivalent resonance stabilised the species more as compared to the normal resonance.
Question 7
4 / -1

A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan⁻¹tan{f(−5) + f(20) + cos⁻¹(f(−10)) + f(17)} is equal to

A
2π − 3

B
3 − 2π

C
2π + 3

D
3 − π

Solution

f(x) is given to be an even, periodic function with period equal to 8.

⇒ f(x + 8) = f(x)

1. f(−5) = f(3) = 2

2. f(20) = f(12) = f(4) = 3

3. f(−10) = f(−2) = f(2) = 1

4. f(17) = f(9) = f(1) = −2

f(−5) + f(20) + cos⁻¹(f(−10)) + f(17) = 2 + 3 + cos⁻¹(1)−2 = 3

tan⁻¹(tan(3)) = tan⁻¹(tan(3 − π)) = 3 − π

(using tan⁻¹(tanx) = x − π if x ∈ (π/2, π))
Question 8
4 / -1

A

B

C

D

Solution
Question 9
4 / -1

A

B

C

D

Solution
Question 10
4 / -1

Let the eccentricity of the hyperbola be the reciprocal of that of the ellipse x² + 4y² = 4. Also, the hyperbola passes through a focus of the ellipse. Then, the equation of the hyperbola is

A
x² - 3y² = 3

B
x² - 3y² = 1

C
3x² - y² = 3

D
3x² - y² = 1

Solution