datavorous/entrance-exam-dataset huggingface dataset:
The correct statement among the following, for a "chromatography" purification method is :

Options:
A Organic compounds run faster than solvent in the thin layer chromatographic plate.
B \(\mathrm{R}_f\) is an integral value.
C \(\mathrm{R}_f\) of a polar compound is smaller than that of a non-polar compound.
D Non-polar compounds are retained at top and polar compounds come down in column chromatography.

The correct answer is:
\(\mathrm{R}_f\) of a polar compound is smaller than that of a non-polar compound.

Solution:
Non polar compounds are having higher value of $\mathrm{R}_{\mathrm{f}}$ than polar compound.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (06 Apr Shift 2)
The correct statement among the following is:

Options:
A SiH33N is planar and less basic than CH33N
B SiH33N is pyramidal and more basic than CH33N
C SiH33N is pyramidal and less basic than CH33N
D SiH33N is planar and more basic than CH33N

The correct answer is:
SiH33N is planar and less basic than CH33N

Solution:
lp of N in NCH33 is free to donate while in NSiH33 lp is involved in back bonding so it is not easy to donate hence it is less basic.

Due to back bonding it is planar and sp2 hybridised.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The correct statement among the following regarding birth control pills is:

Options:
A They contain estrogen only.
B They contain progesterone only.
C They contain a mixture of estrogen and progesterone derivatives.
D Progesterone enhances ovulation.

The correct answer is:
They contain a mixture of estrogen and progesterone derivatives.

Solution:
Birth control pills contain the combination of estrogen and progesterone derivatives which will prevent the ovulation, ie., the release of eggs from ovaries. Birth control pills thickens the cervical mucous and thins the lining of uterus (endometrium). This will not facilitate sperm reaching the egg.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The correct statement for both the processes of physisorption and chemisorption is

Options:
A
both are endothermic

B
chemisorption is endothermic but physisorption is exothermic

C
both are exothermic

D
physisorption is endothermic but chemisorption is exothermic.

The correct answer is:
both are exothermic

Solution:
Both are exothermic.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2012 (12 May Online)
The correct statement for the following addition reaction is

Options:
A (M and O) and (N and P) are two pairs of diasteriomers
B Bromination proceeds through cis-addition in both reaction
C O and P are identical molecules
D (M and O) and (N and P) are two pairs of enantiomers

The correct answer is:
(M and O) and (N and P) are two pairs of diasteriomers

Solution:
Here (M and O) and (N and P) are two pairs of diastereomers.

Topic: Chemistry, Hydrocarbons, JEE Main
The correct statement for the molecule CsI 3 , is :

Options:
A It is a covalent molecule.
B It contains Cs + and I 3 - ions.
C It contains Cs 3 + and I - ions.
D It contains Cs + , I - and lattice I 2 molecule.

The correct answer is:
It contains Cs + and I 3 - ions.

Solution:
I 3 - ion is made by the addition of excess iodine to iodide solution.
I 2 + I - → I 3 -
Cs will always be in +1 oxidation state.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2014 (06 Apr)
The correct statement from the following is

Options:
A A body having zero velocity will not necessarily have zero acceleration
B A body having zero velocity will necessarily have zero acceleration
C A body having uniform speed can have only uniform acceleration
D A body having non-uniform velocity will have zero acceleration

The correct answer is:
A body having zero velocity will not necessarily have zero acceleration

Solution:
When the body is projected vertically upward then at the highest point its velocity is zero but acceleration is not equal to zero $\left(g=9.8 \mathrm{~m} / \mathrm{s}^2\right)$.
\(\Rightarrow\) (1) A body having zero velocity will not necessarily have zero acceleration.

Topic: Physics, Motion In One Dimension, JEE Main
The correct statement in respect of $d$-block elements

Options:
A They are all metals
B They show variable valency
C They form coloured ions and complex salts
D All above statements are correct

The correct answer is:
All above statements are correct

Solution:
All d-block elements are metals with variable oxidation state and form coloured ions and complex salts.
Hence, Option "4" is the correct answer.

