datavorous/entrance-exam-dataset huggingface dataset:
An object takes n times as much time to slide down a 45° rough inclined plane as it takes to slide down a perfectly smooth 45° inclined plane. The coefficient of kinetic friction between the rough plane and the object is

Options:
A n2-1
B 1-1n2
C n2+1
D 1+1n2

The correct answer is:
1-1n2

Solution:
s=ut+12at2,a=gsinθ,u=0

Then, s=0+12at2

or t=2sa=2sgsinθ [for smooth plane]

For the rough plane. the effective value of acceleration along the incline is
a'=gsinθ-μkgcosθ

and t'=2sa=2sgsinθ-μkgcosθ

Now, t't=n=gsinθgsinθ-μkgcosθ

=sin45°sin45°-μkcos45°

=11-μk

⇒ 1n2=1-μk⇒μk=1-1n2

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2018
An organ pipe, open from both end produces 5 beats per second when vibrated with a source of frequency $200 \mathrm{~Hz}$. The second harmonic of the same pipes produces 10 beats per second with a source of frequency $420 \mathrm{~Hz}$. The fundamental frequency of organ pipe is

Options:
A $195 \mathrm{~Hz}$
B $205 \mathrm{~Hz}$
C $190 \mathrm{~Hz}$
D $210 \mathrm{~Hz}$

The correct answer is:
$205 \mathrm{~Hz}$

Solution:
Let the fundamental frequency of organ pipe be $\mathrm{f}$

Case I : $f=200 \pm 5=205 \mathrm{~Hz}$ or $195 \mathrm{~Hz}$

Case II : frequency of 2 nd harmonic of organ pipe $=2 \mathrm{f}$ (as is clear from the second figure) $2 \mathrm{f}=420 \pm 10$ or $\mathrm{f}=210 \pm 5$ or $\mathrm{f}=205$ or 215

Hence fundamental frequency oforgan pipe $=205 \mathrm{~Hz}$

Topic: Physics, Waves and Sound, JEE Main
An organ pipe, open from both end produces 5 beats per second when vibrated with a source of frequency $200 \mathrm{~Hz}$. The second harmonic of the same pipes produces 10 beats per second with a source of frequency $420 \mathrm{~Hz}$. The fundamental frequency of organ pipe is

Options:
A $195 \mathrm{~Hz}$
B $205 \mathrm{~Hz}$
C $190 \mathrm{~Hz}$
D $210 \mathrm{~Hz}$

The correct answer is:
$205 \mathrm{~Hz}$

Solution:
Let the fundamental frequency of organ pipe be $\mathrm{f}$

Case I : $f=200 \pm 5=205 \mathrm{~Hz}$ or $195 \mathrm{~Hz}$

Case II : frequency of 2 nd harmonic of organ pipe $=2 \mathrm{f}$ (as is clear from the second figure) $2 \mathrm{f}=420 \pm 10$ or $\mathrm{f}=210 \pm 5$ or $\mathrm{f}=205$ or 215

Hence fundamental frequency oforgan pipe $=205 \mathrm{~Hz}$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2020
An organic compound $\mathrm{A}\left(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}\right)$ on reaction with Na/diethyl ether gives a hydrocarbon which on monochlorination gives only one chloro derivative, then A is

Options:
A tert-butyl chloride
B sec-butyl chloride
C isobutyl chloride
D $\mathrm{n}$ -butyl chloride

The correct answer is:
tert-butyl chloride

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
An organic compound $\mathrm{A}\left(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}\right)$ on reaction with Na/diethyl ether gives a hydrocarbon which on monochlorination gives only one chloro derivative, then A is

Options:
A tert-butyl chloride
B sec-butyl chloride
C isobutyl chloride
D $\mathrm{n}$ -butyl chloride

The correct answer is:
tert-butyl chloride

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, BITSAT, BITSAT 2014
An organic compound with the formula $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$ forms a yellow crystalline solid with phenylhydrazine and gives a mixture of sorbitol and mannitol when reduced with sodium. Which among the following could be the compound?

