datavorous/entrance-exam-dataset huggingface dataset:
The displacement of a particle is given at time t, by:

$x=A \sin (-2 \omega t)+B \sin ^{2} \omega t \quad$ Then,

Options:
A the motion of the particle is SHM with an amplitude of $\sqrt{A^{2}+\frac{B^{2}}{4}}$
B the motion of the particle is not SHM, but oscillatory with a time period of $T=\pi / \omega$
C the motion of the particle is oscillatory with a time period of $T=\pi / 2 \omega$
D the motion of the particle is a periodic.

The correct answer is:
the motion of the particle is SHM with an amplitude of $\sqrt{A^{2}+\frac{B^{2}}{4}}$

Solution:
The displacement of the particle is given by:

$$

\begin{aligned}

x &=A \sin (-2 \omega t)+B \sin ^{2} \omega t \\

&=-A \sin 2 \omega t+\frac{B}{2}(1-\cos 2 \omega t) \\

&=-\left(A \sin 2 \omega t+\frac{B}{2} \cos 2 \omega t\right)+\frac{B}{2}

\end{aligned}

$$

This motion represents SHM with an amplitude: $\sqrt{A^{2}+\frac{B^{2}}{4}},$ and mean position

$$

\frac{B}{2}

$$

Topic: Physics, Oscillations, BITSAT, BITSAT 2015
The displacement y of a particle in a medium can be expressed as, $y=10^{-6} \sin \left(100 t+20 x+\frac{\pi}{4}\right) m$ where $t$ is in second and $x$ in meter. The speed of the wave is

Options:
A $20 \mathrm{~m} / \mathrm{s}$
B $5 \mathrm{~m} / \mathrm{s}$
C $2000 \mathrm{~m} / \mathrm{s}$
D $5 \pi \mathrm{m} / \mathrm{s}$

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

Solution:
From equation given,

$\begin{array}{l}

\omega=100 \\

\text { and } \mathrm{k}=20 \\

\mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{100}{20}=5 \mathrm{~m} / \mathrm{s}

\end{array}$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2021
The distance between the origin and the normal to curve y=e2x+x2 at x=0 is

Options:
A 23
B 25
C 13
D 15

The correct answer is:
25

Solution:
We have,

y=e2x+x2

at x=0,y=1

dydx=2e2x+2x

dydxx=0=2

∴ Equation of normal at (0,1) is

y-1=-12(x)

⇒ x+2y-2=0

∴ Distance from (0,0) from normal =25

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2019
The distance between the parallel lines $3 x-4 y+7=0$ and $3 x-4 y+5=0$ is $\frac{a}{b}$. Value of $a+b$ is

Options:
A 2
B 5
C 7
D 3

The correct answer is:
7

Solution:
Given parallel lines are

$3 x-4 y+7=0$ and $3 x-4 y+5=0$

Required distance $=\frac{|7-5|}{\sqrt{(3)^{2}+(-4)^{2}}}=\frac{2}{5}$ $\Rightarrow a=2, b=5$

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2021
The distance from the point (3,4,5) to the point where the line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ meets the plane $x+y+z=17$ is

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

The correct answer is:
3

Solution:
Anypoint on the line is $(r+3,2 r+4,2 r+5)$. It lies on the plane $x+y+z=17$ $\therefore(r+3)+(2 r+4)+(2 r+5)=17$ i.e $r=1$

Thus the point of intersection of the plane and the line is (4,6,7) Required distance $=$ distance between (3,4,5) and (4,6,7)

$=\sqrt{\left\{(4-3)^{2}+(6-4)^{2}+(7-5)^{2}\right\}}=3$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2020
The distance of the point \((-1,1)\) from the line \(12(x+6)=5(y-2)\) is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
5

Solution:
The given line is \(12(x+6)=5(y-2)\)
\(\Rightarrow 12 x+72=5 y-10\)
or \(12 x-5 y+72+10=0 \Rightarrow 12 x-5 y+82=0\) The perpendicular distance from \(\left(x_1, y_1\right)\) to the line \(a x+b y+c=0\) is \(\frac{\left(\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}\right)}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\). The point \(\left(x_1, y_1\right)\) is \((-1,1)\), therefore, perpendicular distance from \((-1,1)\) to the line \(12 x-5 y+82=0\) is
\(=\frac{|-12-5+82|}{\sqrt{12^2+(-5)^2}}=\frac{65}{\sqrt{144+25}}=\frac{65}{\sqrt{169}}=5\)

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2011
The distances of the point (1,-5,9) from the plane x+y+z=5 measured along a straight line x=y=z is 23k, then the value of k is

Options:
A 5
B 6
C 3
D 4

The correct answer is:
5

Solution:
Given equation of plane is x+y+z=5.

