datavorous/entrance-exam-dataset huggingface dataset:
The frequency of a sonometer wire is $100 \mathrm{~Hz}$. When the weights producing the tensions are completely immersed in water, the frequency becomes $80 \mathrm{~Hz}$ and on immersing the weights in a certain liquid, the frequency becomes $60 \mathrm{~Hz}$. The specific gravity of the liquid is

Options:
A 1.42
B 1.77
C 1.82
D 1.21

The correct answer is:
1.77

Solution:
As we know, frequency $f \propto \sqrt{m g}$ or $f \propto \sqrt{g}$

In water, $\mathrm{f}_{\mathrm{w}}=0.8 \mathrm{f}_{\text {air }}$

$$

\begin{array}{l}

\frac{\mathrm{g}^{\prime}}{\mathrm{g}}(0.8)^{2}=0.64 \\

\Rightarrow 1-\frac{\rho_{\mathrm{w}}}{\rho_{\mathrm{m}}}=0.64 \\

\Rightarrow \frac{\rho_{\mathrm{w}}}{\rho_{\mathrm{m}}}=0.36

\end{array}

$$

In liquid, $\frac{g^{\prime}}{g}=(0.6)^{2}=0.36$

$$

1-\frac{\rho_{1}}{\rho_{\mathrm{m}}}=0.36 \frac{\rho_{1}}{\rho_{\mathrm{m}}}=0.64

$$

From eq. (1) and (2) $\frac{\rho_{l}}{\rho_{n}}=\frac{0.64}{0.36} \therefore \rho_{l}=1.77$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2015
The frequency of radiation emitted when the electron falls from $\mathrm{n}=4$ to $\mathrm{n}=1$ in a hydrogen atom will be (Given : ionization energy of $\mathrm{H}=2.18 \times 10^{-18} \mathrm{~J}$ atom ${ }^{-1}$ and $\mathrm{h}=6.625 \times 10^{-34} \mathrm{Js}$ )

Options:
A $1.54 \times 10^{15} \mathrm{~s}^{-1}$
B $1.03 \times 10^{15} \mathrm{~s}^{-1}$
C $3.08 \times 10^{15} \mathrm{~s}^{-1}$
D $2.00 \times 10^{15} \mathrm{~s}^{-1}$

The correct answer is:
$3.08 \times 10^{15} \mathrm{~s}^{-1}$

Solution:
$\begin{aligned} \mathrm{v} &=\frac{1}{\mathrm{~h}} \times \mathrm{IE} \times\left[\frac{1}{\mathrm{n}_{1}{ }^{2}}-\frac{1}{\mathrm{n}_{2}{ }^{2}}\right] \\ &=\frac{2.18 \times 10^{-18}}{6.625 \times 10^{-34}} \times\left[\frac{1}{1}-\frac{1}{16}\right]=3.08 \times 10^{15} \mathrm{~s}^{-1} \end{aligned}$

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2021
The frequency of vibration of string is given by

$\mathrm{v}=\frac{\mathrm{p}}{2 l}\left[\frac{\mathrm{F}}{\mathrm{m}}\right]^{1 / 2}$

Here $\mathrm{p}$ is number of segments in the string and $l$ is the length. The dimensional formula for $\mathrm{m}$ will be

Options:
A $\left[\mathrm{M}^{0} \mathrm{LT}^{-1}\right]$
B $\left[\mathrm{ML}^{0} \mathrm{~T}^{-1}\right]$
C $\left[\mathrm{ML}^{-1} \mathrm{~T}^{0}\right]$
D $\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$

The correct answer is:
$\left[\mathrm{ML}^{-1} \mathrm{~T}^{0}\right]$

Solution:
$\mathrm{v}=\frac{\mathrm{p}}{2 \ell}\left[\frac{\mathrm{F}}{\mathrm{m}}\right]^{1 / 2}$

$v^{2}=\frac{p}{4 \ell^{2}} \frac{F}{m} \Rightarrow m=\frac{p^{2} F}{4 \ell^{2} v^{2}}$

Now, dimensional formula of R.H.S.

