datavorous/entrance-exam-dataset huggingface dataset:
$\tan \left[\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right]=$

Options:
A $6 / 17$
B $17/6$
C $7 / 16$
D $16 / 7$

The correct answer is:
$17/6$

Solution:
$\begin{aligned} & \tan \left[\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right] \\ & =\tan \left[\tan ^{-1} \frac{\sqrt{\left(1-\frac{16}{25}\right)}}{\frac{4}{5}}+\tan ^{-1} \frac{2}{3}\right] \\ & =\tan \left[\tan ^{-1}\left(\left.\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right)\right]=\tan \cdot \left(\tan ^{-1} \frac{17}{6}\right)=\frac{17}{6}\right.\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan \left(\cos ^{-1} x\right)$ is equal to

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

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

Solution:
Let $\cos ^{-1} x=\theta$. Then $x=\cos \theta$
$\begin{aligned}
& \Rightarrow \tan \theta=\sqrt{\sec ^2 \theta-1}=\sqrt{\frac{1}{x^2}-1}=\sqrt{\frac{1-x^2}{x}} \\
& \therefore \tan \left(\cos ^{-1} x\right)=\tan \theta=\frac{\sqrt{1-x^2}}{x}
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan y=\frac{2 t}{1-t^2}$ and $\sin x=\frac{2 t}{1+t^2}$, then $\frac{d y}{d x}=$

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

The correct answer is:
1

Solution:
$\tan y=\frac{2 t}{1-t^2}$ $\cdots(i)$
and $\sin x=\frac{2 t}{1+t^2}$ $\cdots(ii)$
From (i), differentiating w.r.t. t of $y$, we get,
$\sec ^2 y \frac{d y}{d t}=\frac{2\left(1+t^2\right)}{\left(1-t^2\right)^2}$
$\frac{d y}{d t}=\frac{2\left(1+t^2\right)}{\left(1-t^2\right)^2} \cdot \frac{1}{\left(1+\tan ^2 y\right)}$
or $\frac{d y}{d t}=\frac{2\left(1+t^2\right)}{\left(1-t^2\right)^2} \cdot \frac{1}{\left[1+\left(\frac{2 t}{1-t^2}\right)^2\right]}=\frac{2}{1+t^2}$ $\cdots(iii)$
and from (i), differentiating w.r.t. $t$ of $x$,$y$
get
$\cos x \frac{d x}{d t}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2}$
or $\frac{d x}{d t}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \frac{1}{\sqrt{1-\frac{(2 t)^2}{\left(1+t^2\right)^2}}}=\frac{2}{1+t^2}$ $\cdots(iv)$
Hence $\frac{d y}{d x}=1$.

Topic: Mathematics, Differentiation, JEE Main
Tangent and normal are drawn at P16,16 on the parabola y2=16x, which intersect the axis of the parabola at A & B, respectively. If C is the center of the circle through the points P, A & B and ∠CPB=θ, then a value of tan⁡θ is:

Options:
A 43
B 12
C 2
D 3

The correct answer is:
2

Solution:
Equation of tangent at x1,y1 is yy1=2ax+x1
Equation of normal at x1,y1 is y-y1=-y12a(x-x1)
Here tangent and normal passes through P(16,16)
∴ Equation of tangent is 2y=x+16
Equation of normal is y+2x=48
Tangent and normal intersect x-axis at A-16,0 and B(24,0)
Since, APB is a right angled triangle, hence AB is the diameter for the circle.

Also mid point of AB is center of the circle.
Hence Centre =4,0
Slope of PC =43
Slope of PB=-2
tan⁡θ=43+21-2×43=10-5=2

Topic: Mathematics, Parabola, JEE Main, JEE Main 2018 (08 Apr)
Tangent at a point \( P \) on \( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \) meets the \( x \)-axis at \( A \) and \( y \)-axis at \( B \). The locus of the midpoint of \( A B \) is \( \frac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=k \), then find \( k \).

Options:

The correct answer is:
4.00

Solution:
xcosθa+ysinθb=1

h=acosθ+02,k=0+bsinθ2
cosθ=a2h, sinθ=b2k
cos2θ+sin2θ=1
a24h2+b24k2=1
∴ locus a2x2+b2y2=4

Topic: Mathematics, Ellipse, JEE Main
Tangent to a curve intersects the y-axis at a point P. A line perpendicular to this tangent through P passes through the point 1,0. The differential equation of the curve is

Options:
A ydydx-xdydx2=1
B xd2ydx2+dydx2=1
C ydxdy+x=1
D None of these

The correct answer is:
ydydx-xdydx2=1

Solution:
Equation of tangent at the point
R x, f x ââis, Yâ f x =fâ² x Xâx
Coordinates of the point are P(0, fx-xfâ²x)
The slope of the perpendicular line through P is
fx-xfâ²x-1=-1fâ²x
ydydx-x dydx2=1 is the differential equation.

