datavorous/entrance-exam-dataset huggingface dataset:
Observe the following statements
I. If $d y+2 x y d x=2 e^{-x^2} d x$, then $y e^{x^2}=2 x+c$
II. If $y e^{-x^2}-2 x=c$, then $d x=\left(2 e^{-x^2}-2 x y\right) d y$ which of the following is a correct statement?

Options:
A Both I and II are true
B Neither I nor II is true
C I is false, but II is true
D I is false, II is true

The correct answer is:
I is false, but II is true

Solution:
I. $d y+2 x y d x=2 e^{-x^2} d x$
$\Rightarrow \quad \frac{d y}{d x}+2 x y=2 e^{-x^2}$
This is a linear differential equation in $y$
Here, $P=2 x, Q=2 e^{-x^2}$
$\therefore \text { I.F. }=e^{\int P d x}=e^{\int 2 x d x}=e^{x^2}$
$\therefore$ Complete solution is
$\begin{aligned}
& y e^{x^2}=2 \int e^{-x^2} e^{x^2} d x+c \\
\Rightarrow \quad y e^{x^2} & =2 x+c
\end{aligned}$
II. $y e^{x^2}-2 x=c$
On differentiating w.r.t $x$, we get
$y e^{x^2} \cdot 2 x+e^{x^2} \frac{d y}{d x}-2=0$
$\begin{aligned}
\Rightarrow & e^{x^2} \frac{d y}{d x} & =2-2 x y e^{x^2} \\
\Rightarrow & \frac{d y}{d x} & =2 e^{-x^2}-2 x y
\end{aligned}$
$\therefore \mathrm{I}$ is true and II is false.

Topic: Mathematics, Differential Equations, JEE Main
Observe the following statements.
I. Sucrose has glycosidic linkage.
II. Cellulose is present in both plants and animals.
III. Lactose contains D-galactose and D-glucose units.
Which of the following statements are correct?

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

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

Solution:
No solution. Refer to answer key.

Topic: Chemistry, Biomolecules, JEE Main
Observe the following statements regarding purification of bauxite
I. During Hall's process, silica is removed as Si (vapour).
II. Bauxite ore contaminated with $\mathrm{Fe}_2 \mathrm{O}_3$ is purified in Baeyer's process.
III. During Serpeck's process, AlN is formed.
The correct answer is

Options:
A I, II and III are correct
B Only I and II are correct
C Only I and III are correct
D Only II and III are correct

The correct answer is:
Only II and III are correct

Solution:
Red bauxite which contains $\mathrm{Fe}_2 \mathrm{O}_3$ as the main impurity, is refined either by Baeyer's process or by Hall's process. White bauxite containing $\mathrm{SiO}_2$ impurity is refined by Serpeck's method. In Serpeck's method, following reactions take place.
$\begin{gathered}
\mathrm{Al}_2 \mathrm{O}_3 \cdot 2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{C}+\mathrm{N}_2 \stackrel{1800^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{AlN}+3 \mathrm{CO}+2 \mathrm{H}_2 \mathrm{O} \\
\mathrm{AlN}+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Al}(\mathrm{OH})_3+\mathrm{NH}_3 \\
2 \mathrm{Al}(\mathrm{OH})_3 \longrightarrow \mathrm{Al}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O}
\end{gathered}$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, AP EAMCET, AP EAMCET 2005
Observe the following statements regarding purification of bauxite
I. During Hall's process, silica is removed as Si (vapour).
II. Bauxite ore contaminated with $\mathrm{Fe}_2 \mathrm{O}_3$ is purified in Baeyer's process.
III. During Serpeck's process, AlN is formed.
The correct answer is

Options:
A I, II and III are correct
B Only I and II are correct
C Only I and III are correct
D Only II and III are correct

The correct answer is:
Only II and III are correct

Solution:
Red bauxite which contains $\mathrm{Fe}_2 \mathrm{O}_3$ as the main impurity, is refined either by Baeyer's process or by Hall's process. White bauxite containing $\mathrm{SiO}_2$ impurity is refined by Serpeck's method. In Serpeck's method, following reactions take place.
$\begin{gathered}
\mathrm{Al}_2 \mathrm{O}_3 \cdot 2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{C}+\mathrm{N}_2 \stackrel{1800^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{AlN}+3 \mathrm{CO}+2 \mathrm{H}_2 \mathrm{O} \\
\mathrm{AlN}+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Al}(\mathrm{OH})_3+\mathrm{NH}_3 \\
2 \mathrm{Al}(\mathrm{OH})_3 \longrightarrow \mathrm{Al}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O}
\end{gathered}$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, TS EAMCET, TS EAMCET 2005
Observe the given diagram of a connective tissue. Identify it. Which of the following statements is incorrect regarding the given tissue?

