datavorous/entrance-exam-dataset huggingface dataset:
A tennis ball (treated as hollow spherical shell) starting from $\mathrm{O}$ rolls down a hill. At point $\mathrm{A}$ the ball becomes air borne leaving at an angle of $30^{\circ}$ with the horizontal. The ball strikes the ground at $\mathrm{B}$. What is the value of the distance $\mathrm{AB}$ ?
(Moment of inertia of a spherical shell of mass $m$ and radius $R$ about its diameter $=\frac{2}{3} m R^2$ )

Options:
A
$1.87 \mathrm{~m}$

B
$2.08 \mathrm{~m}$

C
$1.57 \mathrm{~m}$

D
$1.77 \mathrm{~m}$

The correct answer is:
$2.08 \mathrm{~m}$

Solution:
Velocity of the tennis ball on the surface of the earth or ground
$$
\begin{aligned}
& \mathrm{v}=\sqrt{\frac{2 \mathrm{gh}}{1+\frac{\mathrm{k}^2}{\mathrm{R}^2}}}(\text { where } \mathrm{k}=\text { radius of gyration } \\
& \text { of spherical shell } \left.=\sqrt{\frac{2}{3}} \mathrm{R}\right) \\
& \text { Horizontal range } \mathrm{AB}=\frac{\mathrm{v}^2 \sin 2 \theta}{\mathrm{g}} \\
& =\frac{\left.\sqrt{\frac{2 \mathrm{gh}}{1+\mathrm{k}^2 / \mathrm{R}^2}}\right)^2 \sin \left(2 \times 30^{\circ}\right)}{\mathrm{g}}
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2013 (22 Apr Online)
A tennis ball with (small) mass m2, sits on the top of a basketball with (large) mass m1. The bottom of the basketball is at a height h above the ground and the bottom of the tennis ball is at a height (h+d) above the ground. The balls are dropped from rest. Here all collisions are elastic and m1>>m2. The height from the point of collision up to which the tennis ball bounce is

Options:
A h
B 2h
C 3h
D 9h

The correct answer is:
9h

Solution:
Just before the collision, the velocity of the basketball in the downward direction is 2gh and just after the collision its velocity reverses its direction

So just after the collision, the downward moving tennis ball collides with upward moving basketball. The velocity of the tennis ball after this collision is

v=--22gh+2gh⇒v=32gh

The height up to which the tennis ball will bounce is

h'=32gh22g⇒h'=9h

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A tension of $20 \mathrm{~N}$ is applied to a copper wire of cross sectional area $0.01 \mathrm{~cm}^2$, Young's modulus of copper is $1.1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$ and Poisson's ratio is 0.32 . The decrease in cross sectional area of the wire is

Options:
A $1.16 \times 10^{-6} \mathrm{~cm}^2$
B $1.16 \times 10^{-5} \mathrm{~m}^2$
C $1.16 \times 10^{-4} \mathrm{~m}^2$
D $1.16 \times 10^{-3} \mathrm{~cm}^2$

The correct answer is:
$1.16 \times 10^{-6} \mathrm{~cm}^2$

Solution:
Given, $\sigma=0.32, F=20 \mathrm{~N}$
$$
A=0.01 \mathrm{~cm}^2=0.01 \times 10^{-3} \mathrm{~m}
$$
and $Y=1-1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
We know that
$$
\frac{\Delta l}{l}=\frac{F}{A Y}=\frac{20}{0.01 \times 10^{-3} \times 1.1 \times 10^{11}}=18.1 \times 10^{-7}
$$
and we also known
$$
\begin{gathered}
\sigma=\frac{-\Delta r / r}{\Delta / /} \\
-\frac{\Delta r}{r}=0.32 \times 18.1 \times 10^{-7}=5.79 \times 10^{-7}
\end{gathered}
$$
Hence, decrease in cross reactional area of wire is
$$
\begin{aligned}
\Delta A=2 \frac{\Delta r}{r} \times A & =2 \times 5.79 \times 10^{-7} \times 0.01 \times 10^{-3} \\
& =0.158 \times 10^{-10} \mathrm{~m}^2 \\
& =1.26 \times 10^{-6} \mathrm{~cm}^2
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A terapeptide has −COOH group on alanine. This produces glycine (Gly), valine (Val). Phenyl alamine (Phe) and alanine (Ala) on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) with − NH 2 group attached to a chiral center is.

