datavorous/entrance-exam-dataset huggingface dataset:
An empty tank has concave mirror as its bottom. When sunlight falls normally on the mirror, it is focussed at a height of $32 \mathrm{~cm}$ from the mirror. If the tank is filled with water upto a height of $20 \mathrm{~cm}$, then the sunlight focusses at (refractive index of water $=\frac{4}{3}$ )

Options:
A $16 \mathrm{~cm}$ above water level
B $9 \mathrm{~cm}$ above water level
C $16 \mathrm{~cm}$ below water level
D $9 \mathrm{~cm}$ below water level

The correct answer is:
$9 \mathrm{~cm}$ above water level

Solution:
Sun at infinity, $u=\infty$
Focussed at a height, $\mathrm{v}=-32 \mathrm{~cm}$
Refractive index of water, $\mu_{\mathrm{w}}=\frac{4}{3}$
From the mirror formula
$\begin{aligned}
& \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} \Rightarrow \frac{1}{\mathrm{f}}=\frac{1}{32}+\frac{1}{\infty} \\
& \Rightarrow \mathrm{f}=-32 \mathrm{~cm}
\end{aligned}$
When water is filled in the tank upto a height of $20 \mathrm{~cm}$, the image formed by the mirror will act as virtual object for water surface.
Which will form its image such that
$\begin{aligned} & \frac{\text { Actual height }}{\text { Apparent height }}=\frac{\mu_{\mathrm{w}}}{\mu_{\mathrm{a}}} \\ & \frac{\mathrm{BO}}{\mathrm{BI}}=\frac{4 / 3}{1} \\ & \Rightarrow \mathrm{BI}=\mathrm{BO} \times \frac{3}{4} \\ & =12 \times \frac{3}{4}=9 \mathrm{~cm} \text { above water level }\end{aligned}$

Topic: Physics, Ray Optics, JEE Main
An empty vessel is partially filled with water, then the frequency of vibration of air column in the vessel

Options:
A Remains same
B Decreases
C Increases
D First increases then decreases

The correct answer is:
Increases

Solution:
For closed pipe in general $n=\frac{v}{4 l}(2 N-1) \quad n \propto \frac{1}{l}$
i.e. if length of air column decreases frequency increases.

Topic: Physics, Waves and Sound, JEE Main
An emulsion is a colloidal dispersion of

Options:
A A liquid in a gas
B A liquid in a liquid
C A solid in a liquid
D A gas in a solid

The correct answer is:
A liquid in a liquid

Solution:
Correct option is 2. A liquid in a liquid
An emulsion is a colloidal dispersion of a liquid in either a liquid or a solid. A stable emulsion requires an emulsifying agent to be present.

Topic: Chemistry, Surface Chemistry, JEE Main
An emulsion is a colloidal dispersion of

Options:
A A liquid in a gas
B A liquid in a liquid
C A solid in a liquid
D A gas in a solid

The correct answer is:
A liquid in a liquid

Solution:
Emulsions are colloidal system in which dispersion medium and dispersed phase both are liquids. So, emulsion is dispersion of liquid in liquid.

Topic: Chemistry, Surface Chemistry, JEE Main
An endocrine gland, which secretes melatonin, is located at

Options:
A Dorsal side of forebrain
B Middle side of forebrain
C Dorsal side of midbrain
D Middle side of midbrain

The correct answer is:
Dorsal side of forebrain

Solution:
The Pineal Gland
- Location: Dorsal side of forebrain.
- A hormone secreted: Melatonin.
- Functions of melatonin :
1. Melatonin is very important for the regulation of a 24-hour (diurnal) rhythm of our body.
Example: It helps in maintaining the normal rhythms of the sleep-wake cycle, body temperature, etc.
2. It also affects - metabolism, pigmentation, menstrual cycle, and our immune system.