Topic: Chemistry, d and f Block Elements, JEE Main
The correct statement in respect of protein haemoglobin is that it:

Options:
A function as a catalyst for biological reactions
B maintains blood sugar level
C acts as an oxygen carrier in the blood
D forms antibodies and offers resistance to diseases

The correct answer is:
acts as an oxygen carrier in the blood

Solution:
Haemoglobin in blood carries oxygen from the lungs or gills to the rest of the body (i.e. the tissues). There it releases the oxygen to permit aerobic respiration to provide energy to power the functions of the organism in the process called metabolism.
Related Theory
Four $\mathrm{Fe}^{2+}$ ions of each haemoglobin can bind with 4 molecules of $\mathrm{O}_2$ and it is termed as oxyhaemoglobin.
$$
\mathrm{Hb}_4+4 \mathrm{O}_2 \rightarrow \mathrm{Hb}_4 \mathrm{O}_8
$$

Topic: Chemistry, Biomolecules, NEET, NEET 2004
The correct statement is

Options:
A leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium silicate
B the blistered appearance of copper during the metallurgical process is due to the evolution of CO2
C pig iron is obtained from cast iron
D the Hall-Heroult process is used for the production of aluminium and iron

The correct answer is:
leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium silicate

Solution:
(1) During leaching when bauxite is treated with concentrated NaOH, then sodium aluminate and sodium silicate is formed in the soluble form, whereas Fe2O3 is precipitated.
(2) The blistered appearance of copper during the metallurgical process is due to the evoluation of SO2 .
Cu2S+2Cu2Oâ6Cu+SO2
(3) Cast iron is obtained from pig iron
Pig iron4% âblast of air cast iron3%
(4) Hall-Heroult process is used for production of only aluminium.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The correct statement is:

Options:
A Aniline is a froth stabilizer.
B Zone refining process is used for the refining of titanium.
C Zincite is a carbonate ore.
D Sodium cyanide cannot be used in the metallurgy of silver.

The correct answer is:
Aniline is a froth stabilizer.

Solution:
A Aniline is the forth stabilizer.
B Zone refining process is not used for refining of Ti
C Zincite is ZnO
D Sodium cyanide is used in the metallurgy of silver

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The correct statement is:

Options:
A Pig iron is obtained from cast iron.
B The Hall-Heroult process is used for the production of aluminium and iron.
C leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium silicate.
D the blistered appearance of copper during the metallurgical process is due to the evolution of CO2.

The correct answer is:
leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium silicate.

Solution:
(1) During leaching when bauxite is treated with concentrated NaOH, then sodium aluminate and sodium silicate is formed in the soluble form, whereas Fe2O3 is precipitated
(2) The blistered appearance of copper during the metallurgical process is due to evolution of SO2.
(3) Cast iron is obtained from pig iron.
(4) Hall-Heroult process is used for production of only aluminium.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The correct statement is

Options:
A Potassium dichromate is more soluble than sodium dichromate
B All Cr-O bond lengths in dichromate ion are equal.
C Potassium dichromate is used as a primary standard in volumetric titrations
D All are correct

The correct answer is:
Potassium dichromate is used as a primary standard in volumetric titrations

Solution:
Potassium dichromate is widely used in redox titrations as primary standard.

Topic: Chemistry, d and f Block Elements, JEE Main
The correct statement on the isomerism associated with the following complex ions.

(a) NiH2O5NH32+

(b)NiH2O4NH322+ and

(c) NiH2O3NH332+ is:

Options:
A (a) and (b) show only geometrical isomerism
B (a) and (b) show geometrical and optical isomerism
C (b) and (c) show geometrical and optical isomerism
D (b) and (c) show only geometrical isomerism

The correct answer is:
(b) and (c) show only geometrical isomerism

Solution:
NiH2O5NH32+ does not show any isomerism
NiH2O4NH322+ shows cis-trans (geometrical) isomerism
NiH2O3NH332+ shows fac-mer isomerism

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2015 (10 Apr Online)
The correct statement on the isomerism associated with the following complex ions.

