datavorous/entrance-exam-dataset huggingface dataset:
The length of the latus-rectum of the parabola whose focus is \(\left(\frac{u^2}{2 g} \sin 2 \alpha,-\frac{u^2}{2 g} \cos 2 \alpha\right)\) and directrix is \(y=\frac{u^2}{2 g}\), is

Options:
A \(\frac{u^2}{g} \cos ^2 \alpha\)
B \(\frac{u^2}{g} \cos 2 \alpha\)
C \(\frac{2 u^2}{g} \cos ^2 2 \alpha\)
D \(\frac{2 u^2}{g} \cos ^2 \alpha\)

The correct answer is:
\(\frac{2 u^2}{g} \cos ^2 \alpha\)

Solution:
According to the figure, the length of latus rectum is
\(2(S M)=2 \times \frac{u^2}{2 g}(1+\cos 2 \alpha)=\frac{2 u^2 \cos ^2 \alpha}{g} .\)

Topic: Mathematics, Parabola, JEE Main
The length of the latus-rectum of the parabola whose focus is \(\left(\frac{u^2}{2 g} \sin 2 \alpha,-\frac{u^2}{2 g} \cos 2 \alpha\right)\) and directrix is \(y=\frac{u^2}{2 g}\), is

Options:
A \(\frac{u^2}{g} \cos ^2 \alpha\)
B \(\frac{u^2}{g} \cos 2 \alpha\)
C \(\frac{2 u^2}{g} \cos ^2 2 \alpha\)
D \(\frac{2 u^2}{g} \cos ^2 \alpha\)

The correct answer is:
\(\frac{2 u^2}{g} \cos ^2 \alpha\)

Solution:
According to the figure, the length of latus rectum is
\(2(S M)=2 \times \frac{u^2}{2 g}(1+\cos 2 \alpha)=\frac{2 u^2 \cos ^2 \alpha}{g} .\)

Topic: Mathematics, Parabola, BITSAT, BITSAT 2011
The length of the perpendicular drawn from the point $(3,-1,11)$ to the line $\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ is :

Options:
A $\sqrt{29}$
B $\sqrt{33}$
C $\sqrt{53}$
D $\sqrt{66}$

The correct answer is:
$\sqrt{53}$

Solution:
Let feet of perpendicular is

$(2 \alpha, 3 \alpha+2,4 \alpha+3)$

$\Rightarrow$ Direction ratio of the $\perp$ line is

$2 \alpha-3,3 \alpha+3,4 \alpha-8$. and

$\Rightarrow$ Direction ratio of the line are $2,3,4$

$\Rightarrow 2(2 \alpha-3)+3(3 \alpha+3)+4(4 \alpha-8)=0$

$\Rightarrow \alpha=1$

$\Rightarrow$ Feet of $\perp$ is $(2,5,7)$

$\Rightarrow$ Length $\perp$ is $\sqrt{1^{2}+6^{2}+4^{2}}=\sqrt{53}$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2021
The length of the tangent drawn from any point on the circle $x^{2}+y^{2}+2 f y+\lambda=0$ to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{fy}+\mu=0,$ where $\mu>\lambda>0,$ is

Options:
A $\sqrt{\mu-\lambda}$
B $\sqrt{\mu+\lambda}$
C $\sqrt{\mu^{2}-\lambda^{2}}$
D $\mu+\lambda$

The correct answer is:
$\sqrt{\mu-\lambda}$

Solution:
Let the radius of the first circle be $\mathrm{CT}=\mathrm{r}_{1}$. Also, let the radius of the second circle be $\mathrm{CP}=\mathrm{r}_{2}$
In the triangle $\mathrm{PCT}, \mathrm{T}$ is a right angle

$\begin{aligned} \text { So, } P T &=\sqrt{P C^{2}-C T^{2}}=\sqrt{r_{1}^{2}-r_{2}^{2}} \\ &=\sqrt{\left(f^{2}-\lambda\right)-\left(f^{2}-\mu\right)}=\sqrt{\mu-\lambda} \end{aligned}$

Topic: Mathematics, Circle, JEE Main
The length of the tangent drawn from any point on the circle $x^{2}+y^{2}+2 f y+\lambda=0$ to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{fy}+\mu=0,$ where $\mu>\lambda>0,$ is

