datavorous/entrance-exam-dataset huggingface dataset:
The displacement y of a particle in a medium can be expressed as $y=10^{-6} \sin (110 t+20 x+\pi / 4) m$, where $t$ is in seconds and $x$ in meter. The speed of the wave is

Options:
A
$2000 \mathrm{~m} / \mathrm{s}$

B
$5 \mathrm{~m} / \mathrm{s}$

C
$20 \mathrm{~m} / \mathrm{s}$

D
$5 \pi \mathrm{m} / \mathrm{s}$

The correct answer is:
$5 \mathrm{~m} / \mathrm{s}$

Solution:
$\mathrm{v}=\frac{\omega}{\mathrm{k}}=5 \mathrm{~ms}^{-1}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2004
The displacement y of a particle in a medium can be expressed as y=10-6sin100t-20x+Ï/4 m, where t is in seconds and x in meters. The speed of the wave is

Options:
A 2000 m s-1
B 5 m s-1
C 20 m s-1
D 5π m s-1

The correct answer is:
5 m s-1

Solution:
Given wave equation :
y=10-6sin100t+20x+π4 m
Standard equation : y=a sin ωt+kx+ϕ
Comparing the two, we obtain
∴ ω=100 and k=20
∴ ωk=10020

⇒2πn2π/λ=nλ=v=5
∴ v=5 m s-1

Topic: Physics, Waves and Sound, JEE Main
The displacement $y$ of a wave travelling in the $x$-direction is given by $y=10^{-4} \sin \left(600 t-2 x+\frac{\pi}{3}\right)$ metres where $\mathrm{x}$ is expressed in metres and $\mathrm{t}$ in seconds. The speed of the wave-motion, in $\mathrm{ms}^{-1}$, is

Options:
A
$300$

B
$600$

C
$1200$

D
$200$

The correct answer is:
$600$

Solution:
$y=10^{-4} \sin \left(600 t-2 x+\frac{\pi}{3}\right)$. Comparing it with standard equation
$y=A \sin (v t-k x) ; \quad v=600 m / s$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2003
The displacement $y(t)=A \sin (\omega t+\phi)$ of a pendulum for $\phi=\frac{2 \pi}{3}$ is correctly represented by

Options:
A

B

C

D

The correct answer is:

Solution:
Displacement $y(t)=A \sin (\mathrm{w} t+\phi)$ [Given]
For $\phi=\frac{2 \pi}{3}$
at $t=0 ; y=A \sin \phi=A \sin \frac{2 \pi}{3}$
$$
=A \sin 120^{\circ}=0.87 \mathrm{~A}\left[\because \sin 120^{\circ} \simeq 0.866\right]
$$
Graph (a) depicts $y=0.87 A$ at $t=0$

Topic: Physics, Oscillations, JEE Main, JEE Main 2012 (19 May Online)
The displacements of two particles of same mass executing SHM are represented by the equations
$x_1=4 \sin \left(10 t+\frac{\pi}{6}\right) \text { and } x_2=5 \cos (\omega t) .$
The value of $\omega$ for which the energies of both the particles remain same is

Options:
A 16 unit
B 6 unit
C 4 unit
D 8 unit

The correct answer is:
8 unit

Solution:
The equation of displacement
$x_1=4 \sin \left(10 t+\frac{\pi}{6}\right)$
The energy of this equation
$E_1=\frac{1}{2} m \omega_1^2 a_1^2$
$=\frac{1}{2} m \times 10 \times 10 \times 4 \times 4$
The second equation of displacement
$x_2=5 \cos (\omega t)$
The energy of this equation
$E_2=\frac{1}{2} m \omega_2^2 a_2^2$
$=\frac{1}{2} m \omega_2^2 \times 5 \times 5$
$\because \quad E_1=E_2$
$\therefore \quad \frac{1}{2} m \omega_2^2 \times 5 \times 5$
$=\frac{1}{2} m \times 10 \times 10 \times 4 \times 4$
$\omega_2^2=\frac{10 \times 10 \times 4 \times 4}{5 \times 5}$
$\omega_2=8$ unit