Options:
A It is modified for storage.
B It is a type of loose areolar tissue.
C It is present below the skin as a support framework for epithelium.
D Excess of nutrients are converted to fats in this tissue.

The correct answer is:
It is present below the skin as a support framework for epithelium.

Solution:
Adipose tissue is another type of loose connective tissue located mainly beneath the skin. The cells of this tissue are specialised to store fats. The excess of nutrients which are not used immediately are converted into fats and are stored in this tissue.
Areolar tissue is present beneath the skin. Often it serves as a support framework for epithelium. It contains fibroblasts (cells that produce and secrete fibres), macrophages and mast cells.

Topic: Biology, Structural Organisation in Animals, NEET
Observe the labeling in the diagram and identify A and B

Options:
A A-Corticular nephrons (65-50% of total nephrons)
B-Juxtaglomerular neurons (35-50% of total nephrons)
B A-Juxtaglomerular nephrons (85-90% of total nephrons)
âB-Corticular nephrons (10-15% of total nephrons)
C A-Corticular nephrons (80-85% of total nephrons)
B-Juxtaglomerular neurons (15-20% of total nephrons)
D A-Corticular nephrons (25-30% of total nephrons)
B-Juxtaglomerular neurons (75-80% of total nephrons)

The correct answer is:
A-Corticular nephrons (80-85% of total nephrons)
B-Juxtaglomerular neurons (15-20% of total nephrons)

Solution:
The human kidney has about 85% cortical nephrons and about 15% juxta-medullary nephrons. The cortical neurons are highly coiled structures with major parts present in the cortex region. Juxta-medullary nephrons have a long loop of Henle which runs deep in the medulla, well developed Vasa recta, and are mainly activated during a shortage of water.

Topic: Biology, Excretory Products and their Elimination, NEET
Observe the table depicting interactions between various organism and select the correct option:

Species A	Species B	Name of Interaction

+	+	Mutualism

â	â	A

+	â	B

+	â	Parasitism

+	0	C

â	0	D

(+) Beneficial (-) detrimental (0) Neutral

Options:
A A: Commensalism, B: Predation, C: Amensalism, D: Competition
B A: Predation, B: Parasitism, C: Commensalism, D: Amensalism
C A: Competition, B: Predation, C: Commensalism, D: Amensalism
D A: Competition, B: Predation, C: Amensalism, D: Commensalism

The correct answer is:
A: Competition, B: Predation, C: Commensalism, D: Amensalism

Solution:
Mutualism refers to those species interactions where both the species are benefited from the association and are dependent on each other for survival whereas in competition both species get affected equally.
In predation and parasitism, both only one species that is the predator/parasite is benefitted but in predation, the predator kills the prey whereas in Parasitism the parasite harms but does not kill its host.
The interaction where one species is benefitted and the other is neither benefitted nor harmed is called commensalism.
When one species is harmed and other remains unaffected the type interaction is amensalism.

Topic: Biology, Organisms and Populations, NEET
Oersted is the unit of

Options:
A Magnetic moment
B Intensity of magnetisation
C Magnetic flux
D Magnetic induction

The correct answer is:
Intensity of magnetisation

Solution:
The intensity of magnetic field at a point on the field is defined as the force on a unit north pole placed at that point.
Intensity of magnetization is measured in Oersted as a unit.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds is

Options:
A
n-hexane

B
2, 3-dimethylbutane

C
2,2-dimethylbutane

D
2-methylpentane

The correct answer is:
2, 3-dimethylbutane

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2005
Of the following amines how many can be separated by Hoffmann's mustard oil reaction?

