datavorous/entrance-exam-dataset huggingface dataset:
The major product ' $\mathrm{Y}^{\prime}$ in the following reaction is :

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The mass of $_{7} \mathrm{~N}^{15}$ is $15.00011 \mathrm{amu}$, mass of $_{8} \mathrm{O}^{16}$ is 15.99492 amu and $m_{p}=1.00783$ amu. Determine binding energy of last proton of ${ }_{8} \mathrm{O}^{16}$.

Options:
A $2.13 \mathrm{MeV}$
B $0.13 \mathrm{MeV}$
C $10 \mathrm{MeV}$
D $12.13 \mathrm{MeV}$

The correct answer is:
$12.13 \mathrm{MeV}$

Solution:
$\mathrm{M}\left({ }_{8} \mathrm{O}^{16}\right)=\mathrm{M}\left({ }_{7} \mathrm{~N}^{15}\right)+1 \mathrm{~m}_{\mathrm{p}}$

binding energy of last proton $=M\left(N^{15}\right)+m_{P}-M\left({ }_{1} O^{16}\right)$

$=15.00011+1.00783-15.99492$

$=0.01302 \mathrm{amu}=12.13 \mathrm{MeV}$

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2016
The mass of the substance deposited when one Faraday of charge is passed through its solution is equal to

Options:
A relative equivalent weight
B gram equivalent weight
C specific equivalent weight
D None of the above

The correct answer is:
gram equivalent weight

Solution:
The mass of the substance deposited when one Faraday of charge is passed through its solution is equal to gram equivalent weight.

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2015
The masses of blocks $\mathrm{A}$ and $\mathrm{B}$ are $\mathrm{m}$ and $\mathrm{M}$ respectively. Between $\mathrm{A}$ and $\mathrm{B}$, there is a constant frictional force $\mathrm{F}$ and $\mathrm{B}$ can slide on a smooth horizontal surface. A is set in motion with velocity while $\mathrm{B}$ is at rest. What is the distance moved by A relative to B before they move with the same velocity?

Options:
A $\frac{\mathrm{mMv}_{0}^{2}}{\mathrm{~F}(\mathrm{~m}-\mathrm{M})}$
B $\frac{\mathrm{mMv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{~m}-\mathrm{M})}$
C $\frac{\mathrm{mMv}_{0}^{2}}{\mathrm{~F}(\mathrm{~m}+\mathrm{M})}$
D $\frac{\mathrm{mMv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{M}+\mathrm{m})}$

The correct answer is:
$\frac{\mathrm{mMv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{M}+\mathrm{m})}$

Solution:
For the blocks A and B FBD as shown below

Equations of motion

$$

\begin{array}{l}

\mathrm{a}_{\mathrm{A}}=\frac{\mathrm{F}}{\mathrm{M}}(\text { in }-\mathrm{x} \text { direction }) \\

\mathrm{a}_{\mathrm{B}}=\frac{\mathrm{F}}{\mathrm{M}}(\text { in }+\mathrm{x} \text { direction) }

\end{array}

$$

Relative acceleration, of A w.r.t. B,

$$

\begin{aligned}

\mathrm{a}_{\mathrm{A}, \mathrm{B}} &=\mathrm{a}_{\mathrm{A}}-\mathrm{a}_{\mathrm{B}}=-\frac{\mathrm{F}}{\mathrm{m}}-\frac{\mathrm{F}}{\mathrm{M}} \\

&=-\mathrm{F}\left(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{Mm}}\right)(\text { along }-\mathrm{x} \text { direction })

