datavorous/entrance-exam-dataset huggingface dataset:
The principle of immunization or vaccination is based on which property of the immune system?

Options:
A To produce antibodies
B To retain memory
C To differentiate self and non-self
D All of the above

The correct answer is:
To retain memory

Solution:
There are two types of immunity. innate and Acquired. Innate immunity is the immunity that is present since birth. It is very different from Acquired immunity as it is nonspecific. Acquired immunity is also known as Adaptive immunity. It is very specific and thus can differentiate between self and non self. Another main difference between Innate and Acquired is that Acquired immunity not only protects the body from infections but also prepares the body for further attack. This is due to the formation of memory cells. Acquired immunity is again of two types: AMI (Antibody mediated immunity) and CMI (Cell mediated immunity). Immunity can also be classified as Active and Passive. Active immunity is the introduction of Antigens which induces formation of Antibodies and this can be done Naturally or Artificially. Artificially it is done by use of Vaccines. Vaccines are live or attenuated pathogens which trigger the memory cells and also formation of Antibodies. But the main principle is to retain memory.

Topic: Biology, Human Health and Diseases, NEET
The probabilities of three events A,B and C are given PA=0.6, PB=0.4 and PC=0.5. If PA∪B=0.8, PA∩C=0.3, PA∩B∩C=0.2, PB∩C=β and PA∪B∪C=α, where 0.85≤α≤0.95, then β lies in the interval :

Options:
A 0.35, 0.36
B 0.25, 0.35
C 0.20, 0.25
D 0.36, 0.40

The correct answer is:
0.25, 0.35

Solution:
PA∪B∪C=P(A)+P(B)+P(C)-P(A∩B)-P(B∩C)-P(C∩A)+PA∩B∪C

⇒α=1.4-p(A∩B)-β

⇒α+β=1.4-p(A∩B)....(1)

again

P(A∪B)=P(A)+P(B)-P(A∩B)

⇒P(A∩B)=·2......(2)

by 1 and 2

α=1.2-β

now

0.85≤α≤0.95

⇒0.85≤1.2-β≤0.95⇒β∈0.25,0.35

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The probability distribution of a random variable X is given as

X	â5	â4	â3	â2	â1	0	1	2	3	4	5

P( X )	p	2p	3p	4p	5p	7p	8p	9p	10p	11p	12p

Then, the value of p is

Options:
A 172
B 373
C 572
D 174

The correct answer is:
172

Solution:
Sum of Probabilities =1
â p+2p+3p+4p+5p+7p+8p+9p+10p+11p+12p=1
⇒72p=1⇒p= 1 72

Topic: Mathematics, Probability, JEE Main
The probability distribution of a random variable X is PX=X12: X=1,2,3,4,5,60: otherwise. Then, the conditional probability P32<X<72X>2 is

Options:
A 56
B 518
C 16
D 712

The correct answer is:
16

Solution:
Let, A=32<X<72, B=X>2
A∩B=2<X<72
PAB=PA∩BPB=P3P3+P4+P5+P6
=312312+412+512+612=16

Topic: Mathematics, Probability, JEE Main
The probability distribution of a random variable $\mathrm{X}$ is
Then $\operatorname{Var}(\mathrm{X})=$

Options:
A $\frac{\mathrm{n}^2-1}{12}$
B $\frac{\mathrm{n}^2-\mathrm{n}}{6}$
C $\frac{\mathrm{n}^2-\mathrm{n}}{12}$
D $\frac{\mathrm{n}^2-1}{6}$