Topic: Mathematics, Differential Equations, JEE Main
Tangential acceleration of a particle moving in a circle of radius 1 m varies with time t as shown in figure (initial velocity of the particle is zero). Time after which total acceleration of particle makes an angle of 30Â° with radial acceleration is

Options:
A 4 s

B 43 s

C 223 s

D 2 s

The correct answer is:
223 s

Solution:
Given that a t t = tan 60 o

⇒ at=3t …1
dv dt = 3 t

⇒ ∫ 0 v dv = 3 t dt
v = 3 t 2 2 ...(2)
Given that tan 30o=atac ⇒ 13=atac

⇒ ac=3at
v 2 R = 3 a t … 3
From (1) and (2) 3 2 t 2 2 1 = 3 × 3 t ⇒ 3 4 t 4 = 3 t
t 3 = 4

⇒ t = 2 2 / 3 sec.

Topic: Physics, Laws of Motion, JEE Main
Tangents are drawn at the end points of a normal chord of the parabola y2=4ax. The locus of their point of intersection is

Options:
A x-2ay2+4a3=0
B x-2ay2-4a3=0
C x+2ay2-4a3=0
D x+2ay2+4a3=0

The correct answer is:
x+2ay2+4a3=0

Solution:
Let, Pat12,2at1 and Qat22,2at2 be the end points of a normal chord of the parabola y2=4ax such that PQ is normal at P. Then,
t2=-t1-2t1 …(i)
Let, h,k be the point of intersection of tangents to the parabola at P and Q. Then,
h=at1t2 and k=at1+t2
⇒h=at1-t1-2t1 and k=-2at1. [Using (i)]
⇒h=-at12+2 and t1=-2ak
⇒h=-a4a2k2+2
⇒h+2ak2=-4a3
Hence, the locus of h,k is
x+2ay2=-4a3
or x+2ay2+4a3=0

Topic: Mathematics, Parabola, JEE Main
Tangents are drawn from any point on the directrix of y2=16x to the parabola. If the locus of the midpoint of chords of contact is a parabola, then its length (in units) of the latus rectum is

Options:

The correct answer is:
8

Solution:
Equation of the chord of contact PQ: T=S1

⇒yk-8x+h=k2-16h…….i
According to the property, the chord of contact w.r.t. any point on the directrix always passes through focus 4,0 i.e., 4,0 satisfy the equation i
⇒-84+h=k2-16h
⇒-32-8x=y2-16x
y2=8x-4
⇒ Length of the latus rectum is 8

units

Topic: Mathematics, Parabola, JEE Main
Tangents are drawn from the origin to the curve $\mathrm{y}=\cos \mathrm{x} .$ Their points of contact lie on

Options:
A $x^{2} y^{2}=y^{2}-x^{2}$
B $x^{2} y^{2}=x^{2}+y^{2}$
C $x^{2} y^{2}=x^{2}-y^{2}$
D None of these

The correct answer is:
$x^{2} y^{2}=x^{2}-y^{2}$

Solution:
Let $\left(x_{1}, y_{1}\right)$ be one of the points of contact. Given curve is $y=\cos x$

$$

\begin{array}{l}

\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\sin \mathrm{x} \\

\left.\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)}=-\sin \mathrm{x}_{1}

\end{array}

$$

Now the equation of the tangent at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ is

$$

\begin{aligned}

& y-y_{1}\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}\left(x-x_{1}\right) \\

\Rightarrow & y-y_{1}=-\sin x_{1}\left(0-x_{1}\right)

\end{aligned}

$$

Since, it is given that equation of tangent passes through origin.

$$

\begin{array}{lr}

\therefore & 0-y_{1}=-\sin x_{1}\left(0-x_{1}\right) \\

\Rightarrow & y_{1}=-x_{1} \sin x_{1}

\end{array}

$$

Also, point $\left(x_{1}, y_{1}\right)$ lies on $y=\cos x$

$$

\therefore \quad y_{1}=\cos x_{1}

$$

From Eqs. (i), (ii), we get

$$

\begin{aligned}

& \sin ^{2} x_{1}+\cos ^{2} x_{1}=\frac{y_{1}^{2}}{x_{1}^{2}}+y_{1}^{2}=1 \\

\Rightarrow & x_{1}^{2}=y_{1}^{2}+y_{1}^{2} x_{1}^{2}

\end{aligned}

$$

Hence, the locus of $\left(x_{1}, y_{1}\right)$ is

$$

x^{2}=y^{2}+y^{2} x^{2} \Rightarrow x^{2} y^{2}=x^{2}-y^{2}

$$

Topic: Mathematics, Application of Derivatives, JEE Main
Tangents are drawn from the point $(17,7)$ to the circle $x^2+y^2=169$.
Statement I The tangents are mutually perpendicular.
Statement II The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is $x^2+y^2=338$

