A particle of mass m begins to slide down a fixed smooth sphere from the top as shown. What is its acceleration when it breaks off the sphere?
As the particle of mass m loses contact with the sphere, no normal reaction is exerted by a sphere on the particle. The only force acting on the particle is its weight. Hence its acceleration is g(acceleration due to gravity)
An automobile enters a turn of radius R. If the road is banked at an angle of 450 and the coefficient of friction is 1, the minimum and maximum speed with which the automobile can negotiate the turn without skidding is :
F.B.D. for minimum speed (w.r.t. automobile):
for θ = 45° and μ = 1 vmin
F.B.D. for minimum speed (w.r.t. automobile):
Figure shows the roller coaster track. Each car will start from rest at point A and will roll with negligible friction. It is important that there should be at least some small positive normal force exerted by the track on the car at all points, otherwise the car would leave the track. With the above fact, the minimum safe value for the radius of curvature at point B is (g = 10 m/s2) :
Two masses m1 and m2 which are connected with a light string, are placed over a frictionless pulley. This set up is placed over a weighing machine, as shown. Three combination of masses m1 and m2 are used, in first case m1 = 6 kg and m2 = 2 kg, in second case m1 = 5 kg and m2 = 3kg and in third case m1 = 4 kg and m2 = 4 kg. Masses are held stationary initially and then released. If the readings of the weighing machine after the release in three cases are W1, W2 and W3 respectively then :
Net weight showing in the weighing machine will be equal to the tension in string on both sides i.e. on the side of m1 and m2.
Therefore, W= 2T
Now, from the block diagram,
2T = (m1+m2)g/m1m2
W1=6+ 22 x 6 x 2g=3g
W2=5+ 32 x 5 x 3g=3.75g
W3=4+ 42 x4 x 4g=4g
The ratio of work done by the internal forces of a car in order to change its speed from 0 to V and from V to 2V is (Assume that the car moves on a horizontal road) -
Work done in changing speed from 0 to V is
work done in changing the speed from V to 2V is
A weightless rod of length 2l carries two equal masses 'm', one tied at lower end A and the other at the middle of the rod at B. The rod can rotate in vertical plane about a fixed horizontal axis passing through C. The rod is released from rest in horizontal position. The speed of the mass B at the instant rod, become vertical is :
Let v be the speed of B at lowermost position, the speed of A at lowermost position is 2v. From conservation of energy
Power versus time graph for a given force is given below. Work done by the force upto time t(<t0).
The work done by force from time t = 0 to t = t sec. is given by shaded area in graph below.
Hence as t increases, this area increases.
∴ Work done by force keeps on increasing
A motor car is going due north at a speed of 50 km/h. It makes a 90º left turn without changing the speed. The change in the velocity of the car is about
A body is thrown horizontally with a velocity from the top of a tower of height h. It strikes the level ground through the foot of the tower at a distance x from the tower. The value of x is
A train is standing on a platform , a man inside a compartment of a train drops a stone . At the same instant train starts to move with constant acceleration . The path of the particle as seen by the person who drops the stone is :
Relative to the person in the train, acceleration of the stone is 'g' downward, a (acceleration of train) backwards.
According to him :
P is a point moving with constant speed 10 m/s such that its velocity vector always maintains an angle 60° with line OP as shown in figure (O is a fixed point in space). The initial distance between O and P is 100 m. After what time shall P reach O.
Velocity of approach of P and O is = v cos 60° = 5 m/s
It can be seen that velocity of approach is always constant.
∴ P reaches O after = 100/5 = 20 sec.
A mosquito with 8 legs stands on water surface and each leg makes depression of radius ' a'. If the surface tension and angle of contact are ' T ' and zero respectively, then the weight of mosquito is :
According to the questions the legs of the mosquito landing upon the water surface. Therefore, T.2πa × 8 = Weight of the mosquito.
The work done in increasing the size of a rectangular soap film with dimensions 8 cm × 3.75 cm to 10 cm × 6 cm is 2 × 10–4 J. The surface tension of the film in N/m is :
The property of surface tension is to________
Due to surface tension, the surface area tries to minimize itself.
Two uniform rods of equal length but different masses are rigidly joined to form a L-shaped body, which is then pivoted about O as shown. If in equilibrium the body is in the shown configuration, ratio M/m will be:
In equilibrium, torques of forces mg and Mg about an axis passing through O balance each other.
A large open tank has two small holes in its vertical wall as shown in figure. One is a square hole of side 'L' at a depth '4y' from the top and the other is a circular hole of radius 'R' at a depth ‘y’ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, 'R' is equal to :
Let v1 and v2 be the velocity of efflux from square and circular hole respectively.
S1 and S2 be cross-section areas of square and circular holes.
The volume of water coming out of square and circular hole per second is
∵ Q1 = Q2
A stationary body explodes into two fragments of masses m1 and m2. If momentum of one fragment is p, the energy of explosion is :
By momentum conservation, the momentum of other fragment will be –p Total KE
A shell of mass 2 m projected with a speed ' u ' at an angle θ to the horizontal explodes at the highest point of its motion into two pieces of mass ' m ' each. If one piece whose initial speed is zero, falls vertically, the distance at which the other piece will fall from the gun is given by :
The COM will fall at distance from the initial point, irrespective of explosion. Now one particle is at R/2 the other should be at 3R/2.
A man of 80 kg attempts to jump from a small boat of mass 40 kg on to the shore. He can generate a relative velocity of 6 m/s between himself and boat. His velocity towards the shore is :
Let the velocity of boat be 'V' opposite to the man. By momentum conservation, –40V + 80 (6 – V) = 0
V = 4 m/s
velocity of the man = 6 – 4 = 2m/s
A block of mass m slides along the track with coefficient of kinetic friction μ. A man pulls the block through a rope which makes an angle θ with the horizontal as shown in the figure. The block moves with constant speed V. Power delivered by the man is :
A conical pendulum consists of a simple pendulum moving in a horizontal circle as shown in the figure. C is the pivot, O is the centre of the circle in which the pendulum bob moves and ω the constant angular velocity of the bob. If is the angular momentum about point C, then
The direction of L is perpendicular to the line joining the bob to point C. Since this line keeps changing its orientation in space, direction of L keeps changing however as ω is constant, magnitude of L remain constant.
