datavorous/entrance-exam-dataset huggingface dataset:
The Born-Haber cycle for KCl is evaluated with the following data:
ΔfHΘ for KCl=−436.7kJmol−1;ΔsubHΘ for K=89.2kJmol−1;

Δionization HΘ for K=419.0kJ mol −1;Δelectron gain HΘ for Cl(g)=−348.6kJmol−1

Δbond HΘ for Cl2=243.0kJmol−1
The magnitude of lattice enthalpy of KCl in kJmol-1 is (Nearest integer)

Options:

The correct answer is:
718

Solution:
So, ΔfHΘ=ΔfHΘKCl=Δsub HΘ+Δ1st ionisation HΘK+1/2ΔbondHΘCl2(g)+Δelectron gain HΘ+Δlattice HΘ-436.7=89.2+419+12243+-348.6+Δlattice HΘ

Δlattice HΘ=-717.8 kJ mole-1≈-718 kJ mole-1

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (26 Aug Shift 1)
The π-bounded organometallic compound which has ethylene as one of its component is

Options:
A Dibenzene chromium
B Zeise salt
C Ferrocene
D Tetraethyl tin

The correct answer is:
Zeise salt

Solution:
Zeise salt : K PtCl 3 η 2 - C 2 H 4
π - electrons of ethylene bond by way of π - bonding to Pt.

Topic: Chemistry, Coordination Compounds, JEE Main
The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say bmin ) when:

Options:
A a=λL and bmin=4λL
B a=λ2L and bmin=4λL
C a=λ2L and bmin=2λ2L
D a=λL and bmin=2λ2L

The correct answer is:
a=λL and bmin=4λL

Solution:
In diffraction, first minima, we have

sinθ=λa

b=a+Lλa ...i

Then, minimum size of a spot, we get

∂b∂a=0⇒1-Lλa2=0⇒a=λL ...ii

So,

bmin=λL+λL

=2λL=4λL

Topic: Physics, Wave Optics, JEE Main, JEE Main 2016 (03 Apr)
The brain stem is made up of

Options:
A midbrain, pons, cerebellum.
B midbrain, pons, medulla oblongata.
C diencephalon, medulla oblongata, cerebellum.
D cerebellum, cerebrum, medulla oblongata

The correct answer is:
midbrain, pons, medulla oblongata.

Solution:
The brain stem is constituted by the midbrain, pons and medulla oblongata. It is responsible for connecting the cerebrum to the spinal cord.
The cerebellum is the part of the midbrain which is responsible for maintaining balance and coordination of the body. Diencephalon is the other name for hindbrain.

Topic: Biology, Neural Control and Coordination, NEET
The breakdown of detritus into smaller particles by earthworm is a process called

Options:
A humification
B fragmentation
C mineralisation
D catabolism

The correct answer is:
fragmentation

Solution:
The breakdown of detritus (dead remains of plants and animals) into smaller particle by detrivores like earthworm is called fragmentation. It is the first step in the process of decomposition.

Topic: Biology, Ecosystem, NEET, NEET 2011 (Mains)
The bromination of acetone that occurs in acid solution is represented by this equation
$\mathrm{CH}_3 \mathrm{COCH}_3(\mathrm{aq})+\mathrm{Br}_2(\mathrm{aq}) \rightarrow \mathrm{CH}_3 \mathrm{COCH}_2 \mathrm{Br}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})$
These kinetic data were obtained for given reaction concentrations Initial concentrations, M
Based on these data, the rate equation is

Options:
A $\mathrm{Rate}=\mathrm{k}\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{Br}_2\right]\left[\mathrm{H}^{+}\right]$
B $\mathrm{Rate}=\mathrm{k}\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{H}^{+}\right]$
C $\mathrm{Rate}=\mathrm{k}\left[\mathrm{CH}=\mathrm{COCH}_3\right]\left[\mathrm{Br}_2\right]$
D Rate $=\mathrm{k}\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{Br}_2\right]\left[\mathrm{H}^{+}\right]^2$

