datavorous/entrance-exam-dataset huggingface dataset:
The half-life of a radioactive substance A is 4 days. The probability that a nucleus will decay in two half-lives is

Options:
A 14
B 34
C 12
D 1

The correct answer is:
34

Solution:
After two half lives 14th fraction of nuclei will remain undecayed. Or, 34th fraction will decay. Hence, the probability that a nucleus decays in two half-lives is 34

Topic: Physics, Nuclear Physics, NEET
The half-life of a radioactive substance is 30 minutes. The time taken between $40 \%$ decay and $85 \%$ decay of the same radioactive substance is

Options:
A 15 minutes
B 90 minutes
C 60 minutes
D 30 minutes

The correct answer is:
60 minutes

Solution:
When $40 \%$ nuclei decay, $60 \%$ nuclei remain undecayed Let this number be $\mathrm{N}_0$ When $85 \%$ nuclei decay, $15 \%$ nuclei remain undecayed This number will be $\mathrm{N}=\frac{\mathrm{N}_0}{4}$
The number of nuclei will become one-fourth in two half-lives i.e., 60 minutes.

Topic: Physics, Nuclear Physics, JEE Main
The half-life of a radioactive substance is 30 minutes. The time taken between $40 \%$ decay and $85 \%$ decay of the same radioactive substance is

Options:
A 15 minutes
B 90 minutes
C 60 minutes
D 30 minutes

The correct answer is:
60 minutes

Solution:
When $40 \%$ nuclei decay, $60 \%$ nuclei remain undecayed Let this number be $\mathrm{N}_0$ When $85 \%$ nuclei decay, $15 \%$ nuclei remain undecayed This number will be $\mathrm{N}=\frac{\mathrm{N}_0}{4}$
The number of nuclei will become one-fourth in two half-lives i.e., 60 minutes.

Topic: Physics, Nuclear Physics, MHT CET, MHT CET 2021 (24 Sep Shift 2)
The half-life of a radiocative isotope is $30 \mathrm{~h}$. How long will it take to get reduced to $12.5 \%$ of its initial amount?

Options:
A 120 h
B 90 h
C 60 h
D 50 h

The correct answer is:
90 h

Solution:
Half-life of radioactive isotope,
$T_{1 / 2}=30 \mathrm{~h}$
If $N_0$ be the initial amount of radioactive isotope, then remaining amount $(N)$ in time $t$ is given as
$N=125 \% \text { of } N_0=\frac{125}{100} \times N_0 \Rightarrow N=\frac{N_0}{8}$
We know that,
$\begin{aligned}
N & =N_0\left(\frac{1}{2}\right)^n \Rightarrow \frac{N_0}{8}=N_0\left(\frac{1}{2}\right)^n \\
\Rightarrow \quad\left(\frac{1}{2}\right)^3 & =\left(\frac{1}{2}\right)^n \Rightarrow n=3 \Rightarrow \frac{t}{T_{1 / 2}}=3 \\
\Rightarrow \quad t & =3 \times T_{1 / 2}=3 \times 30=90 \mathrm{~h}
\end{aligned}$

Topic: Physics, Nuclear Physics, TS EAMCET, TS EAMCET 2020 (14 Sep Shift 2)
The half-life of a radio-isotope is 4 h . If initial mass of the isotope was 200 g , then mass remaining after 24 h will be

Options:
A 1.042 g
B 2.084 g
C 3.125 g
D 4.167 g

The correct answer is:
3.125 g

Solution:
$\begin{aligned} & M=M_0\left(\frac{1}{2}\right)^{t / 1 / 2} \\ & \therefore \quad M=200\left(\frac{1}{2}\right)^{24 / 4}=200\left(\frac{1}{2}\right)^6=\frac{200}{64}=3.125 \mathrm{~g}\end{aligned}$

Topic: Physics, Nuclear Physics, JIPMER, JIPMER 2010
The half-life of a substance in a certain enzyme-catalysed reaction is $138 \mathrm{~s}$. The time required for the concentration of the substance to fall from $1.28 \mathrm{mg} \mathrm{L}^{-1}$ to $0.04 \mathrm{mg} \mathrm{L}^{-1}$ is

