datavorous/entrance-exam-dataset huggingface dataset:
An old piece of dicot stem and a dicot root is given to a candidate during practical examination. Which of the following anatomical structures can be used to distinguish between the two?

Options:
A Secondary xylem
B Secondary phloem
C Protoxylem
D Cortical cells

The correct answer is:
Protoxylem

Solution:
The first formed primary xylem elements are called protoxylem and the later formed primary xylem is called metaxylem. In stems, the protoxylem lies towards the centre (pith) and the metaxylem lies towards the periphery of the organ. This type of primary xylem is called endarch. In roots, the protoxylem lies towards periphery and metaxylem lies towards the centre. Such an arrangement of primary xylem is called an exarch.

Topic: Biology, Anatomy of Flowering Plants, NEET
An open air pipe of length $80 \mathrm{~cm}$ has the second harmonic frequency equal to the fundamental frequency of a closed organ air pipe. The length of the closed pipe is

Options:
A $20 \mathrm{~cm}$
B $40 \mathrm{~cm}$
C $60 \mathrm{~cm}$
D $10 \mathrm{~cm}$

The correct answer is:
$20 \mathrm{~cm}$

Solution:
We have
$\left(\mathrm{f}_2\right)_{\text {Open }}=\left(\mathrm{f}_2\right)_{\text {Closed }}$
$\begin{aligned} & \Rightarrow 2 \cdot \frac{\mathrm{V}}{2 l_0}=\frac{\mathrm{V}}{4 l_{\mathrm{C}}} \\ & \Rightarrow \frac{1}{80}=\frac{1}{4 l_{\mathrm{C}}} \quad\left[\because l_0=80 \mathrm{~cm}\right] \\ & \Rightarrow l_{\mathrm{C}}=20 \mathrm{~cm}\end{aligned}$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 1)
An open organ pipe has a fundamental frequency 100 Hz. What frequency will be produced if it's one end is closed?

Options:
A 100,200,300......
B 50,150,250......
C 50,100, 200,300....
D 50,100,150,200......

The correct answer is:
50,150,250......

Solution:
When one end is closed, n1=1002=50 Hz
n2=3n1=150Hz, n3=5n1=250 Hz and so on.

Topic: Physics, Waves and Sound, NEET
An open organ pipe has a fundamental frequency of 300âHz . The first overtone of a closed organ pipe has the same frequency as the first overtone of an open organ pipe. How long is closed organ pipe? speed of sound = 300 m s-1

Options:
A 41.25 cm
B 42.3 cm
C 49.5 cm
D 40.5 cm

The correct answer is:
41.25 cm

Solution:
The fundamental frequency of an organ pipe
f1=v2l0
â´ l0=v2f1=3302Ã300=0.55 m
Given, the first overtone of closed pipe = First overtone of open pipe
Hence, 3v4lC=2v2l0
â lC=34 l0=340.55
â lC=0.4125 m âlC=41.25 cm

Topic: Physics, Waves and Sound, NEET
An open organ pipe having fundamental frequency (n) is in unison with a vibrating string. If the tube is dipped in water so that $75 \%$ of the length of the tube is inside the water then the ratio of fundamental frequency of the air column of dipped tube with that of string will be (Neglect end corrections)

