datavorous/entrance-exam-dataset huggingface dataset:
A solution of (-)-1-chloro-1-phenylethane in toluene recemises slowly in the presence of a small amount of SbCl5 , due to the formation of

Options:
A free radical
B carbene
C carbocation
D carbanion

The correct answer is:
carbocation

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
A solution of 17.1 w\% of sucrose (molar mass $=342 \mathrm{~g} \mathrm{~mol}^1$ ) is isotonic with a $x$ w\% solution of oxalic acid (molar mass $=90 \mathrm{~g}$ mol ${ }^1$ ). Assume the degree of dissociation of oxalic acid as 0.01 . What is $x$ ?

Options:
A 9
B 0.45
C 4.41
D 0.90

The correct answer is:
4.41

Solution:
Given,
Mass of sucrose $\left(w_1\right)=17.1 \mathrm{~g}$
Molar mass of sucrose $\left(M_1\right)=342 \mathrm{~g} \mathrm{~mol}^{-1}$
Mass of oxalic acid $\left(w_2\right)=x g$
Molar mass of oxalic acid $\left(M_2\right)=90 \mathrm{~g} \mathrm{~mol}^{-1}$
Degree of dissociation $(\alpha)$ of oxalic acid $=0.01$
Step I To find the value of van't Hoff factor (i)
$i=(1-\alpha)+n \alpha$
where, $n=$ number of parts in which one molecule of oxalic acid dissociates $(=3)$
$$
\begin{aligned}
& i=(1-0.01)+(3 \times 0.01) \\
& i=0.99+0.03 \\
& i=1.02
\end{aligned}
$$
Step II To find the value of $(x)$
For isotonic solutions :
$\pi$ (sucrose $)=\pi$ (oxalic acid)
$$
[\because \pi=C R T]
$$
or, Molarity (sucrose) = i Molarity (oxalic acid)
$$
\begin{aligned}
& \left.\frac{w_1}{M_1} \text { (sucrose }\right)=i \times\left(\frac{x}{M_2}\right) \text { (oxalic acid) } \\
& x=\frac{w_1}{M_1} \times \frac{M_2}{i}=\frac{171 \times 90}{1.02 \times 342} \\
& x=\frac{1539}{348.84}=4.41 \mathrm{~g}
\end{aligned}
$$
Hence, value of $x=4.41 \mathrm{~g}$
Hence, option (3) is correct.

Topic: Chemistry, Solutions, JEE Main
A solution of 200 mL of 1 M KOH is added to 200 mL of 1 M HCl and the mixture is well shaken. This rise in temperature T1 is noted. The experiment is repeated by using 100 mL of each solution and increase in temperature T2 is again noted. Which of the following is correct?

Options:
A T1=T2
B T2 is twice as large as T1
C T1 is twice as large as T2
D T1 is four times as large as T2

The correct answer is:
T1=T2

Solution:
Heat produced by 200 Meq. of KOH + HCl
=13.7×2001000kcal
This is used to raise the temperature of 400 mL. Solution say T1
Thus, 13.7×2001000=400×S×T1
Now, heat produced by 100 Meq of KOH + HCl
=13.7×1001000
Thus, -13.7×1001000=200×S×T2
It is evident that T1=T2 .

Topic: Chemistry, Thermodynamics (C), JEE Main
A solution of a metal ion when treated with KI gives a red precipitate which dissolves in excess KI to give a colourless solution. Moreover, the solution of metal ion on treatment with a solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion is

Options:
A Pb2+
B Hg2+
C Cu2+
D Co2+

The correct answer is:
Hg2+

Solution:
When a solution of a Hg2+ is treated with KI a red precipitate of (HgI2) is obtained which dissolves in excess KI to give a colourless solution (HgI4)2- . When the solution of Hg2+ is treated with a solution of cobalt (II) thiocyanate a deep blue crystalline precipitate of Hg(SCN)2 is formed. The reactions are as follows
Hg2++2IââHgI2
HgI2(excess)+2Iââ[HgI4]2âSoluble
Hg 2+ +Co (SCN) 2 → Hg (SCN) 2 Blue crystal line ppt.
Here other ion Pb2+ , Hg2+ and Co2+ do not give precipitates formed in these reactions.