The distance measured along the line x=y=z.

Direction ratios of the given line is (1,1,1).

So, the equation of line PQ is

x-11=y+51=z-91=λ

⇒ x=λ+1,y=λ-5,z=λ+9

lies on the plane x+y+z=5

∴ (λ+1)-(-λ+5)+(λ+9)=5

⇒ λ=-10

The coordinate of Q is (-9,-15,-1) and the coordinate of P is (1,-5,9)

PQ=(10)2+(10)2+(10)2=103

∴ 23k=103⇒k=5

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2017
The domain of the function

$f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^{2}}}$

Options:
A [1,2]
B [2,3)
C [1,2]
D [2,3]

The correct answer is:
[2,3)

Solution:
$f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^{2}}}$ is defined

if (i) $-1 \leq x-3 \leq 1 \Rightarrow 2 \leq x \leq 4$ and

(ii) $9-x^{2}>0 \Rightarrow-3
Taking common solution of (i) and (ii), we get $2 \leq x<3$

$\therefore$ Domain $=[2,3)$

Topic: Mathematics, Functions, BITSAT, BITSAT 2015
The domain of the function $f(x)=\log _{2}\left(-\log _{1 / 2}\left(1+\frac{1}{x^{1 / 4}}\right)-1\right)$ is

Options:
A (0,1)
B (0,1]
C $[1, \infty)$
D $(1, \infty)$

The correct answer is:
(0,1)

Solution:
$f(x)$ is defined if

$\begin{array}{l} -\log _{1 / 2}\left(1+\frac{1}{x^{1 / 4}}\right)-1 > 0 \\ \Rightarrow \quad \log _{1 / 2}\left(1+\frac{1}{x^{1 / 4}}\right) < -1 \\ \Rightarrow \quad 1+\frac{1}{x^{1 / 4}} > \left(\frac{1}{2}\right)^{-1} \\ \Rightarrow \quad \frac{1}{x^{1 / 4}} > 1 \\ \Rightarrow \quad 0 < x < 1 \end{array}$

Topic: Mathematics, Functions, BITSAT, BITSAT 2014
The domain of the function $f(x)=\sqrt{x-\sqrt{1-x^2}}$ is

Options:
A $\left[-1,-\frac{1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, 1\right]$
B $[-1,1]$
C $\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{1}{\sqrt{2}},+\infty\right)$
D $\left[\frac{1}{\sqrt{2}}, 1\right]$

The correct answer is:
$\left[\frac{1}{\sqrt{2}}, 1\right]$

Solution:
For $f(x)$ to be defined, we must have
$x-\sqrt{1-x^2} \geq 0 \text { or } x \geq \sqrt{1-x^2}>0 \text { or } x^2 \geq 1-x^2$
or $x^2 \geq \frac{1}{2}$.
Now, $x^2 \geq \frac{1}{2} \Rightarrow\left(x-\frac{1}{\sqrt{2}}\right)\left(x+\frac{1}{\sqrt{2}}\right) \geq 0$
$\Rightarrow x \leq-\frac{1}{\sqrt{2}} \quad \text { or } \quad x \geq \frac{1}{\sqrt{2}}$
Also, $1-x^2 \geq 0$ or $x^2 \leq 1$.
Also, $x^2 \leq 1 \Rightarrow(x-1)(x+1) \leq 0 \Rightarrow-1 \leq x \leq 1$
Thus, $x>0, x^2 \geq \frac{1}{2}$ and $x^2 \leq 1 \Rightarrow x \in\left[\frac{1}{\sqrt{2}}, 1\right]$

Topic: Mathematics, Functions, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The domain of the function $f(x)=\sqrt{x-\sqrt{1-x^{2}}}$ is

Options:
A $\left[-1,-\frac{1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, 1\right]$
B [-1,1]
C $\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{1}{\sqrt{2}},+\infty\right)$
D $\left[\frac{1}{\sqrt{2}}, 1\right]$

The correct answer is:
$\left[\frac{1}{\sqrt{2}}, 1\right]$

Solution:
For $f(x)$ to be defined, we must have $x-\sqrt{1-x^{2}} \geq 0$ or $\quad x \geq \sqrt{1-x^{2}}>0$

$\therefore \quad x^{2} \geq 1-x^{2}$ or $x^{2} \geq \frac{1}{2}$

Also, $1-x^{2} \geq 0$ or $x^{2} \leq 1$.

