datavorous/entrance-exam-dataset huggingface dataset:
If $y=\sqrt{\left(\frac{1+\cos 2 \theta}{1-\cos 2 \theta}\right)}$, then $\frac{d y}{d \theta}$ at $\theta=\frac{3 \pi}{4}$ is :

Options:
A -2
B 2
C $\pm 2$
D None of these

The correct answer is:
-2

Solution:
$y=\sqrt{\frac{1+\cos 2 \theta}{1-\cos 2 \theta}}$
$\Rightarrow y=\sqrt{\frac{2 \cos ^2 \theta}{2 \sin ^2 \theta}}=\sqrt{\cot ^2 \theta}$
$\Rightarrow y=\cot \theta$
Differentiate w.r.t. ' $\theta$ ', we get: $\frac{d y}{d \theta}=-\operatorname{cosec}^2 \theta$
Now, $\left(\frac{\mathrm{dy}}{\mathrm{d} \theta}\right)_{\theta=\frac{3 \pi}{4}}=-\operatorname{cosec}^2\left(\frac{3 \pi}{4}\right)$
$=-\operatorname{cosec}^2\left(\pi-\frac{\pi}{4}\right)=-\operatorname{cosec}^2 \frac{\pi}{4}=-2$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2023 (Memory Based Paper 1)
If $y=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}$, then value of $\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}$ is

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

The correct answer is:
$y$

Solution:
$\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}$

$=\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \cdot \frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha}$

$=\frac{(1+\sin \alpha)^{2}-\cos ^{2} \alpha}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$

$=\frac{\left(1+\sin ^{2} \alpha+2 \sin \alpha\right)-\left(1-\sin ^{2} \alpha\right)}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$

$=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=y$

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2021
If $y=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots . \infty,$ then $\frac{d y}{d x}$ is equal to

Options:
A $x$
B 1
C $y$
D None of these

The correct answer is:
$y$

Solution:
$y=e^{x} \Rightarrow \frac{d y}{d x}=e^{x}=y$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2013
If $y=3 x+6 x^{2}+10 x^{3}+\ldots \ldots \ldots, \infty,$ then

$\frac{1}{3} y-\frac{1.4}{3^{2} 2} y^{2}+\frac{1.4 .7}{3^{2} 3} y^{3}-\ldots \ldots \infty$ is equal to

Options:
A $x$
B $1-x$
C $1+x$
D $x^{x}$

The correct answer is:
$x$

Solution:
We have $y=3 x+6 x^{2}+10 x^{3}+\ldots \ldots$

$\Rightarrow 1+y=1+3 x+6 x^{2}+10 x^{3}+\ldots$

$\Rightarrow 1+y=(1-x)^{-3} \Rightarrow 1-x=(1+y)^{-1 / 3}$

$\Rightarrow x=1-(1+y)^{-1 / 3}$

$=\frac{1}{3} y-\frac{1.4}{3^{2} \cdot 2} y^{2}+\frac{1 \cdot 4 \cdot 7}{3^{2} \cdot 3} y^{3}-\ldots$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2020
If $y=\left(x+\sqrt{1+x^{2}}\right)^{n},$ then $\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}$ is

Options:
A $n^{2} y$
B $-n^{2} y$
C $-y$
D $2 x^{2} y$

The correct answer is:
$n^{2} y$

Solution:
$y=\left(x+\sqrt{1+x^{2}}\right)^{n}$

$$

\begin{array}{c}

\frac{d y}{d x}=n\left(x+\sqrt{1+x^{2}}\right)^{n-1}\left(1+\frac{1}{2}\left(1+x^{2}\right)^{-1 / 2} \cdot 2 x\right) \\

\qquad \begin{array}{c}

\frac{d y}{d x}=n\left(x+\sqrt{1+x^{2}}\right)^{n-1} \frac{\left(\sqrt{1+x^{2}}+x\right)}{\sqrt{1+x^{2}}} \\