Options:
A $1: 1$
B $2: 1$
C $2: 3$
D $3: 2$

The correct answer is:
$2: 1$

Solution:
$\mathrm{n}_{\text {open }}=\frac{\mathrm{v}}{2 \mathrm{~L}}.... (i)$
When dipped in water, pipe becomes closed at one end and open at the other.
Length available for resonance is
$\begin{aligned}
l_1 & =25 \% \times \mathrm{L} \\
& =\mathrm{L} \times \frac{25}{100} \\
& =\mathrm{L} / 4
\end{aligned}$
$\therefore \quad \mathrm{n}_{\text {closed }}=\frac{\mathrm{v}}{4 l_1}=\frac{\mathrm{v}}{4 \times \frac{\mathrm{L}}{4}}=\frac{\mathrm{v}}{\mathrm{L}}.... (ii)$
Comparing (i) and (ii),
$\therefore \quad \frac{\mathrm{n}_{\text {closed }}}{\mathrm{n}_{\text {open }}}=\frac{\left(\frac{\mathrm{v}}{\mathrm{L}}\right)^{\prime}}{\left(\frac{\mathrm{v}}{2 \mathrm{~L}}\right)}=\frac{2}{1}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2023 (10 May Shift 1)
An open organ pipe is closed suddenly with the result that the second overtone of the closed pipe is found to be higher in frequency by 100 than the first overtone of the original pipe. Then the fundamental frequency of the open pipe is

Options:
A 200 s-1
B 100 s-1
C 250 s-1
D 150 s-1

The correct answer is:
200 s-1

Solution:
Frequency of second overtone (fifth harmonic) of close pipe
=5v/4l
Frequency of first overtone (second harmonic) of an open pipe
=2v2l
Accordingly,
5v4l-2v2l=100
Or
v4l=100
Or
v=400l
The fundamental frequency of open pipe
=v2l=400l2l=200 s-1

Topic: Physics, Waves and Sound, NEET
An open organ pipe of length l vibrates in its fundamental mode. The pressure variation is maximum

Options:
A At the two ends
B At the middle of the pipe
C At the distance l 8 inside the ends
D At the distance l 4 inside the ends

The correct answer is:
At the middle of the pipe

Solution:
The first normal mode of vibration is called fundamental mode.
For first normal mode of vibration
l=λ12
Since, the pressure vibration is maximum at node,
Hence, the pressure variation is maximum at the
x=l2

Topic: Physics, Waves and Sound, NEET
An open pipe and a closed pipe have fundamental frequencies $n_1$ and $n_2$ respectively. They are combined to form a closed pipe. Fundamental frequency of the combined pipe is

Options:
A $\mathrm{n}_1+\mathrm{n}_2$
B $\mathrm{n}_1-\mathrm{n}_2$
C $\frac{\mathrm{n}_1 \mathrm{n}_2}{\mathrm{n}_1-2 \mathrm{n}_2}$
D $\frac{\mathrm{n}_1 \mathrm{n}_2}{\mathrm{n}_1+2 \mathrm{n}_2}$

The correct answer is:
$\frac{\mathrm{n}_1 \mathrm{n}_2}{\mathrm{n}_1+2 \mathrm{n}_2}$

Solution:
No solution. Refer to answer key.

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2017 (25 Apr Shift 2)
An open pipe emits a fundamental frequency $n$ when it emits the 3rd harmonic, then the pipe can accommodate

Options:
A 2 nodes, 2 anti-nodes
B 3 nodes, 4 anti-nodes
C 3 nodes, 3 anti-nodes
D 1 node, 2 anti-nodes

The correct answer is:
3 nodes, 4 anti-nodes

Solution:
For third harmonic, $n=3$. So, for an open pipe, the propagation of waves can be shown as
$\therefore$ Nodes $=3$ and anti-nodes $=4$

Topic: Physics, Waves and Sound, COMEDK, COMEDK 2016
An open pipe immersed in water to half its length. The ratio of the fundamental frequency of the pipe before and after immersion in water is

Options:
A $1: 2$
B $1: 1$
C $1: 3$
D $1: 4$

The correct answer is:
$1: 1$

Solution:
Let $L_{o}$ be the length of air column in pipe. After immersion,
$$
L_{c}=\frac{L_{0}}{2}
$$
Fundamental frequency in open pipe, $f_{o}=\frac{v}{2 L_{0}}$
Fundamental frequency in closed pipe, $f_{c}=\frac{v}{4 L_{c}^{r}}$
$$
\therefore \quad \frac{f_{o}}{f_{c}}=\frac{\frac{v}{2 L_{o}}}{\frac{v}{4 L_{c}}}=2\left(\frac{L_{c}}{L_{o}}\right)=2\left(\frac{L_{o}}{2 L_{o}}\right)=1: 1
$$