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

Solution:
Since, the tangents are perpendicular.
So, locus of perpendicular tangents to circle $x^2+y^2=169$ is a director circle having equation
$$
x^2+y^2=338 .
$$

Topic: Mathematics, Circle, JEE Main
Tangents are drawn from the point α, β to the hyperbola 3x2-2y2=6 and are inclined at angles θ and ϕ to the x-axis. If tan⁡θtan⁡ϕ=2, then the value of 2α2-β2 is

Options:

The correct answer is:
7

Solution:
Given hyperbola is 3x2-2y2=6 or x22-y23=1
Slope form of tangent is y=mx±a2m2-b2or y-mx2=a2m2-b2

Tangent from the point α, β is given by,

β-mα2=2m2-3
⇒m2α2-2-2αβm+β2+3=0
m1m2=β2+3α2-2=2=tan⁡θtan⁡ϕ
∴β2+3=2α2-4⇒7=2α2-β2

Topic: Mathematics, Hyperbola, JEE Main
Tangents are drawn to a unit circle with centre at the origin from each point on the line x+y=2. Then, the area (in sq. units) of the locus of the mid-point of the chord of contact is

Options:
A π4
B π16
C π8
D 2π8

The correct answer is:
π8

Solution:
x1,y1 lies on x+y=2
âx1+y1=2â¦i
Chord of contact with respect to x1,y1
xx1+yy1=1
Also equation of chord whose mid point is h,k
h2+k2=hx+ky
â´x1h=y1k=1h2+k2
âx1=hh2+k2, y1=kh2+k2
Substituting in i, we get,
hh2+k2+kh2+k2=2
Hence, the required locus is x2+y2=x2+y2

The area of the circle is Ï 8 sq. units

Topic: Mathematics, Circle, JEE Main
Tangents are drawn to the circle x2+y2=16 at the points where it intersects the circle x2+y2-6x-8y-8=0, then the point of intersection of these tangents is

Options:
A 4,163
B 12,16
C 3,4
D 16,12

The correct answer is:
12,16

Solution:
S1 : x2+y2-16=0
S2 :x2+y2-6x-8y-8=0
In this case, the common chord is the same as the chord of contact AB

So, the equation of AB is 3x+4y-4=0 which is identical with xx1+yy1-16=0
x13=y14=164
⇒ x1=12, y1=16
⇒ Px1,y1=12,16

Topic: Mathematics, Circle, JEE Main
Tangents are drawn to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ at the ends of lattice-rectum the area of the quadrilateral so formed is

Options:
A $27$
B $\frac{27}{4}$
C $\frac{27}{2}$
D $\frac{27}{55}$

The correct answer is:
$27$

Solution:
$\begin{aligned} & \quad \frac{x^2}{5}+\frac{y^2}{5}=1 \\ & e=\sqrt{1-5 / 9} \\ & e=\frac{2}{3} \\ & a^2=5 ; \quad b^2=5 \\ & a=3\end{aligned}$
Equation of tangat at $\left(a e, b^2 / a\right) \equiv\left(2, \frac{1}{3}\right)$
$\begin{aligned} & \frac{x x_1}{9}+\frac{y y_1}{5}=1 \\ & \frac{x \cdot 2}{9}+\frac{y \cdot 5 / 3}{8}=1 \\ & \frac{2 x}{9}+\frac{y}{3}=1\end{aligned}$
Area of Quad $A B C D=\frac{1}{2} \cdot(A C) \cdot(B D)$
$\frac{1}{2} \cdot 9 \cdot 6=27$ sq.units.