=\frac{n\left(\sqrt{1+x^{2}}+x\right)^{n}}{\sqrt{1+x^{2}}}

\end{array}

\end{array}

$$

or $\sqrt{1+x^{2}} \frac{d y}{d x}=n y$ or $\sqrt{1+x^{2}} y_{1}=n y$

$$

\left(y_{1}=\frac{d y}{d x}\right)

$$

Squaring, $\left(1+x^{2}\right) y_{1}^{2}=n^{2} y^{2}$

Differentiating,

$$

\left(1+x^{2}\right) 2 y_{1} y_{2}+y_{1}^{2} \cdot 2 x=n^{2} \cdot 2 y y_{1}

$$

or $\left(1+x^{2}\right) y_{2}+x y_{1}=n^{2} y$

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2015
If $y^x=e^{y-x}$, then $\frac{d y}{d x}$ is equal to

Options:
A $\frac{1+\log y}{y \log y}$
B $\frac{(1+\log y)^2}{y \log y}$
C $\frac{1+\log y}{(\log y)^2}$
D $\frac{(1+\log y)^2}{\log y}$

The correct answer is:
$\frac{(1+\log y)^2}{\log y}$

Solution:
Here, $\mathrm{y}^{\mathrm{x}}=\mathrm{e}^{\mathrm{y}-\mathrm{x}}$
Taking log on both sides, we get
$\begin{aligned}
& \log y^x=\log e^{y-x} \\
& \quad\left(\because \log a^b=b \log a \text { and } \log e=1\right)
\end{aligned}$
$\Rightarrow x \log y=(y-x) \log e \Rightarrow x \log y=y-x \ldots(i)$
On differentiating w.r.t. $x$, we get
$\frac{d}{d x}(x \log y)=\frac{d}{d x}(y-x)$
(using product rule)
$\begin{aligned}
& \Rightarrow x\left(\frac{1}{y}\right) \frac{d y}{d x}+\log y(1)=\frac{d y}{d x}-1 \\
& \Rightarrow \frac{d y}{d x}\left(\frac{x}{y}-1\right)=-1-\log y \\
& \Rightarrow \frac{d y}{d x}\left[\frac{y}{(1+\log y) y}-1\right]=-(1+\log y) \\
& \qquad \quad\left[\because \text { from eq. }(i) \cdot x=\frac{y}{(1+\log y)}\right] \\
& \Rightarrow \frac{d y}{d x}\left[\frac{1-1-\log y}{1+\log y}\right]=-(1+\log y)
\end{aligned}$
$\Rightarrow \frac{d y}{d x}=-\frac{(1+\log y)^2}{-\log y}$
$\Rightarrow \frac{d y}{d x}=\frac{(1+\log y)^2}{\log y}$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If \(y=e^{-x} \cos x\) and \(y_4+k y=0\), where \(y_4=\frac{d^4 y}{d x^4}\), then \(k=\)

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

The correct answer is:
4

Solution:
Let \(y=e^{-x} \cos x\)
\(\begin{aligned}
& \mathrm{y}_1=-e^{-x} \sin x-e^{-x} \cos x=-e^{-x} \sin x-y \\
& y_2=-e^{-x} \cos x+e^{-x} \sin x-y_1 \\
& \Rightarrow y_2=-y-y_1+e^{-x} \sin x=-2\left(y+y_1\right) \\
& \Rightarrow y_3=-2\left(y_1+y_2\right)=-2\left(e^{-x} \sin x-y\right) \\
& \Rightarrow y_4=4 y_1+2 y_2=4 y_1-4 y-4 y_1 \text { or } y_4+4 y=0 \\
& \Rightarrow k=4
\end{aligned}\)

Topic: Mathematics, Functions, BITSAT, BITSAT 2009
If \(y=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}\), then value of \(\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}\) is

Options:
A \(\frac{y}{3}\)
B \(\mathrm{y}\)
C \(2 y\)
D \(\frac{3}{2} y\)

The correct answer is:
\(\mathrm{y}\)