The correct answer is:
$\mathrm{Rate}=\mathrm{k}\left[\mathrm{CH}_3 \mathrm{COCH}_3\right]\left[\mathrm{H}^{+}\right]$

Solution:
In experiment (1) and (2), rate doesn't depend on the concentration of $\mathrm{Br}_2$. So, rate expression will not include $\left[\mathrm{Br}_2\right]$.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2008 (Mains)
The Brownian movement is due to:

Options:
A Enthalpy change during the formation of colloids
B Attractive forces between the colloidal particles and the molecules of dispersion medium
C The impact of molecules of the dispersion medium on the colloidal particles
D The movement of positively charged colloidal particle to negatively charged particle

The correct answer is:
The impact of molecules of the dispersion medium on the colloidal particles

Solution:
The continuous rapid zig-zag motion by a colloidal particle in the dispersion medium is called Brownian movement.

This movement is caused due to interaction of colloid particle with molecules of dispertion medium.

Topic: Chemistry, Surface Chemistry, JEE Main
The buffer system which helps to maintain the $\mathrm{pH}$ of blood between 7.26 to 7.42 , is

Options:
A $\mathrm{H}_2 \mathrm{CO}_3 / \mathrm{HCO}_3^{-}$
B $\mathrm{NH}_4 \mathrm{OH} / \mathrm{NH}_4 \mathrm{Cl}$
C $\mathrm{CH}_3 \mathrm{COOH} / \mathrm{CH}_3 \mathrm{COO}$
D $\mathrm{CH}_3 \mathrm{COONH}_4$

The correct answer is:
$\mathrm{H}_2 \mathrm{CO}_3 / \mathrm{HCO}_3^{-}$

Solution:
$\mathrm{H}_2 \mathrm{CO}_3$ is a weak dibasic acid and dissociates in following two steps :
$$
\begin{aligned}
& \mathrm{H}_2 \mathrm{CO}_3(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{HCO}_3^{-}(\mathrm{aq})+\mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq}) \\
& \mathrm{HCO}_3^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{CO}_3^{2-}(\mathrm{aq})+\mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})
\end{aligned}
$$
$\mathrm{H}_2 \mathrm{CO}_3 / \mathrm{HCO}_3^{-}$buffer system helps to maintain $\mathrm{pH}$ of blood between 7.26 to 7.42 .

Topic: Chemistry, Ionic Equilibrium, TS EAMCET, TS EAMCET 2015
The buffer system which helps to maintain the pH of blood between 7.26 to 7.42 is

Options:
A H2CO3 / HCO3-
B NH4OH / NH4Cl
C CH3COOH / CH3COO-
D CH3COONH4

The correct answer is:
H2CO3 / HCO3-

Solution:
The pH of blood is maintained at ~ 7.4 by the carbonic acid-bicarbonate ion buffering system.

Topic: Chemistry, Ionic Equilibrium, NEET
The bulk moduli of ethanol, mercury and water are given as $0.9,25$ and $2.2$ respectively in units of $10^9 \mathrm{Nm}^{-2}$. For a given value of pressure, the fractional compression in volume is $\frac{\Delta \mathrm{V}}{\mathrm{V}}$. Which of the following statements about $\frac{\Delta V}{V}$ for these three liquids is correct ?

Options:
A
Ethanol $>$ Water $>$ Mercury

B
Water $>$ Ethanol $>$ Mercury

C
Mercury $>$ Ethanol $>$ Water

D
Ethanol $>$ Mercury $>$ Water

The correct answer is:
Ethanol $>$ Water $>$ Mercury

Solution:
Compressibility $=\frac{1}{\text { Bulk modulus }}$
As bulk modulus is least for ethanol (0.9) and maximum for mercury (25) among ehtanol, mercury and water. Hence compression in volume $\frac{\Delta \mathrm{V}}{\mathrm{V}}$ Ethanol $>$ Water $>$ Mercury

