datavorous/entrance-exam-dataset huggingface dataset:
The first overtone in a closed pipe has a frequency

Options:
A Same as the fundamental frequency of an open tube of same length
B Twice the fundamental frequency of an open tube of same length
C Same as that of the first overtone of an open tube of same length
D None of the above

The correct answer is:
None of the above

Solution:
First overtone for closed pipe $=\frac{3 v}{4 l}$
Fundamental frequency for open pipe $=\frac{v}{2 l}$
First overtone for open pipe $=\frac{2 v}{2 l}$.

Topic: Physics, Waves and Sound, JEE Main
The first, second and seventh terms of an arithmetic progression (all the terms are distinct) are in geometric progression and the sum of these three terms is 93. Then, the fourth term of this geometric progression is

Options:
A 21
B 31
C 75
D 375

The correct answer is:
375

Solution:
Let a,a+d,a+6d are in G.P.
âa+d2=aa+6d
âa2+d2+2ad=a2+6ad
âd2=4adâd=4a
Also, given that the sum of the terms is 93.
âa+a+d+a+6d=93
3a+7d=93
3a+74a=93
31a=93âa=3 & d=12
â the terms are 3,15,75
â the fourth term of the GP.. is 75Ã5=375

Topic: Mathematics, Sequences and Series, JEE Main
The first term of a G.P. is 7 , the last term is 448 and sum of all terms is 889 , then the common ratio is

Options:
A $5$
B $4$
C $3$
D $2$

The correct answer is:
$2$

Solution:
$a=7$ and $a r^{n-1}=448$
$S_n=\frac{a\left(r^n-1\right)}{r-1}=889$ $\Rightarrow \frac{\left(a r^{n-1} \cdot r-a\right)}{r-1}=889$
$\Rightarrow \quad \frac{448 r-7}{r-1}=889 \Rightarrow r=2$.

Topic: Mathematics, Sequences and Series, JEE Main
The first term of a G.P. whose second term is 2 and sum to infinity is 8 , will be

Options:
A $6$
B $3$
C $4$
D $1$

The correct answer is:
$4$

Solution:
We have $a r=2$ and $S_{\infty}=8=\frac{a}{1-r}$
$\Rightarrow \quad 8=\frac{2}{r(1-r)}\left( \because a=\frac{2}{r}\right)$
$\Rightarrow \quad 4 r(1-r)=1 \Rightarrow 4 r-4 r^2-1=0$
$\Rightarrow \quad 4 r^2-4 r+1=0 \Rightarrow\left(r-\frac{1}{2}\right)(4 r-2)$ $=0 \Rightarrow r=\frac{1}{2}$
So first term $a=4$

Topic: Mathematics, Sequences and Series, JEE Main
The first term of a harmonic progression is $1 / 7$ and the second term is $1 / 9$. The $12^{\text {th }}$ term is

Options:
A $1 / 19$
B $1 / 29$
C $1 / 17$
D $1 / 27$

The correct answer is:
$1 / 29$

Solution:
Here first term of A.P. be 7 and second be 9 , then $12^{\text {th }}$ term will be $7+11 \times 2=29$.
Hence $12^{\text {th }}$ term of the H.P. be $\frac{1}{29}$

Topic: Mathematics, Sequences and Series, JEE Main
The first three terms of an arithmetic-geometric progression are 3,–1 and –1. The next term of the progression is

Options:
A 2
B –2
C -527
D -59

The correct answer is:
-59

Solution:
Let the first four terms of the A.G.P. be
3, 3+dr, 3+2dr2, 3+3dr3
Hence, 3+dr=-1 and 3+2dr2=-1
3+d2r23+2dr2=1-1
⇒9+d2+6d=-3-2d
⇒d2+8d+12=0⇒d=-2,-6
r=-13+d⇒r=-1 or 13
(i) for d=-2, r=-1; next term is
3+3dr3=3-6-13=3
(ii) for d=-6,r=13; next term is
3+3dr3=3-18133=-1527=-59
So, the next term is 3 or -59

Topic: Mathematics, Sequences and Series, JEE Main
The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is