Topic: Physics, Waves and Sound, COMEDK, COMEDK 2012
An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by 100 Hz than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is

Options:
A 200
B 300
C 240
D 480

The correct answer is:
200

Solution:
Length of the organ pipe is same in both the cases. Fundamental frequency of open pipe is f 1 = v 2 l Frequency of third harmonic of closed pipe will be.
fâ¡ 2 = 3 vâ¡ 4 lâ¡
Given that, fâ¡ 2 = fâ¡ 1 + 1 0 0
or fâ¡ 2 - fâ¡ 1 = 1 0 0
or 3 4 vâ¡ lâ¡ - 1 2 vâ¡ lâ¡ = 1 0 0 â vâ¡ 4 lâ¡ = 1 0 0
â´ vâ¡ 2 lâ¡ or fâ¡ 1 = 2 0 0 Hz
Therefore, fundamental frequency of the open pipe is 200 Hz.

Topic: Physics, Waves and Sound, NEET
An organ pipe closed at one end has fundamental frequency of $1500 \mathrm{~Hz}$. The maximum number of overtones generated by this pipe which a normal person can hear is

Options:
A 14
B 13
C 6
D 9 .

The correct answer is:
6

Solution:
Human ear can hear frequencies upto $20,000 \mathrm{~Hz}$. So for closed pipe,
overtone, $\mathrm{v}=\frac{n v}{4 L}=n \times$ fundamental frequency
$$
\therefore \quad 20,000=n \times 1500 \Rightarrow n \approx 13 \text {. }
$$
Maximum possible harmonics obtained
$$
=1,3,5,7,9,11,13 \ldots \ldots \ldots \ldots
$$
Therefore one can hear maximum upto $13^{\text {th }}$ harmonics. Overtone $=7-1=6$

Topic: Physics, Waves and Sound, AIIMS, AIIMS 2004
An organ pipe filled with a gas at $27^{\circ} \mathrm{C}$ resonates at $400 \mathrm{~Hz}$ in its fundamental mode. If it is filled with the same gas at $90^{\circ} \mathrm{C}$, the resonance frequency at the same mode will be:

Options:
A $420 \mathrm{~Hz}$
B $440 \mathrm{~Hz}$
C $484 \mathrm{~Hz}$
D $512 \mathrm{~Hz}$

The correct answer is:
$440 \mathrm{~Hz}$

Solution:
As per question, we have;
frequency, $f=\frac{v}{4 l}$
Hence,
$\begin{array}{l}
\Rightarrow \frac{4 l}{f_2} =\frac{v_1}{v_2} \\
\Rightarrow \frac{400}{f_2} =\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}} \\
\Rightarrow \frac{400}{f_2} =\sqrt{\frac{300}{363}} \\
\Rightarrow [\because \mathrm{V} \propto \sqrt{\mathrm{T}}] \\
\Rightarrow f_2 =\sqrt{\frac{363}{300}} \times 400 \\
\Rightarrow f_2 =440 \mathrm{~Hz}
\end{array}$

Topic: Physics, Waves and Sound, NEET, NEET 2022 (Phase 2)
An oxygen containing organic compound upon oxidation forms a carboxylic acid as the only organic product with its molecular mass higher by 14 units. The organic compound is

Options:
A an aldehyde
B a primary alcohol
C a secondary alcohol
D a ketone

The correct answer is:
a primary alcohol

Solution:
Because the difference in mass between $-\mathrm{CH}_{2} \mathrm{OH}$ group and - $\mathrm{COOH}$ group is 14 , thus the compound which undergoes oxidation is a primary alcohol. ( $-\mathrm{CH}_{2} \mathrm{OH}$ is the functional group of primary alcohols.)

