datavorous/entrance-exam-dataset huggingface dataset:
If the domain of the function fx=x2−254−x2+log10x2+2x−15 is −∞,α∪β,∞, then α2+β3 is equal to:

Options:
A 140
B 175
C 150
D 125

The correct answer is:
150

Solution:
Given: fx=x2-254-x2+logx2+2x-15

⇒x2-25≥0

⇒x-5x+5≥0

⇒x∈(-∞,-5]∪[5,∞) ...i

4-x2≠0

⇒x≠2,-2 ...ii

⇒x2+2x-15>0

⇒x+5x-3>0

⇒x∈-∞,-5∪3,∞ ...iii

Using i,ii and iii

⇒x∈-∞,-5∪[5,∞)

So, on comparing with x∈-∞,α∪[β,∞) we get,

⇒α2+β3=-52+53

⇒α2+β3=150

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (01 Feb Shift 2)
If the earth be in half of its distance from the Sun, how many days will be in one year?

Options:
A 183
B 730
C 129
D 365

The correct answer is:
129

Solution:
According to Kepler's laws
T2âR3
When R2=R12
T22T12=R123Ã1R13=R138Ã1R13=18
T22T12=18
â T2=T122
=3652Ã1.414
T2=129 days

Topic: Physics, Gravitation, JEE Main
If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh 34 W. The radius of the Earth is 6400 km and g=10 m s-2

Options:
A 0.63×10-3 rad s-1
B 0.28×10-3 rad s-1
C 1.1×10-3 rad s-1
D 0.83×10-3 rad s-1

The correct answer is:
0.63×10-3 rad s-1

Solution:
Due to the rotation of the earth,
Changed gravitational acceleration, g′=g-ω2Rcos2⁡θ
At the equator, angle of inclination, θ=0
3g4=g-ω2R
w2R=g4
w= g4R
= 104×6400×103
=12×8×100
=11600=116×10-2=0.6×10-3 rad s-1.

Topic: Physics, Gravitation, JEE Main
If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh 34 W. The radius of the Earth is 6400 km and g=10 m s-2

Options:
A 0.63×10-3 rad s-1
B 0.28×10-3 rad s-1
C 1.1×10-3 rad s-1
D 0.83×10-3 rad s-1

The correct answer is:
0.63×10-3 rad s-1

Solution:
Due to the rotation of the earth,
Changed gravitational acceleration, g′=g-ω2Rcos2⁡θ
At the equator, angle of inclination, θ=0
3g4=g-ω2R
w2R=g4
w= g4R
= 104×6400×103
=12×8×100
=11600=116×10-2=0.6×10-3 rad s-1.

Topic: Physics, Gravitation, JEE Main, JEE Main 2017 (08 Apr Online)
If the earth moves round the sun in a circular orbit of radius \( 150000000 \mathrm{~km} \) with an angular velocity of about
\( 0.01745 \mathrm{rad} \mathrm{day}^{-1} \). Its linear speed in \( \mathrm{km} \mathrm{s}^{-1} \) is approximately

Options:
A \( 30000000 \)
B \( 1800 \)
C \( 720 \)
D \( 30 \)

The correct answer is:
\( 30 \)

Solution:
Given, angular velocity of earth, ω=0.01745 rad day-1 =0.0174524×60×60 rad day-1 and orbital radius of earth around the sun, R=15×107 km. Let the linear speed of earth be v. We know that, linear speed, v=ωR=0.0174524×60×60×15×107=30 km s-1. Hence, linear speed of earth is 30 km s-1.

Topic: Physics, Motion In Two Dimensions, JEE Main
If the earth rotates faster than its present speed, the weight of an object will

Options:
A Increase at the equator but remain unchanged at the poles
B Decrease at the equator but remain unchanged at the poles
C Remain unchanged at the equator but decrease at the poles
D Remain unchanged at the equator but increase at the poles

The correct answer is:
Decrease at the equator but remain unchanged at the poles

Solution:
$g^{\prime}=g-\omega^2 R \cos ^2 \lambda$
Rotation of the earth results in the decreased weight apparently. This decrease in weight is not felt at the poles as the angle of latitude is $90^{\circ}$.