Options:
A
$-4$

B
$-12$

C
12

D
4

The correct answer is:
$-12$

Solution:
Let $a, a r, a r^2, \ldots$
$a+a r=12$
$a r^2+a r^3=48$
dividing (2) by (1), we have
$$
\begin{aligned}
& \frac{a r^2(1+r)}{a(r+1)}=4 \\
& \Rightarrow r^2=4 \text { if } r \neq-1 \\
& \therefore r=-2 \\
& \text { also, } a=-12 \quad \text { (using (1)). }
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2008
The five successive ionization enthalpies of an element are 800, 2427, 3658, 35024 32824 kJ mol–1 . The number of valence electrons in the element is:

Options:
A 5
B 4
C 3
D 2

The correct answer is:
3

Solution:
As difference in 3rd and 4th ionisation energies is high so atom contains 3 valence electrons.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The flat base of a hemisphere of radius a with no charge inside it lies in a horizontal plane. A uniform electric field $\vec{E}$ is applied at an angle $\frac{\pi}{4}$ with the vertical direction. The electric flux through the curved surface of the hemisphere is

Options:
A
$\pi a^2 E$

B
$\frac{\pi a^2 E}{\sqrt{2}}$

C
$\frac{\pi a^2 E}{2 \sqrt{2}}$

D
$\frac{\left(\pi+2 \pi \phi^2 E\right.}{\left(2 \sqrt{2}{ }^2\right)}$

The correct answer is:
$\frac{\pi a^2 E}{\sqrt{2}}$

Solution:
We know that,
$$
\phi=\oint E \cdot d S=E \oint d S \cos 45^{\circ}
$$
In case of hemisphere
$$
\phi_{\text {curved }}=\phi_{\text {circular }}
$$
Therefore, $\phi_{\text {curved }}=E \pi a^2 \cdot \frac{1}{\sqrt{2}}=\frac{E \pi a^2}{\sqrt{2}}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2012 (19 May Online)
The fleshy receptacle of syconus of fig encloses a number of

Options:
A Mericarps
B Achenes
C Samaras
D Berries

The correct answer is:
Achenes

Solution:
Syconus is fruit type in fig which is composed of many acheneal fruitlets.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2008 (Mains)
The flocculating power of the given ions for the specified colloidal sols will be such that

Options:
A

Arsenic sulphoid sol	Ferric hydroxide sol

FeCN64->PO43->SO42->Cl-	Al3+>Ba2+>Na+

B

Arsenic sulphoid sol	Ferric hydroxide sol

Al3+>Ba2+>Na+	FeCN64->PO43->SO42->Cl-

C

Arsenic sulphoid sol	Ferric hydroxide sol

Na+>Ba2+>Al3+	Cl->SO42->PO43->FeCN64-

D

Arsenic sulphoid sol	Ferric hydroxide sol

Cl->SO42->PO43->FeCN64-	Na+>Ba2+>Al3+

The correct answer is:

Arsenic sulphoid sol	Ferric hydroxide sol

Al3+>Ba2+>Na+	FeCN64->PO43->SO42->Cl-

Solution:
Coagulation power depends on nature of charge and amount of charge
Coagulation power ∝ charge on ion of electrolyte
As Arsenic sulphoid sol is negatively charged so for it cations will cause precipitation of ions and the order is Al3+>Ba2+>Na+
As Ferric hydroxide sol is positively charged so for it anions will cause precipitation of ions and the order is FeCN64->PO43->SO42->Cl-

Topic: Chemistry, Surface Chemistry, JEE Main
The flocculation value of HCl for arsenic sulphide sol is 30 mmolL-1. If H2SO4 is used for the flocculation of arsenic sulphide, the amount, in grams, of H2SO4 in 250ml required for the above purpose is ________
(molecular mass of H2SO4=98 g/mol

Options:

The correct answer is:
0.36

Solution:
For 1L sol 30 mmol of HCl is required
∴ For 1L sol 15 mmol H2SO4 is required
For 250 mL of sol 154×10-3 mmol H2SO4

Mass of H2SO4=3.75 × 10-3×98 = 0.3675 g

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The flocculation value of HCl for arsenic sulphide sol. is 30 m mol L–1. If H2SO4 is used for the flocculation of arsenic sulphide, the amount, in grams, of H2SO4 in 250 ml required for the above purpose is (molecular mass of H2SO4=98 g/mol )

Options:
A 0.1838
B 0.3864
C 0.3675
D 0.1218

The correct answer is:
0.3675

Solution:
→ 1 litre solution of 30 m mol of HCl is required
→ For 1 litre solution 15 m mol H2SO4 is required.
250 ml of solution required: 154×10-3 (m mol) of H2SO4
=0.3675 gm

