datavorous/entrance-exam-dataset huggingface dataset:
A neutron collides elastically with the stationary nucleus of an atom of mass number A. If the collision is perfectly elastic, then after the collision the fraction of the initial kinetic energy retained by the neutron, is

Options:
A 1-AA+12
B A+1A-12
C A-1A2
D A+1A2

The correct answer is:
1-AA+12

Solution:
Let us assume the mass of the neutron is mn=m and mass of the nucleus is mA=Am

Also, let us assume that the initial and the final velocities of the neutron are vi and vf respectively.

Using the formula of final velocity for a perfectly elastic collision we get,

vf=m-Am(m+Am)vi=1-A1+Avi
⇒ vfvi=1-A1+A
⇒KfinalKinitial=1-A1+A2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A neutron moving with a speed 'v' makes a head on collision with a stationary hydrogen atom in ground state. The minimum kinetic energy of the neutron for which perfactly inelastic collision will take place is :

Options:
A 20.4 eV
B 10.2 eV
C 12.1 eV
D 16.8 eV

The correct answer is:
20.4 eV

Solution:
Let velocity before collision v from momentum conservation mv=( m+m ) v 1
v 1 = v 2
Loss in K.E. =12mv2-122mv22
=14 mv2
K.E. lost is used to jump from 1st orbit to 2nd orbit â K.E.=10.2 eV
â 14 mv2=10.2
12 mv2=2Ã10.2=20.4 eV

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2016 (10 Apr Online)
A neutron of mass $m$ is collides elastically with a nucleus of mass $M$, which is initially at rest. Neutron kinetic energy is $\mathrm{k}_0$. Find the maximum kinetic energy it can lose

Options:
A $\frac{2 m M K_0}{(M+m)^2}$
B $\frac{4 m M K_0}{(M+m)^2}$
C $\frac{m M K_0}{(M+m)^2}$
D none of the above

The correct answer is:
$\frac{4 m M K_0}{(M+m)^2}$

Solution:
Maximum energy loss occurs in head on collision
Let $P_0$ be initial momentum
$P_1$ and $P_2$ are final momentum
Applying the law of conservation of Linear momentum
$P_0=P_1+P_2 \cdots$ (1)
Applying the law of conservation of energy
$\frac{P_0^2}{2 m}=\frac{P_1^2}{2 m}+\frac{P_2^2}{2 M} \cdots(2)$
Solving both equation (1) and (2)
$P_1=\frac{m-M}{m+M} P_0$
Final kinetic energy of the Neutron
$\begin{aligned} & =\frac{(m-M)^2 P_0^2}{2 m(M+m)^2} \\ & =\frac{K_0(m-M)^2}{(m+M)^2}\end{aligned}$
Loss in Kinetic energy
$=K_0-\frac{K_0(m-M)^2}{(m+M)^2}=\frac{4 m M K_0}{(M+m)^2}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A new radionucleide is obtained from the radionucleide 90234Th through two successive β -decays followed by one α -decay. The atomic number and the mass number respectively of the resulting radionucleide is

Options:
A 92 and 234
B 94 and 230
C 90 and 230
D 92 and 230

The correct answer is:
90 and 230

Solution:
90Th234→-β 91X234→-β92Y234→-α90Z230

Topic: Chemistry, Chemical Kinetics, JEE Main
A new system of units is proposed in which unit of mass is $\alpha \mathrm{kg}$, unit of length $\beta \mathrm{m}$ and unit of time $\gamma$ s. How much will $5 \mathrm{~J}$ measure in this new system?

