datavorous/entrance-exam-dataset huggingface dataset:
The equation of tangent to the curve \(y=\sin x\) at the point \((\pi, 0)\) is

Options:
A \(x+y=0\)
B \(x+y=\pi\)
C \(x-y=\pi\)
D \(x-y=0\)

The correct answer is:
\(x+y=\pi\)

Solution:
\(y=\sin x \Rightarrow \frac{d y}{d x}=\cos x \Rightarrow\left(\frac{d y}{d x}\right)_{(\pi, 0)}=-1\)
Therefore the equation of tangent at \((\pi, 0)\) is given by
\(y-0=-1(x-\pi) \Rightarrow x+y=\pi\)

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2009
The equation of the circle, which touches the line $y=5$ and passes through $(-1,2)$ and $(1,2)$ is

Options:
A $9 x^{2}+9 y^{2}-60 y+75=0$
B $9 x^{2}+9 y^{2}-60 x-75=0$
C $9 x^{2}+9 y^{2}+60 y-75=0$
D $9 x^{2}+9 y^{2}+60 x+75=0$

The correct answer is:
$9 x^{2}+9 y^{2}-60 y+75=0$

Solution:
The centre of the circle is on the perpendicular bisector of the line joining $(-1,2)$ and $(1,2)$, which is the $y$-axis. The ordinate of the centre is given by

$\begin{array}{l}

(5-y)^{2}=1+(y-2)^{2} \\

\Rightarrow y=\frac{10}{3}

\end{array}$

Hence, eq. of the circle is:

$\begin{array}{l}

x^{2}+\left(y-\frac{10}{3}\right)^{2}=\left(\frac{5}{3}\right)^{2} \\

\Rightarrow 9 x^{2}+9 y^{2}-60 y+75=0

\end{array}$

Topic: Mathematics, Circle, BITSAT, BITSAT 2021
The equation of the circle with centre $(0,2)$ and radius 2 is $x^2+y^2-m y=0$. The value of $m$ is

Options:
A
1
B
2
C
4
D
3

The correct answer is:
4

Solution:
Here $h=0, k=2$ and $r=2$. Therefore, the required equation of the circle is :
$(x-0)^2+(y-2)^2=(2)^2$
or $x^2+y^2-4 y+4=4$ or $x^2+y^2-4 y=0$

Topic: Mathematics, Circle, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The equation of the curve whose slope at any point is equal to y+2x and which passes through the origin is

Options:
A y=2x-1
B y=2ex-x-1
C y=2ex-1
D y=2exx-1

The correct answer is:
y=2ex-x-1

Solution:
We have, dydx=y+2x

⇒dydx-y=2x

Now, IF=e-∫1dx=e-x

y·e-x=∫2xe-xdx+k, k be the constant of integration

=2x∫e-xdx-∫1·-e-xdx+k [using integration by parts]

⇒y·e-x=-2xe-x-2e-x+k ...i

As curve (i) passes through 0,0

∴ 0=0-2+k

⇒k=2

Thus, the curve is

ye-x=-2xe-x-2e-x+2

∴ y=2ex-x-1

Hence, option (b) is correct.

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2018
The equation of the ellipse with focus at \(( \pm 5,0)\) and \(x=\frac{36}{5}\) as one directrix is

Options:
A \(\frac{x^2}{36}+\frac{y^2}{25}=1\)
B \(\frac{x^2}{36}+\frac{y^2}{11}=1\)
C \(\frac{x^2}{25}+\frac{y^2}{11}=1\)
D None of these

The correct answer is:
\(\frac{x^2}{36}+\frac{y^2}{11}=1\)

Solution:
We have ae \(=5\) [Since focus is \(( \pm \mathrm{ae}, 0)\)]
and \(\frac{\mathrm{a}}{\mathrm{e}}=\frac{36}{5} \quad\left[\right.\) since directrix is \(\left.\mathrm{x}= \pm \frac{\mathrm{a}}{\mathrm{e}}\right]\)
On solving we get \(\mathrm{a}=6\)
\(\text { and } \mathrm{e}=\frac{5}{6} \Rightarrow \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=36\left(1-\frac{25}{36}\right)=11\)
Thus, the required equation of the ellipse is
\(\frac{x^2}{36}+\frac{y^2}{11}=1\)