Options:
A $\sqrt{\mu-\lambda}$
B $\sqrt{\mu+\lambda}$
C $\sqrt{\mu^{2}-\lambda^{2}}$
D $\mu+\lambda$

The correct answer is:
$\sqrt{\mu-\lambda}$

Solution:
Let the radius of the first circle be $\mathrm{CT}=\mathrm{r}_{1}$. Also, let the radius of the second circle be $\mathrm{CP}=\mathrm{r}_{2}$
In the triangle $\mathrm{PCT}, \mathrm{T}$ is a right angle

$\begin{aligned} \text { So, } P T &=\sqrt{P C^{2}-C T^{2}}=\sqrt{r_{1}^{2}-r_{2}^{2}} \\ &=\sqrt{\left(f^{2}-\lambda\right)-\left(f^{2}-\mu\right)}=\sqrt{\mu-\lambda} \end{aligned}$

Topic: Mathematics, Circle, BITSAT, BITSAT 2012
The lengths of the tangent drawn from any point

on the circle $15 x^{2}+15 y^{2}-48 x+64 y=0$ to the two circles $5 x^{2}+5 y^{2}-24 x+32 y+75=0$ and $5 x^{2}+5 y^{2}-48 x$ $+64 y+300=0$ are in the ratio of

Options:
A 1: 2
B 2: 3
C 3: 4
D None

The correct answer is:
1: 2

Solution:
Let \(\mathrm{P}(\mathrm{h}, \mathrm{k})\) be a point on the circle

\(15 x^{2}+15 y^{2}-48 x+64 y=0\)

Then the lengths of the tangents from \(\mathrm{P}(\mathrm{h}, \mathrm{k})\) to

\(\begin{aligned} &5 x^{2}+5 y^{2}-24 x+32 y+75=0 \\ &5 x^{2}+5 y^{2}-48 x+64 y+300=0 \text { are } \\ &P T_{1}=\sqrt{h^{2}+k^{2}-\frac{24}{5} h+\frac{32}{5} k+15} \\ &\text { and } P T_{2}=\sqrt{h^{2}+k^{2}-\frac{48}{5} h+\frac{64}{5} k+60} \\ &\text { or } P T_{1}=\sqrt{\frac{48}{15} h-\frac{64}{15} k-\frac{24}{5} h+\frac{32}{5} k+15}=\sqrt{\frac{32}{15} k-\frac{24}{15} h+15} \end{aligned}\)

(Since \((\mathrm{h}, \mathrm{k})\) lies on \(15 \mathrm{x}^{2}-15 \mathrm{y}^{2}-48 \mathrm{x}+64 \mathrm{y}=0\)

\(\begin{aligned} &\left.\therefore \mathrm{h}^{2}+\mathrm{k}^{2}-\frac{48}{15} \mathrm{~h}+\frac{64}{15} \mathrm{k}=0\right) \\ &\text { and } \mathrm{PT}_{2}=\sqrt{\frac{48}{15} \mathrm{~h}-\frac{64}{5} \mathrm{k}-\frac{48}{5} \mathrm{~h}+\frac{64}{5} \mathrm{k}+60} \\ &=\sqrt{-\frac{96}{15} \mathrm{~h}+\frac{128}{15} \mathrm{k}+60}=2 \sqrt{-\frac{24}{15} \mathrm{~h}+\frac{32}{15} \mathrm{k}+15}=2 \mathrm{PT}_{1} \\ &\Rightarrow \mathrm{PT}_{1}: \mathrm{PT}_{2}=1: 2 \end{aligned}\)

Topic: Mathematics, Circle, BITSAT, BITSAT 2016
The letters of the word TOUGH are written in all possible orders and these words are written out as in a dictionary, then the rank of the word TOUGH is

Options:
A 120
B 88
C 89
D 90

The correct answer is:
89

Solution:
\(\begin{aligned}
& \text { Rank }=(4 ! \times 3)+(3 ! \times 2)+(2 ! \times 2)+1 \\
& =72+12+4+1=89
\end{aligned}\)

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2009
The line \(3 x-4 y+7=0\) is rotated through an angle \(\frac{\pi}{4}\) in the clockwise direction about the point \((-1,1)\). The equation of the line in its new position is

