datavorous/entrance-exam-dataset huggingface dataset:
The accompanying graph of position $x$ versus time $t$ represents the motion of a particle. If $p$ and $q$ are both positive constants, the expression that best describes the acceleration $\alpha$ of the particle is

Options:
A $\mathrm{a}=-\mathrm{p}-\mathrm{qt}$
B $a=-p+q t$
C $\mathrm{a}=\mathrm{p}+\mathrm{qt}$
D $a=p-q t$

The correct answer is:
$a=p-q t$

Solution:
From graph $\mathrm{V}$ first increases then decreases
Hence $\mathrm{a}$ is earliar positive then negative
$\mathrm{a=P-q t}$

Topic: Physics, Motion In One Dimension, JEE Main
The accuracy in the measurement of time for 20 oscillations of a pendulum for the following measurements, t1=39.6 s, t2=39.9 s and t3=39.5 s, is

Options:
A ±0.2 s
B ±0.3 s
C ±0.5 s
D ±0.4 s

The correct answer is:
±0.2 s

Solution:
Given, t1 = 39.6 s
t2 = 39.9 s
t3 = 39.5 s
Least count of measuring instrument = 0.1 s
[As measurements have only one decimal place] Precision in the measurement = Least count of the measuring instrument = 0.1 s
Mean value of time for 20 oscillations
t = t 1 + t 2 + t 3 3 = 3 9 . 6 + 3 9 . 9 + 3 9 . 5 3 = 3 9 . 7 s
Absolute errors in the measurements
Î t 1 = t - t 1 = 3 9 . 7 - 3 9 . 6 = 0 . 1 s Î t 2 = t - t 2 = 3 9 . 7 - 3 9 . 9 = - 0 . 2 s Î t 3 = t - t 3 = 3 9 . 7 - 3 9 . 5 = 0 . 2 s
Mean absolute error = Î t 1 + Î t 2 + Î t 3 3
= 0 . 1 + 0 . 2 + 0 . 2 3 = 0 . 5 3 = 0 . 1 7 â 0 . 2 rounding-off upto one decimal place .
â´ Accuracy of measurement = Â± 0 . 2 s

Topic: Physics, Units and Dimensions, JEE Main
The acid that is believed to be mainly responsible for the damage of Taj Mahal is

Options:
A sulfuric acid.
B hydrofluoric acid.
C phosphoric acid.
D hydrochloric acid.

The correct answer is:
sulfuric acid.

Solution:
The air around the city of Agra, where the Taj Mahal is located, contains fairly high levels of sulphur and nitrogen oxides. It is mainly due to a large number of industries and power plants around the area. Use of poor quality of coal, kerosene and firewood as fuel for domestic purposes add up to this problem. The resulting acid rain reacts with marble, CaCO3 of Taj Mahal CaCO3+H2SO4→CaSO4+H2O+CO2 causing damage to this wonderful monument.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2022 (29 Jun Shift 1)
The acid that results when a base accepts a proton is called

Options:
A Conjugate base of the acid
B Conjugate protonated base
C Lewis base
D Conjugate acid of the base

The correct answer is:
Conjugate acid of the base

Solution:
The conjugate bases of weak acids are strong. The weaker an acid, the greater is the strength of its conjugate base. The acid that results when a base accepts a proton is called a conjugate acid of the base.

Topic: Chemistry, Ionic Equilibrium, JEE Main
The acid which contains both -OH and -COOH groups is

Options:
A Phthalic acid
B Adipic acid
C Glutaric acid
D Salicylic acid

The correct answer is:
Salicylic acid

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The acid which reduces Fehling solution is

Options:
A Methanoic acid
B Ethanoic acid
C Butanoic acid
D Propanoic acid

The correct answer is:
Methanoic acid

Solution:
Methanoic acid resemble with aldehyde due to its structure. So it reduce fehling reagent.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The acidic, basic and amphoteric oxides, respectively, are

Options:
A Na2O, SO3, Al2O3
B Cl2O, CaO, P4O10
C N2O3, Li2O, Al2O3
D MgO, Cl2O, Al2O3

The correct answer is:
N2O3, Li2O, Al2O3

Solution:
Non-metal oxides are acidic in nature as they produce acidic solution by reacting with water, so here Cl2O, P4O10, SO3 are acidic oxides which produce HOCl, H3PO4, H2SO4 respectively, with water.

