datavorous/entrance-exam-dataset huggingface dataset:
If the line \(2 x-1=0\) is the directrix of the parabola \(y^2-k x+6=0\), then one of the values of \(k\) is

Options:
A -6
B 6
C \(1 / 4\)
D \(-1 / 4\)

The correct answer is:
-6

Solution:
Given equation of parabola is \(y^2-k x+6=0\)
\(\Rightarrow y^2=k x-6 \Rightarrow y^2=k\left(x-\frac{6}{k}\right)\)
Now, directrix, \(x-\frac{6}{k}=-\frac{k}{4}\)
\(\Rightarrow x=\frac{6}{k}-\frac{k}{4}\) ...(i)
But directrix is given \(\Rightarrow x=\frac{1}{2}\) ...(ii)
\(\begin{aligned}
& \therefore \frac{6}{k}-\frac{k}{4}=\frac{1}{2} \Rightarrow 24-k^2=2 k \\
& \Rightarrow k^2+2 k-24=0 \\
& \Rightarrow(k+6)(k-4)=0 \quad \Rightarrow k=-6, k=4
\end{aligned}\)

Topic: Mathematics, Ellipse, BITSAT, BITSAT 2010
If the line \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\) lies in the plane \(2 x-4 y+z=7\), then the value of \(k\) is

Options:
A 4
B -7
C 7
D No real value

The correct answer is:
7

Solution:
The point \((4,2, k)\) on the line also lies on the plane \(2 x-4 y+z=7\)
So, \(8-8+\mathrm{k}=7 \Rightarrow \mathrm{k}=7\)

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2009
If the line $x \cos \alpha+y \sin \alpha=p$ represents the common chord of the circles $x^{2}+y^{2}=a^{2}$ and $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{b}^{2}(\mathrm{a}>\mathrm{b}),$ where $\mathrm{A}$ and $\mathrm{B}$ lie ont he first circle and $\mathrm{P}$ and $\mathrm{Q}$ lie on the second circle, then AP is equal to

Options:
A $\sqrt{a^{2}+p^{2}}+\sqrt{b^{2}+p^{2}}$
B $\sqrt{a^{2}-p^{2}}+\sqrt{b^{2}-p^{2}}$
C $\sqrt{a^{2}-p^{2}}-\sqrt{b^{2}-p^{2}}$
D $\sqrt{a^{2}+p^{2}}-\sqrt{b^{2}+p^{2}}$

The correct answer is:
$\sqrt{a^{2}-p^{2}}-\sqrt{b^{2}-p^{2}}$

Solution:
The given circles are concentric with centre at (0,0) and the length of the perpendicular from (0,0) on the given line is $\mathrm{p}$. Let $\mathrm{OL}=\mathrm{p}$
then, $\mathrm{AL}=\sqrt{\mathrm{OA}^{2}-\mathrm{OL}^{2}}=\sqrt{\mathrm{a}^{2}-\mathrm{p}^{2}}$
and $\mathrm{PL}=\sqrt{\mathrm{OP}^{2}-\mathrm{OL}^{2}}=\sqrt{\mathrm{b}^{2}-\mathrm{p}^{2}}$
$\Rightarrow \mathrm{AP}=\sqrt{\mathrm{a}^{2}-\mathrm{p}^{2}}-\sqrt{\mathrm{b}^{2}-\mathrm{p}^{2}}$

Topic: Mathematics, Circle, JEE Main
If the line $x \cos \alpha+y \sin \alpha=p$ represents the common chord of the circles $x^{2}+y^{2}=a^{2}$ and $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{b}^{2}(\mathrm{a}>\mathrm{b}),$ where $\mathrm{A}$ and $\mathrm{B}$ lie ont he first circle and $\mathrm{P}$ and $\mathrm{Q}$ lie on the second circle, then AP is equal to

Options:
A $\sqrt{a^{2}+p^{2}}+\sqrt{b^{2}+p^{2}}$
B $\sqrt{a^{2}-p^{2}}+\sqrt{b^{2}-p^{2}}$
C $\sqrt{a^{2}-p^{2}}-\sqrt{b^{2}-p^{2}}$
D $\sqrt{a^{2}+p^{2}}-\sqrt{b^{2}+p^{2}}$

