datavorous/entrance-exam-dataset huggingface dataset:
When phenyl magnesium bromide reacts with $t$-butanol, the product would be

Options:
A Benzene
B Phenol
C $t$-butyl benzene
D $t$-butyl phenyl ether

The correct answer is:
Benzene

Solution:
$\left(\mathrm{CH}_3\right)_3 \mathrm{COH}+\mathrm{PhMgBr} \longrightarrow \mathrm{PhH}$ $+\left(\mathrm{CH}_3\right)_3 \mathrm{COMgBr}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
When phosphate radical react with ammonium molybdate, the colour of precipitate obtained is

Options:
A Green
B Pink
C Canary yellow
D Violet

The correct answer is:
Canary yellow

Solution:
Phosphates give canary yellow precipitate with ammonium molybdate in the presence of conc. HNO3
H 3 PO 4  +  12(NH 4 ) 2 MoO 4  +  21HNO 3 → (NH 4 ) 3 PO 4 .12MoO 3  +  ammonium phosphomolybdate canary yellow ppt. 21NH 4 NO 3  +  12H 2 O

Topic: Chemistry, Practical Chemistry, JEE Main
When photon of energy 4.0 eV strikes the surface of a metal A, the elected photoelectrons have maximum kinetic energy TA eV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectron liberated from another metal B by photon of energy 4.50 eV is TB=TA-1.50 eV. If the de-Broglie wavelength of these photoelectrons λB=2λA, then choose the correct statement(s).

Options:
A The work function of A is 1.50 eV
B The work function of B is 4.0 eV
C TA=3.2 eV
D All of the above

The correct answer is:
The work function of B is 4.0 eV

Solution:
From Einstein photoelectric equation,
E=ϕ0+KEmax

For metal A 4=ϕA+TA....i
For metal B 4.5=ϕB+TA-1.5....ii

From Eq. (i) and (ii), wo got
ϕB-ϕA=2

Now, according to de-Broglie hypothesis,
λA=hmv=h2mTA

Similarly, λB=h2mTB

∴ λAλB=TBTA=TA-1.5TA=1-1.5TA1/2

122=1-1.5TA

On solving, TA=2.0eV

or ϕA=4-TA=4-2=2.0eV

ϕB=6-TA=6-2=4.0eV

Topic: Physics, Dual Nature of Matter, JEE Main
When photon of energy 4.0eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TAeV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50eV is TB=TA-1.5eV. If the de-Broglie wavelength of these photoelectrons λB=2λA, then the work function of metal B is:

Options:
A 4eV
B 2eV
C 1.5eV
D 3eV

The correct answer is:
4eV

Solution:
Relation between de-Broglie wavelength and K.E. is
λ=h2(KE)me⇒λ∝1KE
λAλB=KEBKEA
⇒12=TA-1.5TA⇒TA=2 eV
∴KEB=2-1.5=0.5 eV
ϕB=4.5-0.5=4 eV

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2020 (08 Jan Shift 1)
When photons of energy 4.0 eV fall on the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA (in eV) and a de-Broglie wavelength Î»A. When the same photons fall on the surface of another metal B, the maximum kinetic energy of ejected photoelectrons is TB=TAâ1.5 eV. If the de-Broglie wavelength of these photoelectrons is Î»B=2Î»A, then the work function of metal B is

Options:
A 2 eV
B 3 eV
C 2.5 eV
D 3.5 eV

The correct answer is:
3.5 eV

Solution:
The relation between de-Broglie wavelength and kinetic energy is

λ=h2Kme
λAλB=KBKA⇒12=TA-1.5TA ⇒TA=2 eV

⇒KB=2-1.5=0.5 eV

So, the work-function of B is
ϕB=4.0-0.5=3.5 eV

Topic: Physics, Atomic Physics, JEE Main
When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA (expressed in eV) and de Broglie wavelength Î»A. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20 eV is TB=TA-1.50 eV. If the de Broglie wavelength of these photoelectrons is Î»B=2Î»A, then which is not correct?