Options:

None

Solution:
As we know that
Dimension of energy $=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
Let us consider $\mathrm{M}_1, \mathrm{~L}_1, \mathrm{~T}_1$ and $\mathrm{M}_2, \mathrm{~L}_2, \mathrm{~T}_2$ are units of mass, length and time in given two systems.
$\mathrm{M}_1=1 \mathrm{~kg}, \mathrm{~L}_1=1 \mathrm{~m}, \mathrm{~T}_1=1 \mathrm{~s}, \mathrm{n}_1=$ SI system unit $=5 \mathrm{~J}$
$\mathrm{M}_2=\alpha \mathrm{kg}, \mathrm{L}_2=\beta \mathrm{m}, \mathrm{T}_2=\gamma \mathrm{s}, \mathrm{n}_2=$ New system unit $=$ ?
The magnitude of a physical quantity remains the same, whatever be the system of units of its measurement i.e.,
$$
n_1 u_1=n_2 u_2
$$
So, $n_2=n_1 \frac{u_1}{u_2}=n_1 \frac{\left[\mathrm{M}_1 \mathrm{~L}_1^2 \mathrm{~T}_1^{-2}\right]}{\left[\mathrm{M}_2 \mathrm{~L}_2^2 \mathrm{~T}_2^{-2}\right]}$
$$
\begin{aligned}
&=5\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right] \times\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^2 \times\left[\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right]^{-2} \\
&=5\left[\frac{1}{\alpha} \mathrm{kg}\right] \times\left[\frac{1}{\beta} \mathrm{m}\right]^2 \times\left[\frac{1}{\gamma} \mathrm{s}\right]^{-2} \\
&=5 \times \frac{1}{\alpha} \times \frac{1}{\beta^2} \times \frac{1}{\gamma^{-2}}=\frac{5 \gamma^2}{\alpha \beta^2}
\end{aligned}
$$
$(\because 5$ is constant so neglect it)
So, new system unit of energy will be $\frac{\gamma^2}{\alpha \beta^2}$, or $\left[\alpha^{-1} \beta^{-2} \gamma^2\right]$.

Topic: Physics, Units and Dimensions
A newly discovered plant was found to be a gymnosperm because it

Options:
A lacks ovary but possesses exposed ovules
B possesses xylem vessels
C bears pollen grains
D develops fruits

The correct answer is:
lacks ovary but possesses exposed ovules

Solution:
Gymnosperm word is derived from greek word 'Gymno" means naked and 'sperma' which means naked , thus gymnosperms are naked seed bearing plants in which seed are not present inside fruits, as gymnosperms are non flowering plants in which ovules are borne directly on mega sporophyll. Thus ovules remains exposed before and after fertilization.

Topic: Biology, Plant Kingdom, NEET
A → NH 3 B → KOH/Br 2 CH 3 2 CHNH 2
A is

Options:
A
B
C
D

The correct answer is:

Solution:
In Hoffman degradation reaction one carbon decreases

Topic: Chemistry, Amines, JEE Main
A nine-digit number is formed using the digits 1, 2, 3, 5 and 7. The probability that the product of all digits is always 1920 is

Options:
A 159
B 758
C 7259
D 17!

The correct answer is:
7259

Solution:
Total cases =59
1920=5Ã3Ã27
Favourable cases = no. of ways of arrangement of 5,3,2,2,2,2,2,2,2 in a row =9!7!
hence, required probability =7259

Topic: Mathematics, Probability, JEE Main
A nitrogen containing organic compound gave an oily liquid on heating with bromine and potassium hydroxide solution. On shaking the product with acetic anhydride, an antipyretic drug was obtaine(d) The reactions indicate that the starting compound is

Options:
A acetamide
B nitrobenzene
C aniline
D benzamide

The correct answer is:
benzamide

Solution:
An organic compound gives oily liquid on heating with bromine and potassium hydroxide solution. On shaking with acetic anhydride, an antipyretic drug acetanilide obtained is benzamide.