The correct answer is:
$\frac{\mathrm{n}^2-1}{12}$

Solution:
$\begin{aligned} \therefore \quad \Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} & =\frac{1}{\mathrm{n}}+\frac{2}{\mathrm{n}}+\frac{3}{\mathrm{n}}+\ldots .+\frac{\mathrm{n}}{\mathrm{n}} \\ & =\frac{1+2+3+\ldots \mathrm{n}}{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)}{2(\mathrm{n})}=\frac{\mathrm{n}+1}{2}\end{aligned}$
$\begin{aligned} & \quad \sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=\frac{1}{\mathrm{n}}+\frac{4}{\mathrm{n}}+\frac{9}{\mathrm{n}}+\ldots .+\frac{\mathrm{n}^2}{\mathrm{n}} \\ & \quad \Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=\frac{1}{\mathrm{n}}+\frac{4}{\mathrm{n}}+\frac{9}{\mathrm{n}}+\ldots .+\frac{\mathrm{n}^2}{\mathrm{n}} \\ & \quad=\frac{1+4+9+\ldots \mathrm{n}^2}{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6 \mathrm{n}}=\frac{(\mathrm{n}+1)(2 \mathrm{n}+1)}{6} \\ & \therefore \quad \operatorname{Var}(\mathrm{x})=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-\left(\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^2 \\ & \quad=\frac{(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\left[\frac{(\mathrm{n}+1)}{2}\right]^2=\frac{2 \mathrm{n}^2+3 \mathrm{n}+1}{6}-\frac{\mathrm{n}^2+2 \mathrm{n}+1}{4} \\ & =\frac{4 \mathrm{n}^2+6 \mathrm{n}+2-3 \mathrm{n}^2-6 \mathrm{n}-3}{12}=\frac{\mathrm{n}^2-1}{12}\end{aligned}$

Topic: Mathematics, Probability, JEE Main
The probability distribution of random variable X is given by:

X	1	2	3	4	5

P(X)	K	2K	2K	3K	K

Let p=P(1<X<4∣X<3). If 5p=λK, then λ is equal to

Options:

The correct answer is:
30

Solution:
ΣP(X)=1
⇒K+2K+2K+3K+K=1
K=19
P=P(1<X<4∣X<3)=P(X=2)P(X<3)
P=2K9KK9K+2k9K=23
5P=λK
5×23=λ×19⇒λ=30

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The probability distribution of random variable $X$ with two missing probabilities $p_{1}$ and $p_{2}$ is given below
$\begin{array}{cc}\boldsymbol{X} & \boldsymbol{P}(\boldsymbol{X}) \\ 1 & k \\ 2 & p_{1} \\ 3 & 4 k \\ 4 & p_{2} \\ 5 & 2 k\end{array}$
It is further given that $P(X \leq 2)=0.25$ and $P(X \geq 4)=0.35$. Consider the following statements
$1.$ $p_{1}=p_{2}$
$2.$ $p_{1}+p_{2}=P(X=3)$
Which of the statements given above is/are correct?

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

The correct answer is:
Neither 1 nor 2

Solution:
Let $P(X \leq 2)=0.25$
$\Rightarrow \quad P(X=1)+P(X=2)=0.25$
$\Rightarrow k+p_{1}=0.25($ from the table $)$
$\Rightarrow p_{1}=0.25-k$
and $P(X \geq 4)=0.35$
$\Rightarrow \quad P(X=4)+P(X=5)=0.35$
$\Rightarrow p_{2}+2 k=0.35$ (from the table)
$\Rightarrow p_{2}=0.35-2 k$
From (1) and (2)
$p_{1} \neq p_{2}$
and $p_{1}+p_{2}=0.25-k+0.35-2 k=0.6-3 k$
$\neq P(X=3)$
Hence, neither 1 nor 2 is correct

Topic: Mathematics, Probability, JEE Main
The probability of a bomb hitting a bridge is 23. Two direct hits are needed to destroy the bridge. The minimum number of bombs required such that the probability of bridge being destroyed is greater than 34, is

Options:

The correct answer is:
4

Solution:
Let the number of bombs required =n
Probability of a bomb hitting the target is 23
∴ Probability of atleast 2 hits
=1− 1 3 n − C n 1 2 3 1 1 3 n−1 > 3 4
⇒2n+1313n-1<14
⇒432n+1<3n-1
⇒n≥4

Topic: Mathematics, Probability, JEE Main
The probability of a man hitting a target is 110. The least number of shots required, so that the probability of his hitting the target at least once is greater than 14, is....