\end{aligned}

$$

Initial relative velocity of A w.r.t. B, $\mathrm{u}_{\mathrm{AB}}=\mathrm{v}_{0}$

using equation $v^{2}=u^{2}+2$ as

$$

0=v_{0}^{2}-\frac{2 \mathrm{~F}(\mathrm{~m}+\mathrm{M}) \mathrm{S}}{\mathrm{Mm}} \Rightarrow \mathrm{S}=\frac{\mathrm{Mmv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{~m}+\mathrm{M})}

$$

i.e., Distance moved by A relative to B

$$

\mathrm{S}_{\mathrm{AB}}=\frac{\mathrm{Mmv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{~m}+\mathrm{M})}

$$

Topic: Physics, Laws of Motion, JEE Main
The masses of blocks $\mathrm{A}$ and $\mathrm{B}$ are $\mathrm{m}$ and $\mathrm{M}$ respectively. Between $\mathrm{A}$ and $\mathrm{B}$, there is a constant frictional force $\mathrm{F}$ and $\mathrm{B}$ can slide on a smooth horizontal surface. A is set in motion with velocity while $\mathrm{B}$ is at rest. What is the distance moved by A relative to B before they move with the same velocity?

Options:
A $\frac{\mathrm{mMv}_{0}^{2}}{\mathrm{~F}(\mathrm{~m}-\mathrm{M})}$
B $\frac{\mathrm{mMv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{~m}-\mathrm{M})}$
C $\frac{\mathrm{mMv}_{0}^{2}}{\mathrm{~F}(\mathrm{~m}+\mathrm{M})}$
D $\frac{\mathrm{mMv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{M}+\mathrm{m})}$

The correct answer is:
$\frac{\mathrm{mMv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{M}+\mathrm{m})}$

Solution:
For the blocks A and B FBD as shown below

Equations of motion

$$

\begin{array}{l}

\mathrm{a}_{\mathrm{A}}=\frac{\mathrm{F}}{\mathrm{M}}(\text { in }-\mathrm{x} \text { direction }) \\

\mathrm{a}_{\mathrm{B}}=\frac{\mathrm{F}}{\mathrm{M}}(\text { in }+\mathrm{x} \text { direction) }

\end{array}

$$

Relative acceleration, of A w.r.t. B,

$$

\begin{aligned}

\mathrm{a}_{\mathrm{A}, \mathrm{B}} &=\mathrm{a}_{\mathrm{A}}-\mathrm{a}_{\mathrm{B}}=-\frac{\mathrm{F}}{\mathrm{m}}-\frac{\mathrm{F}}{\mathrm{M}} \\

&=-\mathrm{F}\left(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{Mm}}\right)(\text { along }-\mathrm{x} \text { direction })

\end{aligned}

$$

Initial relative velocity of A w.r.t. B, $\mathrm{u}_{\mathrm{AB}}=\mathrm{v}_{0}$

using equation $v^{2}=u^{2}+2$ as

$$

0=v_{0}^{2}-\frac{2 \mathrm{~F}(\mathrm{~m}+\mathrm{M}) \mathrm{S}}{\mathrm{Mm}} \Rightarrow \mathrm{S}=\frac{\mathrm{Mmv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{~m}+\mathrm{M})}

$$

i.e., Distance moved by A relative to B

$$

\mathrm{S}_{\mathrm{AB}}=\frac{\mathrm{Mmv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{~m}+\mathrm{M})}

$$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2015
The matrix \(A^2+4 A-5 I\), where \(I\) is identity matrix and \(A=\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right]\), equals :

Options:
A \(4\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]\)
B \(4\left[\begin{array}{cc}0 & -1 \\ 2 & 2\end{array}\right]\)
C \(32\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]\)
D \(32\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]\)

The correct answer is:
\(4\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]\)

Solution:
\(\begin{aligned}
& A^2+4 A-5 I=A \times A+4 A-5 I \\
= & {\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right] \times\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right]+4\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right]-5\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] } \\
= & {\left[\begin{array}{cc}
9 & -4 \\
-8 & 17
\end{array}\right]+\left[\begin{array}{cc}
4 & 8 \\
16 & -12
\end{array}\right]-\left[\begin{array}{cc}
5 & 0 \\
0 & 5
\end{array}\right] } \\
= & {\left[\begin{array}{cc}
9+4-5 & -4+8-0 \\
-8+16-0 & 17-12-5
\end{array}\right]=\left[\begin{array}{cc}
8 & 4 \\
8 & 0
\end{array}\right] } \\
= & 4\left[\begin{array}{cc}
2 & 1 \\
2 & 0
\end{array}\right]
\end{aligned}\)