Options:
A $414 \mathrm{~s}$
B $552 \mathrm{~s}$
C $690 \mathrm{~s}$
D $276 \mathrm{~s}$

The correct answer is:
$690 \mathrm{~s}$

Solution:
Enzyme-catalysed reactions follow first order kinetics.
$$
\begin{aligned}
& 1.28 \stackrel{t_{1 / 2}}{\longrightarrow} 0.64 \stackrel{t_{1 / 2}}{\longrightarrow} 0.32 \stackrel{t_{1 / 2}}{\longrightarrow} 0.16 \\
& \stackrel{t_{1 / 2}}{\longrightarrow} 0.08 \stackrel{t_{1 / 2}}{\longrightarrow} 0.04
\end{aligned}
$$
$$
\begin{aligned}
\therefore \text { Time required } & =5 \times t_{1 / 2} \\
& =5 \times 138 \mathrm{~s} \\
& =690 \mathrm{~s} .
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2011 (Mains)
The half-life of Au198 is 2.7 days. The activity of 1.50 mg of Au198, if its atomic weight is 198 g mol-1 is, NA=6×1023 mol-1

Options:
A 240 Ci
B 357 Ci
C 535 Ci
D 252 Ci

The correct answer is:
357 Ci

Solution:
The activity will be

A=λN

Now,

λ=ln2t12 and N=1.5×10-3198×6×1023

Therefore, activity in Ci,

A=ln2t12×1.5×10-3198×6×10233.7×10-10=ln22.7×24×60×60×1.5×10-3198×6×10233.7×10-10=366 Ci≈357 Ci

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The half-life of Au198 is 3 days. If atomic weight of Au198 is 198 g mol-1, then the activity of 2 mg of Au198 is [in disintegration second-1]:

Options:
A 2.67×1012
B 6.06×1018
C 32.36×1012
D 16.18×1012

The correct answer is:
16.18×1012

Solution:
A=λN

λ=ln2t1/2=ln23×24×60×60 sec-1=2.67×10-6 sec-1

N= Number of atoms in 2 mg Au

=2×10-3198×6×1023=6.06×1015

A=λN=1.618×1013=16.18×1012 dps

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The half-life of $\mathrm{Ra}^{226}$ is 1620 years. Then the number of atoms decay in one second in $1 \mathrm{~g}$ of radium (Avogadro number $=6.023 \times 10^{23}$ )

Options:
A $4.23 \times 10^9$
B $3.16 \times 10^{10}$
C $3.61 \times 10^{10}$
D $2.16 \times 10^{10}$

The correct answer is:
$3.61 \times 10^{10}$

Solution:
The number of atoms decay in one second $\frac{d N}{d t}=\lambda N$
$\begin{aligned} & =\frac{0.693}{1620 \times 365 \times 86 \times 400} \times \frac{6.023 \times 10^{23}}{226} \\ & =3.61 \times 10^{10}\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main
The half-life of radio isotope is 4 h . If initial mass of the isotope was 200 g , then mass remaining after 24 h will be

Options:
A 1.042 g
B 2.084 g
C 3.125 g
D 4.167 g

The correct answer is:
3.125 g

Solution:
$\begin{aligned} & M=M_0\left(\frac{1}{2}\right)^{t / t_{1 / 2}} \\ & \therefore \\ & M=200\left(\frac{1}{2}\right)^{24 / 2}=200\left(\frac{1}{2}\right)^6=\frac{200}{64}=3.125 \mathrm{~g}\end{aligned}$

Topic: Physics, Nuclear Physics, JIPMER, JIPMER 2013
The half-life of radioisotopic bromine - 82 is 36 hours. The fraction which remains after one day is _________×10-2.
(Given antilog 0.2006=1.587)

Round off to the nearest integer

Options:

The correct answer is:
63

Solution:
Half life of bromine -82=36 hours

t1/2=0.693K

K=0.69336=0.01925hr-1

K=2.303tlogN0Nt

t=1 day = 24 hr

logN0Nt=24hr×0.01925hr-12.303

=0.2006

Or N0Nt= anti log 0.2006

N0Nt=1.587

Or NtN0=11.587=0.630

Hence, the fraction remain after day =63×10-2

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2024 (29 Jan Shift 2)
The half-life of tritium is 12.5 years. What mass of tritium of initial mass 64 mg will remain undecayed after 50 years?