Options:
A \(7 y+x-6=0\)
B \(7 y-x-6=0\)
C \(7 y+x+6=0\)
D \(7 y-x+6=0\)

The correct answer is:
\(7 y+x-6=0\)

Solution:
As \((-1,1)\) is a point on \(3 x-4 y+7=0\), the rotation is possible. Slope of the given line \(=3 / 4\). Slope of the line in its new position
\(=\frac{\frac{3}{4}-1}{1+\frac{3}{4}}=-\frac{1}{7}\)
The required equation is
\(\mathrm{y}-1=-\frac{1}{7}(\mathrm{x}+1) \text { or } 7 \mathrm{y}+\mathrm{x}-6=0\)

Topic: Mathematics, Parabola, BITSAT, BITSAT 2009
The line \(a x+b y=1\) cuts ellipse \(c x^2+d y^2=1\) only once if

Options:
A \(\mathrm{ca}^2+\mathrm{db}^2=1\)
B \(\frac{\mathrm{c}}{\mathrm{a}^2}+\frac{\mathrm{d}}{\mathrm{b}^2}=1\)
C \(\frac{\mathrm{a}^2}{\mathrm{c}}+\frac{\mathrm{b}^2}{\mathrm{~d}}=1\)
D \(a c^2+b d^2=1\)

The correct answer is:
\(\frac{\mathrm{a}^2}{\mathrm{c}}+\frac{\mathrm{b}^2}{\mathrm{~d}}=1\)

Solution:
Clearly \(a x+b y=1\)
i.e \(y=-\frac{a}{b} x+\frac{1}{b}\) is tangent to
\(\begin{aligned}
& \mathrm{cx}^2+\mathrm{dy}^2=1 \Rightarrow \frac{\mathrm{x}^2}{\frac{1}{c}}+\frac{\mathrm{y}^2}{\frac{1}{d}}=1 \\
& \therefore\left(\frac{1}{b}\right)^2=\left(\frac{1}{c}\right)\left(-\frac{a}{b}\right)^2+\left(\frac{1}{d}\right) \\
& \Rightarrow 1=\frac{a^2}{c}+\frac{b^2}{d}
\end{aligned}\)

Topic: Mathematics, Circle, BITSAT, BITSAT 2010
The line joining (5,0) to $((10 \cos \theta, 10 \sin \theta)$ is divided internally in the ratio 2: 3 at $P$. If $\theta$ varies, then the locus of $\mathrm{P}$ is

Options:
A a pair of straight lines
B a circle
C a straight line
D None of these

The correct answer is:
a circle

Solution:
Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be the point dividing the join of $\mathrm{A}$ and $\mathrm{B}$ in the ratio 2: 3 internally, then

$$

\begin{array}{c}

\mathrm{x}=\frac{20 \cos \theta+15}{5}=4 \cos \theta+3 \Rightarrow \cos \theta=\frac{\mathrm{x}-3}{4} \\

\mathrm{y}=\frac{20 \sin \theta+0}{5}=4 \sin \theta \Rightarrow \sin \theta=\frac{\mathrm{y}}{4} \ldots \text { (ii) }

\end{array}

$$

Squaring and adding (i) and (ii), we get the

required locus $(x-3)^{2}+y^{2}=16,$ which is

a circle.

Topic: Mathematics, Circle, BITSAT, BITSAT 2016
The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x2+9y2=9 meets its auxiliary circle at the point M. Then, the area of the triangle with vertices at A, M and the origin O is

Options:
A 3110
B 2910
C 2110
D 2710

The correct answer is:
2710

Solution:
Equation of auxiliary circle is

x2+y2=9 ..i

∴ Equation of AM is x3+y1=1 ...ii

On solving Eqs. (i) and (ii), we get

M-125,95

Now, area of ΔAOM=12OA×MN=2710 sq units

Hence, option (d) is correct.

Topic: Mathematics, Ellipse, BITSAT, BITSAT 2018
The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x2+9y2=9 meets its auxiliary circle at the point M. Then, the area of the triangle with vertices at A, M and the origin O is

Options:
A 3110
B 2910
C 2110
D 2710

The correct answer is:
2710

Solution:
Equation of auxiliary circle is

x2+y2=9 ..i

∴ Equation of AM is x3+y1=1 ...ii

On solving Eqs. (i) and (ii), we get

M-125,95

Now, area of ΔAOM=12OA×MN=2710 sq units

Hence, option (d) is correct.