Topic: Biology, Chemical Coordination and Integration, NEET
An endothermic reaction $A \rightarrow B$ has an activation energy $15 \mathrm{kcal} /$ mole and energy of reaction $5 \mathrm{kcal} / \mathrm{mole}$. The activation energy of the reaction $B \rightarrow A$ is

Options:
A $20 \mathrm{kcal} / \mathrm{mole}$
B $15 \mathrm{kcal} / \mathrm{mole}$
C $10 \mathrm{kcal} / \mathrm{mole}$
D None of these

The correct answer is:
$10 \mathrm{kcal} / \mathrm{mole}$

Solution:
Its a Endothermic Reaction, So
\(\Delta \mathrm{H}=\left[\mathrm{E}_{\mathrm{a}(\mathrm{f})}\right]-\left[\mathrm{E}_{\mathrm{b}(\mathrm{f})}\right]\)
Given, \(\left[\mathrm{E}_{\mathrm{a}(\mathrm{f})}\right]=15 \mathrm{kcal} / \mathrm{mol}\) and \(\Delta \mathrm{H}=5 \mathrm{kcal} / \mathrm{mol}\)
\(\left[\mathrm{E}_{\mathrm{b}(\mathrm{f})}\right]=15-5=10 \mathrm{kcal} / \mathrm{mol}\)

Topic: Chemistry, Chemical Kinetics, JEE Main
An endothermic reaction with high activation energy for the forward reaction is given by the diagram.

Options:
A
B
C
D

The correct answer is:

Solution:
Endothermic reactions are those which involve absorption of heat. High activation energy means potential energy of product must be much greater than reactants.

Topic: Chemistry, General Organic Chemistry, JEE Main
An energy source will supply a constant current into the load of its internal resistance is

Options:
A
equal to the resistance of the load.

B
very large as compared to the load resistance.

C
zero.

D
non-zero but less than the resistance of the load.

The correct answer is:
very large as compared to the load resistance.

Solution:
$I=\frac{E_0}{R+r} \square \frac{E}{r}$ if $R < r$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2005
An engine approaches a hill with a constant speed. When it is at a distance of $0.9 \mathrm{~km}$, it blows a whistle whose echo is heard by the driver after $5$ seconds. If the speed of sound in air is $330 \mathrm{~m} / \mathrm{s}$, then the speed of the engine is :

Options:
A
$32 \mathrm{~m} / \mathrm{s}$

B
$27.5 \mathrm{~m} / \mathrm{s}$

C
$60 \mathrm{~m} / \mathrm{s}$

D
$30 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$30 \mathrm{~m} / \mathrm{s}$

Solution:
Let after $5 \mathrm{~sec}$ engine at point $\mathrm{C}$
$\begin{aligned}
& \mathrm{t}=\frac{\mathrm{AB}}{330}+\frac{\mathrm{BC}}{330} \\
& 5=\frac{0.9 \times 1000}{330}+\frac{\mathrm{BC}}{330} \\
& \therefore \mathrm{BC}=750 \mathrm{~m}
\end{aligned}$
Distance travelled by engine in $5 \mathrm{~sec}$
$=900 \mathrm{~m}-750 \mathrm{~m}=150 \mathrm{~m}$
Therefore velocity of engine
$=\frac{150 \mathrm{~m}}{5 \mathrm{~sec}}=30 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2013 (09 Apr Online)
An engine develops $10 \mathrm{~kW}$ of power. How much time will it take to lift a mass of $200 \mathrm{~kg}$ to a height of $40 \mathrm{~m}$. $\left(g=10 \mathrm{~m} / \mathrm{sec}^2\right)$

Options:
A 4 sec
B 5 sec
C 8 sec
D 10 sec

The correct answer is:
8 sec

Solution:
$P=\frac{m g h}{t}=10 \times 10^3 \Rightarrow t=\frac{200 \times 40 \times 10}{10 \times 10^3}=8 \mathrm{sec}$

Topic: Physics, Work Power Energy, JEE Main
An engine of a train, moving with uniform acceleration, passes the signal-post with velocity u and the last compartment with velocity v. The velocity with which middle point of the train passes the signal post is :

Options:
A u+v2
B v2+u22
C v-u2
D v2-u22

The correct answer is:
v2+u22

Solution:
v'2=u2+2ad

v2=v'2+2ad

solving, we get

v'=v2+u22

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (25 Feb Shift 1)
An engine of power 58.8 kW pulls a train of mass 2Ã105 kg with a velocity of 36 km h-1. The coefficient of static friction is