Options:

The correct answer is:
3

Solution:
p=110, q=910

P (not hitting in n trials) =910n

∴ P (at least one hit)=1−910n>14

⇒ 910n<34

0.9n<0.75

If n=3 ⇒ 0.729<0.75 which is true.

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (04 Sep Shift 1)
The probability of a man hitting a target is $\frac{2}{5}$. He fires at the target $k$ times $(k$, a given number). Then the minimum $k$, so that the probability of hitting the target at least once is more than $\frac{7}{10}$, is :

Options:
A
3

B
5

C
2

D
4

The correct answer is:
3

Solution:
$$
\begin{aligned}
& \frac{2}{5}+\frac{3}{5} \times \frac{2}{5}+\left(\frac{3}{5}\right)^2 \times \frac{2}{5}+\ldots \ldots+\left(\frac{3}{5}\right)^k \cdot \frac{2}{5}>\frac{7}{10} \\
\Rightarrow & \frac{2}{5}\left[1+\frac{3}{5}+\left(\frac{3}{5}\right)^2+\ldots \ldots+\left(\frac{3}{5}\right)^k\right]>\frac{7}{10} \\
\Rightarrow & \frac{2}{5} \times \frac{1-\left(\frac{3}{5}\right)^k}{1-\frac{3}{5}}>\frac{7}{10} \Rightarrow 1-\left(\frac{3}{5}\right)^k>\frac{7}{10} \\
\Rightarrow & \left(\frac{3}{5}\right)^k < \frac{3}{10} \Rightarrow k \geq 3
\end{aligned}
$$
Hence minimum value of $k=3$

Topic: Mathematics, Probability, JEE Main, JEE Main 2013 (09 Apr Online)
The probability of a problem being solved by 3 students independently are 12, 13 and Î± respectively. If the probability that the problem is solved is PS, then PS lies in the interval (where, Î±â0,1)

Options:
A 0,12
B 13,12
C 23,1
D 13,23

The correct answer is:
23,1

Solution:
Let 3 students are A,B,C and their probability of solving problems is PA,PB,PC respectively.
PA=12,PB=13,PC=α
Probability of the problem is being solved,
PS=1-P(AC∩BC∩CC)
=1-PACPBCP(CC)
=1-1-PA1-PB1-PC
=1-12231-α
=23+13α

Hence, PS∈23,1

Topic: Mathematics, Probability, JEE Main
The probability of a sure event is

Options:
A $0$
B $1$
C $2$
D $\frac{1}{2}$

The correct answer is:
$1$

Solution:
The probability of a sure event is always 1 . Consider there are only red balls and you are going to choose a ball. Here, in any case, the colour of the ball will be only red. Therefore the probability of choosing a red ball will be always 1.

Topic: Mathematics, Probability, JEE Main
The probability of an event A is 45. The probability of an event B, given that the event A occurs is 15. The probability of event A, given that the event B occurs is 23. The probability that neither of the events occurs is

Options:
A 325
B 25
C 125
D 215

The correct answer is:
325

Solution:
PA=45, PBA=PA∩BPA=15⇒PA∩B=425
PAB=PA∩BPB=23⇒PB=425×32=625
PAc∩Bc=PA∪Bc=1-PA∪B=1-PA-PB+PA∩B
=1-45-625+425=325

Topic: Mathematics, Probability, JEE Main
The probability of India winning a test match against Australia is 14. Assuming the matches to be independent events, the probability that in a 7 match series India’s second win occurs at 4th test is P, then 256P is equal to

Options:

The correct answer is:
27

Solution:
2nd win in 4th test ⇒ In the first 3 matches
there must be exactly 1 win
so, required probability =3C114134214
P=27256

Topic: Mathematics, Probability, JEE Main
The probability of selecting integers a∈-5,30 such that x2+2 a+4 x-5a+64>0, for all x∈R, is:

Options:
A 736
B 29
C 16
D 14

The correct answer is:
29

Solution:
We have,

x2+2 a+4 x-5a+64>0

If A>0 and D=B2-4AC<0, then Ax2+Bx+C>0 ∀x∈R.

