datavorous/entrance-exam-dataset huggingface dataset:
Radon- 222 has a half-life of 3.8 days. If one starts with $0.064 \mathrm{kg}$ of radon-222, the quantity of radon- 222 left after 19 days will be

Options:
A $0.002 \mathrm{kg}$
B $0.062 \mathrm{kg}$
C $0.032 \mathrm{kg}$
D $0.024 \mathrm{kg}$

The correct answer is:
$0.002 \mathrm{kg}$

Solution:
Half life of radon $=3 \cdot 8$ days
Total half-lives in 19 days
$\begin{array}{l}
=\frac{19}{3 \cdot 8} \\
=\frac{190}{38}
\end{array}$
$=5$ half-lives
Initially mass of radon $=0.064 \mathrm{kg}$
$0 \cdot 064 \frac{T_{1}}{T_{3}} 0 \cdot 032 \stackrel{T_{2}}{\longrightarrow}$
$0 \cdot 016 \stackrel{T_{4}}{\longrightarrow} 0 \cdot 004 \stackrel{T_{5}}{\longrightarrow} 0 \cdot 002$
So in five half-life periods, the Radon sample will reduce to $0.002 \mathrm{kg}$
$\text {Or } \frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{5}$
$\Rightarrow \quad N=0 \cdot 064 \times \frac{1}{32}$
$=0 \cdot 002 \mathrm{kg}$

Topic: Physics, Nuclear Physics, JEE Main
Raffinose is

Options:
A Trisaccharide
B Monosaccharide
C Disaccharide
D None of these

The correct answer is:
Trisaccharide

Solution:
When carbohydrates on hydrolysis yield more than two to ten monosaccharides, they are known as oligosaccharides. Therefore trisaccharides are basically oligosaccharides.
Raffinose $(\mathrm{C} 18 \mathrm{H} 32 \mathrm{O} 16)$ is a trisaccharide as it contains 3 monosaccharide units.
On hydrolysis it gives glucose+fructose+galactose. It is mostly found in beans, cabbage, broccoli etc. Refer the above figure for the structure of raffinose.

Topic: Chemistry, Biomolecules, JEE Main
Rain, driven by the wind, falls on a railway compartment with a velocity of 20 m s-1, at an angle of 30Â° to the vertical. The train moves, along the direction of wind flow, at a speed of 108 km hr-1. The apparent velocity of rain for a person sitting in the train is

Options:
A 207m s-1
B 107m s-1
C 157m s-1
D 107km h-1

The correct answer is:
107m s-1

Solution:
Speed of train = 1 0 8 Ã 5 1 8 = 3 0 m/s
Let vâ¡ â Râ¡ and vâ¡ â Tâ¡ represent the respective velocities of rain and train.
Now, the relative velocity of rain w.r.t person (train) is given by vâ¡ R, T = vâ¡ â Râ¡ - vâ¡ â Tâ¡ â vâ¡ â Râ¡ + - vâ¡ â Tâ¡
Let OR â and RT â represent the vectors in magnitude and direction.
OT2 = OR2 + RT2 + 2OR Ã RT cos 120o
= 2 0 2 + 3 0 2 - 2 Ã 2 0 Ã 3 0 Ã 1 2
= 4 0 0 + 9 0 0 - 6 0 0 = 7 0 0 = 7 0 0 m/s

Topic: Physics, Motion In Two Dimensions, JEE Main
Rain droplets are falling in vertically downward direction with velocity 5 m/s. A cyclist is moving in northward direction with velocity 10 m/s. The rain droplets will appear to the cyclist to be coming from

Options:
A tan-1â¡2Â above south horizon
B tan-1â¡12 above north horizon
C Vertically downward
D tan-1â¡2Â above north horizon

The correct answer is:
tan-1â¡12 above north horizon

Solution:
tan θ= v R v M
tanâ¡Î¸=510
Î¸=tan-1â¡12 above north horizon.