Topic: Chemistry, General Organic Chemistry, KCET, KCET 2010
An unbiased coin is tossed until the first head appears or until four tosses are completed, whichever happens earlier. Which of the following statements is/are correct?
$1$ The probability that no head is observed is $\frac{1}{16}$.
$2$ The probability that the experiment ends with three
tosses is $\frac{1}{8}$
Select the correct answer using the code given below:

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
$\mathrm{S}=\{\mathrm{H}, \mathrm{TH}, \mathrm{TTH}$, TTTH, TTTT $\}$
$\mathrm{P}(\mathrm{T})=\mathrm{P}(\mathrm{H})=\frac{1}{2}$
[Probability of getting head or tail in a single toss] $\therefore$ Probability that no head is observed $=\mathrm{P}(\mathrm{TTTT})$ $=\mathrm{P}(\mathrm{T}) \mathrm{P}(\mathrm{T}) \mathrm{P}(\mathrm{T}) \mathrm{P}(\mathrm{T})$
$=\frac{1}{2^{4}}=\frac{1}{16}$
And the probability that the experiment ends with 3 tosses
$=\mathrm{P}(\mathrm{TTH})$
$=\mathrm{P}(\mathrm{T}) \mathrm{P}(\mathrm{T}) \mathrm{P}(\mathrm{H})$
$=\frac{1}{8}$
Hence, both statements are correct.

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 2)
An unbiased die is tossed 500 times. The standard deviation of getting sixes in these 500 tosses is

Options:
A $\frac{625}{9}$
B $\sqrt{\frac{250}{3}}$
C $\frac{25}{3}$
D $\frac{250}{3}$

The correct answer is:
$\frac{25}{3}$

Solution:
$\mathrm{n}=500, \mathrm{P}=\frac{1}{6}, \mathrm{q}=\frac{5}{6}$
S.D. $=\sqrt{\mathrm{npq}}=\sqrt{500 \times \frac{1}{6} \times \frac{5}{6}}=\sqrt{\frac{2500}{36}}=\frac{50}{6}=\frac{25}{3}$

Topic: Mathematics, Probability, MHT CET, MHT CET 2022 (06 Aug Shift 1)
An unknown alcohol is treated with the " Lucas reagent " to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism :

Options:
A Secondary alcohol by SN2
B Tertiary alcohol by SN2
C Secondary alcohol by SN1
D Tertiary alcohol by SN1

The correct answer is:
Tertiary alcohol by SN1

Solution:
3Â° alcohol reacts fastest giving white turbidity immediately on adding lucas reagent.
Reaction takes place through carbocation formation involving SN1 mechanism,Rate of reaction in this reaction is directly proportional to the stability of the carbocation formed.
3Â° R-OH(tertiary alcohol) forms 3Â° carbocation and therefore,the reaction is fastest.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
An urn contains one black ball and one green ball. A second urn contains one white and one green ball. One ball is drawn at random from each urn.
What is the probability of getting at least one green ball?

Options:
A $1 / 2$
B $1 / 3$
C $2 / 3$
D $3 / 4$

The correct answer is:
$3 / 4$

Solution:
Total number of balls in urn $-\mathrm{I}=1$ Black $+1$ Green $=2$ Balls Total number of balls in urn $-\mathrm{II}=1$ White $+1$ Green $=2$ Balls
$\begin{aligned} \text { Required prob }=(1 \mathrm{G})_{\mathrm{I}} \times(1 \mathrm{G})_{\mathrm{I}}+(1 \mathrm{G})_{\mathrm{I}} \times(1 \mathrm{~B}) \mathrm{I} \\ &+(1 \mathrm{G})_{\mathrm{II}} \times(1 \mathrm{~W})_{\mathrm{II}} \\=\left(\frac{1}{2} \times \frac{1}{2}\right)+\left(\frac{1}{2} \times \frac{1}{2}\right)+\left(\frac{1}{2} \times \frac{1}{2}\right)=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4} \end{aligned}$

Topic: Mathematics, Probability, NDA, NDA 2012 (Phase 2)
An urn contains one black ball and one green ball. A second urn contains one white and one green ball. One ball is drawn at random from each urn.
What is the probability that both balls are of same colour?

