datavorous/entrance-exam-dataset huggingface dataset:
The anthracene is purified by

Options:
A crystallisation
B filtration
C distillation
D sublimation

The correct answer is:
sublimation

Solution:
Anthracene is purified by sublimation. In sublimation, a solid is converted directly into gaseous state on heating without passing through liquid phase.

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2015
The approximate depth of an ocean is $2700 \mathrm{~m}$. The compressibility of water is $45.4 \times 10^{-11} \mathrm{~Pa}^{-1}$ and density of water is $10^{3} \mathrm{~kg} / \mathrm{m}^{3}$. What fractional compression of water will be obtained at the bottom of the ocean?

Options:
A $1.0 \times 10^{-2}$
B $1.2 \times 10^{-2}$
C $1.4 \times 10^{-2}$
D $0.8 \times 10^{-2}$

The correct answer is:
$1.2 \times 10^{-2}$

Solution:
Compressibility of water, $\mathrm{K}=45.4 \times 10^{-11} \mathrm{~Pa}^{-1}$

density of water $\mathrm{P}=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$

depth of ocean, $\mathrm{h}=2700 \mathrm{~m}$ We have to find $\frac{\Delta \mathrm{V}}{\mathrm{V}}=?$

As we know, compressibility, $\mathrm{K}=\frac{1}{\mathrm{~B}}=\frac{(\Delta \mathrm{V} / \mathrm{V})}{\mathrm{P}}(\mathrm{P}=\mathrm{Pgh})$

So, $(\Delta \mathrm{V} / \mathrm{V})=\mathrm{KPgh}$

$=45.4 \times 10^{-11} \times 10^{3} \times 10 \times 2700$

$=1.2258 \times 10^{-2}$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2016
The area bounded by curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$ is

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

The correct answer is:
$\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)$

Solution:
Given circles are $x^{2}+y^{2}=1 \ldots$ (i) and $(x-1)^{2}+y^{2}=1 \ldots$ (ii)

Centre of (i) is $\mathrm{O}(0,0)$ and radius $=1$

Both these circle are symmetrical about $x$-axis solving (i) and (ii), we get, $-2 x+1=0$

$\Rightarrow x=\frac{1}{2}$

$\begin{array}{l}

\text { then } y^{2}=1-\left(\frac{1}{2}\right)^{2}=34 \\

\Rightarrow y=\frac{\sqrt{3}}{2}

\end{array}$

$\therefore$ The points of intersection are

$\mathrm{P}\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $\mathrm{Q}\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$

It is clear from the figure that the shaded portion in region whose area is required.

$\therefore$ Required area = area OQAPO

$=2 \times$ area of the region OLAP

$=2 \times($ area of the region $O L P O+$ area of $L A P L$ )

$=2\left[\int_{0}^{\frac{1}{2}} \sqrt{1-(x-1)^{2}} d x+\int_{\frac{1}{2}}^{1} \sqrt{1-x^{2}} \mathrm{dx}\right]$

$=2\left[\frac{(x-1) \sqrt{1-(x-1)^{2}}}{2}+\frac{1}{2} \sin ^{-1}(x-1)\right]_{0}^{\frac{1}{2}}+2\left[\frac{x \sqrt{1-x^{2}}}{2}+\frac{1}{2} \sin ^{-1} x\right]_{\frac{1}{2}}^{1}$

$=-\frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\sin ^{-1}\left(\frac{-1}{2}\right)-\sin ^{-1}(-1)+0+\sin ^{-1}(1)-\left(\frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\sin ^{-1}\left(\frac{1}{2}\right)\right)$

$=\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)$ sq. units.

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$ is

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

The correct answer is:
$\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)$

Solution:
Given circles are $x^{2}+y^{2}=1 \ldots$ (i) and $(x-1)^{2}+y^{2}=1 \ldots$ (ii)

Centre of (i) is $\mathrm{O}(0,0)$ and radius $=1$

Both these circle are symmetrical about $x$-axis solving (i) and (ii), we get, $-2 x+1=0$

$\Rightarrow x=\frac{1}{2}$

$\begin{array}{l}

\text { then } y^{2}=1-\left(\frac{1}{2}\right)^{2}=34 \\

\Rightarrow y=\frac{\sqrt{3}}{2}

\end{array}$

$\therefore$ The points of intersection are

$\mathrm{P}\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $\mathrm{Q}\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$

It is clear from the figure that the shaded portion in region whose area is required.

