datavorous/entrance-exam-dataset huggingface dataset:
When a string is divided into three segments of lengths $l_1, l_2$ and $l_3$, the fundamental frequencies of these three segments are $v_1$, $v_2$ and $v_3$ respectively. The original fundamental frequency $(v)$ of the string is

Options:
A $\sqrt{v}=\sqrt{v_1}+\sqrt{v_2}+\sqrt{v_3}$
B $v=v_1+v_2+v_3$
C $\frac{1}{v}=\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}$
D $\frac{1}{\sqrt{v}}=\frac{1}{\sqrt{v_1}}+\frac{1}{\sqrt{v_2}}+\frac{1}{\sqrt{v_3}}$

The correct answer is:
$\frac{1}{v}=\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}$

Solution:
The fundamental frequency of string
$\begin{aligned}
v & =\frac{1}{2 l} \sqrt{\frac{T}{m}} \\
\therefore v_1 l_1 & =v_2 l_2=v_2 l_3=k
\end{aligned}$
From Eq. (i)
$l_1=\frac{k}{v_1}, l_2=\frac{k}{v_2}, l_3=\frac{k}{v_3}$
Original length
$\begin{aligned}
l & =\frac{k}{v} \\
l & =l_1+l_2+l_3 \\
\frac{k}{v} & =\frac{k}{v_1}+\frac{k}{v_2}+\frac{k}{v_3} \\
\frac{1}{v} & =\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}
\end{aligned}$
Here,
$\begin{aligned}
& l=l_1+l_2+l_3 \\
& \frac{k}{v}=\frac{k}{v_1}+\frac{k}{v_2}+\frac{k}{v_3} \\
& \frac{1}{v}=\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}
\end{aligned}$

Topic: Physics, Waves and Sound, JEE Main
When a substance, A reacts with water, it produces a combustible gas B and a solution of a substance, C in water. When another substance, D reacts with this solution of C, it also produces the same gas, B on warming, but D can produce B on reaction with dilute sulphuric acid at room temperature. A imparts a deep golden-yellow colour to a smokeless flame of Bunsen burner. A, B, C and D respectively are :

Options:
A Na,  H2,  NaOH  and  Zn
B K,  H2,  KOH   and  Al
C Ca,  H2,  CaOH2  and  Sn
D CaC2,  C2H2,  CaOH2  and  Sn

The correct answer is:
Na,  H2,  NaOH  and  Zn

Solution:
Na( s )  +   H 2 O( l )  →  NaOH( aq )  +   1 2 H 2 ( g )   [ A ]                                                     [ C ]                             [ B ]
Zn( s )  +  2NaOH( aq )  →   NaZnO 2 ( aq )  +   H 2 ( g )   [ D ]                   [ C ]                                                                             [ B ]
Zn( s )  +   H 2 SO 4   →   ZnSO 4 ( aq )  +   H 2 ( g )                            dilute                                                              [ B ]

Topic: Chemistry, s Block Elements, JEE Main
When a system is taken from state $\mathrm{i}$ to state $\mathrm{f}$ along the path iaf, it is found that $Q=50$ cal and $\mathrm{W}=20$ cal. Along the path $\mathrm{ibf} \mathrm{Q}=36$ cal. $\mathrm{W}$ along the path $\mathrm{ibf}$ is

Options:
A $14 \mathrm{cal}$
B $6 \mathrm{cal}$
C $16 \mathrm{cal}$
D 66 cal

The correct answer is:
$6 \mathrm{cal}$

Solution:
For path iaf,
$$
\begin{array}{l}
\mathrm{Q}=50 \mathrm{cal} \\
\mathrm{W}=20 \mathrm{cal}
\end{array}
$$
By first law of thermodynamics, $\Delta \mathrm{U}=\mathrm{Q}-\mathrm{W}=50-20=30 \mathrm{cal} .$ For path ibf

$$
\begin{array}{l}
\mathrm{Q}^{\prime}=36 \text { cal } \\
\mathrm{W}^{\prime}=? \\
\text { or, } \quad \mathrm{W}^{\prime}=\mathrm{Q}^{\prime}-\Delta \mathrm{U}^{\prime}
\end{array}
$$
Since, the change in internal energy does not depend on the path, therefore $\Delta \mathrm{U}^{\prime}=30 \mathrm{cal}$
$$
\therefore \quad \mathrm{W}^{\prime}=\mathrm{Q}^{\prime}-\Delta \mathrm{U}^{\prime}=36-30=6 \mathrm{cal} .
$$