Topic: Physics, Oscillations, AP EAMCET, AP EAMCET 2010
The displacements of two particles of same mass executing SHM are represented by the equations
$x_1=4 \sin \left(10 t+\frac{\pi}{6}\right) \text { and } x_2=5 \cos (\omega t) .$
The value of $\omega$ for which the energies of both the particles remain same is

Options:
A 16 unit
B 6 unit
C 4 unit
D 8 unit

The correct answer is:
8 unit

Solution:
The equation of displacement
$x_1=4 \sin \left(10 t+\frac{\pi}{6}\right)$
The energy of this equation
$E_1=\frac{1}{2} m \omega_1^2 a_1^2$
$=\frac{1}{2} m \times 10 \times 10 \times 4 \times 4$
The second equation of displacement
$x_2=5 \cos (\omega t)$
The energy of this equation
$E_2=\frac{1}{2} m \omega_2^2 a_2^2$
$=\frac{1}{2} m \omega_2^2 \times 5 \times 5$
$\because \quad E_1=E_2$
$\therefore \quad \frac{1}{2} m \omega_2^2 \times 5 \times 5$
$=\frac{1}{2} m \times 10 \times 10 \times 4 \times 4$
$\omega_2^2=\frac{10 \times 10 \times 4 \times 4}{5 \times 5}$
$\omega_2=8$ unit

Topic: Physics, Oscillations, TS EAMCET, TS EAMCET 2010
The displacement-time graph of two particles \( A \) and \( B \) are straight lines inclined at angles \( 30^{\circ} \) and \( 60^{\circ} \) with time axis. The
ratio of \( v_{A}: v_{B} \) is:

Options:
A \( 1: 2 \)
B \( 1: \sqrt{3} \)
C \( \sqrt{3}: 1 \)
D \( 1: 3 \)

The correct answer is:
\( 1: 3 \)

Solution:
Given the angle of displacement -time graph of the curve va=30°.

The angle of displacement -time graph of the curve Vb=60°.

We know that the slope (tangent of the angle) gives the value of the velocity.

Therefore, The velocity of the first curve is,va=tan 30=13.

The velocity of the second curve,vb=tan 60=3

Therefore,the ratio is.vavb=133=13 or va :Vb=1:3

Topic: Physics, Motion In One Dimension, JEE Main
The disproportionation of MnO42- in acidic medium resulted in the formation of two manganese compounds A and B. If the oxidation state of Mn in B is smaller than that of A, then the spin-only magnetic moment μ value of B in BM is____ . (Nearest integer)

Options:

The correct answer is:
4

Solution:
Potassium permanganate Is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate.

2MnO2+4KOH+O2→2K2MnO4+2H2O

3MnO42-+4H-→2MnO4-+MnO2+2H2O

The oxidation state of Mn in MnO2 is less Mn4+ so, it is B

Mn4+=Ar3d3

Where, n=3

μs=nn+2

⇒33+2

⇒35

⇒15

⇒3.87≈4 B.M.

Therefore, the spin only magnetic moment of Mn4+ ion is approximately 4 B.M.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (28 Jul Shift 1)
The dissociation constant of a weak acid is $1 \times 10^{-4}$. In order to prepare a buffer solution with a pH $=5$, the [Salt]/ [Acid] ratio should be:

Options:
A $4: 5$
B $10: 1$
C $5: 4$
D $1: 10$

The correct answer is:
$10: 1$

Solution:
$\mathrm{pH}=\mathrm{pK}_a+\log \frac{[\text { Salt }]}{\text { Acid }}$
$$
\begin{aligned}
& 5=\log \left[1 \times 10^{-4}\right]+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& 5=4+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& \frac{[\text { Salt }]}{[\text { Acid }]}=10: 1
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The dissociation constant of acetic acid is 1.6 ×10-5 . The degree of dissociation of 0.01 M acetic acid in the presence of 0.1 M HCl is equal to