$\therefore$ Required area = area OQAPO

$=2 \times$ area of the region OLAP

$=2 \times($ area of the region $O L P O+$ area of $L A P L$ )

$=2\left[\int_{0}^{\frac{1}{2}} \sqrt{1-(x-1)^{2}} d x+\int_{\frac{1}{2}}^{1} \sqrt{1-x^{2}} \mathrm{dx}\right]$

$=2\left[\frac{(x-1) \sqrt{1-(x-1)^{2}}}{2}+\frac{1}{2} \sin ^{-1}(x-1)\right]_{0}^{\frac{1}{2}}+2\left[\frac{x \sqrt{1-x^{2}}}{2}+\frac{1}{2} \sin ^{-1} x\right]_{\frac{1}{2}}^{1}$

$=-\frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\sin ^{-1}\left(\frac{-1}{2}\right)-\sin ^{-1}(-1)+0+\sin ^{-1}(1)-\left(\frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\sin ^{-1}\left(\frac{1}{2}\right)\right)$

$=\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)$ sq. units.

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2021
The area bounded by the curve $y=\sin x, x$ -axis and the ordinates $x=0$ and $x=\pi / 2$ is

Options:
A $\pi$
B $\pi / 2$
C 1
D 2

The correct answer is:
1

Solution:
Area $=\int_{0}^{\pi / 2} \mathrm{y} \mathrm{dx}=\int_{0}^{\pi / 2} \sin \mathrm{x} \mathrm{dx}=[-\cos \mathrm{x}]_{0}^{\pi / 2}=1$

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2012
The area bounded by the curves $x+2 y^{2}=0$ and $x+3 y^{2}=1$ is

Options:
A 1 sq. unit
B $\frac{1}{3}$ sq. units
C $\frac{2}{3}$ sq. units
D $\frac{4}{3}$ sq. units

The correct answer is:
$\frac{4}{3}$ sq. units

Solution:
We have, $x+2 y^{2}=0$

$\Rightarrow y^{2}=-\frac{x}{2} \ldots \text { (i) }$

a parabola with vertex $(0,0)$ and $x+3 y^{2}=1$

$\Rightarrow y^{2}=\frac{1-x}{3}=-\left(\frac{x-1}{3}\right) \ldots$ (ii), a parabola with vertex $(1,0)$

Solving (i) and (ii), we get $y=\pm 1$

$\mathrm{A}=\int_{-1}^{1}\left(\left(1-3 y^{2}\right)-\left(-2 y^{2}\right)\right) d y$

$=2 \int_{0}^{1}\left(1-y^{2}\right) d y=2\left[y-\frac{y^{3}}{3}\right]_{0}^{1}$

$=\frac{4}{3}$ sq.units

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2021
The area bounded by the $x$ -axis, the curve $y=f(x)$ and the lines $\mathrm{x}=1, \mathrm{x}=\mathrm{b},$ is equal to $\sqrt{\mathrm{b}^{2}+1}-\sqrt{2}$

for all $b>1,$ then $f(x)$ is

Options:
A $\sqrt{x-1}$
B $\sqrt{x+1}$
C $\sqrt{x^{2}+1}$
D $\frac{x}{\sqrt{1+x^{2}}}$

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

Solution:
Given $\int_{1}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\sqrt{\mathrm{b}^{2}+1}-\sqrt{2}$

Differentiate with respect to $\mathrm{b}$

$$

f(b)=\frac{b}{\sqrt{b^{2}+1}} \Rightarrow f(x)=\frac{x}{\sqrt{x^{2}+1}}

$$

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2014
The area enclosed between the curve $y=\log _e(x+e)$ and the coordinate axes is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
Required area $A=\int_{1-e}^0 y d x=\int_{1-e}^0 \log _e(x+e) d x$
put $x+e=t \Rightarrow d x=d t$ also when $x=1-e, t=$ 1 and when $x=0, t=e$
$\therefore A=\int_1^e \log _e t d t=\left[t \log _e t-t\right]_1^e$
$e-e-0+1=1$

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The area enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ over the interval $\left[0, \frac{\pi}{2}\right]$ is

Options:
A
$4(\sqrt{2}-1)$

B
$2 \sqrt{2}(\sqrt{2}-1)$

C
$2(\sqrt{2}+1)$

D
$2 \sqrt{2}(\sqrt{2}+1)$

The correct answer is:

$2 \sqrt{2}(\sqrt{2}-1)$

Solution:
Given, $y=\sin x+\cos x, x \in[0, \pi / 2]$
$\frac{d y}{d x}=\cos x-\sin x$
$\begin{aligned} y & =|\cos x-\sin x| \\ & =\left[\begin{array}{cc}\cos x-\sin x & x \in[0, \pi / 4] \\ \sin x-\cos x & x \in[\pi / 4, \pi / 2]\end{array}\right]\end{aligned}$

$\therefore$ Required area
$\begin{aligned}
&=\int_0^{\pi / 4}|(\sin x+\cos x)-(\cos x-\sin x)| d x+\int_{\pi / 4}^{\pi / 2}|2 \cos x| d x
\end{aligned}$
$=\int_0^{\pi / 4}|2 \sin x| d x+\int_{\pi / 4}^{\pi / 2}|2 \cos x| d x$
$=2(-\cos x)_0^{\pi / 4}+2(\sin x)_{\pi / 4}^{\pi / 2}$
$=2\left[-\frac{1}{\sqrt{2}}+1+1-\frac{1}{\sqrt{2}}\right]$

$\begin{aligned}= & 2\left(2-\frac{2}{\sqrt{2}}\right) \\ = & 2(2-\sqrt{2}) \\ = & 4-2 \sqrt{2} \\ = & 2 \sqrt{2}(\sqrt{2}-1)\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ over the interval $\left[0, \frac{\pi}{2}\right]$ is

Options:
A
$4(\sqrt{2}-1)$

B
$2 \sqrt{2}(\sqrt{2}-1)$

C
$2(\sqrt{2}+1)$

D
$2 \sqrt{2}(\sqrt{2}+1)$

The correct answer is:

$2 \sqrt{2}(\sqrt{2}-1)$

Solution:
Given, $y=\sin x+\cos x, x \in[0, \pi / 2]$
$\frac{d y}{d x}=\cos x-\sin x$
$\begin{aligned} y & =|\cos x-\sin x| \\ & =\left[\begin{array}{cc}\cos x-\sin x & x \in[0, \pi / 4] \\ \sin x-\cos x & x \in[\pi / 4, \pi / 2]\end{array}\right]\end{aligned}$

$\therefore$ Required area
$\begin{aligned}
&=\int_0^{\pi / 4}|(\sin x+\cos x)-(\cos x-\sin x)| d x+\int_{\pi / 4}^{\pi / 2}|2 \cos x| d x
\end{aligned}$
$=\int_0^{\pi / 4}|2 \sin x| d x+\int_{\pi / 4}^{\pi / 2}|2 \cos x| d x$
$=2(-\cos x)_0^{\pi / 4}+2(\sin x)_{\pi / 4}^{\pi / 2}$
$=2\left[-\frac{1}{\sqrt{2}}+1+1-\frac{1}{\sqrt{2}}\right]$

$\begin{aligned}= & 2\left(2-\frac{2}{\sqrt{2}}\right) \\ = & 2(2-\sqrt{2}) \\ = & 4-2 \sqrt{2} \\ = & 2 \sqrt{2}(\sqrt{2}-1)\end{aligned}$

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2022
The area enclosed by the curves y=x3 and y=x is

Options:
A 53 sq units
B 54 sq units
C 512 sq units
D 125 sq units

The correct answer is:
512 sq units

Solution:
Since, the intersection of two curves

y=x3 and y=x are x=0 and x=1

∴ A=∫01x3-xdx=x44-2x3/2301

=14-23=512 sq units

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2017
The area of the region bounded by the curve \(y=x|x|, x\)-axis and the ordinates \(x=1, x=-1\) is given by :

Options:
A zero
B \(\frac{1}{3}\)
C \(\frac{2}{3}\)
D 1

The correct answer is:
\(\frac{2}{3}\)

Solution:
The area of the region bounded by the curve \(y=f(x)\) and the ordinates \(x=a, x=b\) is given by
\(\text {Area }=\left|\int_a^b y d x\right|\)
According to the question,
\(y=x|x|= \begin{cases}x^2, & x \geq 0 \\ -x^2, & x < 0\end{cases}\)
Required area
\(=\) area of region \(\mathrm{OAB}+\) area of region \(\mathrm{OCD}\)
\(=2 \times\) Area of region \(\mathrm{OAB}\)
\(=2 \int_0^1 x^2 d x=\frac{2}{3} s q\). units