Topic: Mathematics, Matrices, BITSAT, BITSAT 2011
The maximum value of f(x)=x+sin2x,x∈[0,2π] is

Options:
A π2
B 2π
C 3π4
D 3π2

The correct answer is:
2π

Solution:
Given, fx=x+sin2x

⇒ f'(x)=1+2cos2x

For maximum or minimum value, f'x=0

⇒ 1+2cos2x=0⇒cos2x=-12

⇒ 2x=2nπ±2π3

⇒ x=nπ±π3

⇒ x=π3,2π3,4π3,5π3

Find f0, fπ3,f2π3, f4π3, f5π3, f(2π)

⇒ f0=0,fπ3=π3+32,f2π3=2π3-0.8

⇒f4π3=4π3+0.8,f5π3=5π3-0.8

⇒ f2π=2π+0=2π

∴ Maximum value of fx=2π

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2017
The maximum value of Z=4x+2y subject to constraints 2x+3y≤18, x+y≥10 and x,y≥0 is

Options:
A 20
B 36
C 40
D None of these

The correct answer is:
None of these

Solution:
We have,

Z=4x+2y

And,

2x+3y≤18, x+y≥10

Let

L1:2x+3y=18

L2: x+y=10

L3:x=0

L4:y=0

Then the feasible region is

From the figure, it is clear that there is no common area. So, we cannot find maximum value of Z.

Hence, option d is correct.

Topic: Mathematics, Linear Programming, BITSAT, BITSAT 2018
The maximum value of $z=5 x+2 y$ subject to constraints $x+y \leq 7, x+2 y \leq 10, x, y \geq 0$

Options:
A 10
B 26
C 35
D 70

The correct answer is:
35

Solution:
Change the inequalities into equations and draw the graph of lines, thus we get the required feasible region as shown below.
The region bounded by the vertices $A(0,5), B(4,3), C(7,0)$
The objective function is maximum at $\mathrm{C}(7,0)$ and $\operatorname{Max} z=5 \times 7+2 \times 0=35$.

Topic: Mathematics, Linear Programming, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The maximum value of $z=6 x+8 y$ subject to constraints $2 x+y \leq 30, x+2 y \leq 24$ and $x \geq 0, y \geq 0$ is

Options:
A 90
B 120
C 96
D 240

The correct answer is:
120

Solution:
Here, $2 x+y \leq 30, x+2 y \leq 24, x, y \geq 0$

The shaded region represents the feasible region, hence $\mathrm{z}=6 \mathrm{x}+8 \mathrm{y}$. Obviously it is maximum at (12,6) Hence $z=12 \times 6+8 \times 6=120$

Topic: Mathematics, Linear Programming, BITSAT, BITSAT 2020
The maximum value of z=9x+13y subject to constraints 2x+3y≤18,2x+y≤10,x≥0,y≥0 is

Options:
A 130
B 81
C 79
D 99

The correct answer is:
79

Solution:
Given, constraints are2x+3y≤18,2x+y≤10,x≥0,y≥0.

The feasible region is OABCO.

Also, z=9x+13y

At O(0,0),z=0

At A(5,0),z=45

At B(3,4),z=27+52=79

At C(0,6),z=78

∴ Maximum value of z is 79.

