NEET Part Test - 2
180 Questions MCQ Test | NEET Part Test - 2
| 1 Crore+ students have signed up on EduRev. Have you? |
The time taken by a particle performing SHM on a straight line to pass from point A to B where it's velocities are same is 2 seconds. After another 2 seconds it returns to B. The time period of oscillation is (in seconds):
From the given information it can be concluded that points A and B are equidistant from mean position.Hence from diagram it is clear that time period of oscillation is= 2 + 2 x 2 + 2 = 8 second.
For a particle in S.H.M., if the amplitude of displacement is ‘a’ and the amplitude of velocity is ‘v’ the amplitude of acceleration is :
Maximum velocity v = ω a
Maximum acceleeration f = ω2 a ⇒ 
A body performs SHM along the straight line segment ABCDE with C as the mid point of segment AE (A and E are the extreme position for the SHM). Its kinetic energies at B and D are each one fourth of its maximum value. If length of segment AE is 2R, then the distance between B and D is :
When the particle crosses D, its speed is half the maximum speed.
or 
or 
A system is shown in the figure. The time period for small oscillations of the two blocks will be :
Both the spring are in series
∴ 
Time period 
Method II
∴ mx1 = mx2 ⇒ x1 = x2
force equation for first block ;
Put x1 = x2 ⇒ 
∴ 
Two light strings, each of length l, are fixed at points A and B on a fixed horizontal rod xy. A small bob is tied by both strings and in equilibrium, the strings are making angle 45° with the rod. If the bob is slightly displaced normal to the plane of the strings and released then period of the resulting small oscillation will be :
Resulting torque on the bob = 
MI of bob about axis 
For small angle θ
A metre stick swinging in vertical plane about a fixed horizontal axis passing through its one end undergoes small oscillation of frequency f0 as shown in figure. If the bottom half of the stick were cut off, then its new frequency of small oscillation would become:
where, ℓ is distance between point of suspension and centre of mass of the body. Thus, for the stick of length L and mass m :
when bottom half of the stick is cut off
A 25 kg uniform solid sphere with a 20 cm radius is suspended by a vertical wire such that the point of suspension is vertically above the centre of the sphere. A torque of 0.10 N-m is required to rotate the sphere through an angle of 1.0 rad and then maintain the orientation. If the sphere is then released, its time period of the oscillation will be:
τ = -kθ
0.1 = -k(1.0), where k is torsional constant of the wire.
k = 1/10
= 4π second
A rod of length l is in motion such that its ends A and B are moving along x-axis and y-axis respectively. It is given that 
rad/s always. P is a fixed point on the rod as shown in figure. Let M be the projection of P on x-axis. For the time interval in which q changes from 0 to π/2 ,
choose the correct statement :
∴ θ = 2t
Let BP = a ∴ x = OM = a sin θ = a sin (2t)
Hence M executes SHM with in the given time period and its acceleration is opposite to 'x' that means towards left
Which of the following is greatest in SHM ? (assuming potential energy = 0 at mean position)
Average kinetic energy with respect to position 
Average potential energy with respect to position 
Average kinetic energy with respect to time 
Average potential energy with respect to time 
m1 & m2 are connected with a light inextensible string with m1 lying on smooth table and m2 hanging as shown in figure. m1 is also connected to a light spring which is initially unstretched and the system is released from rest :
After the system is released, m2 moves down.
The extention in the spring becomes :
which is the new equilibrium position of the system.
For small 'x' : restoring force on the system is F = kx
⇒ 
(For (m1 + m2 + spring) systeem)
⇒ 
⇒ Angular frequency 
F.B.D. of m1 and m2 just after the system is released :
m1 moves towards right till the total kinetic energy acquired does not converted to potential energy.Hence (D) is also incorrect.Hence (B) is the answer
The potential energy of a particle executing SHM changes from maximum to minimum in 5 sec. Then the time period of SHM is :
P.E. is maximum at extreme position and minimum at mean position.
Time to go from extreme position to mean position is, t = T/4 ; where T is time period of SHM
5 s = T/4 ⇒ T = 20 s.