Topic: Physics, Motion In Two Dimensions, JEE Main
Rain drops fall vertically at a speed of 20 m s-1. At what angle (in degree) do they fall on the wind screen of a car moving with a velocity 15 m s-1, if the wind screen is inclined at an angle of 23Â° to the vertical? [tan-10.75â37Â°]

Options:

The correct answer is:
60

Solution:
tan(90°-θ)=2015
∴cotθ=2015=43
⇒ θ=37°
∴ ϕ=37°+23°=60° (Angle made with the windscreen)

Topic: Physics, Motion In Two Dimensions, JEE Main
Rain is falling vertically with a speed of \( 12 \mathrm{~m} \mathrm{~s}^{-1} \). A woman rides a bicycle with a speed of \( 12 \mathrm{~m} \mathrm{~s}^{-1} \) in east to west direction. What is the direction in which she should hold her umbrella?

Options:
A \( 30^{\circ} \) towards West
B \( 45^{\circ} \) towards West
C \( 30^{\circ} \) towards East
D \( 45^{\circ} \) towards East

The correct answer is:
\( 45^{\circ} \) towards West

Solution:
To protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity of the rain with respect to the woman.
tanθ=vbvr=1

or, tanθ=1

or, θ=45°

Therefore, the direction in which she should hold her umbrella is 45° towards west.

Topic: Physics, Motion In Two Dimensions, JEE Main
Raising the temperature of an equilibrium system

Options:
A Faxeuss the exothermic reaction only
B Faxouss the endothermic reaction only
C Faveurs both the exothermic and endothermic reactions
D Faveurs neither the exothermic nor endothermic reactions

The correct answer is:
Faxouss the endothermic reaction only

Solution:
Favouss the endothermic reaction only

Topic: Chemistry, Chemical Equilibrium, JEE Main
Rajesh has just bought a VCR from Maharashtra Electronics and the shop offers after sales service contract for Rs. 1000 for the next five years. Considering the experience of VCR users, the following distribution of maintenance expenses for the next five years is formed.
The expected value of maintenance cost is

Options:
A Rs. 800
B Rs. 770
C Rs. 700
D Rs. 900

The correct answer is:
Rs. 770

Solution:
$\begin{aligned} & \text { Expected value }=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} \\ & =(0)(0.35)+(500)(0.25)+(1000)(0.15)+(1500)(0.1) \\ & +(2000)(0.08)+(2500)(0.05)+(3000)(0.02) \\ & =0+125+150+150+160+125+60=770\end{aligned}$

Topic: Mathematics, Probability, MHT CET, MHT CET 2021 (20 Sep Shift 1)
Range of $f(x)=\frac{x^2+34 x-71}{x^2+2 x-7}$ is

Options:
A $[5,9]$
B $(-\infty, 5] \cup[9, \infty)$
C $(5,9)$
D None of these

The correct answer is:
$(-\infty, 5] \cup[9, \infty)$

Solution:
$\begin{aligned}
& \text { Let } \frac{x^2+34 x-71}{x^2+2 x-7}=y \\
& \Rightarrow x^2(1-y)+2(17-y) x+(7 y-71)=0
\end{aligned}$
For real value of $x, B^2-4 A C \geq 0$
$\Rightarrow y^2-14 y+45 \geq 0 \Rightarrow y \geq 9, y \leq 5$

Topic: Mathematics, Functions, JEE Main
Range of $f(x)=[x]-x$ is

Options:
A $[0,1]$
B $(-1,0]$
C $R$
D $(-1,1)$

The correct answer is:
$(-1,0]$

Solution:
As shown in graph
$\Rightarrow$ Range is $(-1,0]$

Topic: Mathematics, Functions, JEE Main
Range of the function $\frac{1}{2-\sin 3 x}$ is

Options:
A $[1,3]$
B $\left[\frac{1}{3}, 1\right]$
C $(1,3)$
D $\left(\frac{1}{3}, 1\right)$

The correct answer is:
$\left[\frac{1}{3}, 1\right]$

Solution:
$f(x)=\frac{1}{2-\sin 3 x}, \sin 3 x \in[-1,1]$
Hence $f(x)$ lies in $\left[\frac{1}{3}, 1\right]$.

