datavorous/entrance-exam-dataset huggingface dataset:
Evaluate: $\int \frac{1}{1+3 \sin ^{2} x+8 \cos ^{2} x} \mathrm{dx}$

Options:
A $\frac{1}{6} \tan ^{-1}(2 \tan \mathrm{x})+\mathrm{C}$
B $\tan ^{-1}(2 \tan \mathrm{x})+\mathrm{C}$
C $\frac{1}{6} \tan ^{-1}\left(\frac{2 \tan \mathrm{x}}{3}\right)+\mathrm{C}$
D None of these

The correct answer is:
$\frac{1}{6} \tan ^{-1}\left(\frac{2 \tan \mathrm{x}}{3}\right)+\mathrm{C}$

Solution:
$\mathrm{I}=\int \frac{1}{1+3 \sin ^{2} \mathrm{x}+8 \cos ^{2} \mathrm{x}} \mathrm{dx}$

Dividing the numerator and denominator by $\cos ^{2} \mathrm{x},$ we get

$$

I=\int \frac{\sec ^{2} x}{\sec ^{2} x+3 \tan ^{2} x+8} d x=\int \frac{\sec ^{2} x}{4 \tan ^{2} x+9} d x

$$

Putting $\tan x=t \Rightarrow \sec ^{2} x d x=d t,$ we get

$\mathrm{I}=\int \frac{\mathrm{dt}}{4 \mathrm{t}^{2}+9}=\frac{1}{4} \int \frac{\mathrm{dt}}{\mathrm{t}^{2}+(3 / 2)^{2}}=\frac{1}{4} \times \frac{1}{3 / 2} \tan ^{-1}\left(\frac{\mathrm{t}}{3 / 2}\right)+\mathrm{C}$

$\Rightarrow \mathrm{I}=\frac{1}{6} \tan ^{-1}\left(\frac{2 \mathrm{t}}{3}\right)+\mathrm{C}=\frac{1}{6} \tan ^{-1}\left(\frac{2 \tan \mathrm{x}}{3}\right)+\mathrm{C}$

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2014
Evaluate: $\int \sqrt{\frac{\mathrm{x}}{4-\mathrm{x}^{3}}} \mathrm{dx}$

Options:
A $\frac{2}{3} \sin ^{-1}\left(\frac{x^{3 / 2}}{2}\right)+c$
B $\frac{2}{3} \sin ^{-1}\left(x^{3 / 2}\right)+c$
C $2 \sin ^{-1}\left(\frac{x^{3 / 2}}{2}\right)+c$
D $\frac{1}{3} \sin ^{-1}\left(\frac{\mathrm{x}^{3 / 2}}{2}\right)+\mathrm{c}$

The correct answer is:
$\frac{2}{3} \sin ^{-1}\left(\frac{x^{3 / 2}}{2}\right)+c$

Solution:
$I=\int \sqrt{\frac{x}{4-x^{3}}} d x=\int \frac{\sqrt{x} d x}{\sqrt{4-x^{3}}}$
Here integral of $\sqrt{x}=\frac{2}{3} x^{3 / 2}$ and $4-x^{3}=4-\left(x^{3 / 2}\right)^{2}$
Put $x^{3 / 2}=t \Rightarrow \sqrt{x} d x=\frac{2}{3} d t$
So $I=\frac{2}{3} \int \frac{d t}{\sqrt{4-t^{2}}}=\frac{2}{3} \sin ^{-1}\left(\frac{x^{3 / 2}}{2}\right)+c$

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2012
Evaluate \(\int \frac{x^2}{x^2-1} d x\)

Options:
A \(x-\frac{1}{2} \log \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)+\mathrm{c}\)
B \(x+\frac{1}{2} \log \left(\frac{x+1}{x-1}\right)+c\)
C \(x+\frac{1}{2} \log \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)+\mathrm{c}\)
D None of these

The correct answer is:
\(x+\frac{1}{2} \log \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)+\mathrm{c}\)