A body is executing simple harmonic motion. At a displacement x from mean position, its potential energy is E1 and at a displacement y from mean position, its potential energy is E2. The potential energy E at a displacement (x + y) from mean position is (the potential energy is zero at mean position) :
Therefore 
In the figure shown a block of mass m is attached at ends of two springs. The other ends of the spring are fixed. The mass m is released in the vertical plane when the spring are relaxed. The velocity of the block is maximum when:
Speed of block is maximum at mean position. At mean position upper string is extended and lower spring is compressed.
Figure shows the kinetic energy K of a simple pendulum versus its angle θ from the vertical. The pendulum bob has mass 0.2 kg. The length of the pendulum is equal to (g = 10 m/s2) :
The equation of a wave is given by (all quantity expressed in S.I. units) Y = 5 sin10π (t – 0.01x) along the x-axis. The magnitude of phase difference between the points separated by a distance of 10 m along x- axis is :
The magnitude of phase difference between the points separated by distance 10 metres.
= k x 10 = [10π x 0.01] x 10
= π
A certain transverse sinusoidal wave of wavelength 20 cm is moving in the positive x direction. The transverse velocity of the particle at x = 0 as a function of time is shown. The amplitude of the motion is:
⇒ 
Two vibrating strings of same material stretched under same tension and vibrating with same frequency in the same overtone have radii 2r and r. Then the ratio of their lengths is :
and 
v1 = v2
A chord attached about an end to a vibrating fork divides it into 6 loops, when its tension is 36 N. The tension at which it will vibrate in 4 loops is:
For waves along a string
⇒ 
Now, for 6 loops : 3λ1 = L ⇒ λ1 = L/3
& for 4 loops : 2λ2 = L ⇒ λ2 = L/2
⇒ 
⇒ 
Sinusoidal waves 5.00 cm in amplitude are to be transmitted along a string having a linear mass density equal to 4.00 × 10–2 kg/m. If the source can deliver a maximum power of 90 W and the string is under a tension of 100 N, then the highest frequency at which the source can operate is (take π2 = 10) :
⇒ 
⇒ 
using data f = 30 Hz.
What is the percentage change in the tension necessary in a sonometer of fixed length to produce a note one octave lower (half of original frequency) than before :`
In Sonometer
(Wave length constant for fixed length)
∴ 
⇒ 
∴ %
Figure shown is a graph, at a certain time t, of the displacement function S(x,t) of three sound waves 1,2 and 3 as marked on the curves that travel along x–axis through air. If P1,P2 and P3 represent their pressure amplitudes respectively, then correct relation between them is :
Under similar conditions of temperature and pressure, In which of the following gases the velocity of sound will be largest :
The speed of sound in air is 
of H2 is maximum, hence speed of sound in H2 shall be maximum.
In a sound wave, to increase the intensity by a factor of 10, pressure amplitude must be changed by a factor of :
⇒ 
⇒ 
The two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length LA = 1.5 m and one open end, contains a small sound source that sets up the standing wave with the second lowest resonant frequency of that pipe. Sound from pipe A sets up resonance in pipe B, which has both ends open. The resonance is at the second lowest resonant frequency of pipe B. The length of the pipe B is :
For pipe A, second resonant frequency is third harmonic thus 
For pipe B, second resonant frequency is second harmonic thus 
⇒ 
A source of sound of frequency 256 Hz is moving rapidly towards a wall with a velocity of 5 m/sec. If sound travels at a speed of 330 m/sec, then number of beats per second heard by an observer between the wall and the source is:
For a stationary observer between wall and source, frequency from direct source 
frequency from reflected sound 
So no beats will be heard.
A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:
⇒ 
A source of frequency f is stationary and an observer starts moving towards it at t = 0 with constant small acceleration. Then the variation of observed frequency f ' registered by the observer with time is best represented as :
After a time t, velocity of observer V0 = at
∴ 
Which is a straight line graph of positive slope.
Two persons A and B each carrying a source of frequency 596 Hz and 600 Hz respectively are standing at rest a few metres apart. A starts moving towards B with a velocity of 2 m/s. If the speed of sound is 300 m/s. Which of the following statement is true?
Beats heard per second by A is 8 and B is zero.