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2014 (11 Apr Online)
The bulk modulus of a liquid is 3×1010Nm-2. The pressure required to reduce the volume of liquid by 2% is :

Options:
A 3×108 N m-2
B 6×108 N m-2
C 9×108 N m-2
D 12×108 N m-2

The correct answer is:
6×108 N m-2

Solution:
Bulk modulus B=Volumetric stressVolumetric strain=-∆P∆VV

Then, ∆P=B×-∆VV=3×1010×2100=6×108 N m-2

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2022 (24 Jun Shift 1)
The bulk modulus of a spherical object is B. If it is subjected to uniform pressure p, the fractional decrease in radius is

Options:
A pB
B B3p
C 3pB
D p3B

The correct answer is:
p3B

Solution:
V=Volume of sphere=43πr3
⇒∆VV=3∆rr
also, bulk modulus, B=-p∆VV
-∆VV=pB
3Δrr=pB
Δrr=p3B

Topic: Physics, Mechanical Properties of Solids, JEE Main
The $\mathrm{C}_4$ plants are phoptosynthetically more efficient than $\mathrm{C}_3$ plants because

Options:
A They have more chloroplasts
B The $\mathrm{CO}_2$ compensation point is more
C $\mathrm{CO}_2$ generated during photorespiration is trapped and recycled through PEP carboxylase
D The $\mathrm{CO}_2$ efflux is not prevented

The correct answer is:
They have more chloroplasts

Solution:
$\mathrm{C}_4$ plants do not show photorespiration and have more chloroplast in bundle sheath.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2008 (Mains)
The C-C bond length is maximum in:

Options:
A C70
B C60
C graphite
D diamond

The correct answer is:
diamond

Solution:
In diamond, all carbons are sp3 hybridised and hence bond lengths are similar to carbon-carbon single bonds.
Graphite, C70 and C60 forms of carbon have delocalisation of electrons which leads to double bond characteristics. Hence, their bond lengths are in between carbon-carbon single and double bonds.
Hence, diamond with only single bond character will be have the longest bond.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2019 (12 Apr Shift 2)
The Cl—C—Cl angle in 1, 1, 2, 2-tetrachloroethene and tetrachloromethane respectively will be about

Options:
A 109Â·5Â°and 90Â°
B 120Â°and 109Â·5Â°
C 90Â°and 109Â·5Â°
D 109Â·5Â°and 120Â°

The correct answer is:
120°and 109·5°

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The calculated spin-only magnetic moments BM of the anionic and cationic species of Fe H 2 O 6 +2 and Fe CN64- respectively, are:

Options:
A 0 and 4.9
B 2.84 and 5.92
C 4.9 and 0
D 0 and 5.92

The correct answer is:
0 and 4.9

Solution:
In both cases iron exhibits +2 oxidation state.

Electronic configuration of F e +2 ion =[Ar]3d6

In [Fe(H2O)6]2+, H2O is a weak field ligand and hence, no pairing up.

∴ number of unpaired electron n=4

⇒Spin magnetic moment =4(4+2)=24=4.9 BM

In [Fe(CN)6]4-, CN- is a strong field ligand and hence, pairing will occur.

∴ number of unpaired electron n=0

⇒ spin magnetic moment = zero

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The
can detect the endpoint of a titration involving the reaction of $\mathrm{MnO}_4$ - to Mn2+.

Options:
A Use of oH probe
B Disappearance of the pink colour of the $\mathrm{MnO}_4$
C Addition of an indicator like methyl orange
D Use of red or blue litmus

The correct answer is:
Disappearance of the pink colour of the $\mathrm{MnO}_4$

Solution:
The endpoint of a titration concerning the reaction of $\mathrm{MnO} 4-$, to $\mathrm{Mn} 2+$ can be perceived by the disappearance of pink color.
The pink color disappears in the titration because of the reason that of end point reach. It is an indication that the reaction is attaining its endpoint and the reaction between the sample and the titrant are about to complete.