Options:
A 32 mg

B 8 mg

C 16 mg

D 4 mg

The correct answer is:
4 mg

Solution:
Given, T1/2=12.5 years
Initial mass, N0=64 mg
Number of half-lives, n=5012.5=4
Now, NN0=12n
⇒ N64=124⇒N=64×116=4 mg

Topic: Physics, Nuclear Physics, NEET
The half-life period for catalytic decomposition of AB3 at 50 mm is found to be 4 hrs and at 100 mm it is 2 hrs. The order of the reaction is:

Options:
A 3
B 1
C 2
D 0

The correct answer is:
2

Solution:
t12âa1-n
t1t2= a1a21-n
42= 50100 1-n
n= 2
Therefore,the order of reaction is 2.

Topic: Chemistry, Chemical Kinetics, JEE Main
The half-life period of 234 U is 2.5× 10 5 years. In how much time is the quantity of the isotope reduced to 25% of the original amount-

Options:
A 5Ã 10 5 years
B 4× 10 5 years
C 6× 10 5 years
D 3× 10 5 years

The correct answer is:
5× 10 5 years

Solution:
Initial amount of this isotope N 0 =100
Final amount of the isotope N=25
We know that N= ( 1 2 ) n N 0
So,
25= ( 1 2 ) n ×100 or 25 100 = ( 1 2 ) n
or 1 4 = ( 1 2 ) n or ( 1 2 ) 2 = ( 1 2 ) n
or n=2
Time taken T=n× t 1 2
=2×2.5× 10 5 =5× 10 5 years

Topic: Chemistry, Chemical Kinetics, JEE Main
The half-life period of a first-order reaction is 15 minutes. The amount of substance left after one hour will be

Options:
A 1 1 6 of the original amount
B 1 8 of the original amount.
C 1 3 2 of the original amount.
D 1 4 of the original amount.

The correct answer is:
1 1 6 of the original amount

Solution:
For the first-order reaction
A → Product
[A]t=[A0] 12Δtt1/2
[A]t=A0126015=A0124
∴ [A]t=A016

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2014 (09 Apr Online)
The half-life period of a radio-active element $\mathrm{X}$ is same as the mean life time of another radioactive element $\mathrm{Y}$. Initially they have the same number of atoms. Then

Options:
A
$\mathrm{X}$ will decay faster than $\mathrm{Y}$

B
$\mathrm{Y}$ will decay faster than $\mathrm{X}$

C
$\mathrm{X}$ and $\mathrm{Y}$ have same decay rate initially

D
$\mathrm{X}$ and $\mathrm{Y}$ decay at same rate always.

The correct answer is:
$\mathrm{Y}$ will decay faster than $\mathrm{X}$

Solution:
$\mathrm{t}_{1 / 2}=\frac{\ln 2}{\lambda_{\mathrm{x}}}$
$\tau_{\text {mean }}=\frac{1}{\lambda_{\mathrm{y}}} ; \frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$
$\frac{\ln 2}{\lambda_{\mathrm{x}}}=\frac{1}{\lambda_{\mathrm{y}}} \Rightarrow \lambda_{\mathrm{x}}=\lambda_{\mathrm{y}}(0.6932) \Rightarrow \lambda_{\mathrm{y}}>\lambda_{\mathrm{x}}$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2007
The half-life period of element $\mathrm{X}$ is same as the mean life time of element $Y$. Assume initially $\mathrm{X}$ and $\mathrm{Y}$ have same number of atoms. Then

Options:
A Initially $\mathrm{X}$ and $\mathrm{Y}$ have same decay rates
B Always $\mathrm{X}$ and $\mathrm{Y}$ decay at same rate
C $\mathrm{Y}$ decays faster than $\mathrm{X}$
D $\mathrm{X}$ decays faster than $\mathrm{Y}$

The correct answer is:
$\mathrm{Y}$ decays faster than $\mathrm{X}$

Solution:
For element X,
$\left(\mathrm{t}_{1 / 2}\right)_{\mathrm{x}}=\frac{0.693}{\lambda_{\mathrm{x}}}$
For element $Y$,
$\begin{aligned}
& \left(\mathrm{t}_{\text {moan }}\right)_Y=\frac{1}{\lambda_{\mathrm{y}}} \\
& \frac{0.693}{\lambda_{\mathrm{x}}}=\frac{1}{\lambda_{\mathrm{y}}} \Rightarrow \lambda_{\mathrm{x}}=0.693 \lambda_{\mathrm{y}} \\
& \lambda_{\mathrm{x}} < \lambda_{\mathrm{y}}
\end{aligned}$
Hence, $Y$ will decay faster rate than $X$.