The correct answer is:
$\sqrt{a^{2}-p^{2}}-\sqrt{b^{2}-p^{2}}$

Solution:
The given circles are concentric with centre at (0,0) and the length of the perpendicular from (0,0) on the given line is $\mathrm{p}$. Let $\mathrm{OL}=\mathrm{p}$
then, $\mathrm{AL}=\sqrt{\mathrm{OA}^{2}-\mathrm{OL}^{2}}=\sqrt{\mathrm{a}^{2}-\mathrm{p}^{2}}$
and $\mathrm{PL}=\sqrt{\mathrm{OP}^{2}-\mathrm{OL}^{2}}=\sqrt{\mathrm{b}^{2}-\mathrm{p}^{2}}$
$\Rightarrow \mathrm{AP}=\sqrt{\mathrm{a}^{2}-\mathrm{p}^{2}}-\sqrt{\mathrm{b}^{2}-\mathrm{p}^{2}}$

Topic: Mathematics, Circle, BITSAT, BITSAT 2012
If the lines \(3 x-4 y+4=0\) and \(6 x-8 y-7=0\) are tangents to a circle, then the radius of the circle is

Options:
A \(3 / 2\)
B \(3 / 4\)
C \(1 / 10\)
D \(1 / 20\)

The correct answer is:
\(3 / 4\)

Solution:
The diameter of the circle is perpendicular distance between the parallel lines (tangents) \(3 x-4 y+4=0\) and \(3 x-4 y-\frac{7}{2}=0\) and so it is equal to
\(\frac{4}{\sqrt{9+16}}+\frac{7 / 2}{\sqrt{9+16}}=\frac{3}{2}\). Hence radius is \(\frac{3}{4}\)

Topic: Mathematics, Circle, BITSAT, BITSAT 2011
If the lines $\mathrm{p}_{1} \mathrm{x}+\mathrm{q}_{1} \mathrm{y}=1, \mathrm{p}_{2} \mathrm{x}+\mathrm{q}_{2} \mathrm{y}=1$ and $\mathrm{p}_{3} \mathrm{x}+ \mathrm{q}_{3} \mathrm{y}=1$ be concurrent, then the points $\left(\mathrm{p}_{1}, \mathrm{q}_{1}\right),$

$\left(p_{2}, q_{2}\right)$ and $\left(p_{3}, q_{3}\right)$

Options:
A are collinear
B form an equilateral triangle
C form a sealene triangle
D form a right angled triangle

The correct answer is:
are collinear

Solution:
The equations of the lines are

$$

\begin{array}{l}

\mathrm{p}_{1} \mathrm{x}+\mathrm{q}_{1} \mathrm{y}-1=0 \\

\mathrm{p}_{2} \mathrm{x}+\mathrm{q}_{2} \mathrm{y}-1=0

\end{array}

$$

and $\mathrm{p}_{3} \mathrm{x}+\mathrm{q}_{3} \mathrm{y}-1=0$

As they are concurrent,

$$

\left|\begin{array}{lll}

\mathrm{p}_{1} & \mathrm{q}_{1} & -1 \\

\mathrm{p}_{2} & \mathrm{q}_{2} & -1 \\

\mathrm{p}_{3} & \mathrm{q}_{3} & -1

\end{array}\right|=0 \Rightarrow\left|\begin{array}{lll}

\mathrm{p}_{1} & \mathrm{q}_{1} & 1 \\

\mathrm{p}_{2} & \mathrm{q}_{2} & 1 \\

\mathrm{p}_{3} & \mathrm{q}_{3} & 1

\end{array}\right|=0

$$

This is also the condition for the points $\left(\mathrm{p}_{1}, \mathrm{q}_{1}\right),\left(\mathrm{p}_{2}, \mathrm{q}_{2}\right)$ and $\left(\mathrm{p}_{3}, \mathrm{q}_{3}\right)$ to be collinear.