Topic: Mathematics, Ellipse, BITSAT, BITSAT 2011
The equation of the plane containing the line \(\frac{\mathrm{x}-\mathrm{x}_1}{\ell}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{~m}}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{n}}\) is \(a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0\), then

Options:
A \(\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1=0\)
B \(a \ell+b m+c n=0\)
C \(\frac{\mathrm{a}}{\ell}=\frac{\mathrm{b}}{\mathrm{m}}=\frac{\mathrm{c}}{\mathrm{n}}\)
D \(\ell \mathrm{x}_1+\mathrm{my}_1+\mathrm{nz}_1=0\)

The correct answer is:
\(a \ell+b m+c n=0\)

Solution:
If the given plane contains the given line then the normal to the plane, must be perpendicular to the line and the condition for the same is \(a \ell+b m+c n=0\).

Topic: Mathematics, Probability, BITSAT, BITSAT 2010
The equation of the right bisector plane of the segment joining (2,3,4) and (6,7,8) is

Options:
A $x+y+z+15=0$
B $x+y+z-15=0$
C $x-y+z-15=0$
D None of these

The correct answer is:
$x+y+z-15=0$

Solution:
If the given points be $A(2,3,4)$ and $B$ $(6,7,8),$ then their mid-point $\mathrm{N}(4,5,6)$ must lie on the plane. The direction ratios of $\mathrm{AB}$ are $4,4,4,$ i.e. 1,1,1

$\therefore$ The required plane passes through $\mathrm{N}(4,5,6)$ and is normal to $\mathrm{AB}$. Thus its equation is

$$

1(x-4)+1(y-5)+1(z-6)=0 \Rightarrow x+y+z=15

$$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2014
The equation of the tangent to the curve $y=b e^{-x/a}$ at the point where it crosses the $y$ -axis is

Options:
A $\frac{x}{a}-\frac{y}{b}=1$
B $a x+b y=1$
C $a x-b y=1$
D $\frac{x}{a}+\frac{y}{b}=1$

The correct answer is:
$\frac{x}{a}+\frac{y}{b}=1$

Solution:
$y=b e^{-x / a}$ meets the $y$ -axis at $(0, b)$. Again,

$\frac{d y}{d x}=b e^{-x / a}\left(-\frac{1}{a}\right)$

$\operatorname{At}(0, b), \frac{d y}{d x}=b e^{0}\left(-\frac{1}{a}\right)=-\frac{b}{a}$

Therefore, required tangent is

$y-b=-\frac{b}{a}(x-0) \text { or } \frac{x}{a}+\frac{y}{b}=1$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2020
The equation $\sin ^{4} x-(k+2) \sin ^{2} x-(k+3)=0$ possesses a solution if

Options:
A $k>-3$
B $k<-2$
C $-3 \leq \mathrm{k} \leq-2$
D $\mathrm{k}$ is any positive integer

The correct answer is:
$-3 \leq \mathrm{k} \leq-2$

Solution:
We have, $\sin ^{4} x-(k+2) \sin ^{2} x-(k+3)=0$

$\Rightarrow \sin ^{2} x=\frac{(k+2) \pm \sqrt{(k+2)^{2}+4(k+3)}}{2}$

$=\frac{(k+2) \pm(k+4)}{2}$

$\Rightarrow \sin ^{2} x=k+3\left(\because \sin ^{2} x=-1\right.$ is not possible)

Since $0 \leq \sin ^{2} x \leq 1$

$\therefore 0 \leq \mathrm{k}+3 \leq 1$ or $-3 \leq \mathrm{k} \leq-2$

Topic: Mathematics, Trigonometric Ratios & Identities, BITSAT, BITSAT 2020
The equation $x^{2}-2 \sqrt{3} x y+3 y^{2}-3 x+3 \sqrt{3} y-4=0 \quad$ represents

Options:
A a pair of intersecting lines
B a pair of parallel lines with distance between

them $\frac{5}{2}$
C a pair of parallel lines with distance between them $5 \sqrt{2}$
D a conic section, which is not a pair of straight lines

The correct answer is:
a pair of parallel lines with distance between

them $\frac{5}{2}$

Solution:
We have $a=1, h=-\sqrt{3}, b=3, g=-\frac{3}{2}$, $\mathrm{f}=\frac{3 \sqrt{3}}{2}, \mathrm{c}=-4$

Thus $a b c+2 f g h-a f^{2}-b g^{2}-c h^{2}=0$

Hence the equation represents a pair of straight lines.

