datavorous/entrance-exam-dataset huggingface dataset:
An insulated system contains 4 moles of an ideal diatomic gas at temperature \(T\). When a heat \(Q\) is supplied to the gas, 2 moles of the gas is dissociated into atoms and the temperature remained constant.Then the relation between \(Q\) and \(T\) is
(\(\mathrm{R}=\) universal gas constant.)

Options:
A \(Q=R T\)
B \(Q=2 R T\)
C \(Q=3 R T\)
D \(Q=4 R T\)

The correct answer is:
\(Q=R T\)

Solution:
Given, number of moles of an ideal diatomic gas at the temperature, \(T=4\)
When heat \(Q\) is supplied to the gas, 2 mole of the gas is dissociated into atoms and the temperature remains constant, therefore
heat supplied = change in its internal energy i.e.,
\(Q=\Delta u=\left(u_f-u_i\right)\)
or \(Q=\) (internal energy of 4 moles of a monatomic gas + internal energy of 2 moles of diatomic gas) (internal energy of 4 moles of a diatomic gas)
\(\begin{aligned}
& =\left(4 \times \frac{3}{2} R T+2 \times \frac{5}{2} R T\right)-\left(4 \times \frac{5}{2} R T\right) \\
& =6 R T+5 R T-10 R T=R T
\end{aligned}\)

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2019 (22 Apr Shift 1)
An insulating cylinder contains 4 moles of an ideal diatomic gas. When a heat $\mathrm{Q}$ is supplied, to it, 2 moles of the gas molecules dissociate. If the temperature of the gas remains constant, then the value of $Q$ is
$$
\text { ( } \mathbf{R} \text { - universal gas constant) }
$$

Options:
A 2RT
B RT
C 3RT
D 4RT

The correct answer is:
RT

Solution:
Heat supplied " $Q$ " is given by:
$$
\begin{aligned}
& \mathrm{Q}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}} \\
& \mathrm{U}_{\mathrm{i}}=\mathrm{Initial}_{\text {Internal Energy }} \\
& \mathrm{U}_{\mathrm{f}}=\text { Final Internal Energy }
\end{aligned}
$$
Initially cylinder contains 4 moles of diatomic gas,
$$
\begin{aligned}
& \mathrm{U}_{\mathrm{i}}=\mathrm{n}\left(\frac{\mathrm{f}}{2} \mathrm{RT}\right) \\
& \mathrm{U}=4[-\mathrm{RT}]=10 \mathrm{RT}
\end{aligned}
$$
2 moles of diatomic gas dissociate into monoatomic gas. Hence there are 4 moles of monoatomic gas and 2 moles of diatomic gas.
Final Internal Energy is given by:
$$
\begin{aligned}
& U_f=n\left(\frac{f}{2} R T\right)+n^{\prime}\left(\frac{f^{\prime}}{2}-R T\right) \\
& =4\left(\frac{3}{2} R T\right)+2\left(\frac{5}{2} R T\right) \\
& =11 R T
\end{aligned}
$$
Thus Heat supplied is,
$$
\begin{aligned}
& \mathrm{Q}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}} \\
& =11 \mathrm{RT}-10 \mathrm{RT}=\mathrm{RT} \\
& \mathrm{Q}=\mathrm{RT} .
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, TS EAMCET, TS EAMCET 2023 (12 May Shift 2)
An insulating solid sphere of the radius R is charged in a non-uniform manner such that the volume charge density ρ=Ar, where A is a positive constant and r is the distance from the centre. The potential difference between the centre and surface of the sphere is

Options:
A AR8ε0
B AR4ε0
C ARε0
D AR2Îµ0

The correct answer is:
AR2ε0

Solution:
E Ã 4Ïr2=qÎµ0
EÃ4Ïr2=â«0rArÃ4Ïr2dr
E=A2Îµ0

Since electric field is constant, we get
âV=EÃd
=A2Îµ0ÃR=AR2Îµ0

Topic: Physics, Electrostatics, JEE Main
An insulator container contains 4 moles of an ideal diatomic gas at temperature $T$. Heat $Q$ is supplied to this gas, due to which 2 moles of the gas are dissociated into atoms but temperature of the gas remains constant. Then