In a horizontal spring–mass system, mass m is released after being displaced towards right by some distance at t = 0 on a frictionless surface. The phase angle of the motion in radian when it is first time passing through the equilibrium position is equal to :
Using x = A sin (ωt + φ) and V = Aω cos(ωt + φ) for conditions at t = 0 → x = A and V = 0 then φ = π/2
When it passes equilibrium position for the first time t = T/4
A 75 cm string fixed at both ends produces resonant frequencies 384 Hz and 288 Hz without there being any other resonant frequency between these two. Wave speed for the string is :
Δv = 384 - 288 = 96
Thus 288 and 384 (96 x 3; 96 x 4) are third and fourth harmonics.
For fundamental mode:
λ/2 = 0.75 λ = 1.5 m
η = 96
⇒ v = 96 x 1.5 = 144 m/s.
A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude :
A horizontal rod of mass m and length L is pivoted smoothly at one end. The rod ’s other end is supported by a spring of force constant k. The rod is rotated (in vertical plane) by a small angle q from its horizontal equilibrium position and released. The angular frequency of the
subsequent simple harmonic motion is :
Restoring rorque :
τ = ky L = KL2θ (Since y ≌ Lθ from figure)
⇒ 
⇒ 
(torque due to mg was already balanced so it is not taken in calculation)
A traveling wave y = A sin (kx - ωt + θ) passes from a heavier string to a lighter string. The reflected wave has amplitude 0.5 A. The junction of the strings is at x = 0. The equation of the reflected wave is:
As wave has been reflected from a rarer medium,therefore there is no change in phase.Hence equation for the opposite direction can be written as:
y = 0.5A sin (-kx - ωt + θ) = -0.5A sin (kx + ωt - θ)
A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:
λ = 2ℓ = 3m
Equation of standing wave y = 2A sin kx cos ωt y = A as amplitude is 2A.
A = 2A sin kx
⇒ 
and 
⇒ x2 = 1.25 m ⇒ x2 - x1 = 1m
The average density of Earth’s crust 10 km beneath the surface is 2.7 gm/cm3. The speed of longitudnal seismic waves at that depth is 5.4 km/s. The bulk modulus of Earth’s crust considering its behavior as fluid at that depth, is :
⇒ B = V2ρ
= (5.40 x 103 m/s)2 (2.7 x 103) = 7.9 x 1010 Pa.
The second overtone of an open pipe A and a closed pipe B have the same frequencies at a given temperature. Both pipes contain air. The ratio of fundamental frequency of A to the fundamental frequency of B is:
Second overtone of open pipe 
second overtone of closed pipe 
Since, ratio of frequency are same ∴ 
Now, the ratio of fundamental frequencies : 
A thin uniform rod is suspended in vertical plane as a physical pendulum about point A. The time period of oscillation is T0. Not counting the point A, the number 'n' of other points of suspension on rod such that the time period of oscillation (in vertical plane) is again T0. Then the value of n is : (Since the rod is thin, consider one point for each transverse cross section of rod)
When the point of suspension is at a distance x from centre of length of rod, the time period of oscillation is 
where I is length of the rod.
The time period of oscillation will be same (T0) if the point of suspension is a distance x = λ/2 or x = λ/6 from centre of the rod. Thus there will be three additional points.
Two radio station that are 250m apart emit radio waves of wavelength 100m. Point A is 400m from both station. Point B is 450m from both station. Point C is 400m from one station and 450 m from the other. The radio station emit radio waves in phase. Which of the following statement is true ?
At points A and B, path difference between the waves coming from two radio stations is zero. Hence there will be constructive interference at A and B,
For point C, path difference between the waves is 50 metre i.e. λ/2 so
destructive interference takes places at point C.