Topic: Mathematics, Mathematical Induction, BITSAT, BITSAT 2017
The mean and variance of a random variable $\mathrm{X}$ having binomial distribution are 4 and 2 respectively, then $\mathrm{P}(\mathrm{X}=1)$ is

Options:
A $\frac{1}{4}$
B $\frac{1}{32}$
C $\frac{1}{16}$
D $\frac{1}{8}$

The correct answer is:
$\frac{1}{32}$

Solution:
$\begin{aligned} n p &=4 \\ n p q &=2 \\ \Rightarrow q &=\frac{1}{2}, p=\frac{1}{2}, n=8 \\ \mathrm{P}(\mathrm{X}&=1)={ }^{8} \mathrm{C}_{1}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{7}=8 \cdot \frac{1}{2^{8}}=\frac{1}{2^{5}}=\frac{1}{32} \end{aligned}$

Topic: Mathematics, Probability, BITSAT, BITSAT 2021
The mean of five observations is $4$ and their variance is $5.2$. If three of these observations are $1$,$2$ and $6$ , then the other two are

Options:
A
$2$ and $9$

B
$3$ and $8$

C
$4$ and $7$

D
$5$ and $6$

The correct answer is:

$4$ and $7$

Solution:
Let the two unknown items be $x$ and $y$. Then,
$\operatorname{Mean}=4 \Rightarrow \frac{1+2+6+x+y}{5}=4$
$\Rightarrow \quad x+y=11\dots(i)$
and variance $=5.2$
$\Rightarrow \frac{1^2+2^2+6^2+x^2+y^2}{5}-(\text { mean })^2=5.2$
$\Rightarrow \quad 41+x^2+y^2=5(5.2+16)$
$\begin{array}{ll}\Rightarrow & 41+x^2+y^2=106 \\ \Rightarrow & x^2+y^2=65\end{array}$
On solving Eqs. (i) and (ii), we get
$x=4, y=7 \text { or } x=7, y=4$

Topic: Mathematics, Statistics, BITSAT, BITSAT 2022
The mean square deviation of a set of n observation $\mathrm{x}_{1}, \mathrm{x}_{2^{2}}, \ldots \mathrm{x}_{\mathrm{n}}$ about a point $\mathrm{c}$ is defined as $\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-c\right)^{2}$

The mean square deviations about -2 and 2 are 18 and 10 respectively, the standard deviation of this set of observations is

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

The correct answer is:
3

Solution:
We have $\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}+2\right)^{2}=18$ and

$$

\begin{array}{l}

\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=10 \\

\Rightarrow \sum_{i=1}^{n}\left(x_{i}+2\right)^{2}=18 n \text { and }

\end{array}

$$

$$

\begin{array}{l}

\sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=10 n \\

\Rightarrow \sum_{i=1}^{n}\left(x_{i}+2\right)^{2}+\sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=28 n

\end{array}

$$

and $\sum_{i=1}^{n}\left(x_{i}+2\right)^{2}-\sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=8 n$

$$

\begin{array}{l}

\Rightarrow 2 \sum_{i=1}^{n}\left(x_{i}+4\right)^{2}=28 n \text { and } 2 \sum_{i=1}^{n} 4 x_{i}=8 n \\

\Rightarrow \sum_{i=1}^{n} x_{1}^{2}+4 n=14 n \text { and } \sum_{i=1}^{n} x_{i}=n \\

\Rightarrow \sum_{i=1}^{n} x_{i}^{2}=10 n \text { and } \sum_{i=1}^{n} x_{i}=n \\

\therefore \sigma=\sqrt{\frac{1}{n} \sum_{i=1}^{n} x_{i}^{2}-\left(\frac{1}{n} \sum_{i=1^{\prime}}^{n} x_{i}\right)^{2}}=\sqrt{\frac{10 n}{n}-\left(\frac{n}{n}\right)^{2}}=3

\end{array}

$$

Topic: Mathematics, Statistics, BITSAT, BITSAT 2015
The metal present in vitamin $\mathrm{B}_{12}$ is

Options:
A magnesium
B cobalt
C copper
D zinc

The correct answer is:
cobalt

Solution:
Cobalt is present in vitamin $\mathrm{B}_{12}$.