Topic: Physics, Gravitation, JEE Main
If the earth shrinks in its size (radius) mass remaining the same, the value of $g$ on its surface will

Options:
A increase
B decrease
C remains same
D reduce to zero

The correct answer is:
increase

Solution:
The acceleration due to gravity at earth's surface is given by
$g=\frac{G M}{R^{2}}$
As, mass remains constant.
$\Rightarrow \quad g \propto \frac{1}{R^{2}}$
so, as the radius of earth decreases or the earth shrinks in size, the value of $g$ increases.

Topic: Physics, Gravitation, COMEDK, COMEDK 2016
If the Earth shrinks such that its density becomes 8 times to the present value, then the new duration of the day in hours will be

Options:
A 24
B 12
C 6
D 3

The correct answer is:
6

Solution:
I1ω1=I2ω2
ρ 1 4 3 π r 1 3 = ρ 2 4 3 π r 2 3
ρ 1 r 1 3 = ρ 2 r 2 3
r 2 = ( ρ 1 ρ 2 ) 1 3 r 1
T= 2π ω
⇒ T 1 T 2 = I 1 I 2
T 2 = T 1    I 2 I 1
= T 1 ( ρ 1 ρ 2 ) 2 3
=24(14)=6 h

Topic: Physics, Rotational Motion, JEE Main
If the earth shrinks to 18 of its original volume, while maintaining the same mass, then the duration of the day will be

Options:
A 8 hrs
B 48 hrs
C 6 hrs
D 72 hrs

The correct answer is:
6 hrs

Solution:
Use the concept of conservation of angular momentum the initial angular momentum of the earth should be same as final momentum of the earth, it means,
J1=J2⇒I1ω1=I2ω2, here the value of the moment of inertia for the solid sphere is, I=25mR2 and, the value of ω=2πT,

⇒25mR122πT1=25mR222πT2⇒R12R22=T1T2

When the original volume becomes one by eight of its original volume then, 18×43πR13=43πR23⇒R1=2R2,
⇒2R22R22=24T2⇒T2=6 h.

Topic: Physics, Gravitation, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 2)
If the earth stops rotating, the value of ′g′ at the equator will

Options:
A increase
B remain same
C decrease
D none of the above

The correct answer is:
increase

Solution:
At equator
g'=g-Ï2R
When earth suddenly stops Ï=0
g'=g-02R
g'=g
This mean g will increase.

Topic: Physics, Gravitation, NEET
If the earth suddenly shrinks to 164th of its original volume with its mass remaining the same, the period of rotation of earth becomes 24xh. The value of x is ________

Options:

The correct answer is:
16

Solution:
L= constant; L=Iω

From the conservation of angular momentum,

25MR2ω1=25MR42ω2⇒MR2ω1=MR2ω216⇒ω1ω2=116

Using ω=2πT for each case along with the respective angular frequencies,

T2T1=116

⇒T1T2=161

Using T1=24 h, x=16 by comparing with the given time period in the question.

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (10 Apr Shift 1)
If the earth suddenly shrinks to $\frac{1}{64}$ of its original volume, while keeping the same mass, then the duration of the day will be (Assume earth is a perfect sphere)