Solution:
Given integral \(I=\int\left(1+\frac{1}{x^2-1}\right) d x\)
\(\begin{aligned}
& =\int d x+\int \frac{d x}{(x-1)(x+1)} \\
& =x+\frac{1}{2} \int\left(\frac{1}{x-1}-\frac{1}{x+1}\right) d x=x+\frac{1}{2} \log \left(\frac{x-1}{x+1}\right)+c
\end{aligned}\)

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2011
Evaluate: \(\int x \tan ^{-1} x d x\)

Options:
A \(\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x-\frac{1}{2} x+c\)
B \(\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x+\frac{1}{2} x+c\)
C \(\frac{1}{2}\left(\mathrm{x}^2-1\right) \tan ^{-1} \mathrm{x}-\frac{1}{2} \mathrm{x}+\mathrm{c}\)
D None of these

The correct answer is:
\(\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x-\frac{1}{2} x+c\)

Solution:
\(\begin{aligned}
& \int x \tan ^{-1} x d x=\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^2}{1+x^2} d x \\
& =\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int\left[1-\frac{1}{1+x^2}\right] d x \\
& =\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x-\frac{1}{2} x+c
\end{aligned}\)

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2009
Evaluate : \(\int_0^1 \frac{\mathrm{dx}}{\sqrt{2-\mathrm{x}^2}}\).

Options:
A \(\pi / 4\)
B \(\pi\)
C \(\pi / 2\)
D \(\pi / 3\)

The correct answer is:
\(\pi / 4\)

Solution:
\(\begin{aligned}
& \int \frac{\mathrm{dx}}{\sqrt{2-\mathrm{x}^2}}=\sin ^{-1} \frac{\mathrm{x}}{\sqrt{2}}+\mathrm{c} \\
& \text {So } \int_0^1 \frac{\mathrm{dx}}{\sqrt{2-\mathrm{x}^2}}=\left.\sin ^{-1} \frac{\mathrm{x}}{\sqrt{2}}\right|_0 ^1 \\
= & \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+c-\sin ^{-1}(0)-c=\frac{\pi}{4}-0=\frac{\pi}{4}
\end{aligned}\)

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2009
Evaluate \(\lim _{x \rightarrow \infty} \frac{\sqrt{1+\sin 3 x}-1}{\ln (1+\tan 2 x)}\);

Options:
A \(1 / 2\)
B \(3 / 2\)
C \(3 / 4\)
D \(1 / 4\)

The correct answer is:
\(3 / 4\)

Solution:
\(\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1+\sin 3 x}-1}{\ln (1+\tan 2 x)} \\
= & \lim _{x \rightarrow 0} \frac{(1+\sin 3 x)-1}{\{\sqrt{1+\sin 3 x+1}) \ln (1+\tan 2 x)} \\
= & \lim _{x \rightarrow 0} \frac{1}{(\sqrt{1+\sin 3 x}+1)} \cdot \frac{\sin 3 x}{\ln (1+\tan 2 x)^{\frac{1}{\tan 2 x}}\times \frac{1}{\tan 2 x}}
\end{aligned}\)
\(\begin{aligned} & =\lim _{x \rightarrow 0} \frac{1}{(\sqrt{1+\sin 3 x}+1)} \cdot\left(\frac{\sin 3 x}{3 x}\right)\left(\frac{2 x}{\tan 2 x}\right) \\ & \times \frac{3}{\tan 2 x} \cdot \frac{1}{\ln (1+\tan 2 x)^{\frac{1}{\tan 2 x}}} \\ & =\left(\frac{1}{1+1}\right)(1)(1)\left(\frac{3}{2}\right) \frac{1}{\ln e}=\frac{3}{4} .\end{aligned}\)

Topic: Mathematics, Limits, BITSAT, BITSAT 2011
Experiment to study kinetics of the dissociation of hydrogen peroxide must be performed by group of two or three so that-

Options:
A when one is recording data other should be swirling flask at constant rate.
B experiment can be performed by one student only as outcomes are independent on rate of mixing of mixture 1 and 3 .
C for safety purpose.
D none of these.