Options:
A fructose
B glucose
C mannose
D sucrose

The correct answer is:
fructose

Solution:
Since the compound forms a yellow crystalline solid, i.e. osazone with phenylhydrazine, it may be an aldohexose or a ketohexose. Further, since on reduction, compound forms a mixture of sorbitol and mannitol, it must be a ketohexose, i.e. fructose. Recall that glucose on reduction gives only one alcohol glucitol (Sorbitol)

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2020
An oscillator is nothing but an amplifer with

Options:
A positive feedback
B negative feedback
C large gain
D no feedback

The correct answer is:
positive feedback

Solution:
A positive feedback from output to input in an amplifier provides oscillations of constant amplitude.

Topic: Physics, Semiconductors, BITSAT, BITSAT 2012
An α-particle after passing through potential difference of V volt collides with a nucleus. If the atomic number of the nucleus is Z, then distance of closest approach is

Options:
A 14.4ZV Å
B 14.4ZV m
C 14.4VZ m
D 14.4VZ Å

The correct answer is:
14.4ZV Å

Solution:
Given KE of α-particle =2 eV

r=Zee4πε0·KE

=2Ze×9×1092V

⇒ r=2×Z×16×10-19×9×1092V

=14.4·ZV Å

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2017
An unsaturated hydrocarbon 'X'gives white precipitate with Tollen's reagent. If X is gaseous in nature, the molecular formula of X is

Options:
A C3H6
B C2H4
C C2H2
D C4H8

The correct answer is:
C2H2

Solution:
Tollen's reagent consists of a solution of silver nitrate and ammonia.

Terminal alkynes react with tollen's reagent and form white color precipitate.

With Tollen's reagent (ammoniacal AgNO3), a white precipitate of silver salt is obtained.

H-C≡C-H+2AgNO3+2NH4OH→Ag-C≡C-AgSliver acetylide (white ppt.)+2NH4NO3+H2O

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2018
An X-ray tube is operated at $15 \mathrm{kV}$. Calculate the upper limit of the speed of the electrons striking the target.

Options:
A $7.26 \times 10^{7} \mathrm{~m} / \mathrm{s}$
B $7.62 \times 10^{7} \mathrm{~m} / \mathrm{s}$
C $7.62 \times 10^{7} \mathrm{~cm} / \mathrm{s}$
D $7.26 \times 10^{9} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$7.26 \times 10^{7} \mathrm{~m} / \mathrm{s}$

Solution:
The maximum kinetic energy of an electron accelerated through a potential difference of $\mathrm{V}$ volt is $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$
$$
\begin{array}{l}
\therefore \text { maximum velocity } \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}} \\
\mathrm{v}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 15000}{9.1 \times 10^{-31}}} \\
\mathrm{v}=7.26 \times 10^{7} \mathrm{~m} / \mathrm{s}
\end{array}
$$

Topic: Physics, Electromagnetic Waves, BITSAT, BITSAT 2009
Aniline reacts with nitrous acid to produce

Options:
A phenol
B nitrobenzene
C chlorobenzene
D benzene diazonium chloride

The correct answer is:
benzene diazonium chloride

Solution:
When aniline is treated with nitrous acid in the presence of $\mathrm{HCl}$, then benzene diazonium chloride is obtained.

Topic: Chemistry, Amines, BITSAT, BITSAT 2011
Aniline reacts with phosgene and $\mathrm{KOH}$ to form

Options:
A
B
C
D

The correct answer is:

Solution:
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{COCl}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH.COCl}$

Topic: Chemistry, Amines, BITSAT, BITSAT 2014
Area bounded by two branches of the curve y-x2=x3 and x=1 is

Options:
A 54 sq unit
B 32 sq unit
C 14 sq unit
D 45 sq unit

The correct answer is:
45 sq unit

Solution:
Given, curve is y-x2=x3

⇒ y-x=±xx

⇒ y=x±xx

∴ Area bounded =∫01x+xx-x-xxdx

=2∫01x3/2dx=45 sq unit

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2019
Area intercepted by the curves $y=\cos x$ $\mathrm{x} \in[0, \pi]$ and $\mathrm{y}=\cos 2 \mathrm{x}, \mathrm{x} \in[0, \pi],$ is

Options:
A $\frac{3 \pi}{2}$
B $\frac{3 \sqrt{3}}{2}$
C $\frac{3 \pi}{4}$
D $\frac{3 \sqrt{3}}{4}$