Again $\frac{\mathrm{a}}{\mathrm{h}}=\frac{\mathrm{h}}{\mathrm{b}}=\frac{\mathrm{g}}{\mathrm{f}}=-\frac{1}{\sqrt{3}}$

$\therefore$ the lines are parallel. The distance between them

$$

=2 \sqrt{\frac{g^{2}-a c}{a(a+b)}}=2 \sqrt{\frac{\frac{9}{4}+4}{1(1+3)}}=\frac{5}{2}

$$

Topic: Mathematics, Pair of Lines, BITSAT, BITSAT 2016
The equations $2 x+3 y+4=0 ; 3 x+4 y+6=0$ and $4 x+5 y+8=0$ are

Options:
A consistent with unique solution
B inconsistent
C consistent with infinitely many solutions
D None of the above

The correct answer is:
consistent with unique solution

Solution:
Consider first two equations:
$2 x+3 y=-4 \quad$ and $\quad 3 x+4 y=-6$
We have $\Delta=\left|\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right|=-1 \neq 0$
$$
\Delta_{\mathrm{x}}=\left|\begin{array}{ll}
-4 & 3 \\
-6 & 4
\end{array}\right|=2 \quad \text { and } \quad \Delta_{\mathrm{y}}=\left|\begin{array}{cc}
2 & -4 \\
3 & -6
\end{array}\right|=0
$$
$\therefore x=-2$ and $y=0$
Now this solution satisfies the third, so the equations are consistent with unique solution.

Topic: Mathematics, Determinants, BITSAT, BITSAT 2012
The equilibrium constant for the disproportionation reaction $2 \mathrm{Cu}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{ag})$ at $25^{\circ} \mathrm{C}$

$\left(\mathrm{E}^{\circ} \mathrm{Cu}^{+} / \mathrm{Cu}=0.52 \mathrm{~V}, \mathrm{E}^{\circ} \mathrm{Cu}^{2+} / \mathrm{Cu}=0.16 \mathrm{~V}\right)$ is

Options:
A $6 \times 10^{4}$
B $6 \times 10^{6}$
C $1.2 \times 10^{6}$
D $1.2 \times 10^{-6}$

The correct answer is:
$1.2 \times 10^{6}$

Solution:
The reaction

$$

\begin{array}{l}

2 \mathrm{Cu}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \\

\mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{0.0592}{1} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Cu}^{+}\right]^{2}}

\end{array}

$$

At equilibrium $\mathrm{E}_{\mathrm{cell}}=0$

$$

\therefore \quad \mathrm{E}_{\text {cell }}^{\circ}=0.0592 \log \mathrm{K}_{\mathrm{c}}

$$

or, $\quad \log \mathrm{K}_{\mathrm{c}}=\frac{0.52-0.16}{0.0592}$

$\therefore \mathrm{K}_{\mathrm{c}}=1.2 \times 10^{6}$

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2016
The escape velocity from a planet is $v_{e}$. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be

Options:
A $\mathrm{V}_{\mathrm{e}}$
B $\mathrm{v}_{\mathrm{e}} / \sqrt{2}$
C $\mathrm{v}_{\mathrm{e}} / 2$
D zero

The correct answer is:
$\mathrm{v}_{\mathrm{e}} / \sqrt{2}$

Solution:
$\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}=\frac{\mathrm{GMm}}{\mathrm{R}} ; \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{\mathrm{GMm}}{\mathrm{R}}}=\sqrt{2 \mathrm{gR}}$

In tunnel body will perform SHM at centre $\mathrm{V}_{\mathrm{max}}=\mathrm{A\omega}$ (see chapter on SHM)

$=\frac{\mathrm{R} 2 \pi}{2 \pi \sqrt{\mathrm{R} / \mathrm{g}}}=\sqrt{\mathrm{gR}}=\frac{\mathrm{v}_{\mathrm{e}}}{\sqrt{2}}$

Topic: Physics, Gravitation, BITSAT, BITSAT 2020
The escape velocity from the surface of earth is ve. A body is projected with velocity 3ve. With what constant velocity will it move in the inter planetary space?

