datavorous/entrance-exam-dataset huggingface dataset:
The slope of the tangent to the curve $x=3 t^2+1$, $y=t^3-1$ at $x=1$ is:

Options:
A $\frac{1}{2}$
B
0
C
-2
D $\infty$

The correct answer is:
0

Solution:
Given curve is $x=3 t^2+1$
Second curve is $y=t^3-1$
$\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=3 t^2 \times \frac{1}{6 t}=\frac{t}{2}$
But from (i) when $x=1$
we have $1=3 t^2+1 \Rightarrow 3 t^2=0 \Rightarrow t=0$
$\therefore$ When $\mathrm{x}=1$ then $\mathrm{t}=0 \quad \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathbf{0}$
Hence, slope of the tangent to the curve $=0$

Topic: Mathematics, Differentiation, JEE Main
The slope of the tangent to the curve $x=3 t^2+1$, $y=t^3-1$ at $x=1$ is:

Options:
A $\frac{1}{2}$
B
0
C
-2
D $\infty$

The correct answer is:
0

Solution:
Given curve is $x=3 t^2+1$
Second curve is $y=t^3-1$
$\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=3 t^2 \times \frac{1}{6 t}=\frac{t}{2}$
But from (i) when $x=1$
we have $1=3 t^2+1 \Rightarrow 3 t^2=0 \Rightarrow t=0$
$\therefore$ When $\mathrm{x}=1$ then $\mathrm{t}=0 \quad \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathbf{0}$
Hence, slope of the tangent to the curve $=0$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The slope of the tangent to the curve $y=e^{x}$ cos $x$ is minimum at $x=\alpha, 0 \leq a \leq 2 \pi$, then the value of $\alpha$ is

Options:
A $\underline{0}$
B $\pi$
C $2 \pi$
D $3 \pi 2$

The correct answer is:
$\pi$

Solution:
Let $m$ be the slope of the tangent to the curve

$y=e^{x} \cos x$

Then, $m=\frac{d y}{d x}=e^{x}(\cos x-\sin x)$

Diff. w.r.t 'x'

$$

\begin{array}{r}

\Rightarrow \frac{\mathrm{dm}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}(\cos \mathrm{x}-\sin \mathrm{x})+\mathrm{e}^{\mathrm{x}}(-\sin \mathrm{x}-\cos \mathrm{x}) \\

=-2 \mathrm{e}^{\mathrm{x}} \sin \mathrm{x}

\end{array}

$$

and $\frac{\mathrm{d}^{2} \mathrm{~m}}{\mathrm{dx}^{2}}=-2 \mathrm{e}^{\mathrm{x}}(\sin \mathrm{x}+\cos \mathrm{x})$

Put $\frac{\mathrm{dm}}{\mathrm{dx}}=0 \Rightarrow \sin \mathrm{x}=0 \Rightarrow \mathrm{x}=0, \pi, 2 \pi$

Clearly, $\frac{\mathrm{d}^{2} \mathrm{~m}}{\mathrm{dx}^{2}}>0$ for $\mathrm{x}=\pi$

Thus, $y$ is minimum at $x=\pi$.

Hence the value of $\alpha=\pi$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2015
The slope of the tangent to the hyperbola \(2 x^2-3 y^2=6\) at \((3,2)\) is

Options:
A -1
B 1
C 0
D 2

The correct answer is:
1

Solution:
Differentiating the given equation of the curve
\(\begin{aligned}
& 4 x-6 y \cdot(\mathrm{dy} / \mathrm{dx})=0 \quad \therefore \mathrm{dy} / \mathrm{dx}=2 \mathrm{x} / 3 \mathrm{y} \\
& \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(3,2)}=\frac{2}{3} \cdot \frac{3}{2}=1
\end{aligned}\)

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2010
The smallest positive integral value of $n$ such that
$\left[\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right]^n$ is purely imaginary, is equal to

Options:
A
$4$

B
$3$

C
$2$

D
$8$

The correct answer is:

$4$

Solution:
$\left[\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right]^n=\left[\frac{1+\cos \alpha+i \sin \alpha}{1+\cos \alpha-i \sin \alpha}\right]^n$
$\left[\because \alpha=\frac{\pi}{2}-\frac{\pi}{8}\right]$
$=\left[\frac{2 \cos ^2 \frac{\alpha}{2}+2 i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \cos ^2 \frac{\alpha}{2}-2 i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}\right]^n$
$=\left[\frac{\cos \frac{\alpha}{2}+i \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}-i \sin \frac{\alpha}{2}}\right]_{-}^n=\left(e^{2 i \frac{\alpha}{2}}\right]^n=e^{i n \alpha}$
$=e^{i n\left(\frac{3 \pi}{8}\right)}=\cos \frac{3 n \pi}{8}+i \sin \frac{3 n \pi}{8}$
For $n=4$ we get imaginary part.