Hence,

D<0

⇒ 4a+42-4-5a+64<0

⇒ a2+16+8a+5a-64<0

⇒ a2+13a-48<0

⇒ a+16a-3<0

⇒ a∈-16,3

∴ Possible values for a are -15, -14, …….,2 containing 18 integers.

But given range of a is -5,30, hence a would take values -5, -4, -3, -2, -1, 0, 1, 2, containing 8 integers.

And, -5, 30 has 36 numbers.

∴ Required probability

=836

=29

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The probability of solving a question by three students are $\frac{1}{2}, \frac{1}{4}, \frac{1}{6}$ respectively. Probability of question is being solved will be

Options:
A $\frac{33}{48}$
B $\frac{35}{48}$
C $\frac{31}{48}$
D $\frac{37}{48}$

The correct answer is:
$\frac{33}{48}$

Solution:
(i) This question can also be solved by one student
(ii) This question can be solved by two students simultaneously
(ii) This question can be solved by three students all together.
$P(A)=\frac{1}{2}, P(B)=\frac{1}{4}, P(C)=\frac{1}{6}$
$P(A \cup B \cup C)=P(A)+P(B)+P(C)$
$-[P(A) \cdot P(B)+P(B) \cdot P(C)+P(C) \cdot P(A)]+[P(A) \cdot P(B) \cdot P(C)]$
$=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}-\left[\frac{1}{2} \times \frac{1}{4}+\frac{1}{4} \times \frac{1}{6}+\frac{1}{6} \times \frac{1}{2}\right]+\left[\frac{1}{2} \times \frac{1}{4} \times \frac{1}{6}\right]=\frac{33}{48}$

Topic: Mathematics, Probability, JEE Main
The probability that a married man watches a certain T.V. show is 0.6 and the probability that a married woman watches the show is 0.5. The probability that a man watches the show given that his wife does watch is 0.8. If the probability that a wife watches the show given that her husband does watch is k, then 1k is equal to

Options:

The correct answer is:
1.5

Solution:
Mâ a married man watches the T.V. show
Wâ a married woman watches the T.V. show
PM=0.6,PW=0.5
PMW=0.8
PMW=PMâ©WPW=0.8âPMâ©W=0.4
PWM=PWâ©MPM=0.40.6=23

Topic: Mathematics, Probability, JEE Main
The probability that a non-leap year selected at random will have 53 Sunday is

Options:
A 0
B $1 / 7$
C $2 / 7$
D $3 / 7$

The correct answer is:
$1 / 7$

Solution:
In a non-leap year,total number of days is 365 . Out of them, there are 52 weeks and 1 day extra. Thus, a non-leap year always has 52 Sunday. The remaining 1 day can be Sunday, Monday, Tuenday, Wednesday, Thursday, Friday and Sarurday. Out of these 7 cases, we have Sunday in one
case
$\therefore$ Total number of outcomes = 7
Number of favourable outcomes = 1
Hence, required probability $=\frac{1}{7}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2017
The probability that a particular nucleus of 38Cl will undergo beta decay in any time interval of 4 s is

T1/2 for 38Cl is 37.2 min

Options:
A 3.1×10-4
B 6.2×10-4
C 12.4×10-4
D 24.8×10-4

The correct answer is:
12.4×10-4

Solution:
Probability for a particular nucleus to decay in any time interval dt is

dNN=Î»dt
=0.693T12Ã4
=12.4Ã10-4

Topic: Physics, Atomic Physics, JEE Main
The probability that a randomly chosen 5- digit number is made from exactly two digits is :

Options:
A 135104
B 150104
C 134104
D 121104

The correct answer is:
135104

Solution:
Total number of five-digit numbers are 9104.