Options:
A 0.16
B 0.016
C 0.4
D 0.04

The correct answer is:
0.04

Solution:
For weak acid:
α =  Ka C
α =  1.6 ×  10 −5 10 −2
α = 4× 10 −2

Topic: Chemistry, Ionic Equilibrium, JEE Main
The dissociation constant of two weak acids are K a 1 and K a 2 respectively. Their relative strength of identical concentration is-

Options:
A K a 2 K a 1
B ( K a 1 K a 2 ) 1 2
C K a 1 K a 2
D ( K a 1 × K a 2 K a 1 ) 1 2

The correct answer is:
( K a 1 K a 2 ) 1 2

Solution:
Applying Ostwald's dilution law, α 1 = K a 1 c and α 2 = K a 2 c
Thus, Relative strength = ( strength  of  acid  I ) ( strength  of  acid  II ) = α 1 α 2 = ( K a 1 K a 2 )

Topic: Chemistry, Ionic Equilibrium, JEE Main
The dissociation constants for acetic acid and $\mathrm{HCN}$ at $25^{\circ} \mathrm{C}$ are $1.5 \times 10^{-5}$ and $4.5 \times 10^{-10}$, respectively. The equilibrium constant for the equilibrium -
$$
\mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-}
$$
would be :

Options:
A $3.0 \times 10^4$
B $3.0 \times 10^5$
C $3.0 \times 10^{-5}$
D $3.0 \times 10^{-4}$

The correct answer is:
$3.0 \times 10^4$

Solution:
$$
\begin{aligned}
\mathrm{K}_{\mathrm{c}} & =\mathrm{K}_{\mathrm{a}\left(\mathrm{CH}_3 \mathrm{COOH}\right)} \times \frac{1}{\mathrm{~K}_{\mathrm{a}(\mathrm{HCN})}} \\
& =1.5 \times 10^{-5} \times \frac{1}{4.5 \times 10^{-10}} \\
& \cong 3 \times 10^4
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2009 (Mains)
The dissociation equilibrium of a gas $\mathrm{AB}_2$ can be represented as, $2 \mathrm{AB}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})+\mathrm{B}_2(\mathrm{~g})$. The degree of dissociation is ' $\mathrm{x}$ ' and is small compared to 1. The expression relating the degree of dissociation $(\mathrm{x})$ with equilibrium constant $K_P$ and total pressure $\mathrm{P}$ is

Options:
A $\left(\frac{K_p}{P}\right)$
B $\left(\frac{2 K_P}{P}\right)$
C $\left(\frac{2 K_P}{P}\right)^{1 / 3}$
D $\left(\frac{2 K_P}{P}\right)^{1 / 2}$

The correct answer is:
$\left(\frac{2 K_P}{P}\right)^{1 / 3}$

Solution:
$$
\begin{aligned}
& 2 \mathrm{AB}_2 \rightleftharpoons 2 \mathrm{AB}+\mathrm{B}_2 \\
& 1 \\
& 00 \\
& 1-x \\
& \mathrm{x} \quad \mathrm{x} / 2 \\
&
\end{aligned}
$$
Total mole at equi. $=1+\frac{x}{2}$
$$
\therefore K_P=\frac{\left(\frac{x}{1+x / 2} \times P\right)^2\left(\frac{x / 2}{1+x / 2} \times P\right)}{\left(\frac{1-x}{1+x / 2} \times P\right)^2}
$$
(Here $x$ is degree of dissociation)
or $K_P=\frac{x^3 P}{2}$
or $x^3=\frac{2 K_P}{P}$
or $x=\left(\frac{2 K_P}{P}\right)^{1 / 3}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
The dissociation equilibrium of a gas $\mathrm{AB}_2$ can be represented as, $2 \mathrm{AB}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})+\mathrm{B}_2(\mathrm{~g})$. The degree of dissociation is ' $\mathrm{x}$ ' and is small compared to 1. The expression relating the degree of dissociation $(\mathrm{x})$ with equilibrium constant $K_P$ and total pressure $\mathrm{P}$ is

