datavorous/entrance-exam-dataset huggingface dataset:
The tangential velocity of a body in a non-uniform circular motion varies as $v=7 \mathrm{t} 2-$ $2 \mathrm{v}$ with the radius being equal to $21 \mathrm{~m}$. What is the angular acceleration at $\mathrm{t}=2 \mathrm{~s}$ in $\mathrm{rad} / \mathrm{s} 2 ?$

Options:
A 4/3
B 5/3
C $4$
D $7 / 5$

The correct answer is:
4/3

Solution:
The angular velocity is given as, $\omega=v / r$. Hence, the function for angular velocity is $\omega$ $=(7 t 2-2) / 21$. On differentiating this expression, we will get the angular acceleration as t2/3 On substituting, the suitable values, we will get the angular acceleration as $4 / 3 \mathrm{rad} / \mathrm{s} 2$.

Topic: Physics, Laws of Motion, JEE Main
The tangents at the extremities of a focal chord of a parabola
A. are perpendicular
B. are parallel
C. intersect on the directrix
D. intersect at the vertex

Options:
A intersect on the directrix
B intersect at the vertex
C are parallel
D are perpendicular

The correct answers are:
intersect on the directrix, are perpendicular

Solution:
Let the tangent at $\mathrm{P}$ (at ${ }^2, 2 \mathrm{at}$ ) on the parabola is $t y=x+a t^2$ and at the other end $\left(\frac{a}{t^2}, \frac{-2 a}{t}\right)$ of the focal chord through $\mathrm{P}$
$-\frac{1}{t} y=x+\frac{a}{t^2} \Rightarrow y=-t x-\frac{a}{t}$
Product of the slopes $=\frac{1}{t}(-t)=-1$ and the point of intersection lies on $\left(\frac{1}{t}+\mathrm{t}\right) \mathrm{x}+\mathrm{a}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)=0 \Rightarrow \mathrm{x}=-\mathrm{a}$, the directrix

Topic: Mathematics, Parabola, JEE Main
The tangents at the extremities of the latus rectum of the ellipse 3x2+4y2=12 form a rhombus PQRS. Area (in sq. units) of the rhombus PQRS outside the ellipse is equal to

Options:
A 8-23π
B 12-23π
C 14-π
D 16-23π

The correct answer is:
16-23π

Solution:
One of the end of latus rectum is 1,32
Tangent at 1,32 is x+2y=4
Area of rhombus =412⋅4⋅2=16

Hence, the required area =16-23π sq. units

Topic: Mathematics, Ellipse, JEE Main
The tangents at the points A1,3 and B1,-1 on the parabola y2-2x-2y=1 meet at the point P. Then the area (in unit2) of the triangle PAB is:

Options:
A 4
B 6
C 7
D 8

The correct answer is:
8

Solution:
Given, y2-2x-2y=1

⇒y-12=2x+1

Tangents are drawn at 1,3 & 1,-1, on plotting the diagram we get,

âââââââ

Now, tangent as A: 2y-x-5=0 {using T=0}

Intersection with y=1 is x=-3

Hence, point P is -3,1

We can see diagram has symmetry

So, area of âPAB=2×12×1--3×3-1

=8 sq.units

Topic: Mathematics, Parabola, JEE Main, JEE Main 2022 (25 Jul Shift 2)
The tangents to the curve y=x-22-1 at its points of intersection with the line x-y=3, intersect at the point:

Options:
A 52,1
B 52,-1
C -52,-1
D -52,1

The correct answer is:
52,-1

Solution:
Let coordinates of C be h,k

chord of contact of C w. r. t. y=x-22-1

⇒y=x2-4x+3 is T=0

y+k2=xh+3-2x+h

⇒2h-2x-y=-6-4h-k

compare it with x-y=3

2h-2=1⇒h=52

6-4h-k=-3 ⇒k=-1

∴ C is 52,-1

Topic: Mathematics, Parabola, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The tangents to x2+y2=a2 having inclinations α and β intersect at P. If cotα+cotβ=0, then the locus of P is

