datavorous/entrance-exam-dataset huggingface dataset:
The value of ∑r=110r·CrnCr-1n is equal to

Options:
A 52n-9
B 10n
C 9(n-4)
D None of these

The correct answer is:
52n-9

Solution:
We have,

∑r=110r·CrnCr-1n=∑r=110n-r+1

⇒ ∑r=110r·CrnCr-1n=∑r=110n+1-r

⇒ ∑r=110r·CrnCr-1n=10n+1-∑r=110r

⇒ ∑r=110r·CrnCr-1n=10n+1-55

=10n-45=52n-9

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2017
The value of S=∑n=1∞tan-12nn4+n2+2 is equal to

Options:
A π2
B π
C π4
D None of these

The correct answer is:
π4

Solution:
Given, S=∑n=1∞tan-12nn4+n2+2 ...i

Let n4+n2+1=n22+12+2n21-n2

=n2+12-n2

=n2+n+1n2-n+1 ...ii

Let un=tan-12nn4+n2+2

=tan-12n1+n4+n2+1

=tan-1n2+n+1-n2-n+11+n2+n+1n2-n+1

un=tan-1n2+n+1-tan-1n2-n+1 ...iii

On putting n=1, 2, 3,…… successively in Eq. (iii), we get

u1=tan-13-tan-11

u2=tan-17-tan-13

u3=tan-113-tan-17

un=tan-1n2+n+1-tan-1n2-n+1

On adding vertically, we get

∑n=1∞un=tan-1n2+n+1-tan-11

S=limn→∞∑n=1∞un [from Eq. (i)]

=limn→∞tan-1n2+n+1-tan-11

=π2-π4=π4

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2018
The value of \(\sin \left(4 \tan ^{-1} \frac{1}{3}\right)-\cos \left(2 \tan ^{-1} \frac{1}{7}\right)\) is

Options:
A \(3 / 7\)
B \(7 / 8\)
C \(8 / 21\)
D None of these

The correct answer is:
None of these

Solution:
\(\sin \left(4 \tan ^{-1} \frac{1}{3}\right)\)
\(\begin{aligned}
&= 2 \sin \left(2 \tan ^{-1} \frac{1}{3}\right) \cos \left(2 \tan ^{-1} \frac{1}{3}\right) \\
&=2 \sin \left(\tan ^{-1} \frac{3}{4}\right)=\cos \left(\tan ^{-1} \frac{3}{4}\right)=2 \cdot \frac{3}{5} \cdot \frac{4}{5}=\frac{24}{25} \\
& \cos \left(2 \tan ^{-1} \frac{1}{7}\right)=\cos \left(\tan ^{-1} \frac{7}{24}\right)=\frac{24}{25} .
\end{aligned}\)
The given expression \(=0\)

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2011
The value of $\sin \left[\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}\right], \mathrm{n} \in \mathrm{I}$ is

Options:
A 0
B $\frac{1}{\sqrt{2}}$
C $-\frac{1}{\sqrt{2}}$
D None of these

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

Solution:
$\sin \left[\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}\right]=(-1)^{\mathrm{n}} \sin \left[(-1)^{\mathrm{n}} \frac{\pi}{4}\right]$

$\left[\because \sin (\mathrm{n} \pi+\theta)=(-1)^{\mathrm{n}} \sin \theta\right]$

$\begin{aligned} &=(-1)^{\mathrm{n}}(-1)^{\mathrm{n}} \sin \frac{\pi}{4} \\ \therefore \sin &\left.\left[(-1)^{\mathrm{n}} \theta\right]=(-1)^{\mathrm{n}} \sin \theta\right] \\ &=(-1)^{2 \mathrm{n}} \sin \frac{\pi}{4}=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \end{aligned}$

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2020
The value of the determinant
$\left|\begin{array}{lll}265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181\end{array}\right|$ is

