datavorous/entrance-exam-dataset huggingface dataset:
The ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ and the hyperbola $\frac{x^{2}}{25}-\frac{y^{2}}{16}=1$ have in common

Options:
A centre and vertices only
B centre, foci and vertices
C centre, foci and directrices
D centre only

The correct answer is:
centre and vertices only

Solution:
Equation of ellipse is
$\frac{x^{2}}{25}+\frac{y^{2}}{16}=1, a>b$
and equation of hyperbola is
$\frac{x^{2}}{25}-\frac{y^{2}}{16}=1, a>b$
Let $e$ and $e^{\prime}$ be the eccentricities of the ellipse and hyperbola.
So,
$e=\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}=\sqrt{\frac{25-16}{25}}=\frac{3}{5}$
and $\quad e^{\prime}=\sqrt{\frac{a^{2}+b^{2}}{a^{2}}}=\sqrt{\frac{25+16}{25}}=\frac{\sqrt{41}}{5}$
(i) Centre of ellipse $(0,0)$ and centre of hyperbola is $(0,0)$
(ii) Foci of ellipse are $(\pm a e, 0)$ or $(\pm 3,0)$. Foci of hyperbola are $\left(\pm a e^{\prime}, o\right)$ or $(\pm \sqrt{41}, 0)$.
(iii) Direction of ellipse are
$x=\pm \frac{a}{e} \Rightarrow x=\pm \frac{25}{3}$
and directrices of hyperbola are $x=\pm \frac{a}{e}$
$\Rightarrow \quad x=\pm \frac{25}{\sqrt{41}}$
(iv) Vertices of ellipse are $(\pm a, 0)$ or $(\pm 5,0)$. Vertices of hyperbola are $(\pm a, 0)$ or $(\pm 5,0)$.
From the above discussions, their are common is centre and vertices.

Topic: Mathematics, Hyperbola, JEE Main
The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the straight line $y=m x+c$ intersect in real points only if

Options:
A $a^2 m^2 \lt c^2-b^2$
B $a^2 m^2 \gt c^2-b^2$
C $a^2 m^2 \geq c^2-b^2$
D $c \geq b$

The correct answer is:
$a^2 m^2 \geq c^2-b^2$

Solution:
To cut at real points, $c^2 \leq a^2 m^2+b^2 \Rightarrow$ $a^2 m^2 \geq c^2-b^2$.

Topic: Mathematics, Ellipse, JEE Main
The ellipse E1:x29+y24=1 is inscribed in a rectangle R whose sides are parallel to the coordinate axes. Another ellipse E2 passing through the point 0,4 circumscribes the rectangle R. The length (in units) of the major axis of ellipse E2 is

Options:

The correct answer is:
8

Solution:
Let the ellipse be x2a2+y2b2=1
As it is passing through 0,4 and 3,2. So, b2=16 and 9a2+416=1 or a2=12
So, the length of major axis =2b=8

Topic: Mathematics, Ellipse, JEE Main
The ellipse having its foci $(0, \pm 1)$ and major axis of length $\sqrt{5}$ is

Options:
A $20 x^2+4 y^2=5$
B $36 x^2+20 y^2=45$
C $4 x^2+20 y^2=5$
D $20 x^2+36 y^2=45$

The correct answer is:
$20 x^2+4 y^2=5$

Solution:
Since the foci of given ellipse lie on the $Y$-axis, it is vertical ellipse.
Let the required equation be $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
Where $a^2>b^2$
Foci are $(0, \pm c)=(0, \pm 1) \Rightarrow c=1$
$a=$ length of the semi-major axis $=\frac{1}{2} \times \sqrt{5}=\frac{\sqrt{5}}{2}$
$\therefore \quad a^2=\frac{5}{4}$
Now, $\quad c^2=a^2-b^2 \Rightarrow 1=\left(\frac{\sqrt{5}}{2}\right)^2-b^2$
$\Rightarrow \quad b^2=\frac{5}{4}-1=\frac{1}{4}$
$\therefore$ The required equation is
$\frac{x^2}{1 / 4}+\frac{y^2}{5 / 4}=1 \Rightarrow 4 x^2+\frac{4 y^2}{5}=1$
$\Rightarrow \quad 20 x^2+4 y^2=5$

