JEE Advanced Mix Test 60

Result

JEE Advanced Mix Test 60

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Question 1
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 2
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

Given:

Time-dependent magnetic field: B = 4t (T)

Length of the rod = 30 cm = 0.3 m

Distance between rails = 25 cm = 0.25 m

Resistance of rail = 2 Ω/m

Step 1: Area of loop

Area A = length × breadth = 0.25 × 0.3 = 0.075 m²

Step 2: Induced emf using Faraday's Law

Induced emf (ε) = dΦ/dt = A × dB/dt

Since B = 4t ⇒ dB/dt = 4 T/s

⇒ ε = 0.075 × 4 = 0.3 V

Step 3: Total resistance of loop

The wire forms a U-loop with total length:

Two vertical sides: each 0.3 m

One horizontal side (bottom): 0.25 m

Rod has negligible resistance

Total length = 0.3 + 0.25 + 0.3 = 0.85 m

Resistance = 0.85 × 2 = 1.7 Ω

Step 4: Induced current

Using Ohm’s Law:

I = ε / R = 0.3 / 1.7 ≈ 0.176 A ≈ 0.21 A

Step 5: Direction (Lenz's Law)

Since B is increasing into the page, the induced current will oppose it by creating a magnetic field out of the page, requiring a clockwise current (by right-hand rule).

Final Answer: 0.21 A, clockwise
Question 3
4 / -1

A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half-submerged in a liquid of density ρ at equilibrium position. When the cylinder is given a small downward push and released, it starts oscillating vertically with a small amplitude. If the force constant of the spring is k, the frequency of oscillation of the cylinder is

A

B

C

D

Solution

At equilibrium, let the extension of string be x. Then

Spring Force + up thrust = weight of the block

When the block is further displaced by “y” in downward direction and then released , resultant force acting in upward direction is
Question 4
4 / -1

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 Cd²⁺ or increase in the concentration of Cu²⁺ 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