Daily NEET MCQ- March 31, 2022 - NEET MCQ

Aristotle classified the organisms based on their morphological characteristics. His classification neglected characteristics like nature of the cell wall, evolutionary relationships and method of reproduction.

Aristotle’s classification contained two kingdoms i.e. Plants and Animals. Plants are sub-divided into herbs, shrubs and tree. Animals are further sub-divided into red blooded and non- red blooded animals.

Aristotle classified the organisms into Plants and Animals whereas Carolus Linnaeus classified the organisms into Plantae and Animalia. Whittaker’s 5 kingdom classification contained Monera, Protista, Fungi, Plantae and Animalia.

The kinetic energy of a particle in SHM is given by:
0.5mA²w²cos²(wt + a), where a is the phase constant.
cos²(wt + a) = (1 + cos2(wt+a))/2.
Therefore the time period of oscillation of kinetic energy is 4/2 = 2s.

Let x = Asin(wt).
∴ Kinetic energy = 1/2 mA²w²cos²(wt)
potential energy = 1/2kA²sin²(wt).
According to given condition:
1/2mA²w²cos²(wt) = 1/2A²w²sin²(wt)
using, k = mw², we get:
tan²(wt) = 1.
∴ wt = π/4.
∴ x = Asin(π/4) = A/√2.

1/2ka² = 8 & 1/2kb² = 18.
We have to find value of 1/2k(a + b) ².
∴ 1/2k(a + b)² = 8 + 18 + kab
= 26+(k*4*6/k)
= 26 + 24 = 50J.

F = -dU/dx = 8 – 2x.
Particle starts from x=0, so force will increase its speed till force becomes zero at x=4.
Let acceleration be ‘a’.
a = dv/dt = v dv/dx.
And a = F/m = F/1 = 8-2x.
adx = vdv. Now, integrate both sides. The limits of dx will be 0 to 4 and the limits of dv will be 0 to v.
(8x – x²)₀⁴ = v²/2.
∴ v = 4√2 m/s.

When two or more compounds have the same molecular formula but the different position of functional groups are substituents, they are called positional isomers and the phenomenon is called position isomerism. Here 2-chloropropane and 1-chloropropane differ in position so they exhibit position isomerism.

The compound whose half part of a molecule is a mirror image of the other half, is called meso form. Generally, a meso compound has two or more chiral centers and one plane of symmetry. the compound in meso form is optically inactive due to internal compensation.

The number of optically active isomers for a compound is given as 2ⁿ where, n represents the number of chiral carbons in that particular carbon, here as the compound has 3 chiral carbons n = 3, so the number of optically active isomers equals 2³ = 8.