Topic: Physics, Nuclear Physics, JEE Main
The half-lives of two radioactive nuclides $A$ and $B$ are 1 and $2 \mathrm{~min}$ respectively. Equal weights of $A$ and $B$ are taken separately and allowed to disintegrate for $4 \mathrm{~min}$. What will be the ratio of weights of $A$ and $B$ disintegrated?

Options:
A 1:1
B 5:4
C 1:2
D 1:3

The correct answer is:
5:4

Solution:
For, $A t_{1 / 2}=1 \mathrm{~min}$
$\therefore$ part of $A$ remained after 4 minutes
$\text { (i.e., } 4 \text { half- life) will be }=\frac{1}{16}$
$\therefore$ part of $A$ disintegrated in the same period $=\frac{15}{16}$
For $B, t_{1 / 2}=2 \mathrm{~min}$
$\therefore$ Part of $B$ remained after 4 minutes (i.e., 2 half life)
$=\frac{1}{4}$
$\therefore$ part of $B$ disintegrated in the same time $=\frac{3}{4}$
Hence, ratio of disintegrated weights of $A$ and
$B=\frac{15}{16}: \frac{3}{4}=15: 12=5: 4$

Topic: Chemistry, Chemical Kinetics, AP EAMCET, AP EAMCET 2005
The half-lives of two radioactive nuclides $A$ and $B$ are 1 and $2 \mathrm{~min}$ respectively. Equal weights of $A$ and $B$ are taken separately and allowed to disintegrate for $4 \mathrm{~min}$. What will be the ratio of weights of $A$ and $B$ disintegrated?

Options:
A 1:1
B 5:4
C 1:2
D 1:3

The correct answer is:
5:4

Solution:
For, $A t_{1 / 2}=1 \mathrm{~min}$
$\therefore$ part of $A$ remained after 4 minutes
$\text { (i.e., } 4 \text { half- life) will be }=\frac{1}{16}$
$\therefore$ part of $A$ disintegrated in the same period $=\frac{15}{16}$
For $B, t_{1 / 2}=2 \mathrm{~min}$
$\therefore$ Part of $B$ remained after 4 minutes (i.e., 2 half life)
$=\frac{1}{4}$
$\therefore$ part of $B$ disintegrated in the same time $=\frac{3}{4}$
Hence, ratio of disintegrated weights of $A$ and
$B=\frac{15}{16}: \frac{3}{4}=15: 12=5: 4$

Topic: Chemistry, Chemical Kinetics, TS EAMCET, TS EAMCET 2005
The halides which can react with water or undergo hydrolysis are
PCl5,SiCl4,BCl3,CCl4 and SF6

Options:
A PCl5,SiCl4,BCl3,CCl4 and SF6
B PCl5,SiCl4,BCl3 and CCl4
C PCl5,SiCl4 and BCl3
D PCl5 and SiCl4

The correct answer is:
PCl5,SiCl4 and BCl3

Solution:
Sometimes, when maximum covalency is obtained, the halides become inert to water. Thus, SF6 is stable. This is because SF6 is coordinately saturated and sterically hindered. Thus, SF6 is inert to water, because of kinetic rather than thermodynamic factor. CCl4 cannot undergo hydrolysis as carbon does not have vacant d-orbitals to extend its covalency.
Rest other PCl5,SiCl4 and BCl3 can undergo hydrolysis as follows
PCl5+4H2O→H3PO4+5H+Cl-
SiCl4+4H2O→Si(OH)4+4H+Cl-
BCl3+3H2O→B(OH)3+3H+Cl-