Options:
A $\left(\frac{K_p}{P}\right)$
B $\left(\frac{2 K_P}{P}\right)$
C $\left(\frac{2 K_P}{P}\right)^{1 / 3}$
D $\left(\frac{2 K_P}{P}\right)^{1 / 2}$

The correct answer is:
$\left(\frac{2 K_P}{P}\right)^{1 / 3}$

Solution:
$$
\begin{aligned}
& 2 \mathrm{AB}_2 \rightleftharpoons 2 \mathrm{AB}+\mathrm{B}_2 \\
& 1 \\
& 00 \\
& 1-x \\
& \mathrm{x} \quad \mathrm{x} / 2 \\
&
\end{aligned}
$$
Total mole at equi. $=1+\frac{x}{2}$
$$
\therefore K_P=\frac{\left(\frac{x}{1+x / 2} \times P\right)^2\left(\frac{x / 2}{1+x / 2} \times P\right)}{\left(\frac{1-x}{1+x / 2} \times P\right)^2}
$$
(Here $x$ is degree of dissociation)
or $K_P=\frac{x^3 P}{2}$
or $x^3=\frac{2 K_P}{P}$
or $x=\left(\frac{2 K_P}{P}\right)^{1 / 3}$

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2008 (Mains)
The dissociation equilibrium of a gas AB 2 can be represented as,
2  AB 2 (g)⇌2 AB(g)+ B 2 (g)
The degree of dissociation is x and is small as compared to 1 . The expression relating the degree of dissociation (x) with equilibrium constant K P and total pressure P is:

Options:
A (2K p /P) 1 2
B K p /P​
C 2K p /P​
D (2K p /P) 1 3

The correct answer is:
(2K p /P) 1 3

Solution:
2 ABâ¡ 2 g â 2ABâ¡ g + Bâ¡ 2 gâ¡
At equilibrium 2(1âx) 2x x
Total moles = (2â+âx)
Partial pressures
P AB 2 = 2(1âx) 2â+âx p,â P AB â=â 2x 2â+âx p,â P B 2 = x 2+x p
K p = p AB 2 pB 2 ( pAB 2 ) 2 = [ ( 2 x 2 + x ) p ] 2 [ x p 2 + x ] [ 2 ( 1 - x ) 2 + x p ] 2
K p = 4 x 3 P 4 2+x (1âx) 2
2+x â2 ( x is small)
1âx â1
K P = 4 x 3 p 8 = x 3 p 2
x= 2kp p 3

Topic: Chemistry, Chemical Equilibrium, JEE Main
The dissolution of AlOH3 by a solution of NaOH results in the formation of:

Options:
A AlÂ H2O4OH2+
B AlÂ H2O3OH3
C Al H2O2OH4â
D AlÂ H2O6OH3

The correct answer is:
Al H2O2OH4â

Solution:
AlOH3 dissolve in NaOH solution to give AlOH4â ion which is supposed to have the octahedral complex species AlOH4H2O2â in aqueous solution
Al OH 3 +NaOH aq → Al OH 4 H 2 O 2 – aq + Na + aq

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The distance between $4 x+3 y=11$ and $8 x+6 y=15$, is

Options:
A $\frac{7}{2}$
B 4
C $\frac{7}{10}$
D None of these

The correct answer is:
$\frac{7}{10}$

Solution:
$\begin{aligned}
& 4 x+3 y=11 \text { and } 4 x+3 y=\frac{15}{2} \\
& \qquad D=\left|\frac{11-\frac{15}{2}}{5}\right|=\frac{7}{10} \\
& \text {Therefore, } \quad \mid
\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main
The distance between a light source and photoelectric cell is d. If the distance is decreased to $\frac{d}{2},$ then