(a) NiH2O5NH32+

(b)NiH2O4NH322+ and

(c) NiH2O3NH332+ is:

Options:
A (a) and (b) show only geometrical isomerism
B (a) and (b) show geometrical and optical isomerism
C (b) and (c) show geometrical and optical isomerism
D (b) and (c) show only geometrical isomerism

The correct answer is:
(b) and (c) show only geometrical isomerism

Solution:
NiH2O5NH32+ does not show any isomerism
NiH2O4NH322+ shows cis-trans (geometrical) isomerism
NiH2O3NH332+ shows fac-mer isomerism

Topic: Chemistry, Coordination Compounds, JEE Main
The correct statement regarding B is -

Options:
A It is more reactive than CH3-CHO
B When treated with Con H2SO4+Conc.HNO3/Δ produces para product as a major product
C It gives Benzyl alcohol and Benzoic acid with 50% NaOH/Δ
D It produces Cyanohydrin with KCN/C2H5OH/Δ

The correct answer is:
It gives Benzyl alcohol and Benzoic acid with 50% NaOH/Δ

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is ;

Options:
A Retention occurs in SNl reaction and inversion occurs in SN2 reaction.
B Racemisation occurs in SNl reaction and retention occurs in SN2 reaction.
C Racemisation occurs in both SN1 and SN2 reactions.
D Racemisation occurs in SN1 reaction and inversion occurs in SN2 reaction.

The correct answer is:
Racemisation occurs in SN1 reaction and inversion occurs in SN2 reaction.

Solution:
In the 1st step of SN1 mechanism a carbocation is formed which is planar .Therefore the nucleophile can attack the carbocation from both the sides. When a nucleophile attacks the carbocation from the side where initially halogen was there,the product with retention in configuration is produce and when nucleophile attacks the carbocation from backside, the product with inversion in configuration is produce. Hence 50% of the product will be with retention in configuration and rest 50% will be with inversion in configuration. This produces a racemic mixture which is optically inactive.

In SN2 reaction mechanism requires the attack of nucleophile from the back side of the carbon atom. So the product assumes a stereochemical position opposite to the leaving group originally occupied. This is called inversion of configuration.

SN1 – Racemisation

SN2 – Inversion

Hence option D is the answer.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The correct statement regarding the given Ellingham diagram is:

Options:
A At 1400oC,Al can be used for the extraction of Zn from ZnO .
B At 500oC, coke can be used for the extraction of ZnÂ from ZnO
C At 800oC,Cu can be used for the extraction of ZnÂ from ZnO
D Coke cannot be used for the extraction of CuÂ from Cu2O

The correct answer is:
At 1400oC,Al can be used for the extraction of Zn from ZnO .

Solution:
According to the Ellingham diagram, the species that is going to act as reducing agent should lie below the substance to be reduced at the particular temperature.
As is visible from the graph, at 1400o C, the line of Al lies below the line of ZnO. Hence, Al can be used for reducing ZnO to Zn.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The correct statement with regard to $\mathrm{H}_{2}^{+}$and $\mathrm{H}_{2}^{-}$ is

Options:
A both $\mathrm{H}_{2}^{+}$and $\mathrm{H}_{2}^{-}$are equally stable
B both $\mathrm{H}_{2}^{+}$and $\mathrm{H}_{2}^{-}$do not exist
C $\mathrm{H}_{2}^{-}$is more stable than $\mathrm{H}_{2}^{+}$
D $\mathrm{H}_{2}^{+}$is more stable than $\mathrm{H}_{2}^{-}$

The correct answer is:
$\mathrm{H}_{2}^{+}$is more stable than $\mathrm{H}_{2}^{-}$

Solution:
$\mathrm{H}_{2}^{+}: \sigma 1 \mathrm{~s}^{1}, \stackrel{*}{\sigma} 1 \mathrm{~s}^{0}$
Bond order $=\frac{1}{2}$
$\mathrm{H}_{2}^{-}: \sigma 1 \mathrm{~s}^{2}, \stackrel{*}{\sigma} 1 \mathrm{~s}^{1}$
Bond order $=\frac{2-1}{2}=\frac{1}{2}$
The bond order of $\mathrm{H}_{2}^{+}$and $\mathrm{H}_{2}^{-}$are same but $\mathrm{H}_{2}^{+}$is more stable than $\mathrm{H}_{2}^{-}$. It is due to the presence of one electron in the antibonding molecular orbital in $\mathrm{H}_{2}^{-}$.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The correct statements about H2O2 are :

(A) used in the treatment of effluents.