Solution:
\(\begin{aligned}
& \frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \\
& =\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \cdot \frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha} \\
& =\frac{(1+\sin \alpha)^2-\cos ^2 \alpha}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)} \\
& =\frac{2 \sin \alpha(1+\sin \alpha)}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)} \\
& =\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=y
\end{aligned}\)

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2009
If \(y=x^{x^2}\), then \(\frac{d y}{d x}\) is equal to

Options:
A \((2 \ln x)\)
B \((2 \ln x+1)\)
C \((\ln \ln x+1) x^{x^2}\)
D None of these

The correct answer is:
None of these

Solution:
Consider \(y=x^{x^2} \Rightarrow \ln y=x^2 \ln x\)
\(\begin{aligned}
& \frac{1}{y} \frac{d y}{d x}=2 x \ln x+x^2 \cdot \frac{1}{x}=x(1+2 \ln x) \\
& \frac{d y}{d x}=x^{x^2} \cdot x(1+2 \ln x)=x^{x^2+1}(1+2 \ln x)
\end{aligned}\)

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2011
$\text { If } z_{1}=\sqrt{2}\left[\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right] \text { and } z_{2}=\sqrt{3}\left[\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right]$ then $\left|z_{1} z_{2}\right|$ is equal to $\sqrt{m}$. Value of $m$ is

Options:
A 6
B 3
C 2
D 5

The correct answer is:
6

Solution:
$z_{1}=\sqrt{2}\left[\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right]=\sqrt{2}\left[\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right]=1+i$

$\left|z_{1}\right|=\sqrt{2}$

and $z_{2}=\sqrt{3}\left[\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right]=\sqrt{3}\left[\frac{1}{2}+i \frac{\sqrt{3}}{2}\right]$

$\left|z_{2}\right|=\sqrt{\frac{3}{4}+\frac{9}{4}}=\sqrt{3}$

$\left|z_{1} z_{2}\right|=\left|z_{1}\right|\left|z_{2}\right|=\sqrt{2} \cdot \sqrt{3}=6$

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2021
If $z_{1}=\sqrt{3}+i \sqrt{3}$ and $z_{2}=\sqrt{3}+i$, then the

complex number $\left(\frac{\mathrm{z}_{1}}{\mathrm{z}_{2}}\right)^{50}$ lies in the :

Options:
A first quadrant
B second quadrant
C third quadrant
D fourth quadrant

The correct answer is:
first quadrant

Solution:
$\left(\frac{z_{1}}{z_{2}}\right)^{50}=\left(\frac{\sqrt{3}+i \sqrt{3}}{\sqrt{3}+i}\right)^{50}$

$=\left[\left(\frac{\sqrt{3}(1+i)}{\sqrt{3}+i}\right)^{2}\right]^{25}=\left[\frac{3(2 i)}{3-1+2 \sqrt{3} i}\right]^{25}$

$=\left(\frac{3 i}{1+\sqrt{3} i}\right)^{25}=\frac{3^{25} i^{25}}{\left(-2 \omega^{2}\right)^{25}}$

$$

\begin{array}{l}

=-i . \omega\left(\frac{3}{2}\right)^{25}=-\mathrm{i}\left(\frac{-1+\sqrt{3} \mathrm{i}}{2}\right)\left(\frac{3}{2}\right)^{25} \\

=\left(\frac{3}{2}\right)^{25}\left(\frac{\sqrt{3}}{2}+\frac{1}{2} \mathrm{i}\right)

\end{array}

$$

Hence, $\left(\frac{z_{1}}{z_{2}}\right)^{50}$ lies in the first quadrant as both real and imaginary parts of this number are positive.