Options:
A 1000
B 779
C 679
D 0

The correct answer is:
0

Solution:
Applying $\mathrm{C}_{1}-\mathrm{C}_{2}$ and $\mathrm{C}_{2}-\mathrm{C}_{3},$ we get
$\begin{aligned}
\text { Det. } &=\left|\begin{array}{ccc}
25 & 21 & 219 \\
15 & 27 & 198 \\
21 & 17 & 181
\end{array}\right|=\left|\begin{array}{ccc}
4 & 21 & 9 \\
-12 & 27 & -72 \\
4 & 17 & 11
\end{array}\right| \\
=&\left[\begin{array}{llc}
4 & 21 & 9 \\
0 & 90 & -45 \\
0 & -4 & 2
\end{array} \mid\left[\mathrm{By} \mathrm{R}_{2}+3 \mathrm{R}_{1}, \mathrm{R}_{3}-\mathrm{R}_{1}\right]\right.\\
=4(180-180)=0
\end{aligned}$

Topic: Mathematics, Determinants, BITSAT, BITSAT 2012
The value of $\int \frac{1}{\left[(x-1)^3(x+2)^5\right]^{\frac{1}{4}}} d x$, is

Options:
A
$\frac{4}{3}\left(\frac{x+1}{x-2}\right)^{\frac{1}{4}}+C$

B
$\frac{3}{4}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C$

C
$\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C$

D
$\frac{1}{3}\left(\frac{2 x-1}{4 x-3}\right)^{\frac{1}{4}}+C$

The correct answer is:

$\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C$

Solution:
$I=\int \frac{1}{\left(\frac{x-1}{x+2}\right)^{3 / 4}(x+2)^2} d x$
Let $\frac{x-1}{x+2}=t \Rightarrow \frac{3 d x}{(x+2)^2}=d t$
$$
\begin{aligned}
\Rightarrow \quad I & =\frac{1}{3} \int \frac{1}{t^{3 / 4}} d t=\frac{1}{3}\left(\frac{t^{1 / 4}}{1 / 4}\right)+C \\
& =\frac{4}{3} t^{1 / 4}+C=\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{1 / 4}+C
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2022
The value of $\mathrm{x}$ is maximum for

Options:
A $\mathrm{MgSO}_{4} \cdot \mathrm{x} \mathrm{H}_{2} \mathrm{O}$
B $\mathrm{CaSO}_{4} \cdot \mathrm{x} \mathrm{H}_{2} \mathrm{O}$
C $\mathrm{BaSO}_{4} \cdot \mathrm{x} \mathrm{H}_{2} \mathrm{O}$
D All have the same

The correct answer is:
$\mathrm{MgSO}_{4} \cdot \mathrm{x} \mathrm{H}_{2} \mathrm{O}$

Solution:
Because of smaller size, $\mathrm{Mg}^{2+}$ ions are extensively hydrated.

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2009
The van't Hoff factor i for an electrolyte which undergoes dissociation and association in solvents are respectively

Options:
A greater than 1 and greater than 1
B less than 1 and greater than 1
C less than 1 and less than 1
D greater than 1 and less than 1

The correct answer is:
greater than 1 and less than 1

Solution:
When an electrolyte dissociates van't Hoff factor i is greater than 1 and when it associates the $\mathrm{i}$ is less than $1.$

Topic: Chemistry, Solutions, BITSAT, BITSAT 2012
The vapour density of ozone is

Options:
A 16
B 32
C 24
D 48

The correct answer is:
24

Solution:
We know that, Molecular weight of compound or molecules $=2 \times \mathrm{V}$. $\mathrm{D}$
Vapour density (V. D.) of ozone molecules
$=\frac{\mathrm{M} . \text { wt. of } \mathrm{O}_{3}}{2}=\frac{48}{2}=24$
Hence, $\mathrm{V} . \mathrm{D}$. of $\mathrm{O}_{3}$ is 24

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2010
The vapour pressure of

is higher than

due to

Options:
A Dipole moment
B Dipole- dipole interaction
C $\mathrm{H}$ - bonding
D Lattice structure

The correct answer is:
$\mathrm{H}$ - bonding

Solution:
Ortho-nitrophenol has intramolecular
H bonding

and paranitrophenol has intermolecular H-bonding.