Options:
A $1 / 2$
B $1 / 3$
C $1 / 4$
D $2 / 3$

The correct answer is:
$1 / 4$

Solution:
Total number of balls in urn $-\mathrm{I}=1$ Black $+1$ Green $=2$ Balls Total number of balls in urn $-\mathrm{II}=1$ White $+1$ Green $=2$ Balls
Required prob $=(1 \mathrm{G})_{\mathrm{I}} \times(1 \mathrm{G})_{\mathrm{II}}=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$

Topic: Mathematics, Probability, NDA, NDA 2012 (Phase 2)
Anatomically fairly old dicotyledonous root is distinguished from the dicotyledonous stem by

Options:
A absence of secondary xylem
B absence of secondary phloem
C presence of cortex
D position of protoxylem

The correct answer is:
position of protoxylem

Solution:
In dicotyledonous root, the condition of xylem is exarch as the protoxylem away from the centre and metaxylem towards the centre. In dicotyledonous stem (eg, Cucurbita), the condition of xylen is endarch as the metaxylem away from the centre and protoxylem towards the centre.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2009 (Screening)
Andalusian fowls have two pure forms - black and white. If black forms (BB) and white forms (bb) are crossed, F1 individuals appear blue coloured (Bb), due to incomplete dominance. Which of the following would be an outcome of a cross between black form and blue form?

Options:
A 1 Black: 2 Blue : 1 White
B 2 Black : 1 Blue
C 1 Black : 2 Blue
D 1 Black : 1 Blue

The correct answer is:
1 Black : 1 Blue

Solution:
As Mendel, on crossing homozygous members for contrasting characters, the dominant trait is expressed in the heterozygous or hybrid condition. Incomplete dominance is a deviation of Mendels law of dominance where the heterozygous individual instead of expressing the dominant trait, it expresses a trait which is a mix of the parent traits. On crossing, black form fowl with white form fowl, the offspring is blue form.
When a black form fowl is crossed with blue form fowl we get-

B	B
B	BB	BB
b	Bb	Bb

Genotype of F1 - BB and Bb→1:1
Phenotype of F1 - Black and Blue →1:1

Topic: Biology, Principles of Inheritance and Variation, NEET
Angle $\alpha$ is divided into two parts $\mathrm{A}$ and $\mathrm{B}$ such that $A-B=x$ and $\tan A=\tan B=p: q$. The value of $\sin x$ is equal
to

Options:
A $\frac{(p+q) \sin \alpha}{p-q}$
B $\frac{p \sin \alpha}{p+q}$
C $\frac{p \sin \alpha}{p-q}$
D $\frac{(p-q) \sin \alpha}{p+q}$

The correct answer is:
$\frac{(p-q) \sin \alpha}{p+q}$

Solution:
$A-B=x$ and $\tan A: \tan B=p: q$
Also, given $\alpha=\mathrm{A}+\mathrm{B}$
$\mathrm{A}-\mathrm{B}=\mathrm{x}$ and $\alpha=\mathrm{A}+\mathrm{B}$
$\Rightarrow \mathrm{A}=\frac{\mathrm{x}+\alpha}{2}$ and $\mathrm{B}=\frac{\alpha-\mathrm{x}}{2}$
Now, $\frac{\tan \mathrm{A}}{\tan \mathrm{B}}=\frac{\tan \left(\frac{\mathrm{x}+\alpha}{2}\right)}{\tan \left(\frac{\alpha-\mathrm{x}}{2}\right)}=\frac{\mathrm{p}}{\mathrm{q}} \quad($ Given $)$
$\Rightarrow \frac{2 \sin \left(\frac{\alpha+x}{2}\right) \cos \left(\frac{\alpha-x}{2}\right)}{2 \cos \left(\frac{\alpha+x}{2}\right) \sin \left(\frac{\alpha-x}{2}\right)}=\frac{p}{q}$
$\Rightarrow \frac{\sin \alpha+\sin x}{\sin \alpha-\sin x}=\frac{p}{q}$
$\Rightarrow \frac{\sin \alpha+\sin x+\sin \alpha-\sin x}{\sin \alpha+\sin x-\sin \alpha+\sin x}=\frac{p+q}{p-q}$
$\Rightarrow \frac{2 \sin \alpha}{2 \sin x}=\frac{p+q}{p-q}$
$\Rightarrow \frac{\sin \alpha}{\sin x}=\frac{p+q}{p-q} \Rightarrow \sin x=\frac{(\sin \alpha)(p-q)}{p+q}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2017 (Phase 2)
Annual rings are formed by the activity of

