NEET Test 199

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NEET Test 199

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Weekly Quiz Competition

Question 1
4 / -1

A coil of wire of radius R has 200 turns and self – inductance of 108 mH. The self – inductance of a similar coil of 500 turns will be:

A
375 mH

B
527 mH

C
675 mH

D
None of these

Solution
Question 2
4 / -1

A point source of light is kept at a distance of 15 cm from a converging lens, on its optical axis. The focal length of the lens is 10 cm and its diameter is 3 cm. A screen is placed on the other side of the lens, perpendicular to the axis of the lens, at a distance 20 cm. from it. Then find the area of the illuminated part of the screen?

A

B

C

D

Solution
Question 3
4 / -1

The bond dissociation energy of B − F in BF₃ is 646 kJ per mol whereas that of C − F in CF₄ is 515 kJ mol−1. The correct reason for higher bond dissociation energy of B − F bonds than that of C − F bond is

A

B
Smaller size of B atom than that of C atom

C
Stronger bond between B and F in BF₃ than that between C and F in CF₄

D

Solution

BF₃ is a Lewis acid due to an incomplete octet of Boron. Among BF₃ BCl₃,BBr₃ and so on, BF₃ is the weakest Lewis acid due to back bonding of lone pair electron of F to p-orbital of Boron.

B has vacant available p-orbital in BF₃ and hence it involves pπ - pπ back bonding which is not possible in CF₄ as C does not have any vacant orbital.
Question 4
4 / -1

Which of the following monocyclic ring compounds do not obey Huckel's rule?

P. C₁₂H₁₂^2-

Q. C₁₂H₁₂²⁺

R. C₂₀H₂₀^-

S. C₂₀H₂₀⁺

A
Only P

B
Only Q

C
Q and R

D
R and S

Solution

Huckel's rule (4n + 2 rule): This rule defines that in order to be aromatic a molecule should have a certain number of π electrons (π bonds or lone pair or lone pairs within p orbitals ) within a closed loop of parallel adjacent p- orbitals in this 4n+2π electron where n=0,1,2,3 and so on.

Therefore, in C₁₂H₁₂^2- ,C₁₂H₁₂²⁺, +4n+2π electrons. Therefore Huckel's rule is obeyed and those who don't obey this rule are R and S.
Question 5
4 / -1

In a DNA strand the nucleotides are linked together by:

A
Glycosidic bonds

B
Phosphodiester bonds

C
Peptide bonds

D
Hydrogen bonds

Solution

A nucleotide has three components - nitrogenous base, a pentose sugar ribose in the case of RNA and deoxyribose for DNA), and a phosphate group. There are two types of nitrogenous bases - Purines (Adenine and Guanine) and Pyrimidines (Cytosine and Thymine). A nitrogenous base is linked to the pentose sugar through a N-glycosidic linkage to form a nucleoside. When a phosphate group is linked to 5 -OH of a nucleoside through phosphodiester linkage, a corresponding nucleotide is formed. The nucleotides are linked through 3-5 phosphodiester linkage to form a dinucleotide.
Question 6
4 / -1

How many electrons are present in m shell of an element with atomic number 20:

A
5

B
8

C
15

D
18

Solution

The element having an atomic number 20 is calcium(Ca).

The electronic configuration of Calcium is (2,8,8,2).

K shell - 2 electrons

L shell - 8 electrons

M shell - 8 electrons

N shell - 2 electrons

So, the M shell of an element with an atomic number of 20 has 8 electrons.
Question 7
4 / -1

A rocket is fired from Earth to the Moon. The distance between Earth and the Moon is r and the mass of Earth is 81 times the mass of the Moon. The gravitational force on the rocket will be zero when its distance from the Moon is:

A

B

C

D

Solution

The net force on the rocket is zero.

Because force due to the earth and force due to the moon is the same but in opposite direction.
Question 8
4 / -1

When a body moves in a circular path, no work is done by the force since:

A
Force and displacement are perpendicular to each other

B
The force is always away from the centre

C
There is no displacement

D
There is no net force

Solution
Question 9
4 / -1

Exudation and guttation are due to:

A
Osmotic pressure

B
Root pressure

C
Diffusion pressure

D
transpiration pull

Solution

Osmotic pressure developed in the cells of roots in the root system is termed as root pressure and this supports in the rise of the cell sap towards the leaves. Root pressure results in exudation and as well as guttation.

Fluid emitted via the roots of the plants are referred to as exudates and the process is called as exudation.

Loss of impure water from plant, generally from the margins of leaves is termed as guttation. The water contains a dilute solution of different organic and inorganic substances. Guttation occurs due to low rate of transpiration and more water absorption. This higher rate of absorption develops root pressure and pushes the excess water out of the specialized pores of veins called hydathodes. Guttation is a type of exudation.
Question 10
4 / -1

The decomposition of Cl₂O₇ at 400 K in the gas phase to Cl₂ and O₂ is a first order reaction. After 55 seconds at 400 K, the pressure of Cl₂O₇ falls from 0.062 atm to 0.044 atm. Calculate the pressure of Cl₂O₇, after 100 sec. of decomposition at the temperature.

A
0.033

B
0.044

C
0.055

D
None of the above

Solution