Options:
A x+y=0
B x-y=0
C xy=0
D None of these

The correct answer is:
xy=0

Solution:
Equation of tangent in slope form
y=mxÂ±a1+m2
k-mh2=a2 1+m2
m2 h2-a2-2mhk+k2-a2=0
m1+m2= 2hkh2-a2, m1 m2= k2-a2h2-a2
cotâ¡Î±+cotâ¡Î²=0
1tanâ¡Î±+1tanâ¡Î²=0 â tanâ¡Î±+tanâ¡Î²tanâ¡Î±tanâ¡Î²=0
2hk=0â xy=0

Topic: Mathematics, Circle, JEE Main
The teacher wants to arrange 5 students on the platform such that the boy $\mathrm{B}_1$ occupies second position and the girls $\mathrm{G}_1$ and $\mathrm{G}_2$ are always adjacent to each other, then the number of such arrangements is

Options:
A $24$
B $12$
C $8$
D $16$

The correct answer is:
$8$

Solution:
There are 5 positions. Given that $\mathrm{B}_1$ occupies $2^{\text {nd }}$ position
$\therefore \quad \mathrm{B}_1$ can be arranged in 1 way. As $G_1$ and $G_2$ are always together, none of them can take $1^{\text {st }}$ position.
$\therefore \quad \mathrm{G}_1, \mathrm{G}_2$ and one of the remaining students can be arranged on $3^{\text {rd }}, 4^{\text {th }}$ and $5^{\text {th }}$ position when $\mathrm{G}_1$ and $\mathrm{G}_2$ are always together in $2 ! \times 2$ ! Ways.
And remaining 2 students can be arranged in 2 ! Ways.
$\therefore \quad$ The required number of arrangements
$\begin{aligned}
& =2 ! \times 2 ! \times 2 ! \\
& =8
\end{aligned}$

Topic: Mathematics, Permutation Combination, JEE Main
The teacher wants to arrange 5 students on the platform such that the boy $\mathrm{B}_1$ occupies second position and the girls $\mathrm{G}_1$ and $\mathrm{G}_2$ are always adjacent to each other, then the number of such arrangements is

Options:
A $24$
B $12$
C $8$
D $16$

The correct answer is:
$8$

Solution:
There are 5 positions. Given that $\mathrm{B}_1$ occupies $2^{\text {nd }}$ position
$\therefore \quad \mathrm{B}_1$ can be arranged in 1 way. As $G_1$ and $G_2$ are always together, none of them can take $1^{\text {st }}$ position.
$\therefore \quad \mathrm{G}_1, \mathrm{G}_2$ and one of the remaining students can be arranged on $3^{\text {rd }}, 4^{\text {th }}$ and $5^{\text {th }}$ position when $\mathrm{G}_1$ and $\mathrm{G}_2$ are always together in $2 ! \times 2$ ! Ways.
And remaining 2 students can be arranged in 2 ! Ways.
$\therefore \quad$ The required number of arrangements
$\begin{aligned}
& =2 ! \times 2 ! \times 2 ! \\
& =8
\end{aligned}$

Topic: Mathematics, Permutation Combination, MHT CET, MHT CET 2023 (11 May Shift 2)
The technical term used for the androecium in a flower of China rose (Hibiscus rosa sinensis) is

Options:
A monadelphous
B diadelphous
C polyandrous
D polyadelphous

The correct answer is:
monadelphous

Solution:
In monadelphous condition, all filaments become fused and form a group, while anther remain free, $e g$, China rose Achyranthes, etc. In diadelphous, two separate bundles of united filaments are formed, eg, pea. In polyadelphous, more than two separate bundles of filaments are formed, eg, Ricinus.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2010 (Screening)
The technique called gamete intra fallopian transfer (GIFT) is recommended for those females

Options:
A who cannot produce an ovum
B who cannot retain the foetus inside uterus
C who cervical canal is too narrow to allow passage for the sperms
D who cannot provide suitable environmental for fertilization

The correct answer is:
who cannot produce an ovum

Solution:
The technique called gamete intra fallopian transfer (GIFT) is transfer of an ovum collected from a donor into the fallopian tube of another female. This technique is recommended for those females who can not produce an ovum, but can provide suitable environment for fertilization and further development.