Options:
A ve
B 3ve
C 2ve
D 22ve

The correct answer is:
22ve

Solution:
If v be the velocity in the inter planetary space and Ui, Uf initial and final PE

Then, by the law of conservation of energy,

Ui+Ki=Uf+Kf

-GMemRe+12m3ve2=0+12mv2

-GMeRe+92ve2=12v2

⇒ -2GMeRe+9Ve2=v2

⇒ -2gRe2Re+9ve2=v2 ∵GMe=gRe2

-2gRe+9×2gRe=v2 ∵ve2=2gRe

16 gRe=v2
⇒ v=8·2gRe=22ve

Topic: Physics, Gravitation, BITSAT, BITSAT 2019
The expression ∫0nxdx∫0nxdx, where x and x are integral and fractional part of x and n∈N, is equal to

Options:
A 1n-1
B 1n
C n
D n-1

The correct answer is:
n-1

Solution:
We have, ∫0nxdx

=∫010dx+∫121dx+∫232dx+…+∫n-1nn-1dx

=12-1+23-2+34-3+……+n-1n-n-1

=1+2+3+….+n-1=nn-12

and ∫0nxdx=∫0nx-xdx=n2

∴ ∫0nxdx∫0nxdx=n-1

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2017
The false statement among the followings:

Options:
A The average residence time of $\mathrm{NO}$ is one month.
B Limestone acts as a sink for $\mathrm{SO}_{\mathrm{x}}$
C $\mathrm{SO}_{\mathrm{x}}$ can be removed from flue gases by passing through a solution of citrate ions.
D Ammonia acts as a sink for $\mathrm{NO}_{\mathrm{x}^{-}}$

The correct answer is:
The average residence time of $\mathrm{NO}$ is one month.

Solution:
The average residence time of NO is 4 days.

Topic: Chemistry, Environmental Chemistry, BITSAT, BITSAT 2009
The family of straight lines \((2 a+3 b) x+(a-b) y+2 a-4 b=0\) is concurrent at the point

Options:
A \(\left(\frac{2}{5}, \frac{-14}{5}\right)\)
B \(\left(\frac{-2}{5}, \frac{-14}{5}\right)\)
C \(\left(\frac{-2}{5}, \frac{14}{5}\right)\)
D \(\left(\frac{2}{5}, \frac{14}{5}\right)\)

The correct answer is:
\(\left(\frac{2}{5}, \frac{-14}{5}\right)\)

Solution:
Rewriting the equation \((2 x+y+2) a+(3 x-y-4) b=0\) and for all \(a\), \(\mathrm{b}\) the straight lines pass through the intersection of \(2 x+y+2=0\) and \(3 x-y-4=0\) i.e., the point \(\left(\frac{2}{5},-\frac{14}{5}\right)\).

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2011
The figure shows the graph of pVT versus p for 2×10-4 kg of hydrogen gas at two different temperatures, where p, V and T represents pressure, volume and temperature respectively.Then, the value of pVT, where the curve meet on the vertical axis, is

Options:
A R5
B 2 R3
C R10
D R15

The correct answer is:
R10

Solution:
Number of moles in 2×10-4 kg

μ= Mass of hydrogen in gram Molecular mass

=2×10-4×1032

=0.1 mole

∴ pV=μRT

pVT=μR=0.1×R=R10

Topic: Physics, Thermodynamics, JEE Main
The figure shows the graph of pVT versus p for 2×10-4 kg of hydrogen gas at two different temperatures, where p, V and T represents pressure, volume and temperature respectively.Then, the value of pVT, where the curve meet on the vertical axis, is

Options:
A R5
B 2 R3
C R10
D R15

The correct answer is:
R10

Solution:
Number of moles in 2×10-4 kg

μ= Mass of hydrogen in gram Molecular mass

=2×10-4×1032

=0.1 mole

∴ pV=μRT

pVT=μR=0.1×R=R10

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2019
The final product obtained in the reaction

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2020
The final step for extraction of copper from copper pyrite in bessemer converter involves the reaction