Topic: Mathematics, Ellipse, JEE Main
The ellipse $x^2+4 y^2=4$ is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in another ellipse that passes through the point $(4,0)$. Then the equation of the ellipse is

Options:
A
$x^2+16 y^2=16$

B
$x^2+12 y^2=16$

C
$4 x^2+48 y^2=48$

D
$4 \mathrm{x}^2+64 \mathrm{y}^2=48$

The correct answer is:
$x^2+12 y^2=16$

Solution:
$$
x^2+4 y^2=4 \Rightarrow \frac{x^2}{4}+\frac{y^2}{1}=1 \Rightarrow a=2, b=1 \Rightarrow P=(2,1)
$$
Required Ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \Rightarrow \frac{x^2}{4^2}+\frac{y^2}{b^2}=1$
$(2,1)$ lies on it
$$
\begin{aligned}
& \Rightarrow \frac{4}{16}+\frac{1}{b^2}=1 \Rightarrow \frac{1}{b^2}=1-\frac{1}{4}=\frac{3}{4} \Rightarrow b^2=\frac{4}{3} \\
& \therefore \frac{x^2}{16}+\frac{y^2}{\left(\frac{4}{3}\right)}=1 \Rightarrow \frac{x^2}{16}+\frac{3 y^2}{4}=1 \Rightarrow x^2+12 y^2=16
\end{aligned}
$$

Topic: Mathematics, Ellipse, JEE Main
The ellipse $x^2+4 y^2=4$ is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in another ellipse that passes through the point $(4,0)$. Then the equation of the ellipse is

Options:
A
$x^2+16 y^2=16$

B
$x^2+12 y^2=16$

C
$4 x^2+48 y^2=48$

D
$4 \mathrm{x}^2+64 \mathrm{y}^2=48$

The correct answer is:
$x^2+12 y^2=16$

Solution:
$$
x^2+4 y^2=4 \Rightarrow \frac{x^2}{4}+\frac{y^2}{1}=1 \Rightarrow a=2, b=1 \Rightarrow P=(2,1)
$$
Required Ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \Rightarrow \frac{x^2}{4^2}+\frac{y^2}{b^2}=1$
$(2,1)$ lies on it
$$
\begin{aligned}
& \Rightarrow \frac{4}{16}+\frac{1}{b^2}=1 \Rightarrow \frac{1}{b^2}=1-\frac{1}{4}=\frac{3}{4} \Rightarrow b^2=\frac{4}{3} \\
& \therefore \frac{x^2}{16}+\frac{y^2}{\left(\frac{4}{3}\right)}=1 \Rightarrow \frac{x^2}{16}+\frac{3 y^2}{4}=1 \Rightarrow x^2+12 y^2=16
\end{aligned}
$$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2009
The elongated or tube-like cells, with thick, lignified walls and tapering ends are

Options:
A Tracheids
B Vessels
C Xylem fibre
D Xylem parenchyma

The correct answer is:
Tracheids

Solution:
Xylem is made up of four different kinds of elements in which tracheids are oldest, elongated with thick-walled structures. Tracheids are the main xylem elements in a gymnosperm.
4 types of xylem elements are -
1. Tracheids: Elongated cells with tapering ends. Dead cells (no protoplasm) with lignified cell walls.
2. Vessels (Tracheae): Vessel is tubular structure made up of cells (vessel members) joined together end to end. Dead cells (no protoplasm) with lignified cell walls and a large central cavity. Members are interconnected by their perforated end walls. These are characteristic features of angiosperms (gymnosperms lack vessels).
3. Xylem fibers: Elongated cells with tapering ends and obliterated lumen. Dead cells (no protoplasm) with lignified cell walls. These may be septate or aseptate.
4. Xylem parenchyma: Living cells with a thin cell wall made up of cellulose. These store food (starch, oil), tannin, etc. Ray parenchymatous cells also perform radial conduction of water.