Topic: Chemistry, Practical Chemistry, JEE Main
A solution of acetone in ethanol:

Options:
A obeys Raoult's law
B shows a negative deviation from Raoult's law
C shows a positive deviation from Raoult's law
D behaves like a near ideal solution

The correct answer is:
shows a positive deviation from Raoult's law

Solution:
A solution of acetone in ethanol shows a positive deviation from Raoult's law because of the miscibility of these two liquids with a difference of polarity and the length of a hydrocarbon chain.
Related Theory
In an ideal solution, the solvent-solute interaction is the same as a solvent-solvent or solute - solute interaction. This implies both the solute and the solvent takes the same amount of energy to escape to the vapour phase as when they are in their pure states.
Caution
Raoult's law is apt for describing ideal solutions. Different liquids that are in the mixture do not have the same uniformity in terms of attractive forces, this type of solutions tends to deviate away from the law.

Topic: Chemistry, Solutions, JEE Main
A solution of ammonia in water contains

Options:
A $H^{+}$
B $\mathrm{OH}^{-}$
C Only $\mathrm{NH}_4^{+}$
D $\mathrm{OH}^{-}, \mathrm{NH}_4^{+}$and $\mathrm{NH}_4 \mathrm{OH}$ molecules

The correct answer is:
$\mathrm{OH}^{-}, \mathrm{NH}_4^{+}$and $\mathrm{NH}_4 \mathrm{OH}$ molecules

Solution:
$\mathrm{NH}_3$ when dissolved in water forms
$\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \mathrm{NH}_4 \mathrm{OH}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
A solution of concentration $C g$ equiv/L has a specific resistance $R$. The equivalent conductance of the solution is

Options:
A $\frac{R}{C}$
B $\frac{C}{R}$
C $\frac{1000}{R C}$
D $\frac{1000 R}{C}$

The correct answer is:
$\frac{1000}{R C}$

Solution:
Equivalent conductivity,
$\wedge_{\mathrm{eq}}=\frac{\text { specific conductance } \times 1000}{\text { concentration }(\text { in g equiv } / \mathrm{L})}$
$\because$ Specific conductance
$=\frac{1}{\text { specific resistance }}$
(Given, concentration $=C$ g equiv $/ \mathrm{L})$
$\therefore \quad \wedge_{\mathrm{eq}}=\frac{1}{R} \times \frac{1000}{C}$
or $\quad=\frac{1000}{R C}$

Topic: Chemistry, Electrochemistry, JEE Main
A solution of copper sulphate $\left(\mathrm{CuSO}_4\right)$ is electrolysed for $10$ minutes with a current of $1.5$ amperes. The mass of copper deposited at the cathode (at. mass of $\mathrm{Cu}=63 \mathrm{u}$ ) is :

Options:
A
$0.3892 \mathrm{~g}$

B
$0.2938 \mathrm{~g}$

C
$0.2398 \mathrm{~g}$

D
$0.3928 \mathrm{~g}$

The correct answer is:
$0.2938 \mathrm{~g}$

Solution:
$\mathrm{W}=\mathrm{Zit}$
where $\mathrm{Z}$ = Electrochemical equivalent
Eq. wt. of copper $=\frac{63}{2}=31.5$
$\begin{aligned}
& \mathrm{Z}=\frac{31.5}{96500} \\
& \mathrm{~W}=\mathrm{Zit}=\frac{31.5}{96500} \times 1.5 \times 10 \times 60=0.2938 \mathrm{~g}
\end{aligned}$

Topic: Chemistry, Electrochemistry, JEE Main
A solution of copper sulphate $\left(\mathrm{CuSO}_4\right)$ is electrolysed for $10$ minutes with a current of $1.5$ amperes. The mass of copper deposited at the cathode (at. mass of $\mathrm{Cu}=63 \mathrm{u}$ ) is :