Options:
A Cu2 S+2Cu2O⟶6Cu+SO2
B 4Cu2O+FeS⟶8Cu+FeSO4
C 2Cu2O+FeS⟶4Cu+Fe+SO2
D Cu2 S+2FeO⟶2Cu+Fe+SO2

The correct answer is:
Cu2 S+2Cu2O⟶6Cu+SO2

Solution:
In Bessemer converter, copper-sulphide is partially oxidised to cuprous oxide, which further reacts with the remaining copper sulphide to give copper and sulphur di oxide, i.e. Cu2 S+2Cu2O⟶6Cu+SO2

This reaction represents auto reduction. Cu2O reacts with Cu2S to give copper metal. The solid metal obtained is called as blister copper and is about 99% pure.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, BITSAT, BITSAT 2019
The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at

Options:
A $\frac{9 R}{400} \mathrm{~cm}^{-1}$
B $\frac{7 \mathrm{R}}{144} \mathrm{~cm}^{-1}$
C $\frac{3 \mathrm{R}}{4} \mathrm{~cm}^{-1}$
D $\frac{5 \mathrm{R}}{36} \mathrm{~cm}^{-1}$

The correct answer is:
$\frac{5 \mathrm{R}}{36} \mathrm{~cm}^{-1}$

Solution:
For Balmer $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=3$; $v=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 R}{36} \mathrm{~cm}^{-1}$

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2016
The focal length of a thin convex lens for red and blue rays are 100 cm and 96.8 cm respectively. Then, the dispersive power of the material of the lens is

Options:
A 0.968
B 0.98
C 0.0325
D 0.325

The correct answer is:
0.0325

Solution:
Dispersive power is ratio of angular dispersion and mean deviation.

It can be written in terms of focal length as,

ω=fr-fbfy

Here fr=100 cm

fb=96.8 cm

fy=fb×fr

=96.8×100

≈98.34 cm

∴ Dispersive power

ω=fr-fbfy

=100-96.898.34

≈0.0325
Hence, the dispersive power of the material of the lens is 0.0325.

Topic: Physics, Ray Optics, BITSAT, BITSAT 2017
The following change can be carried out with

Options:
A $\mathrm{NaBH}_{4}$
B $\mathrm{LiAlH}_{4}$
C $\mathrm{H}_{2} / \mathrm{Pt}$
D $\mathrm{PCC} / \mathrm{CH}_{2} \mathrm{Cl}_{2}$

The correct answer is:
$\mathrm{LiAlH}_{4}$

Solution:
$\mathrm{LiAlH}_{4}$ will give the desired compound.

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2020
The following data are for the decomposition of ammonium nitrite in aqueous solution :
The order of reaction is:

Options:
A Zero
B One
C Two
D Three

The correct answer is:
One

Solution:
$\mathrm{NH}_{4} \mathrm{NO}_{2} \longrightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}$
Volume of $\mathrm{N}_{2}$ formed in successive five minutes are $2.75 \mathrm{cc}, 2.40 \mathrm{cc}$ and $2.25 \mathrm{cc}$ which is in decreasing order. So rate of reaction is dependent on concentration of $\mathrm{NH}_{4} \mathrm{NO}_{2}$. As decrease is not very fast so it will be first order reaction.

Topic: Chemistry, Chemical Kinetics, JEE Main
The following data are for the decomposition of ammonium nitrite in aqueous solution :
The order of reaction is:

Options:
A Zero
B One
C Two
D Three

The correct answer is:
One

Solution:
$\mathrm{NH}_{4} \mathrm{NO}_{2} \longrightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}$
Volume of $\mathrm{N}_{2}$ formed in successive five minutes are $2.75 \mathrm{cc}, 2.40 \mathrm{cc}$ and $2.25 \mathrm{cc}$ which is in decreasing order. So rate of reaction is dependent on concentration of $\mathrm{NH}_{4} \mathrm{NO}_{2}$. As decrease is not very fast so it will be first order reaction.

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2010
The following reaction is known as

Options:
A Friedel-Craft's reaction
B Kolbe reaction
C Reimer-Tiemann reaction
D Witting reaction

The correct answer is:
Kolbe reaction

Solution:
At $120-140^{\circ} \mathrm{C}$ temperature and 1.5 atm pressure, sodium phenoxide reacts with $\mathrm{CO}_{2}$ to yield sodium salicylate which on further hydrolysis give to salicylic acid. This reaction is known as Kolbe's reaction.