Aliter : The torque about point is perpendicular to the angular momentum vector about point C. Hence it can only change the direction of L, and not its magnitude.
A sphere of mass m and radius r is projected in a gravity free space with speed v. If coefficient of viscosity is 1/6π, the distance travelled by the body before it stops is :
The only force acting on the body is the viscous force
Two same masses are tied with equal lengths of strings and are suspended at the same fixed point. One mass is suspended freely whereas another is kept in a way that string is horizontal as shown. This mass is given initial velocity u in vertical downward direction. It strikes the freely suspended mass elastically that is just able to complete the circular motion after the collision about point of suspension O. Magnitude of velocity u is :
As collision is elastic, freely suspended mass moves acquiring velocity of colliding mass after the collision. Also acquired velocity must be equal to to complete the circular motion.
A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if
For the ring to move in a circle at constant speed the net force on it should be zero. Here spring force will provide the necessary centripetal force.
∴ kx = mxω2
A force acts on a body due to which its position varies as Work done by this force in initial 2 seconds is :
A particle A is projected with speed VA from a point making an angle 60º with the horizontal. At the same instant, a second particle B is thrown vertically upwards from a point directly below the maximum height point of parabolic path of A , with velocity VB. If the two particles collide then the ratio of VA/VB should be :
A uniform rod of density ρ, and length ‘l’ is having square cross-section of side ‘a’. It is placed in a liquid of equal density ρ vertically along length in a tank having sufficient height of liquid. The surface tension of liquid is ‘T’ and angle of contact is 120º. Then :
A rigid body undergoing uniform pure rolling encounters horizontal tracks AB and BC as shown in the figure AB is a smooth layer of ice and BC is a rough surface with μ = 1. Both AB & BC are rigid tracks. Which of the following statements are CORRECT:
(1) The body will slow down over BC
(2) The body will start slipping on AB
(3) The body remains in pure rolling over the whole stretch AC
(4) The angular velocity of the body remains constant over the whole stretch AC.
For a rigid body in uniform pure rolling, there is no friction, hence irrespective of the nature of surface on which it rolls (as long as the surface is rigid), it will continue in its state of pure rolling with uniform velocity.
A disc of mass m0 rotates freely about a fixed horizontal axis through its centre. A thin cotton pad is fixed to its rim, which can absorb water. The mass of water dripping onto the pad is μ per second. After what time will the angular velocity of the disc get reduced to half of its initial value?
Conservation of angular momentum
A uniform cylinder of mass M and radius R rolls without slipping down a slope of angle θ to the horizontal. The cylinder is connected to a spring constant K while the other end of the spring is connected to a rigid support at P. The cylinder is released when the spring is unstretched. The maximum distance that the cylinder travels is
Potential energy of spring
Workdone by gravity = mg sineθx.
By energy conservations,
A simple pendulum is oscillating in a vertical plane. If resultant acceleration of bob of mass m at a point A is in horizontal direction, find the tangential force at this point in terms of tension T and mg.
When the acceleration of bob is horizontal, net vertical force on the bob will be zero.
T cos θ – mg = 0
The tangential force at that instant is
Hailstones falling vertically with a speed of 10 m/s, hit the wind screen (wind screen makes an angle 30° with the horizontal) of a moving car and rebound elastically. The velocity of the car if the driver finds the hailstones rebound vertically after striking is :
For the driver to observe the rain move vertically upward after the elastic collision, rain should come at an angle 30° with the horizontal (as clear from figure).
Let , velocity of rain w.r.t. car be Vm/c
But since rain fall vertically down.
A water tank stands on the roof of a building as shown. Then the value of 'h' for which the distance covered by the water 'x' is greatest is -
The roots of x are (0,4) and the maximum of x is at h = 2.
The permitted value of h is 0 to 1 clearly h = 1 will give the maximum value of x is this interval.
Aliter : If the column of water itself were from ground upto a height of 4m, h = 2m would give the maximum range x. Farther the hole is from this midpoint, lower the range. Here the nearest point possible to this midpoint is the base of the container. Hence h = 1 m.
A U-tube of base length “l” filled with same volume of two liquids of densities ρ and 2ρ is moving with an acceleration “a” on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height h is given by:
For the given situation, liquid of density 2ρ should be behind that of ρ.
From right limb :
But from left limb :
From (1) and (2) :
A particle A of mass 10/7 kg is moving in the positive direction of x. Its initial position is x = 0 & initial velocity is 1 m/s. The velocity at x = 10 is: (use the graph given)
Area under P –x graph =
from graph ;
∴ v = 4 m/s
ALITER : from graph
P = 0.2 x + 2 or
or mv2 dv = (0.2 x + 2) dx
Now integrate both sides,
⇒ v = 4 m/s
The string of a step rolling wheel is pulled by applying force F with different lines of action in two situations as shown. The wheel starts rolling without slipping due to application of the force :
(Pure rotation about instantaneous point of contact)
Note : If line of action passes through point of contact, it only spins.
Two particles start together from a point O and slide down straight smooth wires inclined at 30º & 60º to the vertical & in the same vertical plane as in figure. The relative acceleration of second with respect to first will be (in magnitude & direction) as :
α = 30°
i.e. Resulting acceleration is in vertical direction.
Two points A & B on a disc have velocities v1 & v2 at some moment. Their directions make angles 60° and 30° respectively with the line of separation as shown in figure. The angular velocity of disc is :
For rigid body separation between two point remains same.
v1 cos60° = v2 cos30°
The spring block system lies on a smooth horizontal surface. The free end of the light spring is being pulled towards right with constant speed v0 = 2m/s. At t = 0 sec, the spring of spring constant k = 100 N/cm is unstretched and the block has a speed 1 m/s to left. The maximum
In the frame (inertial w.rt earth) of free end of spring, the initial velocity of block is 3 m/s to left and the spring unstretched.