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The hardness of a water sample containing 10-3 M MgSO4 expressed as CaCO3 equivalents (in ppm ) is . (molar mass of MgSO4 is 120.37 g/mol )

Options:

The correct answer is:
100

Solution:
10 −3 mole MgSO 4 = 10 −3 mole of MgSO 4 present in 1L solutions.
nCaCO3=nMgSO4
ppm   In term of  CacO 3   = 10 −3 ×100 1000 × 10 6
ppm In term of  CaCO 3 =100  ppm

Hard water is water that has high mineral content. Hard water is formed when water percolates through deposits of limestone, chalk or gypsum which are largely made up of calcium and magnesium carbonates, bicarbonates and sulfates.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The Hardy-Weinberg principle cannot operate if

Options:
A The population is very large
B Frequent mutations occur in the population
C The population has no chance of interaction with other populations
D Free interbreeding occurs among all members of the population

The correct answer is:
Frequent mutations occur in the population

Solution:
The Hardy-Weinberg principle says that allele frequencies in a population are stable and constant from generation to generation as long as
- The population is large enough and changes in allele frequencies due to chance or accident are insignificant
- Mating occurs at random
- Mutation does not occur or if it does occur it must reach a state of equilibrium
- All the members of the population survive and have equal reproductive rates.
Hardy and Weinberg, through this law, have been able to explain why a population as a whole with all its variations and genotype frequencies continue to remain unchanged for many generations.

Topic: Biology, Evolution, NEET
The harmful Ozone is present in

Options:
A Troposphere
B Stratosphere
C Mesosphere
D Thermosphere

The correct answer is:
Troposphere

Solution:
Ozone occurs in two layers of the atmosphere. The layer closest to the Earth's surface is the troposphere. Here, ground-level or 'bad' ozone is an air pollutant that is harmful to breathe and it damages crops, trees and other vegetation. It is the main ingredient of urban smog.

Topic: Biology, Environmental Issues, NEET
The harmonic mean of two positive numbers a and b is 4, their arithmeitc mean is A and the geometric mean is G. If 2A+G2=27, a+b=α and a-b=β, then the value of αβ is equal to

Options:
A 1
B 3
C 52
D 5

The correct answer is:
3

Solution:
Given, harmonic mean H=4

We know that, G2=AH
Since, 2A+G2=27
⇒2A+AH=27
⇒2A+4A=27
⇒A=276=92=a+b2⇒a+b=9
G2=AH=92×4=18
⇒ab=18
a-b=a+b2-4ab=81-4×18=3
⇒α=9, β=3
⇒αβ=3

Topic: Mathematics, Sequences and Series, JEE Main
The heart of a man pumps 5 litres of blood through the arteries per minute at a pressure of 150 mm of mercury. If the density of mercury be 13.6×103 kg m-3 and g=10 m s-2, then the power of heart in watt is:

Options:
A 2.35
B 3.0
C 1.50
D 1.70

The correct answer is:
1.70

Solution:
\(\begin{aligned} & \text { Pressure }=\mathrm{P}=150 \mathrm{~mm} \mathrm{Hg} \\ & \text { Pumping rate }=\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{5 \times 10^{-3}}{60} \mathrm{~m}^3 / \mathrm{s} \\ & \text { Power of heart }=\text { P. } \frac{\mathrm{dV}}{\mathrm{dt}}=\rho \mathrm{gh} \times \frac{\mathrm{dV}}{\mathrm{dt}} \\ & =\left(13.6 \times 10^3\right)(10) \times(0.15) \times \frac{5 \times 10^{-3}}{60} \\ & =\frac{13.6 \times 5 \times 0.15}{6}=1.70 \mathrm{watt}\end{aligned}\)

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The heat absorbed by a system in going through the given cyclic process is :

Options:
A $19.6 \mathrm{~J}$
B $61.6 \mathrm{~J}$
C $616 \mathrm{~J}$
D $431.2 \mathrm{~J}$