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2015
If the lines $\ell \mathrm{x}+\mathrm{my}+\mathrm{n}=0, \mathrm{mx}+\mathrm{ny}+\ell=0$ and

$\mathrm{nx}+\ell \mathrm{y}+\mathrm{m}=0$ are concurrent then

Options:
A $\ell+m+n=0$
B $\ell-m-n=0$
C $\ell+m-n=0$
D $m+n-\ell=0$

The correct answer is:
$\ell+m+n=0$

Solution:
Since the lines are concurrent, so

$$

\Rightarrow \begin{array}{l}

\left|\begin{array}{lll}

\ell & \mathrm{m} & \mathrm{n} \\

\mathrm{m} & \mathrm{n} & \ell \\

\mathrm{n} & \ell & \mathrm{m}

\end{array}\right|=0 \Rightarrow 3 \ell \mathrm{mn}-\ell^{3}-\mathrm{m}^{3}-\mathrm{n}^{3}=0 \\

(\ell+\mathrm{m}+\mathrm{n})\left(\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}-(\mathrm{m}-\mathrm{mn}-\mathrm{n} \ell)=0\right.

\end{array}

$$

$\Rightarrow \quad \ell+\mathrm{m}+\mathrm{n}=0\left[\because \ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}>6 \mathrm{~m}+\mathrm{mn}+\mathrm{n}\right]$

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2013
If the matrix $\left[\begin{array}{ccc}1 & 3 & \lambda+2 \\ 2 & 4 & 8 \\ 3 & 5 & 10\end{array}\right]$ is singular, then $\lambda=$

Options:
A -2
B 4
C 2
D -4

The correct answer is:
4

Solution:
$|\mathrm{A}|=0$ as the matrix $\mathrm{A}$ is singular

$$

\therefore|\mathrm{A}|=\left|\begin{array}{ccc}

1 & 3 & \lambda+2 \\

2 & 4 & 8 \\

3 & 5 & 10

\end{array}\right|=0

$$

Apply $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{1}$

and expand. $-2(4-3 \lambda)+4(4-2 \lambda)=0$

$$

\Rightarrow 8-2 \lambda=0 \Rightarrow \lambda=4

$$

For $\lambda=4,$ the second and the third column are proportional.

Topic: Mathematics, Determinants, BITSAT, BITSAT 2016
If the mean, mode and S.D. of a frequency distribution are 41, 45 and 8 respectively, then its Pearson's coefficient of skewness is

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

The correct answer is:
\(\frac{-1}{2}\)

Solution:
\(-\frac{1}{2}\), We have \(\mathrm{S}_{\mathrm{K}}=\frac{\text { Mean }- \text { Mode }}{\mathrm{S} . \mathrm{D} \text {. }}\)
\(\frac{41-45}{8}=-\frac{1}{2}\)

Topic: Mathematics, Probability, BITSAT, BITSAT 2010
If the middle points of sides $\mathrm{BC}, \mathrm{CA} \& \mathrm{AB}$ of triangle $\mathrm{ABC}$ are respectively $\mathrm{D}, \mathrm{E}, \mathrm{F}$ then position vector of centre of triangle $\mathrm{DEF}$, when position vector of $A, \quad B, \quad C$ are respectively $\hat{i}+\hat{j}, \hat{j}+\hat{k}, \hat{k}+\hat{i}$ is

Options:
A $\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})$
B $(\hat{i}+\hat{j}+\hat{k})$
C $2(\hat{i}+\hat{j}+\hat{k})$
D $\frac{2}{3}(\hat{i}+\hat{j}+\hat{k})$

The correct answer is:
$\frac{2}{3}(\hat{i}+\hat{j}+\hat{k})$

Solution:
The position vector of points $D, E, F$ are respectively $\hat{i}+\hat{j}+\hat{k}, \hat{i}+\frac{\hat{k}+\hat{j}}{2}$ and $\frac{\hat{i}+\hat{k}}{2}+\hat{j}$

So, position vector of centre of $\Delta \mathrm{DEF}$ $=\frac{1}{3}\left[\frac{\hat{i}+\hat{j}}{2}+\hat{k}+\hat{i} \frac{\hat{k}+\hat{j}}{2}+\frac{\hat{i}+\hat{k}}{2}+\hat{j}\right]$

$=\frac{2}{3}[\hat{i}+\hat{j}+\hat{k}]$

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2014
If the middle points of sides \(\mathrm{BC}, \mathrm{CA} ~\&~ \mathrm{AB}\) of triangle \(A B C\) are respectively \(D, E, F\) then position vector of centre of triangle \(\mathrm{DEF}\), when position vector of \(A, B, C\) are respectively \(\hat{i}+\hat{j}\), \(\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{k}}+\hat{\mathrm{i}}\) is -