Topic: Chemistry, Redox Reactions, JEE Main
The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency ' \(\mathrm{A}\) ' \(\times 10^{12}\) hertz and that has a radiant intensity in that direction of \(\frac{1}{'\mathrm{~B}^{\prime}}\) watt per steradian.
'A' and 'B' are respectively

Options:
A 540 and 683
B 450 and 683
C 450 and \(\frac{1}{683}\)
D 540 and \(\frac{1}{683}\)

The correct answer is:
540 and 683

Solution:
The candela is the luminous intensity of a source that emits monochromatic radiation of frequency radiation of frequency $540 \times 10^{12} \mathrm{~Hz}$ and has a radiant intensity in that direction of $\frac{1}{683} \mathrm{w} / \mathrm{sr}$. It is unit of Candela.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The Cannizaro reaction is not given by

Options:
A Trimethyl acetaldehyde
B Acetaldehyde
C Benzaldehyde
D Formaldehyde

The correct answer is:
Acetaldehyde

Solution:
(CH 3 ) 3 C−CHO Trimethyl acetaldehyde CH 3 CHO Acetaldehyde Ph−CHO Benzaldehyde HCHO Formaldehyde
Cannizaro reaction is given by only those aldehydes which does not have Î±-hydrogen atom. As such acetaldehyde has alpha hydrogen so will not give Cannizaro reaction.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The capacitance of a parallel plate capacitor becomes 43 times its original value. If a dielectric slab of thickness t=d2 is inserted between the plates (where d is the distance of separation between the plates), what is the value of the dielectric constant of the slab?

Options:

The correct answer is:
2

Solution:
Let C be the original capacitance.
â´ New, capacitance, Câ²=43C
Thickness, t=d2
â Îµ0Ad-t1-1K=43Îµ0Ad
â 21+1K=43
â K=2

Topic: Physics, Electrostatics, JEE Main
The capacitance of a parallel plate capacitor is 2μF and the charge on its positive plate is 2μC. If the charge on its plates is doubled, the capacitance of the capacitor

Options:
A remains 2μF
B becomes 1μF
C becomes 4μF
D data insufficient

The correct answer is:
remains 2μF

Solution:
The capacity of capacitance is a geometrical quantity which depends on the factors like shape, size, spatial arrangement and the medium present. So if charge is doubled the capacitance would still remain the same.

Topic: Physics, Electrostatics, NEET
The capacitance of a parallel plate capacitor with air as medium is $3 \mu \mathrm{F}$. With the introduction of a dielectric medium between the plates, the capacitance becomes 15 $\mu F$. The permittivity of the medium in SI unit is $\left[\epsilon_{0}=8 \cdot 85 \times 10^{-12}\right.$ SI unit]

Options:
A 15
B $8.845 \times 10^{-11}$
C $0.4425 \times 10^{-10}$
D 44. 5

The correct answer is:
$0.4425 \times 10^{-10}$

Solution:
Capacitance of a paralle plate capacitor $\mathrm{C}=\frac{\varepsilon \mathrm{A}}{\mathrm{d}}$
here $\varepsilon$ is the permitivity of the medium
For air medium $\varepsilon=\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \quad \therefore \mathrm{C}^{\prime}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}} \ldots .1$
whearas for dielectric medium capacitance $\mathrm{C}=\frac{\varepsilon \mathrm{A}}{\mathrm{d}} \ldots 2$
From 1 and 2
$$
\begin{array}{l}
\frac{\mathrm{C}}{\mathrm{C}^{\prime}}=\frac{\varepsilon}{\varepsilon_{0}} \\
\therefore \varepsilon=\frac{15}{3} \times 8.85 \times 10^{-12}=\left(44.25 \times 10^{-12}\right)=0.44 \times 10^{-10} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{2}
\end{array}
$$
Hence the permitivity of the medium $\varepsilon=0.44 \times 10^{-10} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{2}$

Topic: Physics, Capacitance, JEE Main
The capacitance of the capacitors C1, C2 and C3 are 4 μF, 6 μF and 12 μF respectively as shown, and the switch S remains closed for a long time. When the switch S is opened, which of the following statements will be correct about the current flowing through the battery B?