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The floral formula $\bigoplus \underset{+}{\bigcirc} \mathrm{K}_{(5)} \mathrm{C}_{(5)} \mathrm{A}_5 \underset{\mathrm{G}_{(2)}}{ }$ is that of :

Options:
A Tobacco
B Tulip
C Soybean
D Sunnhemp

The correct answer is:
Tobacco

Solution:
No Solution Available

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2009 (Mains)
The fluid coming out of proximal convoluted tubule has a concentration of about

Options:
A 200 mosmol/litre
B 300 mosmol/litre
C 600 mosmol/litre
D 1200 mosmol/litre

The correct answer is:
300 mosmol/litre

Solution:
Mammals have the ability to produce concentrated urine. The capacity of concentrating the urine depends on the length of the loop of Henle. Vasa recta also helps in concentrating the urine. The flow of filtrate in the two limbs of the loop of Henle is in the opposite direction and thus forms a counter current mechanism. The proximity between Henle's loop and vasa recta as well as counter-current in them helps in maintaining an increasing osmolarity towards the inner medullary interstitium i.e. 300mmol/liter in the cortex to 1200 mmol/liter in the medulla. PCT lies in the cortex too hence the concentration is 300mosmol/liter.

Topic: Biology, Excretory Products and their Elimination, NEET
The flux linked with a coil at any instant ' $t$ ' is given by
$\phi=10 t^2-50 t+250$
The induced emf at $t=3 \mathrm{~s}$ is

Options:
A
$190 \mathrm{~V}$

B
$-190 \mathrm{~V}$

C
$-10 \mathrm{~V}$

D
$10 \mathrm{~V}$

The correct answer is:
$-10 \mathrm{~V}$

Solution:
$e=-\frac{d \phi}{d t}=-(20 t-50)=-10$ volt

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2006
The flux of the electric field $\overrightarrow{\mathrm{E}}=24 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}+28 \hat{\mathrm{k}} \mathrm{NC}^{-1}$ through an area of $20 \mathrm{~m}^2$ on the yz plane is

Options:
A $480 \mathrm{Nm}^2 \mathrm{C}^{-1}$
B $600 \mathrm{Nm}^2 \mathrm{C}^{-1}$
C $560 \mathrm{Nm}^2 \mathrm{C}^{-1}$
D $1640 \mathrm{Nm}^2 \mathrm{C}^{-1}$

The correct answer is:
$480 \mathrm{Nm}^2 \mathrm{C}^{-1}$

Solution:
Electric field is given by
$$
\overrightarrow{\mathrm{E}}=24 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}+28 \hat{\mathrm{k}}
$$
Area, $\vec{A}=20 \hat{i} \mathrm{~m}^2$
Electric flux, $\overrightarrow{\mathrm{D}}=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}$
$$
\begin{aligned}
& =(24 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}+28 \hat{\mathrm{k}})(20 \hat{\mathrm{j}}) \\
& =480 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main
The focal chord of the parabola y2=32x touches the ellipse x242+y222=1 in the first quadrant at the point

Options:
A 2,3
B 32,32
C 42,22
D 22,42

The correct answer is:
2,3

Solution:
Let the tangent to the ellipse at 4cosâ¡Î¸,2sinâ¡Î¸ be
xcosâ¡Î¸4+ysinâ¡Î¸2=1
âµ it passes through the focus 8,0
âcosâ¡Î¸=12âÎ¸=Ï3
â point is 42,232â¡2,3

Topic: Mathematics, Ellipse, JEE Main
The focal chord of y2=64x is tangent to x-42+y-22=4, then the square root of the length of this focal chord is equal to

Options:
A 749
B 373
C 743
D 379

The correct answer is:
743

Solution:
Let, the equation of the focal chord is y=mx-16
Since the chord is tangent to the given circle
â´2=4m-16m-21+m2
â41+m2=-12m-22
âm=0 or -1235
âtanâ¡Î¸=1235âcosecâ¡Î¸=3712
â Length of the focal chord, L=64Ã37122 units
â The required value, L=8Ã3712=743

Topic: Mathematics, Parabola, JEE Main
The focal chords of the parabola y2=16x which are tangent to the circle of radius r and centre 6,0 are perpendicular, then the radius r of the circle is

Options:
A 2 units
B 2 units
C 1 units
D 12 units

The correct answer is:
2 units

Solution:
The focus of the given parabola is 4,0

The centre of the circle with two focal chords as tangent is 6,0

From the figure,

the distance between the centre of the circle and the focus is 2 units

Hence, the radius of the circle is 2sin45°=2 units

Topic: Mathematics, Parabola, JEE Main
The focal distance of a point on the parabola $y^{2}=12 x$ is 4 . What is the abscissa of the point?