Topic: Biology, Anatomy of Flowering Plants, NEET
The elongation of a wire on the surface of the earth is 10-4 m. The same wire of same dimensions is elongated by 6×10-5 m on another planet. The acceleration due to gravity on the planet will be _____ m s-2. (Take acceleration due to gravity on the surface of earth =10 m s-2)

Options:

The correct answer is:
6

Solution:
Change in length ∝Force∝mg

mg=kx=YAlx

Thus, x∝g

Then, 10-46×10-5=gg' ⇒ g'=0.6g=6 m s-2

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (26 Jun Shift 1)
The E.M. wave with shortest wavelength among the following is

Options:
A Ultraviolet rays
B X-rays
C Gamma-rays
D Microwaves

The correct answer is:
Gamma-rays

Solution:
If we look at the spectrum of electromagnetic waves then the order of a wavelength is, λgamma<λX-ray<λUV rays<λVisible<λInfrared<λMicrowave<λRadio, so from the above spectrum the wavelength of gamma rays is minimum and then the wavelength of a radio wave is maximum, so the shortest wavelength is for gamma-rays.

Topic: Physics, Electromagnetic Waves, JEE Main
The emf of a cell is 6 V and internal resistance is 0.5 kÎ©. What will be the reading (in volt) of a voltmeter having an internal resistance of 2.5 kÎ©?

Options:

The correct answer is:
5

Solution:
Total resistance=2.5+0.5=3.0 kÎ©=3000 Î©
â´ Current, i=63000 A;
Reading of voltmeter =iÃ2.5Ã1000
=63000Ã2500=5 V

Topic: Physics, Current Electricity, JEE Main
The e.m.f. of a cell is $\mathrm{E}$ volts and internal resistance is $r$ ohm. The resistance in external circuit is also $t$ ohm. The p.d. across the cell will be

Options:
A $E / 2$
B $2 E$
C $4 E$
D $E / 4$

The correct answer is:
$E / 2$

Solution:
Since both the resistors are same, therefore potential difference $V+V=E$
$V=E / 2$

Topic: Physics, Current Electricity, JEE Main
The emf of cell Tl \(\left|\underset{(0.001 \mathrm{M})}{\mathrm{Tl}^{+}}\right|\left|\underset{(0.01 \mathrm{M})}{\mathrm{Cu}^{2+}}\right| \mathrm{Cu}\) is \(0.83 \mathrm{~V}\) at \(298 \mathrm{~K}\). It could be increased by :

Options:
A decreasing concentration of both \(\mathrm{Tl}^{+}\) and \(\mathrm{Cu}^{2+}\) ions
B increasing concentration of \(\mathrm{Cu}^{2+}\) ions
C increasing concentration of \(\mathrm{T1}^{+}\)ions
D increasing concentration of both \(\mathrm{Tl}^{+}\) and \(\mathrm{Cu}^{2+}\) ions

The correct answer is:
increasing concentration of \(\mathrm{Cu}^{2+}\) ions

Solution:
$\begin{array}{ll}
\text{Anodic Reaction} & {\left[\mathrm{T} \ell_{(s)} \rightarrow \mathrm{T}^{+}{ }_{(\mathrm{aq})}+\mathrm{e}^{-}\right] 2} \\
\text{Cathodic Reaction} & \mathrm{Cu}^{+2}{ }_{(\mathrm{aq})}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}_{(\mathrm{s})} \\
\hline \text{Overall Redox Reaction} & 2 \mathrm{~T} \ell_{(\mathrm{s})}+\mathrm{Cu}^{+2}{ }_{(\mathrm{aq})} \rightarrow 2 \mathrm{~T} \ell_{(\mathrm{aq})}^{+}+\mathrm{Cu}_{(\mathrm{s})}
\end{array}$
$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^o-\frac{0.0591}{2} \log \frac{\left[\mathrm{T} \ell^{+}\right]^2}{\left[\mathrm{Cu}^{+2}\right]}$
$\mathrm{E}_{\text {cell }}$ increases by increasing concentration of $\left[\mathrm{Cu}^{+2}\right]$ ions.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (08 Apr Shift 2)
The emf of the cell Zn|Zn2+(0.01M)||Fe2+(0.001M)| Fe at 298 K is 0.2905 volt. Then the value of equilibrium constant for the cell reaction is

Options:
A e0.32/0.0295
B 100.32/0.0295
C 100.26/0.0295
D 100.32/0.0591

The correct answer is:
100.32/0.0295

Solution:
Zn+Fe2+âZn2++Fe(n=2)
E= E o â 0.0591 n logâQ
0.02905=Eo-0.05912log0.010.001
E o =0.2905+0.0295=0.32âvolt
E o = 0.0591 n log  K eq
0.32= 0.0591 2 log  K eq =0.02945 log  K eq
K eq = 10 0.32/0.0295