Options:
A a finite and constant current will flow
B a finite current will flow initially thatwill decrease exponentially with time
C no current will flow
D information is insufficient to predict

The correct answer is:
no current will flow

Solution:
Capacitors are already in steady state.

Topic: Physics, Electrostatics, JEE Main
The capacitor of an oscillatory circuit is enclosed in a container. When the container is evacuated, the resonance frequency of the circuit is $10 \mathrm{kHz}$. When the container is filled with a gas, the resonance frequency changes by $50 \mathrm{~Hz}$. The dielectric constant of the gas is

Options:
A
$1.001$

B
$2.001$

C
$1.01$

D
$3.01$

The correct answer is:
$1.01$

Solution:
The dielectric constant of the gas is $1.01$

Topic: Physics, Capacitance, JEE Main, JEE Main 2012 (26 May Online)
The capacitor shown in the figure is initially uncharged, the battery is ideal. The switch S is closed at time t=0 , then the time after which the energy stored in the capacitor becomes one-fourth of the energy stored in it in steady-state is:

Options:
A RC
B RCln2
C RCln4
D 2RC

The correct answer is:
RCln2

Solution:
U=Umax4 â q22C=14Qmax22C â q=Qmax2
q=CE2 â q=CE1-e-t/RC
CE2=CE1-e-t/RC â 12=1-e-t/RC
e-t/RC=12
â tRC=ln2
ât=RCln2

Topic: Physics, Current Electricity, JEE Main
The capacity of a condenser is $4 \times 10^{-6}$ farad and its potential is 100 volts. The energy released on discharging it fully will be

Options:
A 0.02 Joule
B 0.04 Joule
C $0.025$ Joule
D $0.05$ Joule

The correct answer is:
0.02 Joule

Solution:
$U=\frac{1}{2} C V^2=\frac{1}{2} \times 4 \times 10^{-6} \times(100)^2=0.02 \mathrm{~J}$

Topic: Physics, Capacitance, JEE Main
The capacity of a parallel plate capacitor increases with the

Options:
A Decrease of its area
B Increase of its distance
C Increase of its area
D None of the above

The correct answer is:
Increase of its area

Solution:
Capacity of parallel plate capacitor is directly proportional to area

Topic: Physics, Capacitance, JEE Main
The capacity of a parallel plate condenser is $5 \mu \mathrm{F}$. When a glass plate is placed between the plates of the conductor, its potential becomes $1 / 8$ th of the original value. The value of dielectric constant will be

Options:
A 1.6
B 5
C 8
D 40

The correct answer is:
8

Solution:
$V=\frac{V}{8} \Rightarrow \frac{V}{K}=\frac{V}{8} \Rightarrow K=8$

Topic: Physics, Capacitance, JEE Main
The capacity of a parallel plate condenser is $10 \mu F$ without dielectric. Dielectric of constant 2 is used to fill half the distance between the plates, the new capacitance in $\mu F$ is

Options:
A 10
B 20
C 15
D 13.33

The correct answer is:
13.33

Solution:
$\begin{aligned} & C=\frac{A c_0}{d}=10 \mu F \\ & C_1=\frac{A c_0}{d-t+\frac{t}{k}}=\frac{A c_0}{d-\frac{d}{2}+\frac{d}{2 k}}=\frac{A c_0}{\frac{d}{2}\left(1+\frac{1}{2}\right)}=\frac{4}{3} \cdot \frac{A c_0}{d} \\ & \therefore C_1=\frac{4}{3} \times 10=13.33 \mu F\end{aligned}$

Topic: Physics, Capacitance, JEE Main
The capacity of a parallel plate condenser is $C$. Its capacity when the separation between the plates is halved will be