Options:
A the emission of electron per second will be four times
B maximum kinetic energy of photoelectrons will be four times
C stopping potential will remain same
D the emission of electrons per second will be doubled

The correct answers are:
the emission of electron per second will be four times, stopping potential will remain same

Solution:
The intensity of emitted electrons is
$$
\begin{array}{l}
I=\frac{1}{r^{2}} \\
I_{1}=\frac{1}{d^{2}} \\
I_{2}=\frac{1}{(d / 2)^{2}}=4 \cdot \frac{1}{d^{2}}
\end{array}
$$
and I $\propto$ Number of photons per second $=N$
$\therefore$
$$
N=\frac{1}{r^{2}}
$$
$\therefore$ Number of emithed electrons will become 4 times From Enstein's photo-electric equation
$$
K E_{max}=h v-\phi
$$
where, $\phi=$ work function of surtace
$\mathrm{v}=$ frequency of incident photon
Since, v remains unchanged, thus $KE_{max}$ as well as stopping potential remans unchanged
$\because \mathrm{KE}_{max}=\mathrm{eV}_{s}=\mathrm{v}$
where. $V_{s}$ = stopping potential

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2016
The distance between an object and a screen is 100 cm. A lens can produce real image of the object on the screen for two different positions between the screen and the object. The distance between these two positions is 40 cm. If the power of the lens is close to N100D where N is an integer, the value of N is _________

Options:

The correct answer is:
476

Solution:
f=D2-d24D=1002-4024(100)=(100+40)(100-40)4(100)=21 cm

P=1f=10021=N100

N=476.19

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The distance between charges +q and -q is 2l and between +2q and -2q is 4l. The electrostatic potential at point P at a distance r from centre O is -αqlr2×109 V, where the value of α is ______. (Use 14πε0=9×109 N m2 C-2)

Options:

The correct answer is:
27

Solution:
Considering two dipoles with charges 2q & q, we can write the equivalent setup as shown below:

Therefore, net dipole moment is Pnet=6ql in the leftward direction.

Potential required,

V=Kp→·r→r3=9×1096qlr2cos120o

=−27qlr2×109

Therefore, α=27.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (31 Jan Shift 2)
The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20 cm. The focal length of the lens used is ________ cm.

Options:

The correct answer is:
15

Solution:
The image will be formed on the same side as the object. Therefore, v=3u

Given: v–u=20 cm

⇒2u=20 cm

⇒u=-10 cm (-ve sign as the object is real)

From lens formula, we can write

1(-30)-1(-10)=1f

⇒f=15 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2024 (01 Feb Shift 1)
The distance between object and its two times magnified real image as produced by a convex lens is 45 cm. The focal length of the lens used is ________cm.

Options:

The correct answer is:
10

Solution:
Magnification, m=vu=-2

⇒v=-2u …i

As,

v-u=45 …ii

⇒u=-15 cm

Then, v=30 cm

From lens equation, we can write

1f=1v-1u⇒1f=130-1-15

⇒f=+10 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The distance between Sun and Earth is R. The duration of year if the distance between Sun and Earth becomes 3R will be :

Options:
A 3 yr
B 33 yr
C 3 yr
D 9 yr

The correct answer is:
33 yr

Solution:
According to Kepler's Third Law – The Law of Periods,
T2∝R3
where T is time taken by the planet to go once around the sun and R is the semi-major axis (distance) of the elliptical orbit.
∴ T2=k R3 …(i)
Where k is a constant of proportionality.
When R becomes 3 times, let the time period be T'.
∴ T'2=k(3R)3 …(ii)
∴ T2T'2=127
So, time period will be 33 yr.

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (24 Jun Shift 2)
The distance between the centre of moon and earth is \(D\) and mass of earth is 81 times the mass of moon. At what distance from the centre of the earth, the gravitational force will be zero?