Topic: Chemistry, Electrochemistry, JEE Main
The emission series of hydrogen atom is given by
$$
\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)
$$
where $R$ is the Rydberg constant. For a transition from $\mathrm{n}_{2}$ to $\mathrm{n}_{1}$, the relative change $\Delta \lambda / \lambda$ in the emission wavelength if hydrogen is replaced by deuterium (assume that the mass of proton and neutron are the same and approximately 2000 times larger than that of electrons) is

Options:
A $0.025 \%$
B $0.005 \%$
C $0.0025 \%$
D $0.05 \%$

The correct answer is:
$0.025 \%$

Solution:
$$
\mathrm{R}=\frac{\mathrm{m}_{\mathrm{e}} \mathrm{e}^{4}}{8 \varepsilon_{0}^{2} \mathrm{ch}^{3}}
$$
where $\mathrm{m}_{\mathrm{e}}=$ mass of electron
When we consider mass of nucleus also then we replace $\mathrm{m}$ with reduced mass
$$
\mu=\frac{m_{e} m}{m_{e}+m}
$$
where $\mathrm{m}=$ mass of nucleus
In case of hydrogen atom
$$
\mu_{1}=\frac{\left(\mathrm{m}_{\mathrm{e}}\right)\left(2000 \mathrm{~m}_{\mathrm{e}}\right)}{\mathrm{m}_{\mathrm{e}}+(2000) \mathrm{m}_{\mathrm{e}}}=\frac{2000}{2001} \mathrm{~m}_{\mathrm{e}}
$$
In case of deuterium
$$
\mu_{2}=\frac{\left(\mathrm{m}_{\mathrm{e}}\right)\left(4000 \mathrm{~m}_{\mathrm{e}}\right)}{\mathrm{m}_{\mathrm{e}}+(4000) \mathrm{m}_{\mathrm{e}}}=\frac{4000}{4001} \mathrm{~m}_{\mathrm{e}}
$$
Hence,
$$
\begin{aligned}
\frac{\lambda_{2}}{\lambda_{1}} &=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{\mu_{2}}{\mu_{1}} \\
&=\left(\frac{4000 \mathrm{~m}_{\mathrm{e}}}{4001}\right) \frac{(2001)}{\left(2000 \mathrm{~m}_{\mathrm{e}}\right)} \\
&=\frac{4002}{4001} \\
\Rightarrow \frac{\lambda_{2}}{}^{-\lambda_{1}}{\lambda_{1}}=& \frac{1}{4001} \\
\therefore \quad \frac{\Delta \lambda}{\lambda_{1}} \times 100 &=\frac{1}{4001} \times 100 \\
&=0.025 \%
\end{aligned}
$$

Topic: Physics, Atomic Physics, JEE Main
The emitter-base junction of a transistor is, $\qquad$ biased while the collector-base junction is $\qquad$ biased.

Options:
A reverse,forward
B reverse,reverse
C forward,forward
D forward reverse

The correct answer is:
forward reverse

Solution:
forward, reverse
- From the above figure, it is clear that in order to operate the emitter-base junction of the transistor is forward biased and the collector-base junction is reverse biased.
- Hence option 4 is the answer

Topic: Physics, Semiconductors, JEE Main
The empirical formula and molecular mass of a compound are CH2O and 180 g respectively. The molecular formula of the compound will be

Options:
A C9H18O9
B CH2O
C C6H12O6
D C2H4O2

The correct answer is:
C6H12O6

Solution:
As we know that, the Molar mass of Carbon=12

Molar mass of Hydrogen=1

Molar mass of Oxygen=16

Empirical formula mass =CH2O=12+2+1×16=30

Molecular mass =180

n= Molccular Mass Empirical formula mass =10030=6

∵ Molecular formula =n× empirical formula

∴ Molecular formula =6×CH2O=C6H12O6

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The empirical formula for a compound with a cubic close packed arrangement of anions and with cations occupying all the octahedral sites in Ax B. The value of x is (Integer answer)

Options:

The correct answer is:
1

Solution:
In AxB
Effective number of B atoms =4 (in CCP)
Effective number of A atoms =4 (all O.V.)
So formula of the compound =A4B4=AB
So x=1

Topic: Chemistry, Solid State, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The empirical formula of a hydrocarbon containing \( 80 \% \) carbon and \( 20 \% \) hydrogen is:

Options:
A \( \mathrm{CH} \)
B \( \mathrm{CH}_{2} \)
C \( \mathrm{CH}_{3} \)
D \( \mathrm{CH}_{4} \)

The correct answer is:
\( \mathrm{CH}_{3} \)

Solution:
Elements	Number of moles	Simple ratio

C=80%	8012=6.66	1

H=20%	201=20	3

Hence, empirical formula=CH3 .