Options:
A
$0.3892 \mathrm{~g}$

B
$0.2938 \mathrm{~g}$

C
$0.2398 \mathrm{~g}$

D
$0.3928 \mathrm{~g}$

The correct answer is:
$0.2938 \mathrm{~g}$

Solution:
$\mathrm{W}=\mathrm{Zit}$
where $\mathrm{Z}$ = Electrochemical equivalent
Eq. wt. of copper $=\frac{63}{2}=31.5$
$\begin{aligned}
& \mathrm{Z}=\frac{31.5}{96500} \\
& \mathrm{~W}=\mathrm{Zit}=\frac{31.5}{96500} \times 1.5 \times 10 \times 60=0.2938 \mathrm{~g}
\end{aligned}$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2013 (25 Apr Online)
A solution of $\mathrm{FeCl}_3$ in water acts as acidic due to

Options:
A Hydrolysis of $\mathrm{Fe}^{3+}$
B Acidic impurities
C Dissociation
D Ionization

The correct answer is:
Hydrolysis of $\mathrm{Fe}^{3+}$

Solution:
$\mathrm{FeCl}_3+3 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{Fe}(\mathrm{OH})_3+3 \mathrm{HCl}$. Strong acid and weak base.

Topic: Chemistry, Ionic Equilibrium, JEE Main
A solution of FeCl3 when treated with K4Fe(CN)6 gives a prussian blue precipitate due to the formation of

Options:
A KFe2(CN)6
B FeFe(CN)6
C Fe3Fe(CN)62
D Fe4Fe(CN)63

The correct answer is:
Fe4Fe(CN)63

Solution:
Addition of FeCl3 to K4FeCN6 in dilute and cold solution gives a solution, prussian blue solution Fe4FeCN63.

The reaction is as follows:

4FeCl3 + 3K4[Fe(CN)6] → Fe4 [Fe(CN)6 ]3 + 12KCl

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (01 Feb Shift 1)
A solution of H2SO4 is 31.4% H2SO4 by mass and has a density of 1.25 g/mL. The molarity of the H2SO4 solution is M (nearest integer) [Given molar mass of H2SO4=98 g mol-1]

Options:

The correct answer is:
4

Solution:
Given, Density= 1.25 g/mL,

Molar mass of H2SO4=98 g mol-1

Molarity is defined as no. of moles of solute present in one litre of solution. Thus,

Molarity= No. of moles of soluteVolume of solution in L

No. of moles= mass/ molar mass

M=nsolute V×1000

=31.4981001.25×1000

=4.005≈4

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (29 Jan Shift 1)
A solution of I2 in aqueous KI on reaction with an aqueous solution of Na2S2O3 gets decolourised. The reaction taking place here is

Options:
A Na2S2O3+H2O+I2→Na2S2O4+2HI
B 2Na2S2O3+I2→Na2S4O6+2NaI
C Na2S2O3+2H2O+2I2→Na2S2O5+4HI
D Na2S2O3+2H2O+2I2→Na2S4O8+4HI

The correct answer is:
2Na2S2O3+I2→Na2S4O6+2NaI

Solution:
On running sodium thiosulphate solution into an iodine solution (in KI solution), the reddish-brown colour is gradually discharged through yellow to colourless.
I2+2e-→2I-
2Na2S2O3→Na2S4O6+2Na++2e-
2Na2S2O3+I2→Na2S4O6+2Na+I-

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
A solution of m- chloroaniline, m- chlorophenol and m- chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO3 to give fraction A . The left over organic phase was extracted with dilute NaOH solution to give fraction B . The final organic layer was labelled as fraction C . Fractions A,B and C, contain respectively:

Options:
A m- chlorobenzoic acid, m- chloroaniline and m- chlorophenol
B m- chlorobenzoic acid, m- chlorophenol and m- chloroaniline
C m- chlorophenol, m- chlorobenzoic acid and m- chloroanilinc
D m- chloroaniline, m- chlorobenzoic acid and m- chlorophenol

The correct answer is:
m- chlorobenzoic acid, m- chlorophenol and m- chloroaniline

Solution:
Here are some of the important methods which are commonly employed for the purification of organic compounds:
Filtration.
Crystallisation or Recrystallisation.
Fractional Crystallisation.
Sublimation.
Simple Distillation.
Fractional Distillation.
Distillation under reduced pressure or Vacuum Distillation.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (07 Jan Shift 1)
A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO3 to give fraction A. The left over organic phase was extracted with dilute NaOH solution to give fraction B. The final organic layer was labelled as fraction C. Fraction A, B and C, contain respectively.