Options:
A \((1 / 3)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
B \((\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
C \(2(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
D \((2 / 3)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)

The correct answer is:
\((2 / 3)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)

Solution:
The position vector of points \(D, E, F\) are respectively
\(\frac{\mathrm{i}+\mathrm{j}}{2}+\mathrm{k}, \mathrm{i}+\frac{\mathrm{k}+\mathrm{j}}{2} \text { and } \frac{\mathrm{i}+\mathrm{k}}{2}+\mathrm{j}\)
So, position vector of centre of \(\triangle \mathrm{DEF}\)
\(=\frac{1}{3}\left[\frac{\mathrm{i}+\mathrm{j}}{2}+\mathrm{k}+\mathrm{i} \frac{\mathrm{k}+\mathrm{j}}{2}+\frac{\mathrm{i}+\mathrm{k}}{2}+\mathrm{j}\right]=\frac{2}{3}[\mathrm{i}+\mathrm{j}+\mathrm{k}]\)

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2010
If the normal to the curve $y=f(x)$ at the point (3,4) makes an angle $3 \pi / 4$ with the positive $\mathrm{x}$ -axis, then $\mathrm{f}^{\prime}(3)=$

Options:
A -1
B $-3 / 4$
C $4 / 3$
D 1

The correct answer is:
1

Solution:
Slope of normal to $y=f(x)$ at (3,4) is $\frac{-1}{f^{\prime}(3)}$.
$\text { Thus, } \begin{array}{c}
\frac{-1}{\mathrm{f}^{\prime}(3)}=\tan \left(\frac{3 \pi}{4}\right)=\tan \left(\frac{\pi}{2}+\frac{\pi}{4}\right) \\
=-\cot \frac{\pi}{4}=-1 \Rightarrow \mathrm{f}^{\prime}(3)=1
\end{array}$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2012
If the papers of 4 students can be checked by any one of the 7 teachers, then the probability that all the 4 papers are checked by exactly 2 teachers, is equal to

Options:
A 1249
B 649
C 949
D 1549

The correct answer is:
649

Solution:
Total ways in which papers can be checked is equal 74. Now, two teachers who have to check all the papers can be selected in C27 ways and papers can be checked by them is 24-2 favourable ways.

Thus, required probability

=C2724-274=649

Topic: Mathematics, Probability, BITSAT, BITSAT 2017
If the photon of wavelength 150 pm strikes an atom and one of its inner bound electron is ejected out with a velocity of 1.5×107 m/s, what is the energy with which it is bound to the nucleus?

Options:
A 1.2×102 eV
B 2.15×103 eV
C 7.6×103 eV
D 8.12×103 eV

The correct answer is:
7.6×103 eV

Solution:
Total energy E of any photon is given by the relation : E=hcλ

Where, h= Planck's constant =6.6×10-34 Js

c= Velocity of light =3×108 m/s

λ= wavelength =1.5×10-10 m, i.e, 150 pm

Thus, E=hcλ=6.6×10-34×3×1081.5×10-10; E=1.32×10-15 J and, energy of ejected electron E'

E'=12mv2,

where,

m= mass of electron =9.1×10-31 kg

v= velocity of electron

=1.5×107 m/s

E'=12mv2=12×9.1×10-31×1.5×1072

E'=1.024×10-16

Thus, total energy of photon = Binding energy of electron

B + energy of ejected electron E'

Thus, 1.32×10-15=B+E

∴ B=E-E'=1.32×10-15-1.024×10-16

=1.2176×10-15 J

=1.2176×10-151.6×10-19 eV=7.6×103 eV

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2019
If the plane $3 x+y+2 z+6=0$ is parallel to the line $\frac{3 x-1}{2 b}=3-y=\frac{z-1}{a}$, then the value of $3 a+3 b$ is

Options:
A
$\frac{1}{2}$

B
$\frac{3}{2}$

C
3

D
4

The correct answer is:

$\frac{3}{2}$

Solution:
Given plane $3 x+y+2 z+6=0$
and line $\frac{x-\frac{1}{3}}{\frac{2 b}{3}}=\frac{y-3}{-1}=\frac{z-1}{a}$
Since, plane is parallel to line, then
$$
\begin{aligned}
& 3\left(\frac{2 b}{3}\right)+(1)(-1)+2(a)=0 \\
& 2 b-1+2 a=0 \\
& \Rightarrow \quad a+b=\frac{1}{2} \\
&
\end{aligned}
$$
Now, $3 a+3 b=\frac{3}{2}$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2022
If the planes r→·2i^-λj^+3k^=0 and r→·λi^+5j^-k^=5 are perpendicular to each other, then the value of λ2+λ is

Options:
A 0
B 2
C 1
D 3

The correct answer is:
0

Solution:
We have,

r→·λi^+5j^-k^=5

r→·2i^-λj^+3k^=0

We know that two planes r→·n→1=d1 and r→·n→2=d2 are perpendicular if

n→1·n→2=0

Therefore,

λi^+5j^-k^·2i^-λj^+3k^=0

⇒2λ-5λ-3=0

⇒-3λ-3=0

⇒λ=-1

Therefore, λ2+λ=0

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2018
If the potential of a capacitor having capacity $6 \mu \mathrm{F}$ is increased from $10 \mathrm{~V}$ to $20 \mathrm{~V}$, then increase in its energywill be

Options:
A $4 \times 10^{-4} \mathrm{~J}$
B $4 \times 10^{-6} \mathrm{~J}$
C $9 \times 10^{-4} \mathrm{~J}$
D $12 \times 10^{-6} \mathrm{~J}$

The correct answer is:
$9 \times 10^{-4} \mathrm{~J}$

Solution:
Capacitance of capacitor (C) $=6 \mu \mathrm{F}$ $=6 \times 10^{-6} \mathrm{~F}$; Initial potential $\left(\mathrm{V}_{1}\right)=10 \mathrm{~V}$ and
final potential $\left(\mathrm{V}_{2}\right)=20 \mathrm{~V}$.
The increase in energy $(\Delta \mathrm{U})$
$$
\begin{array}{l}
=\frac{1}{2} \mathrm{C}\left(\mathrm{V}_{2}^{2}-\mathrm{V}_{1}^{2}\right) \\
=\frac{1}{2} \times\left(6 \times 10^{-6}\right) \times\left[(20)^{2}-(10)^{2}\right] \\
=\left(3 \times 10^{-6}\right) \times 300=9 \times 10^{-4} \mathrm{~J} .
\end{array}
$$

Topic: Physics, Capacitance, BITSAT, BITSAT 2011
If the radius of a sphere is measured as $9 \mathrm{~cm}$ with an error of $0.03 \mathrm{~cm}$, then find the approximating error in calculating its volume.

Options:
A $2.46 \pi \mathrm{cm}^{3}$
B $8.62 \pi \mathrm{cm}^{3}$
C $9.72 \pi \mathrm{cm}^{3}$
D $7.6 \pi \mathrm{cm}^{3}$

The correct answer is:
$9.72 \pi \mathrm{cm}^{3}$

Solution:
Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius. Then, $r=9 \mathrm{~cm}$ and $\Delta r=0.03 \mathrm{~cm}$

Let $V$ be the volume of the sphere. Then,

$\begin{array}{l}

\mathrm{V}=\frac{4}{3} \pi r^{3} \\

\Rightarrow \frac{\mathrm{dV}}{\mathrm{dr}}=4 \pi r^{2} \\

\Rightarrow\left(\frac{\mathrm{dV}}{\mathrm{dr}}\right)_{r=9}=4 \pi \times 9^{2}=324 \pi

\end{array}$

Let $\Delta \mathrm{V}$ be the error in $\mathrm{V}$ due to error $\Delta \mathrm{r}$ in $r$.

Then,

$\begin{array}{l}

\Delta \mathrm{V}=\frac{\mathrm{dV}}{\mathrm{dr}} \Delta r \\

\Rightarrow \Delta \mathrm{V}=324 \pi \times 0.03=9.72 \pi \mathrm{cm}^{3}

\end{array}$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2021
If the radius of a star is $R$ and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is $Q$ ?