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2016
If z1 and z¯1 represents adjacent vertices of a regular polygon of n sides and if Imz1Rez1=2-1, then n is equal to

Options:
A 8
B 16
C 18
D 24

The correct answer is:
8

Solution:
Let z1=reiθ

then, z¯1=re-iθ

z1z¯1=e2iθ=e2πi/n

⇒ θ=πn

Also, Imz1Rez1=tanθ=tanπn=2-1

⇒ tanπ8=2-1

∴ n=8

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2019
If z1,z2 and z3 represent the vertices of an equilateral triangle such that z1=z2=z3 then

Options:
A z1+z2=z3
B z1+z2+z3=0
C z1z2=1z3
D z1-z2=z3-z2

The correct answer is:
z1+z2+z3=0

Solution:
Since, z1=z2=z3.

⇒z1-0=z2-0=z3-0.

Thus, the given points are equidistant from origin, and we know that the point which is equidistant from the vertices of a triangle is the circumcentre of the triangle.

∴ 0 is the circumcentre of an equilateral ∆ABC and we also know that for an equilateral triangle the centroid and circumcentre are same.

∴ x1+x2+x33=0=y1+y2+y33

where, z1=x1+iy1.

⇒x1+x2+x33+iy1+y2+y33=0

⇒x1+iy1+x2+iy2+x3+iy3=0

⇒z1+z2+z3=0.

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2017
If $\left|z_{1}\right|=\left|z_{2}\right|=\ldots \ldots \ldots . .\left|z_{n}\right|=1,$ then the

value of $\left|z_{1}+z_{2}+\ldots \ldots . z_{n}\right|-$

$\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\ldots \ldots .+\frac{1}{z_{n}}\right|$ is,

Options:
A 0
B 1
C -1
D None

The correct answer is:
0

Solution:
$z_{1} \bar{z}_{1}=z_{2} \bar{z}_{2}=\ldots .=z_{n} \bar{z}_{n}=1$

$\Rightarrow \overline{\mathrm{z}}_{1}=\frac{1}{\mathrm{z}_{1}}, \overline{\mathrm{z}}_{2}=\frac{1}{\mathrm{z}_{2}}, \overline{\mathrm{z}}_{3}=\frac{1}{\mathrm{z}_{3}}, \ldots \ldots, \overline{\mathrm{z}}_{\mathrm{n}}=\frac{1}{\mathrm{z}_{\mathrm{n}}}$

$\therefore\left|z_{1}+z_{2}+\ldots .+z_{n}\right|-\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\ldots .+\frac{1}{z_{n}}\right|$

$=\left|z_{1}+z_{2}+\ldots+z_{n}\right|-\left|\bar{z}_{1}+\bar{z}_{2}+\ldots .+\bar{z}_{n}\right|=0$

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2020
If z=a+ib satisfies arg(z-1)=arg(z+3i), then (a-1):b=

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

The correct answer is:
1:3

Solution:
We have,

arg(z-1)=arg(z+3i)

⇒arg((a-1)+ib)=arg(a+(b+3)i)

⇒tan-1ba-1=tan-1b+3a

⇒ ba-1=b+3a

⇒ ab=(a-1)(b+3)

⇒ 3(a-1)=b

⇒ (a-1):b=1:3

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2017
If $z$ is a complex number such that $z^{2}=(\bar{z})^{2}$, then

Options:
A $z$ is purely real
B $z$ is purely imaginary
C either $z$ is purely real or purely imaginary
D None of these

The correct answer is:
either $z$ is purely real or purely imaginary

Solution:
Let $z=x+i y$, then its conjugate $\bar{z}=x-i y$

Given that $z^{2}=(\bar{z})^{2}$

$\begin{array}{l}

\Rightarrow x^{2}-y^{2}+2 i x y=x^{2}-y^{2}-2 i x y \\

\Rightarrow 4 i x y=0

\end{array}$

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2021
If $z=x+i y, z^{1 / 3}=a-i b,$ then $\frac{x}{a}-\frac{y}{b}=k\left(a^{2}-b^{2}\right)$

where $k$ is equal to

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
$z^{1 / 3}=a-i b \Rightarrow z=(a-i b)^{3}$