Topic: Chemistry, Amines, JEE Main
A non ideal solution was prepared by mixing $30 \mathrm{ml}$ chloroform and $50 \mathrm{ml}$ acetone. The volume of mixture will be

Options:
A $\gt80 \mathrm{ml}$
B $ \lt 80 \mathrm{ml}$
C $=80 \mathrm{ml}$
D $^3 80 \mathrm{ml}$

The correct answer is:
$ \lt 80 \mathrm{ml}$

Solution:
Chloroform \& acetone form a non-ideal solution, in which A....B $B$ type interaction are more than A...... $A$ \& B..... $B$ $B$ type interaction due to $\mathrm{H}$-bonding. Hence, the solution shows, negative deviation from Raoult's Law i.e.,
$\Delta V_{\text {mix }}=-v e ; \quad \Delta H_{\text {mix }}=-v e$
$\therefore$ total volume of solution $=$ less than $(30+50 \mathrm{~ml} )$
or $ \lt 80 \mathrm{ml}$

Topic: Chemistry, Solutions, JEE Main
A non-conducting cylindrical rod is inserted within two coils of insulated wires, as shown in the given figure. A battery is connected to coil 1, while a galvanometer is connected to coil 2

On switching on the current in coil 1,

Options:
A coil 2 will move towards it
B coil 2 will move away from it
C the pointer of the galvanometer will remain at zero
D the pointer of the galvanometer will show a deflection

The correct answer is:
the pointer of the galvanometer will show a deflection

Solution:
On closing the switch, current begins flowing through coil 1. Hence, a magnetic field will be produced in this coil. The magnetic field varies from zero to a maximum value in a certain time. As a result, the field across coil 2 will change for that time. This change in field will cause current to be induced in coil 2 for that short time. This phenomenon of generation of current in coil 2, caused by the variation in the magnetic field of coil 1, is known as electromagnetic induction. The current induced in coil 2 will deflect the pointer of the galvanometer connected to it.
Why alternative A is wrong:
Coil 2 will not move toward coil 1 because no mechanical force is produced on closing the switch.
Why alternative B is wrong:
Coil 2 will not move away from coil 1 because there is no mechanical force that could repel it.
Why alternative C is wrong:
Since current will be induced in coil 2, the pointer of the galvanometer will not remain at zero.
The correct answer is D.

Topic: Physics, Electromagnetic Induction, NEET
A non-conducting disc of radius R is uniformly charged with surface charge density Ï. A disc of radius R2 is cut from the disc, as shown in the figure. The electric potential at centre C of large disc will be

Options:
A πσR2ε0
B σR2πε0
C σRπ-12πε0
D σRπ-12ε0

The correct answer is:
σRπ-12πε0

Solution:
The electric potential at the centre of the complete disc, V1=ÏR2Îµ0

Let us assume that the potential at the circumference of the smaller disc is V2

r=2R2cosâ¡Î¸2

dr=-2R2sinâ¡Î¸2dÎ¸2

â«dV2=â«Î¸ = 0ÏÏÃrÎ¸dr4ÏÎµ0r=ÏR2ÏÎµ0=V2

â´ Net potential at centre C is,

V=V1âV2=ÏRÏ-12ÏÎµ0

Topic: Physics, Electrostatics, JEE Main
A non-conducting piston of mass m and are S0 divides a non-conducting, closed cylinder as shown in the figure. A piston having mass m is connected with the top wall of the cylinder by a spring of force constant k. The top part is evacuated and the bottom part contains an ideal gas at a pressure P0 in the equilibrium position. Adiabatic exponent γ and in equilibrium length of each part is l. (neglect friction). Find the angular frequency for small oscillation.