Case I : Choosing two non-zero digits i.e. in C29 ways.

So, number of five-digit numbers containing both digits =25-2.

Case II : Choose one non-zero and one zero as digit i.e. in C19 ways.

So, number of five-digit numbers containing one non-zero and one zero both =24-1.

So, favourable number of ways =C2925-2+C1924-1=3630+915=1080+135=1215

Thus, required probability =12159×104=135104.

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The probability that a randomly chosen one-one function from the set a,b,c,d to the set 1,2,3,4,5 satisfied fa+2 fb-fc=fd is

Options:
A 124
B 140
C 130
D 120

The correct answer is:
120

Solution:
Total cases will be 5C4×4!

Now favourable cases for 2fb=fc+fd-fa will be,

Case (I) if fb=1 then fc, fd, fa can take the value 3,4,5 & 4,3,5 respectively

Case (II) if fb=2 then fc, fd, fa can take value 3,5,4 & 5,3,4 respectively

Case (III) if fb=3 then fc, fd, fa can take value 2,5,1 & 5,2,1 respectively

So total favourable case will be 6

So probability=favourable casestotal outcomes=65×4!=120

Topic: Mathematics, Functions, JEE Main, JEE Main 2022 (28 Jun Shift 2)
The probability that a randomly selected 2-digit number belongs to the set n∈N:2n-2 is a multiple of 3} is equal to

Options:
A 16
B 23
C 12
D 13

The correct answer is:
12

Solution:
Total two-digit numbers is 90

So, total number of cases =C190=90

Now, 2n-2=3-1n-2

=C0n3n-C1n·3n-1+…+-1n-1·Cn-1n3+-1n·Cnn-2

=33n-1-n3n-2+…+-1n-1·n+-1n-2

Here, the term -1n-2 should be multiple of 3.

So, 2n-2 is multiple of 3 only when n is odd.

There are total 45 odd two-digit numbers.

Hence, the required probability =C14590=4590=12

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The probability that A speaks truth is $\frac{4}{5}$, while this probability for $B$ is $\frac{3}{4}$. The probability that they contradict each other when asked to speak on a fact is

Options:
A
$\frac{3}{20}$

B
$\frac{1}{5}$

C
$\frac{7}{20}$

D
$\frac{4}{5}$

The correct answer is:
$\frac{7}{20}$

Solution:
$\mathrm{E}_1$ : event denoting that $A$ speaks truth
$\mathrm{E}_2$ : event denoting that $B$ speaks truth
Probability that both contradicts each other $=P\left(E_1 \cap \bar{E}_2\right)+P\left(\bar{E}_1 \cap E_2\right)=\frac{4}{5} \cdot \frac{1}{4}+\frac{1}{5} \cdot \frac{3}{4}=\frac{7}{20}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2004
The probability that a student passes in mathematics is $4 / 9$ and that he passes in physics is $2 / 5$. Assuming that passing in mathematics and physics are independent of each other, what is the probability that he passes in mathematics but fails in physics?

Options:
A $\frac{4}{15}$
B $\frac{8}{45}$
C $\frac{26}{45}$
D $\frac{19}{45}$

The correct answer is:
$\frac{4}{15}$

Solution:
Probability of passing in mathematics $=\frac{4}{9}$
Probability of passing in physics $=\frac{2}{5}$
Probability of failure in physics $=1-\frac{2}{5}=\frac{3}{5}$ Given that both the events are independent. Required probability $=\frac{4}{9} \times \frac{3}{5}=\frac{4}{15}$

Topic: Mathematics, Probability, JEE Main
The probability that an anti-aircraft gun can hit an enemy plane at the first, second and third shot are \( 0.6,0.7 \) and \( 0.1 \)
respectively. The probability that the gun hits the plane is

Options:
A \( 0.108 \)
B \( 0.892 \)
C \( 0.14 \)
D \( 0.91 \)

The correct answer is:
\( 0.892 \)