Options:
A 1
B $-1$
C 2 $\sqrt{3}$
D -2

The correct answer is:
1

Solution:
Focal distance of a point $\left(x_{1}, y_{1}\right)$ on the parabola is $y^{2}=4 a x$ is equal to its distance from directrix $x+a=0$ is
$\mathrm{x}_{1}+\mathrm{a}$
For $y^{2}=12 x$, comparing with $y^{2}=4 a x$.
$4 a=12 \Rightarrow a=3$
so, $x_{1}+3=4$
$\Rightarrow \quad x_{1}=1$

Topic: Mathematics, Parabola, JEE Main
The focal distance of a point on the parabola $y^{2}=16 x$ whose ordinate is twice the abscissa, is

Options:
A 6
B 8
C 10
D 12

The correct answer is:
8

Solution:
Given curve is $y^{2}=16 x$. Let the point be $(h, k)$. But $2 h=k$, then $k^{2}=16 h$
$$
\begin{aligned}
\Rightarrow & & 4 h^{2} &=16 h \\
\Rightarrow & & h &=0, h=4 \\
\Rightarrow & & k &=0, k=8
\end{aligned}
$$
$\therefore$ Points are $(0,0),(4,8)$.
Hence, focal distance are respectively
$$
0+4=4,4+4=8
$$
$(\because$ focal distance $=h-a)$

Topic: Mathematics, Parabola, JEE Main
The focal length f is related to the radius of curvature r of the spherical convex mirror by:

Options:
A f=-12r
B f=+12r
C f=r
D f=-r

The correct answer is:
f=+12r

Solution:
For convex mirror, focus is behind the mirror.

As we can see from the figure that,

⇒f=+r2

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The focal length of a convex mirror is $20 \mathrm{~cm}$ its radius of curvature will be

Options:
A $10 \mathrm{~cm}$
B $20 \mathrm{~cm}$
C $30 \mathrm{~cm}$
D $40 \mathrm{~cm}$

The correct answer is:
$40 \mathrm{~cm}$

Solution:
$f=\frac{R}{2} \Rightarrow R=40 \mathrm{~cm}$

Topic: Physics, Ray Optics, JEE Main
The focal length of a mirror is given by 1f=1u+1 v where u and v represent object and image distances respectively. The maximum relative error in f is

Options:
A ∆ff=∆uu+∆vv
B ∆ff=∆uu+∆vv-∆u+vu+v
C ∆ff=∆uu+∆vv-∆u-vu+v
D ∆ff=∆uu+∆vv+∆uu+v+∆vu+v

The correct answer is:
∆ff=∆uu+∆vv+∆uu+v+∆vu+v

Solution:
f=uvu+v,âff=âuu+âvv+Δ(u+v)u+v

Topic: Physics, Units and Dimensions, JEE Main
The focal length of a mirror is given by 2f=1v-1u. If in the measurement of u and v, the errors are equal to p each, then the relative error in f is

Options:
A p2 1u+1v
B p 1u+1v
C p21u-1v
D p 1u-1v

The correct answer is:
p 1u+1v

Solution:
Given, equation is 2f=1v-1u ....(i)
Differentiating the given equation, we have
-2f2 df=-1v2dv--1u2 du
=-p 1v-1u 1v+1u ∵ dv=du=p
=-2pf1v+1u [using Eq. (i)]
∴ dff=p 1v+1v

Topic: Physics, Units and Dimensions, JEE Main
The focal length of objective and eye lens of a astronomical telescope are respectively $2 \mathrm{~m}$ and $5 \mathrm{~cm}$. Final image is formed at (i) least distance of distinct vision (ii) infinity. The magnifying power in both cases will be