Options:
A $Q=2 R T$
B $Q=R T$
C $Q=3 R T$
D $Q=4 R T$

The correct answer is:
$Q=R T$

Solution:
$Q=\Delta U=U_f-U_i=[$ internal energy of 4 moles of a monoatomic gas + internal energy of 2 moles of a diatomic gas] - [internal energy of 4 moles of a diatomic gas]
$=\left(4 \times \frac{3}{2} R T+2 \times \frac{5}{2} R T\right)-\left(4 \times \frac{5}{2} R T\right)=R T$
Note : (a) 2 moles of diatomic gas becomes 4 moles of a monoatomic gas when gas dissociated into atoms.
(b) Internal energy of $\mu$ moles of an ideal gas of degrees of freedom $F$ is given by $U=\frac{f}{2} \mu R T$

Topic: Physics, Thermodynamics, JEE Main
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident involving a scooter driver, car driver and a truck driver are $0.01,0.03$ and $0.15$ respectively. One of the insured persons meets with an accident. The probability that the person is a scooter driver is

Options:
A $\frac{1}{52}$
B $\frac{3}{52}$
C $\frac{15}{52}$
D $\frac{19}{52}$

The correct answer is:
$\frac{1}{52}$

Solution:
Let $\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3} \& \mathrm{~A}$ be events defined as follows. $\mathrm{E}_{1}=$ person chosen is a scooter driver $\mathrm{E}_{2}=$ person chosen is a car driver. $\mathrm{E}_{3}=$ person chosen is a truck driver $\&$ $\mathrm{A}=$ person meets with an accident $\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{2000}{12000}=\frac{1}{6} ; \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{1}{3} \& \mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{1}{2}$
$\because$ Probability that a person meets with an accident
given that he is a scooter driver $=\mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)=0.01$
$\mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)=0.03 \& \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{3}\right)=0.15$
- the person meets with an accident. $\therefore$ the probability that he was a scooter driver
$=\mathrm{P}\left(\mathrm{E}_{1} / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{3}\right)}$
$=\mathrm{P}\left(\mathrm{E}_{1} / \mathrm{A}\right)=\frac{1 / 6 \times 0.01}{\left(\frac{1}{6} \times 0.01\right)+\left(\frac{1}{3} \times 0.03\right)+\left(\frac{1}{2} \times 0.15\right)}=\frac{1}{52}$

Topic: Mathematics, Probability, JEE Main
An integer is chosen at random from the integers 1 , 2,3, ....., 50. The probability that the chosen integer is a multiple of atleast one of 4,6 and 7 is

Options:
A 825
B 2150
C 950
D 1425

The correct answer is:
2150

Solution:
Let A=1,2,3,4,5,.........,50

Now, the probability of choosing a number which is a multiple of 4 will be P4=1250

Similarly, the probability of choosing a number which is a multiple of 6 will be P6=850

And the probability of choosing a number which is a multiple of 7 will be P7=750

Also, the probability of choosing a number which is a multiple of 4 & 6 will be P4∩6=450

And the probability of choosing a number which is a multiple of 6 & 7 will be P6∩7=150

And the probability of choosing a number which is a multiple of 4 & 7 will be P4∩7=150

And finally, the probability of choosing a number which is a multiple of 4, 6 & 7 will be P4∩6∩7=0

Hence, P4∪6∪7=1250+850+750-450-150-150+0

⇒P4∪6∪7=2150

Topic: Mathematics, Probability, JEE Main, JEE Main 2024 (29 Jan Shift 2)
An integrating factor for the differential equation $\left(1+y^2\right) d x-\left(\tan ^{-1} y-x\right) d y=0$

Options:
A $\tan ^{-1} y$
B $e^{\tan ^{-1} y}$
C $\frac{1}{1+y^2}$
D $\frac{1}{x\left(1+y^2\right)}$

The correct answer is:
$e^{\tan ^{-1} y}$

Solution:
$\begin{aligned} & \left(1+y^2\right) d x-\left(\tan ^{-1} y-x\right) d y=0 \\ & \Rightarrow \frac{d y}{d x}=\frac{1+y^2}{\tan ^{-1} y-x} \Rightarrow \frac{d x}{d y}=\frac{\tan ^{-1} y}{1+y^2}-\frac{x}{1+y^2} \\ & \Rightarrow \frac{d x}{d y}+\frac{x}{1+y^2}=\frac{\tan ^{-1} y}{1+y^2}\end{aligned}$
This is equation of the form $\frac{d x}{d y}+P x=Q$
So, I.F. $=e^{\int P d y}=e^{\int \frac{1}{1+y^2} \cdot d y}=e^{\tan ^{-1} y}$.