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2010
The area of the region bounded by the parabola $(y-2)^2=(x-1)$, the tangent to the parabola at the point $(2,3)$ and the $X$-axis is

Options:
A
3

B
6

C
9

D
12

The correct answer is:

9

Solution:
The equation of the parabola is $y^2-4 y-x+5=0$ The equation of tangent at $(2,3)$ is
$$
\begin{aligned}
3 y-2(y+3)-\frac{x+2}{2}+5 & =0 \\
2 y-x-4 & =0
\end{aligned}
$$

$\therefore$ Required area $A$ is given by
$$
\begin{aligned}
& A=\int_0^3\left(x_2-x_1\right) d y \\
& \Rightarrow \quad A=\int_0^3\left[\left\{(y-2)^2+1\right\}-\{2 y-4\}\right] d y \\
& \Rightarrow \quad A=\int_0^3\left(y^2-6 y+9\right) d y=\int_0^3(3-y)^2 d y \\
& =-\left[\frac{(3-y)^3}{3}\right]_0^3=9 \\
&
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2022
The area of the region included between the curves x2+y2=a2 and x+y=aa>0 is

Options:
A π-23a2 sq units
B 23-πa2 sq units
C 23πa2 sq units
D π+23a2 sq units

The correct answer is:
π-23a2 sq units

Solution:
Given curves are

x+y=a ...i

and x2+y2=a2 ...ii

Now, required areas =4[shaded area in the first quadrant]

=4πa24-∫0aa-x2dx [from Eqs. (i) and (ii)]

=4πa24-∫0aa+x-2axdx

=4πa24-4ax+x22-43ax3/20a

=π-23a2 sq units

Hence, option (a) is correct.

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2018
The area of the region
$\mathbf{R}=\left\{(x, y): 5 x^2 \leq y \leq 2 x^2+9\right\} \text { is : }$

Options:
A $11 \sqrt{3}$ square units
B $12 \sqrt{3}$ square units
C $9 \sqrt{3}$ square units
D $6 \sqrt{3}$ square units

The correct answer is:
$12 \sqrt{3}$ square units

Solution:
$\begin{aligned} & \text { Required area }=2 \int_0^{\sqrt{3}}\left(2 x^2+9-5 x^2\right) d x \\ & =2\left[9 x-x^3\right]_0^{\sqrt{3}}=2[9 \sqrt{3}-3 \sqrt{3}]=12 \sqrt{3} .\end{aligned}$

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The area of the region $\mathrm{R}=\left\{(x, y):|x| \leq|y|\right.$ and $x^{2}$ $\left.+y^{2} \leq 1\right\}$ is

Options:
A $\frac{3 \pi}{8}$ sq units
B $\frac{5 \pi}{8} \mathrm{sq}$ units
C $\frac{\pi}{2}$ sq units
D $\frac{\pi}{8}$ sq unit

The correct answer is:
$\frac{\pi}{2}$ sq units

Solution:
Required area $=4$ (Area of the shaded region in first quadrant)

$$

=4 \int_{0}^{1 \sqrt{2}}\left(y_{1}-y_{2}\right) d x=4 \int_{0}^{1 \sqrt{2}}\left(\sqrt{1-x^{2}}-x\right) d x

$$

$=4\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x-\frac{x^{2}}{2}\right]_{0}^{1 / \sqrt{2}}$

$=4\left[\frac{1}{2 \sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{\pi}{4}-\frac{1}{4}\right]$

$=4\left[\frac{1}{4}+\frac{\pi}{8}-\frac{1}{4}\right]=\frac{4 \pi}{8}=\frac{\pi}{2}$ sq units

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region $\mathrm{R}=\left\{(x, y):|x| \leq|y|\right.$ and $x^{2}$ $\left.+y^{2} \leq 1\right\}$ is

Options:
A $\frac{3 \pi}{8}$ sq units
B $\frac{5 \pi}{8} \mathrm{sq}$ units
C $\frac{\pi}{2}$ sq units
D $\frac{\pi}{8}$ sq unit