$\therefore x+i y=a^{3}+i b^{3}-3 i a^{2} b-3 a b^{2}$

$\Rightarrow x=a^{3}-3 a b^{2} \Rightarrow \frac{x}{a}=a^{2}-3 b^{2}$

and $y=b^{3}-3 a^{2} b \Rightarrow \frac{y}{b}=b^{2}-3 a^{2}$

So, $\frac{x}{a}-\frac{y}{b}=4\left(a^{2}-b^{2}\right)$

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2014
IfM. D. is 12, the value of S.D. will be

Options:
A 15
B 12
C 24
D None of these

The correct answer is:
15

Solution:
We know that Q.D $=\frac{5}{6} \times M \cdot D .=\frac{5}{6} \times 12=10$

$\therefore \mathrm{S} . \mathrm{D}=\frac{3}{2} \times Q \cdot D .=\frac{3}{2} \times 10 \Rightarrow S . D .=15$

Topic: Mathematics, Statistics, BITSAT, BITSAT 2014
In a 1 st order reaction, reactant concentration $C$ varies with time $t$ as :

Options:
A $\frac{1}{\mathrm{C}}$ increases linearly with $t$
B $\log \mathrm{C}$ decreases linearly with $t$
C C decreases with $\frac{1}{t}$
D $\log \mathrm{C}$ decreases with $\frac{1}{t}$

The correct answer is:
$\log \mathrm{C}$ decreases linearly with $t$

Solution:

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2021
In a ∆ABC, D, E, F are the mid-points of the sides BC, CA and AB respectively, the vector AD→ is equal to

Options:
A BE→+CF→
B BE→-CF→
C CF→-BE→
D -BE→-CF→

The correct answer is:
-BE→-CF→

Solution:
AD→=3GD→

=3·12GB→+GC→

=3223EB→+23FC→

=-BE→-CF→

Hence, option (d) is correct.

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2018
In a $\triangle A B C$, the lengths of the two larger sides are 10 and 9 units, respectively. If the angles are in $\mathrm{AP}$, then the length of the third side can be

Options:
A $5 \pm \sqrt{6}$
B $3 \sqrt{3}$
C $\underline{5}$
D None of these

The correct answer is:
$5 \pm \sqrt{6}$

Solution:
Let A, B and $C$ be the three angles of $\Delta A B C$ and Let $a=10$ and $b=9$ It is given that the angles are in $\mathrm{AP}$. $\therefore 2 \mathrm{~B}=\mathrm{A}+\mathrm{C}$ on adding $\mathrm{B}$ both the sides,

we get

$$

\begin{array}{l}

3 \mathrm{~B}=\mathrm{A}+\mathrm{B}+\mathrm{C} \\

\Rightarrow 3 \mathrm{~B}=180^{\circ} \Rightarrow \mathrm{B}=60^{\circ}

\end{array}

$$

Now, we know $\cos \mathrm{B}=\frac{\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2}}{2 \mathrm{ac}}$

$$

\begin{array}{l}

\Rightarrow \cos 60^{\circ}=\frac{10^{2}+\mathrm{c}^{2}-9^{2}}{2 \times 10 \times \mathrm{c}} \\

\Rightarrow \frac{1}{2}=\frac{100+\mathrm{c}^{2}-81}{20 \mathrm{c}} \\

\Rightarrow \mathrm{c}^{2}-10 \mathrm{c}+19=0 \Rightarrow \mathrm{c}=5 \pm \sqrt{6}

\end{array}

$$

Topic: Mathematics, Properties of Triangles, BITSAT, BITSAT 2015
In a $\triangle \mathrm{ABC},$ if $\frac{\cos A}{\mathrm{a}}=\frac{\cos \mathrm{B}}{\mathrm{b}}=\frac{\cos \mathrm{C}}{\mathrm{c}},$ and the side $\mathrm{a}=2,$ then area of the triangle is

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

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

Solution:
$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$

$\Rightarrow \frac{\cos \mathrm{A}}{2 \mathrm{R} \sin \mathrm{A}}=\frac{\cos \mathrm{B}}{2 \mathrm{R} \sin \mathrm{B}}=\frac{\cos \mathrm{C}}{2 \mathrm{R} \sin \mathrm{C}}$

$\Rightarrow \cot A=\cot B=\cot C$

$\Rightarrow \mathrm{A}=\mathrm{B}=\mathrm{C}=60^{\circ} \Rightarrow \Delta \mathrm{ABC}$ is equilateral

Hence, area of $\Delta=\frac{\sqrt{3}}{4} \mathrm{a}^{2}=\sqrt{3}$.