(1/2)*m*v2 = (1/2)*k*A2
The figure shows a hollow cube of side 'a' of volume V. There is a small chamber of volume V/4 in the cube as shown. This chamber is completely filled by m kg of water. Water leaks through a hole H. Then the work done by gravity in this process assuming that the complete water finally lies at the bottom of the cube is :
Let h be the height of water surface, finally
∴ C. M. gets lowered by
∴ Work done by gravity
Two blocks ‘A’ and ‘B’ each of mass ‘m’ are placed on a smooth horizontal surface. Two horizontal force F and 2F are applied on the two blocks ‘A’ and ‘B’ respectively as shown in figure. The block A does not slide on block B. Then the normal reaction acting between the two blocks is :
Acceleration of two mass system is a = F/2m leftward
FBD of block A
A coin is released inside a lift at a height of 2 m from the floor of the lift. The height of the lift is 10 m. The lift is moving with an acceleration of 11 m/s2 downwards. The time after which the coin will strike the lift is :
Relative to lift initial velocity and acceleration of coin are 0 m/s and 1 m/s2 upwards
∴ or t = 4 second
The extension in a uniform rod of length l, mass m, cross section radius r and young’s modulus Y when it is suspended at one of its end is :
Tension at point P in the rod is
Extension of width dx length of rod is
Total extension in road
A spherical soap bubble of radius 1.0 cm is formed inside another of radius 2 cm. If a single soap bubble is formed which maintains the same pressure difference as inside the smaller and outside the larger bubble, the radius of this bubble is -
r = 4/6 cm = 0.0067 m
Block ‘ A ‘ is hanging from a vertical spring and is at rest. Block ‘ B ‘ strikes the block ‘A’ with velocity ‘ v ‘ and sticks to it. Then the value of ‘ v ‘ for which the spring just attains natural length is :
The initial extension in spring is
Just after collision of B with A the speed of combined mass is V/2
For the spring to just attain natural length the combined mass must rise up by and comes
Applying conservation of energy between initial and final states
Solving we get
Alternative solution by SHM
The increase in volume of air, when temperature of 600 ml of it, is increased from 27°C to 47°C under constant pressure, is
so increment = (640 - 600)ml = 40ml
The density of steam at 27°C and 8.314 × 104 pascal is 0.8 Kg m–3. The compressibility factor would be
The velocity corresponding to a gas "X" (MW = 100) corresponding to the maxima of the curve given below at a given temperature is 200 m/s. Then kinetic energy of 300 gram of gas "X" will be :
The velocity corresponding to the maxima is the most probable velocity which is given by the expression.
= 40000 x 50 x = 9 x 103 J = 9 kJ
A 4 : 1 molar mixture of He and CH4 gases is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. The ratio of number of moles of He to CH4 in the mixture effusing out initially will be :
Molar ratio of He and CH4 = 4 : 1
∴ partial pressures
∵ time of diffusion is same hence
20 ml of H2O2 after acidification with dil H2SO4 required 30 ml of N/12 KMnO4 for complete oxidation. The strength of H2O2 solution is [Molar mass of H2O2 = 34]
∴ Strength = N' x equivalent
3 mole of a mixture of FeSO4 and Fe2(SO4)3 required 100 ml of 2M KMnO4 solution in acidic medium. Hence mole fraction of FeSO4 in the mixture is
Arsenic estimation can be done by Bettendorff's process. The reaction is given below :
As4O6 + SnCl2 + HCl → As4 + SnCl4 + H2O
Find out the exact stoichiometric ratio of the reactants : (in the order as given in question)
How many moles of KMnO4 are needed to oxidised a mixture of 1 mole of each FeSO4 and FeC2O4 in acidic medium
Equivalents of KMnO4 = equivalent of FeSO4 + equivalent of FeC2O4
x = 4/5 mole
The correct figure representing isothermal and adiabatic expansions of an ideal gas from a particular initial state is:
pv = constant
pvγ = constant = k
Slope = dp / dv = - γ p / v
as γ > 1
Therefore, graph decreases more faster for adiabatic.
Stops of adiabatic curve is greater than isothermal curve.
An ideal gas with Cv = 3R expands adiabatically into a vacuum thus doubling its volume. The final temperatue is given by :
In free expansion (expansion into vacuum)
q = 0, w = 0 ⇒ ΔU = 0 , ΔT = 0
⇒ T2 = T1
1 mole of an ideal gas at 25°C is subjected to expansion reversibly and adiabatically to ten times of its initial volume. Calculate the change in entropy during expansion (in J k–1 mol–1)
Bond energy of N – H, H – H and N º N bonds are q1, q2 and q3. ΔH of N2 + 3H2 → 2NH3 is:
The heat capacities of A, B and C are in the ratio 1 : 2 : 3. The enthalpy change for the reaction, A + B → C, at temperature T1 is ΔH1. Assuming the heat capacities do not change with temperature, the enthalpy change ΔH2 at temperature T2 will be :
In a homogeneous gaseous reaction, 2.0 mole of ‘A’, 3.0 mole of ‘B’ and 2.0 mole of ‘C’ are placed in a 2.0 L flask and the equilibrium concentration of ‘C’ is 0.5 mole/L. The equilibrium constant (KC) for the reaction is :
(Note that 1 mole of C has reacted to form 1 mole of B and 0.5 mole of A.)
Iron forms a suphide with the approximate formula Fe7S8. Iron exist in both + 2 and + 3 oxidation states. The ratio of Fe(II) atoms to Fe (III) atoms is :
Which of the following solutions would have same pH?
What is the difference in pH for 1/3 and 2/3 stages of neutralisation of 0.1 M CH3COOH with 0.1 M NaOH?
pKa of acetic acid is 4.74, which implies that :
If the solubility of Ag2SO4 in 10–2 M Na2SO4 solution be 2 × 10–8 M then Ksp of Ag2SO4 will be
Which of the following will decrease with dilution at a given temperature?
On dilution of a basic or alkaline solution, concentration of [−OH] ions decrease.
pOH = −log[ −OH]
If [−OH] decrease then pOH increase as pOH = 14−pH or pH = 14−pOH
Thus, on increment of pOH, pH decrease.
Therefore on dilution pH of 10 − 3
M aniline solution decrease.
Option B is correct.
10–2 mole of NaOH was added to 10 litre of water. The pH will change by
The species which has its fifth ionisation potential equal to 340 eV is
For B atom only fifth I.P. = Z2 (13.6 eV) = 25 × 13.6 eV = 340 eV (hydrogen like species).
Consider the ground state of Cr atom (Z = 24). The numbers of electrons with the azimuthal quantum numbers, l = 1 and 2 are, respectively :
ℓ = 1 for p and ℓ = 2 for d.