$=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^{2}\left(\frac{1}{\mathrm{~T}}\right)^{2}}$

[p will have no dimension as it is an integer] $=\mathrm{ML}^{-1} \mathrm{~T}^{0}$

So, dimensions of $\mathrm{m}$ will be $\mathrm{ML}^{-1} \mathrm{~T}^{0}$.

Topic: Physics, Units and Dimensions, BITSAT, BITSAT 2016
The frequency of whistle of an engine appears to be $(4 / 5)^{\text {th }}$ of initial frequency when it crosses a stationary observer. If the velocity of sound is $330 \mathrm{~m} / \mathrm{s}$, then the speed of engine will be

Options:
A $30 \mathrm{~m} / \mathrm{s}$
B $36.6 \mathrm{~m} / \mathrm{s}$
C $40 \mathrm{~m} / \mathrm{s}$
D $330 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$36.6 \mathrm{~m} / \mathrm{s}$

Solution:
$n^{\prime}=\frac{n v}{v-v_{s}} \ldots \ldots \ldots(1)$

$n^{\prime \prime}=\frac{n v}{v+v_{s}} \ldots \ldots . .(2)$

From (1) and (2), $\frac{n^{\prime}}{n^{\prime \prime}}=\frac{v+v_{s}}{v-v_{s}} \ldots \ldots . .(3)$

According to question, $\frac{n^{\prime}}{n^{\prime}}=\frac{5}{4}$

$v_{s}=? v=330 \mathrm{~m} / \mathrm{s} \ldots \ldots \ldots .(4)$

From eq. (3) and (4)

$\frac{5}{4}=\left[\frac{330+v_{s}}{330-v_{s}}\right] \Rightarrow 9 v_{s}=330$

$\therefore v_{s}=36.6 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2021
The frequency $(v)$ of an oscillating liquid drop may depend upon radius $(r)$ of the drop, density $(\rho)$ of liquid and the surface tension $(s)$ of the liquid as :
$v=r^a \rho^b s^c$. The values of $a, b$ and $c$ respectively are

Options:
A $\left(-\frac{3}{2},-\frac{1}{2} ; \frac{1}{2}\right)$
B $\left(\frac{3}{2},-\frac{1}{2}, \frac{1}{2}\right)$
C $\left(\frac{3}{2}, \frac{1}{2} ;-\frac{1}{2}\right)$
D $\left(-\frac{3}{2}, \frac{1}{2}, \frac{1}{2}\right)$

The correct answer is:
$\left(-\frac{3}{2},-\frac{1}{2} ; \frac{1}{2}\right)$

Solution:
$\begin{aligned} & v=\mathrm{r}^{\mathrm{a}} \rho^{\mathrm{b}} \mathrm{s}^{\mathrm{c}} \Rightarrow[v]=[\mathrm{r}]^{\mathrm{a}}[\rho]^{\mathrm{b}}[\mathrm{s}]^{\mathrm{c}} \\ & \Rightarrow \quad\left[\mathrm{T}^{-1}\right]=\left[\mathrm{L}^1\right]^{\mathrm{a}}\left[\mathrm{M}^1 \mathrm{~L}^{-3}\right]^{\mathrm{b}}\left[\frac{\mathrm{MLT}}{\mathrm{L}}\right]^{-2} \\ & \Rightarrow \mathrm{T}^{-1}=\mathrm{M}^{\mathrm{b}+\mathrm{c}} \cdot \mathrm{L}^{\mathrm{a}-3 \mathrm{~b}} \cdot \mathrm{T}^{-2 c} \\ & \Rightarrow \mathrm{b}+\mathrm{c}=0, \mathrm{a}-3 \mathrm{~b}=0,-2 \mathrm{c}=-1 \\ & \Rightarrow \quad \mathrm{c}=\frac{1}{2}, \mathrm{~b}=-\frac{1}{2}, \mathrm{a}-3 \mathrm{~b}=0 ; \\ & \mathrm{a}+\frac{3}{2}=0 \Rightarrow \mathrm{a}=-\frac{3}{2}\end{aligned}$