Options:
A kl+γP0S02nl
B 3kl+γP0S0ml
C km
D kl+γP0S0ml

The correct answer is:
kl+γP0S0ml

Solution:
PVγ=c
dpP+γdVV=0⇒dP= -γP0VdV
Restoring force F=kdx+S0dP
⇒a=-kl+γS0P0mldx
ω=kl+γP0S0ml

Topic: Physics, Thermodynamics, JEE Main
A non-conducting ring having a uniform linear charge density Î» is rotating with an angular speed Ï about an axis passing through its centre and perpendicular to its plane. The magnetic field at its centre is

Options:
A 3μ0λω2π
B μ0λω2
C μ0λωπ
D μ0λω

The correct answer is:
μ0λω2

Solution:
Let the radius of the ring is r.
Charge on the ring,
q=lÎ»=2ÏrÎ» âµl=2Ïr
Current due to the rotation of the ring,
i=qt=2ÏrÎ»2ÏÏâµt=2ÏÏ
=ÏÎ»r
i=ÏÎ»r ...(i)
â´ Magnetic field at the centre of the ring due to the rotation of the charged ring.
B=Î¼02.ir=Î¼02.ÏÎ»r r
=Î¼02.ÏÎ» [Using Eq. (i)]

Topic: Physics, Magnetic Effects of Current, JEE Main
A non-conducting ring of mass m = 4 kg and radius R=10 cm has a charge Q=2 C uniformly distributed over its circumference. The ring is placed on a rough horizontal surface such that the plane of the ring is parallel to the surface. A vertical magnetic field B=4t3 T is switched on at t=0. At t=5 s ring starts to rotate about the vertical axis through the centre. The coefficient of friction between the ring and the surface is found to be k24. Then the value of k is

Options:

The correct answer is:
18

Solution:
The induced electric field at the periphery of the ring is

E=R2dBdt

⇒E=R212t2=6Rt2

When the ring is about to slip,

qER=μmgR

⇒6qRt2=μmg

⇒μ=6qRt2mg=34=1824

k=18

Topic: Physics, Electromagnetic Induction, JEE Main
A nonconducting ring of mass m and radius R is charged as shown. The charged density i.e. charge per unit length is Î». It is then placed on a rough nonconducting horizontal surface plane. At time t = 0, a uniform electric field Eââ=E0i^ is switched on and the ring starts rolling without sliding. The friction force (magnitude and direction) acting on the ring, when it starts moving is

Options:
A λRE0i^
B 3λRE0i^
C 2λRE0i^
D 2λRE0i^

The correct answer is:
λRE0i^

Solution:
As E→ will try to rotate the ring is clockwise direction friction will oppose the rotation and provide translational motion. so frictional force will act forward.
dτ = (dq) × ( E 0 ) 2R sin θ
dτ = (λ Rdθ) 2RE 0 sin θ
τ = 2 R 2 λ E 0 ∫ 0 π / 2 sin θ d θ = 2 R 2 λ E 0 ...(1)
For translational motion, f=ma⇒a = fm ...(2)
For rotational motion , τ- fR = (mR2) (aR) ...(3)
From (2) and (3)
τ=2frR ...(4)
From (1) and (4)
fr=τ2R=2R2λE02R=RλE0⇒f→  r =RλE0i^

Topic: Physics, Electrostatics, JEE Main
A non-conducting ring of radius 0.5 m carries the total charge of 1.11Ã10-10 C, distributed non-uniformly on its circumference, producing an electric field everywhere in space. The value of the line integral â«l=âl=0-E.dl (l=0 being centre of the ring) in volt is

Options:
A +2
B -1
C -2
D Zero

The correct answer is:
+2

Solution:
By definition, ∫l=∞l=0-E→.dl→=V, the potential at the centre of the ring.

V=14πε0.QR=9×109×1.11×10-100.5=2 V
Hence, the value of the given line integral is 2 V

Topic: Physics, Electrostatics, JEE Main
A nonâconducting ring of radius R having uniformly distributed charge Q starts rotating about xâx' axis passing through diameter with an angular acceleration Î±, as shown in the figure. Another small conducting ring having radius a aâªR is kept fixed at the centre of the bigger ring is such a way that axis xx' is passing through its centre and perpendicular to its plane. If the resistance of the small ring is r=1 Î©, find the induced current in it in ampere.
(Given q=16Ã102Î¼0 C, R=1 m, a = 0.1 m, Î±=8 rad s-2)