Options:
A 24 hours
B 1.5 hours
C 16 hours
D 48 hours

The correct answer is:
1.5 hours

Solution:
If earth shrinks to $\frac{1}{64}$ of its original volume;
$\frac{4}{3} \pi r^3=\frac{1}{64}\left(\frac{4}{3} \pi R^3\right)$
Here, $r=$ new radius and $R=$ original radius
$\begin{aligned} \Rightarrow & r^3 & =\frac{R^3}{64} \\ \text { or } & r & =\frac{R}{4}\end{aligned}$
As there is no external torque, angular momentum of earth remains constant
i.e., $\quad I_1 \omega_1=I_2 \omega_2 \quad\left[\because I_{\text {sphere }}=\frac{2}{5} M R^2\right]$
$\begin{aligned} \Rightarrow \quad & \quad \frac{2}{5} M R^2 \cdot \frac{2 \pi}{T_1}=\frac{2}{5} M r^2 \frac{2 \pi}{T_2} \\ & \quad\left(\therefore \omega=\frac{2 \pi}{T}, T=\text { time of revolution about axis }\right)\end{aligned}$
$\Rightarrow \quad T_2=\frac{r^2 T_1}{R^2}$
$\Rightarrow \quad T_2=\frac{\frac{R^2}{16} \times T_1}{R^2} \quad\left[\therefore r=\frac{R}{4}\right]$
$\begin{array}{ll}\Rightarrow & T_2=\frac{T_1}{16}=\frac{24}{16} \mathrm{~h} \\ \text { or } & T_2=1.5 \mathrm{~h}\end{array}$

Topic: Physics, Rotational Motion, AP EAMCET, AP EAMCET 2022 (07 Jul Shift 1)
If the earth were to spin faster, acceleration due to gravity at the poles

Options:
A increases
B decreases
C remains the same
D depends on how fast it spins

The correct answer is:
remains the same

Solution:
The variation of $g$ with angular velocity $\omega$ is given by
$$
g^{\prime}=g-R \omega^2
$$
If earth were to spin faster, that is angular velocity increases, then except at poles, the weight of bodies will decrease at all places.

Topic: Physics, Gravitation, COMEDK, COMEDK 2022
If the Earth were to suddenly contract to \(\frac{1}{n}\) th of its present radius without any change in its mass, the duration of the new day will be nearly

Options:
A \(24 / \mathrm{n} \mathrm{hr}\).
B \(24 \mathrm{n} \mathrm{hr}\).
C \(24 / \mathrm{n}^2 \mathrm{hr}\).
D \(24 n^2 h r\).

The correct answer is:
\(24 / \mathrm{n}^2 \mathrm{hr}\).

Solution:
Hints: \(\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2\)
\(\begin{aligned}
& \frac{2}{5} \mathrm{MR}^2\left(\frac{2 \pi}{\mathrm{T}_1}\right)=\frac{2}{5} \mathrm{M} \cdot \frac{\mathrm{R}^2}{\mathrm{n}^2}\left(\frac{2 \pi}{\mathrm{T}_2}\right) \\
& \mathrm{T}_2=\frac{\mathrm{T}_1}{\mathrm{n}^2}=\frac{24}{\mathrm{n}^2}
\end{aligned}\)

Topic: Physics, Rotational Motion, JEE Main
If the eccentricities of the hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$ be $e$ and $e_1$, then $\frac{1}{e^2}+\frac{1}{e_1^2}=$

Options:
A $1$
B $2$
C $3$
D None of these

The correct answer is:
$1$

Solution:
$\begin{aligned} & e=\sqrt{1+\frac{b^2}{a^2}} \Rightarrow e^2=\frac{a^2+b^2}{a^2} \\ & e_1=\sqrt{1+\frac{a^2}{b^2}} \Rightarrow e_1^2=\frac{b^2+a^2}{b^2} \Rightarrow \frac{1}{e_1^2}+\frac{1}{e^2}=1 .\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
If the eccentricity and length of latus rectum of a hyperbola are $\frac{\sqrt{13}}{3}$ and $\frac{10}{3}$ units respectively, then what is the length of the transverse axis?