Hence former is more volatile than latter.

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2012
The vapour pressure of two pure liquids $A$ and $B$ that form an ideal solution, are 400 and $800 \mathrm{~mm}$ of Hg respectivelyat a temperature $t^{\circ} \mathrm{C}$. The mole fraction of $A$ in a solution of $A$ and $B$ whose boiling point is $t^{\circ} \mathrm{C}$ will be

Options:
A 0.4
B 0.8
C 0.1
D 0.2

The correct answer is:
0.1

Solution:
V.P. of solution at $t^{\circ} \mathrm{C}=760 \mathrm{~mm}$ [at b.p., V.P. of solution =atompheric pressure] Thus $=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{\infty} \mathrm{x}_{\mathrm{B}}$

or $\mathrm{P}=\mathrm{P}_{\mathrm{A}}^{0} \cdot \mathrm{x}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{0}\left(1-\mathrm{x}_{\mathrm{A}}\right)\left[\because \mathrm{x}_{\mathrm{A}}+\mathrm{x}_{\mathrm{B}}=1\right]$

or $760=400 \mathrm{X}_{\mathrm{A}}+800\left(1-\mathrm{X}_{\mathrm{A}}\right)[\because \mathrm{P}=760 \mathrm{mmofHg}]$

or $-800+760=-400 x_{\mathrm{A}}$

or $-40=-400 \mathrm{x}_{\mathrm{A}}$

or $x_{\mathrm{A}}=\frac{40}{400}=0.1$

Thus mole fraction in solution is 0.1

Topic: Chemistry, Solutions, BITSAT, BITSAT 2020
The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) starting from mean position to extreme position (A) is given by

Options:
A

B

C

D

The correct answer is:

Solution:
$\begin{aligned} & T \cdot E=K \cdot E+P . E \\ & \Rightarrow K \cdot E=T \cdot E-K \cdot E \\ & =\frac{1}{2} k^2-\frac{1}{2} k^2 \\ & \Rightarrow K E=\frac{1}{2} m \omega^2\left(A^2-x^2\right) \quad\left[\because k=m \omega^2\right]\end{aligned}$
$\mathrm{KE}$ is maximum at $\mathrm{x}=0$ \& $\mathrm{KE}$ is zero when $\mathrm{x}=\mathrm{A}$ $\therefore \mathrm{K} . \mathrm{E}$ vs $\mathrm{x}$ graph is parabola.

Topic: Physics, Oscillations, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) starting from mean position to extreme position (A) is given by

Options:
A

B

C

D

The correct answer is:

Solution:
$\begin{aligned} & T \cdot E=K \cdot E+P . E \\ & \Rightarrow K \cdot E=T \cdot E-K \cdot E \\ & =\frac{1}{2} k^2-\frac{1}{2} k^2 \\ & \Rightarrow K E=\frac{1}{2} m \omega^2\left(A^2-x^2\right) \quad\left[\because k=m \omega^2\right]\end{aligned}$
$\mathrm{KE}$ is maximum at $\mathrm{x}=0$ \& $\mathrm{KE}$ is zero when $\mathrm{x}=\mathrm{A}$ $\therefore \mathrm{K} . \mathrm{E}$ vs $\mathrm{x}$ graph is parabola.