Topic: Physics, Units and Dimensions, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The fuel charges for running a train are proportional to the square of the speed generated in miles per hour and costs ' 48 per hour at 16 miles per hour. The most economical speed if the fixed charges i.e. salaries etc. amount to 300 per hour is

Options:
A 10
B 20
C 30
D 40

The correct answer is:
40

Solution:
Let the speed of the train be $\mathrm{v}$ and distance to be covered be so that total time taken is $\mathrm{s} / \mathrm{v}$ hours. Cost of fuel per hour $=\mathrm{kv}^{2}(\mathrm{k}$ is constant) Also $48=\mathrm{k} .16^{2}$ by given condition $\therefore \mathrm{k}=\frac{3}{16}$

$\therefore$ Cost to fuel per hour $\frac{3}{16} \mathrm{v}^{2}$. Other charges per hour are 300 . Total running cost,

$$

\begin{array}{r}

C=\left(\frac{3}{16} \mathrm{v}^{2}+300\right) \frac{\mathrm{s}}{\mathrm{v}}=\frac{3 \mathrm{~s}}{16} \mathrm{v}+\frac{300 \mathrm{~s}}{\mathrm{v}} \\

\frac{\mathrm{dC}}{\mathrm{dv}}=\frac{3 \mathrm{~s}}{16}-\frac{300 \mathrm{~s}}{\mathrm{v}^{2}}=0 \Rightarrow \mathrm{v}=40

\end{array}

$$

$\frac{\mathrm{d}^{2} \mathrm{C}}{\mathrm{dv}^{2}}=\frac{600 \mathrm{~s}}{\mathrm{v}^{3}}>0 \therefore \mathrm{v}=40$ results in minimum

running cost

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2014
The function $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}+\frac{2}{\mathrm{x}}$ has a local minimum at

Options:
A $x=2$
B $x=-2$
C $x=0$
D $x=1$

The correct answer is:
$x=1$

Solution:
$\frac{x}{2}+\frac{2}{x}$ is of the form $x+\frac{1}{x} \geq 2$ and equality holds for $x=1$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2015
The function $f(x)=\log (1+x)-\frac{2 x}{2+x}$ is increasing on

Options:
A $(0, \infty)$
B $(-\infty, 0)$
C $(-\infty, \infty)$
D
None of these

The correct answer is:
$(0, \infty)$

Solution:
Given $f(x)=\log (1+x)-\frac{2 x}{2+x}$
$\begin{aligned}
& f^{\prime}(x)=\frac{1}{1+x}-\frac{(2+x)(2)-2 x}{(2+x)^2} \\
& =\frac{1}{1+x}-\frac{4}{(2+x)^2} \\
& =\frac{(2+x)^2-4-4 x}{(1+x)(2+x)^2}=\frac{x^2}{(1+x)(2+x)^2}>0 \text { for all } \\
& x \in(0, \infty)
\end{aligned}$
Thus, given function $\mathrm{f}(\mathrm{x})$ is increasing on $(0$, $\infty)$.

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The function $f(x)=\tan ^{-1}(\sin x+\cos x)$ is an increasing function in

Options:
A $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
B $\left(-\frac{\pi}{2}, \frac{\pi}{4}\right)$
C $\left(0, \frac{\pi}{2}\right)$
D $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

The correct answer is:
$\left(-\frac{\pi}{2}, \frac{\pi}{4}\right)$

Solution:
Since, $f(x)=\tan ^{-1}(\sin x+\cos x)$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{1+(\sin \mathrm{x}+\cos \mathrm{x})^2}(\cos \mathrm{x}-\sin \mathrm{x})$
$=\frac{\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)}{1+(\sin x+\cos x)^2}$
$f(x)$ is increasing if $f^{\prime}(x)>0 \Rightarrow \cos \left(x+\frac{\pi}{4}\right)>0$
$\Rightarrow \quad-\frac{\pi}{2} < x+\frac{\pi}{4} < \frac{\pi}{2} \Rightarrow-\frac{3 \pi}{2} < x+\frac{\pi}{4}$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The function $\mathrm{f}(\mathrm{x})=(\mathrm{x}(\mathrm{x}-2))^{2}$ is increasing in the set