Now 24Cr has configuration 1 s2 2s2 2p6 3s2 3p6 3d5 4s1
Hence there are 12, p-electrons and 5, d-electrons.
The De-broglie wavelength of a tennis ball of mass 66 g moving with the velocity of 10 metre per second is approximately:
The ratio of the “e/m” (specific charge) values of a electron and an α-particle is
The unit of equilibrium constant KC of a reaction is mol–2l2. For this reaction, the product concentration increases by-
Kc = (mole/litre)Δn
where Δn = no of moles on product side – no. of moles on reactant side
Hence Δn = – 2
so moles on reactant side > moles on product side so on increasing pressure reaction will get shifted in forward direction.
At temperature, T, a compound AB2(g) dissociates according to the reaction with a degree of dissociation, x, which is small compared with unity. Deduce the expression for KP, in terms of x and the total pressure, P.
If equilibrium pressure is 6 atm for the above reaction ; Kp will be :
For the chemical reaction,the amount of X3Y at equilibrium is affected by :
Increase in pressure will affect the amount of X3Y at equilibrium. Change in temperature always affects the equilibrium.
For an equilibrium which of the following statements is TRUE.
Volume of solid is more than volume of liquid water. Hence increase in pressure moves the equilibrium forward.
The volume occupied by 2.0 mole of N2 at 200 K and 8.21 atm pressure,
So Volume of 2 moles = 3.6 L.
If 10 gram of V2O5 is dissolved in acid and is reduced to V2+ by zinc metal, how many mole of I2 could be reduced by the resulting solution if it is further oxidised to VO2+ ions ?
[Assume no change in state of Zn2+ions] (V = 51, O = 16, I = 127) :
6e- + 10H+ + V2O5 → 2V2+ + 5 H2O
Zn → Zn2+ + 2e- 3 x 3
V2O5 + 3 Zn + H+ → 3 Zn2+ + 2V2+ + 5 H2O .........(1)
Now H2O + V2+ → VO2+ + 2H+ + 2e-
2e- + I2 → 2I-
V2+ + I2 + H2O → 2I- + VO2+ + 2H+
So we have 1 moles of V2O5 will reduce 2 moles of iodine
So moles of I2 will be reduced by given amount of V2O5 = 0.11 moles of I2
0.5 mole each of two ideal gas are taken in a container and expanded reversibley and adiabatically from V =1 litre to V = 4 litre startily from initial temperature T = 300 K. ΔH for the process (in cal/mol) is.
When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :
Initially solution will have low buffer capacity, will have maximum buffer capacity at half neutralisation and will be close to zero at equivalence point. After equivalence point the buffer capacity will keep on increasing with increase in concentration of base.
One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :
[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].
An electron in a hydrogen like atom makes transition from a state in which its de-Broglie wavelength is λ1 to a state where its de-Broglie wavelength is λ2 then wavelength of photon (λ) generated will be : where m is mass of the electron, c is speed of light in vaccum.
Find out the negative of logaritham of the solubility of solid Zn(OH)2 at 25°C, at pH = 6. Consider Zn(OH)2 makes saturated solution at 25°C.
Dissolved [Zn(OH)2] = [Zn+2]aq + [Zn(OH)+]aq + (Zn(OH)2)aq + [Zn(OH)3-] + [Zn(OH)4]2-
Now, Vn(OH)2]aq = 10-6 M in saturated solution
[Zn(OH)42-] = K5 [Zn(OH)3-] [OH-] = (10-2 M-1) [OH-]2
+ 10-6 + 10-3 x 10-6 + 10-18 = 10-1 + 10-5 + 10-6 + 10-11 = 10-1
The curve of pressure volume (PV) against pressure (P) of the gas at a particular temperature is as shown, according to the graph which of the following is / are incorrect (in the low pressure region):
Hydrogen and Helium show positive deviation whereas CO2, CH4 and O2 first show negative deviation, reach through minimum value and then show positive deviation as the pressure increases.
Hence, the option C is correct.
For 10 minute each, at 0 ºC, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of 4 litre capacity. The resulting pressure is 2.8 atm and the mixture contains 0.4 mole of nitrogen. What is the molar mass of unknown gas?
A 0.200 gm sample containing copper (II) was analysed iodometrically, copper (II) is reduced to
copper (I) by iodine. 2Cu2+ + 4I– ——? 2 CuI + I2
If 20.0mL of 0.10 M Na2S2O3 is required for titration of the liberated iodine then the percentage of copper in the sample will be (Cu = 63.5 g/mole)
2 moles of Cu2+ = 1 mole of I2
= 2 moles of hypo.
so moles of hypo used = 20 x 10-3 x 0.1 = 2 milli moles = milli moles of copper
Hence percentage of copper
An ideal gaseous sample at initial state i (P0 , V0,T0) is allowed to expand to volume 2V0 using two different process; in the first process the equation of process is PV2 = K1 and in second process the equation of the process is PV = K2. Then
Work done in isothermal process will be more than PV2 = const, process whatever be the value of K1 and K2 as is shown in the diagram.
Given that :
ΔGf° (CuO) = –30.4 kcal/mole
ΔGf° (Cu2O) = –34.98 kcal/mole T = 298 K
Now on the basis of above data which of the following predictions will be most appropriate under the standard conditions and reversible reaction.
ΔG°reaction = [2 x (-30.4)] – [–34.98] = – 25.82 kcal and – 25.82 x 103 = 2.303 x 2 x 298 log K
∴ K ≈ 1019, a very high value, hence reaction will be almost complete with a trace of Cu2O.
An acid HA (Ka = 10–5) reacts with NaOH at 298 K. What would be the value of the rate constant of the reverse reaction at the same temperature if the rate constant of the forward reaction is 10–11 mol–1 L sec–1 ?
Equiibrium constant of the backward reaction
∴ Equilibrium constant of the fonvard reaction
In a sample of H-atom electrons make transition from 5th excited state to ground state, producing all possible types of photons, then number of lines in infrared region are
maximum number of lines produced =
out of these 15 lines.
5 lines belong to ultra violet region and 4 lines are in visible region and rest are in infrared region.
When 100 ml of 0.1 M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with volume of HCl added will be :
Initially pH will decrease fast, then slowly due to buffer formation and then will decrease fast as buffer action diminishes.