(B) used as both oxidising and reducing agents.

(C) the two hydroxyl groups lie in the same plane.

(D) miscible with water.

Choose the correct answer from the options given below :

Options:
A (A), (B), (C) and (D)
B (A), (B) and (D) only
C (B), (C) and (D) only
D (A), (C) and (D) only

The correct answer is:
(A), (B) and (D) only

Solution:
Structure of H2O2

(Open book type) → Non-planar

H2O2 is used in the treatment of effluents.

act as both O.A. and R.A.

H2O2 is miscible in water due to hydrogen bonding.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The correct statements among a to d are:
a Saline hydrides produce H2 gas when reacted with H2O.
b Reaction of LiAlH4 with BF3 leads to B2H6 .
c PH3 and CH4 are electron - rich and electron - precise hydrides, respectively.
d HF and CH4 are called as molecular hydrides.

Options:
A a,Â b and c only.
B a,Â c and d only.
C c and d only.
D a, b, c and d

The correct answer is:
a, b, c and d

Solution:
(1) Saline hydrides are ionic hydrides such as LiH, NaH. When reacting with water to form a Base and hydrogen gas.
MHs+H2Oaq ⟶MOHaq+H2
(2) BF3 when reacts with LiAlH4 produces B2H6
(3)
(4) Molecular hydrides are also called covalent hydrides. HF and CH4 are covalent or molecular hydrides.
Hence, all statements are correct.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The correct statements among (a) to (d) regarding $\mathrm{H}_{2}$ as a fuel are :

(i) It produces less pollutants than petrol.

(ii) A cylinder of compressed dihydrogen weighs $\sim 30$ times more than a petrol tank producing the same amount of energy.

(iii) Dihydrogen is stored in tanks of metal alloys like $\mathrm{NaNi}_{5}$

(iv) On combustion, values of energy released per gram of liquid dihydrogen and LPG are 50 and $142 \mathrm{~kJ},$ respectively.

Options:
A $(ii) and (iv) only$
B $(i) and (iii) only$
C $(ii), (iii) and (iv) only$
D $(i), (ii) and (iii) only$

The correct answer is:
$(i), (ii) and (iii) only$

Solution:
option (i), (ii) and (iii) are correct.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The correct statements among I to III are:
(I) Valence bond theory cannot explain the color exhibited by transition metal complexes.
(II) Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes.
(III) Valence bond theory cannot distinguish ligands as weak and strong field ones.

Options:
A (I) and (II) only
B (II) and (III) only
C (I) and (III) only
D (I), (II) and (III)

The correct answer is:
(I) and (II) only

Solution:
Valence Bond Theory cannot explain the colour of transition metal complexes .this theory was also unable to distinguish between weak and strong Fields ligands. Valence Bond Theory does not explain and give quantitative interpretation of magnetic data. Hence, statement (II) is not right.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The correct statements among I to III regrading group 13 element oxides are,
(I) Boron trioxide is acidic.
(II) Oxides of aluminum and gallium are amphoteric.
(III) Oxides of indium and thallium are basic.

Options:
A (I) and (III) only
B (I) and (II) only
C (I), (II) and (III)
D (II) and (III) only

The correct answer is:
(I), (II) and (III)

Solution:
All statements are correct
B2O3→ acidic , B2O3 is an anhydride of Orthoboric Acid.
Al2O3 and Ga2O3 are amphoteric, amphoteric oxides can react with acids as well as bases.
oxides of ln and Tl are basic.
As we move down the group basicity of oxides increases due to increase in metallic character of elements

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2019 (09 Apr Shift 2)
The correct statement(s) among I to III with respect to potassium ions that are abundant within the cell fluids, is/are
I. They activate many enzymes.
II. They participate in the oxidation of glucose to produce ATP.
III. Along with sodium ions, they are responsible for the transmission of nerve signals.