Now, $x^{2} \geq \frac{1}{2} \Rightarrow\left(x-\frac{1}{\sqrt{2}}\right)\left(x+\frac{1}{\sqrt{2}}\right) \geq 0$

$\Rightarrow x \leq-\frac{1}{\sqrt{2}}$ or $x \geq \frac{1}{\sqrt{2}}$

Also, $x^{2} \leq 1 \Rightarrow(x-1)(x+1) \leq 0$

$\Rightarrow-1 \leq x \leq 1$

Thus, $x>0, x^{2} \geq \frac{1}{2}$ and $x^{2} \leq 1$

$\Rightarrow x \in\left[\frac{1}{\sqrt{2}}, 1\right]$

Topic: Mathematics, Functions, BITSAT, BITSAT 2020
The domain of the function $\mathrm{f}(x)=\frac{1}{\sqrt{x^{12}-x^{9}+x^{4}-x+1}}$ is given by

Options:
A $(-\infty,-1)$
B $(1, \infty)$
C $(-1,1)$
D $(-\infty, \infty)$

The correct answer is:
$(-\infty, \infty)$

Solution:
$\mathrm{f}(x)$ is defined for $x^{12}-x^{9}+x^{4}-x+1>0$

$\begin{array}{l}

\Rightarrow x^{4}\left(x^{8}+1\right)-x\left(x^{8}+1\right)+1>0 \\

\Rightarrow\left(x^{8}+1\right) x\left(x^{3}-1\right)+1>0

\end{array}$

If $x \geq 1$ or $x \leq-1$, then the above expres - sion is positive.

If $-1 < x \leq 0$, the above inequality still holds.

If $0 < x < 1$, then

$\begin{array}{l}

x^{12}-x\left(x^{8}+\right)+\left(x^{4}+1\right)>0 \\

\because\left[x^{4}+1>x^{8}+1 \text { and so } x^{4}+1>x\left(x^{8}+1\right)\right]

\end{array}$

The domain of $f=(-\infty, \infty)$

Topic: Mathematics, Functions, BITSAT, BITSAT 2021
The drift velocity of electrons in silver wire with cross-sectional area $3.14 \times 10^{-6} \mathrm{~m}^{2}$ carrying a current of $20 \mathrm{~A}$ is. Given atomic weight of $\mathrm{Ag}=$ $108,$ density of silver $=10.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$

Options:
A $2.798 \times 10^{-4} \mathrm{~m} / \mathrm{sec}$
B $67.98 \times 10^{-4} \mathrm{~m} / \mathrm{sec}$
C $0.67 \times 10^{-4} \mathrm{~m} / \mathrm{sec}$
D $6.798 \times 10^{-4} \mathrm{~m} / \mathrm{sec}$

The correct answer is:
$6.798 \times 10^{-4} \mathrm{~m} / \mathrm{sec}$

Solution:
$\quad$ Number of electrons per kg of silver

$$

=\frac{6.023 \times 10^{26}}{108}

$$

Number of electrons per unit volume of silver

$$

\begin{array}{c}

n=\frac{6.023 \times 10^{26}}{108} \times 10.5 \times 10^{3} \\

\qquad \begin{array}{c}

v_{d}=\frac{I}{n e A} \\

=\frac{20}{6.023 \times 10^{26} \times 10.5 \times 10^{3} \times 1.6 \times 10^{-19} \times 3.14 \times 10^{-6}} \times 108 \\

=6.798 \times 10^{-4} \mathrm{~m} / \mathrm{sec} .

\end{array}

\end{array}

$$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2016
The drug that is antipyretic as well as analgesic is

Options:
A chloroquine
B penicillin
C paracetamol
D chloropromazine hydrochloride

The correct answer is:
paracetamol

Solution:
Acetamidophenol is an antipyretic as well as analgesic drug. It is commonly known by the name of paracetamol. It lowers body temperature as well as is used to get relief from pain.

So, the drug which acts as both antipyretic and analgesic is Acetamidophenol.