Mark the WRONGLY matched-
A muscular sphincter (gastro-oesophageal) regulates the opening of oesophagus into the stomach. The opening of the stomach into the duodenum is guarded by the pyloric sphincter.
The outer pleural membrane of lungs are in close contact with the ______________ and the inner pleural membrane is in contact with the___________.
The parietal pleura is the outer membrane that attaches to and lines the inner surface of the thoracic cavity, covers the upper surface of the diaphragm and is reflected over structures within the middle of the thorax. It separates the pleural cavity from the mediastinum.
Which is CORRECT regarding basophils -
Histamine acts during inflammatory response. It is also an important neurotransmitter and involved in several physiological processes.
Heparin is an anticoagulant and prevents the blood from clotting.
Serotonin is also a compound secreted by basophils and has physiological functions like increasing vascular permeability, dilating capillaries etc.
Consider the following-
(a) In cortical nephron loop of henle is very long and runs deep into medulla
(b) A fall in GFR can activate the JG cells to release renin which can stimulate the Glomerular blood flow and bring GFR normal
When there is a fall in the glomerular filtration rate, it activates the juxtaglomerular cells to release renin. This stimulates the glomerular blood flow, thereby bringing the GFR back to normal. Renin brings the GFR back to normal by the activation of the renin-angiotensin mechanism.
Which is firmly attached to Z-line of myofibril?
At the centre of each I band is an elastic fibre called Z-line to which thin filaments are firmly attached. The thick filaments in the A band are held together by an M line. The portion of the myofibril between two successive Z lines is considered functional and is called a sarcomere.
Medulla have centre that control all except-
This section of the brain helps transfer messages to the spinal cord and the thalamus in the brain from the body and controls breathing, heart function, blood vessel function, digestion, sneezing, and swallowing.
Sensory and motor neurons from the forebrain and midbrain travel through the medulla.
The hypothalamus is a section of the brain responsible for hormone production. The hormones produced by this area of the brain govern body temperature, thirst, hunger, sleep, circadian rhythm, moods, sex drive, and the release of other hormones in the body.
Which is CORRECT regarding hormone?
Hormones are non nutrient chemicals, which act as intracellular messengers and are produced in trace amounts. Endocrine cells are present in different parts of the gastrointestinal tract, e.g., gastrin, secretin, GIP. Atrial wall of our heart secretes a peptide hormone called ANF (Atrial Natriuretic Factor), RH/IH are produced by hypothalamus. Adenohypophysis is not directly under neural control, it is under the control of hypothalamic hormones, brought by portal system.
Which of the following is NOT dense connective tissue?
Adipose tissue, or fat, is an anatomical term for loose connective tissue composed of adipocytes. Its main role is to store energy in the form of fat, although it also cushions and insulates the body.
Which of the following is part of succus entericus?
Succus entericus also called intestinal juice is a fluid that is secreted in small quantity in the small intestine. The secretions of the brush border cells of the mucosa alongwith the secretions of the goblet cells constitute this intestinal juice.
Which of the following is WRONGLY matched-
The partial pressure of oxygen in alveolar air is about 104 mm Hg.
Because PO₂ in the systemic capillaries is greater than the partial pressure in the body cells, oxygen diffuses from the blood and into the cells. Leaving the systemic capillaries, PO₂ = 40 - 50 mmHg.
PaO2 – Partial pressure of oxygen at sea level (160 mmHg in the atmosphere, 21% of standard atmospheric pressure of 760 mmHg) in arterial blood is between 75 mmHg and 100 mmHg.
The partial pressure of oxygen in tissues is low, about 40 mm Hg, because oxygen is continuously used for cellular respiration.
Hence, the correct answer is Option D.
Which condition is NOT Required for erythroblastosis fetalis to develop-
Mother Rh+ve because when mother has Rh +ve then erythroblastosis foetalis is not complete or second progeny is not die. And for erythroblastosis foetalis required condition is mother has Rh-ve and Father has Rh+ve.
Which of the following part involve in selective secretion of potassium and H+ ion to maintain PH and ionic balance?
DCT maintains the pH and ionic balance in the blood by reabsorption of Na+, water, HCO3 and selective secretion of H+, K+ and NH3. Collecting duct plays a role in the maintenance of pH and ionic balance of blood by reabsorption of water, and selective secretion of H+ and K+ ions.
At high level of calcium in muscle, all are true EXCEPT-
Calcium is present in high level in muscle only during contraction and during muscle contraction "size of A band " remains constant.
Ear ossicle attached to oval window of the cochlea is_______
The stapes articulates with the incus through the incudostapedial joint and is attached to the membrane of the fenestra ovalis, the elliptical or oval window or opening between the middle ear and the vestibule of the inner ear.
Consider the following-
(a) Pineal gland is located on the dorsal side of forebrain
(b) These hormone support the process of Red cell formation
How many CORRECT
The pineal gland, also known as the 'pineal body,' is a small endocrine gland. It is located on the back portion of the third cerebral ventricle of the brain, which is a fluid-filled space in the brain.
The hormones of the pineal gland indirectly support the formation of RBCs as it also contains sleep wake cycle, body temperature, pigmentation, menstrual cycle, metabolism and defence mechanism.
Hence, the correct answer is Option C
Maximum absorption of digested food occur in ________ part of Alimentary canal-
Fatty acids and glycerol being insoluble, cannot be absorbed into the blood. Absorption of substances takes place in different parts of the alimentary canal, like mouth, stomach, small intestine and large intestine. However, maximum absorption occurs in the small intestine.
Every 100 ml of __________blood delivers ________ of CO2
4ml of CO2 is delivered to the alveoli by every 100ml of deoxygenated blood.
Which is WRONGLY matched-
Eosinophils are 2 - 3 percent and resist infections and associated with allergic reactions.
Which is WRONG about hormonal control of kidney –
The correct answer is Option C i.e., "ANF can cause vasoconstriction" because release of ANF causes blood vessels dilation (vasodilation)
Fall in GFR releases renin by activating JG cells.
Angiotensin ii releases aldosterone which increases water and sodium ion absorption.
ADH acts on DCT and CD and increases water reabsorption.