Topic: Mathematics, Ellipse, JEE Main
The line which is parallel to X-axis and crosses the curve $y=\sqrt{x}$ at an angle of $45^{\circ}$, is

Options:
A $x=\frac{1}{4}$
B $y=\frac{1}{4}$
C $y=\frac{1}{2}$
D $y=1$

The correct answer is:
$y=\frac{1}{2}$

Solution:
Given equation of a line parallel to X-axis is $\mathrm{y}=\mathrm{k}$

Given equation of the curve is $y=\sqrt{x}$, On solving equation of line with the equation of curve, we get $x=k^{2}$ Thus the intersecting point is $\left(\mathrm{k}^{2}, \mathrm{k}\right)$ It is given that the line $\mathrm{y}=\mathrm{k}$ intersect the curve $y=\sqrt{x}$ at an angle of $\pi / 4$. This means that the slope of the tangent to

$$

\mathrm{y}=\sqrt{\mathrm{x}} \text { at }\left(\mathrm{k}^{2}, \mathrm{k}\right) \text { is } \tan \left(\pm \frac{\pi}{4}\right)=\pm 1

$$

$$

\begin{array}{l}

\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}=\pm 1 \Rightarrow\left(\frac{1}{2 \sqrt{\mathrm{x}}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}=\pm 1 \\

\Rightarrow \mathrm{k}=\pm \frac{1}{2}

\end{array}

$$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2015
The line $y=m x$ bisects the area enclosed by lines $x$ $=0, \mathrm{y}=0$ and $\mathrm{x}=3 / 2$ and the curve $\mathrm{y}=1+4 \mathrm{x}-\mathrm{x}^{2}$

Then the value of $\mathrm{m}$ is

Options:
A $\frac{13}{6}$
B $\frac{13}{2}$
C $\frac{13}{5}$
D $\frac{13}{7}$

The correct answer is:
$\frac{13}{6}$

Solution:
$y=1+4 x-x^{2}=5-(x-2)^{2}$

We have $\int_{0}^{3 / 2}\left(1+4 x-x^{2}\right) \mathrm{dx}=2 \int_{0}^{3 / 2} \operatorname{mx} \mathrm{dx}$

$=\frac{3}{2}+2\left(\frac{9}{4}\right)-\frac{1}{3}\left(\frac{27}{8}\right)=\mathrm{m} \cdot \frac{9}{4}$

On solving we get $\mathrm{m}=\frac{13}{6}$

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2020
The lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar if

Options:
A $\mathrm{k}=3$ or -2
B $\mathrm{k}=0$ or -1
C $\mathrm{k}=1$ or -1
D $\mathrm{k}=0$ or -3

The correct answer is:
$\mathrm{k}=0$ or -3

Solution:
Two planes are coplanar if
$\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{array}\right|=0$
$\left|\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -k \\ k & 2 & 1\end{array}\right|=0$
Applying $\mathrm{C}_2 \rightarrow \mathrm{C}_2+\mathrm{C}_1, \mathrm{C}_3 \rightarrow \mathrm{C}_3+\mathrm{C}_1$
$\Rightarrow\left|\begin{array}{ccc}1 & 0 & 0 \\ 1 & 2 & 1-k \\ k & k+2 & 1+k\end{array}\right|=0$
$\Rightarrow \quad 1[2+2 k-(k+2)(1-k)]=0$
$\Rightarrow \quad 2+2 k-\left(-k^2-k+2\right)=0$
$k^2+3 k=0 \Rightarrow k(k+3)=0$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The load versus elongation graph for four wires is shown. The thinnest wire is

Options:
A $P$
B $Q$
C $R$
D $S$

The correct answer is:
$S$

Solution:
Using \(\quad \mathrm{Y}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \mathrm{L} / \mathrm{L}}\)