Options:
A 0.3
B 0.03
C 0.003
D 0.0003

The correct answer is:
0.003

Solution:
Power of engine,
P=58.8 kW=58.8Ã103 W
Mass of the train, m=2Ã105 kg
Velocity of train, v=36 kmh-1
=36Ã10360Ã60=10 m s-1
Power, P=Fâv
â´ F=Pv=58.8Ã10310
=5880 N
Let the coefficient of friction is Î¼
â´ F=fs=Î¼ mg
â Î¼=Fmg=58802Ã105Ã10
=0.003

Topic: Physics, Work Power Energy, JEE Main
An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close to

Options:

The correct answer is:
19

Solution:
W=2P0v0

Q+=WAB+WBC

=n.cvΔTAB+nCPΔTBC

=32×3P0v0−P0v0+526P0V0−3P0V0

=3P0V0+152P0V0

=212P0V0

η=WQ+=2P0V0212P0V0=421

η%=40021≈19

Topic: Physics, Thermodynamics, JEE Main
An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close to

Options:

The correct answer is:
19

Solution:
W=2P0v0

Q+=WAB+WBC

=n.cvΔTAB+nCPΔTBC

=32×3P0v0−P0v0+526P0V0−3P0V0

=3P0V0+152P0V0

=212P0V0

η=WQ+=2P0V0212P0V0=421

η%=40021≈19

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (06 Sep Shift 2)
An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is:
(Take Cv=1.5R, whereR is gas constant)

Options:
A 0.24
B 0.15
C 0.32
D 0.08

The correct answer is:
0.15

Solution:
Work done in cycle =area under closed curve
W=P0V0
Heat given =QAB+QBC
=nCV ΔTAB+nCp ΔTBC
=32nRTB-nRTA+52nRTC-nRTB
=322P0V0-P0V0+524P0V0-2P0V
=132 P0V0
Thermal efficiency, η=WQabsorbed=213=0.15

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2017 (08 Apr Online)
An engine pumps water continuously through a hose. Water leaves the hose with a velocity $\mathrm{v}$ and $\mathrm{m}$ is the mass per unit length of the water jet. What is the rate which kinetic energy is imparted to water ?

Options:
A $\frac{1}{2} m^2 v^2$
B $\frac{1}{2} \mathrm{mv}^3$
C $\mathrm{mv}^3$
D $\frac{1}{2} \mathrm{mv}^2$

The correct answer is:
$\frac{1}{2} \mathrm{mv}^3$

Solution:
$\mathrm{mv}=\frac{\mathrm{dm}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dt}}=\frac{\mathrm{dm}}{\mathrm{dt}}=$ Rate of flowing mass
$$
\begin{aligned}
& \mathrm{F}_{\mathrm{av}}=\frac{\mathrm{dm}}{\mathrm{dt}} \times \frac{\mathrm{v}}{2}=\frac{(\mathrm{mv}) \mathrm{v}}{2}=\frac{\mathrm{mv}^2}{2} \\
& \mathrm{p}=\frac{\mathrm{dK}}{\mathrm{dt}}=\frac{\mathrm{mv}^2}{2} \times \mathrm{v}=\frac{\mathrm{mv}^3}{3}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
An engine pumps water continuously through a hose. Water leaves the hose with a velocity $\mathrm{v}$ and $\mathrm{m}$ is the mass per unit length of the water jet. What is the rate which kinetic energy is imparted to water ?