Options:
A \(\frac{D}{2}\)
B \(\frac{2 D}{3}\)
C \(\frac{4 D}{3}\)
D \(\frac{9 D}{10}\)

The correct answer is:
\(\frac{9 D}{10}\)

Solution:
Let a unit mass \(m\) is present at a distance \(x\) from the earth, where gravitational force is zero.
\(\frac{G m M_e}{x^2}=\frac{G m M}{(D-x)^2}\)...(i)
where, \(M_e\) is mass of earth and \(M\) is mass of moon.
Given, \(M_e=81 \mathrm{M}\)
\(\therefore\) From Eq. (i), we have
\(\begin{aligned}
& \frac{G m 81 M}{x^2}=\frac{G m M}{(D-x)^2} \Rightarrow \frac{81}{x^2}=\frac{1}{(D-x)^2} \\
\Rightarrow & \left(\frac{9}{x}\right)^2=\frac{1}{(D-x)^2} \Rightarrow \frac{9}{x}=\frac{1}{D-x} \\
\Rightarrow & 9 D-9 x=x \Rightarrow 9 D=10 x \\
\Rightarrow & x= \frac{9 D}{10}
\end{aligned}\)

Topic: Physics, Gravitation, JEE Main
The distance between the chords of contact of the tangents to the circle $x^2+y^2+2 g x+2 f y+c=0$ from the origin and the point $(g, f)$ is

Options:
A $\frac{1}{2}\left(\frac{g^2+f^2-c}{\sqrt{g^2+f^2}}\right)$
B $\left(\frac{g^2+f^2-c}{\sqrt{g^2+f^2}}\right)$
C $\frac{1}{2}\left(\frac{g^2+f^2-c}{g^2+f^2}\right)$
D None of these

The correct answer is:
$\frac{1}{2}\left(\frac{g^2+f^2-c}{\sqrt{g^2+f^2}}\right)$

Solution:
Chord of contact from origin $\equiv g x+f y+c=0$ and from $(g, f) \equiv g x+f y+g(x+g)+f(y+f)+c=0$ or $2 g x+2 f y+g^2+f^2+c=0$
$=\frac{\frac{g^2+f^2+c}{2}-c}{\sqrt{g^2+f^2}}=\frac{g^2+f^2-c}{2 \sqrt{g^2+f^2}}$

Topic: Mathematics, Circle, JEE Main
The distance between the circumcentre and orthocentre of the triangle, whose vertices are \( (0,0),(6,8) \) and \( (-4,3) \), is \( L \), then the value of \( \frac{2}{\sqrt{5}} L \) is

Options:

The correct answer is:
5

Solution:
Given, vertices of the triangle are O(0,0), B(6,8) and C(-4,3)

Clearly, BC2=OB2+OC2

∴ ΔOBC is right-angled at O. Now, circumcentre=mid-point of hypotenuse BC=S1,112 and orthocentre=vertex at the right angle=O(0,0)
∴ Required distance L=OS=1+1214=552 units

Hence, 25L=25×552=5

Topic: Mathematics, Straight Lines, JEE Main
The distance between the directrices of the ellipse $\frac{x^2}{36}+\frac{y^2}{20}=1$ is

Options:
A 8
B 12
C 18
D 24

The correct answer is:
18

Solution:
$\begin{aligned} & a=6, b=2 \sqrt{5} \\ & b^2=a^2\left(1-e^2\right) \Rightarrow \frac{20}{36}=\left(1-e^2\right) \Rightarrow e=\sqrt{\frac{16}{36}}=\frac{2}{3}\end{aligned}$
But directrices are $x= \pm \frac{a}{e}$
Hence distance between them is $2 \cdot \frac{6}{2 / 3}=18$.

Topic: Mathematics, Ellipse, JEE Main
The distance between the foci of a hyperbola is double the distance between its vertices and the length of its conjugate axis is 6 . The equation of the hyperbola referred to its axes as axes of co-ordinates is

Options:
A $3 x^2-y^2=3$
B $x^2-3 y^2=3$
C $3 x^2-y^2=9$
D $x^2-3 y^2=9$

The correct answer is:
$3 x^2-y^2=9$

Solution:
According to given conditions, $2 a e=2.2 a$ or $e=2$ and $2 b=6 \Rightarrow b=3$. Hence, $a=\frac{3}{\sqrt{3}}=\sqrt{3}$
Therefore, equation is $\frac{x^2}{3}-\frac{y^2}{9}=1$ i.e., $3 x^2-y^2=9$.