Options:
A $4 C$
B $2 \mathrm{C}$
C $\frac{\mathrm{c}}{2}$
D $\frac{C}{4}$

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

Solution:
The capacity of a parallel plate condenser is given as,
\(\mathrm{C}=\frac{\varepsilon \mathrm{A}}{\mathrm{d}}\)
When the separation between the plates is halved then the new capacity of the condenser is given as,
\(\begin{aligned}
& \mathrm{C}^{\prime}=\frac{\varepsilon \mathrm{A}}{\frac{\mathrm{d}}{2}} \\
& \mathrm{C}^{\prime}=2 \mathrm{C}
\end{aligned}\)
Thus, the new capacity of parallel plate condenser is \(2 \mathrm{C}\).

Topic: Physics, Capacitance, JEE Main
The capacity of a spherical conductor in MKS system is

Options:
A $\frac{R}{4 \pi \varepsilon_0}$
B $\frac{4 \pi \varepsilon_0}{R}$
C $4 \pi \varepsilon_0 R$
D $4 \pi \varepsilon_0 R^2$

The correct answer is:
$4 \pi \varepsilon_0 R$

Solution:
The potential is given as,
$\mathrm{V}=\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{R}}$
The capacitance is given as,
$\begin{aligned} & C=\frac{q}{V_q} \\ & =\frac{q}{\frac{q}{4 \pi \varepsilon_0 R}} \\ & =4 \pi \varepsilon_0 R\end{aligned}$
Thus, the capacity of a spherical conductor is $4 \pi \varepsilon_0 R$.

Topic: Physics, Capacitance, JEE Main
The capacity of a vessel is 3 L. It contains 6 g oxygen, 8 g nitrogen and 5 g CO2 mixture at 27 Â°C. If R = 8.31 J mol-1 K-1, then the pressure in the vessel in N m-2 will be (approx).

Options:
A 5 × 105

B 5 Ã 104
C 1×106
D 1×105

The correct answer is:
5 Ã 104

Solution:
By Dalton's law
P = P 1 + P 2 + P 3 = μ 1 RT V + μ 2 RT V + μ 3 RT V
= RT V μ 1 + μ 2 + μ 3 = RT V m 1 M 1 + m 2 M 2 + m 3 M 3
= 8.31 × 3 0 0 3 × 1 0 - 3 6 3 2 + 8 2 8 + 5 4 4
=498×103=500×103=5×105 N m-2

Topic: Physics, Thermodynamics, JEE Main
The capacity of an environment to pull on a limited number of individuals is known as

Options:
A Bearing capacity
B Limited capacity
C Environmental resistance
D Carrying capacity

The correct answer is:
Carrying capacity

Solution:
Carrying capacity refers to the size of a population that can be maintained indefinitely. Carrying capacity is the maximum number of individuals of a given species that an area's resources can sustain indefinitely without significantly depleting or degrading those resources.

Topic: Biology, Organisms and Populations, NEET
The capacity of parallel plate condenser depends on

Options:
A The type of metal used
B The thickness of plates
C The potential applied across the plates
D The separation between the plates

The correct answer is:
The separation between the plates

Solution:
The capacity of the parallel plate condenser depends on the separation between the plates.
Explanation:
$C=\frac{K \varepsilon_0 A}{d}$

Topic: Physics, Capacitance, JEE Main
The capacity of parallel plate condenser is $5 \mu \mathrm{F}$. When a glass plate is placed between the plates of the condenser, its potential difference reduces to $1 / 8$ of the original value. The magnitude of relative dielectric constant of glass is

Options:
A 4
B 6
C 7
D 8

The correct answer is:
8

Solution:
Given, $C=5 \mu \mathrm{F}=5 \times 10^{-6} \mathrm{~F}$
Initial potential $=V$
When glass plate of dielectric constant $K$ is introduced between the plates of capacitor, then potential difference is
$$
\begin{aligned}
& V^{\prime} & =\frac{V}{8} \\
\therefore & K & =\frac{V}{V^{\prime}}=\frac{V}{V / 8} \\
\Rightarrow & K & =8
\end{aligned}
$$

Topic: Physics, Capacitance, JEE Main
The carbonate of which of the following cation is insoluble in water?