Options:
A m-chloroaniline, m-chlorobenzoic acid and m-chlorophenol
B m-chlorophenol, m-chlorobenzoic acid and m-chloroaniline
C m-chlorobenzoic acid, m-chloroaniline and m-chlorophenol
D m-chlorobenzoic acid, m-chlorophenol and m-chloroaniline

The correct answer is:
m-chlorobenzoic acid, m-chlorophenol and m-chloroaniline

Solution:
Fraction A, B and C, contain m-chlorobenzoic acid, m-chlorophenol and m-chloroaniline respectively

Topic: Chemistry, General Organic Chemistry, JEE Main
A solution of NiNO32 is electrolyzed between platinum electrode 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode?

Options:
A 0.10
B 0.15
C 0.05
D 0.20

The correct answer is:
0.05

Solution:
Ni+2+2e-→Ni
2 F charge gives = 1 mole of Ni
0.1 F charge gives =12×0.1=0.05  mole  Ni

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A solution of $\mathrm{NH}_{4} \mathrm{Cl}$ and $\mathrm{NH}_{3}$ has $\mathrm{pH}=8.0 .$ Which of the following hydroxides may be precipitated when this solution is mixed with equal volume of 0.2 M of metal ion.

Options:
A $\mathrm{Ba}(\mathrm{OH})_{2}\left(\mathrm{~K}_{\mathrm{sp}}=1.1 \times 10^{-4}\right)$
B $\mathrm{Mg}(\mathrm{OH})_{2}\left(\mathrm{~K}_{\mathrm{sp}}=3.5 \times 10^{-4}\right)$
C $\mathrm{Fe}(\mathrm{OH})_{2}\left(\mathrm{~K}_{\mathrm{sp}}=8.1 \times 10^{-16}\right)$
D $\mathrm{Ca}(\mathrm{OH})_{2}\left(\mathrm{~K}_{\mathrm{sp}}=2.1 \times 10^{-5}\right)$ ).

The correct answer is:
$\mathrm{Fe}(\mathrm{OH})_{2}\left(\mathrm{~K}_{\mathrm{sp}}=8.1 \times 10^{-16}\right)$

Solution:
$\mathrm{pH}=8, \mathrm{pOH}=6 ;\left[\mathrm{OH}^{-}\right]=10^{-6} \mathrm{M}$

Ionic product of $\mathrm{Fe}(\mathrm{OH})_{2}=0.2 \times\left(1 \times 10^{-6}\right)^{2}$

$=2 \times 10^{-13}>\mathrm{K}_{\mathrm{sp}}\left(=8.1 \times 10^{-16}\right)$

Topic: Chemistry, Ionic Equilibrium, JEE Main
A solution of Ni(NO3)2 is electrolyzed between platinum electrodes using a current of 5 amperes for 20 min. What mass of Ni is deposited at the cathode ? (Atomic mass of Ni = 58.7)
[Report your answer by rounding it upto nearset whole number]

Options:

The correct answer is:
2

Solution:
Quantity of electricity used = Current in amperes x Time in second = 5A x (20 x 60) s = 6000 C
Chemical equation for the reaction during electrolysis :
Ni NO 3 2 aq + 2 H + + 2 e - â¶ Ni + 2 HNO 3 2 F 5 8 . 7 g
Charge Q on n moles of electrons is given by : Q = nF
Charge required to deposit 1 mole nickel
= 2 F = 2 x 96500 C = 1.93 x 105 C
Molar mass of nickel = 58.7g mol-1
1.93 x 105 C of charge produces nickel = 58.7g
6000 C of charge produces nickel = 5 8 . 7 g Ã 6 0 0 0 C 1 . 9 3 Ã 1 0 5 C
= 1 . 8 2 g

Topic: Chemistry, Electrochemistry, JEE Main
A solution of phenol in chloroform when treated with aqueous NaOH gives compound Pas a major product. The mass percentage of carbon in P is ______ .

