JEE Advanced Mix Test 43

Result

JEE Advanced Mix Test 43

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Weekly Quiz Competition

Question 1
4 / -1

When liquid medicine of density p 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
Question 2
4 / -1

The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density p(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 p 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

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.

...view full instructions

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

A
oxidising

B
reducing

C
unreactive

D
strong basic

Solution

The oxidation state of Xe in XeF₄ is +4 and in XeF₆ is +6. These oxidation states of Xe are displayed only with elements with high electronegativity, like oxygen and fluorine. Since, Xe will have a tendency to get reduced, the nature of fluorides of xenon is expected to be oxidising
Question 5
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 6
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 7
4 / -1

A

B

C

D
None of these

Solution
Question 8
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
Question 9
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.
Question 10
4 / -1

A
[x]

B
[x + 1]

C
[x + 2]

D
[x + 3]

Solution