Options:
A $\frac{1}{2} m^2 v^2$
B $\frac{1}{2} \mathrm{mv}^3$
C $\mathrm{mv}^3$
D $\frac{1}{2} \mathrm{mv}^2$

The correct answer is:
$\frac{1}{2} \mathrm{mv}^3$

Solution:
$\mathrm{mv}=\frac{\mathrm{dm}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dt}}=\frac{\mathrm{dm}}{\mathrm{dt}}=$ Rate of flowing mass
$$
\begin{aligned}
& \mathrm{F}_{\mathrm{av}}=\frac{\mathrm{dm}}{\mathrm{dt}} \times \frac{\mathrm{v}}{2}=\frac{(\mathrm{mv}) \mathrm{v}}{2}=\frac{\mathrm{mv}^2}{2} \\
& \mathrm{p}=\frac{\mathrm{dK}}{\mathrm{dt}}=\frac{\mathrm{mv}^2}{2} \times \mathrm{v}=\frac{\mathrm{mv}^3}{3}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2009 (Mains)
An engine pumps water continuously through a hose. Water leaves the hose with a velocity $y$ and $m$ is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water ?

Options:
A $\frac{1}{2} \mathrm{mv}^3$
B $\mathrm{mv}^3$
C $\frac{1}{2} \mathrm{mv}^2$
D $\frac{1}{2} m^2 v^2$

The correct answer is:
$\frac{1}{2} \mathrm{mv}^3$

Solution:
Velocity of water is \(v\), mass flowing per unit length is \(m\).
Mass flowing per second \(=\mathrm{mv}\)
\(\therefore\) Rate of kinetic energy or K.E. per second
\(=1 / 2(m v) v^2=1 / 2 m v^3\)

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
An engine pumps water continuously through a hose. Water leaves the hose with velocity $v$ and $m$ is mass per unit length of the water jet. If this jet hits a surface and came to rest instantaneously, the force on the surface is

Options:
A
$m v^3$

B
$m v^2$

C
$\frac{1}{2} m v^2$

D
$\frac{1}{2} m v^3$

The correct answer is:
$m v^2$

Solution:
As power $\mathrm{P}=F . v .=\frac{F \ell}{t}$
$$
\begin{aligned}
& \text { or, } v\left(M \frac{v}{t}\right)=\frac{F . \ell}{t} \\
& \Rightarrow \quad M v^2=F \ell \ell \\
& \Rightarrow \quad F=\frac{M}{\ell} v^2=m v^2 \\
& {\left[\because \frac{M}{\ell}=m(\text { given })\right]}
\end{aligned}
$$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2012 (07 May Online)
An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of $2 \mathrm{~ms}^{-1}$. The mass per unit length of water in the pipe is $100 \mathrm{kgm}^{-1}$. What is the power of the engine?

Options:
A $400 \mathrm{~W}$
B $200 \mathrm{~W}$
C $100 \mathrm{~W}$
D $800 \mathrm{~W}$

The correct answer is:
$800 \mathrm{~W}$

Solution:
$$
\begin{aligned}
\text { Power } & =\mu \mathrm{v}^3=\frac{\mathrm{m}}{\mathrm{l}} \times \mathrm{v}^3 \\
& =100 \times(2)^3=800 \mathrm{~W}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
An engine takes in 5 moles of air at 20 °C and 1 atm, and compresses it adiabatically to 1/10th of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be XkJ. The value of X to the nearest integer is:

Options:

The correct answer is:
46

Solution:
T1V1γ-1=T2V2γ-1

T2=T1V1V2γ-1

=T1(10)75-1

T2=T1(10)2/5

ΔV=52nR;52×5×3102/5-1(293)

6256×1.5×293=461440

≈46ks

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (02 Sep Shift 1)
An equal volume of a reducing agent is titrated separately with 1 M KMnO4 in acid, neutral and alkaline medium. The volumes of KMnO4 required are 20 mL in acid, 33.3 mL in neutral and 100 mL in alkaline media. Find out the volume (mL) of 1M K2Cr2O7 consumed, if the same volume of the reducing agent is titrated in acid medium.
Report your answer by rounding it up to nearest whole number.