Options:

The correct answer is:
8

Solution:
dq=q2ÏR.RdÎ¸=q2Ï.dÎ¸
di=dqT=qdÎ¸Ï2Ï2Ï
di=qÏ4Ï2.dÎ¸
dB=Î¼0di(Rsinâ¡Î¸)22R3
â«dB=â«0ÏÎ¼0sin2â¡Î¸2RqÏ4Ï2dÎ¸
B=Î¼0qÏ16ÏR
Ï=BÏa2
Ï=Ïa2.Î¼0qÏ16ÏR
Ï=Î¼0qÏa216R
|Îµ|=dÏdt
|Îµ|=Î¼0qa216RÎ± .
=8 volt
i=81=8 A.

Topic: Physics, Electromagnetic Induction, JEE Main
A nonâconducting ring of radius R having uniformly distributed charge Q starts rotating about xâx' axis passing through diameter with an angular acceleration Î±, as shown in the figure. Another small conducting ring having radius a aâªR is kept fixed at the centre of bigger ring is such a way that axis xx' is passing through its centre and perpendicular to its plane. If the resistance of small ring is r=1 Î©, find the induced current in it in ampere.
(Given q=16Ã102Î¼0 C, R=1 m, a = 0.1 m, Î±=8 rad s-2)

Options:

The correct answer is:
8

Solution:
dq=q2πR.Rdθ=q2π.dθ
di=dqT=qdθω2π2π
di=qω4π2.dθ

The magnetic field at O due to elementary ring is

dB=μ0di(Rsin⁡θ)22R3
∫dB=∫0πμ0sin2⁡θ2Rqω4π2dθ
B=μ0qω16πR

The flux of the magnetic field through the small ring

ϕ=Bπa2
ϕ=πa2.μ0qω16πR
ϕ=μ0qωa216R
|ε|=dϕdt
|ε|=μ0qa216Rα .
=8 volt
i=81=8 A.

Topic: Physics, Electromagnetic Induction, JEE Main
A non-conducting rod of length L with linear charge density Î»=Î»0x where x is the distance from end A is rotating with constant angular speed Ï about the same end. If the angular velocity of the rod (Ï) is large, then the magnetic dipole moment of the system is ÏÎ»0L4n. What is the value of n?

Options:

The correct answer is:
8

Solution:
dq=Î»0xdxâ i=Ï Î»0xdx2Ï
dm=ÏÎ»0xdx2Ï Ïx2âm=â«dm=â«Ï Î»02x3dx
â´ m=ÏÎ»0L48
n = 8

Topic: Physics, Magnetic Effects of Current, JEE Main
A non-conducting solid sphere has radius $R$ and uniform charge density. A spherical cavity of radius $\frac{R}{4}$ is hollowed out of the sphere. The distance between center of sphere and center of cavity is $\frac{R}{2}$. If the charge of the sphere is $Q$ after the creation of the cavity and the magnitude of electric field at the center of the cavity is $E=K\left(\frac{Q}{4 \pi \in_0 R^2}\right)$, determined the approximate value of $K$.