The correct answer is:
$61.6 \mathrm{~J}$

Solution:
$\begin{aligned} & \Delta \mathrm{U}=0(\text { Cyclic process }) \\ & \Delta \mathrm{Q}=\mathrm{W}=\text { area of } \mathrm{P}-\mathrm{V} \text { curve. } \\ & =\pi \times\left(140 \times 10^3 \mathrm{~Pa}\right) \times\left(140 \times 10^{-6} \mathrm{~m}^3\right) \\ & \Delta \mathrm{Q}=61.6 \mathrm{~J}\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2024 (05 Apr Shift 1)
The heat generated in a circuit is given by $\mathrm{Q}=\mathrm{I}^{2} \mathrm{Rt},$ where I is current, $\mathrm{R}$ is resistance and tis time. If the percentage errors in measuring I. $\mathrm{R}$ and $\mathrm{t}$ are $2 \%, 1 \%$ and $1 \%$ respectively, then the maximum error in measuring heat will be

Options:
A $2 \%$
B $3 \%$
C $4 \%$
D $6 \%$

The correct answer is:
$6 \%$

Solution:
$\begin{aligned} \frac{\Delta Q}{Q} \times 100 &=\frac{2 \Delta I}{I} \times 100+\frac{\Delta R}{R} \times 100+\frac{\Delta t}{t} \times 100 \\ &=2 \times 2 \%+1 \%+1 \%=6 \% \end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
The heat of atomization of methane and ethane are 360 kJ mol-1 and 620 kJ mol-1, respectively. The longest wavelength of light capable of breaking the C-C bond is (Avogadro's number=6.023×1023, h=6.62×10-34 J s)

Options:
A 2.48×104 nm
B 1.49×104 nm
C 2.48×103 nm
D 1.49×103 nm

The correct answer is:
1.49×103 nm

Solution:
4 B.E (C-H) bond=360 kJ mol-1.
B.E (C-H) bond=90 kJ mol-1.
In C2H6, B.EC-C+6×B.EC-H=620 kJ,
B.EC-C bond=620-6×90=80 kJ mol-1,
B.EC-C bond=8096.48=0.83 eV bond-1.
λ(Photon in Å) for rupture of

E=hcλ
The wave length required to break C-C bond=124080.83=14950Å,
=1495 nm,
≈1.49×103 nm.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2015 (10 Apr Online)
The heat of combustion of ethanol into carbon dioxides and water is -327 Kcal at constant pressure. The heat evolved (in cal) at constant volume at 27° C (if all gases behave ideally) is R=2 calmol-1K-1________

Options:

The correct answer is:
326400

Solution:
C2H5OH(ℓ)+O2(g)⟶2O2(g)+3H2O(ℓ);ΔHC=-327Kcal
ΔHc=ΔUC+ΔngRT
-327×103=ΔUC+1×2×300
ΔUC=-326400 cal

So heat evolved as constant volume is 326400cal

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2020 (02 Sep Shift 2)
The heat of combustion of solid benzoic acid at constant volume is \(-321.30 \mathrm{~kJ}\) at \(27^{\circ} \mathrm{C}\). The heat of combustion at constant pressure is \((-321.30-x \mathrm{R}) \mathrm{kJ}\), the value of \(x\) is _______.

Options:

The correct answer is:
150

Solution:
$\begin{aligned} & \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}(\mathrm{S})+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 7 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\ell) \\ & \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\ & =-321.30-\frac{1}{2} \frac{\mathrm{R}}{100} \times 300 \\ & =(-321.30-150 \mathrm{R}) \mathrm{kJ}\end{aligned}$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2024 (05 Apr Shift 1)
The heat of hydrogenation of hexene is $127 \mathrm{~K} \mathrm{~J} /$ mole . When a second double bond is introduced in the molecule, the heat of hydrgenation of the diene is found to be $23 \mathrm{KJ} / \mathrm{mole}$. The diene is most likely to be

Options:
A Hexa -1,2 - diene
B Hexa - 1,3 - diene
C Penta - 1,3 - diene
D Hexa - 1,4 - diene

The correct answer is:
Hexa - 1,3 - diene

Solution:
Hexa - 1, 3 - diene
As the heat of hydrogenation is less than $127 \times 2$; the diene must be stable due to resonance. So it should be hexa - 1,3 - diene.