The correct answer is:
when one is recording data other should be swirling flask at constant rate.

Solution:
No Solution available

Topic: Chemistry, Hydrogen, BITSAT, BITSAT 2010
$\Delta_{f} \mathrm{G}^{-}$ vs $\mathrm{T}$ plot in the Ellingham diagram slopes downward for the reaction

Options:
A $\mathrm{Mg}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{MgO}$
B $2 \mathrm{Ag}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{Ag}_{2} \mathrm{O}$
C $\mathrm{C}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{CO}$
D $\mathrm{CO}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}$

The correct answer is:
$\mathrm{C}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{CO}$

Solution:
$\mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{S}$ increases.

Hence, as the temperature increases, $\mathrm{T} \Delta \mathrm{S}$ increases and hence $\Delta \mathrm{G}(\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S})$ decreases. In other words, the slope of the curve for formation of CO decreases. However, for all other oxides, it increases.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, BITSAT, BITSAT 2020
$f(x)=\left(\frac{e^{2 x}-1}{e^{2 x}+1}\right)$ is

Options:
A an increasing function
B a decreasing function
C an even function
D None of these

The correct answer is:
an increasing function

Solution:
$\begin{array}{l}

\because \mathrm{f}(x)=\left(\frac{e^{2 x}-1}{e^{2 x}+1}\right) \\

\therefore \mathrm{f}(-x)=\frac{e^{-2 x}-1}{e^{-2 x}+1}=\frac{1-e^{2 x}}{1+e^{2 x}} \\

\Rightarrow \mathrm{f}(-x)=\frac{-\left(e^{2 x}-1\right)}{e^{2 x}+1}=-\mathrm{f}(x)

\end{array}$

$\therefore \mathrm{f}(x)$ is an odd function.

Again, $\mathrm{f}^{\prime}(x)=\frac{4 e^{2 x}}{\left(1+e^{2 x}\right)^{2}}>0, \forall x \in R$ $\Rightarrow \mathrm{f}(x)$ is an increasing function.

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2021
$\left[\mathrm{Fe}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]^{2+}+$ en $\rightarrow$ complex $(\mathrm{X})$. The correct statement about the complex (X) is -

Options:
A it is a low spin complex
B it is diamagnetic
C it shows geometrical isomerism
D (a) and (b) both

The correct answer is:
(a) and (b) both

Solution:
Complex $\mathrm{X}$ is $\left[\mathrm{Fe}(\mathrm{en})_{3} \mathrm{right}\right]^{2+} ;$ as 'en' is a strong field ligand pairing of electrons will take place. $\left[\mathrm{Fe}(\mathrm{en})_{3}\right]^{2+}$ :

$d^{2} s p^{3}$ hybridisation

Hence, hybridisation is $d^{2} s p^{3}$ and complex is diamagnetic. As it has 3 bidentate symmetrical 'en' ligands so it will not show geometrical isomerism.

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2021
Figure below shows two paths that may be taken by a gas to go from a state $A$ to a state $(c)$

In process $\mathrm{AB}, 400 \mathrm{~J}$ of heat is added to the system and in process $\mathrm{BC}, 100 \mathrm{~J}$ of heat is added to the system. The heat absorbed by the system in the process AC will be

Options:
A $500 \mathrm{~J}$
B $460 \mathrm{~J}$
C $300 \mathrm{~J}$
D $380 \mathrm{~J}$

The correct answer is:
$460 \mathrm{~J}$

Solution:
In cyclic process ABCA $\mathrm{Q}_{\text {cycle }}=\mathrm{W}_{\text {cycl }}$

$\mathrm{Q}_{\mathrm{AB}}+\mathrm{Q}_{\mathrm{BC}}+\mathrm{Q}_{\mathrm{CA}}=$ ar. of $\Delta \mathrm{ABC}$