Which is correct sequence during muscle contraction -
(a) ATP hydrolysis
(b) decrease in size of sarcomere
(c) release of ADP and Pi
(d) cross bridge broken between actin and myosin
Correct sequence is –
Muscle contraction occurs according to the sliding filanent theory. The active site of actin is masked by troponin. In myosin, the energy for binding of actin is from ATP hydrolysis. Myosin contains ATPase enzyme in its head and is responsible for ATP hydrolysis.
When the signal reaches motor end plate, acetylcholine is released. It binds with the membrane to produce depolarization.
It send neural messages and hence SPR releases Ca2+. This Ca2+ bind with TnC and the masking of actin is removed. At the same time, ATP molecules bind with the head of meromyosin. Therefore the head of meromyosin get energised.
Actin and myosin binds together (cross arm activity) then a fresh ATP is formed and the cross bridge breaks i.e.,ATP hydrolysis lead to the formation of cross bridge.
As a result size of sacromere decreases, then the ADP and iP are released and cross bridge is broken.
Consider the following matched about number of bones-
(1) Thoracic vertebrae - 12
(2) Floating ribs - 11th and 12th
(3) Ankle bones - 8
(4) Phalanges - 15
How many are correctly matched?
Thoracic consists of 12 bones and the bones of last pair (11th and 12th ) are called floating ribs or false bones so 1 and 2 options are right.
options 3 and 4 are wrong because phalanges are 14 and ankle bones are 7.
Consider the following matched-
How many are correctly matched?
The type of joint present between the humerus bone and the pectoral girdle is called the shoulder joint. It is a type of gliding joint.
The carpometacarpal joints of the fingers are synovial plane joints that serve as the articulation between the carpals and the metacarpals and allow the bases of the metacarpal bones to articulate with one another.
The atlantoaxial joint is a joint in the upper part of the neck between the first and second cervical vertebrae; the atlas and axis. It is a pivot joint. The atlantoaxial joint is of a complicated nature.
Gliding joint is the type of joint between the carpal bones.
A suture is a type of fibrous joint that is only found in the skull (cranial suture). The bones are bound together by Sharpey's fibres. A tiny amount of movement is permitted at sutures, which contributes to the compliance and elasticity of the skull. These joints are synarthroses.
Hence, only two out of five statements are correct. The correct option would be Option B.
Which is probably NOT a reason for indigestion?
Lack of digestive enzymes can result in acidity, bloating, gas, constipation and numerous other indigestion problems.
Food-poisoning symptoms—abdominal pain, stomach cramps, and multiple bouts of diarrhea or vomiting—tend to be more severe but shorter-lasting than when it's a stomach bug. If your stomach just feels upset or you have heartburn, bloating, or gas but no vomiting or diarrhea, it's probably indigestion, not an infection.
Enduring periods of stress and anxiety can cause the digestive system to work ineffectively. Acid attack. Stress increases the amount of acid in the stomach, which leads to indigestion and heart burn.
Hence, the correct answer would be Option C.
Activities of intestinal secretion are in control of all except
The anterior pituitary gland produces the following hormones and releases them into the bloodstream: adrenocorticotropic hormone, which stimulates the adrenal glands to secrete steroid hormones, principally cortisol. growth hormone, which regulates growth, metabolism and body composition.
Value of functional residual capacity is -
Functional residual capacity(FRC): volume of the air that will remain in the lungs after a normal expiration this includes expiratory reserve volume ( ERV) + residual volume(RV). ERV = 1000 ml to 1100 ml.
RV = 1100 ml to 1200ml.==> 2100 to 2300.
Which of the following is NOT conducting part in respiratory system?
Bronchioles end in tiny air sacs called alveoli, where the exchange of oxygen and carbon dioxide actually takes place. Each lung houses about 300-400 million alveoli. The lungs also contain elastic tissues that allow them to inflate and deflate without losing shape and are encased by a thin lining called the pleura.
Consider the following statement-
(1) Contraction of Diaphragm increase volume at anterior posterior Axis
(2) maximum gaseous exchange occur in lungs when air which enters comes at body temperature
How many CORRECT?
1) This contraction occurs during the process of inspiration the muscles involved are intercostal muscles and diaphragm muscles.
2) Gas exchange takes place in the millions of alveoli in the lungs and the capillaries that envelop them. As shown below, inhaled oxygen moves from the alveoli to the blood in the capillaries, and carbon dioxide moves from the blood in the capillaries to the air in the alveoli.
Consider the following:
(a) Both SA node and AV node located in right atrium
(b) Number of action potential that could be generated in a minute vary at different parts of nodal system
How many CORRECT?
SA Node: The Sino-Atrial Node (SA Node) is present in the right upper corner of the right atrium.
AV Node: The atrio-ventricular node is present in the lower left corner of the right atrium; close to the atrio-ventricular septum.
The number of action potentials which could be generated per minute varies at different parts of the nodal system. The SA Node can generate the maximum number of action potentials, i.e. 70-75 per minute. The SA Node is responsible for initiating and maintaining the rhythmic contractile activity or beating of the heart. Due to this, the SA Node is called the pacemaker. The human heart normally beats 70-75 times per minute.
Which valve can NOT found open at ventricular diastole?
In this cycle, blood passes through the tricuspid valve first, then the pulmonary valve, mitral valve, and finally the aortic valve. During the diastole phase of the cardiac cycle, the atrioventricular valves are open and semilunar valves closed.
Low pressure in atria during ventricular diastole means –
Diastole: The myocardium is relaxed. The atria and ventricles fill passively. AV valves allow blood to pass from the atria to the ventricles. The aortic and pulmonary artery semilunar valves are closed because the blood in those vessels is at a higher pressure than the ventricles.
An excitatory stimulation is also known as a (an) _________ stimulation
A depolarizing prepulse (DPP) is an electrical stimulus that causes the potential difference measured across a neuronal membrane to become more positive or less negative, and precedes another electrical stimulus.
Myelinated nerve fibre present in –
Cranial nerves are the nerves that emerge directly from the brain (including the brainstem), in contrast to spinal nerves (which emerge from segments of the spinal cord). 10 of the cranial nerves originate in the brainstem.
Nissl’s granules are absent in-
Nissl's granules are the granular structure which is present in the neurons. These are the granules of rough endoplasmic reticulum and are the site of protein synthesis. It is localized in the cyton and dendron. It is visualized by Nissl staining method. The proteins produced by it is used for inter-cellular mechanisms.