Options:

The correct answer is:
4

Solution:
In Hoffmann's mustard oil reaction, the primary amine reacts with alcoholic carbon disulphide and then heated with excess of mercuric chloride and formed the isothiocyanate. Primary amines can only give Hoffmann's mustard oil reaction. Primary amine is the amine in which nitrogen atom is connected only with one carbon atom and two hydrogen atoms.

Topic: Chemistry, Amines, JEE Main
Of the following compounds the most acidic is

Options:
A $\mathrm{As}_2 \mathrm{O}_3$
B $\mathrm{P}_2 \mathrm{O}_5$
C $\mathrm{Sb}_2 \mathrm{O}_3$
D $\mathrm{Bi}_2 \mathrm{O}_3$

The correct answer is:
$\mathrm{P}_2 \mathrm{O}_5$

Solution:
In a group as we go downwards, the oxide basic character increases hence maximum acidic oxide is $\mathrm{P}_2 \mathrm{O}_5$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Of the following four aqueous solutions, the total number of those solutions whose freezing point is lower than that of 0.10MC2H5OH is (Integer answer)
(i) 0.10M Ba3PO42
(ii) 0.10M Na2SO4
(iii) 0.10M KCl
(iv) 0.10M Li3PO4

Options:

The correct answer is:
4

Solution:
ΔTf=ikfm

As m is the same for all solutions so ΔTf depends on Van't Hoff factor (i). Greater is i small is freezing point.

Compound i C2H5OH1Ba3PO425Na2SO43KCl2Li3PO44

So the given solutions have a lower freezing point than ethanol.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Of the following hydrides which has the lowest boiling point

Options:
A $\mathrm{SiH}_4$
B $\mathrm{Ph}_3$
C $\mathrm{HI}$
D $\mathrm{NH}_3$

The correct answer is:
$\mathrm{Ph}_3$

Solution:
$\mathrm{PH}_3$ has the lowest boiling point because it does not form Hydrogen bond.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one of them?

Options:
A
$(n-1) d^8 n s^2$

B
$(n-1) d^5 n s^2$

C
$(n-1) d^3 n s^2$

D
$(n-1) d^5 \mathrm{~ns}^{-1}$

The correct answer is:
$(n-1) d^5 n s^2$

Solution:
No Solution Available

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2004
Of the following pairs, the one containing example of metalloid elements in the periodic table is

Options:
A Sodium and potassium
B Fluorine and chlorine
C Calcium and magnesium
D Boron and silicon

The correct answer is:
Boron and silicon

Solution:
A metalloid is an element with properties similar to both metals and non-metals. The known metalloids are boron, silicon, germanium, arsenic, antimony, tellurium and polonium.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Of the following sets which one does NOT contain isoelectronic species?

Options:
A
$$
\mathrm{PO}_4^{-3}, \mathrm{SO}_4^{-2}, \mathrm{ClO}_4^{-}
$$

B
$\mathrm{CN}, \mathrm{N}_2, \mathrm{C}_2^2$

C
$\mathrm{SO}_3^{-2}, \mathrm{CO}_3^{-2}, \mathrm{NO}_3^{-}$

D
$\mathrm{BO}_3^{-3}, \mathrm{CO}_3^{-2}, \mathrm{NO}_3$

The correct answer is:
$\mathrm{SO}_3^{-2}, \mathrm{CO}_3^{-2}, \mathrm{NO}_3^{-}$

Solution:
No Solution Available

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2005
Of the following sulphides which one is insoluble in dilute acids but soluble in alkalies

Options:
A $\mathrm{PbS}$
B $\mathrm{CaS}$
C $\mathrm{FeS}$
D $\mathrm{Sb}_2 \mathrm{~S}_3$

The correct answer is:
$\mathrm{Sb}_2 \mathrm{~S}_3$

Solution:
It is an acidic salt

Topic: Chemistry, Practical Chemistry, JEE Main
Of the four major causes for the loss of biodiversity (Alien species invasion, habitat loss and fragmentation, overexploitation and co-extinctions) which according to you is the major cause for the loss of biodiversity? Give reasons in support.