Topic: Physics, Thermodynamics, JEE Main
When a system is taken from state $i$ to state $f$ along the path $\mathrm{iaf}$, it is found that $Q=50 \mathrm{~cal}$ and $\mathrm{W}=20 \mathrm{~cal}$. Along the path' $\mathrm{ibf}$ $\mathrm{Q=36 ~cal}$. $\mathrm{W}$ along the path $\mathrm{ibf}$ is

Options:
A
$6 \mathrm{~cal}$

B
$16 \mathrm{~cal}$.

C
$66 \mathrm{~cal}$.

D
$14 \mathrm{~cal}$.

The correct answer is:
$6 \mathrm{~cal}$

Solution:
No Solution Available

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2007
When a train approaches a stationary observer, the apparent frequency of the whistle is nâ² and when the same train recedes away from the observer, the apparent frequency is nâ²â². The apparent frequency n, when the observer moves with the train is

Options:
A n=n′+n2
B n=n′n′′
C n=2nâ²nâ²â²nâ²+nâ²â²
D n=2nâ²nâ²â²nâ²-nâ²â²

The correct answer is:
n=2n′n′′n′+n′′

Solution:
nâ²=vnv-vs,nâ²â²=vnv+vs
â´nnâ²=1-vsv,nnâ²â²=1+vsv
Adding the two, we get
nnâ²+nnâ²â²=2 â´ n=2nâ²nâ²â²nâ²+nâ²â²

Topic: Physics, Waves and Sound, JEE Main
When a transition of electron in He+ takes place from n2 to n1 then wave number in terms of Rydberg constant R will be
(Given n1+n2=4,n2-n1=2)

Options:
A 3R4
B 8R9
C 32R9
D 24R9

The correct answer is:
32R9

Solution:
n1+n2=4

n2−n1=2

2n2=6

n2=3

n1=1

3→1

ν̄=1λ
=RZ21n12-1n22
=R×22112-132
=4R×89
=32R9

Topic: Chemistry, Structure of Atom, JEE Main
When a $U^{238}$ nucleus originally at rest, decays by emitting an alpha particle having a speed ' $u$ ', the recoil speed of the residual nucleus is

Options:
A
$\frac{4 \mathrm{u}}{238}$

B
$-\frac{4 \mathrm{u}}{234}$

C
$\frac{4 \mathrm{u}}{234}$

D
$-\frac{4 \mathrm{u}}{238}$

The correct answer is:
$\frac{4 \mathrm{u}}{238}$

Solution:
Applying the principle of conservation of linear momentum
(4) $(\mathrm{u})=(\mathrm{v})(238) \Rightarrow \mathrm{v}=\frac{4 \mathrm{u}}{238}$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2003
When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of $220 \mathrm{~V}$. mains, the net

Options:
A
input voltage cannot be $\mathrm{AC}$ voltage, but a $\mathrm{DC}$ voltage

B
maximum input voltage is $220 \mathrm{~V}$

C
the meter reads not $\mathrm{v}$ but $\left\langle\mathrm{v}^2>\right.$ and is calibrated to $\operatorname{read} \sqrt{\left\langle\mathrm{v}^2\right\rangle}$

D
The pointer of the meter is stuck by some mechanical defect

The correct answer is:
the meter reads not $\mathrm{v}$ but $\left\langle\mathrm{v}^2>\right.$ and is calibrated to $\operatorname{read} \sqrt{\left\langle\mathrm{v}^2\right\rangle}$

Solution:
As we know that, The voltmeter in $\mathrm{AC}$ reads rms values of voltage
$$
\mathrm{I}_{\mathrm{ms}}=\sqrt{2} \mathrm{I}_0 \text { and } \mathrm{V}_{\mathrm{rms}}=\sqrt{2} \mathrm{v}_0
$$
The voltmeter in AC circuit connected to AC mains reads mean value $\left.\left( < v^2\right\rangle\right)$ and is calibrated in such a way that it gives rms value of $\left\langle v^2\right\rangle$, which is multiplied by form factor $\sqrt{2}$ to give rms value $V_{\text {ms }}$.