Topic: Biology, Reproductive Health, NEET, NEET 2011 (Mains)
The technique of obtaining large number of plantlets by tissue culture method is called:

Options:
A plantlet culture
B organ culture
C micropropagation
D macropropagation.

The correct answer is:
micropropagation

Solution:
Micropropagation or tissue culture is the practice of rapidly multiplying stock plant material to produce many progeny plants, using modern plant tissue culture methods. It facilitates the growth, storage, and maintenance of a large number of plants in small spaces which makes it a costeffective process. Micropropagation is used for germplasm storage and the protection of endangered species.

Topic: Biology, Reproduction in Organisms, NEET, NEET 2004
The technique used for purification of steam volatile water immiscible substance is:

Options:
A Fractional distillation
B Fractional distillation under reduced pressure
C Distillation
D Steam distillation

The correct answer is:
Steam distillation

Solution:
Steam distillation technique is applied to separate substances which are steam volatile and are immiscible with water. It is often used to purify liquids that decompose at boiling point.

The main advantage of steam distillation over conventional distillation is that the steam distillation procedure is carried out at lower temperatures and components can be separated from a mixture below their decomposition point.
The given compound is steam volatile, insoluble in water and decomposes at its boiling point. So, steam distillation techniques will be used for purification.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The technology of biogas production from cow-dung was developed in India largely due to efforts of

Options:
A Gas Authority of India
B Oil and Natural Gas Commission
C Indian Agricultural Research Institute and Khadi and Village Industries Commission
D Indian Oil Corporation.

The correct answer is:
Indian Agricultural Research Institute and Khadi and Village Industries Commission

Solution:
Biogas is a type of biofuel having a mixture of gases which is naturally produced by decomposition of organic matter by anaerobic process. It mainly contains methane and carbon dioxide. Food waste, domestic waste, animal waste form the organic matter used for biogas production. It is produced in a biogas plant where the decomposed organic matter can be used as manure for agriculture purpose. The bacterial group Methanogens are utilized in its productions.
IARI and Khadi and Village Industries commission are credited with the technology for biogas production from cow-dung in India. Biogas is a cheaper renewable source of fuel which can be used for cooking. Biogas is ecofriendly.
GAIL, ONGC and IOC are involved with the fossil fuel and natural gas extraction in India.

Topic: Biology, Microbes in Human Welfare, NEET
The telluric helix was given by

Options:
A De Chan Courtois
B Newlands
C L. Meyer
D Mendeleef

The correct answer is:
De Chan Courtois

Solution:
De Chan Courtois

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The temperature θ at the junction of two insulating sheets, having thermal resistances R1 and R2 as well as top and bottom temperatures θ1 and θ2 (as shown in figure) is given by :

Options:
A θ1R2-θ2R1R2-R1
B θ2R2-θ1R1R2-R1
C θ1R2+θ2R1R1+R2
D θ1R1+θ2R2R1+R2

The correct answer is:
θ1R2+θ2R1R1+R2

Solution:
Heat flow rate will be same through both

∴ θ1-θR1=θ-θ2R2

R2θ1-R2θ=R1θ-R1θ2

θ=R2θ1+R1θ2R1+R2

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The temperature at which a molecule of nitrogen will have the same rms velocity as a molecule of oxygen at 127 Â°C is

Options:
A 457 ℃
B 273 ℃
C 350 ℃
D 77 ℃

The correct answer is:
77 ℃

Solution:
Root mean square velocity
vrms=3RTM
where R is gas constant, T the temperature and M molecular weight.
Given, MN2=28, MO2=32, TO2=127 ℃=127+273=400 K
∴ vO2vN2=TO2MO2×MN2TN2=40032×28TN2=1
⇒ TN2=350 K=77 ℃.