Options:

The correct answer is:
4

Solution:
6 Combinations as last amino acid will be Alanine as it has free -COOH group. Let's represent Alanine as A then the combinations feasible are: DCBA, DBCA, CDBA, BDCA, CBDA, BCDA . But glycine ia achiral thus tetrapeptide with − NH 2 group attached to chiral center is 4.

Topic: Chemistry, Biomolecules, JEE Main
A term is randomly chosen from the expansion of 46+15420. If the probability that it is a rational term is P, then 420P is equal to

Options:

The correct answer is:
40

Solution:
Total number of terms =21
Number of rational terms =2 (9th and 21st)
Required probability P=221

Topic: Mathematics, Probability, JEE Main
A terrestrial animal must be able to:

Options:
A excrete large amounts of water in urine
B conserve water
C actively pump salts out through the skin
D excrete large amounts of salts in urine.

The correct answer is:
conserve water

Solution:
Terrestrial animals (mammals and semi-terrestrial adult amphibians) cannot use water freely for excretion because of less availability in the environment, so ammonia is converted into a less toxic and easily soluble product, urea. Terrestrial animals must osmoregulate to maintain contant internal osmolarity.

Topic: Biology, Ecosystem, NEET, NEET 2004
A tertiary alcohol is obtained when Grignard reagent reacts with

Options:
A pentan \( -2- \) one
B butanone
C propanone
D all of these

The correct answer is:
all of these

Solution:
Grignard reagent reacts with carbonyl compound to form alcohol. Formaldehyde give primary alcohol, other aldehydes give secondary aldehyde whereas ketone give tertiary alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following?

Options:
A +R effect of –CH3 groups
B –R effect of –CH3 groups
C Hyperconjugation
D –I effect of –CH3 groups

The correct answer is:
Hyperconjugation

Solution:
The stability of given carbocations can be explained based on the hyperconjugation effect. The more the number of alpha hydrogen, the more is the stability of carbocations.

Topic: Chemistry, General Organic Chemistry, JEE Main
A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is ___________

Options:

The correct answer is:
135

Solution:
No. of ways of giving wrong answer =3

So, required no. of ways =C46(1)4×(3)2
=15(9)=135

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2020 (04 Sep Shift 2)
A test cross is carried out to

Options:
A determine the genotype of a plant at $\mathrm{F}_2$
B predict whether two traits are linked
C assess the number of alleles of a gene
D determine whether two species or varieties will breed successfully

The correct answer is:
determine the genotype of a plant at $\mathrm{F}_2$

Solution:
When $\mathrm{F}_1$ hybrid is crossed with its recessive parent, it is called test cross. By test cross heterozygosity and homozygosity of the organism can be tested. It gives $1: 1$ ratio in monohybrid and $1: 1: 1: 1$ ratio in dihybrid condition. In presence of linkage the ratio varies from this.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2012 (Mains)
A test particle is moving in a circular orbit in the gravitational field produced by a mass density ρr=Kr2. Identify the current relation between the radius R of the particle’s orbit and its period T:

Options:
A TR is a constant
B T2R3 is a constant
C TR is a constant
D TR2 is a constant

The correct answer is:
TR is a constant

Solution:
M=∫ρ4πr2dr
=∫0RKr24πr2dr
=4πKR
Satellite velocity =GMR=4πK
Time period, T=2πRv=2πR4πK
Now,
TR=2π4πK= constant

Topic: Physics, Gravitation, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A tetrahedron has vertices at $\mathrm{O}(0,0,0), \mathrm{A}(1,2,1) \mathrm{B}(2,1,3)$ and $\mathrm{C}(-1,1,2)$. Then the angle between the faces $\mathrm{OAB}$ and $\mathrm{ABC}$ will be

Options:
A
$90^{\circ}$

B
$cos ^{-1}\left(\frac{19}{35}\right)$

C
$\cos ^{-1}\left(\frac{17}{31}\right)$

D
$30^{\circ}$

The correct answer is:
$cos ^{-1}\left(\frac{19}{35}\right)$

Solution:
Vector perpendicular to the face $\mathrm{OAB}$
$=\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3\end{array}\right|=5 \mathrm{i}-\mathrm{j}-3 \mathrm{k}$
Vector perpendicular to the face $\mathrm{ABC}$
$=\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1\end{array}\right|=\mathrm{i}-5 \mathrm{j}-3 \mathrm{k}$
Angle between the faces $=$ Angle between their normals
$\cos \theta=\left|\frac{5+5+9}{\sqrt{35} \sqrt{35}}\right|=\frac{19}{35}$ or $\theta=\cos ^{-1}\left(\frac{19}{35}\right)$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2003
A tetrahedron has vertices at \( O(0,0,0), A(1,-2,1), B(-2,1,1), C(1,-1,2) \). Then, the angle between the faces \( O A B \) and \( A B C \) will be