The correct answer is:
$\frac{3 \sqrt{3}}{4}$

Solution:
$$

\text { Area }=\int_{0}^{2 \pi / 3}(\cos x-\cos 2 x) d x=\frac{3 \sqrt{3}}{4}

$$

Topic: Mathematics, Area Under Curves, JEE Main
Area intercepted by the curves $y=\cos x$ $\mathrm{x} \in[0, \pi]$ and $\mathrm{y}=\cos 2 \mathrm{x}, \mathrm{x} \in[0, \pi],$ is

Options:
A $\frac{3 \pi}{2}$
B $\frac{3 \sqrt{3}}{2}$
C $\frac{3 \pi}{4}$
D $\frac{3 \sqrt{3}}{4}$

The correct answer is:
$\frac{3 \sqrt{3}}{4}$

Solution:
$$

\text { Area }=\int_{0}^{2 \pi / 3}(\cos x-\cos 2 x) d x=\frac{3 \sqrt{3}}{4}

$$

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2013
Area of the circle in which a chord of length $\sqrt{2}$ makes an angle $\pi / 2$ at the centre, is

Options:
A $\pi / 2$ sq units
B $2 \pi$ sq units
C $\pi$ sq units
D $\pi / 4 \mathrm{sq}$ units

The correct answer is:
$\pi$ sq units

Solution:
Let $\mathrm{AB}$ be the chord of length $\sqrt{2}$. Let $\mathrm{O}$ be the centre of the circle and let $O C$ be the perpendicular from $\mathrm{O}$ on $\mathrm{AB}$.

Then $, \mathrm{AC}=\mathrm{BC}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

In $\Delta \mathrm{OBC},$ we have

$$

\begin{aligned}

\mathrm{OB} &=\mathrm{BC} \operatorname{cosec} 45^{\circ} \\

&=\frac{1}{\sqrt{2}} \times \sqrt{2}=1

\end{aligned}

$$

$\therefore$ Area of the circle $=\pi(\mathrm{OB})^{2}=\pi$ sq units

Topic: Mathematics, Circle, JEE Main
Area of the circle in which a chord of length $\sqrt{2}$ makes an angle $\pi / 2$ at the centre, is

Options:
A $\pi / 2$ sq units
B $2 \pi$ sq units
C $\pi$ sq units
D $\pi / 4 \mathrm{sq}$ units

The correct answer is:
$\pi$ sq units

Solution:
Let $\mathrm{AB}$ be the chord of length $\sqrt{2}$. Let $\mathrm{O}$ be the centre of the circle and let $O C$ be the perpendicular from $\mathrm{O}$ on $\mathrm{AB}$.

Then $, \mathrm{AC}=\mathrm{BC}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

In $\Delta \mathrm{OBC},$ we have

$$

\begin{aligned}

\mathrm{OB} &=\mathrm{BC} \operatorname{cosec} 45^{\circ} \\

&=\frac{1}{\sqrt{2}} \times \sqrt{2}=1

\end{aligned}

$$

$\therefore$ Area of the circle $=\pi(\mathrm{OB})^{2}=\pi$ sq units

Topic: Mathematics, Circle, BITSAT, BITSAT 2015
Area of the triangle formed by the line \(x+y=3\) and angle bisectors of the pair of straight lines \(x^2-y^2+2 y=1\) is

Options:
A 2 sq. units
B 4 sq. units
C 6 sq. units
D 8 sq. units

The correct answer is:
2 sq. units

Solution:
\(\mathrm{x}^2-\mathrm{y}^2+2 \mathrm{y}=1 \Rightarrow \mathrm{x}= \pm(\mathrm{y}-1)\)
Bisectors of above line are \(x=0 ~\&~ y=1\)
So area between \(x=0, y=1 ~\&~ x+y=3\) is shaded Region shown in figure.
Area \(=1 / 2 \times 2 \times 2=2\) sq. units

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2009
Arrange the following ions in the order of decreasing $\mathrm{X}-\mathrm{O}$ bond length, where $\mathrm{X}$ is the central atom