Topic: Physics, Kinetic Theory of Gases, JEE Main
The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300 K is : (Atomic masses : He = 4 u, O = 16 u)

Options:
A 600 K
B 2400 K
C 1200 K
D 300 K

The correct answer is:
2400 K

Solution:
C rms = 3 RT M
C rms He = 3 R × 3 0 0 4 & C rms O 2 = 3 R × T O 2 3 2
∴   ( C rms ) He = ( C rms ) O 2 ⇒   3R×300 4 = 3R× T O 2 32
T O 2 = 3 0 0 × 3 2 4 = 2 4 0 0 K

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2014 (09 Apr Online)
The temperature at which real gases obey the ideal gas laws over a wide range of pressure is called

Options:
A Critical temperature
B Boyle temperature
C Inversion temperature
D Reduced temperature

The correct answer is:
Boyle temperature

Solution:
At Boyle temperature real gas is changed into ideal gas

Topic: Chemistry, States of Matter, JEE Main
The temperature at which the kinetic energy of oxygen molecules becomes double than its value at 27°C is

Options:
A 927°C
B 327°C
C 1227°C
D 627°C

The correct answer is:
327°C

Solution:
The kinetic energy is given by KE=f2kT.

Let the kinetic energy be K1 at 27 °C and the kinetic energy be K' at T' temperature. As given in the question, K'=2K1. Hence,

K'K1=T'T⇒2=T'300⇒T'=600 K

In Celsius the temperature is T'-273=327 °C.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The temperature, at which the root mean square velocity of hydrogen molecules equals their escape their escape velocity from the earth, is closest to:
[ Boltzmann Constant kB=1.38×10-23 J/K
Avogadro number NA=6.02×1026 /kg
Radius of Earth: 6.4×106 m
Gravitational acceleration on Earth =10 ms-2]

Options:
A 3×105K
B 650 K
C 800 K
D 104 K

The correct answer is:
104 K

Solution:
3RTm=11.2 km/s
3×253×T2×10-3=11.2×1032
T=10035.2 K≈104 K

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The temperature at which the second virial coefficient of real gas is zero is called

Options:
A Critical temperature
B Eutetic point
C Boiling point
D Boyle's temperature

The correct answer is:
Boyle's temperature

Solution:
The temperature at which the second virial coefficient of a real gas is zero is called Boyle's temperature.
It is the temperature at which a real gas obeys Boyle's law.
It is given by the expression $\mathrm{TB}=\mathrm{a} / \mathrm{hR}$ where $\mathrm{a}$ and $\mathrm{b}$ are the van der Waal's constants.
Boyle's temperature of a gas is always higher than its critical temperature.

Topic: Chemistry, States of Matter, JEE Main
The temperature at which the speed of sound in air becomes double of its value at $27^{\circ} \mathrm{C}$ is

Options:
A $54^{\circ} \mathrm{C}$
B $327^{\circ} \mathrm{C}$
C $927^{\circ} \mathrm{C}$
D $-123^{\circ} \mathrm{C}$

The correct answer is:
$927^{\circ} \mathrm{C}$

Solution:
$v=\sqrt{\frac{\gamma R T}{M}} \Rightarrow v \propto \sqrt{T}$
i.e. if $v$ is doubled then $T$ becomes four times, hence $T_2=4 T_1=4(273-27)=1200 \mathrm{~K}=927^{\circ} \mathrm{C}$

Topic: Physics, Waves and Sound, JEE Main
The temperature coefficient of e.m.f of a cell can be given by:

Options:
A ∂E∂TP=ΔSnF
B ∂E∂TP=E2-E1T2-T1
C ∂E∂TP=ΔΗnF+E·1T
D All of these

The correct answer is:
All of these

Solution:
ΔG=ΔH+T∂ΔG∂TP (Gibbs-Helmholz equation)
Also, ΔG=ΔH-TΔS; and -ΔG=nF·E
or -TΔS=T∂ΔG∂TP=T-nF∂E∂TP
or ∂E∂TP=ΔSnF, similarly derive for other values.