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The empirical formula of a non-electrolyte is $\mathrm{CH}_{2} \mathrm{O}$. A solution containing $3 \mathrm{~g}$ of the compound exerts the same osmotic pressure as that of $0.05 \mathrm{M}$ glucose solution. The molecular formula of the compound is

Options:
A $\mathrm{CH}_{2} \mathrm{O}$
B $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$
C $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{4}$
D $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}$

The correct answer is:
$\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$

Solution:
For isotonic solution (solution with same osmotic pressure) $C_{1}=C_{2}$
$$
\begin{gathered}
\frac{W_{x}}{M_{X}}=\frac{W_{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}}{M_{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}} \\
\frac{3}{M_{x}}=\frac{9}{180} \\
0.05 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \text { containing } 9 \mathrm{~g} \text { of glucose } \\
M_{x}=\frac{3 \times 180}{9}=60
\end{gathered}
$$
Molecular weight of a non-electrolyte is 60
Molecular formula $=n \times$ empirical formula
Empirical formula $=\mathrm{CH}_{2} \mathrm{O}$
Empirical formula mass $=12+2+16=30$
$$
\begin{aligned}
n &=\frac{\text { molecular mass }}{\text { empirical formula mass }} \\
&=\frac{60}{30}=2
\end{aligned}
$$
Molecular formula $=2 \times \mathrm{CH}_{2} \mathrm{O}=\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$

Topic: Chemistry, Solutions, JEE Main
The empirical formula of a non-electrolyte is $\mathrm{CH}_{2} \mathrm{O}$. A solution containing $6 \mathrm{~g}$ of the compound exerts the same osmotic pressure as that of $0.05 \mathrm{M}$ glucose solution at the same temperature. The molecular formula of the compound is

Options:
A $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$
B $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}$
C $\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}$
D $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{4}$

The correct answer is:
$\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{4}$

Solution:
Solutions having same osmotic pressure, at a given temperature, have same concentration. Concentration of compound $=$ concentration of glucose
$$
\begin{aligned}
&\frac{6}{\mathrm{M} \times 1}=0.05 \\
&\mathrm{M}=\frac{6}{0.05}=120 \\
&\text { Empirical formula mass }\left(\mathrm{CH}_{2} \mathrm{O}\right)=12+2+16 \\
&\mathrm{n}=\frac{\text { molecular mass }}{\text { empirical formula mass }} \\
&=\frac{120}{30}=4
\end{aligned}
$$
Hence, molecular formula $=\left(\mathrm{CH}_{2} \mathrm{O}\right)_{4}=\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{4}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The emulsifying agent in milk is

Options:
A Lactic acid
B Casein
C Lactose
D Fat

The correct answer is:
Casein

Solution:
Correct option is 2. Casein
Casein is the emulsifying agent in milk which keeps two immiscible layers of fat and water together.

Topic: Chemistry, Surface Chemistry, JEE Main
The end A of a rod AB of length 1 m is maintained at 100 Â°C and the end B at 10 Â°C. What is the temperature (in Â°C) at a distance of 60 cm from the end B?

Options:

The correct answer is:
64

Solution:
The situation is given in the figure. Let Î¸ be the temperature at point C.
We know that the rate flow of heat
Qt=KA(Î¸1-Î¸2)d
Here, K = coefficient of thermal conductivity
A = area of cross-section
â KA(100-Î¸)40=KA(Î¸-10)60
100-Î¸2=Î¸-103
300-3Î¸=2Î¸-20
5Î¸=320
Î¸=3205
Î¸=64â

Topic: Physics, Thermal Properties of Matter, JEE Main
The end correction of a resonance column is 1 cm. If the shortest length resonating with the tuning fork is 10 cm, the next resonating length should be