$+400+100+\mathrm{Q}_{\mathrm{C} \rightarrow \mathrm{A}}=\frac{1}{2}\left(2 \times 10^{-3}\right)\left(4 \times 10^{4}\right)$

$\Rightarrow \mathrm{Q}_{\mathrm{C} \rightarrow \mathrm{A}}=-460 \mathrm{~J}$

$\Rightarrow \mathrm{Q}_{\mathrm{A} \rightarrow C}=+460 \mathrm{~J}$

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2016
Figure shows a copper rod joined to a steel rod. The rods have equal length and equal crosssectional area. The free end of the copper rod is kept at $0^{\circ} \mathrm{C}$ and that of steel rod is kept at $100^{\circ} \mathrm{C}$. Find the temperature of the junction of the rod. Conductivity of copper $=390 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}$

Conductivity of steel $=46 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}$

Options:
A $18.01^{\circ} \mathrm{C}$
B $26^{\circ} \mathrm{C}$
C $10.6^{\circ} \mathrm{C}$
D $20^{\circ} \mathrm{C}$

The correct answer is:
$10.6^{\circ} \mathrm{C}$

Solution:
Heat current in first rod (copper)

$$

=\frac{390 \times \mathrm{A}(0-\theta)}{\ell}

$$

Here $\theta$ is temperature of the junction and $\mathrm{A}$ $\& \ell$ are area and length of copper rod. Heat current in second rod (steel)

$$

=\frac{46 \times \mathrm{A}(\theta-100)}{\ell}

$$

In series combination, heat current remains same. So,

$$

\begin{array}{l}

\frac{390 \times \mathrm{A}(0-\theta)}{\ell}=\frac{46 \times \mathrm{A}(\theta-100)}{\ell} \\

-390 \theta==46 \theta-4600 \\

436 \theta=4600 \Rightarrow \theta=10.6^{\circ} \mathrm{C}

\end{array}

$$

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2013
Find out $B$ in the given reactions

Options:
A acetophenone
B benzaldehyde
C cyclohexyl carbaldehyde
D benzoic acid

The correct answer is:
benzaldehyde

Solution:

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Find the A.M. of the first ten odd numbers.

Options:
A 10
B 20
C 15
D 25

The correct answer is:
10

Solution:
First ten odd numbers are 1, 3, 5, 7, 11,13, \(15,17,19\) respectively. So A.M.
\(\begin{aligned}
(\bar{x}) & =\frac{1+3+5+7+9+11+13+15+17+19}{10} \\
& =\frac{100}{10}=10
\end{aligned}\)

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2011
Find the angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$

and the plane $10 \mathrm{x}+2 \mathrm{y}-11 \mathrm{z}=3$

Options:
A $\sin ^{-1}\left(\frac{8}{21}\right)$
B $\sin ^{-1}\left(\frac{5}{21}\right)$
C $\sin ^{-1}\left(\frac{7}{21}\right)$
D $\sin ^{-1}\left(\frac{1}{21}\right)$

The correct answer is:
$\sin ^{-1}\left(\frac{8}{21}\right)$

Solution:
Let $\theta$ be the angle between the line and the normal to the plane. Converting the given equations into vector form, we have $\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+3 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})$ and

$$

\overrightarrow{\mathrm{r}} \cdot(10 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-11 \hat{\mathrm{k}})=3

$$

Here, $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{n}}=10 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-11 \hat{\mathrm{k}}$

$$

\begin{array}{l}

\sin \phi=\left|\frac{(2 \hat{i}+3 \hat{j}+6 \hat{k}) \cdot(10 \hat{i}+2 \hat{j}-11 \hat{k})}{\sqrt{2^{2}+3^{2}+6^{2}} \sqrt{10^{2}+2^{2}+11^{2}}}\right| \\

=\left|\frac{-40}{7 \times 15}\right|=\left|\frac{-8}{21}\right|=\frac{8}{21} \text { or } \phi=\sin ^{-1}\left(\frac{8}{21}\right)