Topic: Chemistry, Thermodynamics (C), JEE Main
The temperature coefficient of resistance for a wire is $0.00125 /{ }^{\circ} \mathrm{C}$. At $300 \mathrm{~K}$ its resistance is $1 \mathrm{ohm}$. The temperature at which the resistance becomes $2 \mathrm{ohm}$ is

Options:
A $1154 \mathrm{~K}$
B $1100 \mathrm{~K}$
C $1400 \mathrm{~K}$
D $1127 \mathrm{~K}$

The correct answer is:
$1127 \mathrm{~K}$

Solution:
$\begin{aligned} & \frac{\rho_1}{\rho_2}=\frac{\left(1+\alpha t_1\right)}{\left(1+\alpha t_2\right)} \Rightarrow \frac{1}{2}=\frac{(1+0.00125 \times 27)}{(1+0.00125 \times t)} \\ & \Rightarrow t=854^{\circ} \mathrm{C} \Rightarrow T=1127 \mathrm{~K}\end{aligned}$

Topic: Physics, Current Electricity, JEE Main
The temperature coefficient of the emf i.e., dE dt =â0.00065 VK â1 for the cell Cd|CdCl21M||AgCls|Ag at 25 o C . Calculate the entropy changes ÎSâ298âK for the cell reaction Cd+2âAgClâ Cd 2+ +2â Cl â +2âAg .

Options:
A â105.5Â JKâ1
B â150.2Â JKâ1
C â75.7Â JKâ1
D â125.5 JKâ1

The correct answer is:
â125.5 JKâ1

Solution:
ÎS=+nFdEdt
=2Ã96500Ãâ0.00065
=–125.5  JK –1

Topic: Chemistry, Electrochemistry, JEE Main
The temperature dependence of rate constant $(k)$ of a chemical reaction is written in terms of Arrhenius equation, $\mathrm{k}=\mathrm{A} \cdot e^{-\mathrm{Ea} / \mathrm{RT}}$. Activation energy $\left(\mathrm{E}_a\right)$ of the reaction can be calculated by ploting:

Options:
A $k$ vs T
B $k$ vs $\frac{1}{\log \mathrm{T}}$
C $\log k$ vs $\frac{1}{\mathrm{T}}$
D $\log k$ vs $\frac{1}{\log \mathrm{T}}$

The correct answer is:
$\log k$ vs $\frac{1}{\mathrm{T}}$

Solution:
We can calculate the Activation Energy by graphing ln k versus 1/T. When the $\ln k$ (rate constant) is plotted versus the inverse of the temperature (kelvin), the slope is a straight line. The value of the slope $(\mathrm{m})$ is equal to $-\mathrm{Ea} / \mathrm{R}$ where $\mathrm{R}$ is a constant equal to $8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$.

Topic: Chemistry, Chemical Kinetics, JEE Main
The temperature dependence of resistance of Cu and undoped Si in the temperature range 300-400 K is best described by

Options:
A linear increase for Cu, exponential decrease for Si
B linear decrease for Cu, linear decrease for Si
C linear increase for Cu, linear increase for Si
D linear increase for Cu, exponential increase for Si

The correct answer is:
linear increase for Cu, exponential decrease for Si

Solution:
Resistance variation with temperature: resistance of a metal increases with increase in temperature linearly and resistance of a semiconductor decreases exponentially with increase in temperature.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2016 (03 Apr)
The temperature gradient in a rod of length $75 \mathrm{~cm}$ is $40^{\circ} \mathrm{C} / \mathrm{m}$. If the temperature of cooler end of the rod is $10^{\circ} \mathrm{C}$, then the temperature of hotter end is