Options:
A I, II and III
B III only
C I and II only
D I and III only

The correct answer is:
I, II and III

Solution:
I. They activate many enzymes.
II. They participate in the oxidation of glucose to produce ATP.
III. Along with sodium ions, they are responsible for the transmission of nerve signals.
All are correct.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The correct structure of product 'A' formed in the following reaction,

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The correct structure of the dipeptide gly-ala is

Options:
A
B
C
D

The correct answer is:

Solution:
The structure of dipeptide gly-ala is

Topic: Chemistry, Biomolecules, JEE Main
The correct truth table for the following logic circuit is :

Options:
A $\begin{array}{|l|l|l|}
\hline A & B & Y \\
\hline 0 & 0 & 1 \\
0 & 1 & 1 \\
1 & 0 & 0 \\
1 & 1 & 1 \\
\hline
\end{array}$
B $\begin{array}{|l|l|l|}
\hline A & B & Y \\
\hline 0 & 0 & 0 \\
0 & 1 & 1 \\
1 & 0 & 0 \\
1 & 1 & 1 \\
\hline
\end{array}$
C $\begin{array}{|l|l|l|}
\hline A & B & Y \\
\hline 0 & 0 & 1 \\
0 & 1 & 1 \\
1 & 0 & 0 \\
1 & 1 & 0 \\
\hline
\end{array}$
D $\begin{array}{|l|l|l|}
\hline A & B & Y \\
\hline 0 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
1 & 1 & 1 \\
\hline
\end{array}$

The correct answer is:
$\begin{array}{|l|l|l|}
\hline A & B & Y \\
\hline 0 & 0 & 1 \\
0 & 1 & 1 \\
1 & 0 & 0 \\
1 & 1 & 1 \\
\hline
\end{array}$

Solution:

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (06 Apr Shift 1)
The cosine of the acute angle between the curves y=x2-1 and y=x2-3 at their points of intersection is

Options:
A 13
B 79
C 1192
D 27

The correct answer is:
79

Solution:
For points of intersection P & Q, solving the curves x2-1=x2-3, we get,
⇒x2-1=3-x2
⇒x=±2⇒P2,1 & Q-2,1
For the slope of tangents at x=2
C1 is y=x2-1
dydx=2x
⇒m1=22
And C2 is y=3-x2
dydx=−2x
⇒m2=-22
∴tan⁡θ=m1-m21+m1m2
=421-8=427
∴cos⁡θ=79

Topic: Mathematics, Application of Derivatives, JEE Main
The cost at \( 5 \) paise / \( \mathrm{kWh} \) of operating an electric motor for \( 8 \) hours which takes \( 15 \) ampere at \( 110 \mathrm{~V} \) is
[Hint: \( \mathrm{W}=\mathrm{I} . \mathrm{V} \) ]

Options:
A Rs \( 66 \)
B \( 66 \) paise
C \( 37 \) paise
D Rs \( 6.60 \)

The correct answer is:
\( 66 \) paise

Solution:
We know that,

Energy = Power×time

and Power = Current (I)×Voltage (V)

Energy consumed during time t hours=I×V×t

Given,

Current, I=15 ampere

Voltage, V=110 V

Time, t=8 hours

Substituting in Energy =I×V×t, we get,

Energy=15×110×8=13200 Wh=13.2 kWh

Cost of operating =5 paise/kWh

Cost of operating 13.2 kWh=5×13.2=66 paise

Topic: Chemistry, Electrochemistry, JEE Main
The cost of running a bus from $A$ to $B$, is $₹\left(a v+\frac{b}{v}\right)$, where $v \mathrm{~km} / \mathrm{h}$ is the average speed of the bus. When the bus travels at $30 \mathrm{~km} / \mathrm{h}$, the cost comes out to be $₹ 75$ while at $40 \mathrm{~km} / \mathrm{h}$, it is $₹ 65$. Then the most economical speed (in $\mathrm{km} / \mathrm{h}$ ) of the bus is :