Options:
A $\frac{7}{2}$ unit
B 12 unit
C $\frac{15}{2}$ unit
D $\frac{15}{4}$ unit

The correct answer is:
$\frac{15}{4}$ unit

Solution:
Length of latus rectum of a hyperbola is $\frac{2 b^2}{a}$ where $a$ is the half of the distance between two vertex of the hyperbola.
Latus rectum $=\frac{2 b^2}{a}=\frac{10}{3}$ or, $b^2=\frac{5 a}{3}$ ...(i)
In case of hyperbola, $\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)$ ...(ii)
Putting value of $b^2$ from equation (i) and $e=$
$\frac{\sqrt{13}}{3}$ in equation (ii), $\frac{5 \mathrm{a}}{3}=\mathrm{a}^2\left(\frac{13}{9}-1\right)$ or,
$\frac{5 a}{3}=\frac{4 a^2}{9}$
$\Rightarrow 4 a^2-15 a=0$ or, $a(4-15 a)=0$
$a \neq 0$, hence, $a=\frac{15}{4}$

Topic: Mathematics, Hyperbola, JEE Main
If the eccentricity of a conic satisfies the equation 2x3+10x-13=0, then that conic is

Options:
A a circle
B a parabola
C an ellipse
D a hyperbola

The correct answer is:
a hyperbola

Solution:
Given equation is 2x3+10x-13=0.

Given that the eccentricity of a conic satisfies the equation i.e., the eccentricity is any one of the roots.

Now, 2x3+10x-13=0.

⇒2x3+10x=13

⇒xx2+5=132

⇒xx2+5=6.5

By verification:

Put x=0⇒0(0+5)=6.5

Put x=1⇒112+5=6.5

⇒6=6.5

Put x=2⇒222+5=6.5

⇒29=6.5

⇒18=6.5

∴ The value of x lies in between 1, 2 i.e., the

eccentricity is greater than one.

∴ The conic is hyperbola.

Topic: Mathematics, Hyperbola, JEE Main
If the eccentricity of a hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$, which passes through $(K, 2)$, is $\frac{\sqrt{13}}{3}$, then the value of $K^2$ is

Options:
A
18

B
8

C
1

D
2

The correct answer is:
18

Solution:
Given hyperbola is
$$
\frac{x^2}{9}-\frac{y^2}{b^2}=1
$$
Since this passes through $(K, 2)$, therefore
$$
\frac{K^2}{9}-\frac{4}{b^2}=1
$$
Also, given $e=\sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{13}}{3}$
$$
\begin{aligned}
& \Rightarrow \sqrt{1+\frac{b^2}{9}}=\frac{\sqrt{13}}{3} \Rightarrow 9+b^2=13 \\
& \Rightarrow b=\pm 2
\end{aligned}
$$
Now, from $\mathrm{eq}^{\mathrm{n}}$ (1), we have
$$
\begin{aligned}
& \frac{K^2}{9}-\frac{4}{4}=1 \quad(\because b=\pm 2) \\
& \Rightarrow \quad K^2=18
\end{aligned}
$$

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2012 (07 May Online)
If the eccentricity of the hyperbola x21+sin⁡θ2-y2cos2⁡θ=1 is 23, then the sum of all the possible values of θ is (where, θ∈0,π)

Options:
A 5π4
B 2π3
C 7π4
D π

The correct answer is:
π

Solution:
e2=1+cos2⁡θ1+sin⁡θ2=43⇒cos2⁡θ1+sin⁡θ2=13
⇒cos⁡θ1+sin⁡θ2=13⇒sin⁡π2-θ1+cos⁡π2-θ2=13
⇒tan2⁡π4-θ2=13
π4-θ2=-π6,π6⇒θ2=π4-π6,π4+π6
θ=π2±π3=π6,5π6

Hence, the required sum =π

Topic: Mathematics, Hyperbola, JEE Main
If the eccentricity of the hyperbola x216-y2b2=-1 is 54, then b2 is equal to