Options:
A $50^{\circ} \mathrm{C}$
B $40^{\circ} \mathrm{C}$
C $35^{\circ} \mathrm{C}$
D $25^{\circ} \mathrm{C}$

The correct answer is:
$40^{\circ} \mathrm{C}$

Solution:
We know
$$
\begin{aligned}
& \mathrm{T}_{\mathrm{g}}=\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{x}} \\
& \Rightarrow \frac{\mathrm{T}_1-10}{0 \cdot 75}=40
\end{aligned}
$$
$\therefore \quad$ Temperature of the hotter end is:
$$
\begin{aligned}
& \mathrm{T}_1=(40 \times 0.75)+10 \\
& \mathrm{~T}_1=40^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
The temperature gradient in the earth's crust is 32 Â°C km-1 and the mean conductivity of earth is 0.008 cal s-1 cm-1 Â°C-1. Considering the earth to be a sphere of radius 6000 km, the loss of heat by the earth every day is about

Options:
A 1030 cal
B 1040 cal
C 1020 cal
D 1018 cal

The correct answer is:
1018 cal

Solution:
∆Q=KA∆T∆x∆t
=0.008102×4 ×2276×1082×32103×86400
≈1018 cal

Topic: Physics, Thermal Properties of Matter, JEE Main
The temperature of 3.00 mol of an ideal diatomic gas is increased by 40.0°C without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is x10. Then the value of x (round off to the nearest integer) is _________.

( GivenR=8.31 J mol-1 K-1

Options:

The correct answer is:
25

Solution:
First, recall the formula of change in internal energy in terms of molar specific heat at constant volume and change in temperature, ∆U=nCVΔT, for diatomic gas, CV=f2R=52R, ⇒∆U=n×52×RΔT

⇒ΔUW=52nRΔTnRΔT=52=2510

So, x=25

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The temperature of 5 moles of a gas at constant volume is changed from 100 Â°C to 120 Â°C. The change in internal energy is 80 J. The total heat capacity of the gas at constant volume (in J K-1) is

Options:
A 8
B 4
C 0.8
D 0.4

The correct answer is:
4

Solution:
Number of moles n=5 mol, T1=100 °C

T2=120 °C, ∆U=80 J

Rise in temperature ∆T=120-100=20 °C
∆U=ms∆T
80=5×M0×s×20, where M0 is the molar mass

The heat capacity is

C=5M0s=4 J K-1

Topic: Physics, Thermodynamics, JEE Main
The temperature of a body falls from 40 Â°C to 36 Â°C in 4 minutes, when placed in a constant ambient temperature of 20 Â°C. The time it takes for the temperature to fall from 36 Â°C to 32 Â°C in the same ambient temperature is

Options:
A 5.1 min
B 6.8 min
C 9 min
D 7.2 min

The correct answer is:
5.1 min

Solution:
Let Î¸ be instantaneous temperature, Î¸
For 40 Â°Câ36 Â°C
36-404=-k(38-20) ...(i)
Î¸avg=38 Â°C
For 36 Â°Câ32 Â°C
32-36t=-k(34-20) ...(ii)
Î¸avg=34 Â°C
(i)(ii) Gives

t=41814ât=5.1 min

Topic: Physics, Thermal Properties of Matter, JEE Main
The temperature of a body on Kelvin scale is found to be $x K$. When it is measured by Fahrenheit thermometer, it is found to be $x^{\circ} F$, then the value of $x$ is

Options:
A $40$
B $313$
C $574.25$
D $301.25$

The correct answer is:
$574.25$

Solution:
We know, $\mathrm{F}=1.8 \mathrm{~T}-459.67$
Where, $\mathrm{F}=$ Fahrenheit Temperature
$\mathrm{T}=$ Kelvin Temperature
Now, $\mathrm{F}=\mathrm{X}^{\circ} \mathrm{F}$ and $\mathrm{T}=\mathrm{XF}$, putting there we get,
$\begin{aligned} & \mathrm{X}=1.8 \mathrm{X}-459.67 \\ & \therefore \mathrm{X}=\frac{459.67}{0.8} \\ & =574.25\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main
The temperature of a gas is $-78^{\circ} \mathrm{C}$ and the average translational kinetic energy of its molecules is $\mathrm{K}$. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2 \mathrm{~K}$ is :