Options:
A Na +
B K +
C NH 4 +
D Ca 2+

The correct answer is:
Ca 2+

Solution:
Carbonates of alkali metal and ammonium are soluble in water.
All other carbonates are insoluble in water.

Topic: Chemistry, s Block Elements, JEE Main
The carbon-carbon bond distance in benzene is

Options:
A Longer than a C-C single bond
B Longer than a C=C double bond
C Shorter than a C=C double bond
D Shorter than a C≡C triple bond

The correct answer is:
Longer than a C=C double bond

Solution:
The bond order of carbon-carbon bond is 1.5 in benzene.
Hence, carbon-carbon bond distance (1.397  A o ) is less than C = C single bond (1.33  A o ). and more than C=C double bond (1.33  A o )

Topic: Chemistry, General Organic Chemistry, JEE Main
The carnot heat engine have an efficiency of 50%. The temperature of sink is maintained at 500 K. To increase the efficiency upto 80%, the increment in the source temperature is

Options:
A 1500 K
B 2500 K
C 500 K
D 2000 K

The correct answer is:
1500 K

Solution:
Recall the formula of efficiency of Carnot's engine in terms of source and sink temperature, η=1-T2T1, here T2=500 K, it means,
0.5=1-500T1 .....1 when the efficiency becomes, 0.8 then,
0.8=1-500T1+x ...2

from the equation one and two,
0.50.2=T1+xT1⇒0.3T1=0.2x⇒x=1.5T1
⇒x=1500 K

Topic: Physics, Kinetic Theory of Gases, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 1)
The carrier frequency generated by a circuit containing 1 nF capacitor and 10 Î¼H inductor is

Options:
A 1592 Hz
B 1592 kHz
C 159.2 Hz
D 15.92 kHz

The correct answer is:
1592 kHz

Solution:
C=1 nF=10-9F, L=10μH=10-5H
f=12πLC=12π10-5×10-9=1072π
=159.2×104 Hz=1592 kHz

Topic: Physics, Semiconductors, JEE Main
The carrier frequency of a transmitter is provided by a tank circuit of a coil of inductance $49 \mu \mathrm{H}$ and a capactiance of $2.5 \mathrm{nF}$. It is modulated by an audio signal of $12 \mathrm{kHz}$. The frequency range occupied by the side bands is:

Options:
A
$18 \mathrm{kHz}-30 \mathrm{kHz}$

B
$63 \mathrm{kHz}-75 \mathrm{kHz}$

C
$442 \mathrm{kHz}-466 \mathrm{kHz}$

D
$13482 \mathrm{kHz}-13494 \mathrm{kHz}$

The correct answer is:
$442 \mathrm{kHz}-466 \mathrm{kHz}$

Solution:
Given: Inductance, $L=49 \mu \mathrm{H}=49 \times 10^{-6} \mathrm{H}$, capacitance $C=2.5 \mathrm{nF}=2.5 \times 10^{-9} \mathrm{~F}$
$$
\begin{aligned}
&\text { Using } \omega=\frac{1}{\sqrt{L C}} \\
&=\frac{1}{\sqrt{49 \times 10^{-6} \times \frac{2.5}{10} \times 10^{-9}}} \\
&=\frac{1}{7 \times 5 \times 10^{-8}}=\frac{10^8}{7 \times 5} \\
&\text { or, } \frac{10^8}{7 \times 5}=2 \pi \times f=2 \times \frac{22}{7} \times f \\
&\text { or, } f=\frac{10^7}{22}=\frac{10^4}{22} \mathrm{kHz}=454.54 \mathrm{kHz}
\end{aligned}
$$
Therefore frequency range $454.54 \pm 12 \mathrm{kHz}$ i.e., $442 \mathrm{kHz}-466 \mathrm{kHz}$

Topic: Physics, Communication System, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)