Topic: Physics, Alternating Current
When a wave travels in a medium, the particle displacement is given by the equation Y=asinâ¡2Ïbt-cx where, a, b and c are constant. The maximum particle velocity will be twice the wave velocity, if

Options:
A ac=1π
B ac=π
C b=ac
D b=1ac

The correct answer is:
ac=1π

Solution:
Given, Y=asinâ¡2Ïbt-cx ...(i)
and standard form Y=Asinâ¡Ït-kx
Thus, we have
Ï=2Ïb and K=2Ïc
or 2ÏÎ»=2Ïcâc=1Î»
â´Â Speed of wave,
v=ÏK=2Ïb2Ïc=bc
Differentiating Equation (i) w.r.t.t, we get
dYdt= Speed of particle,
vp=acosâ¡2Ïbt-cx⋅2Ïb
vp=2Ïabcosâ¡2Ïbt-cx
For vpmax=2Ïab
According to the question's condition,
vpmax=2v
â2Ïab=2⋅bc (Putting values)
âÏa=1c or ac=1Ï

Topic: Physics, Oscillations, JEE Main
When a wave traverses a medium, the displacement of a particle located at $x$ at a time $t$ is given by $y=a \sin (b t-c x)$, where $a, b$ and $c$ are constants of the wave, which of the following is a quantity with dimensions?

Options:
A $\frac{y}{a}$
B $b t$
C $c x$
D $\frac{b}{c}$

The correct answer is:
$\frac{b}{c}$

Solution:
Given, $y=a \sin (b t-c x)$
Comparing the given equation with general wave equation
$y=a \sin \left(\frac{2 \pi t}{T}-\frac{2 \pi x}{\lambda}\right)$
we get
$b=\frac{2 \pi}{T}, c=\frac{2 \pi}{\lambda}$
(a) Dimensions o
$\text { of } \begin{aligned}
\frac{y}{a} & =\frac{\text { metre }}{\text { metre }}=\frac{[\mathrm{L}]}{[\mathrm{L}]} \\
& =\text { Dimensionless }
\end{aligned}$
(b)
$\begin{aligned}
\text { Dimensions of } b t & =\frac{2 \pi}{T} \cdot t=\frac{[\mathrm{T}]}{[\mathrm{T}]} \\
& =\text { Dimensionless }
\end{aligned}$
(c) Dimensions of
$\begin{aligned}
c x & =\frac{2 \pi}{\lambda} \cdot x=\frac{[\mathrm{L}]}{[\mathrm{L}]} \\
& =\text { Dimensionless }
\end{aligned}$
(d) Dimensions of
$\begin{aligned}
\frac{b}{c} & =\frac{2 \pi}{T} / \frac{2 \pi}{\lambda} \\
& =\lambda / T=\left[\mathrm{LT}^{-1}\right]
\end{aligned}$
Thus, option (d) has dimensions.

Topic: Physics, Units and Dimensions, JEE Main
When a white light passes through a hollow prism, then

Options:
A There is no dispersion and no deviation
B Dispersion but no deviation
C Deviation but no dispersion
D There is dispersion and deviation both

The correct answer is:
There is no dispersion and no deviation

Solution:
Effectively there is no deviation or dispersion.

Topic: Physics, Ray Optics, JEE Main
When a wire is stretched and its radius becomes $r / 2$, then its resistance will be

Options:
A $16 R$
B $2 R$
C $4 R$
D 0.