Alkali metal and alkaline earth metal oxides are basic in nature, as they produce metal hydroxide solution with water and metal hydroxides are basic in nature.

Here, Na2O, CaO, Li2O, MgO are basic hydroxides which form NaOH, CaOH2, LiOH, MgOH2 respectively, with water.

Al2O3 is amphoteric oxide as it can react either with acid or base to form salt and water.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The actinoids exhibits more number of oxidation states in general than the lanthanoids. This is because

Options:
A
the $5 \mathrm{f}$ orbitals are more buried than the $4 \mathrm{f}$ orbitals

B
there is a similarity between $4 \mathrm{f}$ and $5 \mathrm{f}$ orbitals in their angular part of the wave function

C
the actinoids are more reactive than the lanthanoids

D
the $5 f$ orbitals extend further from the nucleus than the $4 \mathrm{f}$ orbitals

The correct answer is:
the $5 f$ orbitals extend further from the nucleus than the $4 \mathrm{f}$ orbitals

Solution:
The actinoids exhibit more number of oxidation states in general than the lanthanoids. This is because the $5 \mathrm{f}$ orbitals extend further from the nucleus than the $4 \mathrm{f}$ orbitals.
Hence, (D) is correct.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2007
The activation energy for a reaction which doubles the rate when the temperature is raised from $298 \mathrm{~K}$ to $308 \mathrm{~K}$ is

Options:
A
$59.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$

B
$39.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$

C
$52.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$

D
$29.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$52.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
Activation energy can be calculated from the equation.
$$
\begin{aligned}
& \frac{\log K_2}{\log K_1}=\frac{-E_a}{2.303 R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \\
& \text { Given } \frac{\log K_2}{\log K_1}=2 \quad T_2=308 ; \quad T_1=298 \\
& \therefore \quad \log 2=\frac{-E_a}{2.303 \times 8.314}\left(\frac{1}{308}-\frac{1}{298}\right) \\
& E_a=52.9 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2012 (26 May Online)
The activation energy for a simple chemical reaction $\mathrm{A} \rightarrow \mathrm{B}$ is $\mathrm{E}_a$ in forward direction. The activation energy for reverse reaction:

Options:
A is negative of $\mathrm{E}_a$
B is always less than $\mathrm{E}_a$
C can be less than or more $\mathrm{E}_a$
D is always double of $\mathrm{E}_a$

The correct answer is:
can be less than or more $\mathrm{E}_a$

Solution:
The activation energy for the reverse reaction, $\mathrm{E}_{\mathrm{a}(\mathrm{rev})}$, is the difference between the product energy and transition state at the peak of the diagram. $\mathrm{H}_r$ is the difference between the potential energy of the reactant and the potential energy of the product.
Related Theory
In a reversible reaction of bi-molecular type the activation energy for forward reaction is positive. Whereas, for the backward reaction activation energy is negative.

Topic: Chemistry, Chemical Kinetics, JEE Main
The activation energy for a simple chemical reaction $A \rightarrow B$ is $E_a$ in forward The activation energy for reverse reaction

Options:
A Is always double of $E_a$
B Is negative of $E_a$
C Is always less than $E_a$
D Can be less than or more than $E_a$

The correct answer is:
Can be less than or more than $E_a$

Solution:
In exothermic and endothermic reactions will be more and less than respectively.

Topic: Chemistry, Chemical Kinetics, JEE Main
The activation energy of a reaction is zero. The rate constant of this reaction

Options:
A Increases with increase of temperature
B Decreases with an increase of temperature
C Decreases with decrease of temperature
D Is independent of temperature

The correct answer is:
Is independent of temperature

Solution:
When $E_a=0$, the rate of reaction becomes independent of temperature.
( $E_{\mathrm{a}}=$ Energy of activation).