Topic: Mathematics, Properties of Triangles, BITSAT, BITSAT 2016
In a cell that utilises the reaction
$\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})$
addition of $\mathrm{H}_{2} \mathrm{SO}_{4}$ to cathode compartment, will

Options:
A increase the $\mathrm{E}$ and shift equilibrium to the right.
B lower the $E$ and shift equilibrium to the right.
C lower the $E$ and shift equlibrium to the left.
D increase the $E$ and shift equilibrium to the left.

The correct answer is:
increase the $\mathrm{E}$ and shift equilibrium to the right.

Solution:
$\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}+(\mathrm{aq}) \rightleftharpoons$
$$
\mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})
$$
$$
\mathrm{E}_{\mathrm{cell}}=\mathrm{E}^{\circ} \mathrm{cell}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]\left[\mathrm{H}_{2}\right]}{\left[\mathrm{H}^{+}\right]^{2}}
$$
Addition of $\mathrm{H}_{2} \mathrm{SO}_{4}$ will increase $\left[\mathrm{H}^{+}\right]$ and $\mathrm{E}_{\text {cell }}$ will also increase and the equilibrium will shift towards RHS.

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2009
In a closed system, $\mathrm{A}(\mathrm{s}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+3 \mathrm{C}(\mathrm{g})$, if partial

pressure of $\mathrm{C}$ is doubled, then partial pressure of $\mathrm{B}$ will be

Options:
A $2 \sqrt{2}$ times the original value
B $\frac{1}{2}$ times the original value
C 2 times the original value
D $\frac{1}{2 \sqrt{2}}$ times the original value

The correct answer is:
$\frac{1}{2 \sqrt{2}}$ times the original value

Solution:
$\quad K_{p}=p_{B_{1}}^{2} \times p_{C_{1}}^{3}$

Again, $K_{p}=p_{B_{2}}^{2} \times\left(2 p_{C_{1}}\right)^{3}$

$\therefore \mathrm{p}_{\mathrm{B}_{1}}^{2} \times \mathrm{p}_{\mathrm{C}_{1}}^{3}=\mathrm{P}_{\mathrm{B}_{2}}^{2} \times 8 \mathrm{p}_{\mathrm{C}_{1}}^{3}$

$$

\therefore \frac{\mathrm{p}_{\mathrm{B}_{1}}^{2}}{8}=\mathrm{p}_{\mathrm{B}_{2}}^{2} \text { or, } \frac{\mathrm{p}_{\mathrm{B}_{1}}}{2 \sqrt{2}}=\mathrm{p}_{\mathrm{B}_{2}}

$$

Topic: Chemistry, Chemical Equilibrium, BITSAT, BITSAT 2016
In a coil of resistance 50 Ω, the induced current developed by changing magnetic flux through it, is shown in figure as a function of time. The magnitude of change in flux through the coil is

Options:
A 10 Wb
B 20 Wb
C 32 Wb
D 40 Wb

The correct answer is:
40 Wb

Solution:
According to the given graph I-t induced current (I) is obtained from the slope. Hence, I=1Rdϕdt

⇒dϕ=Idt×R= area of triangle POS×R=8×0.22×50=40 Wb

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2019
In a compound, atoms of elements Y form ccp lattice and those of element X occupy 2/3 rd of tetrahedral voids. The formula of the compound can be

Options:
A X4Y3
B X2Y3
C X2Y
D X3Y4

The correct answer is:
X4Y3

Solution:
Let atoms of Y in ccp structure =100

Then, number of tetrahedral voids =2×100

∵ Number of X atoms =23 rd of tetrahedral voids

∴ Number of X atoms =23×200=4003

Hence, formula of compound can be =X4Y3.