The correct answer is:
$16 R$

Solution:
Initial radius of wire, $r_1=r$ and final radius of wire, $r_2=r / 2=0.5 r$.
Since volume of the wire after stretch remains constant, therefore $l_1 A_1=l_2 A_2$
or
$$
\frac{l_1}{l_2}=\frac{A_2}{A_1}=\frac{r_2^2}{r_1^2}=\left(\frac{0.5 r}{r}\right)^2=\frac{1}{4} .
$$
The resistance $(R)=\rho \frac{l}{A} \propto \frac{l}{A}$.
Therefore $\frac{R_1}{R_2}=\frac{l_1}{l_2} \times \frac{A_2}{A_1}=\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}$ or $\quad R_2=16 R_1=16 R$.

Topic: Physics, Current Electricity, AIIMS, AIIMS 2002
When a wire loop is rotated in a magnetic field, the direction of induced e.m.f. changes once in each

Options:
A $)^{\frac{1}{4}}$ revolution
B ${ }^{\frac{1}{2}}$ revolution
C $1$ revolution
D $2$ revolution

The correct answer is:
${ }^{\frac{1}{2}}$ revolution

Solution:
The induced EMF in a wire loop present in a magnetic field is given as a sinusoidal function.
$\mathrm{E}=\mathrm{E}_0 \sin (\omega \mathrm{t})$
The sinusoidal function has two sign changes in one cycle. Therefore, in every half of its revolution, the direction of the EMF induced will change.

Topic: Physics, Electromagnetic Induction, JEE Main
When a wire of uniform cross section $a$, length $l$ and resistance $R$ is into a complete circle, resistance between any two of diametrically opposite point will be:

Options:
A $\frac{R}{4}$
B $4 R$
C $\frac{R}{8}$
D $\frac{R}{2}$

The correct answer is:
$\frac{R}{4}$

Solution:
Both are in parallel
$$
\begin{aligned}
& \frac{1}{R^{\prime}}=\frac{2}{R}+\frac{2}{R}=\frac{4}{R} \\
& \Rightarrow \quad R^{\prime}=\frac{R}{4}
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main
When a zener diode is used as a regulator with zener voltage of $10 \mathrm{~V}$, nearly five times the load current passes through the zener diode. What should be the series resistance for the zener diode. If load resistance is $2 \mathrm{k} \Omega$ and the unregulated voltage supplied is $16 \mathrm{~V}$.

Options:
A $500 \Omega$
B $100 \Omega$
C $200 \Omega$
D $800 \Omega$

The correct answer is:
$200 \Omega$

Solution:
According to the question, the circuit diagram of zener diode as a voltage regulator is shown below,
Given, voltage supply, $V_s=16 \mathrm{~V}$, voltage across zener diode, $V_Z=10 \mathrm{~V}$, current passes through the zener diede, $I_Z=5 I_L$ and load resistance $R_L=2 \mathrm{k} \Omega$ The current through load resistance is
$$
\begin{aligned}
I_L & =\frac{\operatorname{voltage}\left(V_Z\right)}{\operatorname{resistance}\left(R_L\right)} \\
& =\frac{10}{2 \times 10^3}
\end{aligned}
$$
$$
=5 \times 10^{-3} \mathrm{~A} ...(I)
$$
The current through in the resistance in series
$$
\begin{aligned}
I & =I_Z+I_L \\
& =5 I_L+I_L
\end{aligned}
$$
$$
\begin{aligned}
& =6 I_L \\
& =6 \times 5 \times 10^{-3} \\
& =3 \times 10^{-2} \mathrm{~A}
\end{aligned}
$$
[From Eq. (i)]
Also, $\quad I=\frac{V_S-V_Z}{R_S}$
$$
\Rightarrow \quad R_S=\frac{V_S-V_Z}{I}
$$
substituting the values in Eqs. (ii), we get
$$
R_S=\frac{16-10}{3 \times 10^{-2}}=\frac{6}{3 \times 10^{-2}}=200 \Omega
$$
Thus, the series resistance for zener diode is $200 \Omega$.

Topic: Physics, Semiconductors, JEE Main
When acetaldehyde is heated with Fehling solution, a red precipitate is formed. Which of the following is that?