Options:
A $\mathrm{ClO}_{4}^{-}, \mathrm{SO}_{4}^{2-}, \mathrm{PO}_{4}^{3-}, \mathrm{SiO}_{4}^{-}$
B $\mathrm{SiO}_{4}^{4-}, \mathrm{PO}_{4}^{3-}, \mathrm{SO}_{4}^{2-}, \mathrm{ClO}_{4}^{-}$
C $\mathrm{SiO}_{4}^{4-}, \mathrm{PO}_{4}^{3-}, \mathrm{ClO}_{4}^{-}, \mathrm{SO}_{4}^{2-}$
D $\mathrm{SiO}_{4}^{4-}, \mathrm{SO}_{4}^{2-}, \mathrm{PO}_{4}^{3-}, \mathrm{ClO}_{4}^{-}$

The correct answer is:
$\mathrm{SiO}_{4}^{4-}, \mathrm{PO}_{4}^{3-}, \mathrm{SO}_{4}^{2-}, \mathrm{ClO}_{4}^{-}$

Solution:
More will be the electronegativity of $\mathrm{X}$ lesser will be the bond length of $\underline{X}-\mathrm{O}$ bond.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2020
Arsenic containing medicine used for the treatment of syphilis, is

Options:
A erythromycin
B ofloxacin
C tetracycline
D salvarsan

The correct answer is:
salvarsan

Solution:
No solution required.

Topic: Chemistry, Chemistry in Everyday Life, BITSAT, BITSAT 2021
Aspirin is known as :

Options:
A acetyl salicylic acid
B phenyl salicylate
C acetyl salicylate
D methyl salicylic acid

The correct answer is:
acetyl salicylic acid

Solution:

Topic: Chemistry, Chemistry in Everyday Life, BITSAT, BITSAT 2021
Assuming the diodes to be of silicon with forward resistance zero, the current I in the following circuit is39. Assuming the diodes to be of silicon with forward resistance zero, the current I in the following circuit is

Options:
A 0
B $9.65 \mathrm{~mA}$
C $10 \mathrm{~mA}$
D $10.35 \mathrm{~mA}$

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

Solution:
$I=\frac{V}{R}=\frac{20}{2 \times 10^{3}}=10 \times 10^{-3} \mathrm{~A}=10 \mathrm{~mA}$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2009
At $25^{\circ} \mathrm{C}$ and 1 bar which of the following has a non-zero $\Delta \mathrm{H}_{\mathrm{f}}^{\circ} ?$

Options:
A $\operatorname{Br}_{2}(\ell)$
B C (graphite)
C $\mathrm{O}_{3}(\mathrm{~g})$
D $\mathrm{I}_{2}(\mathrm{~s})$

The correct answer is:
$\mathrm{O}_{3}(\mathrm{~g})$

Solution:
Ozone is allotropic form of oxygen and is of higher energy (by $68 \mathrm{~K}$ Cal mol $^{-1}$ ) than $\mathrm{O}_{2}$. Hence it can not be taken as the reference in standard state.

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2009
At $25^{\circ} \mathrm{C}$, the molar conductance at infinite dilution for the strong electrolytes $\mathrm{NaOH}, \mathrm{NaCl}$ and $\mathrm{BaCl}_{2}$ are $248 \times 10^{-4}, 126 \times 10^{-4}$ and $280 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ respectively. $\Delta_{\mathrm{m}}{ }^{0} \mathrm{Ba}(\mathrm{OH})_{2}$ in $\mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ is

Options:
A $52.4 \times 10^{-4}$
B $524 \times 10^{-4}$
C $402 \times 10^{-4}$
D $262 \times 10^{-4}$