Options:

The correct answer is:
69

Solution:
Reimer-Tiemann formylation reaction :

Molecular formula of product is C7H6O2

Percentage weight of carbon=84122×100=68.85%

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2020 (06 Sep Shift 2)
A solution of sodium sulphate contains 92 g of Na+ ions per kilogram of water. The molality of Na+ ions in that solution in mol kg-1 is:

Options:
A 16
B 4
C 12
D 8

The correct answer is:
4

Solution:
Moles of Na+=Weight of Na+ ionsMolecular weight=9223=4

Molality is number of moles of solute present per Kg of Solvent

4 mole of Na+ are present in 1 kg of water Hence, molality of Na+ ions will be 4

Topic: Chemistry, Solutions, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A solution of sulphur dioxide in water reacts with $\mathrm{H}_2 \mathrm{~S}$ precipitating sulphur. Here sulphur dioxide acts as

Options:
A An oxidising agent
B A reducing agent
C An acid
D A catalyst

The correct answer is:
An oxidising agent

Solution:
$\underset{\text { reducing agent }}{2 \mathrm{H}_2 \mathrm{~S}}+\underset{\text { oxidising agent }}{\mathrm{SO}_2} \rightarrow 2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{~S}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
A solution of two miscible liquids showing negative deviation from Raoult's law will have :

Options:
A increased vapour pressure, increased boiling point
B increased vapour pressure, decreased boiling point
C decreased vapour pressure, decreased boiling point
D decreased vapour pressure, increased boiling point

The correct answer is:
decreased vapour pressure, increased boiling point

Solution:
Raoult's law states that a solvent's partial vapour pressure in a solution (or mixture) is equal or identical to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.

The compounds which show negative deviation from Raoult's law are a mixture of acetone and chloroform. This results in the formation of the hydrogen bonding between atoms of molecules. There is a decrease in the vapour pressure of the mixture that results in a negative deviation from Raoult's law.

Solution with negative deviation has
PT< PA0XA + PB0XB PA< PA0XA PB< PB0XB
If vapour pressure decreases, so the boiling point increases.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (27 Jan Shift 1)
A somatic cell that has just completed the S-phase of its cell cycle, as compared to gamete of the same species has

Options:
A twice the number of chromosomes and twice the amount of DNA
B the same number of chromosomes but twice the amount of DNA
C twice the number of chromosomes and four times the amount of DNA
D four times the number of chromosomes and twice the amount of DNA

The correct answer is:
twice the number of chromosomes and four times the amount of DNA

Solution:
The S phase is interphase where synthesis or replication of DNA takes place. During this phase, the chromosome number remains unchanged while the DNA in each cell doubles. Diploid somatic cells will therefore have twice the number of chromosomes and four times the number of DNA (2n) after the S phase as compared to the gametes which are haploid in number (n).

Topic: Biology, Cell Cycle and Cell Division, NEET
A sonometer wire is vibrating in resonance with a tuning fork. If the tension applied remains constant and the length of the wire is doubled, under what condition would the tuning fork still be in resonance with the wire?

Options:
A The frequency of the tuning fork is halved.
B The frequency of the tuning fork remains the same.
C The frequency of the tuning fork is doubled.
D The frequency of the tuning fork is quadrupled.

The correct answer is:
The frequency of the tuning fork remains the same.

Solution:
Correct Option is : (B)
The frequency of the tuning fork remains the same.

Topic: Chemistry, Waves and Sound, NEET
A sonometer wire of length 1.5m is made of steel. The tension in it produces an elastic strain of 1% . What is the fundamental frequency of steel if density and elasticity of steel are 7.7×103 kg m-3 and 2.2×1011 N m-2 respectively?