Topic: Chemistry, Hydrocarbons, JEE Main
The heat of neutralisation of a strong base and a strong acid is 13.7 kcal. The heat released when 0.7 mole HBr solution is added to 0.35 mole of KOH is

Options:
A 5.425 kcal
B 13.7 kcal
C 4.795 kcal
D 8.795 kcal

The correct answer is:
4.795 kcal

Solution:
For the neutralization between a strong and a strong base, equal number of moles of H+ (from acid) and OÌH (from base) come out. For one mole of each following equation can be drawn
So, 0.35 mole of H2O is produced.
So, heat released =(0.35Ã13.7) kcal=4.795 kcal

Topic: Chemistry, Thermodynamics (C), JEE Main
The heat of solution of anhydrous \(\mathrm{CuSO}_4\) and \(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}\) are \(-70 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(+12 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively.
The heat of hydration of \(\mathrm{CuSO}_4\) to \(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}\) is \(-x \mathrm{~kJ}\). The value of \(x\) is ______ (nearest integer).

Options:

The correct answer is:
82

Solution:
$\begin{gathered}\text { from }(1) \&(2) \\ -70=x+12 \\ x=-82\end{gathered}$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2024 (09 Apr Shift 1)
The heat radiated per unit area in 1 hour by a furnace whose temperature is $3000 \mathrm{~K}$ is $(\sigma=5.7 \times$ $10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}$ )

Options:
A
$1.7 \times 10^{10} \mathrm{~J}$

B
$1.1 \times 10^{12} \mathrm{~J}$

C
$2.8 \times 10^8 \mathrm{~J}$

D
$4.6 \times 10^6 \mathrm{~J}$

The correct answer is:
$1.7 \times 10^{10} \mathrm{~J}$

Solution:
According to Stefan's law
$$
E=\sigma T^4
$$
Heat radiated per unit area in 1 hour ( $3600 \mathrm{~s})$ is
$$
=5.7 \times 10^{-8} \times(300)^4 \times 3600=1.7 \times 10^{10} \mathrm{~J}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2012 (07 May Online)
The heat required to raise the temperature of body by 1 K is called

Options:
A
specific heat

B
thermal capacity

C
water equivalent

D
none of these

The correct answer is:
thermal capacity

Solution:
No Solution Available

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2002
The heating of oxime of acetone in presence of conc. H2SO4 to form N-methyl ethanamide is called

Options:
A Bayer-Villger rearrangement
B Beckmann rearrangement
C Wolf-rearrangement
D Hoffmann reaction

The correct answer is:
Beckmann rearrangement

Solution:
The heating of oxime of acetone in presence of conc. H2SO4 to form N-methyl ethanamide is called Beckmann rearrangement. In Beckmann rearrangement, R-migrates from carbon to nitrogen of an oxime to form N-substituted amide.
Loss of H2O occurs with simultaneous migration of the trans-R'.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The heats of combustion of C x H y , carbon and hydrogen are a, b and c cal respectively. The heat of formation of C x H y , will be:

Options:
A - x b + y c 2 - a 2 cal
B - x b + y c 2 - a cal
C x b - y c 2 + a 2 cal
D x b - y c 2 - a 2 cal

The correct answer is:
- x b + y c 2 - a cal

Solution:
xC +  y 2 H 2 → C x H y            ΔH f CxHy  = ?
ΔH f CxHy  =  ΔH rx n
= ΔH comb . CxHy − xΔH comb . C − y 2 ΔH comb . H 2
= a−x b − y 2 c
= − xb +  yc 2 −a  cal

Topic: Chemistry, Thermodynamics (C), JEE Main
The heats of combustion of carbon and carbon monoxide are -393.5 and-283.5 kJ mol-1, respectively. The heat of formation in kJ of carbon monoxide per mole is:

Options:
A - 676.5
B -110
C 110.5
D 676.5

The correct answer is:
-110

Solution:
Given: Heat of combustion of carbon and carbon monoxide are -393.5 and-283.5 kJ mol-1, respectively.
(i) C+O2→CO2; ∆H1=-393.5 kJ mol-1
(ii) CO+12O2 →CO2; ∆H2=-283.5 kJ mol-1
To find the heat of formation of CO per mole.
C+12O2→CO
Δ H f of CO
From equations (i) and (ii),
=-393.5--283.5
= - 1 1 0 kJ mol-1

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2016 (03 Apr)