Options:
A $\mathrm{Cu}_2 \mathrm{O}$
B $\mathrm{Cu}$
C $\mathrm{CuO}$
D $\mathrm{CuSO}_4$

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

Solution:
When acetaldehyde is heated with Fehling solution, a red precipitate of $\mathrm{Cu}_2 \mathrm{O}$ is obtained.
$\mathrm{CH}_3 \mathrm{CHO}+2 \mathrm{Cu}(\mathrm{OH})_2+\mathrm{NaOH} \longrightarrow$$\mathrm{CH}_3 \cdot \mathrm{COONa}+\mathrm{Cu}_2 \mathrm{O} \downarrow(Red)+3 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Aldehydes and Ketones, JEE Main
When acetaldehyde is heated with Tollen's reagent, following is obtained

Options:
A Methyl alcohol
B Silver acetate
C Silver mirror
D Formaldehyde

The correct answer is:
Silver mirror

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
When acetaldehyde reacts with $\mathrm{PCl}_5$, the resulting compound is

Options:
A Ethyl chloride
B Ethylene chloride
C Ethylidene chloride
D Trichloroacetaldehyde

The correct answer is:
Ethylidene chloride

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
When acetamide is treated with $\mathrm{Br}_{2}$ and caustic soda, the product formed is

Options:
A N-bromamide
B bromoacetic acid
C methanamine
D ethanamine

The correct answer is:
methanamine

Solution:
Amides react with bromine and caustic soda to give their corresponding primary amines. Thus, acetamide gives methanamine. This reaction is known as Hofmann's bromamide degradation reaction.

Topic: Chemistry, Amines, JEE Main
When acetamide is treated with $\mathrm{HNO}_2$, the gas is evolved which of the following?

Options:
A $\mathrm{H}_2$
B $\mathrm{O}_2$
C $\mathrm{N}_2$
D $\mathrm{CH}_4$

The correct answer is:
$\mathrm{N}_2$

Solution:
Explanation: When acetamide is treated with $\mathrm{HNO}_2$, the gas is evolved in nitrogen gas
\(\underset{\text { Acetamide }}{\mathrm{CH}_3-\mathrm{CO}-\mathrm{NH}_2}+\mathrm{HNO}_2 \rightarrow \underset{\text { Acetic acid }}{\mathrm{CH}_3 \mathrm{COOH}}+\mathrm{N}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

Topic: Chemistry, Amines, JEE Main
When acetylene is passed through red hot iron tube, compound $X$ is formed. Which one of the following reactions will yield $X$ as the major product?

Options:
A $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}+\mathrm{Zn} \stackrel{\text { Distillation }}{\longrightarrow}$
B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_3 \mathrm{H}+\mathrm{NaHCO}_3 \longrightarrow$
C $\mathrm{C}_6 \mathrm{H}_{12}+3 \mathrm{H}_2 \stackrel{\mathrm{Ni}}{\longrightarrow}$
D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \stackrel{\Delta}{\longrightarrow}$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}+\mathrm{Zn} \stackrel{\text { Distillation }}{\longrightarrow}$

Solution:
When acetylene is passed through red hot iron tube, benzene is formed as
$3 \mathrm{C}_2 \mathrm{H}_2 \stackrel{\text { red hot tube }}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_6$$(X)$
Reaction (a) also gives $\mathrm{C}_6 \mathrm{H}_6$ (or $X$ ) as
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}+\mathrm{Zn} \stackrel{\text { distillation }}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_6+\mathrm{ZnO}$

Topic: Chemistry, Hydrocarbons, JEE Main
When acetylene is reacted with HBr, we get

Options:
A Methyl bromide
B Ethyl bromide
C Ethylene bromide
D Ethylidene bromide

The correct answer is:
Ethyl bromide

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
When acidulated water (dil. $\mathrm{H}_2 \mathrm{SO}_4$ solution) is electrolysed, will $\mathrm{pH}$ of the solution be affected? Justify your answer.

Options:

None

Solution:
For strong electrolytes, $\Lambda_m$ increases slowly with dilution. Electrolyte ' $B$ ' is stronger than ' $A$ ' because in ' $B$ ' the number of ions remains the same on dilution, but only interionic attraction decreases.
Therefore, $\Lambda_m$ increases only $1.5$ times. While in case of weak electrolyte on dilution, number of constituent ions increases.