Options:
A 0.32
B 0.78
C 0.51
D 0.45

The correct answer is:
0.51

Solution:
A cavity of radius $\frac{R}{4}$ is made inside a non-conducting solid sphere as shown in the figure.
Let $Q_0$ be the charge, present on the sphere when cavity was absent.
Hence, $E=E_{Q_0}-E_{\text {cavity }}$
Where, $E=$ electric field at centre of cavity,
$E_{Q_0}=$ electric field at a distance $\frac{R}{2}$ from centre and
$E_{\text {cavity }}=$ electric field due to cavity itself at its centre.
Hence, $E=\frac{Q_0}{4 \pi \varepsilon_0 R^3}\left(\frac{R}{2}\right)-0$
$\Rightarrow \quad E=\frac{Q_0}{4 \pi \varepsilon_0 R^2}\left(\frac{1}{2}\right)$
Now, it is given that $Q$ be the charge after the creation of the cavity.
So,
$$
\frac{Q}{Q_0}=\frac{\frac{4}{3} \pi\left(R^3-\left(\frac{R}{4}\right)^3\right)}{\frac{4}{3} \pi R^3}
$$
$\because$ Since, for a uniform charged density object, charge $\propto$ volume
$$
\begin{array}{ll}
\Rightarrow & \frac{Q}{Q_0}=\frac{63}{64} \\
\Rightarrow & Q_0=\frac{Q 64}{63}
\end{array}
$$
So, from the Eq. (i) and (ii), we get
$$
E=\frac{Q}{4 \pi \varepsilon_0 R^2}\left(\frac{64}{63} \times \frac{1}{2}\right)
$$
$$
\begin{aligned}
& \qquad=0.5079\left[\frac{Q}{4 \pi \varepsilon_0 R^2}\right) \\
& \Rightarrow \quad E=0.51\left[\frac{Q}{4 \pi \varepsilon_0 R^2}\right] \\
& \text { Hence, } K=0.51 \\
& \therefore \text { So, the correct option is }(\mathrm{3}) .
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main
A non-isotropic solid metal cube has coefficients of linear expansion as: 5× 10 −5 / o C along the x-axis and 5× 10 −6 / o C along the y and the z-axis. If the coefficient of volume expansion of the solid is C× 10 −6 / o C then the value of C is _____________

Options:

The correct answer is:
60

Solution:
V=2Î±2+Î±1
=10Ã10-6+5Ã10-5
=60× 10 −6 / o C

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2020 (07 Jan Shift 1)
A non-reducing sugar A hydrolyses to give two reducing mono saccharides. Sugar A is

Options:
A Fructose
B Galactose
C Glucose
D Sucrose

The correct answer is:
Sucrose

Solution:
Sucrose(Non reducing sugar)⟶H2Oglucose(Reducing sugar)+Fructose(Reducing sugar)

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2021 (18 Mar Shift 1)
A nonuniform rod OM of length l has linear mass density that varies with the distance x from left end of the rod according to Î»= Î» 0 x 3 L 3 ; Where Î» 0 is constant, is kept along x-axis and is rotating about an axis AB , which is perpendicular to the rod as shown in the figure. What is the value of x so that moment of inertia of rod about axis AB I AB is minimum?

Options:
A 7ℓ15
B 2ℓ5
C 8l 15
D 4â5

The correct answer is:
4ℓ5

Solution:
IAB=∫y2dm=λ0ℓ3∫-xl-x(y+x)3y2dy

∫(y+x)3y2dy=y2(y+x)44-∫12y(y+x)4dy

=y2(y+x)24-y10(y+x)5+110(y+x)5dy

=y2(y+x)24-y(y+x)510+(y+x)260

∫-x0(-xy2(y+x)3dx=(ℓ-x)2(ℓ)44-(ℓ-x)ℓ510+ℓ660

IAB=λ0(ℓ-x)24ℓ-110(ℓ-x)ℓ2+160ℓ3

=λ0ℓ(ℓ-x)24ℓ-110(ℓ-x)+160ℓ2

dlABdx=0⇒-(ℓ-x)2+ℓ10=0

⇒5x=4ℓ⇒x=4l5

Topic: Physics, Rotational Motion, JEE Main
A non-uniform rod of length l having mass density Î»(x)=A+Bx2 is placed on -x-axis with its ends at a,0 and a+l, 0. The force it would exert on a point mass m kept at the origin is

Options:
A GmAlaa+2+Bl
B GmAlaa-l-Bl
C GmAlaa+l+Bl
D GmAlaa+l-Bl

The correct answer is:
GmAlaa+l+Bl

Solution:
F=∫aa+lGm dMx2=Gm∫aa+l(A+Bx2)dxx2
=Gm∫aa+lAx2dx+∫aa+lB  dx
=Gm[A[-1x]aa+l+Bl]
=GmA1a-1a+l+Bl