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2022
The solubility of silver bromide in hypo solution involves the formation of:

Options:
A $\mathrm{Ag}_{2} \mathrm{SO}_{3}$
B $\mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$
C $\mathrm{Ag}^{+}$
D $\mathrm{NH}_{4}{ }^{+}$

The correct answer is:
$\mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$

Solution:
Hypo solution is $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ solution which is used in photography for fixing films \& prints. Photographic emulsions are made of AgBr. After developing, the film is put into hypo solution. This forms soluble complex with $\mathrm{Ag}$.

$\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}+\mathrm{Ag} \mathrm{Br} \rightarrow \mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} }{\mathrm{Na}_{5}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{3}\right]$

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2021
The solution of $(2 \cos x-1)(3+2 \cos x)=0$ in the interval $0 \leq x \leq 2 \pi$ is

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{3}, \frac{5 \pi}{3}$
C $\frac{\pi}{3}, \frac{5 \pi}{3}, \cos ^{-1}\left(-\frac{3}{2}\right)$
D None of these

The correct answer is:
$\frac{\pi}{3}, \frac{5 \pi}{3}$

Solution:
We have $(2 \cos x-1)(3+2 \cos x)=0$

If $2 \cos x-1=0,$ then $\cos x=\frac{1}{2}$

$\therefore x=\pi / 3,5 \pi / 3$

If $3+2 \cos x=0,$ the $\cos x=-3 / 2$

which is not possible.

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2014
The solution of dydx=cosx+y+sinx+y, is

Options:
A log1+tanx+y2+C=0
B log1+tanx+y2=x+C
C log1-tanx+y2=x+C
D None of the above

The correct answer is:
log1+tanx+y2=x+C

Solution:
Put x+y=v and 1+dydx=dvdx

Therefore, the differential equation reduces to

dvdx=1+cosv+sinv=2cos2v2+2sinv2cosv2

=2cos2v21+tanv2

⇒ ∫sec2V221+tanv2dv=∫dx

∴ log1+tanx+y2=x+C

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2017
The solution of differential equation \(2 x \frac{d y}{d x}-y=3\) represents a family of

Options:
A circles
B straight lines
C ellipses
D parabola

The correct answer is:
parabola

Solution:
We have, \(2 x \frac{d y}{d x}=y+3 \Rightarrow \frac{2}{y+3} d y=\frac{d x}{x}\) integrating, \(2 \ln (y+3)=\ln x+\ln c=\ln c x\)
\(\Rightarrow \ln (y+3)^2=\ln \mathrm{cx} \Rightarrow(y+3)^2=\mathrm{cx}\)
which is a family of parabolas.

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2010
The solution of the differential equation d2ydx2=sin3x+ex+x2 when y10=1 and y0=0, is

Options:
A -sin3x9+ex+x412+13x-1
B -sin3x9+ex+x412+13x
C -cos3x9+ex+x412+13x+1
D None of the above

The correct answer is:
-sin3x9+ex+x412+13x-1

Solution:
Integrating the given differential equation, we have

dydx=-cos3x3+ex+x33+C1

But y10=1

So, 1=-13+1+C1

⇒C1=13

∴ dydx=-cos3x3+ex+x33+13

Again integrating, we get

y=-sin3x9+ex+x412+13x+C2

But y0=0, so 0=0+1+C2

⇒C2=-1

Thus, y=-sin3x9+ex+x412+13x-1

Hence, option (a) correct.