Options:
A 2569
B 163
C 9
D 3

The correct answer is:
2569

Solution:
y2b2-x216=1

For the hyperbola, e=1+16b2
⇒1+16b2=54⇒b2=2569

Topic: Mathematics, Hyperbola, JEE Main
If the eccentricity of the hyperbola $x^{2}-y^{2} \operatorname{cosec}^{2} \alpha=25$ is $\sqrt{5}$ times the eccentricity of the ellipse $x^{2} \operatorname{cosec}^{2} \alpha+y^{2}=5$ then $\alpha$ is equal to :

Options:
A $\tan ^{-1} \sqrt{2}$
B $\sin ^{-1} \sqrt{\frac{3}{4}}$
C $\tan ^{-1} \sqrt{\frac{2}{5}}$
D $\sin ^{-1} \sqrt{\frac{2}{5}}$

The correct answer is:
$\tan ^{-1} \sqrt{2}$

Solution:
Eccentrcity of $\frac{x^{2}}{25}-\frac{y^{2}}{25 \sin ^{2} \alpha}=1$ is $\sqrt{1+\sin ^{2} \alpha}$

Eccentricity of $\frac{x^{2}}{5 \sin ^{2} \alpha}+\frac{y^{2}}{5}=1 \quad$ is $\sqrt{1-\sin ^{2} \alpha}$
Given, $\sqrt{1+\sin ^{2} \alpha}=\sqrt{5} \sqrt{1-\sin ^{2} \alpha}$
$$
\begin{array}{l}
\Rightarrow \sin ^{2} \alpha=\frac{2}{3} \\
\Rightarrow \alpha=\sin ^{-1} \sqrt{\frac{2}{3}}=\tan ^{-1} \sqrt{2}
\end{array}
$$

Topic: Mathematics, Hyperbola, JEE Main
If the eccentricity of the hyperbola x2-y2sec2⁡α=5 is 3 times the eccentricity of the ellipse x2sec2⁡α+y2=25, then tan2⁡α is equal to

Options:
A 2
B 1
C 3
D 12

The correct answer is:
1

Solution:
Let eccentricity of the hyperbola x25-y25cos2⁡α=1 is
e12=1+b2a2=1+5cos2⁡α5=1+cos2⁡α;
Similarly, eccentricity of the ellipse x225cos2⁡α+y225=1 is

e22=1-25cos2⁡α25=sin2⁡α;

Putting e1=3e2⇒e12=3e22
⇒1+cos2⁡α=3sin2⁡α⇒2=4sin2⁡α
⇒sin⁡α=12⇒tan2⁡α=1

Topic: Mathematics, Hyperbola, JEE Main
If the eccentricity of the standard hyperbola passing through the point (4,6) is 2, then the equation of the tangent to the hyperbola at (4,6) is:

Options:
A 2x-3y+10=0
B x-2y+8=0
C 3x-2y=0
D 2x-y-2=0

The correct answer is:
2x-y-2=0

Solution:
Let us suppose equation of hyperbola is
x2a2-y2b2=1
e=2

∵b2=e2-1a2

⇒b2=3a2
It is passing through 4,6

16a2-36b2=1

⇒a2=4, b2=12
So, equation of tangent is T=0

xx1a2-yy1b2=1

Tangent at point (4,6) is
⇒x-y2=1
⇒2x-y-2=0

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2019 (08 Apr Shift 2)
If the eccentricity of the two ellipse $\frac{x^2}{169}+\frac{y^2}{25}=1$ and $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are equal, then the value of $a / b$ is

Options:
A $5 / 13$
B $6 / 13$
C $13 / 5$
D $13 / 6$

The correct answer is:
$13 / 5$

Solution:
In the first case, eccentricity $e=\sqrt{1-(25 / 169)}$
In the second case, $e^{\prime}=\sqrt{1-\left(b^2 / a^2\right)}$ According to the given condition,
$\begin{aligned} & \sqrt{1-b^2 / a^2}=\sqrt{1-(25 / 169)} \\ & \Rightarrow b / a=5 / 13, \quad(\because a\gt0, b\gt0) \\ & \Rightarrow a / b=13 / 5 .\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
If the electric flux entering and leaving an enclosed surface respectively is $\phi_1$ and $\phi_2$, the electric charge inside the surface will be