Options:
A The work function of A is 2.25 eV
B The work function of B is 3.70 eV
C TA=2.00 eV
D TB=2.75 eV

The correct answer is:
TB=2.75 eV

Solution:
Î»= h mV
V A = h m Î» A ;â V B = h m Î» B â
V A V B = Î» B Î» A = 2 Î» A Î» A =2
T A T B = m V A 2 m V B 2 = V A V B 2 = 2 2 =4
Also TA-TB=1.50
So, TB=0.50
TA=0.50+1.50=2.00 eV
Also, 4.25=hv0A+TA,
So hv0A=4.25-2.00=2.25 eV
4.20=hv0B+TB, hv0B=4.20-0.50=3.70 eV

Topic: Chemistry, Structure of Atom, JEE Main
When photons of wavelength λ1 are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength λ2 are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength λ3 is used then find the stopping potential for this case:

Options:
A hce1λ3+1λ2-1λ1
B hce1λ3+12λ2-1λ1
C hce1λ3-1λ2-1λ1
D hce1λ3+12λ2-32λ1

The correct answer is:
hce1λ3+12λ2-32λ1

Solution:
hcÎ»1=hcÎ»0+eV ...(i)
hcÎ»2=hcÎ»0+3eV ...(ii)
hcÎ»3=hcÎ»0+eVâ² ....(iii)
Equation (i) and (ii)
32Î»1-12Î»2=1Î»0
hcÎ»3-hc32Î»1-12Î»2=eVâ²
hce1Î»3-32Î»1+12Î»2=Vâ²

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2016 (09 Apr Online)
When p-n junction diode is forward biased

Options:
A
the depletion region is reduced and barrier height is increased

B
the depletion region is widened and barrier height is reduced

C
both the depletion region and barrier height reduced

D
both the depletion region and barrier height increased.

The correct answer is:
both the depletion region and barrier height reduced

Solution:
No Solution Available

Topic: Physics, Semiconductors, JEE Main, JEE Main 2004
When potassium is reacted with oxygen, which of the following compound(s) is/are formed?

Options:
A K 2 O
B KO 2
C Both K 2 O and KO 2
D K 2 O 2

The correct answer is:
KO 2

Solution:
Various alkali metals react with oxygen to give different types of compounds. Lithium reacts with oxygen to form simple oxides (Li2O), Sodium reacts to form peroxides (Na2O2) while potassium, rubidium and caesium form superoxides (MO2).
Hence, correct answer will be KO2.

Topic: Chemistry, s Block Elements, JEE Main
When power is drawn from the secondary coil of the transformer, the dynamic resistance

Options:
A increases
B decreases
C remains unchanged
D changes erratically

The correct answer is:
increases

Solution:
When the secondary coil circuit is open, the magnetic flux in the core is produced by the primary current only. When the secondary circuit is closed, the currents in the secondary coil also produce magnetic flux in the core but in opposite direction. This decreases the core flux and hence reduces the back emf produced in the primary coil. The source emf is now in excess of the back emf, more current is drawn in the primary coil. Hence, power factor is no longer zero. The power factor has increased or the phase difference is no longer $90^{\circ}$, ie, phase difference has decreased. Thus, dynamic resistance has increased.

Topic: Physics, Alternating Current, JIPMER, JIPMER 2008
When pressure (P) changes during a process, what condition ensures that the process is reversible?

Options:
A p is always greater than the pressure of the gas
B p is always less than the pressure of the gas
C p is infinitesimally greater than the pressure of the gas
D p remains constant throughout the process

The correct answer is:
p is infinitesimally greater than the pressure of the gas

Solution:
In a reversible process, the external pressure (Pex) is always infinitesimally greater than the pressure of the gas (Pin) during compression, and infinitesimally less during expansion.

Topic: Chemistry, Thermodynamics (C), NEET
When primary amine reacts with chloroform in ethanoic KOH then the product is

Options:
A
an isocyanide

B
an aldehyde

C
a cyanide

D
an alcohol

The correct answer is:
an isocyanide

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2002
When \(\psi_{\mathrm{A}}\) and \(\psi_{\mathrm{B}}\) are the wave functions of atomic orbitals, then \(\sigma^*\) is represented by :

Options:
A \(\psi_{\mathrm{A}}+2 \psi_{\mathrm{B}}\)
B \(\psi_A-\psi_B\)
C \(\psi_{\mathrm{A}}+\psi_{\mathrm{B}}\)
D \(\psi_A-2 \psi_B\)

The correct answer is:
\(\psi_A-\psi_B\)

Solution:
Antibonding molecular orbitals are formed by destructive interference of wave functions.
$(\mathrm{ABMO}) \sigma^*=\psi_{\mathrm{A}}-\psi_{\mathrm{B}}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (08 Apr Shift 2)
When radiation is incident on a photoelectron emitter, the stopping potential is found to be 9 V. If em for the electron is 1.8Ã1011 C Kg-1,the maximum velocity the ejected electron is