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2018
The solution of the differential equation

$(x+1) \frac{d y}{d x}-y=e^{3 x}(x+1)^{2}$ is

Options:
A $y=(x+1) e^{3 x}+c$
B $3 y=(x+1)+e^{3 x}+c$
C $\frac{3 y}{x+1}=e^{3 x}+c$
D $y e^{-3 x}=3(x+1)+c$

The correct answer is:
$\frac{3 y}{x+1}=e^{3 x}+c$

Solution:
The given equation is $\frac{d y}{d x}-\frac{y}{x+1}=e^{3 x}(x+1)$

I.F. $=e^{\int-\frac{1}{x+1} d x}=e^{-\log (x+1)}=\frac{1}{x+1}$

The solution is

$y\left(\frac{1}{x+1}\right)=\int e^{3 x}(x+1) \cdot \frac{1}{x+1} d x+a$

$\Rightarrow \frac{y}{x+1}=\int e^{3 x} d x+a=\frac{e^{3 x}}{3}+a$

$\Rightarrow \frac{3 y}{x+1}=e^{3 x}+c, c=3 a$

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2020
The solution set of $(x-2)^{x^3-6 x+8}>1$ is

Options:
A $(2, \infty)$
B $(2,3) \cup(4, \infty)$
C $(4,5) \cup(5, \infty)$
D $(2,3) \cup(4,5)$

The correct answer is:
$(2,3) \cup(4, \infty)$

Solution:
Clearly $x>2$. Write the given inequality
as
$\begin{aligned}
& \left(x^2-6 x+8\right) \log (x-2)>0 \Leftrightarrow(x-2)(x-4) \log (x-2)> \\
& 0 \\
& \Leftrightarrow(x-4) \log (x-2)>0 \quad[\therefore x>2]
\end{aligned}$

Topic: Mathematics, Basic of Mathematics, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The solution to the differential equation $\frac{d y}{d x}=\frac{y f^{\prime}(x)-y^{2}}{f(x)}$ where $f(x)$ is a given function is

Options:
A $f(x)=y(x+c)$
B $f(x)=c x y$
C $f(x)=c(x+y)$
D $\mathrm{yf}(\mathrm{x})=\mathrm{cx}$

The correct answer is:
$f(x)=y(x+c)$

Solution:
We have $\frac{d y}{d x}=\frac{f^{\prime}(x)}{f(x)} y-\frac{y^{2}}{f(x)}$

$\Rightarrow \frac{d y}{d x}-\frac{f^{\prime}(x)}{f(x)} y=-\frac{y^{2}}{f(x)}$

Divide by $y^{2}: y^{-2} \frac{d y}{d x}-y^{-1} \frac{f^{\prime}(x)}{f(x)}=-\frac{1}{f(x)}$

Put $y^{-1}=z \Rightarrow-y^{-2} \frac{d y}{d x}=\frac{d z}{d x}$

$-\frac{d z}{d x}-\frac{f^{\prime}(x)}{f(x)}(z)=-\frac{1}{f(x)}$

$\Rightarrow \frac{d z}{d x}+\frac{f^{\prime}(x)}{f(x)} z=\frac{1}{f(x)}$

$I F=e^{\int \frac{f^{\prime}(x)}{f(x)} d x}=e^{\log f(x)}=f(x)$

$\therefore$ The solution is $z(f(x))=\int \frac{1}{f(x)}(f(x)) d x+c$

$\Rightarrow y^{-1}(f(x))=x+c \Rightarrow f(x)=y(x+c)$

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2013
The solutions of $(x+y+1) d y=d x$ are

Options:
A $x+y+2=C e^{y}$
B $x+y+4=C \log y$
C $\quad \log (x+y+2)=C y$
D $\log (x+y+2)=C-y$

The correct answer is:
$x+y+2=C e^{y}$

Solution:
Putting $x+y+1=u,$ we have $d u=d x+d y$ and the given equation reduces to $u(d u-d x)=$ $d x$

$\Rightarrow \quad \frac{u d u}{u+1}=d x \quad \Rightarrow \quad u-\log (u+1)=x$

$\Rightarrow \log (x+y+2)=y+$ constant

$\Rightarrow x+y+2=C e^{y}$

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2020
The standard deviation of 5 scores $1,2,3,4,5$ is $\sqrt{a}$. The value of ' $a$ ' is

Options:
A
2
B
3
C
5
D
1

The correct answer is:
2

Solution:
$\begin{aligned} & \text { Mean }(\bar{x})=\frac{1+2+3+4+5}{5}=3 \\ & \text { S.D }=\sigma=\sqrt{\frac{1}{5}(1+4+9+16+25)-9}=\sqrt{11-9} \\ & =\sqrt{2}\end{aligned}$

Topic: Mathematics, Statistics, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The starting material for the manufacture of $\mathrm{KMnO}_4$ is

Options:
A
pyrolusite
B
manganite
C
magnatite
D
haematite

The correct answer is:
pyrolusite

Solution:
Pyrolusite (It is $\mathrm{MnO}_2$ )

Topic: Chemistry, General Principles and Processes of Isolation of Metals, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The statement $p:$ For any real numbers $x, y$ if $x=y$, then $2 x+a=2 y+a$ when $a \in \mathrm{Z}$.