Topic: Mathematics, Functions, JEE Main
Range of the function $f(x)=9-7 \sin x$ is

Options:
A $(2,16)$
B $[2,16]$
C $[-1,1]$
D $(2,16]$

The correct answer is:
$[2,16]$

Solution:
$y=f(x)=9-7 \sin x$. Range $=[2,16]$.

Topic: Mathematics, Functions, JEE Main
Range of the function $f(x)=\sin ^2\left(x^4\right)+\cos ^2\left(x^4\right)$ is

Options:
A $(-\infty, \infty)$
B $\{1\}$
C $(-1,1)$
D $(0,1)$

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

Solution:
$f(x)=\sin ^2\left(x^4\right)+\cos ^2\left(x^4\right)=1$. Hence range $=\{1\}$.

Topic: Mathematics, Functions, JEE Main
Range of the function $f(x)=\frac{x^2}{x^2+1}$ is

Options:
A $(-1,0)$
B $(-1,1)$
C $[0,1)$
D $(1,1)$

The correct answer is:
$[0,1)$

Solution:
Let $y=\frac{x^2}{x^2+1}$
$\Rightarrow(y-1) x^2+0 x+y=1, y \neq 1$ for real values of $x$,
we have $D \geq 0 \Rightarrow-4 y(y-1) \geq 0 \Rightarrow y(y-1) \leq 0 \Rightarrow y \in[0,1)$
$0 \leq \frac{x^2}{x^2+1} \lt 1$

Topic: Mathematics, Functions, JEE Main
Range of the function $f(x)=\frac{x^2+x+2}{x^2+x+1} ; x \in R$ is

Options:
A $(1, \infty)$
B $(1,11 / 7]$
C $(1,7 / 3]$
D $(1,7 / 5]$

The correct answer is:
$(1,7 / 3]$

Solution:
$f(x)=1+\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}} \Rightarrow$ Range $=(1,7 / 3]$

Topic: Mathematics, Functions, JEE Main
Rank of matrix $\left[\begin{array}{cccc}4 & 1 & 0 & 0 \\ 3 & 0 & 1 & 0 \\ 6 & 0 & 2 & 0\end{array}\right]$ is

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

The correct answer is:
$2$

Solution:
$\left[\begin{array}{llll}4 & 1 & 0 & 0 \\ 3 & 0 & 1 & 0 \\ 0 & 0 &
0 & 0\end{array}\right]$ (By $R_3 \rightarrow R_3-2 R_2$ )
$=\left[\begin{array}{llll}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0\end{array}\right]$, (By $C_1 \rightarrow C_1-4 C_2-3 C_3$ )
$=\left[\begin{array}{llll}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\end{array}\right]$,
(Replace $C_1$ by $C_2$ and then Replace $C_2$ by $C_3$ )
Hence rank of matrix is 2 .

Topic: Mathematics, Matrices, JEE Main
Rank the following carbocations in order of stability (1 = most stable)

Options:
A A â 2, B â 1, C â 3, D â 4
B âA → 4, B → 2, C → 1, D → 3
C âA → 4, B → 3, C → 2, D → 1
D âA → 1, B → 2, C → 3, D → 4

The correct answer is:
A â 2, B â 1, C â 3, D â 4

Solution:
A -2 [3o carbocation with cross conjugation]
B - 1 [2o carbocation with extended conjugation ]
C - 3 [3o carbocation with 4 hyperconjugations and +I of 3 alkyl groups]
D - 4 [â2o carbocation with 2 hyperconjugations and +I of 2 alkyl groups]
So order will be
A â 2, B â 1, C â 3, D â 4

Topic: Chemistry, General Organic Chemistry, JEE Main
Rank the following compounds in decreasing order of reactivity in electrophilic aromatic substitution reaction