\end{array}

$$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2014
Find the angle between the two planes $2 \mathrm{x}+\mathrm{y}-2 \mathrm{z}=5$ and $3 x-6 y-2 z=7$

Options:
A $\cos ^{-1}(4 / 21)$
B $\cos ^{-1}(2 / 21)$
C $\cos ^{-1}(1 / 21)$
D $\cos ^{-1}(5 / 21)$

The correct answer is:
$\cos ^{-1}(4 / 21)$

Solution:
The angle between two planes is the angle between their normals. From the equation of the planes, the normal vectors are $\overrightarrow{\mathrm{N}}_{1}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{N}}_{2}=3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$

Therefore, $\cos \theta=\left|\frac{\overrightarrow{\mathrm{N}}_{1} \cdot \overrightarrow{\mathrm{N}}_{2}}{\left|\overrightarrow{\mathrm{N}}_{1}\right|\left|\overrightarrow{\mathrm{N}}_{2}\right|}\right|$

$$

=\left|\frac{(2 \hat{i}+\hat{j}-2 \hat{k}) \cdot(3 \hat{i}-6 \hat{j}-2 \hat{k})}{\sqrt{4+1+4} \sqrt{9+36+4}}\right|=\left(\frac{4}{21}\right)

$$

Hence, $\theta=\cos ^{-1}\left(\frac{4}{21}\right)$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2013
Find the coordinates of the point where the line joining the points (2,-3,1) and (3,-4,-5) cuts the plane $2 x+y+z=7$

Options:
A (1,2,-7)
B (1,-2,7)
C (-1,-2,7)
D (1,2,7)

The correct answer is:
(1,-2,7)

Solution:
The direction ratios of the line are $3-2,-4-(-3),-5-1$ i.e. 1,-1,-6
Hence equation of the line joining the given points is $\frac{x-2}{1}=\frac{y+3}{-1}=\frac{z-1}{-6}=r($ say $)$
Coordinates of any point on this line are $(r+2,-r-3,-6 r+1)$
If this point lies on the given plane $2 x+y+z=7$ then $2(r+2)+(-r-3)+(-6 r+1)=7 \Rightarrow r=-1$
Coordinates of any point on this line are $(-1+2,-(-1)-3,-6(-1)+1)$ i.e. (1,-2,7)

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2012
Find the eccentricity of the conic represented by $x^{2}-y^{2}-4 x+4 y+16=0$

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

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

Solution:
We have $x^{2}-y^{2}-4 x+4 y+16=0$
$\Rightarrow\left(x^{2}-4 x\right)-\left(y^{2}-4 y\right)=16$
$\Rightarrow\left(x^{2}-4 x+4\right)-\left(y^{2}-4 y+4\right)=-16$
$\Rightarrow(x-2)^{2}-(y-2)^{2}=-16$
$\Rightarrow \frac{(x-2)^{2}}{4^{2}}-\frac{(y-2)^{2}}{4^{2}}=1$
This is rectangular hyperbola, whose eccentricity is always $\sqrt{2}$.

Topic: Mathematics, Hyperbola, BITSAT, BITSAT 2012
Find the equation of chord of the circle \(x^2+y^2=8 x\) bisected at the point \((4,3)\)

Options:
A \(y=3\)
B \(y=1\)
C \(y=6\)
D \(y=7\)

The correct answer is:
\(y=3\)

Solution:
\(\begin{aligned}
& \mathrm{T}=\mathrm{S}_1 \Rightarrow \mathrm{x}(4)+\mathrm{y}(3)-4(\mathrm{x}+4)=16+9-32 \\
& \Rightarrow 3 \mathrm{y}-9=0 \Rightarrow \mathrm{y}=3
\end{aligned}\)

Topic: Mathematics, Limits, BITSAT, BITSAT 2010
Find the final intensity of light (I"), if the angle between the axes of two polaroids is $60^{\circ}$.