Solution:
Let the events of hitting the enemy plane at the first, second and third shot are respectively A, B and C. Then as given
PA=0.6, PB=0.7, PC=0.1
Since events A, B, C are independent, so
required probability =P(A+B+C)
=1-PA¯PB¯PC¯
=1-1-0.61-0.7(1-0.1)
=1-0.40.3(0.9)
=1-0.108=0.892

Topic: Mathematics, Probability, JEE Main
The probability that an ordinary or a non-leap year has 53 sunday, is

Options:
A $\frac{2}{7}$
B $\frac{1}{7}$
C $\frac{3}{7}$
D None of these

The correct answer is:
$\frac{1}{7}$

Solution:
In a non-leap year, we have 365 days i.e., 52 weeks and one day. So, we may have any day of seven days. Therefore, 53 Sunday, required probability $=\frac{1}{7}$

Topic: Mathematics, Probability, JEE Main
The probability, that in a randomly selected 3-digit number at least two digits are odd, is

Options:
A 1936
B 1636
C 1933
D 1336

The correct answer is:
1936

Solution:
Total No of 3-digit Number =900

Case -1 When there are 2 odd digit and 1 even digit

(i) Even odd odd =4×5×5=100

{As 0 cannot come at 1st place}

(ii) Odd even odd =5×5×5=125

(iii) Odd odd even =5×5×5=125

Case-2

All 3 odd digit =5×5×5=125

So total number of selecting 3-digit number having atleast 2 odd digit will be addition of both cases which is=100+125+125+125=475

So probability =475900=1936

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The probability that two randomly selected subsets of the set {1,2,3,4,5} have exactly two elements in their intersection, is:

Options:
A 6528
B 6527
C 3527
D 13529

The correct answer is:
13529

Solution:
Total subsets=25=32

Probability=C25×3332×32=10×27210=13529

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The probable reason for the greyish appearance of the cerebral cortex is

Options:
A presence of bundles of axons in this reason
B the cell body of neuron possess myelin sheath
C the cell bodies of neuron are concentrated here
D all of the above

The correct answer is:
the cell bodies of neuron are concentrated here

Solution:
The central nervous system is made up of two types of tissue: the grey matter and the white matter. In the grey matter, these axons are mainly unmyelinated, meaning they are not covered by a whitish-colored, fatty protein called myelin.
The grey matter is mainly composed of neuronal cell bodies and unmyelinated axons

Topic: Biology, Neural Control and Coordination, NEET
The process in which oxidation number increases is known as

Options:
A Oxidation
B Reduction
C Auto-oxidation
D None of the above

The correct answer is:
Oxidation

Solution:
The correct option is 1 Oxidation The increase in oxidation state of an atom, through a chemical reaction, is known as an oxidation

Topic: Chemistry, Redox Reactions, JEE Main
The process of breaking down lumps into small pieces and then into powder is known as

Options:
A Pulverisation
B Concentration
C Powderification
D None of the above

The correct answer is:
Pulverisation

Solution:
ores occur in nature as huge lumps. They are broken to small pieces with the help of crushers or grinders. These pieces are then reduced to fine powder with the help of a ball mill or stamp mill. This process is called pulverisation.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The process of decomposition which is incorrect is

Options:
A Fragmentation: Break down detritus into smaller particles
B Leaching: Water soluble inorganic nutrients go down into the soil horizon
C Catabolism: Enzymes degrade detritus into simpler inorganic substances
D Humification: Release of inorganic nutrients from humus

The correct answer is:
Humification: Release of inorganic nutrients from humus

Solution:
Detritus consisting of remains of dead plant and animals are broken down by process of fragmentation into smaller particles by detrivores like earthworm. Detritus is also degraded into simpler inorganic substances by a bacterial and fungal enzymes by a process called as catabolism.By the process of leaching, watersoluble inorganic nutrients go down into the soil horizon and get precipitated as unavailable salts. There is accumulation of a dark coloured amorphous substance called humus produced by a process of humification which is further degraded by some microbes resulting in release of inorganic nutrients by the process of mineralization.