Options:
A $(-\infty, 0) \cup(2, \infty)$
B $(-\infty, 1)$
C (0,1)$\cup(2, \infty)$
D (1,2)

The correct answer is:
(0,1)$\cup(2, \infty)$

Solution:
$$

\text { } \begin{array}{l}

\text { Here, } \mathrm{f}(\mathrm{x})=(\mathrm{x}(\mathrm{x}-2))^{2} \\

\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}(\mathrm{x}-2)(\mathrm{x}-1)

\end{array}

$$

For $\mathrm{f}(\mathrm{x})$ as increasing, $\mathrm{f}^{\prime}(\mathrm{x})>0$

$$

\text { So, } 4 x(x-1)(x-2)>0 \Rightarrow x(x-1)(x-2)>0

$$

From the above figure required interval is, (0,1)$\cup(2, \infty)$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2013
The function \(f(x)=\sin x-k x-c\), where \(k\) and \(c\) are constants, decreases always when

Options:
A \(k > 1\)
B \(k \geq 1\)
C \(k < 1\)
D \(k \leq 1\)

The correct answer is:
\(k \geq 1\)

Solution:
Let \(f(x)=\sin x-k x-c\) where \(k\) and \(c\) are constants. \(f^{\prime}(x)=\cos x-k\)
\(\therefore f\) decreases if \(\cos x \leq k\)
Thus, \(f(x)=\sin x-k x-c\) decrease always when \(k \geq 1\).

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2011
The function \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^3-3 \mathrm{x}^2-12 \mathrm{x}+4\), has

Options:
A two points of local maximum
B two points of local minimum
C one maxima and one minima
D nomaxima or minima

The correct answer is:
one maxima and one minima

Solution:
\(\begin{aligned}
& f(x)=2 x^3-3 x^2-12 x+4 \\
& \Rightarrow \quad f^{\prime}(x)=6 x^2-6 x-12=6\left(x^2-x-2\right) \\
& =6(x-2)(x+1)
\end{aligned}\)
For maxima and minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\)
\(\therefore \quad 6(\mathrm{x}-2)(\mathrm{x}+1)=0 \Rightarrow \mathrm{x}=2,-1\)
Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}-6\)
\(\begin{aligned}
& \text { At } \mathrm{x}=2 ; \mathrm{f}^{\prime \prime}(\mathrm{x})=24-6=18 > 0 \\
& \therefore \quad \mathrm{x}=2, \text { local min. point } \\
& \text { At } \mathrm{x}=-1 ; \mathrm{f}^{\prime \prime}(\mathrm{x})=12(-1)-6=-18 < 0 \\
& \therefore \mathrm{x}=-1 \text { local max. point }
\end{aligned}\)

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2011
The function \(\mathrm{f}(\mathrm{x})=(\mathrm{x}-1) \sqrt{|\ell n \mathrm{x}|}\) is at \(\mathrm{x}=1\)

Options:
A discontinuous
B continuous but not differentiable
C differentiable with \(f^{\prime}(1)=0\)
D differentiable with \(\mathrm{f}^{\prime}(1) \neq 0\)

The correct answer is:
differentiable with \(f^{\prime}(1)=0\)

Solution:
Continuous as well as differentiable, so \( f^{\prime}(1)=0\)

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2011
The function \(\mathrm{f}(\mathrm{x})=\tan \mathrm{x}-4 \mathrm{x}\) is strictly decreasing on

Options:
A \(\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)\)
B \(\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\)
C \(\left(-\frac{\pi}{3}, \frac{\pi}{2}\right)\)
D \(\left(\frac{\pi}{2}, \pi\right)\)

The correct answer is:
\(\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)\)