=GmAlaa+l+Bl

Topic: Physics, Gravitation, JEE Main
A normal is drawn to the ellipse x29+y2=1 at the point 3cos⁡θ,sin⁡θ where 0<θ<π2. If N is the foot of the perpendicular from the origin O to the normal such that ON=2, then θ is equal to

Options:
A π4
B π12
C π3
D π6

The correct answer is:
π6

Solution:
Equation of the normal at 3cosθ,sinθ is 3xcos⁡θ-ysin⁡θ=8
So, ON=-89sec2⁡θ+cosec2⁡θ=2 (given)
⇒16=91+tan2⁡θ+1+cot2⁡θ⇒6=6+3tan⁡θ-cot⁡θ2

⇒3tan⁡θ-cot⁡θ=0
Hence, tan2⁡θ=13⇒θ=π6

Topic: Mathematics, Ellipse, JEE Main
A normal line with positive direction cosines to the plane P makes equal angles with the coordinate axes. The distance of the point A1,2,3 from the line x-11=y+21=z-32 measured parallel to the plane P is equal to

Options:
A 3 units
B 13 units
C 14 units
D 25 units

The correct answer is:
14 units

Solution:
Direction ratios of the normal to the plane n→=<1,1,1>
Any general point on the given line is λ+1,λ-2,2λ+3
From the diagram
AB→=λ,λ-4,2λ
AB→.n→=0⇒λ+λ-4+2λ=0
⇒λ=1
So, coordinates of the point B are 2,-1,5
Hence, AB=1+9+4=14 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A normal to the hyperbola, $4 x^2-9 y^2=36$ meets the co-ordinate axes $x$ and $y$ at $A$ and $B$, respectively. If the parallelogram $O A B P(O$ being the origin) is formed, then the locus of $P$ is

Options:
A
$4 x^2-9 y^2=121$

B
$4 x^2+9 y^2=121$

C
$9 x^2-4 y^2=169$

D
$9 x^2+4 y^2=169$

The correct answer is:
$9 x^2-4 y^2=169$

Solution:
Given, $4 x^2-9 y^2=36$
After differentiating w.r.t. $x$, we get
$$
\begin{aligned}
&\text { 4.2.x-9.2.y. } \frac{d y}{d x}=0 \\
&\Rightarrow \text { Slope of tangent }=\frac{d y}{d x}=\frac{4 x}{9 y}
\end{aligned}
$$
So, slope of normal $=\frac{-9 y}{4 x}$
Now, equation of normal at point $\left(x_0, y_0\right)$ is given by
$$
y-y_0=\frac{-9 y_0}{4 x_0}\left(x-x_0\right)
$$
As normal intersects $\mathrm{X}$ axis at $A$, Then
$$
A \equiv\left(\frac{13 x_0}{9}, 0\right)
$$
and $B \equiv\left(0, \frac{13 y_0}{4}\right)$
As $O A B P$ is a parallelogram
$\therefore$ midpoint of $O B \equiv\left(0, \frac{13 y_0}{8}\right) \equiv$ Midpoint of $A P$
So, $P(x, y) \equiv\left(\frac{-13 x_0}{9}, \frac{13 y_0}{4}\right)$
$\because\left(x_0, y_0\right)$ lies on hyperbola, therefore $4\left(x_0\right)^2-9\left(y_0\right)^2=36$
From equation (i): $x_0=\frac{-9 x}{13}$ and $y_0=\frac{4 y}{13}$
From equation (ii), we get $9 x^2-4 y^2=169$
Hence, locus of point $P$ is : $9 x^2-4 y^2=169$