Options:
A 200.5 Hz
B 770 Hz
C 188.5 Hz
D 178.2 Hz

The correct answer is:
178.2 Hz

Solution:
f = V 2 â = 1 2 â T Î¼ = 1 2 â T Ad
Also, Y = T â A Î â â T A = Y Î â â â f = 1 2 â Y Î â â d
â = 1.5 m, Î â â = 0.01, d = 7.7 Ã 1 0 3 kg / m 3
Y = 2.2 Ã 1 0 1 1 N / m 2
After solving
f = 2 7 Ã 1 0 3 3 Hz
f â 178.2 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2013 (07 Apr)
A sonometer wire of length $114 \mathrm{~cm}$ is fixed at both the ends. Where should the two bridges be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio $1: 3: 4$ ?

Options:
A
At $36 \mathrm{~cm}$ and $84 \mathrm{~cm}$ from one end

B
At $24 \mathrm{~cm}$ and $72 \mathrm{~cm}$ from one end

C
At $48 \mathrm{~cm}$ and $96 \mathrm{~cm}$ from one end

D
At $72 \mathrm{~cm}$ and $96 \mathrm{~cm}$ from one end

The correct answer is:
At $72 \mathrm{~cm}$ and $96 \mathrm{~cm}$ from one end

Solution:
Total length of the wire, $\mathrm{L}=114 \mathrm{~cm}$ $\mathrm{n}_1: \mathrm{n}_2: \mathrm{n}_3=1: 3: 4$
Let $\mathrm{L}_1, \mathrm{~L}_2$ and $\mathrm{L}_3$ be the lengths of the three parts
As $\mathrm{n} \propto \frac{1}{\mathrm{~L}}$
$$
\begin{aligned}
& \therefore \quad \mathrm{L}_1: \mathrm{L}_2: \mathrm{L}_3=\frac{1}{1}: \frac{1}{3}: \frac{1}{4}=12: 4: 3 \\
& \therefore \quad \mathrm{L}_1=72 \mathrm{~cm}\left(\frac{12}{12+4+3} \times 114\right) \\
& \mathrm{L}_2=24 \mathrm{~cm}\left(\frac{4}{19} \times 114\right) \\
& \text { and } \quad \mathrm{L}_3=18 \mathrm{~cm}\left(\frac{3}{19} \times 114\right)
\end{aligned}
$$
Hence the bridges should be placed at $72 \mathrm{~cm}$ and $72+24=96 \mathrm{~cm}$ from one end.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2013 (23 Apr Online)
A sonometer wire of resonating length $90 \mathrm{~cm}$ has a fundamental frequency of $400 \mathrm{~Hz}$ when kept under some tension. The resonating length of the wire with fundamental frequency of $600 \mathrm{~Hz}$ under same tension _____$\mathrm{cm}$.

Options:

The correct answer is:
60

Solution:
$\begin{aligned} & \mathrm{f}_0=400 \mathrm{~Hz} ; \mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}}=\text { constant } \\ & \frac{\lambda}{2}=\mathrm{L} ; \quad \mathrm{v}=\mathrm{f}_0 \lambda\end{aligned}$
$\begin{aligned} & \frac{\mathrm{v}}{2 \mathrm{f}_0}=\mathrm{L} \Rightarrow \mathrm{v}=2 \mathrm{Lf}_0 \\ & \mathrm{~L}^{\prime}=\frac{\mathrm{v}}{2 \mathrm{f}^{\prime}}=\frac{2 \mathrm{Lf}_0}{2 \mathrm{f}^{\prime}} \\ & =\frac{\mathrm{Lf}_0}{\mathrm{f}^{\prime}}=\frac{90 \times 400}{600}=60\end{aligned}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2024 (05 Apr Shift 2)
A sound absorber attenuates the sound level by $20 \mathrm{~dB}$. The intensity decreases by a factor of

Options:
A
$1000$

B
$10000$

C
$10$

D
$100$

The correct answer is:
$100$

Solution:
$\begin{aligned}
& \mathrm{B}_1=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}_0}\right) \\
& \mathrm{B}_2=\log \left(\frac{\mathrm{I}^{\prime}}{\mathrm{I}_0}\right)
\end{aligned}$
given $B_2-B_1=20$
$\begin{aligned}
& 20=10 \log \left(\frac{\mathrm{I^{\prime}}}{\mathrm{I}}\right) \\
& \mathrm{I}^{\prime}=100 \mathrm{I}
\end{aligned}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2007
A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of