Options:
A Cambium
B Secondary xylem
C Phellogen
D Xylem and phloem

The correct answer is:
Cambium

Solution:
During secondary growth, vascular cambium divides to form secondary xylem which forms wood. The activity of cambium varies in the spring and autumn season. For plants growing in the temperate region, more quantity of light coloured wood is formed in the spring season and less quantity of dark coloured wood is formed in the autumn season. These concentric light and dark coloured wood constitutes an annual ring.

Topic: Biology, Anatomy of Flowering Plants, NEET
Answer the following :
(a) The triple point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$ respectively. On the absolute scale, one of the fixed points is the triple point of water, which on the Kelvin absolute scale is assigned the number $273.16 \mathrm{~K}$. What is the other fixed point on this (Kelvin) scale?
(c) The absolute temperature (kelvin scale) $T$ is related to temperature $t_c$ on the Celsius scale by $t_c=T-273.15$. Why do we have $273.15$ in this relation, 0 and not $273.16$ ?
(d) What is the temperature of the triple point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Options:

None

Solution:
(a) Triple point of water has a unique value $(273.16 \mathrm{~K}$ ) at fixed values of pressure and volume. But melting point of ice and boiling point of water do not have unique value as they depend both on pressure and temperature.
(b) The other fixed point is the absolute zero.
(c) On Celsius scale, $0^{\circ} \mathrm{C}$ corresponds to melting point of ice at normal pressure and the same temperature in Kelvin scale is $273.15 \mathrm{~K}$. The triple point of water in Kelvin scale is $273.16 \mathrm{~K}$.
$\therefore \quad$ By the relation, $t_c=T-273.15$,
the triple point of water on Celsius scale $=273.16$ $-273.15=0.01^{\circ} \mathrm{C}$
(d) Relation between temperature in Fahrenheit scale and absolute scale is
$\frac{\mathrm{F}-32}{180}=\frac{\mathrm{K}-273.15}{100}$
For another set of temperature, $\frac{\mathrm{F}^{\prime}-32}{180}=\frac{\mathrm{K}^{\prime}-273.15}{100}$
Subtracting (i) from (ii), we get
$$
\begin{aligned}
& \frac{F^{\prime}-F}{180}=\frac{K^{\prime}-K}{100} \\
\Rightarrow F^{\prime}-F &=\frac{180}{100} \times\left(K^{\prime}-K\right)
\end{aligned}
$$
If $K^{\prime}-K=1 K$, then
$$
\mathrm{F}^{\prime}-\mathrm{F}=\frac{180}{100} \times 1=\frac{9}{5}
$$
$\therefore \quad$ The triple point of water $(273.16 \mathrm{~K})$ in the new scale is $273.16 \times \frac{9}{5}=491.69$.

Topic: Physics, Thermal Properties of Matter
Answer the following :
(a) Which part of the ear determines the pitch of a sound?
(b) Which part of the human brain is the most developed ?
(c) Which part of our central neural system acts as a master clock?