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2015
The middle term in the expansion of $\left(\frac{10}{x}+\frac{x}{10}\right)^{10}$ is

Options:
A ${ }^{10} \mathrm{C}_5$
B ${ }^{10} \mathrm{C}_6$
C ${ }^{10} \mathrm{C}_5 \frac{1}{x^{10}}$
D ${ }^{10} \mathrm{C}_5 \mathrm{x}^{10}$

The correct answer is:
${ }^{10} \mathrm{C}_5$

Solution:
General term $=T_{r+1}$
$={ }^{10} \mathrm{C}_r\left(\frac{10}{x}\right)^{10-r}\left(\frac{x}{10}\right)^r$
Here $n=10$, which is an even number.
Now, $\left[\frac{10}{2}+1\right]^{\text {th }}$ term i.e. $6^{\text {th }}$ term is the middle term.
Hence, middle term $=T_6$
$\mathrm{T}_{5+1}={ }^{10} \mathrm{C}_5\left(\frac{10}{x}\right)^{10-5}\left(\frac{x}{10}\right)^5$
$={ }^{10} \mathrm{C}_5\left(\frac{10}{x}\right)^5\left(\frac{x}{10}\right)^5={ }^{10} \mathrm{C}_5$.

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The minimum value of \(\frac{x^4+y^4+z^2}{x y z}\) for positive real number \(x, y, z\) is

Options:
A \(\sqrt{2}\)
B \(2 \sqrt{2}\)
C \(4 \sqrt{2}\)
D \(8 \sqrt{2}\)

The correct answer is:
\(2 \sqrt{2}\)

Solution:
By A.M. \(\geq\) G.M.
\(x^4+y^4 \geq 2 x^2 y^2\) and \(2 x^2 y^2+z^2 \geq \sqrt{8} x y z\).
\(\Rightarrow \frac{x^4+y^4+z^2}{x y z} \geq \sqrt{8}\)

Topic: Mathematics, Limits, BITSAT, BITSAT 2010
The minimum value of \(f(x)=\sin ^4 x+\cos ^4 x\) in the interval \(\left(0, \frac{\pi}{2}\right)\) is

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

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

Solution:
Let \(y=\sin ^4 x+\cos ^4 x\)
\(\begin{aligned}
& \frac{d y}{d x}=4 \sin ^3 x \cos x+4 \cos ^3 x(-\sin x) \\
& =4 \sin x \operatorname{cox}\left(\sin ^2 x-\cos ^2 x\right) \\
& =(2 \sin 2 x)(-\cos 2 x)=-\sin 4 x \\
& \therefore \frac{d y}{d x}=0 \Rightarrow \sin 4 x=0 \Rightarrow 4 x=0, \pi, 2 \pi, 3 \pi
\end{aligned}\)
or \(\mathrm{x}=0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \ldots \ldots \ldots, \Rightarrow \mathrm{x}=\frac{\pi}{4}\)

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2011
The minimum value of the function \(y=x^4-2 x^2+1\) in the interval \(\left[\frac{1}{2}, 2\right]\) is

Options:
A 0
B 2
C 8
D 9

The correct answer is:
0

Solution:
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^4-2 \mathrm{x}^2+1\right)=4 \mathrm{x}\left(\mathrm{x}^2-1\right)\)
For \(\max\). or \(\min , \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
\(4 \mathrm{x}\left(\mathrm{x}^2-1\right)=0 ;\), either \(\mathrm{x}=0\) or \(\mathrm{x}= \pm 1\)
\(x=0\) and \(x=-1\) does not belong to \(\left[\frac{1}{2}, 2\right]\)
\(\frac{d^2 y}{d x^2}=12 x^2-4 \therefore\left(\frac{d^2 y}{d x^2}\right)_{x=1}\)
\(=12(1)^2-4=8 > 0\)
\(\therefore \quad\) there is minimum value of function at \(\mathrm{x}=1\)
\(\therefore \quad\) minimum value is
\(y(1)=1^4-2(1)^2+1=1-2+1=0\)