Solution:
\(f(x)=\tan x-4 x \Rightarrow f^{\prime}(x)=\sec ^2 x-4\)
When \(\frac{-\pi}{3} < \mathrm{x} < \frac{\pi}{3}, 1 < \sec \mathrm{x} < 2\)
Therefore, \(1 < \sec ^2 x < 4\)
\(\Rightarrow-3 < \left(\sec ^2 x-4\right) < 0\)
Thus, for \(\frac{-\pi}{3} < \mathrm{x} < \frac{\pi}{3}, \mathrm{f}^{\prime}(\mathrm{x}) < 0\)
Hence, \(\mathrm{f}\) is strictly decreasing on \(\left(\frac{-\pi}{3}, \frac{\pi}{3}\right)\)

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2010
The fundamental frequency of an open organ pipe is $300 \mathrm{~Hz}$. The first overtone of this pipe has same frequency as first overtone of a closed organ pipe. If speed of sound is $330 \mathrm{~m} / \mathrm{s}$, then the length of closed organ pipe is

Options:
A $41 \mathrm{~cm}$
B $37 \mathrm{~cm}$
C $31 \mathrm{~cm}$
D $80 \mathrm{~cm}$

The correct answer is:
$41 \mathrm{~cm}$

Solution:
For open pipe, $\mathrm{n}=\frac{\mathrm{v}}{2 \ell},$ where $\mathrm{n}_{0}$ is the
fundamental frequency of open pipe. $\therefore \ell=\frac{\mathrm{v}}{2 \mathrm{n}}=\frac{330}{2 \times 300}=\frac{11}{20}$
As freq. of 1 st overtone of open pipe $=$ freq. of 1 st overtone of closed pipe
$\begin{aligned} \therefore \quad 2 \frac{\mathrm{v}}{2 \ell}=3 \frac{\mathrm{v}}{4 \ell^{\prime}} \Rightarrow \ell^{\prime} &=\frac{3 \ell}{4}=\frac{3}{4} \times \frac{11}{20} \\ &=41.25 \mathrm{~cm} \end{aligned}$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2011
The general solution of \(8 \tan ^2 \frac{x}{2}=1+\sec x\) is

Options:
A \(2 n \pi \pm \cos ^{-1}\left(\frac{-1}{3}\right)\)
B \(2 \mathrm{n} \pi \pm \frac{\pi}{6}\)
C \(2 \mathrm{n} \pi \pm \cos ^{-1}\left(\frac{1}{3}\right)\)
D None of these

The correct answer is:
\(2 \mathrm{n} \pi \pm \cos ^{-1}\left(\frac{1}{3}\right)\)

Solution:
We have, \(8 \tan ^2 \frac{x}{2}=1+\sec x\)
\(\begin{aligned}
& \Rightarrow 8\left(\frac{1-\cos x}{1+\cos x}\right)=1+\frac{1}{\cos x}=\frac{1+\cos x}{\cos x} \\
& \Rightarrow 8 \cos x-8 \cos ^2 x=(1+\cos x)^2 \\
& \Rightarrow 9 \cos ^2 \mathrm{x}-6 \cos \mathrm{x}+1=0 \\
& \Rightarrow(3 \cos \mathrm{x}-1)^2=0 \Rightarrow 3 \cos \mathrm{x}-1=0 \\
& \Rightarrow \cos \mathrm{x}=\frac{1}{3}=\cos \alpha \text { (say) } \Rightarrow \mathrm{x}=2 \mathrm{n} \pi \pm \alpha \\
& \therefore \mathrm{x}=2 \mathrm{n} \pi \pm \cos ^{-1}\left(\frac{1}{3}\right) \text { where } \mathrm{n} \in \mathrm{I}
\end{aligned}\)

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2009
The general solution of sinx-3sin2x+sin3x=cosx-3cos2x+cos3x is

Options:
A nπ+π8
B nπ2+π8
C -1nnπ2+π8
D 2hπ+cos-132

The correct answer is:
nπ2+π8

Solution:
We have,

sinx-3sin2x+sin3x=cosx-3cos2x+cos3x

⇒sinx+sin3x-3sin2x=cosx+cos3x-3cos2x

⇒2sin2xcosx-3sin2x-2cos2xcosx+3cos2x=0

⇒sin2x(2cosx-3)-cos2x(2cosx-3)=0

⇒(sin2x-cos2x)(2cosx-3)=0

⇒ sin2x=cos2x [∵cosx≠3/2]