Options:
A \( \cos ^{-1}\left(\frac{1}{2}\right) \)
B \( \cos ^{-1}\left(\frac{-1}{6}\right) \)
C \( \cos ^{-1}\left(\frac{-1}{3}\right) \)
D \( \cos ^{-1}\left(\frac{1}{4}\right) \)

The correct answer is:
\( \cos ^{-1}\left(\frac{-1}{3}\right) \)

Solution:
Vector perpendicular to face OAB=n1→

=OA→×OB→

=i^-2j^+k^×-2i^+j^+k^

=-2-1i^+-2-1j^+1-4k^

=-3i^-3j^-3k^

Vector perpendicular to face ABC=n2→

=AB→×AC→

=-3i^+3j^×j^+k^

=3-0i^+3+0j^+-3-0k^

=3i^+3j^-3k^

Since, angle between faces is equal to angle between their normals.

∴ cosθ=n→1·n→2n1n2

=-33+-33+-3-39+9+99+9+9

=-9-9+92727=-13

⇒θ=cos-1-13

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A tetrahedron has vertices P1, 2, 1, Q2, 1, 3, R-1, 1, 2 and O0, 0, 0. The angle between the faces OPQ and PQR is

Options:
A cos-1⁡731
B cos-11731
C cos-11935
D cos-1935

The correct answer is:
cos-11935

Solution:
Vector perpendicular to face OPQ=i^j^k^121213=5i^-j^-3k^

Vector perpendicular to face PQR=i^j^k^21-11-12=i^-5j^-3k^

Angle between two faces cosθ=5+5+93535=1935

⇒θ=cos-11935

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (12 Jan Shift 1)
'A' tetramerises in water upto \( 80 \% \). A solution of \( 2.5 \mathrm{~g} \) of \( \mathrm{A} \) in \( 100 \mathrm{~g} \) of water lowers the freezing point by \( 0.3^{\circ} \mathrm{C} \). The molar mass of \( \mathrm{A} \) is \( \left(\mathrm{K}_{\mathrm{f}}\right. \) for \( \mathrm{H}_{2} \mathrm{O} \) is \( \left.1.86\right) \)

Options:
A \( 31 \)
B \( 62 \)
C \( 122 \)
D \( 244 \)

The correct answer is:
\( 62 \)

Solution:
α=1-i1-1n
0.8=1-i1-14
i=0.4
∵ ∆T=i×kf×m
0.3=0.4×1.86×wA×1000MA×wH2O
0.3=0.4×1.86×2.5×1000MA×100
MA=0.4×1.86×2.5×10000.3×100=62 u

Topic: Chemistry, Solutions, JEE Main
A tetrapeptide has -COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) with -NH2 group attached to a chiral center is

Options:

The correct answer is:
4

Solution:
No. of possible sequences (primary structures) with -NH2 group attached to a chiral center is 4.
Since -COOH group is attached to alanine, one terminal of the tetrapeptide contains alanine. Thus, its position is fixed.
Glycine is achiral whereas valine and phenylalanine are chiral. Thus, -NH 2
group is attached to either valine or phenylalanine.
Thus following are the four possible combinations of the tetrapeptide.
H 2 ​N−Val−Gly−Phe−Ala−COOH
H 2 ​N−Val−Phe−Gly−Ala−COOH
H 2 ​N−Phe−Gly−Val−Ala−COOH
H 2 ​N−Phe−Val−Gly−Ala−COOH

Topic: Chemistry, Biomolecules, JEE Main
A. The drift velocity of electrons decreases with the increase in the temperature of conductor.
B. The drift velocity is inversely proportional to the area of cross-section of given conductor.
C. The drift velocity does not depend on the applied potential difference to the conductor.
D. The drift velocity of electron is inversely proportional to the length of the conductor.
E. The drift velocity increases with the increase in the temperature of conductor.