Topic: Chemistry, Electrochemistry
When acidulated water (dil. H2SO4solution) is electrolysed, will the pH of the solution be affected?

Options:
A Yes, the pH will increase (become more alkaline).
B Yes, the pH will decrease (become more acidic).
C It depends on the concentration of H2SO4used.
D pH of the solution will not be affected as [H+] remains constant.

The correct answer is:
pH of the solution will not be affected as [H+] remains constant.

Solution:
Correct Option is : (D)
pH of the solution will not be affected as [H+] remains constant.

Topic: Chemistry, Electrochemistry, NEET
When AgCl is dissolved in excess of each of NH3 , KCN and Na2S2O3 separately, the complex ions produced in each case are respectively

Options:
A AgNH342+,AgCN23-, and Ag2S2O322-
B AgNH32+,AgCN2-, and AgS2O323-
C AgNH32+,AgCN23-, and Ag2S2O322-
D AgNH322+,AgCN23- and Ag2S2O322-

The correct answer is:
AgNH32+,AgCN2-, and AgS2O323-

Solution:
When AgCl is dissolved in excess of each of NH3, KCN and Na2S2O3 separately, the complex ions produced in each case are AgNH32+,AgCN2-, and AgS2O323- respectively as follows
AgCl+2NH4OH→AgNH32Cl+2H2O
AgCl+2KCN→KAgCN2+KCl

Topic: Chemistry, d and f Block Elements, JEE Main
When $\mathrm{AgCl}$ is treated with $\mathrm{KCN}$

Options:
A $\mathrm{Ag}$ is precipitated
B a complex ion is formed
C double decomposition takes place
D no reaction takes place

The correct answer is:
a complex ion is formed

Solution:
Hints: $\mathrm{AgCl}+2 \mathrm{KCN} \rightarrow \mathrm{K}\left[\mathrm{Ag}(\mathrm{CN})_2\right]+\mathrm{KCl}$

Topic: Chemistry, Coordination Compounds, JEE Main
When $\mathrm{AgNO}_3$ solution is added in excess to $1 \mathrm{M}$ solution of $\mathrm{CoCl}_3 . X \mathrm{NH}_3$, one mole of $\mathrm{AgCl}$ is formed. What is the value of ' $X$ '?

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

The correct answer is:
$4$

Solution:
It means that there is only one counter ion, i.e., the possible complex is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$. Hence, $X$ is 4 .

Topic: Chemistry, Coordination Compounds, JEE Main
When air is pumped into a cycle tyre, resulting in an increase in both the volume and pressure of the air in the tyre, how does this scenario align with Boyle's law?

Options:
A It contradicts Boyle's law, as the law applies when the number of molecules remains constant.
B It confirms Boyle's law, as the volume and pressure of the air change proportionally.
C It demonstrates a modification of Boyle's law, accommodating changes in the number of molecules.
D It suggests an extension of Boyle's law, allowing for situations involving the addition of more molecules.

The correct answer is:
It contradicts Boyle's law, as the law applies when the number of molecules remains constant.

Solution:
Correct Option is : (A)
It contradicts Boyle's law, as the law applies when the number of molecules remains constant.

Topic: Chemistry, Kinetic Theory of Gases, NEET
When air is pumped into a cycle tyre the volume and pressure of the air in the tyre both are increased. What about Boyle's law in this case?

Options:

None

Solution:
According to Boyle's law, at a constant temperature the volume of gas is inversely proportional to pressure i.e. it is valid only for constant mass of gas.
When air is pumped, more molecules are pumped and Boyle's law is stated for situation where number of molecules remains constant.
In that case, as the number of air molecules keep increasing.
Hence this is a case of variable mass. Boyle's law (and even Charle's law) is only applicable in situations, where number of gas molecules remains fixed.
Hence, Boyle's law is not applicable in this case.

Topic: Physics, Kinetic Theory of Gases
When aluminium is bombarded with fast neutrons, it changes into sodium with emission of particle x according to the equation Al1327+01n→1124Na+x
What is x?