⇒ 2x=2nπ±π2-2x

⇒ x=nπ2+π8 [∵ neglect -ve sign]

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2017
The general solution of the differential equation $\frac{d y}{d x}+\sin (x+y)=\sin (x-y)$ is

Options:
A $\log \tan y+\sin x=\mathrm{C}$
B $\quad \log \tan \frac{y}{2}+2 \sin x=\mathrm{C}$
C $\tan \frac{y}{2}+\log \sin x=\mathrm{C}$
D None of these

The correct answer is:
$\quad \log \tan \frac{y}{2}+2 \sin x=\mathrm{C}$

Solution:
The equation is,

$\frac{d y}{d x}=\sin (x-y)-\sin (x+y)=2 \cos x \sin (-y)$

$\Rightarrow \frac{d y}{\sin y}+2 \cos x d x=0$

$\Rightarrow \int \cos e c y d y+2 \int \cos x d x=\mathrm{C}$

$\Rightarrow \log \tan \frac{y}{2}+2 \sin x=\mathrm{C}$

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2013
The general solution of the differential equation $\left(\tan ^{-1} y-x\right) d y=\left(1+y^{2}\right) d x$ is

Options:
A $x=\left(\tan ^{-1} y+1\right)+\mathrm{C} e^{-\tan ^{-1} y}$
B $x=\left(\tan ^{-1} y-1\right)+\mathrm{C} e^{-\tan ^{-1} y}$
C $x=\left(\tan ^{-1} x-1\right)+\mathrm{C} e^{-\tan ^{-1} x}$
D $x=\left(\tan ^{-1} x+1\right)+\mathrm{Ce}^{-\tan ^{-1} x}$

The correct answer is:
$x=\left(\tan ^{-1} y-1\right)+\mathrm{C} e^{-\tan ^{-1} y}$

Solution:
The given differential equation can be written as $\frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{\tan ^{-1} y}{1+y^{2}} \ldots$ (i)

Now, eq. (i) is a linear differential equation of the form $\frac{d x}{d y}+\mathrm{P}_{1} x=\mathrm{Q}_{1}$

where $\mathrm{P}_{1}=\frac{1}{1+y^{2}}$ and $\mathrm{Q}_{1}=\frac{\tan ^{-1} y}{1+y^{2}}$

Therefore, I.F $=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y}$

Thus, the solution of the given differential equation is given by $x e^{\tan ^{-1} y}=\int\left(\frac{\tan ^{-1} y}{1+y^{2}}\right) e^{\tan ^{-1} y} \mathrm{dy}+\mathrm{C} \ldots$ (ii)

Let $\mathrm{I}=\int\left(\frac{\tan ^{-1} y}{1+y^{2}}\right) e^{\tan ^{-1} y} \mathrm{dy}$

On substituting $\tan ^{-1} y=t$, so that

$\left(\frac{1}{1+y^{2}}\right) \mathrm{dy}=\mathrm{dt}$, we get

$I=\int t e^{t} d t=t e^{t}-\int 1 \cdot e^{t} d t=t e^{t}-e^{t}=e^{t}(t-1)$

or $\mathrm{I}=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)$

On substituting the value of 1 in equation (i1), we get

$x \cdot e^{\tan ^{-1} y}=\cdot e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)+\mathrm{C}$

or $x=\left(\tan ^{-1} y-1\right)+\mathrm{C} e^{\tan ^{-1} y}$

which is the general solution of the given differntial equation.