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

The correct answer is:
3 > 1 > 4 > 2

Solution:
Rate of electrophilic aromatic substitution is effected by group/s already present. If they -O-,-NH-,-CH2- are ring activating and ortho, para directing due to +R and +H effects reactivity is more but if they -CO-,-NO2,-CN are ring deactivating and meta directing groups due to -R and -H effects reactivity decreases.
Here reactivity order is 3 > 1 > 4 > 2.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Rate constant for a reaction $\mathrm{H}_2+\mathrm{I}_2 \rightarrow 2 \mathrm{HI}$ is 49 , then rate constant for reaction $2 \mathrm{HI} \rightarrow \mathrm{H}_2+\mathrm{I}_2$ is

Options:
A 7
B 1/49
C 49
D 21

The correct answer is:
1/49

Solution:
For reversible reaction rate constant is also reverse.

Topic: Chemistry, Chemical Kinetics, JEE Main
Rate law equation for a reaction is $R=k[x][y]$, rate of reaction doubles when

Options:
A conc. of $\mathrm{x}$ is kept constant and conc. of $\mathrm{y}$ is halved
B conc. of both $\mathrm{x}$ and $\mathrm{y}$ is doubled
C conc. of $\mathrm{x}$ is doubled and conc. of $\mathrm{y}$ is kept constant
D conc. of $\mathrm{y}$ is doubled and conc. of $\mathrm{x}$ is halved

The correct answer is:
conc. of $\mathrm{x}$ is doubled and conc. of $\mathrm{y}$ is kept constant

Solution:
$\mathrm{r}=\mathrm{k}(\mathrm{x})(\mathrm{y})$
Of $\mathrm{x}$ is doubled and $\mathrm{y}$ remains constant the rate is doubled.

Topic: Chemistry, Chemical Kinetics, MHT CET, MHT CET 2022 (06 Aug Shift 1)
Rate of a reaction can be expressed by Arrhenius equation as:
$$
\mathrm{k}=\mathrm{Ae}^{-\mathrm{E} / \mathrm{RT}}
$$
In this equation, E represents

Options:
A
the energy above which all the colliding molecules will react

B
the energy below which colliding molecules will not react

C
the total energy of the reacting molecules at a temperature, $T$

D
the fraction of molecules with energy greater than the activation energy of the reaction

The correct answer is:
the energy below which colliding molecules will not react

Solution:
No Solution Available

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2006
Rate of biomass production is expressed as:

Options:
A kcalÂ m-2Â Â yr-1
B kcal g m2 yr-1
C g m3 yr
D g m-2 yr

The correct answer is:
kcal m-2 yr-1

Solution:
The amount of biomass produced per unit area over a period of time is called primary productivity expressed in weight (g–2) or energy kcal m–2 . The rate at which this biomass is produced is called productivity. To compare the productivity of different ecosystems it is expressed by units g–2 yr –1 or kcal m–2 yr –1 . It is divided into gross primary productivity and net primary productivity. The rate at which organic matter is produced during photosynthesis is gross primary productivity and the remaining biomass after the respiratory loss is net primary productivity.

Topic: Biology, Ecosystem, NEET
Rate of dehydration of alcohols follows the order:

Options:
A
$2^{\circ}>1^{\circ}>\mathrm{CH}_3 \mathrm{OH}>3^{\circ}$

B
$3^{\circ}>2^{\circ}>1^{\circ}>\mathrm{CH}_3 \mathrm{OH}$

C
$2^{\circ}>3^{\circ}>1^{\circ}>\mathrm{CH}_3 \mathrm{OH}$

D
$\mathrm{CH}_3 \mathrm{OH}>1^{\circ}>2^{\circ}>3^{\circ}$

The correct answer is:
$3^{\circ}>2^{\circ}>1^{\circ}>\mathrm{CH}_3 \mathrm{OH}$

Solution:
The order of dehydration among three type of alcohols is $3^{\circ}>2^{\circ}>1^{\circ}>\mathrm{CH}_3 \mathrm{OH}$. This behaviour is related to the relative stabilities of carbocations $\left(3^{\circ}>2^{\circ}>1^{\circ}\right)$.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2013 (09 Apr Online)
Rate of physisorption increases with

Options:
A Decrease in temperature
B Increase in temperature
C Decrease in pressure
D Decrease in surface area

The correct answer is:
Decrease in temperature

Solution:
It is an exothermic process, according to Le-Chatelier’s principle, lowering temperature drive the process in forward direction.