The correct answer is:
$\frac{\pi}{2}$ sq units

Solution:
Required area $=4$ (Area of the shaded region in first quadrant)

$$

=4 \int_{0}^{1 \sqrt{2}}\left(y_{1}-y_{2}\right) d x=4 \int_{0}^{1 \sqrt{2}}\left(\sqrt{1-x^{2}}-x\right) d x

$$

$=4\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x-\frac{x^{2}}{2}\right]_{0}^{1 / \sqrt{2}}$

$=4\left[\frac{1}{2 \sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{\pi}{4}-\frac{1}{4}\right]$

$=4\left[\frac{1}{4}+\frac{\pi}{8}-\frac{1}{4}\right]=\frac{4 \pi}{8}=\frac{\pi}{2}$ sq units

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2015
The area of the triangle formed by joining the origin to the point of intersection of the line x5+2y=35 and circle x2+y2=10 is

Options:
A 3
B 4
C 5
D 6

The correct answer is:
5

Solution:
Length of perpendicular from origin to the line x5+2y=35 is

OL=3552+22=359=5

Radius of the given circle =10=OQ=OP

PQ=2QL

=2OQ2-OL2

=210-5=25

Thus, area of ΔOPQ=12×PQ×OL

=12×25×5=5 sq units

Hence, option (c) is correct.

Topic: Mathematics, Circle, BITSAT, BITSAT 2018
The area under the curve y=cosx-sinx,0≤x≤π2 and above X-axis, is

Options:
A 22
B 22-2
C 22+2
D 0

The correct answer is:
22-2

Solution:
Required area =∫0π/2cosx-sinxdx

=∫0π/4cosx-sinxdx+∫π/4π/2sinx-cosxdx

=sinx+cosx0π/4+-cosx-sinxπ/4π/2

=2-1-1-2=22-2

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2017
The arithmetic mean of a set of observation is X¯. If each observation is divided by α and increased by 10, then the mean of the new series is

Options:
A X¯α
B X¯+10α
C X¯+10αα
D αX¯+10

The correct answer is:
X¯+10αα

Solution:
Let x1, x2,…,xn be n observations.

Then, X¯=1nΣxi

Let yi=xiα+10

Then. 1n∑i=1nyi=1α1nΣxi+1n10n

⇒X¯new=1αX¯+10=X¯+10αα

Hence option (c) is correct.

Topic: Mathematics, Statistics, BITSAT, BITSAT 2018
The arithmetic mean of numbers a, $\mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}$ is $\mathrm{M}$. What is the value of $(a-M)+(b-M)+(c-M)+$ $(\mathrm{d}-\mathrm{M})+(\mathrm{e}-\mathrm{M}) ?$

Options:
A $\mathrm{M}$
B $a+b+c+d+e$
C 0
D $5 \mathrm{M}$

The correct answer is:
0

Solution:
Given $\mathrm{M}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}}{5}$

$\Rightarrow a+b+c+d+e=5 M$

$\Rightarrow a+b+c+d+e-5 m=0$

$$

\begin{array}{l}

\Rightarrow(a-M)+(b-M)+(c-M)+(d-M) \\

+(e-M)=0

\end{array}

$$

Hence, required value $=0$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2016
The arithmetic mean of the data $0,1,2, \ldots \ldots, n$ with frequencies $1,{ }^{4} \mathrm{C}_{1}{ }^{4} \mathrm{C}_{2}, \ldots . .,{ }^{\infty} \mathrm{C}_{\mathrm{a}}$ is

Options:
A $\mathrm{n}$
B $\frac{2^{n}}{n}$
C $\mathrm{n}+1$
D $\frac{\mathrm{n}}{2}$

The correct answer is:
$\frac{\mathrm{n}}{2}$

Solution:
Since, Mean $=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$ where $\mathrm{x}_{\mathrm{i}}$ are

observations with frequencies $\mathrm{f}_{\mathrm{i}}, \mathrm{i}=1,2,$ $\ldots \ldots \ldots$

The required mean is given by

$$

\begin{aligned}

\overline{\mathrm{X}} &=\frac{0.1+1 .{ }^{\mathrm{n}} \mathrm{C}_{1}+2{ }^{\mathrm{n}} \mathrm{C}_{2}+\ldots \ldots+\mathrm{n} .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}}{1+{ }^{\mathrm{n}} \mathrm{C}_{1}+{ }^{\mathrm{n}} \mathrm{C}_{2}+\ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}} \\

=& \frac{\sum_{\mathrm{r}=0}^{\mathrm{n}} \mathrm{r}^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r} \cdot \frac{\mathrm{n}}{\mathrm{r}}{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}}{\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}} \\

