NEET Unit Test - 2 - NEET MCQ

Assume initial speed of man = u and kinetic energy = K
Final speed of man = (u+2) then kinetic energy = 2K. Also,

r=0.5m, t=15s
n₁=5rpm=5/60 rps
n₂=25rpm=25/60rps
… ω₁=2π(5) rad/s
… ω₂=2π(25) rad/s
As,
a=r∝          [∝=ω₂- ω₁/t]
so, a=0.5[2π(n₂-n₁)/t]
a=0.5x6.28x20/60x15
a=6.28x2/60x3
a=6.28/90
a=0.697 m/s²
a≈0.7 m/s²

For a system to be in equilibrium torques on the system must be equal to zero
This is true only if the torques are taken about the centre of mass as the whole mass of the object seems to be situated at centre of mass.

  (orbital speed υ₀ of a satellite)

Near the earth’s surface is equal to 1/√2
times the escape velocity of a particle on earth’s surface) Now from conservation of mechanical energy: Decrease in kinetic energy = increase in potential energy

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. Let M be the mass of the planet R its radius and m the mass of a particle on its surface. Then

Given,
 l=2m
m=1kg
ω=15rd//s
K.E.=1/2I ω
=1/2x(ml²/3) ω²
=1/2x(1x4/3)(15)²
=150J

For ring 1
(MOI)1 = MOI about diameter + PQ²
MOI about diameter = ½ PQ² 
(MOI)1 = 3/2 PQ²
Similarly, (MOI)2 = 3/2 PQ²
(MOI)3 = ½ PQ²
Total MOI = 7/2 PQ²

The difference in the potential of the body when it rolls down through a vertical height h, is mgh.
As the KE at the top point is zero and let say KE at bottom is ½ mv² + ½ Iw²
Where m is its mass, I is its moment of inertia, I = ½ mr²
Where r is its radius, v is its gained translational speed and w is its gained angular speed.
w = v/r
Hence equating PE and KE gives
mgh = ½ mv² + ½ Iw²
That is mgh = ½ mv² + ½ mr².(v/r)²
We get mgh = ½  mv² + ¼  mv² 
Thus we get v = √gh/3

In compound X, all the bond angles are exactly 109^∘28′X is carbon tetrachloride.
CCl has sp³ - hybridization with 4 bond pairs and 0 lone pairs of electrons giving regular tetrahedron geometry as all the four C−Cl bonds are equivalent. In others the geometry is little distorted inspite of sp³ - hybridization due to different atoms on the vertices of tetrahedron.

In (i), the Ip are atequatorial position so there are less Ip - bp repulsions as compared to other positions.

Hence, T-shape is moststable.

The Phosphorus Pentafluoride (PF₅) is a nonpolar molecule because the configuration of PF₅ has trigonal bipyramidal, and PF₅ consists of a central phosphorus atom, surrounded by five fluorine atoms. So the three planar fluorine’s electronegativity cancels out each other, which resulting PF₅ is a nonpolar molecule.

similarly, in PCl₅ case chlorine's electronegativity cancels out each other.

 In case of PCl_3​F₂​, the molecule is non-polar, even though the fluorine atoms are slightly more electro-negative than the chlorine. This could be explain in a couple ways. The more intuitive approach is to consider how the presence of two fluorine atoms would create symmetry. The less intuitive, but more accurate, process would be to realize that the structure of the molecule is divided in to two non-polar structure(trigonal planer and linear). With other molecules, like PF₃​Cl₂​ it's not possible to divide it in to non-polar shapes. The repulsion between the fluorine atom is greater than that between the chlorine. Thus PF_3​Cl_2​ is polar.

No lone pair ofelectrons is available in BCI₃

One lone pair of electrons is available on N atom, occupies a corner in the tetrahedral arrangement. Therefore, NCI₃ appears pyramidal in shape.

Phyllode is a modification of leaf or petiole. It is green in colour and is able to carry out photosynthesis. It is a xerophytic adapation. It reduces the rate of transpiration in xerophytic plants like Opuntia. Thus, the correct answer is 'Petiole.'

Reticulate venation is a characteristic in Dicotyledonous leaves. In reticulate venation, veins are interconnected and form a web-like network. There is a prominent vein called the midrib from which many small veins arise which finally form a net-like structure in the lamina. The veins vary in thickness in the reticulate venation of the dicot leaves.