A particle performs SHM with a time period T and amplitude 'a'. The magnitude of average velocity of the particle over the time interval during which it travels a distance a/2 from the extreme position is :
The magnitude of displacement in the given time interval = a/2
Time taken by the particle to cover a distance a/2 starting from rest = T/6
Hence the magnitude of average velocity over given time interval is 
A simple pendulum 50 cm long is suspended from the roof of a cart accelerating in the horizontal direction with constant acceleration √3 g m/s2. The period of small oscillations of the pendulum about its equilibrium position is (g = π2 m/s2) :
With respect to the cart, equilibrium position of the pendulum is shown. If displaced by small angle θ from this position, then it will execute SHM about this equilibrium position, time period of which is given by :
; 
⇒ geff = 2g ⇒ T = 1.0 second
A particle is subjected to two simple harmonic motions along x and y directions according to, x = 3 sin 100 πt; y = 4 sin 100 πt :
x = 3 sin 100 πt
y = 4 sin 100 πt
Equation of path is
which is equation of a straight line having slope = 4/3
Equation of resulting motion is 
Amplitude is 
When a wave pulse traveling in a string is reflected from a rigid wall to which string is tied as shown in figure. For this situation two statements are given below :
(1) The reflected pulse will be in same orientation of incident pulse due to a phase change of p radians
(2) During reflection the wall exert a force on string in upward direction
For the above given two statements choose the correct option given below :
Reflected pulse will be inverted as it is reflected by a denser medium. The wall exerts force in downward direction.
The equation of displacement due to a sound wave is s = s0 sin2 (ωt - kx). If the bulk modulus of the medium is B, then the equation of pressure variation due to that sound is :
The equation of pressure variation due to sound is
A point source of power 50π watts is producing sound waves of frequency 1875Hz. The velocity of sound is 330m/s, atmospheric pressure is 1.0 x 105 Nm-2, density of air is 1.0 kgm-3. Then pressure amplitude at r = 
m from the point source is (using π = 22/7) :
where P, P0 , V are power, pressure amplitude and velocity respectively.
⇒ 
An organ pipe of length L is open at one end and closed at other end. The wavelengths of the three lowest resonating frequencies that can be produced by this pipe are :
when attached to a ring, it is named as carboxamide.
has higher priority than – C ≡ N
(3) – C ≡ N has perfix cyano
are homologue
Conditions for geometrical isomerism is not fulfilled in C.
I & II have change in position of π-bond.
II & III have different geometrical orientations.
I & III also have change in position of π-bond.
'm' monochloro structural isomeric products
The value of m and n are respectively :
'm' no. of products 
'n' no. of products 
reacts with carbonyls (Aldehyde/ketones) to give yellow ppt.
A mixture of benzaldehyde and formaldehyde is heated with 2,4 DNP. How many fractions of organic products will obtained after fractional distillation of product mixture.
Benzaldehyde is unsymmetrical, gives 2 product but formaldehyde is symmetrical ,gives only one product.
Compare the physical properties (μ, b.p, m.p, &stability) in the geometrical isomers CH3 - CH = CH - CNChoose from the following options:
In trans isomer there is linear addition of bond dipoles hence μ isgreater and its B.P will be more. Packing in trans is better hence m.p is more.
An organic compound X(C5H10) on ozonolysis gives Y & Z. The product mixture Y and Z on reaction with NH2 – OH gives four oximes. The structure of X is
-I group increases acidic strength and I effects depend upon distance.
Which of the following structures are superimposable (identical) :
Both in (I) & (III) configuration at lower chiral carbon is 'R' and upper chiral carbon is 'S'.
If optical rotation is produced by 
is + 36o then that is produced by 
Rotation of the compound in question is zero because it is a meso compound.
(single stereoisomers)
(Stable due to intramolecular H bonding)
Electron donating ability of group order :
-NH2 > -N(CH3)2 > - NH - CHO > -NH2
A Natural Drug was having [a] = 40.3º in polarimeter experiment in clockwise direction what can be the structure of this drug.
is optically active compound so it rotate the plane of ppl.
Lucas reagent can not be used to distinguished Acetaldehyde and acetylene.
Negative charge on more electronegative atom is more stable.
In all other cases positive charge is not in conjugation.
Which of the following pairs of structures do not represent resonating structures ?
Structures given in option(A) are tautomers.
Catalytic hydrogenation of which of the following, gives a product that gives two monophotochlorination products (considering structural isomers only)?
No of hyperconjugation structures in I ,III ,IV and II are respectively 3,6,6 and 6.Trans is more stable than cis isomer.
Which of the following can not give H2 gas with sodium metal ?