Choose the correct answer from the options given below:

Options:
A A and B only
B A and D only
C B and E only
D B and C only

The correct answer is:
A and D only

Solution:
Drift velocity =eτmE

vd=eτmΔVl

ΔV= Potential difference applied across the wire As temperature increases, relaxation time decreases, hence Vd decreases.
As per formula, Vd∝1l

vd=I neA , as it is not mentioned that current is at steady state neither it is mentioned that n is constant for given conductor. So it can't be said that vd is inversely proportional to A.

I=neAvd=VR=VρlA

vd=eEτm

τ decrease with temperature increase.

First and fourth statements are correct.

vd=Vρlne E=Vl

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (27 Jul Shift 2)
A. The physical association between the two genes is termed as ____a____.

B. The generation of non-parental gene combination is termed as ____b____ .

Options:
A a - recombination, b - crossing over
B a - recombination, b - linkage
C a - linkage, b - recombination
D a - linkage, b - mutation

The correct answer is:
a - linkage, b - recombination

Solution:
Linkage: It is the tendency of genes to remain together even during the formation of gametes. This is due to the physical association or linkage between the 2 genes. In such cases, parental gene combinations remain higher than the non-parental gene combinations (recombinants).
Recombination: It is the tendency of the generation of non-parental gene combinations. In such cases, non-parental gene combinations (recombinants) remain higher than the parental gene combinations.

Topic: Biology, Principles of Inheritance and Variation, NEET
A. The pollen grain is generally spherical measuring about 25-50 cm in diameter.
B. In over 60% of angiosperms, pollen grains are shed at 2 celled stage.

Select the statement that holds true for the above statement.

Options:
A Both 'A' and 'B' are correct
B Only 'B' is correct
C Only 'A' is correct
D Both 'A' and 'B' are incorrect

The correct answer is:
Only 'B' is correct

Solution:
The pollen grain is generally spherical measuring about 25-50 micrometre in diameter. In 60% of angiosperms, pollen grains are shed at 2 celled stages, the smaller cell is generative cell and the larger cell is a vegetative cell. In remaining 40% of angiosperms, generative cell divides to form 2 male gametes, thus pollen grains are liberated at 3 celled stage.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
A 'thermacole' icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side $30 \mathrm{~cm}$ has a thickness of $5.0 \mathrm{~cm}$. If $4.0 \mathrm{~kg}$ of ice is put in the box, estimate the amount of ice remaining after $6 \mathrm{~h}$. The outside temperature is $45^{\circ} \mathrm{C}$, and co-efficient of thermal conductivity of thermacole is $0.01 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$. [Heat of fusion of water $=335 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1}$ ]

Options:

None

Solution:
Length of each side $=l=30 \mathrm{~cm}=0.3 \mathrm{~m}$, Thickness $=\Delta x=5 \mathrm{~cm}=0.05 \mathrm{~m}$
Total surface area of the box, $A=6 l^2=6 \times 0.3 \times 0.3$,
Temperature difference $=\Delta T=45-0=45^{\circ} \mathrm{C}$ $k=0.01 \mathrm{~J} / \mathrm{s} / \mathrm{m} /{ }^{\circ} \mathrm{C}$,
Time $=\Delta t=6 \mathrm{hrs}=6 \times 60 \times 60 \mathrm{~s}$
Latent heat of fusion $=\mathrm{L}=335 \times 10^3 \mathrm{~J} / \mathrm{kg}$
If $m$ is the mass of ice melted, then $\mathrm{mL}=k A\left(\frac{\Delta T}{\Delta x}\right) \Delta t=\mathrm{m}=\frac{\mathrm{KA}}{\mathrm{L}}\left(\frac{\Delta \mathrm{T}}{\Delta x}\right) \Delta t$ $=\frac{0.01 \times 0.54 \times 45 \times 6 \times 60 \times 60}{0.05 \times 335 \times 10^3}=0.313 \mathrm{~kg}$
$\therefore$ Mass of ice left $=4-0.313 \mathrm{~kg}=3.687 \mathrm{~kg}$

Topic: Physics, Thermal Properties of Matter
A thermally insulated container is divided into two parts by a screen. In one part the pressure and temperature are P and T for an ideal gas-filled. In the second part, it is vacuum. If now a small hole is created in the screen, then the temperature of the gas will

Options:
A Decrease
B Increase
C Remain same
D None of these

The correct answer is:
Remain same

Solution:
In second part there is a vacuum i.e., P = 0. So work done in expansion = P Δ V = 0 . Also Δ Q = 0 . From the first law of thermodynamics, Δ U = 0 i.e., temperature of an ideal gas remains same due to free expansion.