Topic: Chemistry, Surface Chemistry, JEE Main
Rate of reaction (r) is plotted against temperature (T) for an enzyme catalysed reaction. Which of the following is correct representation?

Options:
A
B
C
D

The correct answer is:

Solution:
For enzyme catalyst reaction, upto optimum temperature rate increases and then decreases.

Topic: Chemistry, Surface Chemistry, JEE Main
Rate of the reaction

Options:
A
$\mathrm{Cl}$

B
$\mathrm{OCOCH}_3$

C
$\mathrm{OC}_2 \mathrm{H}_5$

D
$\mathrm{NH}_2$

The correct answer is:
$\mathrm{Cl}$

Solution:
Rate of reaction will be fastest when $\mathrm{Z}$ is $\mathrm{Cl}$ because it is a weakest base

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2004
Ratio among linear expansion coefficient ($\alpha$), areal expansion coefficient $(\beta)$ and volume expansion coefficient $(\gamma)$ is

Options:
A $1: 2: 3$
B $3: 2: 1$
C $4: 3: 2$
D None of these

The correct answer is:
$1: 2: 3$

Solution:
As $\alpha=\frac{\beta}{2}=\frac{\gamma}{3} \Rightarrow \alpha: \beta: \gamma=1: 2: 3$

Topic: Physics, Thermal Properties of Matter, JEE Main
Ratio of amplitude of interfering waves is $3: 4$. Now ratio of their intensities will be

Options:
A $\frac{16}{9}$
B $49: 1$
C $\frac{9}{16}$
D None of these

The correct answer is:
$\frac{9}{16}$

Solution:
$I \propto a^2 \Rightarrow \frac{I_1}{I_2}=\left(\frac{a_1}{a_2}\right)^2=\left(\frac{3}{4}\right)^2=\frac{9}{16}$

Topic: Physics, Wave Optics, JEE Main
Ratio of $C_p$ and $C_v$ of a gas ' $X$ is 1.4. The number of atoms of the gas ' $X$ present in 11.2 litrers of it at N.T.P. is

Options:
A $6.02 \times 10^{23}$
B $1.2 \times 10^{24}$
C $3.01 \times 10^{23}$
D $2.01 \times 10^{23}$

The correct answer is:
$6.02 \times 10^{23}$

Solution:
Since $\frac{C_p}{C_V}=1.4$, the gas should be diatomic.
If volume is 11.2
then, no. of moles $=\frac{1}{2}$
$\therefore$ no. of molecules $=\frac{1}{2} \times$ Avagadro's No.
no. of atoms $=2 \times$ no. of molecules
$2 \times \frac{1}{2} \times$ Avagadro's No.
$=6.0223 \times 10^{23}$

Topic: Chemistry, States of Matter, JEE Main
Ratio of centripetal acceleration for an electron revolving in $3^{\text {rd }}$ orbit to $5^{\text {th }}$ Bohr orbit
of hydrogen atom is

Options:
A $\frac{424}{21}$
B $\frac{625}{81}$
C $\frac{125}{4}$
D $\frac{775}{61}$

The correct answer is:
$\frac{625}{81}$

Solution:
Centripetal acceleration $\mathrm{a}=\frac{\mathrm{v}^{2}}{\mathrm{r}}$
$v \propto \frac{1}{n}$ and $r \alpha n^{2}$
$\therefore a=\frac{1}{n^{4}}$
$\therefore \frac{a_{5}}{a_{3}}=\frac{(3)^{4}}{(5)^{4}}=\frac{81}{125}$

