JEE Advanced Mix Test 80

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JEE Advanced Mix Test 80

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Weekly Quiz Competition

Question 1
4 / -1

When liquid medicine of density ρ is to be put in the eye it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius R (assuming r << R)

A
2πrT

B
2πRT

C
2πr²T/R

D
2πr²T/r

Solution

The vertical force due to the surface tension on the drop is,
Question 2
4 / -1

The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density ρ(r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The electric field is only along the radial direction.

For a = 0, the value of d (maximum value of ρ as shown in the figure) is

A

B

C

D

Solution
Question 3
4 / -1

Two small equally charged spheres, each of mass m, are suspended from the same point by silk threads of length l. The distance between the spheres x << l. Find the rate dq/dt with which the charge leaks off each sphere if their approach velocity varies as v = a/√x , where a is a constant.

A

B

C

D

Solution
Question 4
4 / -1

A composite wire of uniform diameter 3.0 mm consists of a copper wire of length 2.2 m and a steel wire of length 1.6 m which is stretched under a load by 0.7 mm. Calculate the load, given that the Young's modulus of elasticity for copper is 1.1 × 10¹¹ N m^–2 and that for steel is 2 × 10¹¹ N m^–2.

A
1.8 × 10² N

B
1.5 × 10^–2 N

C
2.8 × 10² N

D
1.0 × 10² N

Solution
Question 5
4 / -1

Directions For Questions

Directions: The following question is based on the paragraph given below.

The noble gases have closed-shell electronic configuration and are monatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.

The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF₄ reacts violently with water to give XeO₃. The compound can also be prepared using XeF₆ as the starting compound. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.

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The chemical nature of the compounds XeF₄ and XeF₆ is expected to be

A
oxidising

B
reducing

C
unreactive

D
strong basic

Solution

XeF₄ (Xenon tetrafluoride) and XeF₆ (Xenon hexafluoride) are fluorides of xenon, where xenon is in a high oxidation state (+4 in XeF₄ and +6 in XeF₆).

Elements in high oxidation states tend to gain electrons to return to lower, more stable states — this makes them oxidising agents.

Additionally, it is mentioned that XeF₄ reacts violently with water to form XeO₃, indicating a chemical transformation where xenon remains in a high oxidation state, again showing oxidising character.

These compounds oxidise other substances during reactions (such as water), hence are oxidising in nature.
Question 6
4 / -1

(1R, 3S)-Cis-1-Bromo-3-methyl cyclohexane. The product formed in the reaction is

A
(1R, 3S)-Cis-3-methyl cyclohexanol

B
(1S, 3S)-Cis-3-methyl cyclohexanol

C
(1S, 3S)-Trans-3-methyl cyclohexanol

D
(1R, 3R)-Trans-3-methyl cyclohexanol

Solution

In the presence of a polar aprotic solvent, like acetone, the mechanism followed will be SN².

Walden inversion takes place at C₁, where -Br is substituted by -OH.

Hence, the product formed will be (1S, 3S)-Trans-3-methyl cyclohexanol
Question 7
4 / -1

What is the magnetic moment of coordination compound formed during brown ring test?

A
3.87 BM

B
4.92 BM

C
5.92 BM

D
2.83 BM

Solution

Brown ring test When a freshly prepared FeSO₄ solution is added to aqueous solution of NO⁻₃ ion followed by addition of concentrated H₂SO₄ the brown ring is observed at junction between two liquids. This colour is due to charge transfer oxidation state of iron in this complex is + l. In this state Fe⁺, has 3 unpaired electrons, and hence, the magnetic moment will be: √3(3+2) = 3.87 BM
Question 8
4 / -1

A

B

C

D
None of these

Solution
Question 9
4 / -1

Out of 3n consecutive integers, three are selected at random. Find the probability that their sum is divisible by 3.

A

B

C

D

Solution

Suppose the sequence of 3n consecutive integers begin with integer m.

Then the 3n consecutive integers are:

m, m + 1, m + 2,…………, m + (3n - 1)

Out of these integers, 3 integers can be chosen in ³ⁿC₃ ways.

Let us divide these 3n consecutive integers in the groups G1, G2 and G3 as follows:

G1: m, m + 3. m + 6, …….., m + (3n - 3)

G2: m + 1, m + 4, m + 7, ……., m + (3n - 2)

G3: m + 2, m + 5, m + 8, …….., m + (3n - 1)
Question 10
4 / -1

If SK be the perpendicular from the focus S on the tangent at any point P on the ellipse then locus of the foot of the perpendicular K is equal to

A
a²x²+ b²y²= (ax − by)²

B
x² + y² = a²

C
x²+ y²= b²

D
x²+ y.= a²+ b²

Solution

By eliminating θ from both the eqs. (i) and (ii) [by squaring and adding (i) & (ii)], we get the locus of K as x²+ y²= a² i.e., the auxiliary circle of the ellipse.