Options:
A $\frac{3 \mathrm{I}_{0}}{2}$
B $\frac{\mathrm{I}_{0}}{2}$
C $\frac{\mathrm{I}_{0}}{4}$
D $\frac{\mathrm{I}_{0}}{8}$

The correct answer is:
$\frac{\mathrm{I}_{0}}{8}$

Solution:
From the first polaroid

$$

\mathrm{I}^{\prime}=\frac{\mathrm{I}_{0}}{2}

$$

From second polaroid

$$

\begin{aligned}

\mathrm{I}^{\prime \prime} &=\mathrm{I}^{\prime} \cos ^{2} \theta=\frac{\mathrm{I}_{0}}{2} \cos ^{2}\left(60^{\circ}\right) \\

&=\frac{\mathrm{I}_{0}}{2} \times\left(\frac{1}{2}\right)^{2}=\frac{\mathrm{I}_{0}}{8}

\end{aligned}

$$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2014
Find the mean deviation about the mean for the data $4,7,8,9,10,12,13,17$

Options:
A 3
B 24
C 10
D 8

The correct answer is:
3

Solution:
Arithmetic mean $\bar{x}$ of $4,7,8,9,10,12,13$, 17 is
$\bar{x}=$ $\frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}=10$
$\Sigma\left|x_i-\bar{x}\right|=6+3+2+1+0+2+3+7=24$
$\therefore$ Mean deviation about mean
$=\text { M.D. }(\bar{x})=\frac{\sum\left|x_i-\bar{x}\right|}{n}=\frac{24}{8}=3$

Topic: Mathematics, Statistics, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Find the number of photon emitted per second by a 25 watt source of monochromatic light of wavelength $6600 Å$. What is the photoelectric current assuming $3 \%$ efficiency for photoelectric effect?

Options:
A $\frac{25}{3} \times 10^{19} \mathrm{~J}, 0.4 \mathrm{amp}$
B $\frac{25}{4} \times 10^{19} \mathrm{~J}, 6.2 \mathrm{amp}$
C $\frac{25}{2} \times 10^{19} \mathrm{~J}, 0.8 \mathrm{amp}$
D None of these

The correct answer is:
$\frac{25}{3} \times 10^{19} \mathrm{~J}, 0.4 \mathrm{amp}$

Solution:
$P_{\text {in }}=25 \mathrm{~W}, \lambda=6600 \AA=6600 \times 10^{-10} \mathrm{~m}$

$n h v=P$

Number of photons emitted/sec $\begin{aligned} n &=\frac{P}{\frac{h c}{\lambda}}=\frac{P \lambda}{h c}=\frac{25 \times 6600 \times 10^{-10}}{6.64 \times 10^{-34} \times 3 \times 10^{8}} \\ &=8.28 \times 10^{19}=\frac{25}{3} \times 10^{19} \end{aligned}$

$3 \%$ of emitted photons are producing current

$\begin{aligned} \therefore \quad & I=\frac{3}{100} \times n e \\ &=\frac{3}{100} \times \frac{25}{3} \times 10^{19} \times 1.6 \times 10^{-19}=0.4 \mathrm{~A} \end{aligned}$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2016
Find the value of
\(2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{2}+2 \tan ^{-1} \frac{1}{8}\)

Options:
A \(\pi / 4\)
B \(\pi / 2\)
C \(3 \pi / 4\)
D None of these

The correct answer is:
\(\pi / 4\)

Solution:
\(\begin{aligned}
& 2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{2}+2 \tan ^{-1} \frac{1}{8} \\
& =2 \tan ^{-1} \frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5} \cdot \frac{1}{8}}+\tan ^{-1} \frac{1}{7} \\
& =2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{2 \cdot \frac{1}{3}}{1-\frac{1}{9}}+\tan ^{-1} \frac{1}{7} \\
& =\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4} \cdot \frac{1}{7}} \\
& =\tan ^{-1} \frac{25}{25}=\tan ^{-1} 1=45^{\circ}=\frac{\pi}{4}
\end{aligned}\)

Topic: Mathematics, Matrices, BITSAT, BITSAT 2009
Find the value of $a$ such that the sum of the squares of the roots of the equation $x^{2}-(a-2) x-(a+1)=0$ is least.