Ketones do not give H2 gas with sodium metal.
Select the structure with correct numbering for IUPAC name of the compound.
OH is senior than SH.
Find the acid strength order:
(I) O2N - CH2 - COOH
(II) F - CH2 - COOH
(III) H3CO - CH2 - COOH
(IV) CH3 - CH2 - COOH
(Acidity ∝ -I strength of groups attached)
-I strength is in order -NO2 > -F> -OCH3 & -CH2CH3 is +I group
-NO2 at para position increases acidity at the maximum extent due to its strong -m and -I effect.
Consider the set of three compounds, which is true about basicity of these compounds in water ?
(1) MeNH2
(2) Me2NH and
(3) Me3N
In RNH2, R2NH, R3N ; the order of base strength is 2° > 1° > 3° when R = Me and the order is 2° > 3° > 1° when R = Et.
A saturated hydrocarbon 'X' having molecular formula (C6H12) will form only one monochloro product on reaction with Cl2 in presence of ultravoilet light. Calculate the number of dichloroproducts (structural only) of the same compound 'X'.
Which resonating structure is more contributor to the actual structure ?
The structures with the least separation of formal charges are more stable. In structure C the charges are closer together making it more stable.
The correct structure of tautomer of the following compound 
can be:
The corresponding anion
because negative charges is stablised by two -m groups.
∴ The enolate ion 
is easily formed
Acidic strength of marked hydrogen of following compound in decreasing order is
is parent chain and 
(2- hydroxybutyl) group is substituent.
A hydrocarbon (P) on ozonolysis in presence of zinc gives only one dicarbonyl compound, which gives both Tollen’s and iodoform test. Identify the structure of (P).
Product gives both tollen's & iodoform test.
Consider the true/false of the following statements:
S1: The most stable resonating structure of p-nitrophenol (not having aromatic ring) is 
.
S2 : In 
all C–O bonds are of equal length.
S3 : CH3COONa is more resonance stabilised than the protonated acid CH3COOH.
S4 : Benzene ring is more electron dense in phenol than phenoxide.
S1 : Most stable resonating structure is 
group has greater +m effect, so it makes phenoxide ion more electron dense.
Only in (D) the l.p of N atom is not involved in resonance with benzene ring.
A hydrocarbon ‘X’ C7H10 is catalytically hydrogenated to C7H14 ‘Y’. 'Y' gives six monochloro products after photochemical chlorination. The structure of 'X' is -
C7H10 DU = 3
six monochloro products
Salicylic acid is more acidic than p-hydroxy benzoic acid.
The compound (I) does not racemise during enolisation, because the carbon atom with α-H is not asysmmetric carbon atom.
For the given reaction the correct reactivity order of H-atoms is :
Reactivity depends upon stability of free radicals generated at y,zx and w respectively.
In the given sequence of reactions which of the following is the correct structure of compound A.
Which of the following is not correctly ordered for resonance stability
Follow the rules for stability of resonating structures (structure with more number of p-bonds is more stable).
In acid (C), the group at ortho position is a –M group.
Sulphonic acid will be strongest acid.
Amongst the following compounds select the correct acidic strength order
–M effect increases acidic strength more as compare to –I effect.
Arrange the following carbocations in increasing order of stability:
The human body possesses four types of tissues, namely muscle, epithelial, connective and nervous. Connective tissue is further divided into six types, blood, lymph, bone, cartilage, adipose tissue, tendons and ligaments.
Few features are given-
(I) hypodermal sclerenchyma
(II) Conjoint Vascular bundle
(III ) unequal size Vascular bundle
(IV) parenchymatous endodermis
(V) presence of protoxylem in vascular bundle
How many features can belong to dicot stem-
Dicot stem have conjoint vascular bundle and endodermis is made up of parenchyma and also protoxylem present in primary xylem
Which of the following option have all cell/tissue is present as a ground tissue in plant organ-
Transverse section – epidernmis – vascular bundle = ground tissue
Meristematic cells are: thin walled, dense cytoplasm, in an active state of division, conspicuous nuclei, small size, isodiametric or polyhedral in shape, possess only primary cell wall, no lignin/suberin deposition, plasmodesmata present.
pericycle of dicot stem- sclerenchyma and hypodermis of monocot stem sclerenchyma
Secondary xylem is the type of xylem formed fromsecondary growth. In comparison, the primary xylemforms during primary growth. Because of this, thesecondary xylem is associated with lateral growth rather than vertical growth as in the primary xylem.