Options:
A is true
B is false
C its contrapositive is not true
D None of these

The correct answer is:
is true

Solution:
We prove the statement $p$ is true by contrapositive method and by direct method.

Direct method: For any real number $x$ and $y$,

$\begin{array}{l}

x=y \\

\Rightarrow 2 x=2 y \\

\Rightarrow 2 x+a=2 y+a \text { a for some } a \in Z

\end{array}$

Contrapositive method: The contrapositive statement of $p$ is "For any real numbers $x, y$ if $2 \mathrm{x}+\mathrm{a} \neq 2 \mathrm{y}+\mathrm{a}$, where $\mathrm{a} \in \mathrm{Z}$, then $\mathrm{x} \neq \mathrm{y}$."

Given, $2 x+a \neq 2 y+a$

$\begin{array}{l}

\Rightarrow 2 x \neq 2 y \\

\Rightarrow x \neq y

\end{array}$

Hence, the given statement is true.

Topic: Mathematics, Mathematical Reasoning, BITSAT, BITSAT 2021
The statement which is not correct about control of particulate pollution is:

Options:
A In electrostatic precipitator, the particulates are made to acquire positive charge which are then attracted by the negative electrode and removed.
B Gravity settling chamber removes larger particles from the air.
C Cyclone collector removes fine particls in the diameter range $5-20$ microns.
D Wet scrubbers are used to wash away all types of particulates.

The correct answer is:
In electrostatic precipitator, the particulates are made to acquire positive charge which are then attracted by the negative electrode and removed.

Solution:
Particulates acquire negative charge and are attracted by the positive electrode.

Topic: Chemistry, Environmental Chemistry, BITSAT, BITSAT 2009
The stopping potential $\left(V_0\right)$ versus frequency $(v)$ of a graph for photoelectric effect in a metal. From the graph, the planck's constant $(h)$ is.

Options:
A
$6.60 \times 10^{-34} \mathrm{~J}-\mathrm{s}$

B
$6.69 \times 10^{-34} \mathrm{~J}-\mathrm{s}$

C
$6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}$

D
$6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}$

The correct answer is:

$6.60 \times 10^{-34} \mathrm{~J}-\mathrm{s}$

Solution:
$$
\begin{aligned}
e V_0=h\left(v-v_0\right) & =h v-h v_0 \\
V_0 & =\left(\frac{h}{e}\right) v-\left(\frac{h}{e}\right) v_0 \quad \quad \ldots (i)
\end{aligned}
$$
From Eq. (i), it follows that the slope of the graph is $\frac{h}{e}$. Therefore,
$$
\begin{aligned}
h=e \times \text { slope } & =1.6 \times 10^{-19} \times \frac{165-0}{(8-4) \times 10^{15}} \\
& =\frac{1.6 \times 10^{-19} \times 16.5}{4 \times 10^{15}}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s} .
\end{aligned}
$$

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2022
The structural feature which distinguishes proline from natural $\alpha$ -amino acids?

Options:
A Proline is optically inactive.
B Proline contains aromatic group.
C Proline is a dicarboxylic acid.
D Proline is a secondary amine.

The correct answer is:
Proline is a secondary amine.

Solution:

Topic: Chemistry, Biomolecules, JEE Main
The structural feature which distinguishes proline from natural $\alpha$ -amino acids?

Options:
A Proline is optically inactive.
B Proline contains aromatic group.
C Proline is a dicarboxylic acid.
D Proline is a secondary amine.

The correct answer is:
Proline is a secondary amine.

Solution:

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2011
The structure of $\mathrm{H}_2 \mathrm{O}_2$ is

Options:
A
planar, linear

B
non-planar, linear

C
planar, non-linear

D
non-planar, non-linear

The correct answer is:

non-planar, non-linear

Solution:
Hydrogen peroxide $\left(\mathrm{H}_2 \mathrm{O}_2\right)$ is a non-planar molecule with twisted symmetrical structure. It has non-linear structure with an open book structure.