Options:
A 6Ã105 m s-1
B 8Ã105 m s-1
C 1.8Ã106 m s-1
D 1.8Ã105 m s-1

The correct answer is:
1.8Ã106 m s-1

Solution:
12mvmax2=eV0
⇒ vmax=2emV0
=2×1.8×1011×9
= 18×105 m s-1
= 1.8×106 m s-1

Topic: Physics, Dual Nature of Matter, JEE Main
When radiation of wavelength A is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3A, the stopping potential is V4. If the threshold wavelength for the metallic surface is nλ then value of n will be :

Options:

The correct answer is:
9

Solution:
hcλ=ϕ+eV ...........(1)

hc3λ=ϕ+eV4 ---------(2)

from (1) & (2)

hcλ1-13=34ev

hcλ23=34eV

eV=89hcλ

hcλ=ϕ+89hcλ

ϕ=hc9λ=hcλth

λth=9λ

∴k=9

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2020 (02 Sep Shift 1)
When radiation of wavelength λ is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is:

Options:
A 2λ
B 4λ
C 8λ
D 6λ

The correct answer is:
4λ

Solution:
VS=hv-ϕ

4.8=hcλ-ϕ ...(i)

1.6=hc2λ-ϕ ...(ii)

Using above equation (i) -(ii)

3.2=hcλ-hc2λ

3.2=hc2λ ...(iii)

λ=hc6.4

Put in equation (ii)

ϕ=1.6

hcλth=1.6

λth=hc1.6

=hc6.4×4=4λ

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (25 Jul Shift 2)
When rain is accompanied by a thunderstorm, the collected rain water will have a $\mathrm{pH}$ value

Options:
A
slightly higher than that when the thunderstorm is not there

B
uninfluenced by occurence of thunderstorm

C
which depends on the amount of dust in air

D
slightly lower than that of rain water without thunderstorm

The correct answer is:
slightly lower than that of rain water without thunderstorm

Solution:
The rain water after thunderstorm contains dissolved acid and therefore the $\mathrm{pH}$ is less than rain water without thunderstorm.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2003
When resonance is produced in a series LCR circuit, then which of the following is not correct?

Options:
A
Current in the circuit is in phase with the applied voltage.

B
Inductive and capacitive reactances are equal.

C
If $\mathrm{R}$ is reduced, the voltage across capacitor will increase.

D
Impedance of the circuit is maximum.

The correct answer is:
Impedance of the circuit is maximum.

Solution:
Impedance $(Z)$ of the series LCR circuit is
$Z=\sqrt{R^2+\left(X_L-X_C\right)^2}$
At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$
Therefore, $Z_{\text {minimum }}=R$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2013 (25 Apr Online)
When salt is added to water, which of the following statement is true?

Options:
A Boiling point decreases
B Boiling point increases
C Boiling point remain constant
D Freezing point increases

The correct answer is:
Boiling point increases

Solution:
This question is based on colligative properties of solution. According to this, These are the properties of solution which depend only on the number of solute particle but not the nature of the solute are called colligative properties.

When non-volatile solute (salt) mixed in water (solvent) then the boiling point of solution is higher than boiling point of pure solvent (water).

Topic: Chemistry, Solutions, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 2)
When same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide solution, the ratio of volumes of hydrogen evolved is:

Options:
A 1: 1
B 1: 2
C 2: 1
D 9: 4

The correct answer is:
1: 1

Solution:
$\mathrm{Zn}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{ZnSO}_{4}+\mathrm{H}_{2}$

$\mathrm{Zn}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2}+\mathrm{H}_{2}$

$\therefore$ Ratio of volumes of $\mathrm{H}_{2}$ evolved is 1: 1

Topic: Chemistry, d and f Block Elements, JEE Main
When $\mathrm{SCN}^{-}$is added to an aqueous solution containing $\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}$, the complex ion produced is

Options:
A $\left[\mathrm{Fe}\left(\mathrm{OH}_{2}\right)_{2}\left(\mathrm{SCN}^{-}\right)\right]^{2+}$
B $\left[\mathrm{Fe}\left(\mathrm{OH}_{2}\right)_{5}\left(\mathrm{SCN}^{-}\right)\right]^{2+}$
C $\left[\mathrm{Fe}\left(\mathrm{OH}_{2}\right)_{8}\left(\mathrm{SCN}^{-}\right)\right]^{2+}$
D $\left[\mathrm{Fe}\left(\mathrm{OH}_{2}\right)\left(\mathrm{SCN}^{-}\right)\right]^{6+}$