Topic: Chemistry, Alcohols Phenols and Ethers, BITSAT, BITSAT 2012
The following two compounds are

Options:
A enantiomers
B diastereomers
C identical
D epimers

The correct answer is:
enantiomers

Solution:
The given two structures are optical isomers but as these are mirror image of each other, hence they represent enantiomers of each other.

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2012
The following two compounds are

Options:
A enantiomers
B diastereomers
C identical
D epimers

The correct answer is:
enantiomers

Solution:
The given two structures are optical isomers but as these are mirror image of each other, hence they represent enantiomers of each other.

Topic: Chemistry, General Organic Chemistry, JEE Main
The foot of the perpendicular from the point (7,14,5) to the plane $2 \mathrm{x}+4 \mathrm{y}-\mathrm{z}=2$ are

Options:
A (1,2,8)
B (3,2,8)
C (5,10,6)
D (9,18,4)

The correct answer is:
(1,2,8)

Solution:
We know that the length of the perpendicular from the point $\left(x_{1}, y_{1}, z_{1}\right)$ to the plane $a x+b y+c z+d=0$ is $\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$
and the co-ordinate $(\alpha, \beta, \gamma)$ of the foot of the $\perp$ are given by $\frac{\alpha-x_{1}}{a}=\frac{\beta-y_{1}}{b}=\frac{\gamma-z_{1}}{c}$
$=-\left(\frac{a x_{1}+b y_{1}+c z_{1}+d}{a^{2}+b^{2}+c^{2}}\right)$ ...(1)
In the given ques, $x_{1}=7, y_{1}=14, z_{1}=5$,
$a=2 b=4, c=-1, d=-2$
By putting these values in (1), we get $\frac{\alpha-7}{2}=\frac{\beta-14}{4}=\frac{\gamma-5}{-1}=-\frac{63}{21}$
$\Rightarrow \alpha=1, \beta=2$ and $\gamma=8$
Hence, foot of $\perp$ is (1,2,8)

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2012
The force between two small charged spheres having charges of $1 \times 10^{-7} \mathrm{C}$ and $2 \times 10^{-7} \mathrm{C}$ placed $20 \mathrm{~cm}$ apart in air is

Options:
A $4.5 \times 10^{-2} \mathrm{~N}$
B $5.4 \times 10^{-2} \mathrm{~N}$
C $4.5 \times 10^{-3} \mathrm{~N}$
D $5.4 \times 10^{-3} \mathrm{~N}$

The correct answer is:
$4.5 \times 10^{-3} \mathrm{~N}$

Solution:
Here, $q_1=1 \times 10^{-7} \mathrm{C}, q_2$ and $2 \times 10^{-7} \mathrm{C}$, $r=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$
$F=\frac{q_1 q_2}{4 \pi \varepsilon_0 r^2}=\frac{9 \times 10^9 \times 1 \times 10^{-7} \times 2 \times 10^{-7}}{\left(20 \times 10^{-2}\right)^2}$
$=4.5 \times 10^{-3} N$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The force on a body of mass 1 kg is 20 i^+10 j^ N. If it starts from rest, then the position of the body at time t=2 s, is

Options:
A -20 i^-40 j^
B 20 i^-40 j^
C 40 i^-20 j^
D 40 i^+20 j^

The correct answer is:
40 i^+20 j^

Solution:
F=20 i^+10 j^

∴ Fx=20 N and Fy=10 N

∴ Acceleration,

ax=Fxm=201
ax=20 m s-2

and ay=Fym=101=10 m s-2

∴ Displacement of body in x -direction in 2 s.

x=ux t+12ax t2

=0+12×20×22
=40 m

Displacement of body in y-direction in 2 s.

y=uy t+12 ay t2

=0+12×10×22

=20 m

∴ Position of the body at t=2 s

r=x i^+y j^
r=40 i^+20 j^

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2019
The formation of $\mathrm{CO}$ and $\mathrm{CO}_{2}$ illustrates the law of

Options:
A reciprocal proportion
B conservation of mass
C multiple proportion
D constant composition

The correct answer is:
multiple proportion

Solution:
Formation of $\mathrm{CO}$ and $\mathrm{CO}_{2}$ illustrates the law of multiple proportion that is constant mass of $\mathrm{C}$ reacts with different masses of oxygen. These masses here bears simple ratio of 1: 2 .