Topic: Physics, Oscillations, JEE Main
The velocity and acceleration vectors of a particle undergoing circular motion are $\overrightarrow{\mathrm{v}}=2 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}$ and $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}^{2}$ respectively at an instant of time. The radius of the circle is -

Options:
A $1 \mathrm{~m}$
B $2 \mathrm{~m}$
C $3 \mathrm{~m}$
D $4 \mathrm{~m}$

The correct answer is:
$1 \mathrm{~m}$

Solution:
It can be observed that component of acceleration perpendicular to velocity is $\mathrm{a}=4 \mathrm{~m} / \mathrm{s}^{2}$

$$

\therefore \text { radius }=\frac{\mathrm{v}^{2}}{\mathrm{a}_{\mathrm{c}}}=\frac{(2)^{2}}{4}=1 \text { metre }

$$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2013
The velocity of a projectile at the initial point A is 2i^-3j^ m s-1. Its velocity (in m s-1) at point B is

Options:
A -2i^-3j^
B -2i^+3j^
C 2i^-3j^
D 2i^+3j^

The correct answer is:
2i^+3j^

Solution:
From the figure, the X-component remains unchanged while the Y-component is reverse. Then, the velocity at point B is 2i^+3j^ m s-1.

Topic: Physics, Mathematics in Physics, JEE Main
The velocity of a projectile at the initial point A is 2i^-3j^ m s-1. Its velocity (in m s-1) at point B is

Options:
A -2i^-3j^
B -2i^+3j^
C 2i^-3j^
D 2i^+3j^

The correct answer is:
2i^+3j^

Solution:
From the figure, the X-component remains unchanged while the Y-component is reverse. Then, the velocity at point B is 2i^+3j^ m s-1.

Topic: Physics, Mathematics in Physics, BITSAT, BITSAT 2017
The vertices of the hyperbola $9 x^2-16 y^2-36 x+96 y-252=0$ are

Options:
A $(6,3),(-2,3)$
B $(6,3),(-6,3)$
C $(-6,3),(-6,-3)$
D $(2,3),(-2,3)$

The correct answer is:
$(6,3),(-2,3)$

Solution:
Given hyperbola is
$\begin{aligned}
& 9 x^2-16 y^2-36 x+96 y-252=0 \\
& \Rightarrow 9\left(x^2-4 x\right)-16\left(y^2-6 y\right)=252 \\
& \Rightarrow 9\left(x^2-4 x+4\right)-16\left(y^2-6 y+9\right) \\
& =252+36-144 \\
& \Rightarrow 9(x-2)^2-16(y-3)^2=144
\end{aligned}$
Put $x=X+2$ and $y=Y+3$
$\therefore$ Equation (i) becomes $\frac{X^2}{16}-\frac{Y^2}{9}=1$
Now, vertices are $X= \pm a$ wehre $a=4$ and $Y=0$
Hence, vertices are $(6,3),(-2,3)$.

Topic: Mathematics, Hyperbola, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The wave number of the limiting line in Lyman series of hydrogen is $109678 \mathrm{~cm}^{-1}$. The wave number of the limiting line in Balmer series of $\mathrm{He}^{+}$ would be :

Options:
A $54839 \mathrm{~cm}^{-1}$
B $109678 \mathrm{~cm}^{-1}$
C $219356 \mathrm{~cm}^{-1}$
D $438712 \mathrm{~cm}^{-1}$

The correct answer is:
$109678 \mathrm{~cm}^{-1}$

Solution:
$\mathrm{R}_{\mathrm{H}}=109678 \mathrm{~cm}^{-1}$

Wave number of the limiting line in Balmer series of $\mathrm{He}^{+}$

$$

\begin{array}{l}

=\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^{2}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right] \\

=109678 \times(2)^{2}\left[\frac{1}{[2)^{2}}-\frac{1}{\infty}\right] \\

=109678 \mathrm{~cm}^{-1}

\end{array}

$$

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2014
The wavelength corresponding to maximum energy for hydrogen is 91.2 nm. The corresponding wavelength for He+ ion is

Options:
A 2.28 nm
B 22.8 nm
C 182.4 nm
D 364.8 nm

The correct answer is:
22.8 nm

Solution:
Accordinq to Rydberg's equation

v¯=1λ=RZ21nL2-1nH2

where,

v¯= wave number

λ= wavelength

R= Rydberg constant =109677 cm-1

Z= atomic number

nL= lower energy state

nH= higher energy state.

i.e. 1λ∝Z2

Thus, λHe+λH=Z2HZ2He+

To find λHe+=?