Cross-section area of the wire \(\quad \mathrm{A}=\pi \mathrm{R}^{2}\) where \(\quad \mathrm{R}\) is the radius of the wire \(\therefore \quad \mathrm{Y}=\frac{\mathrm{F}}{\pi \mathrm{R}^{2}} \times \frac{\mathrm{L}}{\Delta \mathrm{L}}\)

\(\Longrightarrow \mathrm{F}=\frac{\pi \mathrm{R}^{2} \mathrm{Y}}{\mathrm{L}} \Delta \mathrm{L}\)

Thus Slope of load vs elongation graph, \(\mathrm{s}=\frac{\pi \mathrm{R}^{2} \mathrm{Y}}{\mathrm{L}}\)

\(\Longrightarrow \quad \mathrm{s} \propto \mathrm{R}^{2}\)

Now from the graph shown, OA has the least slope and hence the least radius. Thus \(\mathrm{OA}\) represents the thinnest wire.

Topic: Physics, Mechanical Properties of Solids, JEE Main
The load versus elongation graph for four wires is shown. The thinnest wire is

Options:
A $P$
B $Q$
C $R$
D $S$

The correct answer is:
$S$

Solution:
Using \(\quad \mathrm{Y}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \mathrm{L} / \mathrm{L}}\)

Cross-section area of the wire \(\quad \mathrm{A}=\pi \mathrm{R}^{2}\) where \(\quad \mathrm{R}\) is the radius of the wire \(\therefore \quad \mathrm{Y}=\frac{\mathrm{F}}{\pi \mathrm{R}^{2}} \times \frac{\mathrm{L}}{\Delta \mathrm{L}}\)

\(\Longrightarrow \mathrm{F}=\frac{\pi \mathrm{R}^{2} \mathrm{Y}}{\mathrm{L}} \Delta \mathrm{L}\)

Thus Slope of load vs elongation graph, \(\mathrm{s}=\frac{\pi \mathrm{R}^{2} \mathrm{Y}}{\mathrm{L}}\)

\(\Longrightarrow \quad \mathrm{s} \propto \mathrm{R}^{2}\)

Now from the graph shown, OA has the least slope and hence the least radius. Thus \(\mathrm{OA}\) represents the thinnest wire.

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2014
The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2=6 on any tangent to it is

Options:
A x2-y22=6x2+2y2
B x2-y22=6x2-2y2
C x2+y22=6x2+2y2
D x2+y22=6x2-2y2

The correct answer is:
x2+y22=6x2+2y2

Solution:
We have,

x2+3y2=6

⇒x26+y22=1

Centre of an ellipse is 0,0.

Let the foot of perpendicular be (h,k).

Equation of tangent in slope form to the standard ellipse x2a2+y2b2=1 is

y=mx±a2m2+b2.

Hence, equation of tangent with slope m passing (h,k) is

y=mx±6m2+2

⇒k=mh±6m2+2, where m=-hk.

⇒k=h-hk±6m2+2

⇒6h2k2+2=h2+k2k

⇒6h2+2k2=h2+k22

So, required locus is

6x2+2y2=x2+y22.

Topic: Mathematics, Ellipse, BITSAT, BITSAT 2019
The locus of the mid-point of the chord if contact of tangents drawn from points lying on the straight line $4 x-5 y=20$ to the circle $x^2+y^2=9$ is

Options:
A
$20\left(x^2+y^2\right)-36 x+45 y=0$

B
$20\left(x^2+y^2\right)+36 x-45 y=0$

C
$36\left(x^2+y^2\right)-20 x+45 y=0$

D
$36\left(x^2+y^2\right)+20 x-45 y=0$

The correct answer is:

$20\left(x^2+y^2\right)-36 x+45 y=0$

Solution:
Let $P\left(t, \frac{4 t-20}{5}\right)$ be a point on the line
$4 x-5 y=20$. Then the chord of contact of tangents drawn from $P$ to the circle $x^2+y^2=9$ is
$t x+\left(\frac{4 t-20}{5}\right) y=9\quad \quad \dots(i)$
Let $\theta(h, k)$ be the mid-point of this chord of contact, then its equation is also $h x+k y=h^2+k^2\quad \quad \dots(ii)$
$\left[\right.$ using $\left.T=S^{\prime}\right]$
Clearly Eqs. (i) and (ii) represent the same line
$\begin{aligned}
& \therefore \frac{t}{h}=\frac{4 t-20}{5 k}=\frac{9}{h^2+k^2} \\
& \Rightarrow \frac{t}{h}=\frac{9}{h^2+k^2} \text { and } \frac{t}{h}=\frac{4 t-20}{5 k} \\
& \Rightarrow t=\frac{9 h}{h^2+k^2} \text { and } t=\frac{20 h}{4 h-5 k} \\
& \Rightarrow \frac{9 h}{h^2+k^2}=\frac{20 h}{4 h-5 k} \\
& \Rightarrow h\left\{20\left(h^2+k^2\right)-36 h+45 k\right\}=0 \\
& x=0, \text { or }\left[20\left(x^2+y^2\right)-36 x+45 y\right]=0
\end{aligned}$

Topic: Mathematics, Circle, BITSAT, BITSAT 2022
The locus of the point of intersection of two tangents to the parabola $y^{2}=4 a x,$ which are at right angle to one another is

Options:
A $x^{2}+y^{2}=a^{2}$
B $a y^{2}=x$
C $x+a=0$
D $x+y \pm a=0$

The correct answer is:
$x+a=0$

Solution:
Let the two tangents to the parabola \(y^{2}=4 a x\) be \(P T\) and \(Q T\) which are at right angle to one another at \(\mathrm{T}(\mathrm{h}, \mathrm{k})\).

Then we have a find the locus of \(\mathrm{T}(\mathrm{h}, \mathrm{k})\).

We know that \(y=m x+\frac{a}{m}\), where \(m\) is the slope is the equation of tangent to the parabola \(\mathrm{y}^{2}=4 \mathrm{ax}\) for all \(\mathrm{m}\).

Since this tangent to the parabola will pass through \(\mathrm{T}(\mathrm{h}, \mathrm{k})\), so

\(\mathrm{k}=\mathrm{mh}+\frac{\mathrm{a}}{\mathrm{m}} ; \text { or } \mathrm{m}^{2} \mathrm{~h}-\mathrm{mk}+\mathrm{a}=0\)

This is a quadratic equation in \(m\), so will have two roots, say \(m_{1}\) and \(m_{2}\), then \(\mathrm{m}_{1}+\mathrm{m}_{2}=\frac{\mathrm{k}}{\mathrm{h}}\), and \(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=\frac{\mathrm{a}}{\mathrm{h}}\)

Given that the two tangents intersect at right angle so

\(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1 \text { or } \frac{\mathrm{a}}{\mathrm{h}}=-1 \text { or } \mathrm{h}+\mathrm{a}=0\)

The locus of \(\mathrm{T}(\mathrm{h}, \mathrm{k})\) is \(\mathrm{x}+\mathrm{a}=0\), which is the equation of directrix.

Topic: Mathematics, Parabola, JEE Main
The locus of the point of intersection of two tangents to the parabola $y^{2}=4 a x,$ which are at right angle to one another is

Options:
A $x^{2}+y^{2}=a^{2}$
B $a y^{2}=x$
C $x+a=0$
D $x+y \pm a=0$

The correct answer is:
$x+a=0$

Solution:
Let the two tangents to the parabola \(y^{2}=4 a x\) be \(P T\) and \(Q T\) which are at right angle to one another at \(\mathrm{T}(\mathrm{h}, \mathrm{k})\).

Then we have a find the locus of \(\mathrm{T}(\mathrm{h}, \mathrm{k})\).

We know that \(y=m x+\frac{a}{m}\), where \(m\) is the slope is the equation of tangent to the parabola \(\mathrm{y}^{2}=4 \mathrm{ax}\) for all \(\mathrm{m}\).

Since this tangent to the parabola will pass through \(\mathrm{T}(\mathrm{h}, \mathrm{k})\), so

\(\mathrm{k}=\mathrm{mh}+\frac{\mathrm{a}}{\mathrm{m}} ; \text { or } \mathrm{m}^{2} \mathrm{~h}-\mathrm{mk}+\mathrm{a}=0\)

This is a quadratic equation in \(m\), so will have two roots, say \(m_{1}\) and \(m_{2}\), then \(\mathrm{m}_{1}+\mathrm{m}_{2}=\frac{\mathrm{k}}{\mathrm{h}}\), and \(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=\frac{\mathrm{a}}{\mathrm{h}}\)

Given that the two tangents intersect at right angle so

\(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1 \text { or } \frac{\mathrm{a}}{\mathrm{h}}=-1 \text { or } \mathrm{h}+\mathrm{a}=0\)

The locus of \(\mathrm{T}(\mathrm{h}, \mathrm{k})\) is \(\mathrm{x}+\mathrm{a}=0\), which is the equation of directrix.