Topic: Mathematics, Differential Equations, JEE Main
An interaction in which one organism is benefitted and other is unaffected

Options:
A predation
B commensalism
C mutualism
D parasitism

The correct answer is:
commensalism

Solution:
Commensalism is a type of interaction where one organism is benefitted and other remain unaffected. In predation, one organism feeds on another organism.
In parasitism, one organism (parasite) gets the benefit at the expense of the host.
Mutualism is a type of positive interaction in which all individuals are benefitted by one another.

Topic: Biology, Organisms and Populations, JIPMER, JIPMER 2019
An interference is observed due to two coherent sources A and B separated by a distance 4Î» along Y-axis where Î» is the wavelength of the light source used. A detector D is moved along the positive X-axis. Find the total number of maxima points are observed on the X-axis excluding the points x=0 and x=â?

Options:

The correct answer is:
3

Solution:
To obtain maxima the path difference
AD-BD=nλ
AD-xn=nλ
AD=nλ+xn ...(i)
Form ∆le ABD,
AD2=AB2+BD2
nλ+xn2=16λ2+xn2
n2λ2+xn2+2nλxn=16λ2+xn2
2nλxn=16-n2λ2
xn=16-n2λ2n
For n=0⇒x0=∞
n=1⇒x1=16-12λ=15λ2=7.5λ
n=2⇒x2=16-12λ=12λ4=3λ
n=3⇒x3=16-96λ=7λ6
n=4⇒x4=16-162×4λ=0
So the total no. of maxima points observed are 3.

Topic: Physics, Wave Optics, JEE Main
An interference pattern was made by using red light. If the red light changes with blue light, the fringes will become

Options:
A Wider
B Narrower
C Fainter
D Brighter

The correct answer is:
Narrower

Solution:
$\beta=\frac{\lambda D}{d} \Rightarrow \beta \propto \lambda$

Topic: Physics, Wave Optics, JEE Main
An inverted bell, lying at the bottom of lake 47.6 m deep, has 50 cm3 of air trapped in it. The bell is brought to the surface of the lake, then the volume of the trapped air will become [atmospheric pressure = 70 cm of Hg and density of Hg=13.6 g cm-3]

Options:
A 600 cm3
B 300 cm3
C 250 cm3
D 200 cm3

The correct answer is:
300 cm3

Solution:
According to Boyle's law, pressure and volume are inversely proportional to each other i.e. p â 1 v

â P 1 V 1 = P 2 V 2
â ( P 0 + h Ï w g ) V 1 = P 0 V 2
â V 2 = ( 1 + h Ï w g P 0 ) V 1
V 2 =(1+ 47.6*1*1000*10 70* 10 â2 *13.6*1000*10 ) [Asâ P 2 = P 0 =70ââcmââofâHg=70* 10 â2 *13.6*1000*10]
â V 2 = 1 + 5 5 0 cm 3 = 3 0 0 cm 3

Topic: Physics, Thermodynamics, JEE Main
An inverted pear-shaped organ attached to the pelvic wall involved in reproduction is

Options:
A Womb
B Uterus
C Breast
D Both (a) and (b)

The correct answer is:
Both (a) and (b)

Solution:
The uterus is a female reproductive organ located between the bladder and rectum, in the pelvic area, and is an inverted pear-shaped structure. The uterus is connected to the fallopian tubes, the cervix, and the vagina. The main purpose of the uterus is to nourish a foetus prior to birth. In menstruating females, the ovaries release eggs that travel via the fallopian tubes to the uterus. If fertilized, the eggs bind themselves to the wall of the uterus (womb) and the foetus develops.

Topic: Biology, Human Reproduction, NEET
An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of the faces of the cube. The empirical formula for this compound would be

Options:
A
$\mathrm{A}\mathrm{~B}$

B
$\mathrm{A}_2 \mathrm{~B}$

C
$\mathrm{A}\mathrm{~B}_3$

D
$\mathrm{A}_3 \mathrm{~B}$

The correct answer is:
$\mathrm{A}\mathrm{~B}_3$

Solution:
$$
A=\frac{1}{8} \times 8=1
$$
(Corner)
$\mathrm{B}=\frac{1}{2} \times 6=3$
(Face centre)
$\therefore \mathrm{AB}_3$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2005
An ionic compound is generally a

Options:
A Good electrolyte
B Weak electrolyte
C Non-electrolyte
D Neutral

The correct answer is:
Good electrolyte

Solution:
Ionic compounds are good conductors of heat and electricity so they are good electrolyte.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
An iron ball and a wooden ball of the same radius are released from a height ' $h$ ' in vacuum. The time taken by both of them to reach the ground is

Options:
A Unequal
B Exactly equal
C Roughly equal
D Zero

The correct answer is:
Exactly equal

Solution:
Time of decent $t=\sqrt{\frac{2 h}{g}}$. In vacuum no other force works except gravity so time period will be exactly equal.