The correct answer is:
$524 \times 10^{-4}$

Solution:
$\Delta_{\mathrm{Na}^{+}}^{\circ}+\Delta_{\mathrm{OH}^{-}}^{\circ}=248 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ $\Delta_{\mathrm{Na}^{+}}^{\circ}+\Delta_{\mathrm{Cl}^{-}}^{\circ}=126 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ $\Delta_{\mathrm{Ba}}^{\circ}{ }^{2+}+\Delta{ }_{2 \mathrm{Cl}^{-}}^{\circ}=280 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ $\mathrm{NoW}$, $\Delta_{\mathrm{Ba}(\mathrm{OH})_{2}}^{\circ}=\Delta_{\mathrm{BaCl}_{2}}^{\circ}+2 \Delta{ }^{\circ}{ }_{\mathrm{NaOH}}^{\circ}-2 \Delta{ }_{\mathrm{NaC} 1}^{\circ}$ $\Delta{ }_{\mathrm{Ba}(\mathrm{OH})_{2}}^{\circ}=280 \times 10^{-4}+2 \times 248 \times 10^{-4}-2 \times 126 \times 10^{-4}$ $\Delta_{\mathrm{Ba}(\mathrm{OH})_{2}}^{\circ}=524 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2021
At a certain temperature, the number density of charge carriers in a semiconductor is n. When an electric field is applied to it, the charge carriers drift with an average speed v. If the temperature of the semiconductor is raised.

Options:
A n will increase but v will decrease
B n will decrease but v will increase
C Both n and v will increase
D Both n and v will decrease

The correct answer is:
n will increase but v will decrease

Solution:
On raising the temperature of the semiconductor, covalent bonds start breaking up and thus more charge carriers release. This will increase the value of n and as a result the rate of collision of charge carriers will increase. This will decrease the drift speed v.

Topic: Physics, Semiconductors, BITSAT, BITSAT 2018
At a constant volume the specific heat of a gas is 0.075 and its molecular weight is $40 .$ The gas is:

Options:
A Monoatomic
B Diatomic
C Triatomic
D None of these

The correct answer is:
Monoatomic

Solution:
$\begin{aligned} & q=\frac{c}{m} \\ & \Rightarrow C_{v}=q \times m \\ &=0.075 \times 40 \\ &=3.0 \mathrm{cal} \\ & C_{P}-C_{v}=R \\ & C_{P}=C_{V}+R \\ &=3+2=5 \\ & \frac{C_{P}}{C_{V}}=\frac{5}{3}=1.66 \end{aligned}$
Monoatomic gas

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2010
At an extreme point of a function $f(x),$ the tangent to the curve is

Options:
A parallel to the $\mathrm{x}$ -axis
B perpendicular to the $\mathrm{x}$ -axis
C inclined at an angle $45^{\circ}$ to the $\mathrm{x}$ -axis
D inclined at an angle $60^{\circ}$ to the $x$ -axis

The correct answer is:
parallel to the $\mathrm{x}$ -axis

Solution:
At an extreme point of a function $f(x)$, slope is always zero. Thus, At an extreme point of a function $f(x)$, the tangent to the curve is parallel to the x-axis.

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2016
At any instant the velocity of a particle of mass $500 \mathrm{~g}$ is $\left(2 t \hat{i}+3 t^2 \hat{j}\right) \mathrm{ms}^{-1}$. If the force acting on the particle at $t=1 s$ is $(\hat{i}+x \hat{j}) N$. Then the value of $\mathrm{x}$ will be:

Options:
A 3
B 4
C 6
D 2

The correct answer is:
3

Solution:
Mass of particle, $\mathrm{m}=500 \mathrm{~g}=0.5 \mathrm{~kg}$
velocity of a particle, $\vec{v}=2 t \hat{i}+3 t^2 \hat{j}$
$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}=2 \hat{\mathrm{i}}+6 \mathrm{t} \hat{\mathrm{j}}$
at $\mathrm{t}=1, \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}$
Force acting on the particle,
$\overrightarrow{\mathrm{F}}=\mathrm{ma}=0.5(2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}})=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}$
$\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+x \hat{\mathrm{j}} \quad$ Hence $x=3$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2023 (Memory Based Paper 1)
At high pressure, the entire surface gets covered by a monomolecular layer of the gas follows

Options:
A three-halved order
B second-order
C first-order
D zero-order

The correct answer is:
zero-order

Solution:
At high pressure the extent of adsorption follows zero order kinetics.

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2013
At the corners of an equilateral triangle of side a (1 metre), three point charges are placed (each of $0.1 \mathrm{C}$ ). If this system is supplied energy at the rate of $1 \mathrm{kw},$ then calculate the time required to move one of the mid-point of the line joining the other two.