Topic: Mathematics, Hyperbola, JEE Main
A normal to the hyperbola, $4 x^2-9 y^2=36$ meets the co-ordinate axes $x$ and $y$ at $A$ and $B$, respectively. If the parallelogram $O A B P(O$ being the origin) is formed, then the locus of $P$ is

Options:
A
$4 x^2-9 y^2=121$

B
$4 x^2+9 y^2=121$

C
$9 x^2-4 y^2=169$

D
$9 x^2+4 y^2=169$

The correct answer is:
$9 x^2-4 y^2=169$

Solution:
Given, $4 x^2-9 y^2=36$
After differentiating w.r.t. $x$, we get
$$
\begin{aligned}
&\text { 4.2.x-9.2.y. } \frac{d y}{d x}=0 \\
&\Rightarrow \text { Slope of tangent }=\frac{d y}{d x}=\frac{4 x}{9 y}
\end{aligned}
$$
So, slope of normal $=\frac{-9 y}{4 x}$
Now, equation of normal at point $\left(x_0, y_0\right)$ is given by
$$
y-y_0=\frac{-9 y_0}{4 x_0}\left(x-x_0\right)
$$
As normal intersects $\mathrm{X}$ axis at $A$, Then
$$
A \equiv\left(\frac{13 x_0}{9}, 0\right)
$$
and $B \equiv\left(0, \frac{13 y_0}{4}\right)$
As $O A B P$ is a parallelogram
$\therefore$ midpoint of $O B \equiv\left(0, \frac{13 y_0}{8}\right) \equiv$ Midpoint of $A P$
So, $P(x, y) \equiv\left(\frac{-13 x_0}{9}, \frac{13 y_0}{4}\right)$
$\because\left(x_0, y_0\right)$ lies on hyperbola, therefore $4\left(x_0\right)^2-9\left(y_0\right)^2=36$
From equation (i): $x_0=\frac{-9 x}{13}$ and $y_0=\frac{4 y}{13}$
From equation (ii), we get $9 x^2-4 y^2=169$
Hence, locus of point $P$ is : $9 x^2-4 y^2=169$

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A nozzle throws a stream of gas against a wall with a velocity v much larger than the thermal agitation of the molecules. After collision of the molecules with wall the magnitude of their velocity remains same. Also assume that the force exerted on the wall by the molecules is perpendicular to wall. (This is not strictly true for a rough wall.) Find the pressure exerted on the wall. ( n= number of molecules per unit volume, m=mass of a gas molecule )

Options:
A 2n⁡m⁡v⁡2cos2θ
B 3n⁡m⁡v⁡2cos2θ
C n⁡m⁡v⁡2cos2θ
D 2n⁡m⁡v⁡2sin2θ

The correct answer is:
2n⁡m⁡v⁡2cos2θ

Solution:
Since, the molecules rebound from the wall, the component of velocity perpendicular to wall is reversed, while its velocity parallel to wall does not change. The change in velocity of molecules is parallel to normal N. The magnitude of charge is
Î v â = 2 vâ¡ cos Î¸
The change in momentum of a molecule is
Î p â = mâ¡ Î v â
= 2 mâ¡ vâ¡ cos Î¸
in the direction of normal N. Let n be the number of molecules per unit volume.The number of molecules arriving at an area A of the wall per unit time is the number in a slanted cylinder whose length is equal to the velocity v and whose cross-section is A cos Î¸
Number of molecules =nAv cos Î¸.
Each molecule suffers a change of momentum 2mv cos Î¸. Change of momentum of the stream of gas in a direction perpendicular to wall is equal to nAv cos Î¸Ã2mv cosÎ¸=2 An mv2cos2Î¸.
Hence, force exerted by stream of gas on the wall.
Fâ¡=2Anmv2cos2Î¸
This is also the force exerted by gas molecules on the wall.
Pressure = Normal force Area
=Fâ¡Aâ¡
=2nâ¡mâ¡vâ¡2cos2Î¸

Topic: Physics, Laws of Motion, JEE Main