Topic: Physics, Gravitation, JEE Main
An iron bar of length $10 \mathrm{~m}$ is heated from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$. If the coefficient of linear thermal expansion of iron is $10 \times 10^{-6} /{ }^{\circ} \mathrm{C}$, the increase in the length of bar is

Options:
A $0.5 \mathrm{~cm}$
B $1.0 \mathrm{~cm}$
C $1.5 \mathrm{~cm}$
D $2.0 \mathrm{~cm}$

The correct answer is:
$1.0 \mathrm{~cm}$

Solution:
Increase in length $\Delta L=L_0 \propto \Delta \theta$
$=10 \times 10 \times 10^{-6} \times(100-0)=10^{-2} \mathrm{~m}=1 \mathrm{~cm}$

Topic: Physics, Thermal Properties of Matter, JEE Main
An iron load of $2 \mathrm{~kg}$ is suspended in air from the free end of a sonometer wire of length $1 \mathrm{~m}$. A tuning fork of frequency $256 \mathrm{~Hz}$, is in resonance with $\frac{1}{\sqrt{7}}$ times the length of the sonometer wire. If the load is immensed in water, the length of the wire in metre that will be in resonance with the same tuning fork is
$($ Specific gravity of iron $=8)$

Options:
A $\sqrt{8}$
B $\sqrt{6}$
C $\frac{1}{\sqrt{6}}$
D $\frac{1}{\sqrt{8}}$

The correct answer is:
$\frac{1}{\sqrt{8}}$

Solution:
We know $I \propto \sqrt{T}$
$$
\therefore \quad \frac{I_{\text {air }}}{I_{\text {water }}}=\sqrt{\frac{T_{\text {air }}}{T_{\text {water }}}}
$$ $\begin{array}{llrl}\text { But specific gravity } 8 & =\frac{T_{\text {air }}}{T_{\text {air }}-T_{\text {water }}} \\ \Rightarrow & T_{\text {water }} =\frac{7}{8} T_{\text {air }} \\ \therefore & \frac{l_{\text {air }}}{l_{\text {water }}} =\sqrt{\frac{8}{7}} \\ \text { But } & l_{\text {air }} =\frac{1}{\sqrt{7}} l \text { (given) } \\ \therefore & l_{\text {water }} =\frac{1}{\sqrt{8}} l\end{array}$

Topic: Physics, Waves and Sound, JEE Main
An iron load of 2 kg is suspended in the air from the free end of a sonometer wire of length 1 m. A tuning fork of frequency 256 Hz is in resonance with 1 7 times the length of the sonometer wire. If the load is immersed in water, the fraction of length of the wire that will be in resonance with the same tuning fork, will be (specific gravity of iron = 8) (assume both cases in same harmonic)

Options:
A 8
B 6
C 16
D 18

The correct answer is:
18

Solution:
We know lâT
â´Î¹airÎ¹water=TairTwater
But specific gravity x = 8
Tâ¡airâ¡=mâ¡gâ¡, Tâ¡water=mg1-1x
= 7 8 mg
â Twater=78Tair
â´ lairlwater=87
But
lair=17l (Given)
â´ lwater=18l

Topic: Physics, Waves and Sound, JEE Main
An iron rocket fragment initially at -100 Â°C enters the earth's atmosphere almost horizontally and quickly fuses completely in atmospheric friction. Specific heat of iron is 0.11 kcal kg-1 Â°C-1, its melting point is 1535 Â°C and the latent heat of fusion is 3 kcal kg-1. The minimum velocity with which the fragment must have entered the atmosphere is

Options:
A 0.45 km s-1
B 1.32 km s-1

C 2.32 km s-1

D Zero

The correct answer is:
1.32 km s-1

Solution:
The kinetic energy of the fragment is changed to heat. Expressing mass m in kilograms everywher, we have
12mv2=m30+m0.111535∘+100∘4184
v2=836830+180≃1.76×106
Hence
v=1.76×103=1.32 km s-1

Topic: Physics, Kinetic Theory of Gases, JEE Main
An iron rod is placed parallel to magnetic field of intensity 2000 A m. The magnetic flux through the rod is 6Ã10-4 Wb and its cross-sectional area is 3 cm2. The magnetic permeability of the rod in Wb A-1m-1 is