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2021
The general solution of the equation $\sin 2 x+2 \sin x+2 \cos x+1=0$ is

Options:
A $3 \mathrm{n} \pi-\frac{\pi}{4}$
B $2 \mathrm{n} \pi+\frac{\pi}{4}$
C $2 \mathrm{n} \pi+(-1)^{\mathrm{n}} \sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
D $\mathrm{n} \pi-\frac{\pi}{4}$

The correct answer is:
$\mathrm{n} \pi-\frac{\pi}{4}$

Solution:
Given, $\sin 2 \mathrm{x}+2 \sin \mathrm{x}+2 \cos \mathrm{x}+1=0$

$\Rightarrow 1+\sin 2 \mathrm{x}+2(\sin \mathrm{x}+\cos \mathrm{x})=0$

$\Rightarrow(\sin x+\cos x)^{2}+2(\sin x+\cos x)=0$

$\Rightarrow(\sin x+\cos x)(\sin x+\cos x+2)=0$

$\therefore \sin x+\cos x=0$ or $\sin x+\cos x=-2$

But, $\sin x+\cos x=-2$ is inadmissible.

Since, $|\sin x| \leq 1,|\cos x| \leq 1$

$\therefore \sin x+\cos x=0 \Rightarrow \sin \left(x+\frac{\pi}{4}\right)=0$

$\Rightarrow x+\frac{\pi}{4}=n \pi \Rightarrow x=n \pi-\frac{\pi}{4}$

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2016
The given graph represents the variation of compressibility factor $(Z)=\frac{p V}{n R T}$, for three real gases $A, B$ and $C$. Identify the only incorrect statement.

Options:
A
For the gas $A, a=0$ and its dependence on $p$ is linear at all pressure.

B
For the gas $B, b=0$ and its dependence on $p$ is linear at all pressure.

C
For the gas $C$, which is typical real gas for which neither. $a$ nor $b=0$. By knowing the minima and point of the intersection, with $Z=1, a$ and $b$ can be calculated.

D
At high pressure the slope is positive for all real gases.

The correct answer is:

For the gas $B, b=0$ and its dependence on $p$ is linear at all pressure.

Solution:
From the graph it is clear that, the value of ' $Z$ ' decreases with increase of pressure. We can explain as follows on the basis of van der Waals' equation.
At high pressure, when ' $p$ ' is large, $V$ will be small and one cannot ignore ' $b$ ' in comparison to $V$. However, the team $a / V^2$ may be considered negligible in comparison to ' $p$ ' in van der Waals' equation.
$\left(p+\frac{a}{V^2}\right)(V-b)=n R T$
$p(V-b)=n R T$
$\Rightarrow \quad p V-p b=n R T$
or $\frac{p V}{n R T}=1+\frac{p b}{n R T}$
or $Z=1+\frac{p b}{n R T}$
Thus, $Z$ is greater than 1. As pressure is increased (at constant $T$ ), the factor $\frac{p b}{n R T}$ increases. This explains why after minima in the curves, $Z$ increase continuously with pressure. Hence, the only incorrect statement is (b).

Topic: Chemistry, States of Matter, BITSAT, BITSAT 2022
The given lens is broken into four parts and rearranged as shown. If the initial focal length is $f$ then after rearrangement the equivalent focal length is-

Options:
A $f$
B $f / 2$
C $f / 4$
D $4 f$

The correct answer is:
$f / 2$

Solution:
Cutting a lens in transverse direction doubles their focal length i.e. $2 f$. Using the formula of equivalent focal length, $\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{1}{f_{3}}+\frac{1}{f_{4}}$

We get equivalent focal length as $f / 2$

Topic: Physics, Ray Optics, BITSAT, BITSAT 2013
The greatest and least values of $\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}$ are respectively

Options:
A $\frac{\pi^{2}}{4}$ and 0
B $\frac{\pi}{2}$ and $\frac{-\pi}{2}$
C $\frac{5 \pi^{2}}{4}$ and $\frac{\pi^{2}}{8}$
D $\frac{\pi^{2}}{4}$ and $\frac{-\pi^{2}}{4}$

The correct answer is:
$\frac{5 \pi^{2}}{4}$ and $\frac{\pi^{2}}{8}$

Solution:
We have, $\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}$

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

$=\frac{\pi^{2}}{4}-2 \sin ^{-1} x\left(\frac{\pi}{2}-\sin ^{-1} x\right)$

$=\frac{\pi^{2}}{4}-\pi \sin ^{-1} x+2\left(\sin ^{-1} x\right)^{2}$

$=2\left[\left(\sin ^{-1} x\right)^{2}-\frac{\pi}{2} \sin ^{-1} x+\frac{\pi^{2}}{8}\right]$