The presence of interfasicular cambium and Vascular bundle is shown in the diagrammatic representation of dicotyledonous stem above.
- Phloem parenchyma is ordinary living elongated parenchyma cells having abundant plasmodesmata. They store food, resins, latex, mucilage, etc. Phloem parenchyma is absent in most of the monocots and some herbaceous dicots.
- Cambium is a thin strip of primary meristem present between the xylem and phloem in dicot stems. Cambium is absent in the monocotyledons.
- Phloem parenchyma is found in both primary and secondary phloem. It is a part of the phloem elements. These are found in dicot roots, leaves, and stems but are absent in monocot plants.
Spring wood have less amount of xylem fibres that are present in secondary xylem and due to this reason density of spring wood is less.
Cork cambium is product of dedifferentiation
Interfasicular cambium is product of dedifferentiation.
The sclerenchymatous hypodermis is present in a monocotyledonous stem. The hypodermis is two to three-layered thick and lies below the epidermis. It is made up of thick-walled lignified sclerenchyma fibers.
Lenticels and collenchymatous hypodermis both can be present in-
Lenticels are present in dicot stem and dicot root.
In Given, how many are present in monocot plant – Lenticels, stomata, phloem parenchyma, phloem fibre , cuticle, endodermis, root hair, phellogen ,secondary medullary rays
lenticels, secondary medullary rays are product of secondary growth Phloem fibre and phloem parenchyma absent in monocot and phellogen also absent in monocot.
The functions of parenchyma tissues are storage, photosynthesis, and to help the plant float on water. Collenchyma- Are similar to parenchyma cells with thicker cell walls. They are meant to provide mechanical support to the plant structure in parts such as petiole of the leaf.
Sclereids are a reduced form of sclerenchyma cells with highly thickened, lignified cellular walls that form small bundles of durable layers of tissue in most plants. The presence of numerous sclereids form the cores of apples and produce the gritty texture of guavas.
In Given Examples, how many are triploblastic-
Echinus, Cucumaria, pleurobranchia , Ctenoplana , Gorgonia , Ascidia and Salpa
- Triploblastic animals start from playthelminthis to the chordate, hence phylum is not triploblastic.
- Pleurobranchia,ctenoplana,gorgonia are diploblastic because they're the members of phylum.
Clarias batrachus, a benthic species, has both a suprabranchial chamber (SC) and a Weberian apparatus-linked, encapsulated gas bladder (WGB).
In mosses, as in liverworts and hornworts, the leafy shoots belong to the gametophytic phase and produce sex organs when they mature. The leafy shoots (often called gametophores, because they bear the sex organs) arise from a preliminary phase called the protonema, the direct product of spore germination.
The fucus has a diplontic life cycle. Both the male and female fuse in the ostiole of the bump.
Consider the following statement-
(a) Marchantia have dioceious sporophyte
(b) homosporus Pteridophye have both sporophye and gametophyte freeliving
(c) gemmae cup develop on thallus produce sexual bud
Mark the correct option-
- Marchentia (a liverwort) have diocious gametophyte not sporophyte
- In pteridophyte both sporophyte and gametophyte are free living.
- Gemma buds are asexual.
- Hence, only statement b is correct.
Consider the following statement-
(a) Zygote of algae always undergo reduction division except in diplontic and haplo-diplontic algae
(b) Both algae and bryophyte is homosporus
How many statements are correct -
Red algae is non motile.
Match the following-
Which is/are correctly matched-
(1) Heterosporus - selaginella and salvinia
(2) Gymnosperm female gametophyte - archegonia and embryo sac
(3) Air pollinated - Cedrus, ficus, pinus
(4) Prothallus - Dryopteris and Pteridium
How many are the correct -
Gymnosperm lack embryosac and prothallus is present in homosporus pteridophyte like fern.
Which of following features matched with most of the bryophytes-
(1) spore stage undergo mitosis.