Topic: Chemistry, Chemical Kinetics, JEE Main
The activation energy of one of the reactions in a biochemical process is 532611 J mol-1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300=x×10-3k310°. The value of x is
[Given: ln10=2.3 R=8.3 J K-1 mol-1]

Options:

The correct answer is:
1

Solution:
lnK2 K1=EaR1 T1-1 T2

lnK2 K1=5326118.3×10310×300

where K2 is at 310 K & K1 is at 300 K

lnK2 K1=6.9

=3×ln10

lnK2 K1=ln103

K2=K1×103

K1=K2×103

So

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2022 (29 Jun Shift 1)
The activation energy, rate-constant and Arrhenius parameters of a chemical reaction at 25oC are 104.4 KJ mol-1, 3.0 x 10-4 s-1, 6.0 x 1014 s-1 respectively. The value of rate-constant in s â1 at T â â is (write your answer by dividing by 1014)

Options:

The correct answer is:
6

Solution:
k = A Â· e - Ea / RT (Arrhenius equation)
As T â â
k â A
i.e., k = A = 6 Ã 1 0 1 4 sec -1

Topic: Chemistry, Coordination Compounds, JEE Main
The active mass of $64 \mathrm{gm}$ of $\mathrm{HI}$ in a two litre flask would be

Options:
A 2
B 1
C 5
D 0.25

The correct answer is:
0.25

Solution:
$[\mathrm{HI}]=\frac{64 \mathrm{gm}}{128 \times 2 \text { litre }}=0.25$
Active mass is the concentration in moles/litre,

Topic: Chemistry, Chemical Equilibrium, JEE Main
The activity of a freshly prepared radioactive sample is $10^{10}$ disintegrations per second, whose mean life is $10^{-9} \mathrm{~s}$. The mass of an atom of this radioisotope is $10^{-25} \mathrm{~kg}$. The mass (in $\mathrm{mg}$ ) of the radioactive sample is

Options:

The correct answer is:
1

Solution:
Activity $\left(-\frac{d N}{d t}\right)=\lambda N=\left(\frac{1}{t_{\text {mean }}}\right) \times N$
$$
\begin{aligned}
& \therefore \quad N=\left(-\frac{d N}{d t}\right) \times t_{\text {mean }} \\
& \quad=\text { Total number of atoms } \\
& \text { Mass of one atom is } 10^{-25} \mathrm{~kg}=m \text { (say) } \\
& \therefore \quad \text { Total mass of radioactive substance } \\
& =(\text { number of atoms) } \\
& \quad \times(\text { mass of one atom) } \\
& =\left(-\frac{d N}{d t}\right)\left(t_{\text {mean }}\right)(m)
\end{aligned}
$$
Substituting the values, we get
Total mass of radioactive substance
$$
=1 \mathrm{mg}
$$
$\therefore$ Answer is 1 .
Analysis of Question
Question is very simple. Again in my opinion one can solve it.

Topic: Physics, Nuclear Physics, JEE Main
The activity of a radioactive element decreases in 10 years to 1/5 of initial activity A0. After further next 10 years, its activity will be

Options:
A A04
B A010
C A015
D A025

The correct answer is:
A025

Solution:
∵ The activity of a radioactive sample,
A = A0(12)tt12
Where, A0= initial activity.
In the first case,
A05 = A0(12)10t12
15 = (12)10t12 ...(i)
In the second case,
A = A0(12)20t12 ...(ii)
From Equations. (i) and (ii), we get
1/5A = (12)20t12A0(12)20t12
⇒15A=1A01210t12
⇒15A=1Ao5 [from Equation (i)]
∴A=A05×5=A025

Topic: Physics, Nuclear Physics, JEE Main
The activity of a radioactive material is 2.56×10-3Ci. If the half life of the material is 5 days, after how many days the activity will become 2×10-5Ci?