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2014
The fourth term of an A.P. is three times of the first term and the seventh term exceeds the twice of the third term by one, then the common difference of the progression is

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

The correct answer is:
2

Solution:
Let the progression be $a$, $a+d$, a +2 d,

Then $x_{4}=3 x_{1} \Rightarrow a+3 d=3 a \Rightarrow 3 d=2 a \ldots$ (i)

Again $x_{7}=2 x_{3}+1$

$\Rightarrow a+6 d=2(a+2 d)+1 \Rightarrow 2 d=a+1 \quad \ldots$ (ii)

Solving (i) and (ii), we get $\mathrm{a}=3, \mathrm{~d}=2$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2016
The fraction of total volume occupied by the atoms present in a simple cube is

Options:
A $\frac{\pi}{3 \sqrt{2}}$
B $\frac{\pi}{4 \sqrt{2}}$
C $\frac{\pi}{4}$
D $\frac{\pi}{6}$

The correct answer is:
$\frac{\pi}{6}$

Solution:
Number of atoms per unit cell $=1$ Atoms touch each other along edges. Hence $r=\frac{a}{2}$
$(\mathrm{r}=$ radius of atom and $\mathrm{a}=$ edge length)
Therefore $\%$ $=\frac{\frac{4}{3} \pi r^{3}}{(2 r)^{3}}=\frac{\pi}{6}$

Topic: Chemistry, Solid State, BITSAT, BITSAT 2011
The free space inside a current carrying toroid is filled with a material of susceptibility $2 \times 10^{-2}$. The percentage increase in the value of magnetic field inside the toroid will be

Options:
A $2 \%$
B $0.2 \%$
C $0.1 \%$
D $1 \%$

The correct answer is:
$2 \%$

Solution:
Given,
Susceptibility of material, $\chi_m=2 \times 10^{-2}$
Using $\mu_{\mathrm{r}}=1+\chi_{\mathrm{m}}=1+0.02=1.02$
$\mathrm{B}_{\text {final }}=\mu_{\mathrm{r}} \mathrm{B}_0$ (here, $\mathrm{B}_0=$ initial magnetic field)
$\%$ increase in magnetic field
$=\frac{\mathrm{B}_{\text {final }}-\mathrm{B}_0}{\mathrm{~B}_0} \times 100=\frac{\mu_{\mathrm{r}} \mathrm{B}_0-\mathrm{B}_0 \times 100}{\mathrm{~B}_0}$
$=\frac{(\chi+1)-1 \times 100}{1}=0.02 \times 100=2 \%$

Topic: Physics, Magnetic Properties of Matter, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The frequency and the intensity of incident beam of light falling on the surface of a photoelectric material is increased by a factor of two. This will

Options:
A increase the maximum kinetic energy of the ejected photoelectrons by a factor of more than two and would increase the photoelectric current by a factor of two.
B increase the maximum kinetic energy of the photo electron and would increase the photoelectric current both by a factor of two.
C increase the maximum kinetic energy of the photoelectron by a factor of two and will have no effect on the magnitude of the photoelectric current produced.
D not produce any effect on the kinetic energy of the emitted photoelectrons but will increase the photoelectric current by a factor of two.

The correct answer is:
increase the maximum kinetic energy of the ejected photoelectrons by a factor of more than two and would increase the photoelectric current by a factor of two.

Solution:
Using Einstein's photoelectric equations,

hv=ϕ+eVs

eVs=hv-ϕ....(i)

Thus, when the frequency of incident light radiation (photons) v increases to 2v, the stopping potential Vs changes to Vs

eVs'=h(2v)-ϕ....ii

On dividing Eq. (ii) by Eq. (ii), we get

Vs'Vs=2hv-ϕhv-ϕ=2hv-2ϕ+ϕhv-ϕ

=2(hv-ϕ)+ϕhv-ϕ=2+ϕhv-ϕ

∴ V'sVs>2⇒Vs'>2Vs

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2018