(2) diploid stage is multicellular and undergo meiosis
(3) diploid stage is represented by one cell stage
(4) no free living sporophyte
How many are the correct -
Diploid sporophyte is multicellular in bryophyte which is dependent and show meiosis.
Consider the following statements -
(1) In cycas male sporangia and female sporangia are borne on different tree
(2) branched stem is present in pinus and cedrus
(3) In cycas corraloid root is present
(4) In gymnosperm ovule are exposed both before and after fertilization
How many are the correct –
Cycas show coralloid root, and exposed ovule is present before fertilization and after fertilization it form seed.
Polysiphonia and porphyra is red algae and non motile.
Spirogyra is non-motile anisogamou , gelidium and porphyra is non-motile but oogamous.
Which of the following are oogamous and non-motile male gamets-
Bryophyte and pteridophyte have motile male gamete and gymnosperm and angiosperm have non-motile gamets.
Pteridophyte, gymnosperm and angiosperm have vascular sporophyte.
Consdier the following feature-
Internal fertilization, open circulatory system, malphigian tubule, triploblastic, nephridia, organ system-
How many feature are common in annelida and arthropoda-
Annelida and arthropoda share dissimilarities on the mode of fertilisation, circulatory system and organ system.
Sensory tentacle located on anterior head region.
- Ophiura or the serpent star is a species of the brittle star in the order Ophiurida belonging to the phylum Echinodermata and class Ophiuroidea.
- Cucumaria is a genus of sea cucumbers belonging to the phylum Echinodermata and class Holothuroidea.
The body of amphibians is divisible into head and trunk.
Mesoglea layer present in diploblastic organism.
Sponges and Coelentrata reproduce both sexually and asexually.
Echinodermata and Aschehelminthes are mostly dioceious.
The mouth of Osteichthyes is apical in position.
Consider the following statements-
(1) acoelomates organism can be triploblastic
(2) some acoelomate organism also show organ system level of organization
How many statement are correct-
Platyhelminthes are acoelomate and triploblastic but Achehelminthes show organ system but are pseudocoelomate.
- Ascaris (Round worm) belong to phylum Aschelminthes. Their fertilisation is internal and development may be direct or indirect.
- Locusta (Locust) is a gregarious pest belonging to largest phylum Arthropoda. They usually have internal fertilisation.
Flat worm (taenia) lack organ system level of organization.
Dentallium is an example of phylum mollusca and Asterias, Ophiura, Antedon are examples of phylum echinodermata which posses water vascular system.
Water vascular system is present in all except one of the function
- The water vascular system is present in the organisms which belong to the group of Echinoderms.
- The system is a hydraulic system. This system is used by the organisms for respiration, food circulation, waste removal, excretion and locomotion.
- It does not, however, play role in acting as sense organs.
Most reptiles reproduce sexually and have internal fertilisation. Calotes are an example of internal fertilisation.
Birds have direct development.
All four families –fabaceae, solanaceae,liiaceae and brassicaceae are hypogynous along with mustard, chilli and brinjal.
Asparagus is fibrous root modification and pneumatophore develop from prop root
Modified stems or leaves known as tendrils are seen in Vitis and passiflora. These tendrils develop from either the axillary bud or the terminal bud of the stem. Specifically, In Passiflora, the tendrils develop from the axillary bud.
Which region in root is responsible for Growth of root in length-
Length increase by only zone of elongation
In Australian Accacia plant, leaves are small and long lived.
The seed coat is thin membranous in groundnut, gram and maize seeds.
Few Examples are given-
Pea, bean, belladonna, Aloe, Asparagus, mustard, indigofera, Gloriosa, groundnut and makoi
How many have zygomophic flower
Examples of fabaceae are zygomorphic
Few features are given-
Endospermous, Actinomorphic, basal leaf, epipetalous, bicarpellary, superior ovary, reticulate venation and glabrous
How many belong to liliaceae-
Epipetalous, bicarpellary, reticulate venation and glabrous are feature of solanaceae
Among given feature-
Axile placentation, apocarpous, diadelphous, twisted aestivation, monodelphous, alternate phyllotaxy
How many belong to china rose-
China rose given separately in example of axile placentation, monodelphous, alternate phyllotaxy and twisted aestivation
petunia is bicarpellary as it belong to solanaceae
Papilionaceous corolla has five free petals which resemble the butterfly. The posterior superior petal is larger than the others, two anterior inferior petals are more or less united forming a boat-shaped structure and are called carina or keel, two lateral petals called wings are present.