Options:
A 30 days
B 35 days
C 40 days
D 25 days

The correct answer is:
35 days

Solution:
The activity of a radioactive sample at any time t is given by,

A=A02-tT12⇒2×10-5=2.56×10-3×2-t5⇒2t5=128⇒t5=7⇒t=35 days

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2022 (29 Jun Shift 1)
The activity of a radioactive material is 6.4×10-4 curie. Its half life is 5 days. The activity will become 5×10-6 curie after

Options:
A 7 days
B 15 days
C 25 days
D 35 days

The correct answer is:
35 days

Solution:
Given here, A0=6.4×10-4 Curie and A=5×10−6 Curie

Half life, T12=5 days

The activity of a radioactive sample at time t is given by,

A=A012tT12

⇒5×10-66.4×10-4=12t5

⇒7=t5⇒t=35 days

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2022 (27 Jul Shift 2)
The activity of a radioactive sample decreases to 13rd of its original value in 3 days. Then, in 9 days its activity will become

Options:
A 127 of the original value
B 19 of the original value
C 18 of the original value
D 13 of the original value

The correct answer is:
127 of the original value

Solution:
R=R0e-Î»t
â 13=e-Î» Ã 3=e-3Î» â¦(i)
Again R'R0=e-Î» Ã 9=e-9Î»=(e-3Î»)3
=133 [From Eq.i]
=127
â Râ = R027
Hence, in 9 days activity will become 127 of the original value.

Topic: Physics, Nuclear Physics, JEE Main
The activity of a radioactive sample falls from 700s-1 to 500s-1 in 30 minutes. Its half life is close to:

Options:
A 72min
B 62min
C 66min
D 52min

The correct answer is:
62min

Solution:
From equation of radioactivity. At=A0e-λt

lnA0At=λt

⇒ ln700500=λ(30min) ------1

also

⇒ ln2=λt1/2 --------2

(ii)/(i)

⇒ ln2ln(7/5)=t1/2(30min) ⇒ (2.06004)30=t1/2=61.8min

≈ 62min

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The activity of a radioactive sample is measured as \( 9750 \) counts per minute at \( t=0 \) and as \( 975 \) counts per minute at \( t=5 \) minutes. The decay constant is approximately,

Options:
A \( 0.922 \) per minute.
B \( 0.691 \) per minute.
C \( 0.461 \) per minute.
D \( 0.230 \) per minute.

The correct answer is:
\( 0.461 \) per minute.

Solution:
According to radioactivity decay law, the number of decays per second is directly proportional to the number of active samples.

dNdt=λN

9750=λN0 …. (i)

975=λN … (ii)

Dividing (i) by (ii),

N N 0 = 1 10 .

We know that,

λ=2.303tlogN0N

=2.3035log10

=0.4606=0.461 per minute.

Topic: Physics, Nuclear Physics, JEE Main
The activity of a radioactive sample is measured as $\mathrm{N}_0$ counts per minute at $\mathrm{t}=0$ and $\mathrm{N}_0 / \mathrm{e}$ counts per minute at $\mathrm{t}=5 \mathrm{~min}$. The time (in minute) at which the activity reduces to half its value is

Options:
A $\log _e 2 / 5$
B $\frac{5}{\log _e 2}$
C $5 \log _{10} 2$
D $5 \log _{\mathrm{e}} 2$

The correct answer is:
$5 \log _{\mathrm{e}} 2$

Solution:
Fraction remains after $n$ half lives
$$
\frac{\mathrm{N}}{\mathrm{N}_0}=\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}}
$$
Given $\mathrm{N}=\frac{\mathrm{N}_0}{\mathrm{e}} \Rightarrow \frac{\mathrm{N}_0}{\mathrm{eN}_0}=\left(\frac{1}{2}\right)^{5 / \mathrm{T}}$
or $\frac{1}{\mathrm{e}}=\left(\frac{1}{2}\right)^{5 / T}$
Taking $\log$ on both sides, we get
$$
\begin{aligned}
\log 1-\log \mathrm{e} & =\frac{5}{\mathrm{~T}} \log \frac{1}{2} \\
-1 & =\frac{5}{\mathrm{~T}}(-\log 2) \\
\Rightarrow \quad \mathrm{T} & =5 \log _{\mathrm{e}} 2
\end{aligned}
$$
Now, let $\mathrm{t}^{\prime}$ be the time after which activity reduces to half
$$
\begin{aligned}
\Rightarrow \quad\left(\frac{1}{2}\right) & =\left(\frac{1}{2}\right)^{\mathrm{t}^{\prime} / 5 \log _{\mathrm{e}} 2} \\
\Rightarrow \quad \mathrm{t}^{\prime} & =5 \log _{\mathrm{e}} 2
\end{aligned}
$$