Options:

None

Solution:
(a) Pitch represents the perceived fundamental frequency of a sound. Pitch is a subjective sensation in which a listener assigns perceived tones to relative positions on a musical scale based primarily on the frequency of vibration. As sound is finally perceived by the temporal lobe of the cerebral cortex so it can be said that cerebral cortex perceives the pitch of the sound.
(b) Cerebral cortex
(c) Hindbrain

Topic: Biology, Neural Control and Coordination
Antibodies present in colostrum which protect the new born from certain diseases is of

Options:
A
IgG type

B
IgA type

C
IgD type

D
IgE type

The correct answer is:
IgA type

Solution:
- During the initial days of lactation the yellowish fluid wlostrum is secreted by mother. It has abundant antibodies (IgA) that protects the infant from several diseases. IgG immunoglobulin (antibody) is most abundant (appro. \( 80 \% \) antibody in human and found in serum IgA is second abundant (approx. 10-15\%) antibody. It is found in saliva and tear also. IgD and IgE constitute 2-3\% of total antibodies which are found in most cells and serum.

Topic: Biology, Human Health and Diseases
Antibodies present in colostrum, which protects the newborn from certain diseases, are of

Options:
A IgG type
B Ig A type
C Ig D type
D Ig E type

The correct answer is:
Ig A type

Solution:
The yellowish fluid colostrum secreted by mother during the initial days of lactation has abundant antibodies (IgA) to protect the infant.

Topic: Biology, Human Health and Diseases, NEET
Apis, Bombyx, Culex and Limulus belongs to

Options:
A the largest phylum of animalia
B the smallest group of animalia
C the second-largest animal phylum
D the animals have an endoskeleton of calcareous ossicles

The correct answer is:
the largest phylum of animalia

Solution:
An "arthropod" is an invertebrate animal that has an exoskeleton, a segmented body, and jointed appendages. The following families of organisms are all examples of arthropods: Insects such as ants, dragonflies, and bees; Arachnids such as spiders and scorpions; Myriapods (a term which means "many feet") such as centipedes and millipedes; Crustaceans such as crabs, lobsters, and shrimp.

Topic: Biology, Animal Kingdom, NEET
Apomixis is

Options:
A Formation of seeds by fusion of gametes
B Formations of seeds without syngamy and meiosis
C Formation of seeds with syngamy but no meiosis
D None of the given options

The correct answer is:
Formations of seeds without syngamy and meiosis

Solution:
Apomixis (asexual seed formation) is the result of a plant gaining the ability to bypass the most fundamental aspects of sexual reproduction: meiosis and fertilization. Without the need for fertilization involving the male gamete, the resulting seed germinates a plant that develops as a maternal clone.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Are fruits more dependable for reproduction than seeds?

Options:
A Yes
B No
C It depends on the species
D It depends on the habitat

The correct answer is:
No

Solution:
Fruits are less dependable for reproduction compared to seeds as seed formation is independent of water, whereas fruits may depend on water availability for fertilization.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately $10^{-15} \mathrm{~m}$.

Options:

None

Solution:
To detect the properties of nucleons inside the nucleus the wavelength of particle which may detect nucleons or each particle (neutron and proton) present inside the nucleus is called nucleon.
Given, wavelength $\lambda=10^{-15} \mathrm{~m}$
To detect separate parts inside a nucleon the electron must have wavelength less than $10^{-15} \mathrm{~m}$.
We know that energy of free electron
$$
\mathrm{KE}=\mathrm{PE}
$$
Energy $\mathrm{E}=\mathrm{hv}=\frac{\mathrm{hc}}{\lambda}$
$$
=\frac{\mathrm{hc}}{\lambda}
$$
From Eq. (i) and Eq. (ii),
kinetic energy of electron
$$
\begin{aligned}
\mathrm{PE} &=\frac{\mathrm{hc}}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{10^{-15} \times 1.6 \times 10^{-19}} \mathrm{eV} \\
&=1.24 \times 10^9 \mathrm{eV} \\
\mathrm{KE} &=1.24 \times 10^9 \mathrm{eV}
\end{aligned}
$$

Topic: Physics, Nuclear Physics