Topic: Chemistry, Solid State, BITSAT, BITSAT 2019
In a compound C, H and N are present in 9:1:3.5 by weight. If molecular weight of compound is 108, the molecular formula of compound is

Options:
A C2H6 N2
B C3H4 N
C C6H8 N2
D C9H12 N3

The correct answer is:
C6H8 N2

Solution:
Element	Ratio (by weight)	Atomic weight	Mol
Ratio	Simplest Mol Ratio

C	9	12	912=0.75	0.750.25=3

H	1	1	11=1.00	1.000.25=4

N	3.5	14	3.514=0.25	0.250.25=1

∴ Empirical formula =C3H4 N

The empirical formula mass =3×12+4×1+14=54 g

Given, molecular mass =108 g

Thus, n=10854=2

Hence, molecular formula = (Empirical formula) ×2

C3H4 N×2=C6H8 N2

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2019
In a cuboid of dimension $2 \mathrm{~L} \times 2 \mathrm{~L} \times \mathrm{L}$, a charge $q$ is placed at the centre of the surface ' $S$ ' having area of $4 \mathrm{~L}^2$. The flux through the opposite surface to ' $\mathrm{S}$ ' is given by

Options:
A $\frac{\mathrm{q}}{12 \varepsilon_0}$
B $\frac{\mathrm{q}}{3 \varepsilon_0}$
C $\frac{\mathrm{q}}{2 \varepsilon_0}$
D $\frac{q}{6 \varepsilon_0}$

The correct answer is:
$\frac{q}{6 \varepsilon_0}$

Solution:
After placing similar cubic at the bottom, we get cube of side $2 \mathrm{~L}$ having change $\mathrm{q}$ at its centre.
From the Gauss's law electric flux through whole
Flux passing through shaded face
$\phi=\frac{\mathrm{q} / \varepsilon_0}{6}=\frac{\mathrm{q}}{6 \varepsilon_0}$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2023 (Memory Based Paper 2)
In a cuboid of dimension $2 \mathrm{~L} \times 2 \mathrm{~L} \times \mathrm{L}$, a charge $q$ is placed at the centre of the surface ' $S$ ' having area of $4 \mathrm{~L}^2$. The flux through the opposite surface to ' $\mathrm{S}$ ' is given by

Options:
A $\frac{\mathrm{q}}{12 \varepsilon_0}$
B $\frac{\mathrm{q}}{3 \varepsilon_0}$
C $\frac{\mathrm{q}}{2 \varepsilon_0}$
D $\frac{q}{6 \varepsilon_0}$

The correct answer is:
$\frac{q}{6 \varepsilon_0}$

Solution:
After placing similar cubic at the bottom, we get cube of side $2 \mathrm{~L}$ having change $\mathrm{q}$ at its centre.
From the Gauss's law electric flux through whole
Flux passing through shaded face
$\phi=\frac{\mathrm{q} / \varepsilon_0}{6}=\frac{\mathrm{q}}{6 \varepsilon_0}$

Topic: Physics, Electrostatics, JEE Main
In a first order reaction with time the concentration of the reactant decreases

Options:
A linearly
B exponentially
C no change
D None of these

The correct answer is:
exponentially

Solution:
For first order reaction $[\mathrm{A}]=\left[\mathrm{A}_{0}\right] \mathrm{e}^{\mathrm{k}}$

$\therefore$ The concentration of reactants will exponentially decreases with time.

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2015
In a given network the equivalent capacitance between $A$ and $B$ is $C_{1}=C_{4}=1 \mu F, C_{2}=C_{3}=2 \mu F$.