Topic: Physics, Nuclear Physics, NEET, NEET 2010 (Screening)
The activity of a radioactive sample is measured as N0 counts per minute at t=0 and N0e counts per minute at t=5 min. The time (in minutes) at which the activity reduces to half its value is

Options:
A loge25
B 5loge2
C 5log102
D 5loge2

The correct answer is:
5loge2

Solution:
Fraction remains after n half-lives

NN0=12n=12tT

Given, N=N0e⇒N0eN0=125T

⇒ 1e=125T

Taking log on both sides, we get

log1-loge=5Tlog12

⇒ -1=5T-log2

⇒ T=5loge2

Now, let t be the time after which activity reduces to half

12=12t/5loge2

⇒ t'=5loge2

Topic: Physics, Nuclear Physics, JEE Main
The activity of a radioactive sample is measured as N0 counts per minute at t=0 and N0/e counts per minute at t=5 min. The time, (in minutes) at which the activity reduces to half its value, is

Options:
A loge⁡25
B 5loge⁡2
C 5log10⁡2
D 5loge⁡2

The correct answer is:
5loge⁡2

Solution:
Fraction remains after n half-lives
NN0=12n=12+tT
t =5
N=N0e
1e=12+5T
Taking log on both sides, we get
logâ¡1-logâ¡e=5Tlogâ¡12
-1=5T-logâ¡2
T=5logeâ¡2
Now, let tâ² be the time after which activity reduces to half
12=12tâ²/5logeâ¡2
tâ²=5logeâ¡2

Topic: Physics, Nuclear Physics, NEET
The Î²- activity of a sample of CO2 prepared from a contemporary wood gave a count rate of 25.5 counts per minute (cpm). The same mass of CO2 from an ancient wooden statue gave a count rate of 20.5 cpm under the same conditions. If the half life of 14C is 5770 years, then the age of the statue is close to [Take log10255205â0.095]

Options:
A 1822 years
B 182 years
C 822 years
D 18220 years

The correct answer is:
1822 years

Solution:
r=20.5 cpm, r0=25.5 cpm
âµ r0âN0 and râN
â´r0r=N0N
Also, t=2.303Î»logâ¡N0N=2.303Î»logâ¡r0r
t=2.303Ã57700.693logâ¡25.520.5=1822 years

Topic: Physics, Nuclear Physics, JEE Main
The actual value of resistance R , shown in the figure is 30Ω. This is measured in an experiment as shown using the standard formula R=VI , where V and I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is:

Options:
A 35Ω
B 600Ω
C 570Ω
D 350Ω

The correct answer is:
570Ω

Solution:
Measured value of R=5% less than
actual value of R i.e., R'=30-(5% of 30)=30-5100×30

R'=28.5 Ω ...i

Now, let us assume that internal resistance of voltmeter RV. Replacing voltmeter with its internal resistance, we get following circuit.

It is clear that the measured value, R' should be equal to parallel combination of R and RV. Mathematically,
R'=R×RVR+RV=28.5 Ω

Given. R=30 Ω

⇒ 30 RV30+RV=28.5

⇒ 30 RV=28.5×30+28.5 RV

⇒ 1.5 RV=28.5×30

⇒ RV=28.5×301.5=19×30

or RV=570 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The acute angle between the lines $y=3$ and $y=\sqrt{3} x+9$ is

Options:
A $30^{\circ}$
B $60^{\circ}$
C $45^{\circ}$
D $90^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
It is clear from figure that angle $=60^{\circ}$

Topic: Mathematics, Straight Lines, JEE Main
The acute angle between the pair of tangents drawn to the ellipse 2x2+3y2=5 from the point 1,3 is