Mark the WRONG statement –
There is a small gap between the axon terminal of the presynaptic neuron and the membrane of the postsynaptic cell, and this gap is called the synaptic cleft. Across the synaptic cleft, there is the post-synaptic cell surface covered in receptors (ligand-gated ion channels) for the neurotransmitter.
Adrenaline show affect all except-
Just as a cup of coffee can give you a temporary boost of energy, adrenaline increases alertness and awareness and can improve memory and cognitive functioning.
In stressful situations, cold sweats are usually mediated by adrenaline. This is because the adrenaline causes the blood vessels to narrow and a few sweat glands to become active – producing a drop in skin temperature and a cold sweat.
Epinephrine (EP), also known as adrenaline, is an important hormone and neurotransmitter that plays a role in the fight/flight responses, regulating blood sugar, and pupil dilation.
Fright causes the brain to send signals to the renal glands which start pumping large amounts of adrenalin into the bloodstream. This increases the heart and breathing rate in preparation for the ensuing action.
Hence, the correct answer is Option D.
Pace-maker is implanted when one of these is defective
Impulse for contraction is generated in the SA node hence if this node is defective then the pace maker is implanted in position of SA node.
Pulse pressure is difference of diastole and systole and is equal to ______ mm Hg.
Pulse pressure is the difference between the systolic and diastolic blood pressure. It is measured in millimeters of mercury (mmHg). It represents the force that the heart generates each time it contracts. For example, if resting blood pressure is 120/80 mmHg, then the pulse pressure is 40 mmHg.
Which of the following is hormone of pars nervosa-
Neurohypophysis is known also as the pars nervosa.Pars nervosa doesn't synthesise any hormone. Its function is to store and release Vasopressin (Anti diuretic Hormone) and oxytocin into blood.
Mark the INCORRECT –
Erythropoietin hormone is not steroid hormone it is produced mainly in kidney ,which maintenans the production of red blood cells.
Consider the following-
(a) Androgen can act on CNS
(b) Glucagon mainly act on liver cells
How many CORRECT?
The mechanisms of action of androgens on the CNS are extremely complex and multi-faceted. Much data has been accumulated in recent years that suggest a beneficial role of androgens in several aspects of CNS function. These include positive effects on mood, cognition, memory, and libido.
Glucagon is an extremely potent hormone released by drops in blood glucose. Glucagon acts on the liver to elevate plasma glucose, an action opposite to that of insulin. Glucagon promotes hepatic glycogenolysis and increases hepatic gluconeogenesis.
Mark the INCORRECT statement-
The most important physiological role of glucagon is to promote hyperglycemia in response to a hypoglycemia. Glucagon acts only on liver glycogen, unlike epinephrine, which acts on both liver and muscle glycogen. Like most hormones,glucagon is first bound to surface receptors on a cell, in this case, the hepatocyte.
Which of the following is NOT role of glucocorticoids-
Cortisol and epinephrine facilitate the movement of immune cells from the bloodstream and storage organs, such as the spleen, into tissue where they are needed to defend against infection. Glucocorticoids do more than help the body respond to stress. They also help the body respond to environmental change.
Consider the following statement-
(A) Collagen is a part of dense regular connective tissue but not dense irregular connective tissue.
(B) Osteocytes are located in spaces called lacunae.
Mark the CORRECT statement-
Between the rings of matrix, the bone cells (osteocytes) are located in spaces called lacunae. Small channels (canaliculi) radiate from the lacunae to the osteonic (haversian) canal to provide passageways through the hard matrix.
Dense regular connective tissue is mainly made up of type I collagen fibers. It is found in areas of the body where large amounts of tensile strength are required, like in ligaments, tendons and aponeurosis. The collagen fibers are densely packed together and arranged in parallel to each other. Dense irregular connective tissue consists of mostly collagen fibers. It has less ground substance than loose connective tissue.
Fibroblasts are the predominant cell type, scattered sparsely across the tissue.
Hence, the correct option is Option B.
Type of epithelium act as a diffusion boundary-
Squamous epithelium: This type of epithelium is often permeable and occurs where small molecules need to pass quickly through membranes via filtration or diffusion.
Which is absent in female cockroach but present in male cockroach-
Gonapophysis is an organ of the anal region of an insect that serves in copulation, oviposition, or stinging. Some chitinous asymmetrical structures are found surrounding the male gonopore at the end of the abdomen.
In cockroach Grinding and incising region present in _________ mouth part.
The mandible has grinding and incision regions present on it, thus helping in breaking the food into smaller particles.
Spiracles present in cockroach is-
In cockroach, the respiratory system has a network of the trachea. They open through 10 pairs of spiracles that are present on the lateral side of the body. Thin tubes carry oxygen from the air to all the parts of the body. The spiracles are regulated by the sphincters.
Consider the following statement-
(A) Smooth muscles lack striation and are fusiform
(B) Cardiac cell function together as cell junctions fuse the plasma membrane
Mark the CORRECT statement.
Smooth muscle fibers are long, spindle-shaped (fusiform) cells. Note the single and centrally placed nucleus in each smooth muscle cell. The absence of striation is also characteristic of the smooth muscle cells.
Which is NOT true about myasthenia gravis?
Myasthenia gravis is a neuromuscular disorder that is usually caused by an autoimmune problem. In this condition, antibodies attack the neuromuscular junctions of own body mistakenly. damage to neuromuscular junctions reduces communication b/w nerve cells & muscles. this results in muscle weakness also causes droopy eyelids and mouth that may lead to Paralysis, but degeneration of muscle does not occurs.
Which is INCORRECT?
Macrophage show ameoboied movement while platelets does not show ameoboied movement.
Fascicles lined by common
The Three Connective Tissue Layers. Bundles of muscle fibers, called fascicles, are covered by the perimysium. Muscle fibers are covered by the endomysium. In other places, the mysia may fuse with a broad, tendon-like sheet called an aponeurosis, or to fascia, the connective tissue between skin and bones.
Consider the following-
(a) Striation is due to distribution pattern of actin and myosin
(b) Each dark band have both actin and myosin
How many CORRECT?
a) The myofibril appears striated as they are composed of actin and myosin filaments, repeated in units called sarcomere, which is the basic functional unit of the muscle fiber. The sarcomere is responsible for the striated appearance of muscle.The actin is thin filament and myosin is thick filaments.