Its structure is as follows

Topic: Chemistry, Chemistry in Everyday Life, BITSAT, BITSAT 2019
The drug used as an antidepressant is

Options:
A Luminol
B Tofranil
C Mescaline
D Sulphadiazine

The correct answer is:
Tofranil

Solution:
Tofranil is used for the treatment of antidepressant.

Topic: Chemistry, Chemistry in Everyday Life, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The eccentricity of an ellipse, with its centre at the origin, is $1 / 2$. If one of the directrices is $\mathrm{x}=4$ then the equation of the ellipse is:

Options:
A $4 x^{2}+3 y^{2}=1$
B $3 x^{2}+4 y^{2}=12$
C $4 x^{2}+3 y^{2}=12$
D $3 x^{2}+4 y^{2}=1$

The correct answer is:
$3 x^{2}+4 y^{2}=12$

Solution:
$\mathrm{e}=\frac{1}{2} .$ Directrix $, \mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=4$

$$

\therefore \mathrm{a}=4 \times \frac{1}{2}=2 \quad \therefore \mathrm{b}=2 \sqrt{1-\frac{1}{4}}=\sqrt{3}

$$

Equation of ellispe is

$$

\frac{x^{2}}{4}+\frac{y^{2}}{3}=1 \Rightarrow 3 x^{2}+4 y^{2}=12

$$

Topic: Mathematics, Ellipse, BITSAT, BITSAT 2015
The eccentricity of the ellipse whose major axis is three times the minor axis is:

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

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

Solution:
Let a be the major axis and $b$, the minor axis of the ellipse, then 3 minor axis = major axis.

$\Rightarrow 3 b=a$

Eccentricity is given by:

$\begin{array}{l}

b^{2}=a^{2}\left(1-e^{2}\right) \\

\Rightarrow b^{2}=9 b^{2}\left(1-e^{2}\right) \\

\Rightarrow \frac{1}{9}=\left(1-e^{2}\right) \\

\Rightarrow e^{2}=\frac{8}{9} \\

\Rightarrow e=\frac{2 \sqrt{2}}{3}

\end{array}$

Topic: Mathematics, Ellipse, BITSAT, BITSAT 2021
The effect of increase in temperature on the number of electrons in conduction band $\left(\mathrm{n}_{\mathrm{e}}\right)$ and resistance of a semiconductor will be as:

Options:
A Both $\mathrm{n}_{\mathrm{e}}$ and resistance decrease
B Both $\mathrm{n}_{\mathrm{e}}$ and resistance increase
C $\mathrm{n}_{\mathrm{e}}$ increases, resistance decreases
D $n_e$ decreases, resistance increases

The correct answer is:
$\mathrm{n}_{\mathrm{e}}$ increases, resistance decreases

Solution:
When temperature increases, more electrons excite to conduction band and hence conductivity increases, therefore resistance decreases.

Topic: Physics, Semiconductors, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The efficiency of fuel cell is given by the expression, $\eta$ is

Options:
A $\eta=-\frac{\mathrm{nFE}_{\text {cell }}}{\Delta \mathrm{H}} \times 100$
B $\eta=-\frac{\mathrm{nFE}_{\text {cell }}}{\Delta \mathrm{S}} \times 100$
C $\eta=-\frac{\mathrm{nFE}_{\text {cell }}}{\Delta \mathrm{A}} \times 100$
D None of the above

The correct answer is:
$\eta=-\frac{\mathrm{nFE}_{\text {cell }}}{\Delta \mathrm{H}} \times 100$

Solution:
Efficiency of fuel cell is:

$\eta=\frac{-n F E_{\text {cell }}}{\Delta H} \times 100$

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2015
The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (given mass and charge of an electron respectively are $9.1 \times 10^{-31} \mathrm{~kg}$ and $1.6 \times$ $\left.10^{-19} \mathrm{C} .\right)$

Options:
A $-5.6 \times 10^{-11} \mathrm{~N} / \mathrm{C}$
B $-4.8 \times 10^{-15} \mathrm{~N} / \mathrm{C}$
C $-1.6 \times 10^{-19} \mathrm{~N} / \mathrm{C}$
D $-3.2 \times 10^{-19} \mathrm{~N} / \mathrm{C}$

The correct answer is:
$-5.6 \times 10^{-11} \mathrm{~N} / \mathrm{C}$

Solution:
$-\mathrm{eE}=\mathrm{mg}$
$\overrightarrow{\mathrm{E}}=-\frac{9.1 \times 10^{-31} \times 10}{1.6 \times 10^{-19}}=-5.6 \times 10^{-11} \mathrm{~N} / \mathrm{C}$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2012
The electric potential $V$ is given as a function of distance $\times($ metre $)$ by $V=\left(5 x^{2}+10 x-4\right)$ volt.