Options:
A $50 \mathrm{~h}$
B $60 \mathrm{~h}$
C $48 \mathrm{~h}$
D $54 \mathrm{~h}$

The correct answer is:
$50 \mathrm{~h}$

Solution:
Initial potential energy of the system

$$

\begin{array}{l}

=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}^{2}}{\mathrm{a}}+\frac{\mathrm{q}^{2}}{\mathrm{a}}+\frac{\mathrm{q}^{2}}{\mathrm{a}}\right]=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{3 \mathrm{q}^{2}}{\mathrm{a}}\right) \\

=9 \times 10^{9}\left(3 \times \frac{(0.1)^{2}}{1}\right)=27 \times 10^{7} \mathrm{~J}

\end{array}

$$

Let charge at $\mathrm{A}$ is moved to mid-point $\mathrm{O}$ Then final potential energy of thhe system

$$

\begin{aligned}

\mathrm{U}_{\mathrm{f}} &=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{2 \mathrm{q}^{2}}{(\mathrm{a} / 2)}+\frac{\mathrm{q}^{2}}{\mathrm{a}}\right] \\

&=5 \times \frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}^{2}}{\mathrm{a}^{2}}\right)=45 \times 10^{7} \mathrm{~J}

\end{aligned}

$$

Work done $=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=18 \times 10^{7} \mathrm{~J}$

Also, energy supplied per sec $=1000 \mathrm{~J}$ (given) Time required to move one of the mid-point of the line joining the other two

$$

\mathrm{t}=\frac{18 \times 10^{7}}{1000}=18 \times 10^{4} \mathrm{~s}=50 \mathrm{~h}

$$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2015
At the magnetic north pole of the earth, the value of horizontal component of earth's magnetic field and angle of dip are, respectively

Options:
A zero, maximum
B maximum, minimum
C maximum, maximum
D minimum, minimum

The correct answer is:
zero, maximum

Solution:
At the magnetic north pole, the magnetic needle will point vertically. There is no component of earth's magnetic field in the horizontal direction and the angle of dip (the angle that the resultant magnetic field at the place makes with the horizontal) is $90^{\circ}$.
$$
\mathrm{H}=0, \delta=90^{\circ}(\text { maximum })
$$

Topic: Physics, Magnetic Properties of Matter, BITSAT, BITSAT 2009
At what angle θ to the horizontal should an object is projected so that the maximum height reached is equal to the horizontal range?

Options:
A tan-12
B tan-14
C tan-1(23)
D tan-13

The correct answer is:
tan-14

Solution:
Given, H=R

or u2sin2θ2g=u2sin2θg

sinθ×sinθ2=2sinθ×cosθ

⇒tanθ=4

⇒ θ=tan-14

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2017
At what temperature, the rate of effusion of $\mathrm{N}_{2}$ would be 1.625 times than that of $\mathrm{SO}_{2}$ at $50^{\circ} \mathrm{C}$ ?

Options:
A $110 \mathrm{~K}$
B $173 \mathrm{~K}$
C $373 \mathrm{~K}$
D $273 \mathrm{~K}$

The correct answer is:
$373 \mathrm{~K}$

Solution:
$\quad r \propto U$ and $U=\sqrt{\frac{3 R T}{\text { M }}}$

$\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\sqrt{\frac{\mathrm{T}_{1} \mathrm{M}_{2}}{\mathrm{~T}_{2} \mathrm{M}_{1}}}$ or $\frac{\mathrm{r}_{\mathrm{N}_{2}}}{\mathrm{r}_{\mathrm{SO}_{2}}}=\sqrt{\frac{\mathrm{T}_{1} \times 64}{323 \times 28}}=1.625$

or $\mathrm{T}_{2}=373 \mathrm{~K}$

Topic: Chemistry, States of Matter, BITSAT, BITSAT 2016
$$
\text { If } A=\left[\begin{array}{ccc}
0 & c & -b \\
-c & 0 & a \\
b & -a & 0
\end{array}\right] \text { and } B=\left[\begin{array}{ccc}
a^2 & a b & a c \\
a b & b^2 & b c \\
a c & b c & c^2
\end{array}\right] \text {, }
$$
then $A B$ is equal to

