Monthly Mock Test (August 31) - NEET MCQ

e = | v₁ – v₂ |/ | u₁ – u₂ | which is 1 for a perfectly elastic collision.

Here, P = 2kW = 2000 W..
W = Mgh = M × 10 × 10 = 100 M and t = 60 s.
This gives, M = 1200 kg Its volume = 1200 litre as 1 litre of water contains 1 kg of its mass.

Initial momentum (p₁) = p; Final momentum

(p₂) = 1.5 p and initial kinetic energy (K₁) = K.

 Kinetic energy

or, K₂ = 2.25 K.
Therefore, increase in kinetic energy is 2.25 K – K = 1.25 K or 125%.

Applying law of conservation of momentum,

Applying conservation of momen tum, 

Here, v₁ = 36 km/hr = 10 m/s m₁ = 2 kg, m₂ = 3 kg

K.E. (initial) =

K.E. (Final) =

Loss in K.E. = 100 – 40 = 60 J

As  ΔH = ΔE + Δn_gRT if n p < nr; Δ n_g = n_p – n_r = – v_e.
Hence Δ H < Δ E.

During isothermal expansion of ideal gas,
ΔT = 0.   Now H = E + PV
∵ ΔH = ΔE + Δ(PV)
∴ ΔH = ΔE + Δ(nRTT));
Thus if ΔT = 0., ΔH = ΔE i.e., remain unaffected

If more tran s-2-pentene is added, then its concentration in right hand side will increase. But in order to maintain the K constant, concentration of cis-2-pentene will also increase. Therefore more cis-2-pentene will be formed.

Enthalpy of formation of C₂H₄ , CO₂ and H₂O are 52 , – 394 and – 2 86 k J/ mol respectively. (Given) The reaction is
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O.
change in enthalpy, (ΔH) = ΔH_products - ΔH_reactants
= 2 x (-394) + 2 x (-286) - (52 + 0)
= – 1412 kJ/ mol.

The relation between free energy change and equilibrium constant is given by Nernst equation

At equilibrium,  E_c = 0 and Q = K_C

................(i)

Again ΔGº = nFEº (ii)

put in (i)

Entropy change ΔS =  ΔS_product - ΔS_reactant  
= 2 (186.7) – (223 + 130.6)
 = 373.4 – 353.6
 = 19.8 JK^–1 mol^–1

For a cyclic process

ΔE = 0, ΔH = 0 & ΔG = 0 .

As all depend upon final state and initial state,w doesn’t depend on path followed.

= 9.77 J K^–1 mol^–1

 = –98.2 + 298.2 = 200 kJ/Mole

In the alveoli, where there is high pO₂, low pCO₂, lesser H⁺ concentration and lower temperature, the factors are all favourable for the formation of oxyhaemoglobin, whereas in the tissues, where low pO₂, high pCO₂, high H⁺ concentration and higher temperature exist, the conditions are favourable for dissociation of oxygen from the oxyhaemoglobin.

Expiratory capacity is the total volume of air a person can expire after normal inspiration. It includes tidal volume and expiratory reserve volume.
EC = TV + ERV = 500 + 1000 = 1500 mL

Residual volume is the volume of air which remains in the lungs after the most forceful expiration. This residual air enables the lungs to continue exchange of gases even after maximum exhalation. Due to this, lungs do not collapse even after forceful expiration.

Frog has lungs as its main respiratory organs but during hibernation & aestivation and during its habitat in water it respires through skin

CO₂ from the respiratory tissues to the lungs is transported by the blood in 3 ways :

(i) In dissolved state or as a physical solution : Very small amount physically dissolved in plasma (7% i.e. @ 0.3 ml of CO₂ by each 100 ml of blood)

(ii) Bicarbonate ions : @ 70% (i.e. @ 2.5 ml per 100 ml of blood) CO₂ diffuses in plasma & then into RBCs where it (in the  presence of carbonic anhydrase) combines with H₂O to form carbonic acid which is almost spontaneously dissociated into hydrogen ion & bicarbonate ions.

(iii) Carbaminohaemoglobin : @ 23% (i.e. @ 1 ml of CO₂ per 100 ml of blood) combines with haemoglobin forming an unstable compound

CO₂ enters RBC and reacts with water to form carbonic acid. Carbonic acid dissociates to form bicarbonate and hydrogen ions. Some bicarbonate ions are transported in erythrocytes while some diffuse into the blood plasma. Exit of bicarbonate ions change the ionic balance between the plasma and erythrocytes. To restore this balance chloride ions diffuse from plasma into erythrocytes. Due to this the pH of blood is maintained.

At the altitude of 5400 meters the low atmospheric pressure of O₂ will be too low so the solubility of oxygen in the blood will be very less hence the oxygen carried by each RBC will be too less. But to fulfill the oxygen requirement of the body blood has to carry more oxygen to the body tissue and this is done by the increased no. of RBCs.

Gaseous exchange between blood (dissolved phase) and alveolar air (gaseous phase) across respiratory membrane occurs by simple diffusion. Obviously, it depends upon the concentration gradient (=partial pressure) of the concerned gases in blood and alveolar air.

The pulse making impulse travels in the heart in the order of SA node→AV node→bundle of His→Purkinje fibres→heart muscles.

White blood cells are of two types : (i) Granulocytes (with granular cytoplasm) - neutrophils, basophils and eosinophils. (ii) Agranulocytes (with clear cytoplasm)  - Lymphocytes and monocytes

Histamine is a potant vasodilator  formed by decarboxylation of the amino acid histidine and released by mast cells in response to appropriate antigens.
Mast cells are especially prevalent in the connective tissue of the skin and respiratory tract and in surrounding blood vessels.

SA is the site of generation of the rhythmic cardiac impulse. AV node is a compact mass of  myogenic fibres which receives impulse from SA node and transmits it to ventricles. Purkinje fibres are in contact with the contractile muscles of the ventricular walls. Chordae tendinae are the muscles which keep the AV valves  in position.

Leucocytes  are colourless, nucleated amoeboid cells found in blood which are devoid of  haemoglobin and are  capable of coming out of blood capillaries through the process of diapedesis. Fall of  WBC count is called leucopenia, and occurs due to folic acid deficiency and AIDS etc. 

Sickle cell anaemia is a hereditary disorder of autosomal nature caused  by mutation of the gene controlling b-chain of haemoglobin. It involves substitution of glutamine by valine.

S.A node is called the pacemaker because it initiates the cardiac impulse AV node receives the impulse from the S.A  node and transmits it to the ventricles