Topic: Physics, Atomic Physics, JEE Main
Ratio of radii of second and first Bohr orbits of $\mathrm{H}$ atom

Options:
A 2
B 4
C 3
D 5

The correct answer is:
4

Solution:
Bohr radius $=\frac{r_2}{r_1}=\frac{(2)^2}{(1)^2}=4$

Topic: Chemistry, Structure of Atom, JEE Main
Ratio of Î T b K b of 10 g AB2 and 14 g A2B per 100 g of solvent in their respective, solution (AB2 and A2B both are non-electroytes) is 1 mol/kg in both cases. Hence, atomic wt. of A and B are respectively :

Options:
A 100, 40
B 60, 20
C 20, 60
D None of these

The correct answer is:
60, 20

Solution:
Δ T b = K b w solute W solvent × 1 M solute × 1 0 0 0
1 = 1 0 0 0 × 1 0 1 0 0 × M AB 2 ⇒ M AB 2 = 1 0 0
Similarly M A 2 B = 1 4 0
1 0 0 = M A + 2 · M B
and 1 4 0 = 2 · M A + M B ;
M A = 6 0 and M B = 2 0

Alternative solution

Δ T b K b = 1 = 1 0 M A + 2 M B 0.1
M A + 2 M B = 1 0 0 ........ (1)
Δ T b K b = 1 = 1 4 2M A + M B 0.1
2 M A + M B = 1 4 0 ........ (2)
M A = 60, M B = 2 0

Topic: Chemistry, Solutions, JEE Main
Ratio of thermal energy released in two resistor R and 3R connected in parallel in an electric circuit is :

Options:
A 3:1
B 1:1
C 1:3
D 1:27

The correct answer is:
3:1

Solution:
In parallel connection, potential difference across resistors remains the same.

Thermal energy released in resistor is given by H=V2R×t

So, the ratio is H1H2=V2tRV2t3R=3RR=3:1.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Ravi and Rashmi are each holding 2 red cards and 2 black cards (all four red and all four black cards are identical). Ravi picks a card at random from Rashmi and then Rashmi picks a card to random from Ravi. This process is repeated a second time. Let p be the probability that both have all 4 cards of the same colour. Then, p satisfies

Options:
A p≤5%
B 5%<p≤10%
C 10%<p≤15%
D 15%<p

The correct answer is:
p≤5%

Solution:
It is given that, Ravi and Rashmi are each holding 2 red cards and 2 black cards (all four red and all four black cards are identical).

Let Ravi picks a Red card from Rashmi and Rashmi picks black card from Ravi and in second time also Ravi picks a red card from Rashmi and Rashmi picks black card from Ravi, then finally Ravi have four red cards and Rashmi have four black cards, then probability of such event is

C12C14×C12C15×C11C14×C11C15=4400=1100

Similarly, the probability if at the end Ravi have four black and Rashmi have four red cards is 1100.

So, required probability =150=2%=p≤5%

Topic: Mathematics, Probability, JEE Main
Ray diverging from a point source form a wave front that is

Options:
A Cylindrical
B Spherical
C Plane
D Cubical

The correct answer is:
Spherical

Solution:

Topic: Physics, Wave Optics, JEE Main
Rayon is

Options:
A Natural silk
B Artificial silk
C Regenerated fibre
D Synthetic fibre

The correct answer is:
Regenerated fibre

Solution:
Rayon is regenerated fibre.

Topic: Chemistry, Polymers, JEE Main
RBC and a plant cell (with thick cell wall) are placed in distilled water. The solute concentration is the same in both the cells. What changes would be observed in them?