The correct answer is:
$\left[\mathrm{Fe}\left(\mathrm{OH}_{2}\right)_{5}\left(\mathrm{SCN}^{-}\right)\right]^{2+}$

Solution:
Fe(III) ion from ferric nitrate will react with thiocyanate, $\mathrm{SCN}^{-}$ion to form a blood red complex i.e. $\mathrm{FeSCN}^{2+} .$ But in presence of water it forms a complex containing five water molecules. i.e. $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\left(\mathrm{SCN}^{-}\right)\right]^{2+}$

Topic: Chemistry, Coordination Compounds, JEE Main
When shunt of 4 Î© is attached to a galvanometer, the deflection in the galvanometer is reduced to 15th. If an additional shunt of 2 Î© is also connected. The deflection of galvanometer will be reduced to

Options:
A 110th of initial
B 113th of initial
C remains the same
D none of these

The correct answer is:
113th of initial

Solution:
Intial condition
When shunt of 4Î© is used, i5ÃG=45iÃ4

âG=16Î©
When additional shunt of 2Î© is attached,
i'Ã16=iâi'43
12i'=iâi'
âi'=113
It will reduce to 113of the initial value.

Topic: Physics, Current Electricity, NEET
When sin9θcos27θ+sin3θcos9θ+sinθcos3θ=ktan27θ-tanθ is defined, then k=

Options:
A π2
B -12
C 12
D π4

The correct answer is:
12

Solution:
sin9θcos27θ+sin3θcos9θ+sinθcos3θ=ktan27θ-tanθ

Substitute θ=π3 in the above expression,

⇒sin9π3cos27π3+sin3π3cos9π3+sinπ3cos3π3=ktan27π3-tanπ3

⇒0+0+32-1=ktan9π-tan60°

⇒-3k=-32

∴ k=12

Topic: Mathematics, Trigonometric Equations, JEE Main
When $\mathrm{SO}_{2}$ is passed through acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{0}$, the process that takes place is

Options:
A the solution turns blue
B the solution is decolourised
C $\mathrm{SO}_{2}$ is reduced
D green $\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ is formed

The correct answer is:
green $\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ is formed

Solution:
When $\mathrm{SO}_{2}$ is passed through acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ solution. $\mathrm{SO}_{2}$ gas turns acidified potassium dichromate solution from orange to green reduced chromium $+4$ to $+3$.
$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{H}_{2} \mathrm{SO}_{4}+3 \mathrm{SO}_{2} \rightarrow 2 \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}$

Topic: Chemistry, Redox Reactions, JEE Main
When sodium dicarbonate is heated strongly for calcined in a kiln, it forms

Options:
A $\mathrm{Na}$
B $\mathrm{Na}_2 \mathrm{CO}_3$
C $\mathrm{NaCO}_3$
D $\mathrm{NaHCO}_3$

The correct answer is:
$\mathrm{Na}_2 \mathrm{CO}_3$

Solution:
On heating sodium bicarbonate strongly it decomposes to give carbon dioxide, water vapour, and sodium carbonate.
Reaction :
$2 \mathrm{NaHCO}_3(\mathrm{~s}) \rightarrow \mathrm{Na}_2 \mathrm{CO}_3(\mathrm{~s})+\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})$
Hence, Option "2" is the correct answer.

Topic: Chemistry, s Block Elements, JEE Main
When sodium dichromate is treated with \( \mathrm{KCl} \), it produces orange crystals of compound [A] which is treated with conc. \( \mathrm{H}_{2} \mathrm{SO}_{4} \) and soluble \( \mathrm{KCl} \) produces deep red vapour of [B]. Identify [A] and [B].

Options:
A [A] is \( \mathrm{K}_{2} \mathrm{CrO}_{4} \) and [B] is \( \mathrm{CrO}_{2} \mathrm{Cl}_{2} \)
B \( [\mathrm{A}] \) is \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) and \( [\mathrm{B}] \) is \( \mathrm{CrO}_{3} \)
C \( [\mathrm{A}] \) is \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) and [B] is \( \mathrm{CrO}_{2} \mathrm{Cl}_{2} \)
D \( [\mathrm{A}] \) is \( \mathrm{K}_{2} \mathrm{CrO}_{4} \) and \( [\mathrm{B}] \) is \( \mathrm{CrO}_{3} \)