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2009
The minimum value of the function \(z=4 x+3 y\) subject to the constraints \(3 x+2 y \geq 160,5 x+2 y \geq 200\), \(x+2 y \geq 80, x \geq 0, y \geq 0\) is

Options:
A 320
B 300
C 220
D 200

The correct answer is:
220

Solution:
No Solution Available

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2009
The minimum velocity (in $\mathrm{ms}^{-1}$ ) with which a car driver must traverse a flat curve of radius $150 \mathrm{~m}$ and coefficient of friction 0.6 to avoid skidding is

Options:
A 60
B 30
C 15
D 25

The correct answer is:
30

Solution:
The condition to avoid skidding,

$$

\mathrm{v}=\sqrt{\mu \mathrm{rg}}=\sqrt{0.6 \times 150 \times 10}=30 \mathrm{~m} / \mathrm{s}

$$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2013
The modulus of $\frac{(1+i \sqrt{3})(2+2 i)}{(\sqrt{3}-i)}$ is

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

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

Solution:
$\begin{aligned} & \frac{(1+i \sqrt{3})(2+2 i)}{\sqrt{3}-i}=\frac{2+2 \sqrt{3} i+2 i-2 \sqrt{3}}{\sqrt{3}-i} \\ & =\frac{(2-2 \sqrt{3})+(2 \sqrt{3}+2) i}{\sqrt{3}-i} \times \frac{\sqrt{3}+i}{\sqrt{3}+i} \\ & =\frac{2 \sqrt{3}-6+2 i-2 \sqrt{3} i+6 i+2 \sqrt{3} i-2 \sqrt{3}-2}{3+1} \\ & =\frac{8 i-8}{4}=-2+2 i \\ & \therefore \text { Modulus }=\sqrt{(-2)^2+(2)^2}=2 \sqrt{2}\end{aligned}$

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The molecules of a given mass of a gas have

r.m.s. velocity of $200 \mathrm{~ms}^{-1}$ at $27^{\circ} \mathrm{C}$ and $1.0 \times 10^{5}$ $\mathrm{Nm}^{-2}$ pressure. When the temperature and pressure of the gas are respectively, $127^{\circ} \mathrm{C}$ and $0.05 \times 10^{5} \mathrm{Nm}^{-2},$ the r.m.s. velocity of its molecules in $\mathrm{ms}^{-1}$ is :

Options:
A $100 \sqrt{2}$
B $\frac{400}{\sqrt{3}}$
C $\frac{100 \sqrt{2}}{3}$
D $\frac{100}{3}$

The correct answer is:
$\frac{400}{\sqrt{3}}$

Solution:
Here $v_{1}=200 \mathrm{~m} / \mathrm{s}$; temperature $\mathrm{T}_{1}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{k}$

temperature $\mathrm{T}_{2}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{k}$

$\mathrm{V}=?$

R.M.S. Velocity, $\mathrm{V} \propto \sqrt{\mathrm{T}}$

$$

\begin{array}{l}

\Rightarrow \quad \frac{v}{200}=\sqrt{\frac{400}{300}} \\

\Rightarrow \quad v=\frac{200 \times 2}{\sqrt{3}} \mathrm{~m} / \mathrm{s} \\

\Rightarrow \quad \mathrm{v}=\frac{400}{\sqrt{3}} \mathrm{~m} / \mathrm{s}

\end{array}

$$

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2016
The molecules of a given mass of gas have a root mean square velocity of $200 \mathrm{~m} \mathrm{~s}^{-1}$ at $27^{\circ} \mathrm{C}$ and $1.0 \times 10^{5} \mathrm{Nm}^{-2}$ pressure. When the temperature is $127^{\circ} \mathrm{C}$ and the pressure $0.5 \times 10^{5} \mathrm{Nm}^{-2},$ the root mean square velocity in $\mathrm{ms}^{-1}$, is