Options:
A $-48,-40$
B $-40,-48$
C $-40,48$
D $-48,40$

The correct answer is:
$-48,-40$

Solution:
When the final image is at the least distance of distinct vision, then
$m=-\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=\frac{200}{5}\left(1+\frac{5}{25}\right)=\frac{200 \times 6}{5 \times 5}=-48$
When the final image is at infinity, then
$m=\frac{-f_o}{f_e}=\frac{200}{5}=-40$

Topic: Physics, Ray Optics, JEE Main
The focal length of objective and eye lens of a microscope are $4 \mathrm{~cm}$ and $8 \mathrm{~cm}$ respectively. If the least distance of distinct vision is $24 \mathrm{~cm}$ and object distance is $4.5 \mathrm{~cm}$ from the objective lens, then the magnifying power of the microscope will be

Options:
A 18
B 32
C 64
D 20

The correct answer is:
32

Solution:
For objective lens $\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}$
$\Rightarrow \frac{1}{(+4)}=\frac{1}{v_o}-\frac{1}{(-4.5)} \Rightarrow v_o=36 \mathrm{~cm}$
$\therefore\left|m_D\right|=\frac{v_o}{u_o}\left(1+\frac{D}{f_e}\right)=\frac{36}{4.5}\left(1+\frac{24}{8}\right)=32$

Topic: Physics, Ray Optics, JEE Main
The focal length of the objective and the eyepiece of a telescope are $50 \mathrm{~cm}$ and $5 \mathrm{~cm}$ respectively. If the telescope is focussed for distinct vision on a scale distant $2 \mathrm{~m}$ from its objective, then its magnifying power will be :

Options:
A
$-4$

B
$-8$

C
$+8$

D
$-2$

The correct answer is:
$-2$

Solution:
Given: $\mathrm{f}_0=50 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=5 \mathrm{~cm}$
$\mathrm{d}=25 \mathrm{~cm}, \mathrm{u}_0=-200 \mathrm{~cm}$
Magnification $\mathrm{M}=$ ?
As $\frac{1}{\mathrm{v}_0}-\frac{1}{\mathrm{u}_0}=\frac{1}{\mathrm{f}_0}$
$$
\Rightarrow \frac{1}{\mathrm{v}_0}=\frac{1}{\mathrm{f}_0}+\frac{1}{\mathrm{u}_0}=\frac{1}{50}-\frac{1}{200}=\frac{4-1}{200}=\frac{3}{200}
$$
or $\mathrm{v}_0=\frac{200}{3} \mathrm{~cm}$
Now $\mathrm{v}_{\mathrm{e}}=\mathrm{d}=-25 \mathrm{~cm}$
From, $\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$
$$
\begin{aligned}
& -\frac{1}{u_e}=\frac{1}{f_e}-\frac{1}{v_e} \\
& =\frac{1}{5}+\frac{1}{25}=\frac{6}{25}
\end{aligned}
$$
or, $\mathrm{v}_{\mathrm{e}}=\frac{-25}{6} \mathrm{~cm}$
$$
\begin{aligned}
& \text { Magnification } \mathrm{M}=\mathrm{M}_0 \times \mathrm{M}_{\mathrm{e}} \\
& =\frac{\mathrm{v}_0}{\mathrm{u}_0} \times \frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{u}_{\mathrm{e}}}=\frac{-200 / 3}{200} \times \frac{-25}{-25 / 6} \\
& =-\frac{1}{3} \times 6=-2
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2013 (22 Apr Online)
The focal length of the objective lens of a compound microscope is

Options:
A Equal to the focal length of its eye piece
B Less than the focal length of eye piece
C Greater than the focal length of eye piece
D Any of the above three

The correct answer is:
Less than the focal length of eye piece

Solution:
For a compound microscope $\lt f_{\text {objective }} \lt f_{\text {eye piece }}$

Topic: Physics, Ray Optics, JEE Main
The focal length of the objective of a terrestrial telescope is $80 \mathrm{~cm}$ and it is adjusted for parallel rays, then its magnifying power is 20 . If the focal length of erecting lens is $20 \mathrm{~cm}$, then full length of telescope will be

Options:
A $84 \mathrm{~cm}$
B $100 \mathrm{~cm}$
C $124 \mathrm{~cm}$
D $164 \mathrm{~cm}$