Question: Value of electric field at $x=1 \mathrm{~m}$ is

Options:
A $-23 \mathrm{~V} / \mathrm{m}$
B $11 \mathrm{~V} / \mathrm{m}$
C $6 \mathrm{~V} / \mathrm{m}$
D $-20 \mathrm{~V} / \mathrm{m}$

The correct answer is:
$-20 \mathrm{~V} / \mathrm{m}$

Solution:
$\mathrm{V}=5 x^{2}+10 x-4$

$\mathrm{E}=\frac{-\mathrm{dV}}{\mathrm{dx}}=-(10 x+10)$

At $x=1 m, \mathrm{E}=-20 \mathrm{~V} / m$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2021
The electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. The ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells will be

Options:
A 3: 1
B 2: 1
C 1: 1
D 3: 2

The correct answer is:
3: 2

Solution:
At cathodes: $\mathrm{Fe}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} ;$

$$

\begin{array}{l}

\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} \\

\left(\mathrm{E}_{\mathrm{Fe}}\right)_{1}=\frac{\text { At. wt. }}{2} ;\left(\mathrm{E}_{\mathrm{Fe}}\right)_{2}=\frac{\text { At. wt. }}{3}

\end{array}

$$

Hence, $\frac{\left(\mathrm{E}_{\mathrm{Fe}}\right)_{1}}{\left(\mathrm{E}_{\mathrm{Fe}}\right)_{2}}=\frac{3}{2}$

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2013
The element, with atomic number $118,$ will be

Options:
A alkali
B noble gas
C lanthanide
D transition element

The correct answer is:
noble gas

Solution:
Electronic configuration of element with atomic number 118 will be $[\mathrm{Rn}] 5 \mathrm{f}^{14} 6 \mathrm{~d}^{10}$ $7 \mathrm{~s}^{2} 7 \mathrm{p}^{6} .$ Since its elctronic configuration in the outer moot or bit (ns $^{2} n p^{6}$ ) resemble with that of inert or noble gases, therefore it will be noble gas element.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2015
The EMF of the cell $\mathrm{Tl} / \mathrm{Tl}^{+}(0.001 \mathrm{M}) \| \mathrm{Cu}^{2+}$

$(0.01 \mathrm{M}) / \mathrm{Cu}$ is $0.83 .$ The cell EMF can be increased by

Options:
A Increasing the concentration of $\mathrm{Tl}^{+}$ ions.
B Increasing the concentration of $\mathrm{Cu}^{2+}$ ions.
C Increasing the concentration of $\mathrm{Tl}^{+}$ and $\mathrm{Cu}^{2+}$ ions.
D None of these

The correct answer is:
Increasing the concentration of $\mathrm{Cu}^{2+}$ ions.

Solution:
The oxidation potential

$\propto \frac{1}{\text { Concentration of ions }}$ and reduction

potential $\propto$ concentration of ions. The cell voltage can be increased by decreasing the concentration of ions around anode or by increasing the concentration of ions around cathode

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2016
The empirical formula and molecular mass of a compound are CH2O and 180 g respectively. The molecular formula of the compound will be

Options:
A C9H18O9
B CH2O
C C6H12O6
D C2H4O2

The correct answer is:
C6H12O6

Solution:
As we know that, the Molar mass of Carbon=12

Molar mass of Hydrogen=1

Molar mass of Oxygen=16

Empirical formula mass =CH2O=12+2+1×16=30

Molecular mass =180

n= Molccular Mass Empirical formula mass =10030=6

∵ Molecular formula =n× empirical formula

∴ Molecular formula =6×CH2O=C6H12O6

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2018
The energies of energy levels $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ for a given atom are in the sequence $\mathrm{E}_{\mathrm{A}}<\mathrm{E}_{\mathrm{B}}<\mathrm{E}_{\mathrm{C}}$. If
the radiations of wavelengths $\lambda_{1}, \lambda_{2}$ and $\lambda_{3}$ are emitted due to the atomic transitions $\mathrm{C}$ to $\mathrm{B}, \mathrm{B}$ to A and $C$ to A respectively then which of the following relations is correct?