Topic: Mathematics, Hyperbola, JEE Main
The distance between the foci of an ellipse is 16 and eccentricity is $\frac{1}{2}$. Length of the major axis of the ellipse is

Options:
A 8
B 64
C 16
D 32

The correct answer is:
32

Solution:
Given, distance between the foci $=2 a e=16$ and eccentricity of ellipse $(e)=1 / 2$. We know that length of the major axis of the ellipse
$=2 a=\frac{2 a e}{e}=\frac{16}{1 / 2}=32$.

Topic: Mathematics, Ellipse, JEE Main
The distance between the foci of the ellipse $3 x^2+4 y^2=48$ is

Options:
A 2
B 4
C 6
D 8

The correct answer is:
4

Solution:
$\begin{aligned} & \frac{x^2}{(48 / 3)}+\frac{y^2}{(48 / 4)}=1 \\ & a^2=16, b^2=12 \Rightarrow e=\sqrt{1-\frac{b^2}{a^2}}=\frac{1}{2}\end{aligned}$
Distance is $2 a e=2 \cdot 4 \cdot \frac{1}{2}=4$.

Topic: Mathematics, Ellipse, JEE Main
The distance between the foci of the hyperbola $x^2-3 y^2-4 x-6 y-11=0$ is

Options:
A $4$
B $6$
C $8$
D $10$

The correct answer is:
$8$

Solution:
Given, equation of hyperbola is
$$
\begin{gathered}
x^2-3 y^2-4 x-6 y-11=0 \\
\Rightarrow \quad\left(x^2-4 x+4\right)-3\left(y^2+2 y+1\right)-11 \\
=4-3 \\
\Rightarrow \quad(x-2)^2-3(y+1)^2=12 \\
\Rightarrow \quad \frac{(x-2)^2}{12}-\frac{(y+1)^2}{4}=1 \\
\text { Now, } \quad e=\sqrt{1+\frac{4}{12}}=\frac{2}{\sqrt{3}}
\end{gathered}
$$
Now,
$$
e=\sqrt{1+\frac{4}{12}}=\frac{2}{\sqrt{3}}
$$
$\therefore$ Distance between foci
$$
=2 a e=2 \times \sqrt{12} \times \frac{2}{\sqrt{3}}=8
$$

Topic: Mathematics, Hyperbola, JEE Main
The distance between the focus and directrix of the parabola 3x-y2=48x+3y is

Options:

The correct answer is:
12

Solution:
Given, the equation of parabola is 3x-y22=24x+3y2
⇒ length of the latus rectum =4a=24
⇒ distance between the focus and directrix is equal to 2a=12

Topic: Mathematics, Parabola, JEE Main
The distance between the line $\vec{r}=2 \hat{i}-2 \hat{j}+3 \hat{k}+\lambda(\hat{i}-\hat{j}+4 \hat{k})$ and the plane $\vec{r} \cdot(\hat{i}+5 \hat{j}+\hat{k})=5$ is

Options:
A
$\frac{10}{9}$

B
$\frac{10}{3 \sqrt{3}}$

C
$\frac{3}{10}$

D
$\frac{10}{3}$

The correct answer is:
$\frac{10}{3 \sqrt{3}}$

Solution:
Distance between the line
$\vec{r}=2 \hat{i}-2 \hat{j}+3 \hat{k}+\lambda(\hat{i}-\hat{j}+4 \hat{k})$ and the plane $\vec{r} \cdot(\hat{i}+5 \hat{j}+\hat{k})=5$ is
equation of plane is $x+5 y+z=5$
$\therefore$ Distance of line from this plane
= perpendicular distance of point $(2,-2,3)$ from the plane
i.e. $\left|\frac{2-10+3-5}{\sqrt{1+5^2+1}}\right|=\frac{10}{3 \sqrt{3}}$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2005
The distance between the lines $3 x+4 y=9$ and $6 x+8 y=15$ is