The correct answer is:
\( [\mathrm{A}] \) is \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) and [B] is \( \mathrm{CrO}_{2} \mathrm{Cl}_{2} \)

Solution:
This problem includes, mixing of preparation and properties of K2Cr2O7 involving chromyl chloride test.
Preparation of K2Cr2O7 When sodium dichromate Na2Cr2O7 is treated with KCl it converts into K2Cr2O7. K2Cr2O7 is more stable than Na2Cr2O7. The chemical reaction occurs during this transformation is as follows.
Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl
Properties of K2Cr2O7 - K2Cr2O7 on reaction with conc. H2SO4 and water soluble inorganic compounds containing chloride ion produces deep red vapours of CrO2Cl2. This process is used to identification of chloride ion in inorganic salt. this test is know as chromyl chloride test.
The reaction occurs during chromyl chloride test is as follows.
K 2 Cr 2 O 7 + 2 H 2 SO 4 conc. → Δ 2 KHSO 4 + 2 CrO 3 red + H 2 O
NaCl + H 2 SO 4 conc. → NaHSO 4 + HCl
CrO 3 + 2 HCl → Δ CrO 2 Cl 2 chromyl chloride deep red vapour + H 2 O

Topic: Chemistry, d and f Block Elements, JEE Main
When sodium is heated with moist air, then the product obtained is

Options:
A $\mathrm{Na}_2 \mathrm{O}$
B $\mathrm{NaOH}$
C $\mathrm{Na}_2 \mathrm{CO}_3$
D $\mathrm{Na}_2 \mathrm{O}_2$

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

Solution:
$\begin{aligned} & 2 \mathrm{Na}+2 \mathrm{NH}_3 \xrightarrow{\text { heat }} 2 \mathrm{NaNH}_2+\mathrm{H}_2 \\ & (a, b) 2 \mathrm{Na}+\frac{1}{2} \mathrm{O}_2 \xrightarrow{\text { moistair }} \mathrm{Na}_2 \mathrm{O} \\ & \mathrm{Na}_2 \mathrm{O}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \text {. } \\ & \end{aligned}$

Topic: Chemistry, s Block Elements, JEE Main
When some amount of sodium acetate is further added to a mixture of acetic acid and sodium acetate, then pH of the solution

Options:
A increases
B decreases
C remains same
D none of these can be predicted from given information

The correct answer is:
increases

Solution:
When some amount of sodium acetate is further added to a mixture of acetic acid and sodium acetate, then pH of the solution increases
pH=pKa+log[Salt][Acid]
when the concentration of salt is increased, the value of pH will increase.

Topic: Chemistry, Ionic Equilibrium, NEET
When sound waves travel from air to water which one of the following remains constant?

Options:
A Time period
B Frequency
C Velocity
D Wavelength

The correct answer is:
Frequency

Solution:
Frequency of a sound wave remains independent of the medium through which it travels.Wavelength, velocity and phase change with the change in media.

Topic: Physics, Waves and Sound, JIPMER, JIPMER 2011
When spectral lines are passed through magnetic field, splitting is observed. This effect is known as

Options:
A Stark effect
B Photoelectric effect
C Zeeman effect
D Ramn effect

The correct answer is:
Zeeman effect

Solution:
The splitting of spectral line by the magnetic field is called Zeeman effect while by electric field is known as Stark effect.

Topic: Chemistry, Structure of Atom, JEE Main
When sucrose is heated with conc. $\mathrm{HNO}_3$ the product is

Options:
A Sucrose nitrate
B Formic acid
C Oxalic acid
D Citric acid

The correct answer is:
Oxalic acid

Solution:

Topic: Chemistry, Biomolecules, JEE Main
When sugar 'X' is boiled with dilute H2SO4 in alcoholic solution, two isomers 'A' and 'B' are formed. 'A' on oxidation with HNO3 yields saccharic acid where as 'B' is laevorotatory. The compound 'X' is

Options:
A Maltose
B Sucrose
C Lactose
D Strach

The correct answer is:
Sucrose

Solution:
Compound X is Sucrose. which on acidic hydrolysis gives Glucose and fructose.