Options:
A $\frac{400}{\sqrt{3}}$
B $100 \sqrt{2}$
C $\frac{100 \sqrt{2}}{3}$
D $\frac{100}{3}$

The correct answer is:
$\frac{400}{\sqrt{3}}$

Solution:
$\frac{c_{2}}{c_{1}}=\sqrt{\frac{400}{300}}=\frac{2}{\sqrt{3}}$
$$
\Rightarrow c_{2}=\frac{2}{\sqrt{3}} \times 200=\frac{400}{\sqrt{3}} \mathrm{~ms}^{-1}
$$

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2011
The most acidic oxide for nitrogen is

Options:
A NO2
B N2O
C NO
D N2O5

The correct answer is:
N2O5

Solution:
The acidic nature of the oxides of nitrogen can be determined by the strength of the acid formed by the nitrogen oxide in water.

More higher be the (+) ve oxidation number of N-atom in its oxide more acidic be the nature of oxide. Thus N2O5, is most acidic oxide.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2017
The most likely acid-catalysed aldol condensation products of each of the two aldehydes I and II will respectively be

Options:
A
B
C
D

The correct answer is:

Solution:
\(\left[F e F_{6}\right]^{3-}\) complex ion possesses sp \(^{3} d^{2}\) hybridization.

The electronic configurations of \(C r, F e, C o\) and \(T i\) in the complexes are \((3 d)^{3},(3 d)^{5},(3 d)^{6}\) and \((3 d)^{1}\), respectively.

These involve \(d^{2} s p^{3}, s p^{3} d^{2}, d^{2} s p^{3}\) and \(d^{2} s p^{3}\) hybridization respectively.

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2016
The most reactive amine towards reactions with dil. HCl is

Options:
A CH3-NH2
B (CH3)2NH
C CH33·N
D PhNH2

The correct answer is:
(CH3)2NH

Solution:
More be the ability of N-atom of -NH2 group to donate lone pair of electrons to HCl (dil.), more be the reactivity of amine.

In (B) two alkyl groups are attached to the -NH2 (having electron releasing nature) which increases the electron donating ability of N-atom of -NH2. In aqueous medium, secondary amine is most basic. Hence, it is most reactive. In (D), due to resonance, lone pair of electrons get delocalised and its basic strength is very less.

Topic: Chemistry, Amines, BITSAT, BITSAT 2017
The most stable carbanion among the following is:

Options:
A
B
C
D

The correct answer is:

Solution:
$-\mathrm{NO}_{2}$ group, being strong electron withdrawing, disperses the -ve charge, hence stabilizes the concerned carbanion.

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2013
The most stable ion is

Options:
A $\left[\mathrm{Fe}(\mathrm{OH})_{3}\right]^{3-}$
B $\left[\mathrm{FeCl}_{6}\right]^{3-}$
C $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$
D $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$

The correct answer is:
$\left[\mathrm{FeCl}_{6}\right]^{3-}$

Solution:
A more basic ligand forms stable bond with metal ion, $\mathrm{Cl}^{-}$ is most basic amongst all.

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2011
The nth roots of unity are in

Options:
A AP
B GP
C HP
D None of these

The correct answer is:
GP

Solution:
Let x=1

=cos0°+isin0°

⇒ x1/n=ei2πr/n; r=0, 1, 2,…

Then, the roots are 1, e2πin, e4πin,… which are clearly on GP with common ratio e2πi/n

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2018
The nearest point on the line $3 x+4 y=12$ from the origin is

Options:
A $\left(\frac{36}{25}, \frac{48}{25}\right)$
B $\left(3, \frac{3}{4}\right)$
C $\left(2, \frac{3}{2}\right)$
D None of these