Options:

The correct answer is:
17

Solution:
Let equal volumes of R.A in each case be X Meqs
MnO 4 - + 8 H + + 5 e - â Mn 2 + + 4 H 2 O Acid Z = 5 MnO 4 - + 2 H 2 O + 3 e - â MnO 2 + 4 OH - Neutral Z = 3 MnO 4 - + e - â MnO 4 2 - Alkaline Z = 1
X milliââeqs. of R.A = 20Ã1 Ã5 = 33.3Ã1 Ã3 = 100Ã1 Ã1 =100 milliâeqs
100 milli eqs. of R.A = 100 milli eqs. of K2 Cr2O7
Cr 2 O 7 2 - + 1 4 H + + 6 e - â 2 Cr 3 + + 7 H 2 O
100 milli eqs. of Cr 2 O 7 2 - = 6 Ã 1 Ã Volume mL,
Vol = 16.7 mL

Topic: Chemistry, Redox Reactions, JEE Main
An equation of a plane parallel to the plane $x-2 y+2 z-5=0$ and at a unit distance from the origin is

Options:
A
$x-2 y+2 z-3=0$

B
$x-2 y+2 z+1=0$

C
$x-2 y+2 z-1=0$

D
$x-2 y+2 z+5=0$

The correct answer is:
$x-2 y+2 z-3=0$

Solution:
Equation of plane parallel to $x-2 y+2 z-5=0$ is $x-2 y+2 z+k=0 \dots \dots(1)$
perpendicular distance from $\mathrm{O}(0,0,0)$ to $(1)$ is $1$
$\frac{|k|}{\sqrt{1+4+4}}=1 \quad \Rightarrow|k|=3 \quad \Rightarrow k=\pm 3 \quad \therefore x-2 y+2 z-3=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2012 (Offline)
An equation of the tangent to the curve $y=x^4$ from the point $(2,0)$ not on the curve is

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

The correct answer is:
$y=0$

Solution:
Let the point of contact be $(h, k)$, where $k=h^4$.
Tangent is $y-k=4 h^3(x-h),\left[0 \frac{d y}{d x}=4 x^3\right]$
It passes through $(2,0), \therefore-k=4 h^3(2-h)$ $\Rightarrow h=0$ or $8 / 3, \therefore k=0$ or $(8 / 3)$
$\therefore$ Points of contact are $(0,0)$ and $\left(\frac{8}{3},\left(\frac{8}{3}\right)^4\right)$
$\therefore$ Equation of tangents are
$y=0 \text { and } y-\left(\frac{8}{3}\right)^4=4\left(\frac{8}{3}\right)^3\left(x-\frac{8}{3}\right)$

Topic: Mathematics, Application of Derivatives, JEE Main
An equi-convex lens has power $P$ it is cut into two symmetrical halves by a plane containing the principal axis. The power of one part will be

Options:
A 0
B $\frac{P}{2}$
C $\frac{P}{4}$
D $P$

The correct answer is:
$P$

Solution:
When the lens is cut along its principal axis, the focal length of the two halves will remain same because the radius of curvature of both the surfaces are still same. So, the power also remains same as

Topic: Physics, Ray Optics, JEE Main
An equi-convex lens has power $P$ it is cut into two symmetrical halves by a plane containing the principal axis. The power of one part will be

Options:
A 0
B $\frac{P}{2}$
C $\frac{P}{4}$
D $P$

The correct answer is:
$P$

Solution:
When the lens is cut along its principal axis, the focal length of the two halves will remain same because the radius of curvature of both the surfaces are still same. So, the power also remains same as

Topic: Physics, Ray Optics, NEET, NEET 2019 (Odisha)
An equiconvex lens is cut two halves along (i) XOX' and (ii) YOY' as shown in the figure. Let $f, f^{\prime}, f^{\prime \prime}$ be the focal length of the complete lens, of each half in case (i) and each half in case (ii), respectively.
Choose the correct statement from the following:

Options:
A $f^{\prime}=f, f^{\prime \prime}=2 f$
B $f^{\prime}=2 f, f^{\prime \prime}=f$
C $f^{\prime}=f, f^{\prime \prime}=f$
D $f=2 f, f^{\prime \prime}=2 f$