Options:
A $127^{\circ} \mathrm{C}$
B $117^{\circ} \mathrm{C}$
C $-39^{\circ} \mathrm{C}$
D $-78^{\circ} \mathrm{C}$

The correct answer is:
$117^{\circ} \mathrm{C}$

Solution:
$\begin{aligned}& \mathrm{K} . \mathrm{E}=\frac{\mathrm{nf}_1 \mathrm{RT}}{2} \\& \mathrm{~T}_{\mathrm{i}}=-78^{\circ} \mathrm{C} \rightarrow 273+\left[-78^{\circ} \mathrm{C}\right]=195 \mathrm{~K}
\end{aligned}$
K.E $\alpha \mathrm{T}$
To double the K.E energy temp also become double
$\begin{aligned}& \mathrm{T}_{\mathrm{f}}=390 \mathrm{~K} \\
& \mathrm{~T}_{\mathrm{f}}=117^{\circ} \mathrm{C}\end{aligned}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The temperature of a gas is raised from $27^{\circ} \mathrm{C}$ to $927^{\circ} \mathrm{C}$. The root mean square speed

Options:
A gets halved
B gets doubled
C is $\sqrt{\left(\frac{927}{27}\right)}$ times the earlier value
D remains the same

The correct answer is:
gets doubled

Solution:
Given, $\begin{aligned} T_1 & =27^{\circ} \mathrm{C}=300 \mathrm{~K} \\ T_2 & =927^{\circ} \mathrm{C}=1200 \mathrm{~K}\end{aligned}$
$v_{\mathrm{rms}}=\sqrt{\left(\frac{3 k T}{m}\right)}$
$\begin{aligned} & \Rightarrow \quad v_{\mathrm{rms}} \propto \sqrt{T} \\ & \therefore \quad \frac{\left(v_{\text {rms }}\right)_1}{\left(v_{\text {rms }}\right)_2}=\sqrt{\frac{T_1}{T_2}} \\ & \frac{\left(v_{\text {rms }}\right)_1}{\left(v_{\text {rms }}\right)_2}=\sqrt{\frac{300}{1200}}=\frac{1}{2} \\ & \Rightarrow \quad\left(v_{\mathrm{rms}}\right)_2=2\left(v_{\mathrm{rms}}\right)_1 \\ & \end{aligned}$

Topic: Physics, Kinetic Theory of Gases, JIPMER, JIPMER 2016
The temperature of a gas is raised while its volume remains constant, the pressure exerted by the gas on the walls of the container increases because its molecules

Options:
A Lose more kinetic energy to the wall
B Are in contact with the wall for a shorter time
C Strike the wall more often with higher velocities
D Collide with each other with less frequency

The correct answer is:
Strike the wall more often with higher velocities

Solution:
Due to increase in temperature root-mean-square velocity of gas molecules increases. So they strike the wall more often with higher velocity. Hence the pressure exerted by a gas on the walls of the container increase.

Topic: Physics, Thermodynamics, NEET
The temperature of a liquid drops from 365 K to 361 K in 2 minutes. The time during which temperature of the liquid drops from 344 K to 342 K is (Room temperature is 293 K) -

Options:
A 84 s
B 72 s
C 66 s
D 60 s

The correct answer is:
84 s

Solution:
From Newton's law of cooling,
365-3612=K365+3612-293 ..... (1)
344-342t=K344+3422-293 .... (2)
Equation 1Ã·(2)
t=365+361-586344+342-586=140100 min
=140Ã60100 s=84 s

Topic: Physics, Thermal Properties of Matter, JEE Main