Options:
A Electron
B Proton
C Neutron
D Alpha- particle

The correct answer is:
Alpha- particle

Solution:
The mass number of x=27+1-24=4 and its atomic number =13+0-11=2.
Hence particle x is the helium nucleus, which is called an alpha particle

Topic: Physics, Nuclear Physics, JEE Main
When an $A C$ source of emf $E$ with frequency $\omega=100 \mathrm{~Hz}$ is connected across a circuit, the phase difference between
$E$ and current I in the circuit is observed to be $\frac{\pi}{4}$ as shown in the figure. If the circuit consist of only $\mathrm{RC}$ or $\mathrm{RL}$ in series, then

Options:
A $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=5 \mu \mathrm{F}$
B $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{L}=10 \mathrm{H}$
C $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{L}=1 \mathrm{H}$
D $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=10 \mu \mathrm{F}$

The correct answer is:
$\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=10 \mu \mathrm{F}$

Solution:
.. Since supply voltage lags the current $\operatorname{By}(\phi=\pi / 4)$
$$
\begin{aligned}
&\tan \phi=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}=1, \mathrm{X}_{\mathrm{C}}=\mathrm{R} \ldots \ldots(1) \\
&\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 10 \times 10^{-6}}=1 \mathrm{k} \Omega=\mathrm{R}
\end{aligned}
$$

Topic: Physics, Alternating Current, JEE Main
When an AC voltage is applied to a L-C-R circuit, which of the following is true?

Options:
A \( I \) and \( V \) are out of phase with each other in \( R \)
B \( I \) and \( V \) are in phase in \( L \) while in \( C \), they are out of phase
C \( I \) and \( V \) are out of phase in both, \( C \) and \( L \)
D \( I \) and \( V \) are out of phase in \( L \) and in phase in \( C \)

The correct answer is:
\( I \) and \( V \) are out of phase in both, \( C \) and \( L \)

Solution:
Now in this phasor diagram VR is in phase with I but VL leads I by 90° while VC lags behind I by 90°.

Topic: Physics, Alternating Current, JEE Main
When an acid or alkali is mixed with buffer solution, then $\mathrm{pH}$ of buffer solution

Options:
A Not changes
B Changes slightly
C Increases
D Decreases

The correct answer is:
Not changes

Solution:
Because the $\mathrm{pH}$ of buffer are not changed.

Topic: Chemistry, Ionic Equilibrium, JEE Main
When an acyl chloride is heated with $\mathrm{Na}$ salt of a carboxylic acid, the product is

Options:
A An ester
B An anhydride
C An alkene
D An aldehyde

The correct answer is:
An anhydride

Solution:
When an acyl halide is heated with acid salt, anhydrides are formed
$\mathrm{CH}_3 \mathrm{COONa}+\mathrm{CH}_3 \mathrm{COCl} \xrightarrow{4} \underset{\text { acetic anhydride }}{\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}}+\mathrm{NaCl}$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
When an air bubble of radius $r$ rises from the bottom to the surface of a lake, its radius becomes $\frac{5 \mathrm{r}}{4}$. Taking the atmospheric pressure to be equal to $10 \mathrm{~m}$ height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature):

Options:
A
$10.5 \mathrm{~m}$

B
$8.7 \mathrm{~m}$

C
$11.2 \mathrm{~m}$

D
$9.5 \mathrm{~m}$

The correct answer is:
$9.5 \mathrm{~m}$

Solution:
Using $\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2$
$$
\begin{aligned}
&\left(P_1\right) \frac{4}{3} \pi r^3=\left(P_2\right) \frac{4}{3} \pi \frac{125 r^3}{64} \\
&\frac{\rho g(10)+\rho g h}{\rho g(10)}=\frac{125}{64} \\
&640+64 h=1250
\end{aligned}
$$
On solving we get $h=9.5 \mathrm{~m}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
When an alkyl halide is allowed to react with a sodium alkoxide the product most likely is

Options:
A An aldehyde
B A ketone
C An ether
D A carboxylic acid

The correct answer is:
An ether

Solution:
RX+RONa→R-O-REther+NaX
It is a Williamson's synthesis reaction.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main