Topic: Physics, Thermodynamics, NEET
A thermally insulated piece of metal is heated under atmospheric pressure by an electric current so that it receives electric energy at constant power P. This leads to an increase in the absolute temperature T of the metal with time t as Tt= T01 +at - t014, where a, t0 and T0 are constants. The heat capacity CPT of the metal is

Options:
A 4PaT0
B 4PT3aT04
C 2PT3aT04
D 2PaT0

The correct answer is:
4PT3aT04

Solution:
Heat given to the metal
dQ = Pdt = CP(t) dT ...(i)
At constant pressure in time interval at
Given, T = T0 [1 + a(t - t0)]1/4
dT dt = T 0 4 1 + a t - t 0 - 3 / 4 Ã a ...(ii)
From Eqs. (i) and (ii)
C P T = P dT dt = 4 P 1 + a t - t 0 3 / 4 T 0 a = 4 PT 3 a T 0 4

Topic: Physics, Thermal Properties of Matter, JEE Main
A thermally insulated rigid container contains an ideal gas heated by a filament of resistance $100 \Omega$ through a current of $1 A$ for $5 \mathrm{~min}$ then change in internal energy is

Options:
A $0 \mathrm{~kJ}$
B $10 \mathrm{~kJ}$
C $20 \mathrm{~kJ}$
D $30 \mathrm{~kJ}$

The correct answer is:
$30 \mathrm{~kJ}$

Solution:
Volume of the ideal gas is constant so $W=P \Delta V=0$ using FLOT $\Delta Q=\Delta U \Rightarrow$
$\begin{aligned} & \Delta U=i^2 R t=1^2 \times 100 \times 5 \times 60 \\ & =30 \times 10^3=30 K J\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
A thermally insulated vessel contains 150 g of water at 0°C . Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be closest to: (Latent heat of vaporization of water =2.10×106 J kg-1 and Latent heat of Fusion of water =3.36×105 J kg-1 )

Options:
A 35 g
B 20 g
C 130 g
D 150 g

The correct answer is:
20 g

Solution:
Suppose m grams of water evaporate. Then, heat required
ΔQreq=mLv
Mass that converts into ice =150-m
Heat released in the process,
ΔQrel=150-mLf
Now, ΔQrel=ΔQreq
150-mLf=mLv
mLf+Lv=150 Lf
m=150LfLf+Lv
m=20 g

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2019 (08 Apr Shift 1)
A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats 1.4. Vessel is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by :
(R= universal gas constant)

Options:
A Mv27R
B Mv25R
C 2Mv27R
D 7Mv25R

The correct answer is:
Mv25R

Solution:
KE of the vessel =12mv2
When the vessel is suddenly stopped, the ordered motion of the gas molecules is converted into disordered motion of the molecules increasing thereby the internal energy of the gas. Thus,
∆U=nCv∆T=12mv2=12nMv2
Where n is number of moles of the gas in the vessel and M is molecular weight of the gas.
∴ ∆T=Mv22Cv
As Cv=Rγ-1
∴ ∆T=Mv2(γ-1)2R=Mv2(1.4-1)2R=Mv25R

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (26 Jun Shift 1)
A thermally insulated vessel contains an ideal gas of molecular mass $M$ and ratio of specific heats $\gamma$. It is moving with speed $v$ and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by :

Options:
A
$\frac{(\gamma-1)}{2 \gamma \mathrm{R}} \mathrm{Mv}^2 \mathrm{~K}$

B
$\frac{\gamma M v^2}{2 \mathrm{R}} \mathrm{K}$

C
$\frac{(\gamma-1)}{2 \mathrm{R}} \mathrm{Mv}^2 \mathrm{~K}$

D
$\frac{(\gamma-1)}{2(\gamma+1) R} \mathrm{Mv}^2 \mathrm{~K}$

The correct answer is:
$\frac{(\gamma-1)}{2 \mathrm{R}} \mathrm{Mv}^2 \mathrm{~K}$