Options:
A 4
B 2
C 1
D 3

The correct answer is:
1

Solution:
Let $\alpha, \beta$ be the roots of the equation.

$\therefore \alpha+\beta=a-2$ and $\alpha \beta=-(a+1)$

Now, $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$

$=(a-2)^{2}+2(a+1)=(a-1)^{2}+5$

$\therefore \alpha^{2}+\beta^{2}$ will be minimum if $(a-1)^{2}=0$,

i.e. $a=1$.

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2021
Find the value of \(\int_0^{4 \pi}|\sin x| d x\)

Options:
A 8
B 6
C 4
D 2

The correct answer is:
8

Solution:
We know that \(|\sin x|\) is a periodic function of \(\pi\)
\(\text { Hence } \begin{array}{r}
\int_0^{4 \pi}|\sin x| d x=4 \int_0^\pi|\sin x| d x=4 \int_0^\pi \sin x d x \\
=4[-\cos x]_0^\pi=8
\end{array}\)

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2011
Find the value of \(\lim _{x \rightarrow 0} \frac{\sqrt{\left(1+x^2\right)}-\sqrt{1-x^2}}{x^2}\)

Options:
A 1
B 2
C 3
D 5

The correct answer is:
1

Solution:
\(\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x^2} \cdot \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}} \\
\Rightarrow & \lim _{x \rightarrow 0} \frac{1+x^2-1+x^2}{x^2\left(\sqrt{1+x^2}+\sqrt{1-x^2}\right)} \\
\Rightarrow & \lim _{x \rightarrow 0} \frac{2 x^2}{x^2\left(\sqrt{1+x^2}+\sqrt{1-x^2}\right)}=\frac{2}{\sqrt{1}+\sqrt{1}}=\frac{2}{2}=1
\end{aligned}\)

Topic: Mathematics, Statistics, BITSAT, BITSAT 2010
Find the value of $\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]$

Options:
A $-1 / 3$
B $-7 / 17$
C $-1 / 2$
D $-1 / 4$

The correct answer is:
$-7 / 17$

Solution:
$\begin{aligned} & 2 \tan ^{-1} \frac{1}{5}=\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{5} \\ &=\tan ^{-1} \frac{\frac{1}{5}+\frac{1}{5}}{1-\frac{1}{5} \cdot \frac{1}{5}}=\tan ^{-1} \frac{2 / 5}{24 / 25}=\tan ^{-1} \frac{5}{12} \\ &=\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)=\tan \left(\tan ^{-1} \frac{5}{12}-\frac{\pi}{4}\right) \\ &=\frac{\tan \left(\tan ^{-1} \frac{5}{12}\right)-\tan \frac{\pi}{4}}{1+\tan \left(\tan ^{-1} \frac{5}{12}\right) \tan \frac{\pi}{4}}=\frac{-\frac{7}{12}}{\frac{17}{12}}=-\frac{7}{17} \end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2014
Find the variance of the data given below

Options:
A 1.29
B 2.19
C 1.32
D None of these

The correct answer is:
1.32

Solution:
Variance $\sigma^{2}=\frac{\Sigma f_{i}\left(x_{i}-\bar{x}\right)^{2}}{\Sigma f_{i}}=\frac{286.49}{217}$

$=1.32$

Topic: Mathematics, Statistics, BITSAT, BITSAT 2013
Find the vertex of the parabola \(x^2-8 y-x+19=0\).