The correct answer is:
$164 \mathrm{~cm}$

Solution:
$m=\frac{f_o}{f_e} \Rightarrow \frac{80}{f_e}=20 \Rightarrow f_e=4 \mathrm{~cm}$
Hence length of terrestrial telescope
$=f_o+f_e+4 f=80+4+4 \times 20=164 \mathrm{~cm}$
Hence length of terrestrial telescope
$=f_o+f_e+4 f=80+4+4 \times 20=164 \mathrm{~cm}$

Topic: Physics, Ray Optics, JEE Main
The focal lengths of convex lens for red and blue light are $100 \mathrm{~cm}$ and $96.8 \mathrm{~cm}$ respectively. The dispersive power of material of lens is

Options:
A 0.325
B 0.0325
C 0.98
D 0.968

The correct answer is:
0.0325

Solution:
$\omega=\frac{f_R-f_V}{f_y}=\frac{f_R-f_V}{\sqrt{f_V f_R}}$
Putting value of $f_V$ and $f_R$ we get $\omega=0.0325$.

Topic: Physics, Ray Optics, JEE Main
The focal lengths of objective lens and eye lens of a Galilean Telescope are respectively 30 cm and 3.0 cm. Telescope produces virtual, erect image of an object situated far away from it at least distance of distinct vision from the eye lens. In this condition the Magnifying Power of the Galilean Telescope should be :

Options:
A +8.8
B -11.2
C +11.2
D -8.8

The correct answer is:
+11.2

Solution:
MP = - fâ¡ 0 fâ¡ e 1 + fâ¡ e D
= - 3 0 - 3 1 + - 3 2 5
= + 11.2

Topic: Physics, Ray Optics, JEE Main, JEE Main 2014 (09 Apr Online)
The focal lengths of the objective and eye-lens of a microscope are $1 \mathrm{~cm}$ and $5 \mathrm{~cm}$ respectively. If the magnifying power for the relaxed eye is 45 , then the length of the tube is

Options:
A $30 \mathrm{~cm}$
B $25 \mathrm{~cm}$
C $15 \mathrm{~cm}$
D $12 \mathrm{~cm}$

The correct answer is:
$15 \mathrm{~cm}$

Solution:
by using $m_{\infty}=\frac{\left(L_{\infty}-f_o-f_e\right) \cdot D}{f_o f_e}$
$\Rightarrow 45=\frac{\left(L_{\infty}-1-5\right) \times 25}{1 \times 5} \Rightarrow L_{\infty}=15 \mathrm{~cm}$

Topic: Physics, Ray Optics, JEE Main
The foci of a hyperbola are (±2,0) and its eccentricity is 32. A tangent, perpendicular to the line 2x+3y=6, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the x- and y-axes are a and b respectively, then |6a|+|5b| is equal to

Options:

The correct answer is:
12

Solution:
The given equation of hyperbola is x2a2-y2b2=1,ae=2,e=32

⇒a=43

⇒b2=a2e2-a2=4-169=209

The tangent is perpendicular to 2x+3y=6.

Slope of tangent is m=-1-23=32

Equation of tangent, m=32

Equation of tangent is y=mx-a2m2-b2
⇒y=32x-169×94-209

( ∵ Tangent is in 1st quadrant⇒C<0

⇒y=32x-43

Converting the equation into intercept form:

⇒x89+y-43=1

⇒6a+5b=6×89+5×-43=12

Hence this is the required option.

Topic: Mathematics, Hyperbola, JEE Main
The foci of a hyperbola are (±2,0) and its eccentricity is 32. A tangent, perpendicular to the line 2x+3y=6, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the x- and y-axes are a and b respectively, then |6a|+|5b| is equal to

Options:

The correct answer is:
12

Solution:
The given equation of hyperbola is x2a2-y2b2=1,ae=2,e=32

⇒a=43

⇒b2=a2e2-a2=4-169=209

The tangent is perpendicular to 2x+3y=6.