Topic: Biology, Ecosystem, NEET
The process of industrial manufacturing of sodium carbonate is known as

Options:
A Castner process
B Haber's process
C Le-blanc process
D Chamber process

The correct answer is:
Le-blanc process

Solution:
Although this process was effective, it could not be used to produce large volumes. The first process that allowed production of significant amounts of sodium carbonate was a synthetic process known as the LeBlanc process. Nowadays solvay's process is used to manufacture sodium carbonate .

Topic: Chemistry, s Block Elements, JEE Main
The process of migration of chloride ions from plasma to RBC and of carbonate ions from RBC to plasma is

Options:
A Chloride shift
B Ionic shift
C Atomic shift
D Na+ pump

The correct answer is:
Chloride shift

Solution:
During the transport of CO2 through the blood, bicarbonate ions diffuse out of RBCs while chloride ions from plasma enter the RBCs to maintain ionic equilibrium. This is called the chloride shift.

Topic: Biology, Breathing and Exchange of Gases, NEET
The process of migration of chloride ions from plasma to RBC and that of carbonate ions from RBC to plasma is

Options:
A

Chloride shift

B

Ionic shift

C

Atomic shift

D

Na+ pump

The correct answer is:

Chloride shift

Solution:
During the transport of CO2 through the blood, bicarbonate ions diffuse out of RBCs, whereas chloride ions from plasma enter the RBCs to maintain ionic equilibrium. This is called as chloride shift or Hamburger's phenomenon.

Topic: Biology, Breathing and Exchange of Gases, NEET
The process of mineralisation by microorganisms helps in the release of

Options:
A
inorganic nutrients from humus

B
both organic and inorganic nutrients from detritus

C
organic nutrients from humus

D
inorganic nutrients from detritus and formation of humus.

The correct answer is:
inorganic nutrients from humus

Solution:
During the process of decomposition five important steps occured i.e fragmentation, leaching, catabolism humification and mineralisation. Humus and mineral (inorganic nutrients) are obtained by humification and mineralisaton.
Mineralisation is also responsible for release of inorganic substances, both minerals ( $\mathrm{K}+, \mathrm{Mg}++, \mathrm{Ca}++$ and $)$ and nonminerals like $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{CO}_2$ from dead organic matter of plant's and animal's remains.

Topic: Biology, Ecosystem
The process of sperm penetrating the ovum is mainly

Options:
A Mechanical
B Chemical
C Electrostatical
D Thermal

The correct answer is:
Chemical

Solution:
At the time of fertilization, when sperm comes in contact with the ovum, an acrosomal reaction takes place with the release of an enzyme hyaluronidase, and acrosin. These enzymes then begin to digest the zona pellucida and allow the sperm to tunnel toward the egg's plasma membrane by dissolving the actin filaments through biochemical reactions.

Topic: Biology, Human Reproduction, NEET
The process that involves the removal of sulphur from the ores is :

Options:
A Smelting
B Roasting
C Leaching
D Refining

The correct answer is:
Roasting

Solution:
In roasting process, metal sulphide ore are converted into metal oxide and sulphur is remove in the form of SO2 gas.

2MS+3O2⟶Δ2MO+2SO2↑

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The process that is NOT endothermic in nature is :

Options:
A Ar(g)+e-→Arg-
B H(g)+e-→Hg-
C O(g)-+e-→O(g)2-
D Na(g)→Nag++e-

The correct answer is:
H(g)+e-→Hg-

Solution:
H(g)+e- →extremicH-(g)	ΔHeg=-72KJ/mol

O-(g)+e-⟶ →endeterm c O2-(g)	ΔHeg=+744 KJ/mol

Ar(g)+e- →endeterm c Ar-(g)	ΔHeg=+96 KJ/mol

Na(g)⟶ endotherm c Na+(g)+e-	IE =495.8KJ/mol

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main