A muscle fascicle is a bundle of skeletal muscle fibers surrounded by perimysium.
Troponin is a globular protein complex involved in muscle contraction.It occurs with tropomyosin in the thin filament of muscle fiber.
Meromyosin are subunits of myosin, the thick filament.
b) The ‘A’ and ‘I’ bands are arranged alternately throughout the length of the myofibrils. The light bands contain actin(thin filament) and are called I-band or Isotropic band, whereas the dark band called ‘A’ or Anisotropic band contains myosin(thick filament).
In a muscle fibre five dark band and no light band visible this implies-
As the myofibrils contract the muscle cell contracts.The arrangement of the thick myosin filaments across the myofibrils and the cell causes them to refract light and produce a dark band known as the A Band. In between the A bands is a light area where there are no thick myofilaments, only thin actin filaments.
Mucosa have all except
Mucosa is made up of connective tissue and not mesothelium & it has goblet cells,villi and irregular folds rugae in stomach. Serosa is made up of mesothelium.
Consider the following-
(a) Both stomach and liver situated in abdominal cavity
(b) Glisson’s capsule is connective tissue cover hepatic lobule
How many CORRECT?
ECG is graphical representation of -
Graphic record of the electrocardiography is called electrocardiogram (ECG, EKG). It is based on recording of electric potential generated by heart on body surface.
Consider the following-
(a) Cardiac output depend upon stroke volume
(b) SA node generate action potential after closure of semilunar valve
How many CORRECT?
The cardiac output (CO) is dependent on the stroke volume (SV) and the heart rate (HR): CO (L/min) = SV (ml/beat)Ⅹ HR (beats/min) = 4.9 L/min (normally).
Pacemaker of heart is Sinoatrial node located in right atrium. It generates impulses which lead to contraction of atria.
Consider the statements regarding the angiosperm.
A. Pollen exine is highly thickened due to sporopollenin.
B. Three celled pollen liberation shows vegetative cell, generative cell and male gamete nucleus.
C. Cleistogamous flowers are seen in Commelina, Viola and Oxalis plants.
D In maize and bamboo, insects are acting as the pollinating agent.
3-celled stage of pollen liberation takes place by vegetative (tube) cell and 2 male gametic cells (nuclei).
Consider the following-
(a) Neural signals through the sympathetic nerves can increase cardiac output
(b) Parasympathetic neural signal decrease heart beat
How many CORRECT?
Normally the parasympathetic nervous system is active keeping heart rate & stock volume at normal level. Exercises stimulate the sympathetic nervous system which increases the heart rate & makes the heart contract with a greater force, which increase stock volume & then the Cardiac output increases Parasympathetic activity increase after a crisis has passed. This reduces the heart rate & stock volume from their high level , bringing cardiac output back to normal. Hence, Option B is correct.
ANS have no impact on -
The action potential is conducted along the axon membrane by contiguous conduction and by salitatory conduction. This will begin to depolarise this area of the membrane until it reaches threshold potential. Once threshold is reached, the action potential is fired and this is the area of the AP.
The medullary gradient is mainly caused by-
This osmotic gradient is formed by the accumulation of solutes, primarily NaCl and urea, in the cells, interstitium, tubules, and vessels of the medulla.
Nacl is transported by the ascending limb of Henle’s loop which is exchanged with-
The vasa recta capillaries are long, hairpin-shaped blood vessels that run parallel to the loops of Henle. The hairpin turns slow the rate of blood flow, which helps maintain the osmotic gradient required for water reabsorption. The Henle’s loop and vasa recta play a significant role in this.
The flow of filtrate in the two limbs of Henle’s loop is in opposite directions and thus forms a counter current.
The flow of blood through the two limbs of vasa recta is also in a countercurrent pattern.
The proximity between the Henle’s loop and vasa recta, as well as the countercurrent in them, help in maintaining an increasing osmolarity towards the inner medullary interstitium, i.e., from 300 mOsmolL–1 in the cortex to about 1200 mOsmolL–1 in the inner medulla. This gradient is mainly caused by NaCl and urea.
NaCl is transported by the ascending limb of Henle’s loop which is exchanged with the descending limb of vasa recta.
NaCl is returned to the interstitium by the ascending portion of vasa recta. Similarly, small amounts of urea enter the thin segment of the ascending limb of Henle’s loop which is transported back to the interstitium by the collecting tubule.
The above-described transport of substances facilitated by the special arrangement of Henle’s loop and vasa recta is called the countercurrent mechanism. This mechanism helps to maintain a concentration gradient.
The reabsorption of water in the kidneys is under the control of a hormones –
Antidiuretic hormone binds to receptors on cells in the collecting ducts of the kidney and promotes reabsorption of water back into the circulation. In the absense of antidiuretic hormone, the collecting ducts are virtually impermiable to water, and it flows out as urine.
Which of the following factor can activate the JG cells to release renin?
JG cells which releases renin to maintain GFR, actually GFR is regulated by a mechanism that is juxta glomerular apparatus which has JG cells. Fall in any of this that is GBP, GBF, and GFR directly affects GFR that is the filterate being formed and thus release of renin works on it and GFR comes back to be normal.
The GFR/day in a healthy adult is-
GFR is 125 ml/min.
So in a day there are 24 * 60 = 1440 minutes.
Therefore in 1440 min, GFR will be 125 * 1440 = 180,000 ml or 180 litres.
Inner to hilum of kidney is a broad funnel shaped structure called-
Inner Structure: Inner to the hilum is a broad funnel shaped space called the renal pelvis with projections called calyces. The outer layer of kidney is a tough capsule. Inside the kidney, there are two zones, an outer cortex and an inner medulla.
Which of the following statement about proximal convulated tubule( PCT) is FALSE?
The proximal convoluted tubule (PCT) is lined with cuboidal epithelium with brush border. It increases the surface area for reabsorption. Almost all essential nutrients and 70-80 % of electrolytes and water are reabsorbed by PCT. The PCT is responsible for selective reabsorption of glucose, amino acids, NaCl and H2O.
Hence, the correct answer is Option C.
If loop of Henle were absent in mammalian nephrons, which of the following is to be expected?
The main function of the Henle's loop is to absorb water from the tubular lumen, thus making the urine concentrated. If loop of Henle becomes absent then the urine becomes more dilute.
Which does NOT occur in DCT-