=& \frac{\mathrm{n} \sum{\sum}^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}}{\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{\mathrm{n} .2^{\mathrm{n}-1}}{2^{\mathrm{n}}}=\frac{\mathrm{n}}{2}

\end{aligned}

$$

Topic: Mathematics, Statistics, BITSAT, BITSAT 2015
The atomic radius of atom is r. Total volume of atoms present in a $f c c$ unit cell of an element is

Options:
A $\frac{24}{3} \pi r^{3}$
B $\frac{12}{3} \pi r^{3}$
C $\frac{16}{3} \pi r^{3}$
D None

The correct answer is:
$\frac{16}{3} \pi r^{3}$

Solution:
4 atom are present in $f c c$.
$$
\mathrm{So}, \mathrm{V}=4\left[\frac{4}{3} \pi \mathrm{r}^{3}\right]=\frac{16}{3} \pi \mathrm{r}^{3}
$$

Topic: Chemistry, Solid State, BITSAT, BITSAT 2009
The average kinetic energy of an ideal gas per molecule in SI unit at $25^{\circ} \mathrm{C}$ will be

Options:
A $6.17 \times 10^{-21} \mathrm{~kJ}$
B $6.17 \times 10^{-21} \mathrm{~J}$
C $6.17 \times 10^{-20} \mathrm{~J}$
D $7.16 \times 10^{-20} \mathrm{~J}$

The correct answer is:
$6.17 \times 10^{-21} \mathrm{~J}$

Solution:
$\overline{\mathrm{KE}}=\frac{3}{2} \mathrm{kT}=\frac{3}{2} \times \frac{8.313}{6.023 \times 10^{23}} \times 298=6.17 \times 10^{-21} \mathrm{~J}$

(Average Kinetic energy $\left.\overline{\mathrm{KE}}=\frac{3}{2} \mathrm{k} \mathrm{T}=\frac{3}{2} \frac{\mathrm{R}}{\mathrm{N}} \mathrm{T}\right)$

Topic: Chemistry, States of Matter, BITSAT, BITSAT 2016
The average translational kinetic energy of $\mathrm{O}_{2}$ (molar mass 32) molecules at a particular temperature is $0.048 \mathrm{eV}$. The translational kinetic energy of $\mathrm{N}_{2}$ (molar mass 28 ) molecules in $\mathrm{eV}$ at the same temperature is

Options:
A 0.0015
B 0.003
C 0.048
D 0.768

The correct answer is:
0.048

Solution:
Given,
\(\mathrm{E}_{\mathrm{O}_{2}}=0.048 \mathrm{eV}\)
We know that,
Translational kinetic energy,
\(\mathrm{E}=\frac{3}{2} \mathrm{KT}\)
where, \(\mathrm{k}=\) Boltzmann constant
\(\mathrm{T}=\) Temperature
\(\therefore \mathrm{E} \propto \mathrm{T}\)
Since, the temperature is same for both oxygen and nitrogen
therefore, \(\mathrm{E}_{\mathrm{O}_{2}}=\mathrm{E}_{\mathrm{N}_{2}}\)
\(\therefore \mathrm{E}_{\mathrm{N}_{2}}=0.048 \mathrm{eV}\)

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2012
The back emf induced in a coil, when current changes from 1 ampere to zero in one milli-second, is 4 volts, the self inductance of the coil is

Options:
A 1 henry
B 4 henry
C $10^{-3}$ henry
D $4 \times 10^{-3}$ henry

The correct answer is:
$4 \times 10^{-3}$ henry

Solution:
$\mathrm{e}=-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$
But $\mathrm{e}=4 \mathrm{~V}$ and $\frac{\mathrm{di}}{\mathrm{dt}}=\frac{0-1}{10^{-3}}=-1 / 10^{-3}$
$\therefore \frac{-1}{10^{-3}}(-\mathrm{L})=4 \Rightarrow \mathrm{L}=4 \times 10^{-3}$ henry

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2010
The beam of light has three wavelengths $4144 Ã, 4972 Ã$ and $6216 Ã$ with a total intensity of $3.6 \times 10^{-3} \mathrm{Wm}^{2}$ equally distributed amongst the three wavelengths. The beam falls normally on the area $1 \mathrm{~cm}^{2}$ of a clean metallic surface of work function $2.3 \mathrm{eV}$. Assume that there is $\mathrm{n} 0$ loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $2 \mathrm{~s}$