The correct answer is:
$f^{\prime}=f, f^{\prime \prime}=2 f$

Solution:
Since the lens is equiconvex, the radius of curvature of each half is same.
Using lens maker formula
$$
\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
$$
Here $R_1=R$
$R_2=-R$ by convection
$$
\therefore \frac{1}{f}=(\mu-1) \frac{2}{\mathrm{R}}(\mu-1) \frac{1}{\mathrm{R}}=\frac{1}{2 f} \ldots(\mathrm{i})
$$
If we cut the lens along $\mathrm{XOX}^{\prime}$ then two halves of the lens will be having the same radii of curvalur and
$$
\begin{aligned}
& R_1=R, R_2=-R \\
& \therefore \frac{1}{f^{\prime}}=(\mu-1)\left(\frac{2}{R}\right) \\
& \Rightarrow \frac{1}{f^{\prime}}=\frac{2}{2 f}=\frac{1}{f}
\end{aligned}
$$
From (i) it is clear that $f^{\prime}=f$
But when we cut along $\mathrm{YOY}$ ' then
$$
\begin{aligned}
& R_1=R \text { and } R_2=\infty \\
& \therefore \infty \frac{1}{f^{\prime \prime}}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right) \\
& =(\mu-1) \frac{1}{R}=\frac{1}{2 f} \\
& \Rightarrow \quad f^{\prime \prime}=2 f
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main
An equilateral prism of mass m rests on a rough horizontal surface, with a coefficient of friction Î¼. A horizontal force F is applied to the prism, as shown in the figure. If the coefficient of friction is sufficiently high, so that the prism does not slide before toppling, then the minimum force required to topple the prism is

Options:
A mg3
B mg4
C μmg3
D μmg4

The correct answer is:
mg3

Solution:
The prism will topple after normal reaction shifts to the rightmost part of the prism.

Hence, at that time, equating torque about the toppling point ( rightmost point of the prism) due to force F and force due to gravity mg

FÃa32=mga2

âF=mg3

Topic: Physics, Rotational Motion, JEE Main
An equilateral triangle ABC is cut from a thin solid sheet of wood. (See figure) D, E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I0 . If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then:

Options:
A I=34I0
B I=1516I0
C I=916I0
D I=I04

The correct answer is:
I=1516I0

Solution:
Dimension analysis
Io=KMa2
Now for small lamina
I′=KM4a22=kma216
I′=Io16
So moment of Inertia of remaining part
IL=Io-I′
=Io-Io16
IL=15Io16

Topic: Physics, Rotational Motion, JEE Main
An equilateral triangle is inscribed in the ellipse whose equation is x2+4y2=4. One vertex of the triangle is 0,1 and one altitude is contained in the y-axis. If the length of each side is k3 units, then k is

Options:
A 1613
B 813
C 1316
D 138

The correct answer is:
1613

Solution:
x24+y21=1
Equation of AB is y=3x+1
x2+43x2+23x+1=4
⇒13x2+83x=0⇒x=-8313
BC=16313 units

Topic: Mathematics, Ellipse, JEE Main
An equilateral triangle is made by uniform wires $A B, B C, C A .$ A current $I$ enters at $A$ and leaves from the mid point of $B C$. If the lengths of each side of the triangle is $L,$ the magnetic field $B$ at the centroid $O$ of the triangle is

Options:
A $\frac{\mu_{0}}{4 \pi}\left(\frac{4 L}{L}\right)$
B $\frac{\mu_{0}}{2 \pi}\left(\frac{4 L}{L}\right)$
C $\frac{\mu_{0}}{4 \pi}\left(\frac{2 L}{L}\right)$
D zero

The correct answer is:
zero

Solution:
The given situation can be shown as

The magnetic field at centroid $O$ = Magnetic field due to left part + Magnetic field due to right part
$\therefore$
$$
\mathrm{B}=\mathrm{B}_{1}+\mathrm{B}_{2}=0
$$
(: direction of $B$ due to both parts is different of $O$ )

Topic: Physics, Magnetic Effects of Current, JEE Main
An equilateral triangleâs sides increase at the rate of 2 cm/sec. If the area of its incircle increases at a rate of k cm2/sec (when the length of the side is 6Ïcm ), then the value of k is

Options:

The correct answer is:
2

Solution:
ra2=tan30o
2ra=13
âr=a23
â´ Area of incircle, A=Ïr2=Ïa212
â´dAdt=2Ïa12dadt
=Ï6â6Ïâ2
=2cm2/sec.