The correct answer is:
$\left(\frac{36}{25}, \frac{48}{25}\right)$

Solution:
If ' $\mathrm{D}$ ' be the foot of altitude, drawn from origin to the given line, then ${ }^{4} \mathrm{D}^{\prime}$ is the required point. Let $\angle \mathrm{OBA}=\theta$
$\Rightarrow \tan \theta=4 / 3$
$\Rightarrow \angle \mathrm{DOA}=\theta$
$\begin{aligned}
&\text { we have } \quad \mathrm{OD}=12 / 5 \text { . }\\
&\text { If } D \text { is }(h, k) \text { then } h=O D \cos \theta, k=O D \sin \theta\\
&\Rightarrow \mathrm{h}=36 / 25, \mathrm{k}=48 / 25 .
\end{aligned}$

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2012
The negation of the Boolean expression $((\sim q)$ $\wedge p) \Rightarrow((\sim p) \vee q)$ is logically equivalent to

Options:
A $p \Rightarrow q$
B $q \Rightarrow p$
C $\sim(p \Rightarrow q)$
D $\sim(q \Rightarrow p)$

The correct answer is:
$\sim(p \Rightarrow q)$

Solution:
Given boolen expression are
$\begin{aligned}
& \sim p \vee q \equiv p \rightarrow q \text { and } \sim q \wedge p \equiv \sim(p \rightarrow q) \\
& \text { Negation of } \sim(p \rightarrow q) \rightarrow(p \rightarrow q) \\
& \text { is } \sim(p \rightarrow q) \wedge(\sim(p \rightarrow q)) \text { i.e., } \sim(p \rightarrow q)
\end{aligned}$

Topic: Mathematics, Mathematical Reasoning, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The night-blindness is developed due to deficiency of vitamin

Options:
A B6
B C
C B12
D A

The correct answer is:
A

Solution:
Night blindness is developed due to deficiency of vitamin-A.

Vitamin A deficiency leads to a disease known as night blindness. People with night blindness have a normal vision when adequate light is present but poor vision in the dark.

Vitamin A is necessary for the production of retinal, a component of rhodopsin.

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2017
The non stoichiometric compound $\mathrm{Fe}_{0.94} \mathrm{O}$ is formed when $\mathrm{x} \%$ of $\mathrm{Fe}^{2+}$ ions are replaced by as many $\frac{2}{3} \mathrm{Fe}^{3+}$ ions, $\mathrm{x}$ is

Options:
A 18
B 12
C 15
D 6

The correct answer is:
18

Solution:
The number of $\mathrm{Fe}^{3+}$ ions replacing $\mathrm{x} \mathrm{Fe}^{2+}$ ions $=\frac{2 \mathrm{x}}{3}$ vacancies of cations

$$

\begin{array}{l}

=x-\frac{2 x}{3}=x / 3 \\

\text { But } x / 3=1-0.94=0.06, x=0.06 \times 3 \\

=0.18=18 \%

\end{array}

$$

Topic: Chemistry, Solid State, BITSAT, BITSAT 2016
The normality of $26 \%$ (wt/vol) solution of ammonia (density $=0.855$ ) is approximately:

Options:
A 1.5
B 0.4
C 15.3
D 4

The correct answer is:
15.3

Solution:
Wt. of $\mathrm{NH}_{3}=26 \mathrm{~g}=\frac{26}{17} \mathrm{~g} \mathrm{eq}=1.53 \mathrm{~g} \mathrm{eq}$

Vol. of soln. $=100 \mathrm{mc}=0.1 \mathrm{~L}$

$\therefore$ Normality $=\frac{1.53}{0.1}=15.3 \mathrm{~N}$

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2016
The number of $3 \mathrm{C}-2 e^{-}$bonds present in diborane is

Options:
A
1

B
2

C
3

D
4

The correct answer is:

2

Solution:
The structure of diborane is

In this structure, there are two $3 \mathrm{C}-2 e^{-}$bonds. 3 atoms, $\quad \mathrm{B}-\mathrm{H}-\mathrm{B}$ share two electrons and form an angular geometry, leading bent bond, also known as banana bond.

Topic: Chemistry, p Block Elements (Group 13 & 14), BITSAT, BITSAT 2022