Options:
A 10-1
B 10-2
C 10-3
D 10-4

The correct answer is:
10-3

Solution:
μ=ϕA×1H=6×10−43×10−4×12×103=10−3 Wb A-1m-1

Topic: Physics, Magnetic Properties of Matter, JEE Main
An iron rod of length $2 \mathrm{~m}$ and cross-sectional area of $50 \mathrm{~mm}^{2}$ stretched by $0.5 \mathrm{~mm}$, when a mass of $250 \mathrm{~kg}$ is hung from its lower end. Young's modulus of iron rod is

Options:
A $19.6 \times 10^{20} \mathrm{~N} / \mathrm{m}^{2}$
B $19.6 \times 10^{18} \mathrm{~N} / \mathrm{m}^{2}$
C $19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$
D $19.6 \times 10^{15} \mathrm{~N} / \mathrm{m}^{2}$

The correct answer is:
$19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$

Solution:
$Y=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\frac{250 \times 9.8}{50 \times 10^{-6}}}{\frac{0.5 \times 10^{-3}}{2}}$
$=\frac{250 \times 9.8}{50 \times 10^{-6}} \times \frac{2}{0.5 \times 10^{-3}} \Rightarrow 19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
An iron rod of length $2 \mathrm{~m}$ and cross-sectional area of $50 \mathrm{~mm}^{2}$ stretched by $0.5 \mathrm{~mm}$, when a mass of $250 \mathrm{~kg}$ is hung from its lower end. Young's modulus of iron rod is

Options:
A $19.6 \times 10^{20} \mathrm{~N} / \mathrm{m}^{2}$
B $19.6 \times 10^{18} \mathrm{~N} / \mathrm{m}^{2}$
C $19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$
D $19.6 \times 10^{15} \mathrm{~N} / \mathrm{m}^{2}$

The correct answer is:
$19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$

Solution:
$\begin{aligned}
\mathrm{Y}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\frac{250 \times 9.8}{50 \times 10^{-6}}}{\frac{0.5 \times 10^{-3}}{2}} \\
=\frac{250 \times 9.8}{50 \times 10^{-6}} \times \frac{2}{0.5 \times 10^{-3}} \Rightarrow 19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}
\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
An iron rod of volume 10-3 m3 and relative permeability 1000 is placed as core in a solenoid with10 turns cm-1 . If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be :

Options:
A 50×102Am2
B 5×102Am2
C 500×102Am2
D 0.5×102Am2

The correct answer is:
5×102Am2

Solution:
Given,

Volume of iron rod is, V=10-3 m3

Relative permeability, μr=1000

No of turns per unit length of solenoid is, n=Nl=10 turns cm-1=1000 turns m-1

Current through the solenoid is, i=0.5 A

Magnetic moment of an iron core solenoid is given by,

M=(μr-1) NiA

⇒M=μr-1NiVl=μr-1NliV

⇒M=1000-1×1000×0.5×10-3

⇒M=499.5≈500 A m2

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (05 Sep Shift 2)
An iron wire AB has diameter of 0.6 mm and length 3 m at 0 â C .The wire is now stretched between the opposite walls of a brass casing at 0 â C . What is the extra tension that will be set up in the wire when the temperature of the system is raised to 40 â C ?

Given Î± brass =18Ã 10 â6 /K

Î± iron =12Ã 10 â6 /K

Y iron =21Ã 10 10 N/m 2

Options:
A 14.2â N
B 13.8 N
C 16.3 N
D 21.7 N

The correct answer is:
14.2â N

Solution:
Increase in length of the iron wire = L α Δ t = L × 12 × 10 - 6 × 40 = 48L × 10 - 5 m
Increase in length of the brass tube = L α Δ t = L × 18 × 10 - 6 × 40 = 72L × 10 - 5 m
Since the increase in length of the brass tube is greater than that of the iron wire, the latter will be tightened. Let T be the tension in the wire. Then
Stress = T⁡ π r⁡ 2 = T⁡ π 3 × 1 0 - 4 2 = T⁡ 9 π × 1 0 - 8
Strain = 7 2 L ⁡ - 4 8 L⁡ × 1 0 - 5 L⁡ = 2 4 × 1 0 - 5
Since stress=Y×strain,T⁡9π×10-8=21×1010×24×10-5
T = 2 1 × 1 0 10 × 2 4 × 1 0 - 5 × 9 π × 1 0 - 8 = 14.2 N

Topic: Physics, Mechanical Properties of Solids, JEE Main
An irreversible process occurring isothermally in an isolated system leads to

Options:
A Zero entropy
B An increase in the total entropy of the system
C A decrease in the total entropy of the system
D None of these

The correct answer is:
An increase in the total entropy of the system

Solution:
Entropy of the system increases as the process occur irreversibly and isothermally in an isolated system.