Topic: Mathematics, Ellipse, JEE Main
Tangents are drawn to the hyperbola 4x2-y2=36 at the points P and Q. If these tangents intersect at the point T0, 3 then the area (in sq. units) of ΔPTQ is:

Options:
A 365
B 455
C 543
D 603

The correct answer is:
455

Solution:
Given hyperbola 4x2-y2=36
⇒x29-y236=1
Equation of chord of contact w.r.t. point T(0, 3) : xx19-yy136=1
x.09-y.336=1
⇒ y= -12
This intersect the hyperbola x29-14436=1
at x=±35
P35, -12 and Q-35, -12
Hence, area of ∆=12×65×15=455

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2018 (08 Apr)
Tangents at the extremities of any focal chord of a parabola intersect

Options:
A At right angles
B On the directrix
C On the tangents at vertex
D None of these

The correct answers are:
At right angles, On the directrix

Solution:
Let the tangent at \(\mathrm{P}\) (at \(\mathrm{t}^2\), 2at) on the parabola is ty \(=\mathrm{x}+\mathrm{at}^2\) and at the other end \(\left(\frac{a}{t^2}, \frac{-2 a}{t}\right)\) of the focal chord through \(P_i\)
\(-\frac{1}{\mathrm{t}} \mathrm{y}=\mathrm{x}+\frac{\mathrm{a}}{\mathrm{t}^2} \Rightarrow \mathrm{y}=-\mathrm{tx}-\frac{\mathrm{a}}{\mathrm{t}}\)
Product of the slopes \(=\frac{1}{t}(-t)=-1\) and the point of intersection lies on
\(\left(\frac{1}{t}+\mathrm{t}\right) \mathrm{x}+\mathrm{a}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)=\mathrm{o} \Rightarrow \mathrm{x}=-\mathrm{a}\), the directrix

Topic: Mathematics, Parabola, JEE Main
Tangents drawn from the point $(-8,0)$ to the parabola $y^2=8 x$ touch the parabola at $P$ and $Q$. If $F$ is the focus of the parabola, then the area of the triangle $P F Q$ (in sq. units) is equal to

Options:
A
48

B
32

C
24

D
64

The correct answer is:
48

Solution:
Equation of the chord of contact $P Q$ is given by:
$$
T=0
$$
or $T \equiv y y_1-4\left(x+x_1\right)$,
where $\left(x_1, y_1\right) \equiv(-8,0)$
$\therefore$ Equation becomes: $x=8$
$\&$ Chord of contact is $x=8$
$\therefore$ Coordinates of point $P$ and $Q$ are $(8,8)$ and $(8,-8)$
and focus of the parabola is $F(2,0)$
$\therefore$ Area of triangle $P Q F=\frac{1}{2} \times(8-2) \times(8+8)$ $=48$ sq. units

Topic: Mathematics, Parabola, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Tangents \( P A \) and \( P B \) are drawn to the circle \( x^{2}+y^{2}=8 \) from any arbitrary point \( P \) on the line \( x+y=4 \). The locus of the mid-point of chord of contact \( A B \) is

Options:
A \( x^{2}+y^{2}-2 x-2 y=0 \)
B \( x^{2}+y^{2}+2 x+2 y=0 \)
C \( x^{2}+y^{2}-2 x+2 y=0 \)
D \( x^{2}+y^{2}+2 x-2 y=0 \)

The correct answer is:
\( x^{2}+y^{2}-2 x-2 y=0 \)

Solution:
Let Pa, 4-a be the arbitrary point on the given line x+y=4
The equation of chord of contact AB is T=0
xa+y4-a=8 ....i

Also, using T=S1
The equation of chord AB whose mid point is Mh,k is
hx+ky=h2+k2 ....ii
Since i & ii are identical, on comparing their coefficients, we get,
ah=4-ak=8h2+k2=a+4-ah+k=4h+k

⇒4h2+k2=8h+k

Hence, the locus of Mh,k is x2+y2-2x-2y=0

Topic: Mathematics, Circle, JEE Main
Tangents to a circle at points $P$ and $Q$ on the circle intersect at a point $R$. If $P Q=6$ and $P R=5$ then the radius of the circle is

Options:
A $\frac{13}{3}$
B $4$
C $\frac{15}{4}$
D $\frac{16}{5}$

The correct answer is:
$\frac{15}{4}$

Solution:
In $\Delta R C P \Rightarrow \cos \theta=\frac{4}{5}$
In $\Delta \mathrm{PCO} \Rightarrow \cos \theta=\frac{3}{\mathrm{r}}$

Topic: Mathematics, Circle, JEE Main
Tangents to the parabola y2=4ax at Pat12,2at1 and Qat22,2at2 meet at T. If â³PTQ is right-angled at T, then 1PS+1QS is equal to (where, S is the focus of the given parabola)

Options:
A 1a
B 2a
C 12a
D 14a

The correct answer is:
1a

Solution:
T lies on the directrix

and PQ is the focal chord.