Options:
A $Q / 4 \pi R^{2} \sigma$
B $\left(\mathrm{Q} / 4 \pi R^{2} \sigma\right)^{-1 / 2}$
C $\left(4 \pi R^{2} Q / \sigma\right)^{1 / 4}$
D $\left(\mathrm{Q} / 4 \pi R^{2} \sigma\right)^{1 / 4}$

The correct answer is:
$\left(\mathrm{Q} / 4 \pi R^{2} \sigma\right)^{1 / 4}$

Solution:
Stefan's law for black bodyradiation $Q=\sigma e A T^{4}$

$$

T=\left[\frac{Q}{\sigma\left(4 \pi R^{2}\right)}\right]^{1 / 4}

$$

Here $e=1$

$$

A=4 \pi R^{2}

$$

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2013
If the radius of $\mathrm{H}$ is $0.53 Ã…$, then what will be the radius of ${ }_{3} \mathrm{Li}^{2+}$?

Options:
A $0.17 Ã…$
B $0.36 Ã…$
C $0.53 Ã…$
D $0.59 Ã…$

The correct answer is:
$0.17 Ã…$

Solution:
Radius of orbit $=\frac{\mathrm{n}^{2} \mathrm{a}_{\mathrm{o}}}{\mathrm{z}}\left(\mathrm{a}_{\mathrm{o}}=0.529 \mathrm{~A}\right)$
Radius of $\mathrm{H}=\frac{(1)^{2} \times 0.529 \mathrm{~A}}{1}=0.53 \mathrm{~A}$
Thus, the radius of ${ }_{3} \mathrm{Li}^{2+}$ will be $=\frac{(1)^{2} \times 0.529}{3}=0.17 Ã…$

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2012
If the radius of the $3 \mathrm{rd}$ Bohr's orbit of hydrogen atom is $r_3$ and the radius of $4^{\text {th }}$ Bohr's orbit is $r_4$. Then:

Options:
A $r_4=\frac{9}{16} r_3$
B $r_4=\frac{16}{9} r_3$
C $r_4=\frac{3}{4} r_3$
D $r_4=\frac{4}{3} r_3$

The correct answer is:
$r_4=\frac{16}{9} r_3$

Solution:
$\begin{aligned} & r=0.529 \times \frac{n^2}{Z} Å \\ & r_3=0.529 \times \frac{3^2}{1} \Rightarrow r_4=0.529 \times \frac{4^2}{1} \\ & \frac{r_4}{r_3}=\frac{4^2}{3^2}=\frac{16}{9} \Rightarrow r_4=\frac{16 r_3}{9}\end{aligned}$

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If the radius of the earth were to shrink by one per cent, its mass remaining the same, the value of g on the earth's surface would

Options:
A increase by $0.5 \%$
B increase by $2 \%$
C decrease by $0.5 \%$
D decrease by $2 \%$

The correct answer is:
increase by $2 \%$

Solution:
$\quad \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \Rightarrow \frac{\mathrm{dg}}{\mathrm{g}}=-2 \frac{\mathrm{dR}}{\mathrm{R}}$
$\frac{\mathrm{dR}}{\mathrm{R}}=-1 \% \Rightarrow \frac{\mathrm{dg}}{\mathrm{g}}=2 \%$

Topic: Physics, Gravitation, BITSAT, BITSAT 2012
If the real part of $\frac{\bar{z}+2}{\bar{z}-1}$ is $4, z \neq 1,$ then the locus of the point representing $z$ in the complex plane is

Options:
A a straight line parallel to $x$ -axis
B a straight line equally inclined to axes
C a circle with radius 2
D a circle with radius $\frac{1}{2}$

The correct answer is:
a circle with radius $\frac{1}{2}$

Solution:
Real part of $\frac{\bar{z}+2}{\bar{z}-1}$ is given by

$$

\begin{array}{l}

\frac{1}{2}\left[\frac{\bar{z}+2}{\bar{z}-1}+\left(\frac{\bar{z}+2}{\bar{z}-1}\right)\right]=4 \\