Options:
A $3 \mu \mathrm{F}$
B $6 \mu \mathrm{F}$
C $4.5 \mu \mathrm{F}$
D $2.5 \mu \mathrm{F}$

The correct answer is:
$3 \mu \mathrm{F}$

Solution:
The equivalent circuit is shown in figure.

$\mathrm{C}_{\mathrm{AB}}=1+\frac{2 \times 2}{2+2}+1$

$\mathrm{C}_{\mathrm{AB}}=3 \mu \mathrm{F}$

Topic: Physics, Capacitance, BITSAT, BITSAT 2021
In a homonuclear molecule, which of the following set of orbitals is degenerated?

Options:
A σ1s and σ2s
B π2px and π2py
C π*2px and σ*2pz
D σ2px and σ*2pz

The correct answer is:
π2px and π2py

Solution:
Orbitals of equal energy are called degenerated orbitals. Homonuclear molecules are the molecules composed of only one type of element and contain following set of degenerated orbitals:

As, energy of,

σ2s>σ1s

π2px≈2py

π*2px<σ*2pz

σ2px<σ*2pz

Thus, only π2px and π2py are degenerate orbitals.

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2019
In a monoclinic unit cell, the relation of sides and angles are respectively:

Options:
A $a=b \neq c$ and $\alpha=\beta=\gamma=90^{\circ}$
B $a \neq b \neq c$ and $\alpha=\beta=\gamma=90^{\circ}$
C $a \neq b \neq c$ and $\beta=\gamma=90^{\circ} \neq \alpha$
D $a \neq b \neq c$ and $\alpha \neq \beta \neq \gamma \neq 90^{\circ}$

The correct answer is:
$a \neq b \neq c$ and $\beta=\gamma=90^{\circ} \neq \alpha$

Solution:
No Solution

Topic: Chemistry, Solid State, BITSAT, BITSAT 2020
In a photoelectric experiment, with light of wavelength $\lambda,$ the fastest electron has speed $v$. If the exciting wavelength is changed to $3 \lambda / 4$, the speed of the fastest emitted electron will become

Options:
A $v \sqrt{\frac{3}{4}}$
B $v \sqrt{\frac{4}{3}}$
C less than $v \sqrt{\frac{4}{3}}$
D greater than $v \sqrt{\frac{4}{3}}$

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

Solution:
$\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{hc}}{\lambda}-\phi$
$$
\begin{array}{l}
\frac{1}{2} \mathrm{~m}^{\prime}=\frac{\mathrm{hc}}{(3 \lambda / 4)}-\phi=\frac{4 \mathrm{hc}}{3 \lambda}-\phi \\
\text { Clearly, } \mathrm{v}^{\prime}>\sqrt{\frac{4}{3} \mathrm{v}}
\end{array}
$$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2009
In a polygon no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon be 70 then the number of diagonals of the polygon is

Options:
A 20
B 28
C 8
D None of these

The correct answer is:
20

Solution:
A combination of four vertices is equivalent to one interior point of intersection of diagonals.
\(\therefore\) No. of interior points of intersection
\(\begin{aligned}
& ={ }^{\mathrm{n}} \mathrm{C}_4=70 \\
& \Rightarrow \mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)=5.6 .7 .8 \\
& \therefore \mathrm{n}=8
\end{aligned}\)
So, number of diagonals \(={ }^8 \mathrm{C}_2-8=20\)

Topic: Mathematics, Permutation Combination, JEE Main
In a polygon no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon be 70 then the number of diagonals of the polygon is

Options:
A 20
B 28
C 8
D None of these

The correct answer is:
20

Solution:
A combination of four vertices is equivalent to one interior point of intersection of diagonals.
\(\therefore\) No. of interior points of intersection
\(\begin{aligned}
& ={ }^{\mathrm{n}} \mathrm{C}_4=70 \\
& \Rightarrow \mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)=5.6 .7 .8 \\
& \therefore \mathrm{n}=8
\end{aligned}\)
So, number of diagonals \(={ }^8 \mathrm{C}_2-8=20\)

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2010