Options:
A 32 cm
B 40 cm
C 28 cm
D 36 cm

The correct answer is:
32 cm

Solution:
Given: e=1 cm. For first resonance, λ4= l1+e=11 cm. For second resonance, 3λ4= l2+e ⇒ l2=3 ×11-1=32 cm.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2018 (16 Apr Online)
The end of two rods of different materials with their thermal conductivities, area of cross-section and lengths all are in the ratio 1 : 2 are maintained at the same temperature difference. If the rate of flow of heat in the first rod is 4 cal s-1, then in the second rod rate of heat flow in cal/s will be

Options:
A 1
B 2
C 8
D 16

The correct answer is:
8

Solution:
For the first rod, KAθ1-θ2L=4 cal s-1
For the second rod, 2K2Aθ1-θ22L=x cal s-1
Which gives x=8 cal s-1

Topic: Physics, Thermal Properties of Matter, NEET
The end product $B$ in the sequence of reactions
$\mathrm{R}-\mathrm{X} \xrightarrow{\mathrm{CN}^{-}} A \xrightarrow{\mathrm{NaOH}} B$ is

Options:
A An alkane
B A carboxylic acid
C Sodium salt of carboxylic acid
D A ketone

The correct answer is:
Sodium salt of carboxylic acid

Solution:
$\mathrm{R}-\mathrm{X} \xrightarrow{\mathrm{KCN}} \mathrm{R}-\mathrm{CN} \xrightarrow[\mathrm{H}_2 \mathrm{O}]{\mathrm{NaOH}} R-\mathrm{COONa}+\mathrm{NH}_3$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The end product in the following sequence of reaction is
$\mathrm{HC} \equiv \mathrm{CH} \xrightarrow[2. \mathrm{H}_2 \mathrm{SO}_4]{1 \mathrm{H}_8 \mathrm{SO}_4} A \xrightarrow{\mathrm{CH}_3 \mathrm{MeX}} B \xrightarrow{[0]}$

Options:
A Acetic acid
B Isopropyl alcohol
C Acetone
D Ethanol

The correct answer is:
Acetone

Solution:
\(\mathrm{HC} \equiv \mathrm{CH} \xrightarrow[20 \mathrm{HgSO}_4]{1 \% \mathrm{H}_4} \underset{\text { (A) }}{\mathrm{CH}_3 \mathrm{CHO}} \xrightarrow[\mathrm{H}_2 \mathrm{O}]{\mathrm{CH}_3 \mathrm{Mex}} \underset{\text { (A) }}{\mathrm{CH}_3 \mathrm{CHOHCH}_3} \xrightarrow{[\mathrm{O}]} \underset{\text{Acetone}}{\mathrm{CH}_3 \mathrm{COCH}_3}\)

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The end product of the reactions is
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 \xrightarrow{\mathrm{HNO}_2} A \xrightarrow{\mathrm{PCl}_5} B \xrightarrow{\mathrm{H}_{\mathrm{H}} \mathrm{NH}_2} C$

Options:
A Ethyl cyanide
B Ethyl amine
C Methyl amine
D Acetamide

The correct answer is:
Ethyl amine

Solution:

Topic: Chemistry, Amines, JEE Main
The end products of the following sequence of reactions are

Options:
A yellow precipitate of CHΙ3 ,
B yellow precipitate of CHΙ3 ,
C yellow precipitate of CHΙ3 ,
D yellow precipitate of CHΙ3 ,

The correct answer is:
yellow precipitate of CHΙ3 ,

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The endpoints of a conducting string (shaped as a circular loop) of constant length are being pulled at a constant velocity v as shown in the figure. There exists a uniform magnetic field B in space which is perpendicular to the circular loop. If the loop always remains circular during the motion of its endpoints, then the emf induced in the loop at t=ÏR2v is

Options:
A 2BRv
B BRv
C 4BRv
D 0.5BRv

The correct answer is:
BRv

Solution:
At any instant, the radius of the loop r, will satisfy the following relation

2πr+2vt=2πR

⇒r=R-vtπ

The flux of the magnetic field at any instant, will be

ϕB=πr2B

So, the induced emf is

Ε=-dϕdt=2πBrdrdt

⇒Ε=2πBR-vtπv

⇒Ε=BRv

Topic: Physics, Electromagnetic Induction, NEET
The ends of a rod of length l and mass m are attached to two identical springs as shown in the figure. The rod is free to rotate about its centre O. The rod is depressed slightly at end A and released. The time period of the oscillation is