Given, λH=91.2 nm

ZH=1

ZHe=2

∴ λHe+=1×91.24=22.8 nm

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2019
The wavelength of radiation emitted by a body depends upon

Options:
A the nature of its surface
B the area of its surface
C the temperature of its surface
D All of the above

The correct answer is:
All of the above

Solution:
No Solution

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2014
The work done in blowing a soap bubble of surface tension $0.06 \times \mathrm{Nm}^{-1}$ from $2 \mathrm{~cm}$ radius to $5 \mathrm{~cm}$ radius is

Options:
A $0.004168 \mathrm{~J}$
B $0.003168 \mathrm{~J}$
C $0.003158 \mathrm{~J}$
D $0.004568 \mathrm{~J}$

The correct answer is:
$0.003168 \mathrm{~J}$

Solution:
$\begin{aligned} \Delta \mathrm{W} &=\mathrm{S} \times \Delta \mathrm{A}=0.06 \times 4 \pi\left(\mathrm{r}_{2}^{2}-\mathrm{r}_{1}^{2}\right) \\ &=0.003168 \mathrm{~J} \end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2014
The work done in placing a charge of $8 \times 10^{-18}$ coulomb on a condenser of capacity 100 microfarad is

Options:
A $3.1 \times 10^{-26}$ joule
B $4 \times 10^{-10}$ joule
C $32 \times 10^{-32}$ joule
D $16 \times 10^{-32}$ joule

The correct answer is:
$32 \times 10^{-32}$ joule

Solution:
Work done $=\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{C}}=\frac{\left(8 \times 10^{-18}\right)^2}{2 \times 100 \times 10^{-6}}$
$=32 \times 10^{-32} \mathrm{~J}$

Topic: Physics, Capacitance, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The Young's modulus of a perfectly rigid body is

Options:
A unity
B zero
C infinity
D some finite non-zero constant

The correct answer is:
infinity

Solution:
For a perfectly rigid body strain produced is zero for the given force applied, so $\mathrm{Y}=$ stress/strain $=\infty$

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2012
There are four numbers of which the first three are in G.P. and the last three are in A.P., whose common difference is 6 . If the first and the last numbers are equal then two other numbers are

Options:
A -2,4
B -4,2
C 2,6
D none

The correct answer is:
-4,2

Solution:
Let the last three numbers in A.P. be a, a +6 , a $+12,$ then the first term is alsoa +12 . But $a+12, a, a+6$ are in G.P.

$\therefore \mathrm{a}^{2}=(\mathrm{a}+12)(\mathrm{a}+6) \Rightarrow \mathrm{a}^{2}=\mathrm{a}^{2}+18 \mathrm{a}+72$

$\therefore \mathrm{a}=-4$

$\therefore$ The numbers are 8,-4,2,8.

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2020
There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be

Options:
A 1: 1
B 2: 1
C 1: 2
D 4: 1

The correct answer is:
4: 1

Solution:
$\because$ Both wires are same materials so both will have same Young's modulus, and let it be Y. $Y=\frac{\text { stress }}{\text { strain }}=\frac{F}{\text { A. }(\Delta L / L)}, F=$ applied force

$\mathrm{A}=$ area of cross-section of wire

Now,

$\mathrm{Y}_{1}=\mathrm{Y}_{2} \Rightarrow \frac{\mathrm{FL}}{\left(\mathrm{A}_{1}\right)\left(\Delta \mathrm{L}_{1}\right)}=\frac{\mathrm{FL}}{\left(\mathrm{A}_{2}\right)\left(\Delta \mathrm{L}_{2}\right)}$

Since load and length are same for both

$\begin{array}{l}

\Rightarrow \mathrm{r}_{1}^{2} \Delta \mathrm{L}_{1}=\mathrm{r}_{2}^{2} \Delta \mathrm{L}_{2},\left(\frac{\Delta \mathrm{L}_{1}}{\Delta \mathrm{L}_{2}}\right)=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2}=4 \\