In non- endospermic seeds like orchid, pea, lupin, the endosperm is completely consumed by the embryo.
In Solanaceae, the gynoecium is bicarpellary and the seeds are endospermic. Flowers of Solanaceae family are called as hypogynous flower as the sepals, petals, stamens are successively and freely inserted below the ovary and ovary is termed as superior.
Consider the following statement-
(1) Cassia and gulmohur have imbricate aestivation
(2) Superior ovary present in soyabean
How many are correct?
- In imbricate aestivation, one petal is completely inside and one is completely outside. The remaining perianth members show one edge inside and the other edge outside and are divided into two types descending imbricate and ascending imbricate. The example is Gulmohar and Cassia.
- Gynoecium of soyabean is monocarpellary, unilocular with ovules on marginal placentation and superior ovary.
In epigynous flower, the thalamus grows around the ovary, fusing with its wall. The other floral parts are present above the ovary. Hence the ovary is said to be inferior. So, peach is not epigynous.
Formed from six tepals arranged into two separate whorls of three parts each,Superior ovary, syncarpous, with three connate (fused) carpels and is trilocular or unilocular, actinomorphic (radially symmetric) or slightly zygomorphic (bilaterally symmetric), generally large and showy but may be inconspicuous.
Consider the following example-
Diatom, euglenoids, dinoflagellate, Trypanosoma, Agaricus, Ustilago, Albugo
How many can have motile asexual stage-
euglenoids, dinoflagellates, trypanosome and Albugo can have motile stage
Dikaryon not visible in phycomycetes.
Few structure are given –
Basidiocarp, ascocarp, conidiophores, prothallus, sporangium, Asci,
Zygospore and conidia
How many are diploid-
Conidiophore, prothallus and conidia are haploid.
Which of the following group of structure are genetically identical-
- Conidiophore - Erect fungal filament on which conidia are formed is called conidiophore.
- Conidia - Conidia are formed in chain. Each conidium forms new fungal filament (mycelium) by germination.
Hence, conidia and conidiophores are genetically identical.
Consider the following statement -
a) bacteria is chemosynthetic, photoautotrophic and heterotrophic
b) mostly bacteria is heterotrophic
c) bacteria show most variation in their metabolic diversity
Mark the correct option-
- Bacterias can be chemosynthetic, photoautotrophic and heterotrophic. Most bacteria are heterotrophs while some are photo or chemosynthetic autotrophs.
- The bacteria is known to have the most extensive metabolic diversity because it is able to extract energy from any carbohydrate which is in its reduced form like sugar.
Which is true about viroids, how many are correct-
1) high molecular weight RNA molecule
2) have protein coat
3) cause potato spindle tuber disease
Mark the correct option-
- Viroids consist solely of a low molecular weight RNA of about 75,000–130,000 and lack protein coat.
- Potato spindle tuber disease is caused by a small, naked, single-stranded, circular molecule of infectious RNA, which is called a viroid. Viroids have been found to be the cause of several dozen plant diseases.
- Hence, only statement 3 is correct.
Sexual spores are mainly produced by fungi. Agaricus and puccinia belong to the kingdom fungi.
Which of the following example have endogenously produced asexual spores-
- Mucor are commonly known as sac-fungi which produce asexual spores known as conidia. These spores are produced outside the mycelium which is known as conidiospores.
- Albugo is endogenously produced asexual spore which is non-motile.
Mark the correct statements-
1) basidiomycetes lack asexual spores
2) basidiomycetes lack sex organs
3) basidiomycetes have rust and smut fungi
Mark the correct statements-
In basidiomycetes, asexual reproduction by asexual spores is absent except in rusts and smut. In basidiomycetes, sex organs are absent because secular reproduction is by somatogamy.
Which of the following not give much help in correct identification –
Zoological park is for keeping animal in human care.