Options:
A 2πm2k
B 2π2mk
C π2m3k
D π3m2k

The correct answer is:
π2m3k

Solution:
Let the rod be depressed by a small amount x (in the figure). Both the springs are compressed by x. When the rod is released, the restoring torque is given by
Ï=kxÃl2+kxÃ12=kxl
Now tanÎ¸=xl/2=2xl.
Since Î¸ is small, tanÎ¸âÎ¸, where Î¸ is expressed in radian.
Thus Î¸=2xl or x=Î¸l2
Ï=kÎ¸l2Ãl=kÎ¸l22
If I is the moment of inertia of the rod about O, then
Id2Î¸dt2=-kl22Î¸
Or d2Î¸dt2=-kl22IÎ¸
Since d2Î¸dt2â-Î¸, the motion is simple harmonic whose angular frequency is given by
Ï=kl22I
Now, Ï=2ÏT and I=ml212. Therefore, we have
2ÏT=kl22Ã12ml2=6km
Or T=Ï2m3k, which is choice Ï2m3k

Topic: Physics, Oscillations, JEE Main
The ends of a stretched wire of length l are fixed at x=0 and x=l. In one experiment, the displacement of the wire is y1=asinâ¡Ïxlsinâ¡Ït and energy E1 and in another experiment, its displacement is y2=asinâ¡3Ïxlsinâ¡3Ït and energy is E2 then

Options:
A E2=E1
B E2=3E1
C E2=E13
D E2=9E1

The correct answer is:
E2=9E1

Solution:
In first case freq. âÏ1=Ï
In second case freq. Ï2=3Ï
âµEnergy Î±freq.2
E1E2=Ï2(3Ï)2
âE2=9E1

Topic: Physics, Waves and Sound, JEE Main
The energies of activation for forward and reverse reactions for $A_2+B_2 \rightleftharpoons 2 A B$ are $180 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The enthalpy change of the reaction $\left(\mathrm{A}_2+\mathrm{B}_2 \longrightarrow 2 \mathrm{AB}\right)$ in the presence of catalyst will be (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ )

Options:
A
$300$

B
$120$

C
$280$

D
$20$

The correct answer is:
$20$

Solution:
So, $\Delta H_{\text {Reaction }}=E_f-E_b$
$=80-100=-20$
Hence, (D) is correct.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2007
The energy absorbed by each molecule $\left(\mathrm{A}_2\right)$ of a substance is $4.4 \times 10^{-19} \mathrm{~J}$ and bond energy per molecule is $4.0 \times 10^{-19} \mathrm{~J}$. The kinetic energy of the molecule per atom will be :

Options:
A $4.0 \times 10^{-20} \mathrm{~J}$
B $2.0 \times 10^{-20} \mathrm{~J}$
C $2.2 \times 10^{-19} \mathrm{~J}$
D $4 \times 10^{-19} \mathrm{~J}$

The correct answer is:
$2.0 \times 10^{-20} \mathrm{~J}$

Solution:
$$
\begin{gathered}
\mathrm{KE}=4.4 \times 10^{-19}-4.0 \times 10^{-19} \\
\mathrm{KE} / \text { molecule }=0.4 \times 10^{-19} \\
\mathrm{KE} / \text { atom }=\frac{0.4 \times 10^{-19}}{2} \\
=2 \times 10^{-20} \mathrm{~J}
\end{gathered}
$$

Topic: Chemistry, States of Matter, NEET, NEET 2009 (Mains)
The energy associated with electric field is UE and with magnetic field is UB for an electromagnetic wave in free space. Then:

Options:
A UE>UB
B UE=UB
C UE=UB2
D UE<UB

The correct answer is:
UE=UB

Solution:
The average energy density of electric field is given by

uE=12εE02

Here, E0 is the maximum value of electric field.

The average energy density of magnetic field is given by

uB=B022μ

Here, B0 is the maximum value of magnetic field.

The speed of light is given by c2=1με

Also, E0=cB0

So, uE=12c2μcB02=B022μ=uB

As the energy density of electric field and magnetic field are equal, the energy associated for equal volume will also be equal.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2019 (09 Jan Shift 2)