Options:
A $\lambda_{1}+\lambda_{2}+\lambda_{3}=0$
B $\lambda_{3}=\lambda_{1}^{2}+\lambda_{2}$
C $\lambda_{3}=\lambda_{1}+\lambda_{2}$
D $\lambda_{3}=\frac{\lambda_{1} \lambda_{2}}{\lambda_{1}+\lambda_{2}}$

The correct answer is:
$\lambda_{3}=\frac{\lambda_{1} \lambda_{2}}{\lambda_{1}+\lambda_{2}}$

Solution:
No Solution available

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2011
The energy levels of an atom is shown is figure. Which one of these transitions will result in the emission of a photon of wavelength $124.1 \mathrm{~nm}$ ? Given $\left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{Js}\right)$

Options:
A B
B A
C C
D D

The correct answer is:
D

Solution:
As $\mathrm{E}(\mathrm{eV})=\frac{1240}{\lambda(\mathrm{nm})}=\frac{1240}{124.1} \simeq 10 \mathrm{eV}$
Only is transition (D), the energy gap is $10 \mathrm{eV}$ So, option (d) is correct

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The energy of electron in the nth orbit of hydrogen atom is expressed as $\mathrm{E}_{\mathrm{n}}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}$.

The shortest and longest wavelength of Lyman series will be

Options:
A $910 Ã, 1213 Ã$
B $5463 Ã, 7858 Ã$
C $1315 Ã, 1530 Ã$
D None of these

The correct answer is:
$910 Ã, 1213 Ã$

Solution:
$\frac{1}{\lambda_{\max }}=\mathrm{R}\left[\frac{1}{(\mathrm{l})^{2}}-\frac{1}{(2)^{2}}\right]$

$\Rightarrow \lambda_{\mathrm{max}}=\frac{4}{3 \mathrm{R}} \approx 1213 Ã…$

and

$\frac{1}{\lambda_{\min }}=\mathrm{R}\left[\frac{1}{(1)^{2}}-\frac{1}{\infty}\right] \Rightarrow \lambda_{\min }=\frac{1}{\mathrm{R}} \approx 910 Ã…$

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2020
The engine of a train moving with speed $10 \mathrm{~ms}^{-1}$ towards a platform sounds a whistle at frequency $400 \mathrm{~Hz}$. The frequency heard by a passenger inside the train is (neglect air speed. Speed of sound in air $330 \mathrm{~ms}^{-1}$ )

Options:
A $200 \mathrm{~Hz}$
B $400 \mathrm{~Hz}$
C $412 \mathrm{~Hz}$
D $388 \mathrm{~Hz}$

The correct answer is:
$400 \mathrm{~Hz}$

Solution:
The relative velocity of a passenger with respect to source of sound (engine of train) is 0 . So there will be no Doppler's effect. So frequency heard is $400 \mathrm{~Hz}$.

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The engine of a truck moving along a straight road delivers constant power. The distance travelled by the truck in time $t$ is proportional to

Options:
A $t$
B $t^{2}$
C $\sqrt{\mathrm{t}}$
D $t^{3 / 2}$

The correct answer is:
$t^{3 / 2}$

Solution:
$\frac{\mathrm{mv}^{2}}{\mathrm{t}}=\frac{\mathrm{ms}^{2}}{\mathrm{t}^{3}}$

since both $\mathrm{P}$ and $\mathrm{m}$ are constants

$\therefore \frac{\mathrm{s}^{2}}{\mathrm{t}^{3}}=$ constant

Topic: Physics, Units and Dimensions, BITSAT, BITSAT 2020
The enthalpy of combustion of 2 moles of benzene at $27^{\circ} \mathrm{C}$ differs from the value determined in bomb calorimeter by

Options:
A $-2.494 \mathrm{~kJ}$
B $2.494 \mathrm{~kJ}$
C $-7.483 \mathrm{~kJ}$
D $7.483 \mathrm{~kJ}$

The correct answer is:
$-7.483 \mathrm{~kJ}$

Solution:
By bomb calorimeter we get $\Delta \mathrm{E}$.

$$

\begin{array}{l}

2 \mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{l})+15 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \\