Options:
A \(\left(\frac{1}{2}, \frac{75}{32}\right)\)
B \(\left(\frac{1}{5}, \frac{65}{32}\right)\)
C \(\left(\frac{1}{3}, \frac{65}{22}\right)\)
D \(\left(\frac{1}{3}, \frac{35}{12}\right)\)

The correct answer is:
\(\left(\frac{1}{2}, \frac{75}{32}\right)\)

Solution:
The given equation of Parabola can be written as
\(\begin{aligned}
&\left(x-\frac{1}{2}\right)^2-8 y+19-\frac{1}{4}=0 \\
&\left(x-\frac{1}{2}\right)^2=8 y-\frac{76-1}{4} \\
& \Rightarrow\left(x-\frac{1}{2}\right)^2=8\left(y-\frac{75}{32}\right) \therefore \text { vertex }=\left(\frac{1}{2}, \frac{75}{32}\right)
\end{aligned}\)

Topic: Mathematics, Functions, BITSAT, BITSAT 2009
For a CE transistor amplifier, the audio signal voltage across the collector resistance of $2.0 \mathrm{k} \Omega$ is $2.0 \mathrm{~V}$. Suppose the current amplification factor of the transistor is 100 , What should be the value of $\mathrm{R}_{\mathrm{B}}$ in series with $\mathrm{V}_{\mathrm{BB}}$ supply of $2.0 \mathrm{~V}$ if the dc base current has to be 10 times the signal current?

Options:
A $14 \mathrm{k} \Omega$
B $18 \mathrm{k} \Omega$
C $10 \mathrm{k} \Omega$
D $5 \mathrm{k} \Omega$

The correct answer is:
$14 \mathrm{k} \Omega$

Solution:
The output ac voltage is $2.0 \mathrm{~V}$. So, the ac collector current $\mathrm{i}_{\mathrm{C}}=2.0 / 2000=1.0 \mathrm{~mA}$.

The signal current through the base is, therefore given by $\mathrm{i}_{\mathrm{B}}=\mathrm{i}_{\mathrm{C}} / \beta=1.0 \mathrm{~mA} / 100=0.010 \mathrm{~mA}$

The dc base current has to be $10 \times 0.010$ $=0.10 \mathrm{~mA}$

$$

\mathrm{R}_{\mathrm{B}}=\left(\mathrm{V}_{\mathrm{BB}}-\mathrm{V}_{\mathrm{BE}}\right) / \mathrm{I}_{\mathrm{B}}

$$

Assuming $\mathrm{V}_{\mathrm{BE}}=0.6 \mathrm{~V}, \mathrm{R}_{\mathrm{B}}=(2.0 .0 .6) / 0.10$

$=14 \mathrm{k} \Omega$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2014
For a gas if ratio of specific heats at constant pressure and volume is $\gamma$ then value of degrees of freedom is

Options:
A $\frac{3 \gamma-1}{2 \gamma-1}$
B $\frac{2}{\gamma-1}$
C $\frac{9}{2}(\gamma-1)$
D $\frac{25}{2}(\gamma-1)$

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

Solution:
$\begin{aligned} & \gamma=1+\frac{2}{f}, \Rightarrow \gamma-1=\frac{2}{f} \\ & \Rightarrow \frac{f}{2}=\frac{1}{\gamma-1} \Rightarrow f=\frac{2}{\gamma-1} \end{aligned}$

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2012
For a particular reversible reaction at temperature $T, \Delta H$ and $\Delta S$ were found to be both +ve. If $T_{e}$ is the temperature at equilibrium, the reaction would be spontaneous when

Options:
A $T_{e}>T$
B $T>T_{e}$
C $T_{e}$ is 5 times $T$
D $T=T_{e}$

The correct answer is:
$T>T_{e}$

Solution:
At equilibrium $\Delta G=0$

$$

\begin{array}{l}

\text { Hence, } \Delta G=\Delta H-T_{e} \Delta S=0 \\

\therefore \Delta H=T_{e} \Delta S \text { or } T_{e}=\frac{\Delta H}{\Delta S}

\end{array}

$$

For a spontaneous reaction $\Delta G$ must be negative which is possible only if $\Delta H-T \Delta S<0$

$\therefore \Delta H
or $T>\frac{\Delta H}{\Delta S} ; T_{e}

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Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2016