\Rightarrow \frac{\bar{z}+2}{\bar{z}-1}+\frac{z+2}{z-1}=8 \\

\Rightarrow z \bar{z}-\bar{z}+2 z-2+z \bar{z}+2 \bar{z}-z-2 \\

=8(z \bar{z}-\bar{z}-z+1) \\

\Rightarrow z \bar{z}-\frac{3}{2} z-\frac{3}{2} \bar{z}+2=0

\end{array}

$$

Comparing with the equation $z \bar{z}+\bar{a} z+a \bar{z}+b=0,$ we get $a=-\frac{3}{2}$ and $b$

$=2$. Thus, the locus of $z$ given by the equation

(i) is a circle with centre $=\frac{1}{2}$

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2013
If the series limit wavelength of Lyman series for the hydrogen atom is $912 Å,$ then the series limit wavelength for Balmer series of hydrogen atoms is

Options:
A $912 Å$
B $912 \times 2 Å$
C $912 \times 4 Å$
D $\frac{912}{2} Å$

The correct answer is:
$912 \times 4 Å$

Solution:
$\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$

For limiting wavelength of Lyman series

$$

n_{1}=1, n_{2}=\infty \quad \frac{1}{\lambda_{L}}=R

$$

For limiting wavelength of Balmer series $n_{1}=2, n_{2}=\infty$

$$

\begin{array}{l}

\frac{1}{\lambda_{B}}=R\left(\frac{1}{4}\right) \Rightarrow \lambda_{B}=\frac{4}{R} \\

\therefore \lambda_{B}=4 \lambda_{L}=4 \times 912 \AA

\end{array}

$$

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2016
If the slope of the tangent to the curve at any point $\mathrm{P}(x, y)$ is $\frac{y}{x}-\cos ^{2} \frac{y}{x}$, then the equation of a curve passing through $\left(1, \frac{\pi}{4}\right)$ is

Options:
A $\tan \left(\frac{y}{x}\right)+\log x=1$
B $\tan \left(\frac{y}{x}\right)+\log y=1$
C $\tan \left(\frac{x}{y}\right)+\log x=1$
D $\tan \left(\frac{x}{y}\right)+\log y=1$

The correct answer is:
$\tan \left(\frac{y}{x}\right)+\log x=1$

Solution:
According to the condition,

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y}{x}-\cos ^{2} \frac{y}{x} \ldots \text { (i) }$

This is a homogeneous differential equation Substituting $y=v x$, we get

$\begin{array}{l}

v+x \frac{\mathrm{dv}}{\mathrm{dx}}=v-\cos ^{2} v \\

\Rightarrow x \frac{\mathrm{dv}}{\mathrm{dx}}=-\cos ^{2} v \\

\Rightarrow \int \sec ^{2} \mathrm{vdv}=-\int \frac{\mathrm{dx}}{\mathrm{x}} \\

\Rightarrow \tan v=-\log x+\mathrm{C} \\

\Rightarrow \tan \frac{y}{x}+\log x=\mathrm{C}

\end{array}$

Substituting $\mathrm{x}=1, y=\frac{\pi}{4}$, we get $\mathrm{C}=1$

Thus, we get

$\tan \left(\frac{y}{x}\right)+\log x=1$

which is the required solution.

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2021
If the straight line $y=m x+c$ touches the parabola $y^2-4 a x+4 a^3=0$, then $c$ is

Options:
A
$a m+\frac{a}{m}$

B
$a m-\frac{a}{m}$

C
$\frac{a}{m}+a^2 m$

D
$\frac{a}{m}-a^2 m$

The correct answer is:

$\frac{a}{m}-a^2 m$

Solution:
Solving the given equations,
$$
\begin{aligned}
&(m x+c)^2=4 a x-4 a^3 \\
& \Rightarrow m^2 x^2+2 m c \cdot x+c^2=4 a x-4 a^3 \\
& \Rightarrow m^2 x^2+(2 m c-4 a) x+c^2+4 a^3=0
\end{aligned}
$$
Since the straight line touches the parabola at a point, so, the discriminant $=0$
$$
\begin{aligned}
& \Rightarrow \quad(2 m c-4 a)^2-4 m^2\left(c^2+4 a^3\right)=0 \\
& \Rightarrow 4 m^2 c^2-16 a m c+16 a^2-4 m^2 c^2-16 a^3 m^2=0 \\
& \Rightarrow-m c+a-a^2 m^2=0 \\
& \Rightarrow m c=a-a^2 m^2 \Rightarrow c=\frac{a}{m}-a^2 m
\end{aligned}
$$