Topic: Chemistry, Hydrogen, BITSAT, BITSAT 2022
The sum of all odd numbers between 1 and 1000 which are divisible by 3 is

Options:
A 83667
B 90000
C 83660
D None of these

The correct answer is:
83667

Solution:
Sum of odd numbers between 1 and 1000 , which is divisible by $3=3+9+15+21+27+\ldots+999=\mathrm{S}$ (let)

$\therefore$ Let $\mathrm{n}$ be the number of terms in series and $\mathrm{a}$ is first term.

$\therefore l=a+(n-1) d$,

where $l$ is last term and $d$ is is common difference.

$\begin{array}{l}

999=3+(n-1) \times 6 \\

n-1=\frac{999-3}{6}=\frac{996}{6} \\

\Rightarrow n-1=166 \\

\Rightarrow n=167 \\

\therefore \mathrm{S}=\frac{n}{2}[2 a+(n-1) d] \\

=\frac{167}{2}[2 \times 3+(167-1) \times 6] \\

=\frac{167}{2}[1002]=167 \times 501=83667

\end{array}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2021
The sum of all the solution of the equation $\cos \theta \cos \left(\frac{\pi}{3}+\theta\right) \cos \left(\frac{\pi}{3}-\theta\right)=\frac{1}{4}, \theta \in[0,6 \pi]$

Options:
A
$15 \pi$

B
$30 \pi$

C
$\frac{100 \pi}{3}$

D
None of these

The correct answer is:

$30 \pi$

Solution:
$$
\begin{aligned}
& 2 \cos \theta\left[\cos 120^{\circ}+\cos 2 \theta\right]=1 \\
\Rightarrow & 2 \cos \theta\left(-\frac{1}{2}+2 \cos ^2 \theta-1\right)=1 \\
\Rightarrow & 4 \cos ^3 \theta-3 \cos \theta-1=0 \\
\Rightarrow & \cos 3 \theta=1=\cos 0 \\
\Rightarrow & 3 \theta=2 n \pi \text { or } \theta=\frac{2 n \pi}{3}, n \in z
\end{aligned}
$$
Given the values so that $2 n$ does not exceed 18 .
Hence, the sum $=\frac{2 \pi}{3} \sum_1^9 n=\frac{2 \pi}{3} \times \frac{9(9+1)}{2}=30 \pi$

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2022
The sum of mean and variance of a Binomial distribution BD→ for 3 trials is 2.7, then the binomial distribution BD→ is given by

Options:
A 0.2+0.85
B 0.3+0.75
C 0.4+0.65
D None of these

The correct answer is:
0.3+0.75

Solution:
Mean m for BD=np

and variance σ2 for BD=npq

Given, np+npq=2.7 and n=3

∴ np1+q=2.7⇒p1+q=2.73

⇒p1+q=0.9

⇒1-q1-q=0.9 [∵ p+q=1]

⇒1-q2=0.9

⇒q2=0.1⇒q=±0.3

∴ q=0.3 [∵ q=-0.3 can't possible as 0≤q≤1]

⇒p=0.7 and q=0.3

∴ BD is given by 0.3+0.75

Topic: Mathematics, Probability, BITSAT, BITSAT 2018
The sum of the infinite series $\frac{2^{2}}{2 !}+\frac{2^{4}}{4 !}+\frac{2^{6}}{6 !}+\ldots$

is equal to

Options:
A $\frac{\mathrm{e}^{2}+1}{2 \mathrm{e}}$
B $\frac{\mathrm{e}^{4}+1}{2 \mathrm{e}^{2}}$
C $\frac{\left(\mathrm{e}^{2}-1\right)^{2}}{2 \mathrm{e}^{2}}$
D $\frac{\left(\mathrm{e}^{2}+1\right)^{2}}{2 \mathrm{e}^{2}}$

The correct answer is:
$\frac{\left(\mathrm{e}^{2}-1\right)^{2}}{2 \mathrm{e}^{2}}$

Solution:
We know that

$\frac{e^{x}+e^{-x}}{2}=1+\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+\frac{x^{6}}{6 !}+\ldots$

keeping $\mathrm{x}=2$, we get

Expression $=\frac{1}{2}\left[\frac{e^{2}+e^{-2}}{2}\right]-1=\frac{\left(e^{2}-1\right)^{2}}{2 e^{2}}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2020
The sum of the series $1+2.2+3.2^{2}+4.2^{3}+\ldots .+100.2^{99}$ is

Options:
A $100 \cdot 2^{100}+1$
B $99.2^{100}+1$
C $99.2^{100}-1$
D $100.2^{100}+1$

The correct answer is:
$99.2^{100}+1$

Solution:
Let $S=1+2.2+3.2^{2}+4.2^{2}+\ldots .+100.2^{99}$

It is an arithmetico - geometric series. On multiplying Eq.