12 \mathrm{CO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(1) \\

\Delta \mathrm{H}-\Delta \mathrm{E}=\Delta \mathrm{nRT} \\

=(12-15) \times 8.314 \times 300=-7.483 \mathrm{~kJ}

\end{array}

$$

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2013
The enthalpy of sublimation of aluminium is 330 $\mathrm{kJ} / \mathrm{mol}$. Its Ist, II $^{\text {nd }}$ and III $^{\text {rd }}$ ionization enthalpies are 580,1820 and $2740 \mathrm{~kJ}$ respectively. How much heat has too be supplied (in $\mathrm{kJ}$ ) to convert $13.5 \mathrm{~g}$ of aluminium into $\mathrm{Al}^{3+}$ ions and electrons at $298 \mathrm{k}$

Options:
A 5470
B 2735
C 4105
D 3765

The correct answer is:
2735

Solution:
Heat needed too be supplied

$$

\text { per } m o l=330+580+1820+2740

$$

$=5470 \mathrm{~kJ}$

No. of mols of Al taken $=\frac{13.5}{27}=0.5 \mathrm{~mol}$

Heat required $=0.5 \times 5470 \mathrm{~kJ}=2735 \mathrm{~kJ}$

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2014
The equation of a common tangent to the parabolas $y=x^2$ and $y=-(x-2)^2$ is

Options:
A $y=4(x-2)$
B $y=4(x-1)$
C $y=4(x+1)$
D $y=4(x+2)$

The correct answer is:
$y=4(x-1)$

Solution:
Equation of tangent of parabola $y=x^2$ be
$t x=y+a t^2$ ...(i)
$y=t x-\frac{t^2}{4}$
Solve with $\mathrm{y}=-(\mathrm{x}-2)^2$
$t x-\frac{t^2}{4}=-(x-2)^2$
$x^2+x(t-4)-\frac{t^2}{4}+4=0$
Here, Discriminant $=0$.
$(t-4)^2-4 \cdot\left(4-\frac{t^2}{4}\right)=0 \Rightarrow t^2-4 t=0 \Rightarrow t=0$
or $\mathrm{t}=4$
Put value of $t$ in eq. (i), then $y=4(x-1)$.

Topic: Mathematics, Parabola, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The equation of a projectile is $y=\sqrt{3} x-\frac{g x^{2}}{2}$ The angle of projection is given by

Options:
A $\tan \theta=\frac{1}{\sqrt{3}}$
B $\tan \theta=\sqrt{3}$
C $\frac{\pi}{2}$
D zero.

The correct answer is:
$\tan \theta=\sqrt{3}$

Solution:
Comparing the given equation with

$\mathrm{y}=\mathrm{x} \tan \theta-\frac{\mathrm{gx}^{2}}{2 \mathrm{u}^{2} \cos ^{2} \theta}, \text { we get }$

$\tan \theta=\sqrt{3}$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2020
The equation of all lines having slope 2 which are tangent to the curve $y=\frac{1}{x-3}, x \neq 3,$ is

Options:
A $y=2$
B $y=2 x$
C $y=2 x+3$
D None of these

The correct answer is:
None of these

Solution:
The equation of the given curve is $\mathrm{y}=\frac{1}{\mathrm{x}-3}, \mathrm{x} \neq 3$

The slope of the tangent to the given curve

at any point $(\mathrm{x}, \mathrm{y})$ is given by $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{(\mathrm{x}-3)^{2}}$

For tangent having slope $2,$ we must have

$2=\frac{-1}{(x-3)^{2}}$

$$

\Rightarrow 2(x-3)^{2}=-1 \Rightarrow(x-3)^{2}=-\frac{1}{2}

$$

which is not possible as square of a real number cannot be negative. Hence, there is no tangent to the given curve having slope 2 .

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2013
The equation of plane containing line x-y=1,z=1 and parallel to x2-z3=1,y=3 is

Options:
A 3x+3y-2z=1
B 3x-3y-2z=1
C 3x+3y+2z=1
D 3x+3y+2z=-1

The correct answer is:
3x-3y-2z=1

Solution:
Equation of plane containing line x-y-1=0 and z-1=0 is (x-y-1)+λ(z-1)=0

⇒ x-y+λz-1-λ=0 ...(i)

Since, this plane is parallel to line x2-z3=1 and y=3

x-22=y-30=z3 ...(ii)

∴ 2(1)+(0)(-1)+3(λ)=0

⇒ λ=-2/3

Putting the value of λ in Eq. (i), we get

x-y-23z-1+23=0⇒3x-3y-2z=1

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2019