Options:
A tan-11675
B tan-12475
C tan-13275
D tan-13+8535

The correct answer is:
tan-12475

Solution:
Let equation of tangents to the ellipse x252+y253=1 be y=mx±52 m2+53

Since the tangents pass through 1,3, so

3=m±15m2+106

⇒9m2+36m-44=0

Let θ be the angle between the tangents, so

tanθ=m1-m21+m1 m2

i.e. tanθ=42+44491-449=24535

⇒θ=tan-12475

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2022 (26 Jul Shift 2)
The acute angle between the planes P1 and P2, when P1 and P2 are the planes passing through the intersection of the planes 5x+8y+13z-29=0 and 8x-7y+z-20=0 and the points 2,1,3 and 0,1,2, respectively, is

Options:
A π3
B π4
C π6
D π12

The correct answer is:
π3

Solution:
Equation of plane passing through the intersection of two planes 5x+8y+13z-29=0 and 8x-7y+z-20=0 can be written as 5x+8y+13z-29+λ8x-7y+z-20=0 and if it is passing through 2,1,3 then

10+8+39-29+λ16-7+3-20=0

i.e. λ=72.

P1: Equation of plane through intersection of 5x+8y+13z-29=0 and 8x-7y+z-20=0

and the point 2,1,3 is 5x+8y+13z-29+728x-7y+z-20=0

⇒2x-y+z=6

Similarly P2: Equation of plane through intersection of

5x+8y+13z-29=0 and 8x-7y+z-20=0 and the point 0,1,2 is

5x+8y+13z-29+μ8x-7y+z-20=0

Solving we get μ=15

So P2: x+y+2z=5

Angle between planes =θ=cos-12-1+266=cos-112=π3

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The acute angle between two lines such that the direction cosines $l, m, n$, of each of them satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is :

Options:
A
$15^{\circ}$

B
$30^{\circ}$

C
$60^{\circ}$

D
$45^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
Let $l_1, m_1, n_1$ and $l_2, m_2, n_2$ be the d.c of line 1 and 2 respectively, then as given $l_1+m_1+n_1=0$
and $l_2+m_2+n_2=0$
and $l_1{ }^2+m_1{ }^2-n_1^2=0$ and $l_2^2+m_2^2-n_2^2=0$
$\left(\because l+m+n=0\right.$ and $\left.l^2+m^2-n^2=0\right)$
Angle between lines, $\theta$ is
$$
\cos \theta=l_1 l_2+m_1 m_2+n_1 n_2
$$
As given $l^2+m^2=n^2$ and $l+m=-n$ $\Rightarrow(-n)^2-2 l m=n^2 \Rightarrow 2 l m=0$ or $l m=0$
So $l_1 m_1=0, l_2 m_2=0$
If $l_1=0, m_1 \neq 0$ then $l_1 m_2=0$
If $m_1=0, l_1 \neq 0$ then $l_2 m_1=0$
If $l_2=0, m_2 \neq 0$ then $l_2 m_1=0$
If $m_2=0, l_2 \neq 0$ then $l_1 m_2=0$
Also $l_1 l_2=0$ and $m_1 m_2=0$
$$
\begin{aligned}
& l^2+m^2-n^2=l^2+m^2+n^2-2 n^2=0 \\
& \Rightarrow 1-2 n^2=0 \Rightarrow n=\pm \frac{1}{\sqrt{2}} \\
& \therefore n_1=\pm \frac{1}{\sqrt{2}}, n_2=\pm \frac{1}{\sqrt{2}} \\
& \therefore \cos \theta=\frac{1}{2} \theta=60^{\circ} \text { (acute angle) }
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2013 (22 Apr Online)
The acute angle of intersection of the curves x2y=1 and y=x2 in the first quadrant is Î¸, then tanâ¡Î¸ is equal to

Options:
A 1
B 23
C 34
D 43

The correct answer is:
43

Solution:
Solving the equation of the curves, we get
1x2=x2⇒x=±1
∴ Intersection point is 1,1 in the first quadrant.
Now, x2y=1⇒y=1x2
⇒dydx=-2x3
⇒m1=-2 (by putting x=1)
And y=x2
⇒dydx=2x
⇒m2=2
∴tan⁡θ=m1-m21+m1m2
=43

Topic: Mathematics, Application of Derivatives, JEE Main
The acute angle \( x \), satisfying the equation \( \sin x-3 \sin 2 x+\sin 3 x=\cos x-3 \cos 2 x+\cos 3 x \) is \( \frac{\pi}{n} \), then find the value of 'n'?