Topic: Mathematics, Parabola, BITSAT, BITSAT 2016
The logic operations performed by the given digital circuit is equivalent to:

Options:
A
AND
B
NOR
C
OR
D
NAND

The correct answer is:
AND

Solution:
$\begin{aligned} & \mathrm{Y}=\overline{\mathrm{Z}}=(\mathrm{A}+\mathrm{B}) \cdot(\mathrm{A} \cdot \mathrm{B})=\mathrm{A} \cdot \mathrm{A} \cdot \mathrm{B}+\mathrm{AB} \cdot \mathrm{B} \\ & \mathrm{Y}=\mathrm{AB}+\mathrm{AB}=\mathrm{A} \cdot \mathrm{B} \\ & \therefore \text { It is an } \mathrm{AND} \text { Gate. }\end{aligned}$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The magnetic field at the centre O of the current-carrying square loop shown in the figure is

Options:
A 12×10-5 T
B 4×10-5 T
C 42×10-5 T
D 22×10-5 T

The correct answer is:
42×10-5 T

Solution:
Given, current through the square loop is 2 A.

I=2 A

The given situation is shown in the figure. Magnetic field due to wire AB at the centre O is

BAB=μ04π·Irsin45°+sin45°

=10-7×22×10-212+12

=2×10-5 T [Downward to the plane of square loop]

Since, magnetic field due to side AB, BC, CD and DA is same in magnitude and direction.

Hence, net magnetic field at O

B=4×BAB=42×10-5 T

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2019
The magnetic field at the origin due to the current flowing in the wire is $-$

Options:
A $-\frac{\mu_{0} \mathrm{I}}{8 \pi \mathrm{a}}(\hat{\mathrm{i}}+\hat{\mathrm{k}})$
B $\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{a}}(\hat{\mathrm{i}}+\hat{\mathrm{k}})$
C $\frac{\mu_{0} \mathrm{I}}{8 \pi \mathrm{a}}(-\hat{\mathrm{i}}+\hat{\mathrm{k}})$
D $\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a} \sqrt{2}}(\hat{\mathrm{i}}-\hat{\mathrm{k}})$

The correct answer is:
$\frac{\mu_{0} \mathrm{I}}{8 \pi \mathrm{a}}(-\hat{\mathrm{i}}+\hat{\mathrm{k}})$

Solution:
$\mathrm{B}_{\mathrm{OD}}=0, \mathrm{~B}_{\mathrm{OB}}=0$
$\mathrm{~B}_{\mathrm{AB}}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a} \sqrt{2}}\left[\cos 45^{\circ}(-\hat{\mathrm{i}})+\cos 45^{\circ} \hat{\mathrm{k}}\right]$
$=\frac{\mu_{0} \mathrm{I}}{8 \pi \mathrm{a}}(-\hat{\mathrm{i}}+\hat{\mathrm{k}})$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2010
The magnetic field in a travelling electromagnetic wave has a peak value of $20 \mathrm{nT}$. The peak value of electric field strength is

Options:
A $3 \mathrm{~V} / \mathrm{m}$
B $6 \mathrm{~V} / \mathrm{m}$
C $9 \mathrm{~V} / \mathrm{m}$
D $12 \mathrm{~V} / \mathrm{m}$

The correct answer is:
$6 \mathrm{~V} / \mathrm{m}$

Solution:
$\overrightarrow{\mathrm{E}}_{0}=\overrightarrow{\mathrm{B}}_{0} \times \overrightarrow{\mathrm{C}}$
$$
\left|\overrightarrow{\mathrm{E}}_{0}\right|=\left|\mathrm{B}_{0}\right| \cdot|\overrightarrow{\mathrm{C}}|=20 \times 10^{-9} \times 3 \times 10^{8}=6 \mathrm{~V} / \mathrm{m}
$$