Options:
A
$\left(\phi_2-\phi_1\right) \varepsilon_0$

B
$\left(\phi_1+\phi_2\right) / \varepsilon_0$

C
$\left(\phi_2-\phi_1\right) / \varepsilon_0$

D
$\left(\phi_1+\phi_2\right) \varepsilon_0$

The correct answer is:
$\left(\phi_2-\phi_1\right) \varepsilon_0$

Solution:
According to Gauss's Law
$$
\begin{aligned}
& \int(\text { E.dA })=\mathrm{q}_0 / \varepsilon_0 \Rightarrow \mathrm{q}=\varepsilon_0\left(\phi_2-\phi_1\right) \\
& {\left[\text { since } \phi=\int \mathrm{E} \cdot \mathrm{dA}\right]}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main
If the electric flux entering and leaving an enclosed surface respectively is $\phi_1$ and $\phi_2$, the electric charge inside the surface will be

Options:
A
$\left(\phi_2-\phi_1\right) \varepsilon_0$

B
$\left(\phi_1+\phi_2\right) / \varepsilon_0$

C
$\left(\phi_2-\phi_1\right) / \varepsilon_0$

D
$\left(\phi_1+\phi_2\right) \varepsilon_0$

The correct answer is:
$\left(\phi_2-\phi_1\right) \varepsilon_0$

Solution:
According to Gauss's Law
$$
\begin{aligned}
& \int(\text { E.dA })=\mathrm{q}_0 / \varepsilon_0 \Rightarrow \mathrm{q}=\varepsilon_0\left(\phi_2-\phi_1\right) \\
& {\left[\text { since } \phi=\int \mathrm{E} \cdot \mathrm{dA}\right]}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2003
If the electron in a hydrogen atom jumps from an orbit with level $\mathrm{n}_1=2$ to an orbit with level $\mathrm{n}_2=1$ the emitted radiation has a wavelength given by

Options:
A $\lambda=5 / 3 \mathrm{R}$
B $\lambda=4 / 3 \mathrm{R}$
C $\lambda=\mathrm{R} / 4$
D $\lambda=3 \mathrm{R} / 4$

The correct answer is:
$\lambda=4 / 3 \mathrm{R}$

Solution:
$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3 R}{4}$
$\Rightarrow \lambda=\frac{4}{3 R}$

Topic: Physics, Atomic Physics, JEE Main
If the electron in a hydrogen atom moves from ground state orbit to $5^{\text {th }}$ orbit, then the potential energy of the electron

Options:
A is increased
B is zero
C is decreased
D remains unchanged

The correct answer is:
is increased

Solution:
The potential energy of electron in hydrogen atom is given by
$$
\mathrm{U}=-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{e}^2}{\mathrm{r}}
$$
Hence as $r$ increase, potential energy will increase. (It will become less negative)

Topic: Physics, Atomic Physics, MHT CET, MHT CET 2021 (20 Sep Shift 2)
If the electron in the hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation in term of Rydberg constant is

Options:
A $\frac{6}{5 R}$
B $\frac{36}{5 R}$
C $\frac{64}{7 R}$
D None of these

The correct answer is:
$\frac{36}{5 R}$

Solution:
As we know $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)$ $\quad \frac{1}{\lambda}=\mathrm{R}\left(\frac{9-4}{36}\right)=\frac{5 \mathrm{R}}{36}$ $\therefore \quad \lambda=\frac{36}{5 \mathrm{R}}$

Topic: Physics, Atomic Physics, JEE Main
If the elevation in boiling point of a solution of $10 \mathrm{~g}$ of solute $(\mathrm{mol} . \mathrm{wt}=100)$ in $100 \mathrm{~g}$ of water is $\Delta T b$, the ebullioscopic constant of water is