Glucose on oxidation with HNO3 yields saccharic acid where as fructose is laevorotatory.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2022 (28 Jun Shift 2)
When sulphur is boiled with $\mathrm{Na}_2 \mathrm{SO}_3$ solution, the compound formed is

Options:
A Sodium sulphide
B Sodium sulphate
C Sodium persulphate
D Sodium thiosulphate

The correct answer is:
Sodium thiosulphate

Solution:
$\mathrm{Na}_2 \mathrm{SO}_3+\mathrm{S} \rightarrow N a_2 \mathrm{~S}_2 \mathrm{O}_3$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
When temperature increases, the frequency of a tuning fork

Options:
A
increases

B
decreases

C
remains same

D
increases or decreases depending on the material

The correct answer is:
decreases

Solution:
No Solution Available

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2002
When the absolute temperature of the source of a Carnot heat engine is increased by $25 \%$, its efficiency increases by $80 \%$. The new efficiency of the engine is

Options:
A $12 \%$
B $24 \%$
C $48 \%$
D $36 \%$

The correct answer is:
$36 \%$

Solution:
According to the question, initially
$\begin{aligned}
& \eta=\frac{100-T_2}{100} \\
& =\eta+\eta \times \frac{80}{100}=1.8 \eta=\frac{125-T_2}{125} \\
& \frac{\eta}{\eta^{\prime}}=\frac{100-T_2}{100} \times \frac{125}{125-T_2} \\
& \Rightarrow \quad \frac{1}{18}=\frac{100-T_2}{100} \times \frac{125}{125-T_2} \\
& \text { or, } \frac{5}{9}=\frac{100-T_2}{100} \times \frac{5}{\left(125-T_2\right)} \\
& \text { or, } 9\left(100-\mathrm{T}_2\right)=500-4 \mathrm{~T}_2 \\
& \text { or, } 900-9 \mathrm{~T}_2=500-4 \mathrm{~T}_2 \\
& \text { or, } 400=5 \mathrm{~T}_2 \\
& \text { or, } \mathrm{T}_2=\frac{400}{5}=80
\end{aligned}$
The new efficiency,
$\begin{aligned}
& \eta^{\prime \prime}=1-\frac{T_2}{T_1}=1-\frac{80}{125} \\
& \frac{125-80}{125}=\frac{45}{125}=\frac{45}{125} \times 100=36 \%
\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
When the angle of incidence on a transparent material is 60°, the reflected light is completely polarised. The velocity of refracted ray inside the material is

Options:
A 3 Ã 1 0 8 m/s
B 3 Ã 1 0 8 m/s
C 3 2 × 1 0 8 m/s
D 1 3 × 1 0 8 m/s

The correct answer is:
3 × 1 0 8 m/s

Solution:
Î¼=tanip
â´cv=tanip
â´v=3Ã1083=3Ã108 m/s

Topic: Physics, Wave Optics, JEE Main
When the area of cross-section of a stretched wire is halved and tension is doubled, the speed of propagation of transverse waves along it becomes $k$ times the initial speed. Then, $k$

Options:
A 1
B 4
C 2
D 8

The correct answer is:
2

Solution:
Speed of transverse wave in stretched wire is given as
$$
v=\sqrt{\frac{T}{m}}
$$
where, $T=$ tension in the wire
and $\quad m=$ mass per unit length of wire.
$$
\begin{aligned}
& \therefore \quad v=\sqrt{\frac{T}{M / l}} \quad\left[\because m=\frac{M}{l}\right] \\
& =\sqrt{\frac{T l}{M}}=\sqrt{\frac{T l}{V \cdot \rho}} \\
& {[\because M=V \cdot \rho]} \\
& =\sqrt{\frac{T l}{A \cdot l \cdot \rho}} \\
& {[\because V=A \cdot l]} \\
& v=\sqrt{\frac{T}{A \rho}} \Rightarrow v \propto \sqrt{\frac{T}{A}} \\
& \Rightarrow \quad \frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1} \times \frac{A_1}{A_2}} \\
& \text { Given, } \quad A_2=\frac{A_1}{2}, T_2=2 T_1 \\
&
\end{aligned}
$$
$\therefore$ From Eq. (i), we get
$$
\begin{aligned}
& \frac{v_2}{v_1}=\sqrt{\frac{2 T_1}{T_1} \times \frac{A_1}{\frac{A_1}{2}}}=\sqrt{4} \\
& v_2=2 v_1 \Rightarrow K=2
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main
When the azimuthal quantum number has the value of 2 , the number of orbitals possible are

Options:
A 7
B 5
C 3
D 0

The correct answer is:
5

Solution:
Given, azimuthal quantum number $(1)=2$
Number of orbitals $=(2 l+1)$
$$
=(2 \times 2+1)=4+1=5
$$

Topic: Chemistry, Structure of Atom, JEE Main