$=2\left[\left(\sin ^{-1} x-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{16}\right]$

Thus, the least value is $2\left(\frac{\pi^{2}}{16}\right)$ i.e. $\frac{\pi^{2}}{8}$

and the greatest value is $2\left[\left(\frac{-\pi}{2}-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{16}\right]$ i.e. $\frac{5 \pi^{2}}{4} .$

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2013
The greatest positive integer, which divides

$\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)$ for all $\mathrm{n} \in \mathbf{N},$ is

Options:
A 2
B 6
C 24
D 120

The correct answer is:
24

Solution:
The product of $\mathrm{r}$ consecutive integers is divisible byr! . Thus $n(n+1)(n+2)(n+3)$ is divisible by $4 !=24$

Topic: Mathematics, Basic of Mathematics, BITSAT, BITSAT 2014
The greatest term in the expansion of 31+1320 is

Options:
A 268409
B 248409
C 258409
D None of these

The correct answer is:
258409

Solution:
Tr+1=3·20Cr13r

and Tr=3·20Cr-113r-1

Now, Tr+1Tr=20-r+1r13

Since, Tr+1≥Tr

⇒20-r+1≥3r

⇒r≤213+1=212.73

⇒r≤7.692⇒r=7

∴ The greatest term is T8=3·C720137=258409

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2018
The greenhouse effect is because of the

Options:
A presence of gases, which in general are strong infrared absorbers, in the atmosphere
B presence of $\mathrm{CO}_{2}$, only in the atmosphere
C pressure of $\mathrm{O}_{3}$ and $\mathrm{CH}_{4}$ in the atmosphere
D $\mathrm{N}_{2} \mathrm{O}$ and chlorofluorohydrocarbons in the atmosphere

The correct answer is:
presence of gases, which in general are strong infrared absorbers, in the atmosphere

Solution:
Green house gases such as $\mathrm{CO}_{2}$, ozone, methane, the chlorofluorocarbon compounds and water vapour form a thick cover around the earth which prevents the IR rays emitted by the earth to escape. It gradually leads to increase in temperature of atmosphere.

Topic: Chemistry, Environmental Chemistry, BITSAT, BITSAT 2020
The greenhouse effect is because of the

Options:
A presence of gases, which in general are strong infrared absorbers, in the atmosphere.
B presence of $\mathrm{CO}_{2}$, only in the atmosphere.
C presence of $\mathrm{O}_{3}$ and $\mathrm{CH}_{4}$ in the atmosphere.
D $\mathrm{N}_{2} \mathrm{O}$ and chlorofluorohydrocarbons in the atmosphere.

The correct answer is:
presence of gases, which in general are strong infrared absorbers, in the atmosphere.

Solution:
Green house gases such as $\mathrm{CO}_{2}$, ozone, methane, the chlorofluoro carbon compounds and water vapour form a thick cover around the earth which prevents the IR rays emitted by the earth to escape. It gradually leads to increase in temperature of atmosphere.

Topic: Chemistry, Environmental Chemistry, BITSAT, BITSAT 2013
The heat generated in a circuit is given by $\mathrm{Q}=\mathrm{I}^{2} \mathrm{Rt},$ where I is current, $\mathrm{R}$ is resistance and tis time. If the percentage errors in measuring I. $\mathrm{R}$ and $\mathrm{t}$ are $2 \%, 1 \%$ and $1 \%$ respectively, then the maximum error in measuring heat will be

Options:
A $2 \%$
B $3 \%$
C $4 \%$
D $6 \%$

The correct answer is:
$6 \%$

Solution:
$\begin{aligned} \frac{\Delta Q}{Q} \times 100 &=\frac{2 \Delta I}{I} \times 100+\frac{\Delta R}{R} \times 100+\frac{\Delta t}{t} \times 100 \\ &=2 \times 2 \%+1 \%+1 \%=6 \% \end{aligned}$

Topic: Physics, Mathematics in Physics, BITSAT, BITSAT 2020