(i) by 2 and then subtracting it from Eq. (i), we get

$S=1+2.2+3.2^{2}+4.2^{3}+\ldots+100.2^{99}$

$2 S=1.2+2.2^{2} \pm 3.2^{3}+\ldots+99.2^{99}+100.2^{100}$

$\begin{aligned}-S &=1+2+2^{2}+2^{3} \ldots+2^{99}-100.2^{100} \\ & \Rightarrow-S=\frac{1\left(2^{100}-1\right)}{2-1}-100.2^{100} \\ & \Rightarrow-S=2^{100}-1-100.2^{100} \\ & \Rightarrow-S=-1-99.2^{100} \\ & \Rightarrow S=99.2^{100}+1 \end{aligned}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2013
The sum to infinite term of the series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\cdots$ is

Options:
A
3
B
4
C
6
D
2

The correct answer is:
3

Solution:
We have
Multiplying both sides by $\frac{1}{3}$, we get
Subtracting eqn. (ii) from eqn. (i), we get
$\begin{aligned} & \frac{2}{3} S=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \ldots \ldots \infty \\ & \Rightarrow \frac{2}{3} S=\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \ldots \ldots \infty \\ & \Rightarrow \frac{2}{3} S=\frac{\frac{4}{3}}{1-\frac{1}{3}}=\frac{4}{3} \times \frac{3}{2} \Rightarrow S=3\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main
The sum to infinite term of the series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\cdots$ is

Options:
A
3
B
4
C
6
D
2

The correct answer is:
3

Solution:
We have
Multiplying both sides by $\frac{1}{3}$, we get
Subtracting eqn. (ii) from eqn. (i), we get
$\begin{aligned} & \frac{2}{3} S=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \ldots \ldots \infty \\ & \Rightarrow \frac{2}{3} S=\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \ldots \ldots \infty \\ & \Rightarrow \frac{2}{3} S=\frac{\frac{4}{3}}{1-\frac{1}{3}}=\frac{4}{3} \times \frac{3}{2} \Rightarrow S=3\end{aligned}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The sum to $\mathrm{n}$ terms of the series

$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots \ldots \ldots \ldots$ is

Options:
A $n-1-2^{-n}$
B 1
C $n-1+2^{-n}$
D $1+2^{-n}$

The correct answer is:
$n-1+2^{-n}$

Solution:
$\begin{aligned} \frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots \\=&\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+\left(1-\frac{1}{16}\right)+\ldots . \\=& \mathrm{n}-\frac{\frac{1}{2}\left\{1-\frac{1}{2^{\mathrm{n}}}\right\}}{1-\frac{1}{2}}=\mathrm{n}-1+2^{-\mathrm{n}} \end{aligned}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2016
The surface charge density of a thin charged disc of radius $\mathrm{R}$ is $\sigma$. The value of the electric field at the centre of the disc is $\frac{\sigma}{2 \epsilon_{0}}$. With respect to the field at the centre, the electric field along the axis at a distance $\mathrm{R}$ from the centre of the $\operatorname{disc}$

Options:
A reduces by $70.7 \%$
B reduces by $29.3 \%$
C reduces by $9.7 \%$
D reduces by $14.6 \%$

The correct answer is:
reduces by $70.7 \%$

Solution:
Electric field intensity at the centre of the disc.

