JEE Advanced Mix Test 64

Result

JEE Advanced Mix Test 64

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Weekly Quiz Competition

Question 1
4 / -1

A point Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity . This can be considered equivalent to a loop carrying a steady current Qω/2π . A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant y.

The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is BR/t. Find the value of t.

A
2

B
5

C
3

D
1

Solution
Question 2
4 / -1

A point Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity . This can be considered equivalent to a loop carrying a steady current Qω/2π . A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant y.

The change in the magnetic dipole moment associated with the orbit, at the end of time interval of the magnetic field change, is Find the value of p.

A
5

B
4

C
3

D
2

Solution
Question 3
4 / -1

A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 22 seconds). The later ball is thrown at a velocity of half the first. At t = 2 s, both the balls reach their maximum heights. At this time the vertical gap between first and second ball is +15m.

The speed of the first ball is

A
20 m s⁻¹

B
10 m s⁻¹

C
5 m s⁻¹

D
15 m s⁻¹

Solution

Let the speeds of the two balls, 1 and 2, be v₁ and v₂, respectively. Since the speed of the second ball is half of the first, let

v₁ = 2v ...(1)

v₂ = v ...(2)

If y₁ and y₂ are the maximum heights reached by the balls 1 and 2, respectively, then
Question 4
4 / -1

A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 22 seconds). The later ball is thrown at a velocity of half the first. At t = 2 s, both the balls reach their maximum heights. At this time the vertical gap between first and second ball is +15m.

The time interval between the throw of balls is

A
1.2 sec

B
0.5 sec

C
0.8 sec

D
1 sec

Solution

Let the speeds of the two balls, 1 and 2, be v₁ and v₂, respectively. Since the speed of the second ball is half of the first, let

v₁ = 2v ...(1)

v₂ = v ...(2)

If y₁ and y₂ are the maximum heights reached by the balls 1 and 2, respectively, then

Since t₁ (time taken by ball 1 to cover distance of 20 m) is 2 s, time interval between the two throws,

= t₁ − t₂ = 2 − 1 = 1 s
Question 5
4 / -1

Treatment of compound O with KMnO₄/H⁺ gave P, which on heating with ammonia gave Q. The compound Q on treatment with Br₂/NaOH produced R. On strong heating, Q gave S, which on further treatment with ethyl 2-bromopropanoate in the presence of KOH followed by acidification, gave a compound T.

The sum of N- and O-atoms present in compound T is

A
5

B
4

C
3

D
2

Solution

Oxidation of o-dipropylbenzene (O) yields phthalic acid (P), which reacts with NH3 to form phthalamide (Q).

(Q) on heating forms phthalimide (S) with the elimination of NH3.

(S) on further treatment with ethyl 2-bromopropanoate followed by hydrolysis yields alanine (T).
Question 6
4 / -1

A definite amount of reducing agent is oxidised by 20 mL of 1 M KMnO₄ in acid medium, then the same amount of reducing agent is oxidised to the same state by how many mL of 1 M KMnO4 in neutral medium itself changing to Mn⁴⁺ state?

A
3 mL

B
33.3 mL

C
12 mL

D
24 mL

Solution

∴ Milliequivalents of KMnO₄ in acid medium = Milliequivalents of KMnO₄ in neutral medium.

1 × 5 × 20 = 1 × 3 × V

∴ V = 33.3mL
Question 7
4 / -1

Name the end product in the following series of reaction.

A
CH₄

B
CH₃OH

C
CH₃CN

D
CH₃COONH₄

Solution
Question 8
4 / -1

Let F₁(x₁, 0) and F₂(x₂, 0), for x₁ < 0 and x₂ > 0, be the foci of the ellipse

Suppose a parabola having vertex at the origin and focus at F₂ intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant.

If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x-axis at Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF1NF2 is m : n. Find the sum of m and n.

A
12

B
13

C
14

D
15

Solution
Question 9
4 / -1

The number of real values of x for which the equation has solution is

A
1

B
2

C
3

D
Infinite

Solution
Question 10
4 / -1

If x takes the values for which the equation has a solution, then the number of values of a∈[0,100] is

A
2

B
1

C
5

D
3

Solution

For a ∈ [0, 100], there are exactly 3 values of a i.e. a = 3π, 15π and 27π.