\Delta \mathrm{L}_{1}: \Delta \mathrm{L}_{2}=4: 1

\end{array}$

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2020
There is a shell of mass $\mathrm{M}$ and density of the shell is uniform. The work done to take a point mass from point $\mathrm{A}$ to $\mathrm{B}$ is $[\mathrm{AB}=\mathrm{r}]$

Options:
A $\frac{\mathrm{GmM}}{\mathrm{r}}$
B $\frac{\mathrm{GmM}}{\mathrm{R}}$
C $-\frac{\mathrm{GmM}}{\mathrm{r}}$
D zero

The correct answer is:
zero

Solution:
Gravitaional field inside the shell is zero, so no work required.

Topic: Physics, Gravitation, JEE Main
There is a shell of mass $\mathrm{M}$ and density of the shell is uniform. The work done to take a point mass from point $\mathrm{A}$ to $\mathrm{B}$ is $[\mathrm{AB}=\mathrm{r}]$

Options:
A $\frac{\mathrm{GmM}}{\mathrm{r}}$
B $\frac{\mathrm{GmM}}{\mathrm{R}}$
C $-\frac{\mathrm{GmM}}{\mathrm{r}}$
D zero

The correct answer is:
zero

Solution:
Gravitaional field inside the shell is zero, so no work required.

Topic: Physics, Gravitation, BITSAT, BITSAT 2013
Thermosetting polymer bakelite is formed by the reaction of phenol with

Options:
A CH3·CH2·CHO
B CH3.CHO
C H.CHO
D HCOOH

The correct answer is:
H.CHO

Solution:
Bakelite is a thermosetting plastic which is formed by the reaction of phenol with H.CHO in the presence of H2SO4 (conc).

The reaction occurs as follows:

A cross-linked condensation polymer (at o-and p-position) bakelite is thus produced.

Topic: Chemistry, Polymers, BITSAT, BITSAT 2019
Ther.m.s. velocity of oxygen molecule at $16^{\circ} \mathrm{C}$ is $474 \mathrm{~m} / \mathrm{sec}$. The r.m.s. velocity in $\mathrm{m} / \mathrm{s}$ of hydrogen molecule at $127^{\circ} \mathrm{C}$ is

Options:
A 1603
B 1896
C 2230.59
D 2730

The correct answer is:
2230.59

Solution:
$v_{\operatorname{oxg} .}=\sqrt{\frac{3 R \times 289}{32}}\left(v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{\mathrm{M}}}\right)$

$\mathrm{v}_{\mathrm{H}}=\sqrt{\frac{3 \mathrm{R} \times 400}{2}}$ so $\mathrm{v}_{\mathrm{H}}=2230.59 \mathrm{~m} / \mathrm{sec}$

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2020
Thiamine and pyridoxine are also known respectively as :

Options:
A
Vitamin $B_2$ and Vitamin $E$
B
Vitamin $E$ and Vitamin $B_2$
C
Vitamin $B_6$ and Vitamin $B_2$
D
Vitamin $B_1$ and Vitamin $B_6$

The correct answer is:
Vitamin $B_1$ and Vitamin $B_6$

Solution:
Vitamin $B_1$ - Thiamin Vitamin
$\mathrm{B}_6$ - Pyridoxine

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Thomas slag is

Options:
A $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$
B $\mathrm{CaSiO}_{3}$
C Mixture of (a) and (b)
D $\mathrm{FeSiO}_{3}$

The correct answer is:
Mixture of (a) and (b)

Solution:
Calcium silicophosphate (a mixture of $\left.\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \& \mathrm{Ca}_{2} \mathrm{SiO}_{4}\right)$ is called Thomas slag.