\(\mathrm{E}=\frac{\sigma}{2 \epsilon_{0}} \text { (given) }\)

Electric field along the axis at any distance \(x\) from the centre of the disc

\(\mathrm{E}^{\prime}=\frac{\sigma}{2 \epsilon_{0}}\left(1-\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}}-\mathrm{R}^{2}}\right)\)

From question, \(\mathrm{x}=\mathrm{R}\) (radius of disc)

\(\begin{aligned}

&\therefore E^{\prime}=\frac{\sigma}{2 \epsilon_{0}}\left(1-\frac{R}{\sqrt{R^{2}+R^{2}}}\right) \\

&=\frac{\sigma}{2 \epsilon_{0}}\left(\frac{\sqrt{2} R-R}{\sqrt{2} R}\right) \\

&=\frac{4}{14} E

\end{aligned}\)

\(\therefore\) \% reduction in the value of electric field

\(=\frac{\left(\mathrm{E}-\frac{4}{14} \mathrm{E}\right) \times 100}{\mathrm{E}}=\frac{1000}{14} \%=70.7 \%\)

Topic: Physics, Electrostatics, BITSAT, BITSAT 2016
The susceptibility of a magnetism at $300 \mathrm{~K}$ is $1.2 \times 10^{-5}$. The temperature at which the susceptibility increases to $1.8 \times 10^{-5}$ is

Options:
A $150 \mathrm{~K}$
B $200 \mathrm{~K}$
C $250 \mathrm{~K}$
D $20 \mathrm{~K}$

The correct answer is:
$200 \mathrm{~K}$

Solution:
$\begin{aligned} & \chi=\frac{\mathrm{C}}{\mathrm{T}} \\ & \Rightarrow \frac{\chi_{1}}{\chi_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} \Rightarrow \frac{1.2 \times 10^{-5}}{1.8 \times 10^{-5}}=\frac{\mathrm{T}_{2}}{300} \\ & \Rightarrow \mathrm{T}_{2} \Rightarrow \frac{12}{18} \times 300=200 \mathrm{~K} \end{aligned}$

Topic: Physics, Magnetic Properties of Matter, BITSAT, BITSAT 2014
The temperature of an ideal gas is increased from $27^{\circ} \mathrm{C}$ to $127^{\circ} \mathrm{C}$, then percentage increase in $\mathrm{v}_{\mathrm{rms}}$

is

Options:
A $37 \%$
B $11 \%$
C $33 \%$
D $15.5 \%$

The correct answer is:
$15.5 \%$

Solution:
We know, $\mathrm{V}_{\mathrm{mms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$

$\Rightarrow \%$ increase in

$$

\begin{array}{l}

\mathrm{V}_{\mathrm{rms}}=\frac{\sqrt{\frac{3 \mathrm{R} \mathrm{T}_{2}}{\mathrm{M}}}-\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}}}}{\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}}}} \times 100 \\

=\frac{\sqrt{\mathrm{T}_{2}}-\sqrt{\mathrm{T}_{1}}}{\sqrt{\mathrm{T}_{1}}} \times 100 \\

=\frac{\sqrt{400}-\sqrt{300}}{\sqrt{300}} \times 100 \\

=\frac{20-17.32}{17.32} \times 100=15.5 \%

\end{array}

$$

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2013
The temperature of an ideal gas is increased from $120 \mathrm{~K}$ to $480 \mathrm{~K}$. If at $120 \mathrm{~K}$, the root mean square velocity of the gas molecules is $v$, at $480 \mathrm{~K}$ it becomes

Options:
A $4 u$
B $2 u$
C $\frac{u}{2}$
D $\frac{u}{4}$

The correct answer is:
$2 u$

Solution:
\(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}}}}{\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}}}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}} \therefore \mathrm{v}_{2}=2 \mathrm{v}_{1}\)

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2021
The term independent of \(x\) in the expansion of \(\left(9 \mathrm{x}-\frac{1}{3 \sqrt{\mathrm{x}}}\right)^{18}, \mathrm{x} > 0\), is a times the corresponding binomial coefficient. Then a is

Options:
A 3
B \(1 / 3\)
C \(-1 / 3\)
D None of these

The correct answer is:
None of these

Solution:
\(\begin{aligned}
\mathrm{T}_{\mathrm{r}+1} & ={ }^{18} \mathrm{C}_{\mathrm{r}}(9 \mathrm{x})^{18-\mathrm{r}}\left(-\frac{1}{3 \sqrt{\mathrm{x}}}\right)^{\mathrm{r}} \\
& =(-\mathrm{r})^{\mathrm{r}}{ }^{18} \mathrm{C}_{\mathrm{r}} 9^{18-\frac{3 \mathrm{r}}{2}} \mathrm{x}^{18-\frac{3 \mathrm{r}}{2}}
\end{aligned}\)
is independent of \(\mathrm{x}\) provided \(\mathrm{r}=12\) and then \(\mathrm{a}=1\).

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2011