Topic: Physics, Electromagnetic Waves, BITSAT, BITSAT 2009
The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by:

Options:
A $\vec{\mu}_L=\frac{e \vec{L}}{2 m}$
B $\vec{\mu}_L=-\frac{e \vec{L}}{2 m}$
C $\vec{\mu}_l=-\frac{e \vec{L}}{m}$
D $\vec{\mu}_l=\frac{2 e \vec{L}}{m}$

The correct answer is:
$\vec{\mu}_L=\frac{e \vec{L}}{2 m}$

Solution:
As $\overrightarrow{\mathrm{M}}=\mathrm{IA}$
$\Rightarrow|\overrightarrow{\mathrm{M}}|=\frac{\mathrm{e}}{\frac{2 \pi \mathrm{R}}{\mathrm{v}}} \pi \mathrm{R}^2$ $\left[\because \mathrm{I}=\frac{\mathrm{Q}}{\mathrm{T}}=\frac{\mathrm{e}}{\frac{2 \pi \mathrm{R}}{\mathrm{V}}}\right]$
$\Rightarrow|\vec{M}|=\frac{1}{2} e v R \Rightarrow|\vec{M}|=\frac{\mathrm{mvR}}{1} \cdot \frac{\mathrm{e}}{2 \mathrm{~m}}$
$\Rightarrow|\overrightarrow{\mathrm{M}}|=\frac{\mathrm{eL}}{2 \mathrm{~m}} \Rightarrow|\overrightarrow{\mathrm{M}}|=-\frac{\mathrm{e} \overrightarrow{\mathrm{L}}}{2 \mathrm{~m}}$
[ $\because$ Here $\overrightarrow{\mathrm{M}}$ and $\overrightarrow{\mathrm{L}}$ will always be opposite]

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The magnetic moments associated with two closely wound circular coils A and B of radius $r_A=10 \mathrm{~cm}$ and $r_B=20 \mathrm{~cm}$ respectively are equal if: (Where $\mathrm{N}_A, \mathrm{I}_A$ and $\mathrm{N}_{\mathrm{B}}, \mathrm{I}_{\mathrm{B}}$ are number of turn and current of $A$ and $B$ respectively)

Options:
A $2 \mathrm{~N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}$
B $\mathrm{N}_{\mathrm{A}}=2 \mathrm{~N}_{\mathrm{B}}$
C $\mathrm{N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=4 \mathrm{~N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}$
D $4 \mathrm{~N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}$

The correct answer is:
$\mathrm{N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=4 \mathrm{~N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}$

Solution:
The magnetic moment associated with circular coil,
$\begin{aligned}
& M=N I A \\
& M_A=M_B \\
& \therefore N_A I_A A_A=N_B I_B A_B \\
& \therefore N_A I_A \pi(0.1)^2=N_B I_B \pi(0.2)^2\left(\because A=\pi r^2\right) \\
& \therefore N_A I_A=4 N_B I_B
\end{aligned}$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The magnifying power of a telescope is $9 .$ When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be $20 \mathrm{~cm}$. The focal length of lenses are

Options:
A $18 \mathrm{~cm}, 2 \mathrm{~cm}$
B $11 \mathrm{~cm}, 9 \mathrm{~cm}$
C $10 \mathrm{~cm}, 10 \mathrm{~cm}$
D $15 \mathrm{~cm}, 5 \mathrm{~cm}$

The correct answer is:
$18 \mathrm{~cm}, 2 \mathrm{~cm}$

Solution:
$\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}=9, \quad \therefore \mathrm{f}_{0}=9 \mathrm{f}_{\mathrm{e}}$

Also $\mathrm{f}_{0}+\mathrm{f}_{\mathrm{e}}=20(\because$ final image is at infinity $)$ $9 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=20, \mathrm{f}_{\mathrm{e}}=2 \mathrm{~cm}, \quad \therefore \mathrm{f}_{0}=18 \mathrm{~cm}$

Topic: Physics, Ray Optics, BITSAT, BITSAT 2014
The major product ' $\mathrm{Y}^{\prime}$ in the following reaction is :

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, BITSAT, BITSAT 2021