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2015
Three blocks of masses $\mathrm{m}_{1}, \mathrm{~m}_{2}$ and $\mathrm{m}_{3}$ are connected by massless strings, as shown, on a frictionless table. They are pulled with a force $\mathrm{T}_{3}$ $=40 \mathrm{~N}$. If $m_{1}=10 \mathrm{~kg}, \mathrm{~m}_{2}=6 \mathrm{~kg}$ and $\mathrm{m}_{3}=4 \mathrm{~kg},$ the
tension $\mathrm{T}_{2}$ will be

Options:
A $20 \mathrm{~N}$
B $40 \mathrm{~N}$
C $10 \mathrm{~N}$
D $32 \mathrm{~N}$

The correct answer is:
$32 \mathrm{~N}$

Solution:
For equilibrium of all 3 masses,
$$
a=\frac{\mathrm{T}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}
$$
For equilibrium of $\mathrm{m}_{1} \& \mathrm{~m}_{2}$ $\mathrm{T}_{2}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \cdot \mathrm{a}$ or, $\mathrm{T}_{2}=\frac{\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{T}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}$
Given $m_{1}=10 \mathrm{~kg}, \mathrm{~m}_{2}=6 \mathrm{~kg}, \mathrm{~m}_{3}=4 \mathrm{~kg}$
$\mathrm{T}_{3}=40 \mathrm{~N}$
$\therefore \mathrm{T}_{2}=\frac{(10+6) \cdot 40}{10+6+4}=32 \mathrm{~N}$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2011
Three bricks each of length $L$ and mass $M$ are arranged as shown from the wall. The distance of the centre of mass of the system from the wall is

Options:
A $\frac{\mathrm{L}}{4}$
B $\frac{\mathrm{L}}{2}$
C $\left(\frac{3}{2}\right) \mathrm{L}$
D $\left(\frac{11}{12}\right) \mathrm{L}$

The correct answer is:
$\left(\frac{11}{12}\right) \mathrm{L}$

Solution:
$x_{1}=\frac{\mathrm{L}}{2}, x_{2}=\mathrm{L}, x_{3}=\frac{5 \mathrm{~L}}{4}$

$\therefore x_{\mathrm{CM}}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}}$

$=\frac{\mathrm{M} \times \frac{\mathrm{L}}{2}+\mathrm{M} \times \mathrm{L}+\mathrm{M} \times \frac{5 \mathrm{~L}}{4}}{\mathrm{M}+\mathrm{M}+\mathrm{M}}=\frac{11 \mathrm{~L}}{12}$

Topic: Physics, Center of Mass Momentum and Collision, BITSAT, BITSAT 2021
Three capacitors $\mathrm{C}_{1}=1 \mu \mathrm{F}, \mathrm{C}_{2}=2 \mu \mathrm{F}$ and $\mathrm{C}_{3}=3$

$\mu \mathrm{F}$ are connected as shown in figure, then the equivalent capacitance between points A and $\mathrm{B}$ is

Options:
A $3 \mu \mathrm{F}$
B $4 \mu \mathrm{F}$
C $5 \mu \mathrm{F}$
D $6 \mu \mathrm{F}$

The correct answer is:
$6 \mu \mathrm{F}$

Solution:
$C_{\mathrm{eq}}=(1+2+3) \mu \mathrm{F}=6 \mu \mathrm{F}$

Topic: Physics, Capacitance, JEE Main
Three capacitors $\mathrm{C}_{1}=1 \mu \mathrm{F}, \mathrm{C}_{2}=2 \mu \mathrm{F}$ and $\mathrm{C}_{3}=3$

$\mu \mathrm{F}$ are connected as shown in figure, then the equivalent capacitance between points A and $\mathrm{B}$ is

Options:
A $3 \mu \mathrm{F}$
B $4 \mu \mathrm{F}$
C $5 \mu \mathrm{F}$
D $6 \mu \mathrm{F}$

The correct answer is:
$6 \mu \mathrm{F}$

Solution:
$C_{\mathrm{eq}}=(1+2+3) \mu \mathrm{F}=6 \mu \mathrm{F}$

Topic: Physics, Capacitance, BITSAT, BITSAT 2014
Three charge q, Q and 4q are placed in a straight line of length $l$ at points distant $0, \frac{1}{2}$ and $l$ respectively from one end. In order to make the net froce on q zero, the charge Q must be equal to