Solution: Meristematic tissue consists of immature cells that are in a constant state of division, contributing to the growth of plants. This tissue is found in specialized regions such as the tips of roots and shoots, where growth is most active. Unlike other tissues that might specialize in support, transport, or photosynthesis, the primary role of meristematic tissue is to facilitate growth through cell division

Exarch xylem is the arrangement in which the protoxylem (the first formed primary xylem cells in a plant shoot/root) is directed towards the periphery and metaxylem (later formed primary xylem cells in a plant shoot/root) is directed towards the centre. Its development follows a centripetal pattern and is characteristic of roots. 

Thus the correct answer is option D.

• The respiratory or gas-exchange surface consists of millions of small sacs, or alveoli, lined by a simple squamous epithelium.
• This epithelium is exceedingly thin to facilitate the diffusion of oxygen and CO₂. 
• The alveolar walls also contain cuboidal surfactant-secreting cells.
  

Hence, the correct option is A

NCERT Reference: Topic- EXCHANGE OF GASES” of chapter "Breathing and Exchange of Gases" of NCERT 

• Graph is plotted between oxygen concentration and percentage saturation of haemoglobin with this curve is called dissociation curve.
• Dissociation curve is sigmoid in shape.

Hence, the correct option is D.

NCERT Reference: Topic-  Transport of Oxygen” of chapter "Breathing and Exchange of Gases" of NCERT

• Workers in a mill or industrial setting may be at risk of developing occupational respiratory disorders.
• These disorders can be caused by exposure to various airborne pollutants, such as dust, fumes, chemicals, and particulate matter, which are commonly found in industrial and manufacturing environments.
• In industries, produced dust can’t be cope up by the defense system of human body.
• Long and continuous exposure gives rise to inflammation leading to fibrosis and thus leading to severe lung damage.

Hence, the correct option is D

NCERT Reference: Topic -"Respiratory Disorders"  of chapter "Breathing and Exchange of Gases" of NCERT.

Silicosis and asbestosis are known as occupational disorder as they occur due to one's occuption. Workers working in industries where silica and asbestos are used can suffer from these respiratory disorders.

All the daughter cells of multicellular organism have the same number and type of chromosomes as parent cells due to equitable distribution of all chromosomes in mitosis. This helps in ensuring genetic continuity.

SAN (sino-atrial node) is located in the right upper y corner of right atrium. AV bundle continues from AVN (Atrio-ventricular node). Purkinje fibres are modified cardiac muscle fibres that originate from AV node and spread into two ventricles. The nodal musculature has the ability to generate action potential without any external stimuli i.e. it is autoexcitable. AVN is located in the lower left corner of the right atrium close to the atrio-ventricular septum.

Coronary arteries supply blood to the heart muscle. Like all other tissues in the body, the heart muscle needs oxygen-rich blood to function.

Systemic circulation is the movement of the blood between heart and rest of the body (tissues) and the back to the heart. The oxygenated blood moves from the left auricle into the left ventricle. From here the blood moves into the aorta which ultimately divides into arteries that carry oxygenated blood to various tissues in the body. From the tissues deoxygenated blood is carried by the veins into the right atrium via superior vena cava and inferior vena cava.

The cardiac output of a person having 72 heart beats per minute and a stroke volume of 50 mL would be 72 x 50 = 3600 mL. Thus, the correct answer is 3600 mL, i.e Option D.

As the ventricle is completely divided in birds, mammals and some reptiles (Crocodiles alligator) the left and right pans of the heart function as air tight conduits for pure and impure blood. The right part receives impure blood from whole body and sends it to the lungs for oxygenation. The left part receives purified blood from the lungs and supplies it to the whole body Thus the right and left parts of the heart respectively serve as completely separated pulmonary and systamic hearts. This is known as double heart circuit. In man the rate of heart beat (double circulation) is about 75 times per minute

Coronary heart disease (CHD) is a disease in which a waxy substance called plaque builds up inside the coronary arteries. These arteries supply oxygen-rich blood to your heart muscle. When plaque builds up in the arteries, the condition is called atherosclerosis. The buildup of plaque occurs over many years.

Rate of breathing is mainly controlled by CO₂ level in the blood.
Excess CO₂ decrease the pH of the blood and it mainly stimulate the respiratory centre of the brain (i.e., Medulla oblongata) and increase the inspiratory and expiratory signals to the respiratory muscles.

Normal blood pressure is 120/80 mm Hg. Blood pressure of 190/110 mm Hg is regarded as high blood pressure. It leads to heart diseases and also affects vital organs like brain and kidney.