Slope of tangent is m=-1-23=32

Equation of tangent, m=32

Equation of tangent is y=mx-a2m2-b2
⇒y=32x-169×94-209

( ∵ Tangent is in 1st quadrant⇒C<0

⇒y=32x-43

Converting the equation into intercept form:

⇒x89+y-43=1

⇒6a+5b=6×89+5×-43=12

Hence this is the required option.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The foci of the basic terms of conics \( 25 x^{2}+16 y^{2}-150 x=175 \) are

Options:
A \( (0, \pm 3) \)
B \( (0, \pm 2) \)
C \( (3, \pm 3) \)
D \( (0, \pm 1) \)

The correct answer is:
\( (3, \pm 3) \)

Solution:
Given equation,

25x2+16y2-150x=175

⇒25x2-6x+9+16y2=175+225

⇒25x-32+16y2=400
⇒x-3216+y225=1
Here, a2=16 and b2=25

It is a vertical ellipse.
∴e=1-a2b2=1-1625=35
Hence, the foci are 0,±be

⇒x-3=0 & y=±be

⇒x=3 & y=±3

≡3,±3.

Topic: Mathematics, Ellipse, JEE Main
The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide. Then the value of $b^2$ is

Options:
A
9

B
1

C
5

D
7

The correct answer is:
7

Solution:
$\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$
$\mathrm{a}=\sqrt{\frac{144}{25}}, \mathrm{~b}=\sqrt{\frac{81}{25}}, \mathrm{e}=\sqrt{1+\frac{81}{144}}=\frac{15}{12}=\frac{5}{4}$
Foci $=(3,0)$, focus of ellipse $=(3,0) \Rightarrow \mathrm{e}=\frac{3}{4}$
$b^2=16\left(1-\frac{9}{16}\right)=7$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2003
The foci of the ellipse $25(x+1)^2+9(y+2)^2=225$ are at

Options:
A $(-1,2)$ and $(-1,-6)$
B $(-1,2)$ and $(6,1)$
C $(1,-2)$ and $(1,-6)$
D $(-1,-2)$ and $(1,6)$

The correct answer is:
$(-1,2)$ and $(-1,-6)$

Solution:
$\begin{aligned} & \frac{(x+1)^2}{\frac{225}{25}}+\frac{(y+2)^2}{\frac{225}{9}}=1 \\ & a=\sqrt{\frac{225}{25}}=\frac{15}{5}, b=\sqrt{\frac{225}{9}}=\frac{15}{3} \Rightarrow e=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\end{aligned}$
Focus $=\left(-1,-2 \pm \frac{15}{3} \cdot \frac{4}{5}\right)=(-1,-2 \pm 4)=(-1,2) ; \quad(-1,-6)$

Topic: Mathematics, Ellipse, JEE Main
The foci of the hyperbola $9 x^2-16 y^2=144$ are

Options:
A $( \pm 4,0)$
B $(0, \pm 4)$
C $( \pm 5,0)$
D $(0, \pm 5)$

The correct answer is:
$( \pm 5,0)$

Solution:
The equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=1$
Now $b^2=a^2\left(e^2-1\right) \Rightarrow e=\frac{5}{4}$
Hence foci are $( \pm a e, 0) \Rightarrow\left( \pm 4 \cdot \frac{5}{4}, 0\right)$ i.e., $( \pm 5,0)$

Topic: Mathematics, Hyperbola, JEE Main
The focus and corresponding directrix of an ellipse are 3,4 and x+y-1=0 respectively. If the eccentricity of the ellipse is 12, then the coordinates of the centre of the ellipse are

Options:
A 2,3
B 4,5
C 8,9
D 1,2

The correct answer is:
4,5

Solution:
Let, the lengths of the semi-major axis and semi-minor axis of the ellipse are a & b respectively, then the distance of the focus 3,4 from the corresponding directrix x+y-1=0 is equal to ae-ae
âa12-a12=3+4-12=32
â3a2=32âa=22
Distance between the focus and centre =ae=2
The slope of the axis of the ellipse is 1=tanâ¡Î¸
âcosâ¡Î¸,sinâ¡Î¸=12,12 or -12,-12
The points on the axis of the ellipse at a distance 2 units from 3,4 are 3Â±212,4Â±212
â Points are 2,3 or 4,5 , but the centre is 4,5 because it is far from the directrix as compared to 2,3.

Topic: Mathematics, Ellipse, JEE Main
The focus of the conic represented parametrically by the equation y=t2+3,x=2t-1 is

Options:
A 1,4
B 4,1
C -1,3
D -1,4

The correct answer is:
-1,4

Solution:
Given y=t2+3, x=2t-1

Eliminating t, we get,

y=x+122+3
⇒x+12=4y-3
Hence, vertex: -1,3 and focus: -1,4

Topic: Mathematics, Parabola, JEE Main