Options:
A $\frac{\Delta T_{b}}{10}$
B $\Delta T_{b}$
C $10 \Delta T_{b}$
D $100 \Delta T_{b}$

The correct answer is:
$\Delta T_{b}$

Solution:
$\mathrm{m}=\frac{1000 \times \mathrm{k}_{b} \times \mathrm{w}}{W \times \Delta T_{b}}$
or $\quad \begin{aligned} k_{\mathrm{b}} &=\frac{\mathrm{m} \times W \times \Delta T_{b}}{1000 \times \mathrm{w}}=\frac{100 \times 100 \times \Delta T_{b}}{1000 \times 10} \\ &=\Delta T_{b} \end{aligned}$

Topic: Chemistry, Solutions, JEE Main
If the ellipse x2a2+y2b2=1 meets the line x7+y26=1 on the x-axis and the line x7-y26=1 on the y-axis, then the eccentricity of the ellipse is

Options:
A 57
B 267
C 37
D 257

The correct answer is:
57

Solution:
Given,

x2a2+y2b2=1 meets the line x7+y26=1 on the x-axis, so at x-axis y=0 putting in both equation and comparing we get, a=7

And x2a2+y2b2=1 meets the line x7-y26=1 on the y-axis where x=0 so putting in both equation and comparing we get, b=26

Therefore, e2=1-b2a2=1-2449

⇒e=2549=57

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2022 (25 Jul Shift 2)
If the energy of a photon is $10 \mathrm{eV}$, then its momentum is

Options:
A $5.33 \times 10^{-23} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
B $5.33 \times 10^{-25} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
C $5.33 \times 10^{-29} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
D $5.33 \times 10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$5.33 \times 10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

Solution:
Momentum of a photon $\propto \frac{\mathrm{E}}{\mathrm{c}}$
$$
=5.33 \times 10^{-27} \mathrm{~kg} \mathrm{~ms}^{-1}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
If the energy of photons corresponding to the wavelength of \( 6000 \mathrm{~A} \) is \( 3.2 \times 10^{-19} \mathrm{~J} \), the photon energy for a wavelength
of \( 4000 Å \) will be

Options:
A \( 1.11 \times 10^{-19} \mathrm{~J} \)
B \( 2.22 \times 10^{-19} \mathrm{~J} \)
C \( 4.40 \times 10^{-19} \mathrm{~J} \)
D \( 4.80 \times 10^{-19} \mathrm{~J} \)

The correct answer is:
\( 4.80 \times 10^{-19} \mathrm{~J} \)

Solution:
Given for photons wavelength λ1=6000 A∘ and energy E1=3.2×10-19 J and λ2=4000 A∘, let energy for second photons is E2.

We know energy of a photon is E=hcλ, then we can write

E2E1=λ1λ2 or E2=λ1λ2×E1, using given data,

E2=60004000×3.2×10-19, on solving we get,

E2=4.80×10-19 J

Topic: Physics, Dual Nature of Matter, JEE Main
If the enolate ion combines with carbonyl group of ester, we get

Options:
A Aldol
B $\alpha, \beta$-unsaturated ester
C $\quad \beta$-keto aldehyde
D Acid

The correct answer is:
$\quad \beta$-keto aldehyde

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
If the equal sides AB and AC (each equal to 5 units) of a right-angled isosceles triangle ABC are produced to P and Q such that BP.CQ=AB2, then the line PQ always passes through the fixed point (where A is the origin and AB, AC lie along the positive x and positive y-axis respectively)

Options:
A 7,6
B 6,5
C 5,5
D 6,6

The correct answer is:
5,5

Solution:
Let, AP=h, AQ=k
Equation of the line PQ is xh+yk=1 … 1
Given, BP.CQ=AB2
⇒h-5k-5=25
or, 5h+5k=hk
or, 5h+5k=1
From 2, it follows that line 1, i.e., PQ passes through the fixed point 5,5