Options:
A $3 / 2$
B $3 / 10$
C 6
D None of these

The correct answer is:
$3 / 10$

Solution:
Here the lines are, $3 x+4 y-9=0$
and
$6 x+8 y-15=0$
Now distance from origin of both the lines are
$\frac{-9}{\sqrt{3^2+4^2}}=-\frac{9}{5} \text { and } \frac{-15}{\sqrt{6^2+8^2}}=-\frac{15}{10}$
Hence distance between both the lines are
$\left|-\frac{9}{5}-\left(-\frac{15}{10}\right)\right|=\frac{3}{10}$
Ailter: Put $y=0$ in the first equation, we get $x=3$ therefore, the point $(3,0)$ lies on it. So the required distance between these two lines is the perpendicular length of the line $6 x+8 y=15$ from
$\frac{6 \times 3-15}{\sqrt{6^2+8^2}}=\frac{3}{10}$

Topic: Mathematics, Straight Lines, JEE Main
The distance between the nearest node and antinode in a stationary wave is

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

The correct answer is:
$\frac{\lambda}{4}$

Solution:
The distance between the nearest node and antinode in a stationary wave is $\frac{\lambda}{4}$

Topic: Physics, Waves and Sound, JEE Main
The distance between the planes $x+2 y+3 z+7=0$ and $2 x+4 y+6 z+7=0$ is

Options:
A $\frac{\sqrt{7}}{2 \sqrt{2}}$
B $\frac{7}{2}$
C $\frac{\sqrt{7}}{2}$
D $\frac{7}{2 \sqrt{2}}$

The correct answer is:
$\frac{\sqrt{7}}{2 \sqrt{2}}$

Solution:
Required distance $=\frac{7-\frac{7}{2}}{\sqrt{1+4+9}}=\frac{\sqrt{7}}{2 \sqrt{2}}$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The distance between the point -1,-5,-10 and the point of intersection of the line x-23=y+14=z-212 with the plane x-y+z=5 is 13t, then t equals to

Options:

The correct answer is:
1

Solution:
Any point on the line is 3r+2, 4r –1, 12r+2
If it lies on the given plane, then 3r+2-4r-1+12r+2=5
⇒ 11r=0 ⇒ r=0
∴ Co-ordinates of the point of intersection of the given line and given plane are 2, -1, 2 and thus the required distance is 2+12+-1+52+2+102=13

⇒13=13t ⇒ t=1

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The distance between the tangent lines to the hyperbola $x^2-2 y^2=18$ which are perpendicular to the line $y=x$ is

Options:
A 6
B $2 \sqrt{3}$
C $3 \sqrt{2}$
D 0

The correct answer is:
$2 \sqrt{3}$

Solution:
The line perpendicular to line $y=x$ is
$$
y=-x+c \text { or } x+y-c=0
$$
Given, hyperbola is $x^2-2 y^2=18$
or $\quad \frac{x^2}{18}-\frac{y^2}{9}=1$
Here, $a^2=18, b^2=9$
Condition for tangency of hyperbola is
$$
\begin{aligned}
& c^2=a^2 m^2-b^2 \\
& c^2=18 \times(-1)^2-9 \\
& c^2=9 \Rightarrow c=+3
\end{aligned}
$$
Equation of tangents are
$$
x+y \pm 3=0
$$
$$
\begin{aligned}
\text { Distance between tangent } & =\frac{|6|}{\sqrt{1^2+1^2}} \\
& =\frac{6}{\sqrt{2}} \text { or } 3 \sqrt{2} \text { units }
\end{aligned}
$$

Topic: Mathematics, Hyperbola, JEE Main