Options:

None

Solution:
Loss of habitat and fragmentation could be regarded the main cause out of the four major causes for the loss of biodiversity. This is because
(i) Habitat loss and fragmentation caused by clearing and over-exploitation of forest areas for agriculture, urbanisation and industrialisation, results in destruction of natural habitats.
(ii) Increasing human population has overburdened the forest resources and have destroyed forest land, which means loss of habitat for several species.
(iii) In addition, large habitats are broken up into small fragments, because of which mammals and birds requiring large territories and with migratory habits are adversely affected, leading to decline in population.

Topic: Biology, Biodiversity and its Conservation
Of the metals $B e, \quad M g, \quad C a$ and $S r$ of group II A. In the periodic table the least ionic chloride would be formed by

Options:
A $\mathrm{Be}$
B $\mathrm{Mg}$
C $\mathrm{Ca}$
D $\mathrm{Sr}$

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

Solution:
Because of small atomic size and high I.E. Be forms covalent chloride.

Topic: Chemistry, s Block Elements, JEE Main
Of the two slits producing interference in Young's experiment, one is covered with glass so that light intensity passing is reduced to $50 \%$. Which of the following is correct?

Options:
A Intensity of fringes remains unaltered
B Intensity of bright fringe decreases and that of dark fringe increases
C Intensity of bright fringe increases and that of dark fringe decreases
D Intensity of both bright and dark fringes decreases

The correct answer is:
Intensity of bright fringe decreases and that of dark fringe increases

Solution:
$\therefore \quad$ The intensity of bright fringe decreases and dark fringe increases when one slit is covered. The reason behind this is that the bright fringes are produced when the amplitudes from both slits are in phase and added constructively.

Topic: Physics, Wave Optics, MHT CET, MHT CET 2023 (13 May Shift 2)
____________ of volume of neural tissue in our body is contributed by neuroglia.

Options:
A < 25%
B > 50%
C > 75%
D 90%

The correct answer is:
> 50%

Solution:
Neural tissue consists of two types of cells :
1. Non-excitatory cells : These are also known as neuroglial or glial cells forming more than half of the nervous tissue.These non-neural cells provide support and nutrition to neurons and maintain homeostasis.
2. Excitatory cells or neurons :Neuron is the longest cell of human body (of Animal kingdom as well).
It is the structural and functional unit of nervous system.

Topic: Biology, Structural Organisation in Animals, NEET
Ohm's law is true

Options:
A For metallic conductors at low temperature
B For metallic conductors at high temperature
C For electrolytes when current passes through them
D For diode when current flows

The correct answer is:
For metallic conductors at low temperature

Solution:
Because with rise in temperature resistance of conductor increase, so graph between $V$ and $i$ becomes non linear.

Topic: Physics, Current Electricity, JEE Main
Oil of winter green is

Options:
A

B

C

D

The correct answer is:

Solution:
Methyl salicylate is known as oil of winter green. It is the methyl ester of salicyclic acid.
It is colourless, viscous liquid with a sweet, fruity odour.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
On a frictionless surfaces, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle θ to its initial direction and has a speed v3 . The second block's speed after the collision is:

Options:
A 34v
B 32v
C 32v
D 223v

The correct answer is:
223v

Solution:
In elastic collision energy of system remains same. so, \((\mathrm{KE})_{\text {before collision }}=(\mathrm{KE})_{\text {after collision }}\)
Let speed of second body after collision is \(\mathrm{v}^{\prime}\).
\(\begin{aligned}
& \frac{1}{2} \mathrm{Mv}^2+0=\frac{1}{2} \mathrm{M}\left(\frac{\mathrm{v}}{3}\right)^2+\frac{1}{2} \mathrm{M}\left(\mathrm{v}^{\prime}\right)^2 \\
& \Rightarrow \mathrm{v}^{\prime}=\frac{2 \sqrt{2}}{3} \mathrm{v}
\end{aligned}\)

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2015 (Phase 2)
On a hot summer night, the refractive index of air is the smallest near the ground and increases with a height from the ground. When a light beam is directed horizontally, the Huygens' principle leads us to conclude that as it travels, the light beam,

Options:
A bends upward.
B becomes narrower.
C goes horizontally without any deflection.
D bends downward.