Options:
A $2 \times 10^{9}$
B $1.075 \times 10^{12}$
C $9 \times 10^{8}$
D $3.75 \times 10^{6}$

The correct answer is:
$1.075 \times 10^{12}$

Solution:
As we know, threshold wavelength

$$

\begin{array}{c}

\left(\lambda_{0}\right)=\frac{\mathrm{hc}}{\phi} \\

\Rightarrow \lambda_{0}=\frac{\left(6.63 \times 10^{-34}\right) \times 3 \times 10^{8}}{2.3 \times\left(1.6 \times 10^{-19}\right)}=5.404 \times 10^{-7} \mathrm{~m}

\end{array}

$$

$\Rightarrow \lambda_{0}=5404 Ã…$

Hence, wavelength $4144 Ã…$ and $4972 Ã…$ will emit electron from the metal surface. For each wavelength energy incident on the surface per unit time = intensity of each $\times$ area of the surface wavelength

$$

=\frac{3.6 \times 10^{-3}}{3} \times(\mathrm{lcm})^{2}=1.2 \times 10^{-7} \text {joule }

$$

Therefore, energy incident on the surface for each wavelength in $2 \mathrm{~s}$

$$

\mathrm{E}=\left(1.2 \times 10^{-7}\right) \times 2=2.4 \times 10^{-7} \mathrm{~J}

$$

Number of photons $\mathrm{n}_{1}$ due to wavelength $4144 Ã…$

$$

n_{1}=\frac{\left(2.4 \times 10^{-7}\right)\left(4144 \times 10^{-10}\right)}{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}=0.5 \times 10^{12}

$$

Number of photons $\mathrm{n}_{2}$ due to the wavelength $4972 Ã…$

$$

\mathrm{n}_{2}=\frac{\left(2.4 \times 10^{-7}\right)\left(4972 \times 10^{-10}\right)}{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}=0.572 \times 10^{12}

$$

Therefore total number of photoelectrons liberated in $2 \mathrm{~s}$

$$

\begin{aligned}

\mathrm{N} &=\mathrm{n}_{1}+\mathrm{n}_{2} \\

&=0.5 \times 10^{12}+0.575 \times 10^{12} \\

&=1.075 \times 10^{12}

\end{aligned}

$$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2015
The bob of simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob get suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

Options:
A first increase and then decrease to the original value
B first decrease and then increase to the original value
C remain unchanged
D increase towards a saturation value

The correct answer is:
first increase and then decrease to the original value

Solution:
The bob filled completely with water has its centre of mass at its centre. The time period of oscillation is

T=2πlg

As the water starts coming out of the bob, its centre of mass shifts vertically downward and as a result, effective length of the pendulum increases and hence its period also increases. When the bob is empty, again its centre of mass appears at its centre and as a result, the period of oscillation again reaches to its original value.

Topic: Physics, Oscillations, BITSAT, BITSAT 2018
The Bohr orbit radius for the hydrogen atom $(n=1)$ is approximately $0.530 Å$. The radius for the first excited state $(n=2)$ orbit is (in $Å$ )

Options:
A 0.13
B 1.06
C 4.77
D 2.12

The correct answer is:
2.12

Solution:
Given : Radius of hydrogen atom $=0.530$
$Å$, Number of excited state $(n)=2$ and atomic number of hydrogen atom $(Z)=1$. We know that the Bohr radius
$(r)=\frac{n^2}{Z} \times \text { radius of atom }=\frac{(2)^2}{1} \times 0.530$
$=4 \times 0.530=2.12 Å$

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The Bohr orbit radius for the hydrogen atom $(n=1)$ is approximately $0.530 Ã$. The radius for the first excited state $(n=2)$ orbit is $(\operatorname{in} Ã)$

Options:
A 0.13
B 1.06
C 4.77
D 2.12

The correct answer is:
2.12

Solution:
Radius of hydrogen atom $=0.530 Ã…$, Number of excited state $(n)=2$ and atomic number of hydrogen atom $(Z)=1$. We know that the Bohr radius.

$$

(r)=\frac{n^{2}}{Z} \times \text { Radius of atom }=\frac{(2)^{2}}{1} \times 0.530

$$

$=4 \times 0.530=2.12 Ã…$

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2013