Options:
A 100
B 1000
C 10000
D 10

The correct answer is:
100

Solution:
L 1 = 10 log I 1 I 0 ; L 2 = 10 log I 2 I 0
∴ L 1 - L 2 = 10 log I 1 I 0 - 10 log I 2 I 0
or Δ L = 10 log I 1 I 2 or 20 dB = 10 log I 1 I 2
or 10 2 = I 1 I 2 or I 2 = I 1 100 .

Topic: Physics, Waves and Sound, JEE Main
A sound source emitting sound of frequency 450 Hz is approaching a stationary observer with velocity 30 m s-1 and another identical source is going away from the observer with the same velocity. If the velocity of sound is 330 m s-1, then the difference of frequencies heard by observer is

Options:
A 103.5 Hz

B 82.5 Hz
C 33.5 Hz
D 92.5 Hz

The correct answer is:
82.5 Hz

Solution:
f1=vv-vsf0=330330-30Ã450
f1=330300Ã450=11Ã45
f1=495 Hz
f2=330360Ã450
f2=412.5 Hz
Îf=82.5 Hz

Topic: Physics, Waves and Sound, JEE Main
A sound source S, emitting a sound of frequency 400 Hz and a receiver R of mass m are at the same point. R is performing SHM with the help of a spring of force constant K. At a time t=0, R is at the mean position and moving towards the right, as shown. At the same time, the source starts moving away from R with some acceleration a. The frequency registered by the receiver at a time t=10 s is 250 Hz. What is the time (in seconds) at which the corresponding registered frequency of 250 Hz was emitted by the source?
[Given that mK=25Ï2 and amplitude of oscillation =100Ï m, vsound=320 m s-1]

Options:

The correct answer is:
8

Solution:
The time period of oscillation T=2πmK=10 s

So at t = 10 s, the receiver passes through mean position towards the right with a speed

vR=Aω=100π×π5=20 m s-1

Using doppler’s formula

fapp=f0v-vRv+vS

250=400 320-20320+vS ⇒ vS=160 m s-1

So the velocity of the source is vS=160 m s-1 at the time of emission of the wave which was received by the receiver at t=10 s

Let's say the time instant when the wave was emitted is T, then the distance of the source from the receiver is 12 at2. This distance is travelled by the wave in a time 10-T. So,

12aT2=v10-T

vs=aT

⇒12vsTT2=v10-T

⇒12×160×T=320×10-T

⇒T=40-4T

⇒T=8 s

Topic: Physics, Waves and Sound, JEE Main
A sound wave is passing through air column in the form of compression and rarefaction. In consecutive compressions and rarefactions,

Options:
A
density remains constant

B
Boyle's law is obeyed

C
bulk modulus of air oscillates

D
there is no transfer of heat

The correct answer is:
there is no transfer of heat

Solution:
(a) The density of medium changes due to compression and rarefraction. At compressed regions density is maximum and at rarefactions density is minimum.
(b) As density is changing very rapidly so temperature of medium increases. Hence Boyle's law is not obeyed.
(c) Bulk modulus of air remains same or constant.
(d) The time of compressions and rarefraction is very small i.e., we can assume adiabatic process, i.e. no transfer of heat from surroundings.

Topic: Physics, Waves and Sound
A sound wave of frequency 245 Hz travels with the speed of 300 m s-1 along the positive x-axis. Each point of the wave moves to and fro through a total distance of 6 cm. What will be the mathematical expression of this travelling wave?

Options:
A Y(x,t)=0.03sin5.1x-0.2×103t
B Y(x,t)=0.06sin5.1x-1.5×103t
C Y(x,t)=0.06sin0.8x-0.5×103t
D Y(x,t)=0.03sin5.1x-1.5×103t

The correct answer is:
Y(x,t)=0.03sin5.1x-1.5×103t

Solution:
ω=2πf

=1.5×103

A=62=3 cm=0.03 m

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (17 Mar Shift 2)