Options:
A
45

B
50

C
60

D
40

The correct answer is:
60

Solution:
Let $\operatorname{cost} \mathrm{C}=a v+\frac{b}{v}$
According to given question,
$$
\begin{aligned}
& 30 a+\frac{b}{30}=75 \\
& 40 a+\frac{b}{40}=65
\end{aligned}
$$
On solving (i) and (ii), we get $a=\frac{1}{2}$ and $b=1800$
Now, $C=a v+\frac{b}{v}$
$$
\begin{aligned}
& \Rightarrow \frac{d C}{d v}=a-\frac{b}{v^2} \\
& \frac{d C}{d v}=0 \Rightarrow a-\frac{b}{v^2}=0 \\
& \Rightarrow v=\sqrt{\frac{b}{a}}=\sqrt{3600} \\
& \Rightarrow v=60 \mathrm{kmph}
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2013 (23 Apr Online)
The counter current mechanism involves the transport of ___________ to maintain the concentration of _________ in medullary interstitium.

Options:
A NaCl and urea
B KCl and glucose
C NaCl and water
D urea and water

The correct answer is:
NaCl and urea

Solution:
Concentrating urine is very important because this will prevent huge water loss and salt imbalance through our bodies. This is done through a counter-current mechanism that operates in the kidneys. This takes place because the loop of Henle and vasa recta are parallel to each other but the flow within them are in opposite direction. This opposite flow is counter current. This causes movement of Nacl and urea and thus water and salt balance is maintained.

Topic: Biology, Excretory Products and their Elimination, NEET
The counting rate observed from a radioactive source at $t=0$ was 1600 counts $\mathrm{s}^{-1}$, and $t=8 \mathrm{~s}$, it was 100 counts $\mathrm{s}^{-1}$. The counting rate observed as counts $\mathrm{s}^{-1}$ at $t=6 \mathrm{~s}$ will be

Options:
A
250

B
400

C
300

D
200

The correct answer is:
200

Solution:
As we know,
$$
\begin{aligned}
& {\left[\frac{N}{N_0}\right]=\left[\frac{1}{2}\right]^n} \\
& n=\text { no. of half life } \\
& N \text { - no. of atoms left } \\
& N_0 \text { - initial no. of atoms } \\
& \text { By radioactive decay law, } \\
& \frac{d N}{d \mathrm{t}}=k N \\
& k-\operatorname{disin} t e g r a t i o n \text { constant } \\
& \therefore \frac{\frac{d N}{d t}}{\frac{d N_0}{d t}}=\frac{N}{N_0}
\end{aligned}
$$
From (i) and (ii) we get
$$
\frac{\frac{d N}{d t}}{\frac{d N_0}{d t}}=\left[\frac{1}{2}\right]^n
$$
or, $\left[\frac{100}{1600}\right]=\left[\frac{1}{2}\right]^n \Rightarrow\left[\frac{1}{2}\right]^4=\left[\frac{1}{2}\right]^n$
$\therefore n=4$, Therefore, in 8 seconds 4 half life had occurred in which counting rate reduces to 100 counts $\mathrm{s}^{-1}$.
$\therefore$ Half life, $\frac{T_1}{2}=2 \mathrm{sec}$
In $6 \mathrm{sec}, 3$ half life will occur
$$
\begin{aligned}
& \therefore\left[\frac{\frac{d \mathrm{~N}}{d t}}{1600}\right]=\left[\frac{1}{2}\right]^3 \\
& \Rightarrow \frac{d \mathrm{~N}}{d t}=200 \text { counts } \mathrm{s}^{-1}
\end{aligned}
$$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2012 (26 May Online)
The covalency and oxidation state respectively of boron in BF4-, are

Options:
A 3 and 5
B 3 and 4
C 4 and 4
D 4 and 3

The correct answer is:
4 and 3

Solution:
Number of covalent bond formed by Boron is 4.

BF4- Covalency =4

The oxidation state of an element represents the charge it would have if all the bonding electrons were assigned to the more electronegative atom in the bond.

Oxidation number of fluorine is -1

Then,

B + 4 x (-1) = -1B-4=-1B=+3

Oxidation state =+3 for Boron

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (13 Apr Shift 2)