Topic: Chemistry, Thermodynamics (C), JEE Main
An isolated particle of mass m is moving in the horizontal plane x, y along the x-axis, at a certain height above the ground. It suddenly explodes into two fragments of masses m4 and 3m4. An instant later, the smaller fragment is at y=+15 cm. The larger fragment at this instant is at

Options:
A y=-5 cm
B y=+20 cm
C y=+5 cm
D y=-20 cm

The correct answer is:
y=-5 cm

Solution:
Before the explosion, the particle was moving along x-axis i.e., it has no y-component of velocity. Therefore, the centre of mass will not move in y-direction or we can say ycm=0.
Now, ycm=m1y1+m2y2m1+m2
Therefore, 0=m/4+15+3m/4ym/4+3m/4
or, y=-5 cm

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
An isolated soap bubble is kept in a container in which the pressure is maintained at P0. The bubble is given some charge due to which its radius increases to r and the pressure inside the bubble reduces to P0. If the surface tension of the soap bubble is T, then the charge given to the bubble is

Options:
A Q=8πrrTε0
B Q=8πr2rTε0
C Q=4πr2rTε0
D Q=4πrrTε0

The correct answer is:
Q=8πr2rTε0

Solution:
The excess pressure inside a soap bubble due to surface tension is 4Tr.

In the given problem the pressure of the air inside and outside the bubble is the same, hence the excess pressure in this situation should be due to the electrostatic pressure.

4Tr=σ22ε0

⇒4Tr=12ε0Q4πr22⇒Q=8πr2rTε0

Topic: Physics, Electrostatics, JEE Main
An isomer of C6H14 forms three monochloro derivatives. The isomer may be -
(Excluding stereo isomer)

Options:
A neo-pentane
B n-hexane
C 2, 3-dimethylbutane
D iso-hexane

The correct answer is:
n-hexane

Solution:
In n-hexane, there is a symmetry of the structure at third and fourth carbon. So, chlorine can attach to first, second, and third carbon to give 3 monochloro derivatives. These monochloro derivatives are:

Topic: Chemistry, General Organic Chemistry, JEE Main
An isosceles triangle made of wood of base 10 feet and height 8 feet is placed vertically with its base on the ground and vertex directly above it. If the triangle faces the sun whose altitude is 30o, then the tangent of the angle at the apex of the shadow is

Options:
A 803
B 803167
C 8923
D 803217

The correct answer is:
803167

Solution:
Let, ABC is the triangle and BCE is its shadow on the ground
Height of the triangular shadow =DE=8cot⁡30o=83 feet
∴ tan⁡α=583 from ΔBDE
⇒tan⁡2α=2tan⁡α1-tan2α
=25831-25192=543×192167=803167

Topic: Mathematics, Heights and Distances, JEE Main
An isosceles triangle of wood of base 10 feet and height 83 feet is placed vertically with its base on the ground and vertex directly above. The triangle faces the sun whose altitude is 30o. Then, the tangent of the angle at the apex of the shadow is

Options:
A 80
B 8039
C 892
D 80217

The correct answer is:
8039

Solution:
Let, ABC is the triangle and triangle BCE is its shadow on the ground
Height of the triangular shadow =DE=83cot⁡30o=8 feet
∴tan⁡α=58 (from ∆BDE )
tan⁡2α=2tan⁡α1-tan2⁡α
=2×581-2564=54×6439=8039

Topic: Mathematics, Heights and Distances, JEE Main
An isotope of the original nucleus can be formed in a radioactive decay, with the emission of following particles.

Options:
A one $\alpha$ and one $\beta$
B one $\alpha$ and two $\beta$
C one $\alpha$ and four $\beta$
D four $\alpha$ and one $\beta$

The correct answer is:
one $\alpha$ and two $\beta$

Solution:
${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}} \stackrel{\alpha}{\longrightarrow}{ }_{\mathrm{Z}-2} \mathrm{Y}^{\mathrm{A}-4} \stackrel{2 \beta^{-}}{\longrightarrow}{ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}-4}$

Topic: Physics, Nuclear Physics, JEE Main