Topic: Mathematics, Parabola, BITSAT, BITSAT 2022
If the straight lines $\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2}$ intersect at a point, then the integer $\mathrm{k}$ is equal to

Options:
A
-5
B
5
C
2
D
-2

The correct answer is:
-5

Solution:
When the two lines intersect then shortes distance between them is zero i.e.
$\begin{aligned}
& \frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot \vec{b}_1 \times \vec{b}_2}{\left|\vec{b}_1 \times \vec{b}_2\right|}=0 \\
& \Rightarrow\left(\vec{a}_2-\vec{a}_1\right) \cdot \vec{b}_1 \times \vec{b}_2=0
\end{aligned}$
where $\vec{a}_1=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}_1=k \hat{i}+2 \hat{j}+3 \hat{k}$
$\begin{aligned}
& \vec{a}_2=2 \hat{i}+3 \hat{j}+\hat{k}, \hat{b}_2=3 \hat{i}+k \hat{j}+2 \hat{k} \\
& \Rightarrow\left|\begin{array}{ccc}
1 & 1 & -2 \\
k & 2 & 3 \\
3 & k & 2
\end{array}\right|=0 \\
& \Rightarrow \quad 1(4-3 k)-1(2 k-9)-2\left(k^2-6\right)=0 \\
& \Rightarrow \quad-2 k^2-5 k+25=0 \Rightarrow k=-5 \text { or } \frac{5}{2}
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If the sum of a certain number of terms of the A.P. $25,22,19, \ldots$ is 116 then the last term is

Options:
A 0
B 2
C 4
D 6

The correct answer is:
4

Solution:
$a=25, d=22-25=-3 \text {. }$

Let $n$ be the no. of terms

$\begin{array}{l}

\text { Sum }=116 ; \\

\text { Sum }=\frac{n}{2}[2 a+(n-1) d] \\

116=\frac{n}{2}[50+(n-1)(-3)] \\

\text { or } 232=n[50-3 n+3]=n[53-3 n]=-3 n^{2}+53 n \\

\Rightarrow 3 n^{2}-53+232=0 \\

\Rightarrow(n-8)(3 n-29)=0 \\

\Rightarrow n=8 \text { or } n=\frac{29}{3}, \\

n \neq \frac{29}{3} \\

\therefore n=8 \\

\therefore \text { Now, } \mathrm{T}_{8}=a+(8-1) d=25+7 \times(-3)=25-21=4 \\

\therefore \text { Last term }=4

\end{array}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2021
If the sum of an infinite GP $a, a r, a r^2, a r^3, \ldots$ is 15 and the sum of the squares of its each term is 150 , then the sum of $a r^2, a r^4, a r^6, \ldots$ is :

Options:
A $\frac{5}{2}$
B $\frac{1}{2}$
C $\frac{25}{2}$
D $\frac{9}{2}$

The correct answer is:
$\frac{1}{2}$

Solution:
Sum of infinite terms of series a $+a r+a r^2+$ $=15$
$\therefore \frac{a}{1-r}=15$ ...(i)
Sequence formed by square of terms: $a^2, a^2 r^2, a^2 r^4, a^2 r^6 \ldots$
$a^2+a^2 r^2+\cdots=150$
Sum $=\frac{a^2}{1-r^2}=150$
$\Rightarrow \frac{a}{1-r} \cdot \frac{a}{1+r}=15 \Rightarrow 15 \cdot \frac{a}{1+r}=150$ $\left(\because \frac{\mathrm{a}}{1-\mathrm{r}}=15\right)$
$\Rightarrow \frac{a}{1+r}=10$ ...(ii)
On dividing equation (i) by (ii)
$\frac{1+\mathrm{r}}{1-\mathrm{r}}=\frac{15}{10}$
or $r=\frac{1}{5} \Rightarrow a=12$
Now series : $a r^2+a r^4+a r^6+\ldots$
Sum $=\frac{a^2}{1-r^2}=\frac{12 \cdot\left(\frac{1}{25}\right)}{1-\frac{1}{25}}=\frac{1}{2}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2023 (Memory Based Paper 1)