The correct answer is:
bends upward.

Solution:
Every point on the horizontal ray of light behaves as a source of light. Thus, the secondary rays which are directed towards denser medium would bend towards the normal. These phenomena cause the bending of the rays upwards.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2015 (04 Apr)
On a linear temperature scale $\mathrm{Y}$, water freezes at $-160^{\circ} \mathrm{Y}$ and boils at $-50^{\circ} \mathrm{Y}$. On this $\mathrm{Y}$ scale, a temperature of $340 \mathrm{~K}$ would be read as : (water freezes at $273 \mathrm{~K}$ and boils at $373 \mathrm{~K}$ )

Options:
A
$-73.7^{\circ} \mathrm{Y}$

B
$-233.7^{\circ} \mathrm{Y}$

C
$-86.3^{\circ} \mathrm{Y}$

D
$-106.3^{\circ} \mathrm{Y}$

The correct answer is:
$-86.3^{\circ} \mathrm{Y}$

Solution:
Reading on any scale - LFP
UFP - LFP
$=$ constant for all scales
$\frac{340-273}{373-273}=\frac{{ }^{\circ} \mathrm{y}-(-160)}{-50-(-160)}$
$\Rightarrow \frac{67}{100}=\frac{\mathrm{y}+160}{110}$
$\therefore \quad y=-86.3^{\circ} \mathrm{y}$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2013 (09 Apr Online)
On a long horizontally moving belt, a child runs to and fro with a speed of $9 \mathrm{~km} / \mathrm{h}$ (w.r.t. belt) between his father and mother located $50 \mathrm{~m}$ apart on the moving belt. The belt moves with a speed of $4 \mathrm{kmh}^{-1}$. For an observer on a stationary platform outside, what is the
(a) speed of the child running in the direction of motion of the belt?
(b) speed of the child running opposite to the direction of motion of the belt? and
(c) time taken by the child in case (a) and (b)? Which of the answers alter if motion is viewed by one of the parents ?

Options:

None

Solution:
Let us consider left to right to be the positive direction of $x$-axis.
(a) Here, velocity of belt, $v_{\mathrm{B}}=+4 \mathrm{~km} / \mathrm{h}$
Speed of the child w.r.t. belt, $v_{\mathrm{C}}=+9 \mathrm{~km} / \mathrm{h}$
Speed of the child w.r.t. stationary observer, while running in direction of motion of the belt $v_C^{\prime}=v_C+\mathrm{v}_B=9+4=13 \mathrm{~km} / \mathrm{h}$
(b) Here, $v_B=+4 \mathrm{~km} / \mathrm{h} ; \quad v_C=-9 \mathrm{~km} / \mathrm{h}$
Speed of the child w.r.t. stationary observer, while running in opposite direction of motion of belt $v_C^{\prime}=v_C+v_B=-9+4=5 \mathrm{~km} / \mathrm{h}$
$\therefore \quad$ Velocity of the child in any direction (from father to mother or from mother to father) will be $9 \mathrm{kmh}^{-1}$
or $\frac{5}{2} \mathrm{~ms}^{-1}$.
$\therefore \quad$ Time taken by the child in each case (a) and (b) each is $\mathrm{t}=50 /(5 / 2)=20 \mathrm{~s}$
If the motion is observed by one of the parents, answer to case (a) or case (b) will get altered. It is so because speed of child w.r.t. either of mother or father is $9 \mathrm{~km} / \mathrm{h}$
But answer (c) remains unaltered due to the fact that parents and child are on the same belt and as such all are equally affected by the motion of belt.

Topic: Physics, Motion In One Dimension
On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are $39^{\circ} \mathrm{W}$ and $239^{\circ} \mathrm{W}$ respectively. What will be the temperature on the new scale, corresponding to a temperature of $39^{\circ} \mathrm{C}$ on the Celsius scale?