Options:
A $-\mathrm{q}$
B $-2 q$
C $\frac{-\mathrm{q}}{2}$
D q

The correct answer is:
$-\mathrm{q}$

Solution:
$\left(\mathrm{F}_{\text {net }}\right)_{\mathrm{q}}=0$

$\Rightarrow \quad \mathrm{k} \frac{\mathrm{Qq}}{\left(\frac{\ell}{2}\right)^{2}}+\mathrm{k} \frac{4 \mathrm{q}^{2}}{\ell^{2}}=0$

$\begin{array}{l}

\text { where } \mathrm{k}=\frac{1}{4 \pi \varepsilon_{0}} \\

\Rightarrow 4 \mathrm{Qq}+4 \mathrm{q}^{2}=0 \quad \Rightarrow \mathrm{Q}=-\mathrm{q}

\end{array}$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2020
Three charges are placed at the three vertices of an equilateral triangle of side a as shown in the figure. The force experienced by the charge placed at the vertex A in a direction normal to BC is

Options:
A Q24πε0a2
B -Q24πε0a2
C zero
D Q22πε0a2

The correct answer is:
zero

Solution:
Resultant force

F'=F2+F2+2FFcos120°=F ∵cos120°=-12

Now, from figure

F=F'2+F''2+2F'F''cos90°

Now, the force normal to BC at vertex A is

F''=F2-F'2=0 ∵F'=F

Topic: Physics, Electrostatics, BITSAT, BITSAT 2017
Three equal charges (q) are placed at corners of an equilateral triangle of side $a$. The force on any charge is

Options:
A zero
B $\sqrt{3} \frac{\mathrm{Kq}^{2}}{\mathrm{a}^{2}}$
C $\frac{\mathrm{Kq}^{2}}{\sqrt{3} \mathrm{a}^{2}}$
D $3 \sqrt{3} \frac{\mathrm{Kq}^{2}}{\mathrm{a}^{2}}$

The correct answer is:
$\sqrt{3} \frac{\mathrm{Kq}^{2}}{\mathrm{a}^{2}}$

Solution:
$\mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}^{2}+\mathrm{F}^{2}+2 \mathrm{~F}^{2} \cos 60^{\circ}}=\sqrt{3} \mathrm{~F}$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2013
Three numbers are chosen at random without replacement from {1,2,3,……,8}. The probability that their minimum is 3, given that their maximum is 6, is

Options:
A 38
B 15
C 14
D 25

The correct answer is:
15

Solution:
Let A be the event that maximum is 6.

B be the event that minimum is 3.

PA=C25C38 (number less than 6 is 5)

PB=C25C38 (number greater than 3 is 5)

PA∩B=C12C38

∴ Required probability

=PB/A=PA∩BPA=C12C25=210=15

Topic: Mathematics, Probability, BITSAT, BITSAT 2019
Three particles, each of mass $m$ gram, are situated at the vertices of an equilateral triangle $A B C$ of side $\ell \mathrm{~cm}$ (as shown in the figure). The moment of inertia of the system about a line $A X$ perpendicular to $A B$ and in the plane of $A B C$, in gram $-\mathrm{cm}^2$ units will be

Options:
A $\frac{3}{2} m l^{2}$
B $\frac{3}{4} m l^{2}$
C $2 m l^{2}$
D $\frac{5}{4} m l^{2}$

The correct answer is:
$\frac{5}{4} m l^{2}$

Solution:
$\mathrm{I}_{\mathrm{AX}}=\mathrm{m}(\mathrm{AB})^{2}+\mathrm{m}(\mathrm{OC})^{2}=\mathrm{m} l^{2}+\mathrm{m}\left(l \cos 60^{\circ}\right)^{2}=m l^{2}+\frac{\mathrm{m} l^{2}}{4}=\frac{5}{4} \mathrm{~m} l^{2}$

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2021