Topic: Mathematics, Straight Lines, JEE Main
If the equation $a_n x^n+a_{n-1} x^{n-1}+\ldots .+a_1 x=0, \quad a_1 \neq 0$,$n \geq 2$, has a positive root $x=\alpha$, then the equation $n a_n x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots .+a_1=0$ has a positive root, which is

Options:
A Greater than or equal to $\alpha$
B Equal to $\alpha$
C Greater than $\alpha$
D Smaller than $\alpha$

The correct answer is:
Smaller than $\alpha$

Solution:
Let
$f(x)=a_n x^n+a_{n-1} x^{n-1}+\ldots .+a_1 x$
$f(0)=0 ; \quad f(\alpha)=0$
$\Rightarrow f^{\prime}(x)=0$, has atleast one root between $(0, \alpha)$
i.e., equation $n a_n x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots .+a_1=0$
has a positive root smaller than $\alpha$.

Topic: Mathematics, Quadratic Equation, JEE Main
If the equation
$a_n x^n+a_{n-1} x^{n-1}+\ldots \ldots+a_1 x=0, a_1 \neq 0, n \geq 2$, has a positive root $x=\alpha$, then the equation $n a_n x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots . .+a_1=0$ has a positive root, which is

Options:
A
greater than $\alpha$

B
smaller than $\alpha$

C
greater than or equal to $\alpha$

D
equal to $\alpha$

The correct answer is:
smaller than $\alpha$

Solution:
$f(0)=0, f(\alpha)=0$
$\Rightarrow f^{\prime}(k)=0$ for some $k \in(0, \alpha)$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2005
If the equation ax2+2bx-3c=0 has no real roots and 3c4<a+b, then

Options:
A c<0
B c>0
C c=0
D a+2b-3c<0

The correct answer is:
c<0

Solution:
ax2+2bx-3c=0 has no real roots

f2=4a+4b-3c
=4a+b-3c4>0 (given)
So, fx>0 ∀x
⇒f0>0⇒-3c>0⇒c<0

Topic: Mathematics, Quadratic Equation, JEE Main
If the equation a|z|2+α¯z+αz¯¯+d=0 represents a circle where a,d are real constants then which of the following condition is correct?

Options:
A |α|2-ad≠0
B |α|2-ad>0 and a∈R-{0}
C |α|2-ad≥0 and a∈R
D α=0,a,d∈R+

The correct answer is:
|α|2-ad>0 and a∈R-{0}

Solution:
Given a|z|2+α¯z+αz¯¯+d=0

∵z2=zz¯ & α¯z+αz¯¯=α¯z¯+αz¯¯=αz¯+α¯z

Then, azz¯+αz¯+α¯z+d=0 or zz¯+αaz¯+α¯az+da=0 is the equation of circle.

Centre =-αa, radius, r=αα¯a2-da=αα¯-ada2

αa2-da≥0 for the equation to represent circle.

So, |α|2-ad>0 and a∈R-{0}

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (18 Mar Shift 1)
If the equation \( a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x=0, a_{1} \neq 0, n \geq 2 \), has a positive root \( x=\alpha \), then the equation \( n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_{1}=0 \) has a positive root, which is

Options:
A Smaller then \( \alpha \)
B Greater then \( \alpha \)
C Equal to \( \alpha \)
D Greater then or equal \( \alpha \)

The correct answer is:
Smaller then \( \alpha \)

Solution:
Let fx=anxn+an-1xn-1+...+a1x ...1

∴ f⁡ 0 = 0
Now fα=0 Since, x=α is a root of differentiate equation 1 w.r.t. x, we get
f⁡ ' x = n a n x n - 1 + n - 1 a n - 1 x n - 2 + ... + a 1 = 0
Since, fα=f0=0.

So, according to Rolle's theorem f'x=0 somewhere in between 0 & α.
Hence, f'x=0 will have a root smaller than α.

Topic: Mathematics, Quadratic Equation, JEE Main