Options:
A $139^{\circ} \mathrm{W}$
B $78^{\circ} \mathrm{W}$
C $117^{\circ} \mathrm{W}$
D $200^{\circ} \mathrm{W}$

The correct answer is:
$117^{\circ} \mathrm{W}$

Solution:
$$
\begin{aligned}
& \frac{39-0}{100-0}=\frac{x-39}{239-39} \\
& \Rightarrow \quad x=117^{\circ} \mathrm{W}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are $39^{\circ} \mathrm{W}$ and $239^{\circ} \mathrm{W}$ respectively. What will be the temperature on the new scale, corresponding to a temperature of $39^{\circ} \mathrm{C}$ on the Celsius scale?

Options:
A $139^{\circ} \mathrm{W}$
B $78^{\circ} \mathrm{W}$
C $117^{\circ} \mathrm{W}$
D $200^{\circ} \mathrm{W}$

The correct answer is:
$117^{\circ} \mathrm{W}$

Solution:
$$
\begin{aligned}
& \frac{39-0}{100-0}=\frac{x-39}{239-39} \\
& \Rightarrow \quad x=117^{\circ} \mathrm{W}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2008 (Mains)
On a photosensitive material, when the frequency of incident radiation is increased by 30%, the kinetic energy of emitted photoelectrons increases from 0.4 eV to 0.9 eV. The work function of the surface is

Options:
A 1 eV
B 1.267 eV

C 1.4 eV

D 1.8 eV

The correct answer is:
1.267 eV

Solution:
hv=0.4+W0 ....... (i)
1.3 hv=0.9+W0 ....... (ii)
(ii) - 1.3 (i) gives
0.3 W0=0.9-1.3 (0.4)
â´ W0=0.380.3=1.267 eV

Topic: Physics, Dual Nature of Matter, JEE Main
On a photosensitive surface, if the intensity of incident radiation is increased, the stopping potential

Options:
A first increases and then decreases
B increases
C decreases
D remains unchanged

The correct answer is:
remains unchanged

Solution:
Stopping potential does not depend on the intensity of light.

Topic: Physics, Dual Nature of Matter, MHT CET, MHT CET 2021 (21 Sep Shift 1)
On a rectangular hyperbola x2-y2=a2,a>0, three points A, B, C are taken as follows : A=-a, 0 ; B and C are placed symmetrically with respect to the X-axis on the branch of the hyperbola not containing A. Suppose that the ΔABC is equilateral. If the side length of the ΔABC is ka, then k lies in the interval

Options:
A (0,2]
B (2,4]
C (4,6]
D (6,8]

The correct answer is:
(2,4]

Solution:
We have rectangular hyperbola

x2-y2=a2

Given ABC is an equilateral triangle.

∴AB=BC=AC

AB2=BC2

a2secθ+12+a2tan2θ=4a2tan2θ

⇒secθ+12=3tan2θ

⇒secθ+12=3sec2θ-1

⇒secθ+12=3secθ+1secθ-1

⇒secθ+1=3secθ-3

⇒secθ=2

⇒θ=60°

∵ Side BC=2atanθ

=2atan60°=2a3

But side of triangle is ka.

∴ka=2a3

k=23

Hence, k∈(2,4]

Topic: Mathematics, Hyperbola, JEE Main
On a temperature scale 'X', the boiling point of water is 65°X and the freezing point is -15°X. Assuming that the X scale is linear. The equivalent temperature corresponding to -95°X on the Fahrenheit scale would be

Options:
A -112°F
B -48°F
C -148°F
D -63°F

The correct answer is:
-148°F

Solution:
The relation between two temperature scales can be written as follows

65-x65--15=212-F180 ...1

Substitute the values of the known parameters in to equation (1) and sole to calculate the required temperature in Fahrenheit scale.

65+9580=212-F180⇒2×180=212-F⇒F=212-360=-148°F

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2023 (11 Apr Shift 1)
On a thin layer chromatographic plate, an organic compound moved by 3.5 cm, while the solvent moved by 5 cm. The retardation factor of the organic compound is _____________×10-1

Options:

The correct answer is:
07

Solution:
Given, an organic compound moved by 3.5 cm

Solvent moved by 5 cm

The Rf (retardation factor) value is the ratio of the solute’s distance travelled to the solvent’s distance travelled.

Retardation factor =Distance travelled bysample/organic compoundDistance travelled by solvent

By putting the values in the above formula,

Rf=3.55=7×10-1

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (30 Jan Shift 1)