Solution:
$\mathrm{W}=\Delta \mathrm{U}$
$\frac{1}{2} \mathrm{mv}^2=\mathrm{nC}_{\mathrm{v}} \mathrm{dT}$
$=\frac{\mathrm{m}}{\mathrm{M}} \frac{\mathrm{R}}{\gamma-1} \mathrm{dT}$
$\mathrm{dT}=\frac{\mathrm{M}(\gamma-1) \mathrm{v}^2}{2 \mathrm{R}} \mathrm{K}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2011
A thermally insulated vessel with nitrogen gas at $27^{\circ} \mathrm{C}$ is moving with a velocity of $100 \mathrm{~ms}^{-1}$. If the vessel is stopped suddenly, then the percentage change in the pressure of the gas is nearly
(assume entire loss in $\mathrm{KE}$ of the gas is given as heat to gas and $R=8.3 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ )

Options:
A 1.1
B 0.93
C 0.5
D 2.25

The correct answer is:
2.25

Solution:
Let there are $n$ moles of $N_2$ gas in the cylinder.
Assuming all of $K, E$ appears in form of heat, $n\left(\frac{1}{2} M v^2\right)=\frac{f}{2} n R \Delta T$
Here, $M=28 \mathrm{~g}=28 \times 10^{-3} \mathrm{~kg}, f=5$
Increment in pressure due to change of temperature is
Also, $\Delta p=\frac{n R \Delta T}{V}$
So, $\frac{\Delta p}{p}=\frac{\left(\frac{n R \Delta T}{V}\right)}{\left(\frac{n R T}{V}\right)}=\frac{n R \Delta T}{n R T} \Rightarrow \frac{\Delta p}{p}=\frac{n M v^2}{f n R T}=\frac{M v^2}{f R T}$
So, percentage change in pressure is,
$$
\therefore \frac{\Delta p}{p} \times 100=\frac{28 \times 10^{-3} \times 100 \times 100}{5 \times 8.3 \times 300} \times 100=2.25 \%
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A thermocol box has a total wall area (including the lid) of $1.0 \mathrm{~m}^2$ and wall thickness of $3 \mathrm{~cm}$. It is filled with ice at $0^{\circ} \mathrm{C}$. If the average temperature outside the box is $30^{\circ} \mathrm{C}$ throughout the day, the amount of ice that melts in one day is
$$
\begin{aligned}
& \text { IUse } K_{\text {thermosol }}=0.03 \mathrm{~W} / \mathrm{mK}, \\
& L_{\text {fusion (ire) }}=3.00 \times 10^5 \mathrm{~J} / \mathrm{kg} \text { ] }
\end{aligned}
$$

Options:
A $1 \mathrm{~kg}$
B $2.88 \mathrm{~kg}$
C $25.92 \mathrm{~kg}$
D $8.64 \mathrm{~kg}$

The correct answer is:
$8.64 \mathrm{~kg}$

Solution:
Given,
Total wall area (including the lid $\rangle\langle A\rangle=1.0 \mathrm{~cm}^2$
Thickness of wall $(l)=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$
Average temperature outside the box $=30^{\circ} \mathrm{C}$
$$
\begin{aligned}
\Delta \theta & =30-0=30^{\circ} \mathrm{C} \\
L_{\text {fusion (ice) }} & =3 \times 10^5 \mathrm{~J} / \mathrm{kg} \\
K_{\text {thermocol }} & =0.03 \mathrm{~W} / \mathrm{m} \mathrm{K}
\end{aligned}
$$
We know that,
$$
\frac{Q}{t}=\frac{K A}{l} \Delta \theta
$$
For one day, $t=24 \times 60 \times 60 \mathrm{~s}$
$$
\begin{aligned}
\frac{m \times L_{\text {fusion (ice) }}}{t} & =\frac{K A}{l} \Delta Q \\
\frac{m \times 3 \times 10^5}{24 \times 60 \times 60} & =\frac{0.03 \times 1}{3 \times 10^{-2}} \times 30 \\
m & =\frac{0.03 \times 1 \times 30 \times 24 \times 60 \times 60}{3 \times 10^{-2} \times 3 \times 10^5} \\
m & =\frac{77760}{9000}=8.64 \mathrm{~kg}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold then, an electric current will

Options:
A
flow from Antimony to Bismuth at the cold junction

B
flow from Antimony to Bismuth at the hot junction

C
flow from Bismuth to Antimony at the cold junction

D
not flow through the thermocouple

The correct answer is:
flow from Antimony to Bismuth at the cold junction

Solution:
Flow from Antimony to Bismuth at cold junction

Topic: Physics, Current Electricity, JEE Main, JEE Main 2006