Topic: Mathematics, Application of Derivatives, JEE Main
An equilateral triangular loop ADC of uniform specific resistivity having some resistance is pulled with a constant velocity v out of a uniform magnetic field directed into the paper. At time t=0, side DC of the loop is at the edge of the magnetic field. The induced current I versus time t graph will be as :

Options:
A
B
C
D

The correct answer is:

Solution:
Let 2a be the side of the triangle and b the length AE.

âAHG and âAEC are similar triangles.
AHAE=GHEC
â´GH=AHAEEC
=b-vtb.a=a-abvt
â´FG=2GH=2a-abvt
Induced e.m.f., e=BvFG=2Bva-abvt
â´ Induced current, I=eR=2BvRa-abvt
or I=k1-k2t
Thus, the I-t graph is a straight line with a negative slope and positive intercept.

Topic: Physics, Electromagnetic Induction, JEE Main
An equilateral triangular loop ADC having some resistance is pulled with a constant velocity v out of a uniform magnetic field directed into the screen. At time t=0, side DC of the loop is at edge of the magnetic field. The induced current i versus time t graph will be as

Options:
A
B
C
D

The correct answer is:

Solution:
Let 2a be the side of the triangle and b the length AE.

AH AE = GH EC
â´ GH = AH AE EC
or GH = b - vt b Â· a
= a - a b vt
â´ FG = 2 GH = 2 a - a b vt
Induced emf e = Bv FG = 2 Bv a - a b vt
â´ Induced current i = e R = 2 Bv R a - a b vt
or i = k1 - k2t
Thus, i-t graph is a straight line with negative slope and positive intercept.

Topic: Physics, Electromagnetic Induction, JEE Main
An equilibrium mixture in a vessel of capacity 100 litre contain 1 mol N2, 2 mol O2 and 3 mol NO . Find number of moles of O2 to be added, so that at new equilibrium the concentration of NO is 0.04 mol/lit. .

Options:
A 10118
B 1019
C 2029
D None of these

The correct answer is:
10118

Solution:
N2+O2⇌2NO
K=322×1 …(i)
N21-z+O22+y-z⇌2NO2z+3
K=(3+2z)2(2+y-z)(1-z) …(ii)
Given, 3+2z=4
z=12
(i) = (ii)
y=10118

Topic: Chemistry, Chemical Equilibrium, JEE Main
An ester $(A)$ with molecular formula $\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}_{2}$ was treated with excess of $\mathrm{CH}_{3} \mathrm{MgBr}$ and the complex so formed was treated with $\mathrm{H}_{2} \mathrm{SO}_{4}$ to give an olefin $(B)$. Ozonolysis of $(B)$ gave $a$ ketone with molecular formula $\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}$ which shows positive iodoform test. The structure of $(A)$ is

Options:
A $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOC}_{2} \mathrm{H}_{5}$
B $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOC}_{6} \mathrm{H}_{5}$
C $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOCH}_{3}$
D

The correct answer is:
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOC}_{2} \mathrm{H}_{5}$

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
An ester A(C 4 H 8 O 2 ), on treatment with excess methyl magnesium chloride followed by acidification, gives an alcohol B as the sole organic product. Alcohol B, on oxidation with NaOCl followed by acidification, gives acetic acid. Ester A is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
An ester \( \mathrm{A}\left(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\right) \), on treatment with excess methyl magnesium chloride followed by acidification, gives an alcohol \( \mathrm{B} \) as the sole organic product. Alcohol \( \mathrm{B} \), on oxidation with \( \mathrm{NaOCl} \) followed by acidification, gives acetic acid. Ester A is:

Options:
A
B
C
D

The correct answer is:

Solution:
The given ester reacts with excess of Grignard reagent forms 2-propanol as a single alcohol product. 2-propanol on oxidation with NaOCl, gives acetic acid and chloroform(It is a haloform reaction). The reaction are shown below,

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main