Options:
A
B
B
A
C
O
D
I

The correct answer is:
O

Solution:
$\begin{aligned} &\quad A B=\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right]\left[\begin{array}{ccc}a^2 & a b & a c \\ a b & b^2 & b c \\ a c & b c & c^2\end{array}\right] \\ & A B=\left[\begin{array}{ccc}a b c-a b c & b^2 c-b^2 c & b c^2-b c^2 \\ -a^2 c+a^2 c & -a b c+a b c & -a c+a c \\ a^2 b-a^2 b & a b^2-a b^2 & a b c-a b c\end{array}\right] \\ & =\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=\mathrm{O} \\ & \end{aligned}$

Topic: Mathematics, Matrices, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Bag $P$ contains 6 red and 4 blue balls and bag $Q$ contains 5 red and 6 blue balls. A ball is transferred from bag $P$ to bag $Q$ and then a ball is drawn from bag $Q$. What is the probability that the ball drawn is blue?

Options:
A $\frac{7}{15}$
B $\frac{8}{15}$
C $\frac{4}{19}$
D $\frac{8}{19}$

The correct answer is:
$\frac{8}{15}$

Solution:
Let $E_1, E_2$ and $A$ be the events defined as follows:
$\mathrm{E}_1=$ red ball is transferred from bag $\mathrm{P}$ to bag $\mathrm{Q}$
$\mathrm{E}_2=$ blue ball is transferred from bag $\mathrm{P}$ to bag $\mathrm{Q}$
$\mathrm{A}=$ the ball drawn from bag $\mathrm{Q}$ is blue
As the bag $\mathrm{P}$ contains 6 red and 4 blue balls,
$P\left(E_1\right)=\frac{6}{10}=\frac{3}{5} \text { and } P\left(E_2\right)=\frac{4}{10}=\frac{2}{5}$
Note that $E_1$ and $E_2$ are mutually exclusive and exhaustive events.
When $E_1$ has occurred i.e., a red ball has already been transferred from bag $P$ to $Q$, then bag $Q$ will contain 6 red and 6 blue balls, So, $P\left(A \mid E_1\right)=$$\frac{6}{12}=\frac{1}{2}$
When $E_2$ has occurred i.e., a blue ball has already been transferred from bag $P$ to $Q$, then bag $Q$ will contain 5 red and 7 blue balls, So, $P\left(A \mid E_2\right)=$$\frac{7}{12}$
By using law of total probability, we get
$P(A)=P\left(E_1\right) P\left(A \mid E_1\right)+P\left(E_2\right) P\left(A \mid E_2\right)$
$=\frac{3}{5} \times \frac{1}{2}+\frac{2}{5} \times \frac{7}{12}=\frac{8}{15}$

Topic: Mathematics, Probability, BITSAT, BITSAT 2023 (Memory Based Paper 1)
$\mathrm{BeF}_2,$ is soluble in water whereas fluorides of other alkaline earth metals are insoluble because of

Options:
A ionic nature of $\mathrm{BeF}_{2}$.
B covalent nature of $\mathrm{BeF}_{2}$
C greater hydration energy of $\mathrm{Be}^{2+}$ ion as compared to its lattice energy.
D none of these.

The correct answer is:
greater hydration energy of $\mathrm{Be}^{2+}$ ion as compared to its lattice energy.

Solution:
$\mathrm{Be}^{2+}$ being small in size is heavily hydrated and heat of hydration exceeds the lattice energy. Hence $\mathrm{BeF}_{2}$ is soluble in water.

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2020
Beryllium and aluminium exhibit many properties which are similar. But, the two elements differ in

Options:
A forming covalent halides
B forming polymeric hydrides
C exhibiting maximum covalency in compounds
D exhibiting amphoteric nature in their oxides

The correct answer is:
exhibiting maximum covalency in compounds

Solution:
The valency of beryllium is +2 while that of aluminium is +3

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2011
Beryllium resembles much with :

Options:
A $\mathrm{Zn}$
B $\mathrm{Al}$
C $\mathrm{Li}$
D $\mathrm{Ra}$

The correct answer is:
$\mathrm{Al}$

Solution:
Beryllium resembles with aluminium due to similarity in the size of ions and similarity in electropositive character. This type of resemblance between first element of a group in second period with second element of the next group is termed as diagonal relationship.

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2009