Options:

The correct answer is:
8

Solution:
We have

sinx-3sin2x+sin3x=cosx-3cos2x+cos3x

⇒sinx+sin3x-3sin2x=cosx+cos3x-3cos2x

∴ sinC+sinD= 2sinC+D2cosC-D2∴ cosC+cosD=2cosC+D2cosC-D2

⇒2sinx+3x2cosx-3x2-3sin2x=2cosx+3x2cosx-3x2-3cos2x

⇒2sin2xcos-x-3sin2x=2cos2xcos-x-3cos2x

⇒2sin2xcosx-3sin2x=2cos2xcosx-3cos2x

⇒2sin2xcosx-3sin2x-2cos2xcosx+3cos2x=0

⇒2cosx(sin2x-cos2x)-3(sin2x-cos2x)=0

⇒(sin2x-cos2x)(2cosx-3)=0

⇒2cosx-3=0 & sin2x-cos2x=0

⇒ cosx=32 which is not possible ascosx∈-1, 1 & tan2x=1

⇒2x=tan-1(1)

⇒2x=λπ+π4; λ∈I

⇒x=λπ2+π8; λ∈I

So the solution is n=8.

Topic: Mathematics, Trigonometric Equations, JEE Main
The acute angles between the curves y=2x2-x and y2=x at 0,0 and 1,1 are α and β respectively, then

Options:
A α-β=0
B α+β=0
C α>β
D α<β

The correct answer is:
α-β=0

Solution:
y=2x2-x
⇒dydx=4x-1
Let, m1=4x-1
y2=x
⇒2ydydx=1
Let, m2=dydx=12y
At P0,0, m1=-1, m2→not defined
Angle α=π4
At Q1,1, m1=3, m2=12
β=tan-13-121+32=π4

Topic: Mathematics, Application of Derivatives, JEE Main
The addition of $\mathrm{HCN}$ to carbonyl compounds is an example of

Options:
A Nucleophilic substitution
B Electrophilic addition
C Nucleophilic addition
D Electrophilic substitution

The correct answer is:
Nucleophilic addition

Solution:
Addition of $\mathrm{HCN}$ to carbonyl compounds is an example of nucleophilic addition

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The addition of $\mathrm{HI}$ in the presence of peroxide catalyst does not follow anti-Markovnikov's rule because :

Options:
A
$\mathrm{HI}$ is a strong reducing agent.

B
$\mathrm{H-I}$ bond is too strong to be broken homolytically.

C
$\mathrm{I}$ atom combines with $\mathrm{H}$ atom to give back $\mathrm{HI}$.

D
Iodine atom is not reactive enough to add across a double bond.

The correct answer is:
Iodine atom is not reactive enough to add across a double bond.

Solution:
$\mathrm{HI}$ does not exhibit peroxide effect. $\mathrm{HI}$ bond although dissociates easily into iodine radicals, they being bigger in size are not much reactive but recombine together to form iodine molecule.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2013 (09 Apr Online)
The addition of NaOH to Cr3+ solution produces the precipitate of

Options:
A Cr(OH)3
B CrO3
C Cr2O3(H2O)n
D Cr2O3

The correct answer is:
Cr2O3(H2O)n

Solution:
Cr(OH)3 is not precipitated, but the hydrated oxide, Cr2O3(H2O)n is precipitated.
Cr3++3OH-→Cr(OH)3→Cr2O3(H2O)n

Topic: Chemistry, d and f Block Elements, JEE Main
The addition of silica during the extraction of copper from its sulphide ore :-

Options:
A converts copper sulphide into copper silicate
B converts iron oxide into iron silicate
C reduces copper sulphide into metallic copper
D reduces the melting point of the reaction mixture

The correct answer is:
converts iron oxide into iron silicate

Solution:
Silica is used to remove FeO impurity from the ore of copper

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (27 Jul Shift 2)