Options:
A Both plant cell and RBC would not undergo any change
B The RBC would increase in size and burst, while the plant cell would remain about the same size
C The plant cell would increase in size and burst, while the RBC would remain about the same size
D Both plant cell and RBC would decrease in size and collapse

The correct answer is:
The RBC would increase in size and burst, while the plant cell would remain about the same size

Solution:
When RBC and a plant cell are placed in distilled water, endosmosis takes place as a result of which RBC would increase in size and burst, while the plant cell would remain about the same size because of the presence of rigid cell wall.

Topic: Biology, Transport in Plants, NEET
$R C O O H$ on treatment with $\mathrm{PCl}_s$ and $\mathrm{KCN}$, is subjected to hydrolysis followed by Clemmensen's reduction, the product obtained is

Options:
A $\mathrm{RCH}_2-\mathrm{COCl}$
B $\mathrm{RCH}_2-\mathrm{COOH}$
C $\mathrm{RCOCN}$
D $\mathrm{RCN}$

The correct answer is:
$\mathrm{RCH}_2-\mathrm{COOH}$

Solution:
\(\mathrm{RCOOH}+\mathrm{PCl}_5 \rightarrow \mathrm{RCOCl}+\mathrm{POCl}_3+\mathrm{HCl} \xrightarrow{\mathrm{KCN}} \mathrm{RCOCN}+\mathrm{KCl} \xrightarrow{2 \mathrm{H}_2 \mathrm{O}} \mathrm{RCOCOOH}+2 \mathrm{NH}_3 \xrightarrow[\mathrm{Zn}-\mathrm{Hg} / \text { Conc. } \mathrm{HCl}]{\text { Clemenson reduction }} \mathrm{R} \mathrm{CH}_2 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O}\)

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\mathrm{RCOOH} \longrightarrow \mathrm{RCH}_2 \mathrm{OH}$
This mode of reduction of an acid to alcohol can be affected only by

Options:
A $\mathrm{Zn} / \mathrm{HCl}$
B $N a$-alcohol
C Aluminium isopropoxide and isopropyl alcohol
D $\mathrm{LiAlH}_4$

The correct answer is:
$\mathrm{LiAlH}_4$

Solution:
$\mathrm{RCOOH} \xrightarrow{\mathrm{LIAlH}_4} \mathrm{RCH}_2 \mathrm{OH}$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\operatorname{Re} \frac{(1+i)^2}{3-i}=$

Options:
A $-1 / 5$
B $1 / 5$
C $1 / 10$
D $-1 / 10$

The correct answer is:
$-1 / 5$

Solution:
$\begin{aligned} & \operatorname{Re}\left[\frac{(1+i)^2}{3-i}\right]=\operatorname{Re}\left[\left(\frac{2 i}{3-i}\right)\left(\frac{3+i}{3+i}\right)\right] \\ & \operatorname{Re}\left[\frac{(1+i)^2}{3-i}\right]=\operatorname{Re}\left[\left(\frac{2 i}{3-i}\right)\left(\frac{3+i}{3+i}\right)\right] \\ & =\operatorname{Re}\left[\frac{6 i-2}{9+1}\right]=\operatorname{Re}\left[-\frac{2}{10}+\frac{6}{10} i\right]=-\frac{1}{5}\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
Reactant A forms two products:
\( \mathrm{A} \stackrel{\mathrm{k}_{1}}{\rightarrow} \mathrm{B} \), Activation energy \( \mathrm{E}_{\mathrm{a}_{1}} ; \mathrm{A} \stackrel{\mathrm{k}_{2}}{\rightarrow} \mathrm{C} \), Activation energy \( \mathrm{E}_{\mathrm{a}_{2}} \)
\( \mathrm{E}_{\mathrm{a}_{2}}=2 \mathrm{E}_{\mathrm{a}_{1}} \), then \( \mathrm{k}_{1} \) and \( \mathrm{k}_{2} \) will be related as:

Options:
A \( \mathrm{k}_{2}=\mathrm{k}_{1} \mathrm{e}^{-\mathrm{E}_{\mathrm{a}_{1}} / \mathrm{RT}} \)
B \( \mathrm{k}_{2}=\mathrm{k}_{1} \mathrm{e}^{-\mathrm{E}_{\mathrm{a}_{2}} / \mathrm{RT}} \)
C \( \mathrm{k}_{1}=\mathrm{k}_{2} \mathrm{e}^{-\mathrm{E}_{\mathrm{a}_{1}} / \mathrm{RT}} \)
D \( \mathrm{k}_{1}=2 \mathrm{k}_{2} \mathrm{e}^{\mathrm{E}_{\mathrm{a}_{2}} / \mathrm{RT}} \)

The correct answer is:
\( \mathrm{k}_{2}=\mathrm{k}_{1} \mathrm{e}^{-\mathrm{E}_{\mathrm{a}_{1}} / \mathrm{RT}} \)

Solution:
k2=Ae-Ea2/RT
k 1 = Ae - E a 1 / RT
k2k1=e-Ea2+Ea1/RT Since,Ea2=2Ea1
k 2 k 1 = e - E a 1 / RT
or k 2 = k 1 e - E a 1 / RT

Topic: Chemistry, Chemical Kinetics, JEE Main
Reaction
${ }_1 D^2+{ }_1 \mathrm{~T}^3 \rightarrow{ }_2 \mathrm{He}^4+{ }_0 n^1+$ energy is an example of

Options:
A Nuclear fission
B Nuclear fusion
C Artificial radioactivity
D Radioactive disintegration

The correct answer is:
Nuclear fusion

Solution:
Nuclear fusion is a type of nuclear reaction where two or more atomic nuclei combine to form a single, heavier nucleus. This process is accompanied by the release or absorption of energy depending on the masses of the nuclei involved.
The reaction mentioned, ${ }^1 \mathrm{D}_2+{ }^1 \mathrm{~T}_3 \rightarrow{ }^2 \mathrm{He}_4+{ }^0 \mathrm{n}_1+$ energy, is a nuclear fusion reaction.

Topic: Physics, Nuclear Physics, JEE Main
Reaction $\mathrm{A} \rightarrow \mathrm{B}$ follows first order kinetics. The time taken for 0.8 mole of $A$ to produce 0.6 mole of $B$ is 1 hour. What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of $\mathrm{B}$ ?

Options:
A 1 hour
B 0.5 hour
C 0.25 hour
D 2 hour

The correct answer is:
1 hour

Solution:
For first order reaction,
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \ldots \ldots .(1)\)
Where \(k\) is rate constant
when 0.8 mol of \(A\) produces 0.6 mol of \(B\) then, \(0.8-0.6=0.2\) moles of A is left .
final moles \(\mathrm{A}=\mathrm{a}-\mathrm{x}=0.2\)
initial moles \(\mathrm{a}=0.8\)
\(\begin{aligned}
& \mathrm{t}=\frac{2.303}{\mathrm{k}} \log \frac{0.8}{\mathrm{o} .2} \\
& \mathrm{k}=\frac{2.303}{1} \log [4] \ldots \ldots . . (2) \\
& \mathrm{k}=2.303 \times 0.6020 \\
& \mathrm{k}=1.3865 \ldots \ldots . .(3)
\end{aligned}\)
In second case, 0.9 moles of A produces 0.675 moles of \(B\) so remaining moles of \(A\) are \(0.9-0.675=0.225\)
final moles of \(\mathrm{A}=\mathrm{a}-\mathrm{x}=0.225\)
initial mole \(=0.9\)
By using first order equation,
\(\begin{aligned}
\mathrm{t} & =\frac{2.303}{\mathrm{k}} \log \frac{0.9}{\mathrm{o} .225} \\
\mathrm{t} & =\frac{2.303}{1.3865} \log (4) \ldots . . (4) \\
\mathrm{t} & =1.661 \times 0.6020 \\
\mathrm{t} & =0.999=1 \mathrm{hr}
\end{aligned}\)

Topic: Chemistry, Chemical Kinetics, JEE Main