ât1t2=-1
1PS=1a1+t12, 1QS=1a1+t22=t12a1+t12
So, 1PS+1QS=1a

Topic: Mathematics, Parabola, JEE Main
Tautomerism is exhibited by

Options:
A $\left(\mathrm{CH}_3\right)_3 \mathrm{CNO}$
B $\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$
C $\mathrm{R}_3 \mathrm{CNO}_2$
D $\mathrm{RCH}_2 \mathrm{NO}_2$

The correct answer is:
$\mathrm{RCH}_2 \mathrm{NO}_2$

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Taxonomic studies consider a group of individual organisms with fundamental similarities as a

Options:
A genus
B species
C order
D phylum

The correct answer is:
species

Solution:
The classification of all living organism has been done by dividing them into different levels based on their features and characteristics. There is a hierarchy in this classification and each step represents a rank or category known as taxonomic category. There are around seven main taxonomic ranks or categories - species, genus, family, order, class, phylum/division and kingdom. Group of individuals having similar fundamental qualities are called species. When the group of related species have more common characters among them they are referred to as genus. Families which exhibit few similar characters are placed under the same Order. All the fishes, amphibians, reptiles, birds along with mammals constitute the taxonomic level phylum.

Topic: Biology, The Living World, NEET
Team A'' consists of 7 boys and n girls and Team B'' has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then n is equal to:

Options:
A 5
B 2
C 4
D 6

The correct answer is:
4

Solution:
Total matches between boys of both team=C17×C14=28

Total matches between girls of both team=C1n×C16=6n

Given, 28+6n=52

⇒n=4

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (17 Mar Shift 1)
Tears from eyes and interferons are included under which of the following barriers?

Options:
A Physiological and physical
B Physical and cytokine
C Physiological and cytokine
D Cellular and cytokine

The correct answer is:
Physiological and cytokine

Solution:
Innate immunity consists of 4 types of barriers:
I. Physical barriers
- Skin: It is the main barrier that prevents entry of the pathogens.
- Mucus: It covers the epithelium lining of respiratory, gastrointestinal and urogenital tracts so that the microbes may not enter our body.
II. Physiological barriers
These prevent microbial growth and include acid in the stomach, saliva in the mouth and tears.
III. Cellular barriers
Phagocytosis and destruction of microbes by certain types of WBCs and body cells,
A. Neutrophils (Polymorpho-nuclear leucocytes or PMNL), B. Monocytes, C. Natural killer (a type of lymphocytes), D. Tissue macrophages.
IV. Cytokine barriers
Interferons are proteins secreted by the virus-infected cells to protect the spread of infection to healthy or non-infected cells.

Topic: Biology, Human Health and Diseases, NEET
Television signals are

Options:
A Amplitude modulated
B Frequency modulated
C Phase modulated
D Both frequency and amplitude modulated

The correct answer is:
Both frequency and amplitude modulated

Solution:
Television signals are both frequency and amplitude modulated.

Topic: Physics, Communication System, JEE Main
Temperature and volume curves are drawn for two thermodynamic processes. For the first process, it is a straight line and for the second, it is a rectangular hyperbola. The ratio of work done in the first process to the work done in the second process is

Options:
A 2:1
B 3:1
C 4:1
D 3:2

The correct answer is:
2:1

Solution:
The first process is a constant pressure process, hence
w1 = nR(2T0 - T0) = nRT0
The equation for the second process is
T = c V
Hence, P = nRT V = nRc V 2
â´ w = â« V 0 2 V 0 PdV = nRT 0 2
â´ w 1 w 2 = 2 : 1
â´ 2 : 1

Topic: Physics, Thermodynamics, JEE Main
Temperature coefficient of a reaction is $2$. When temperature is increased from $30^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$, rate of the reaction increases by

Options:
A $128$ times
B $100$ times
C $500$ times
D $250$ times

The correct answer is:
$128$ times

Solution:
For every $10^{\circ} \mathrm{C}$ rise in temperature, rate increase by 2 times i.e.,
$$
\begin{array}{ll}
& r_{1}=2^{n} r \\
\text { where, } \quad & n=\frac{\Delta t}{10}=\frac{100-30}{10}=7 \\
\therefore \quad & r_{1}=2^{7} r \Rightarrow r_{1}=128 r
\end{array}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main
Temperature difference of 120oC is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB and length 3L2, is connected across AB (See figure). In steady state, temperature difference between P and Q will be close to:

Options:
A 45oC
B 35oC
C 60oC
D 75oC

The correct answer is:
45oC

Solution:
T1-T2=120oC
We define thermal resistance
R=LKA
Given circuit can be reduced to

Equivalent resistance Req=8R5
Thermal current i=T1-T28R5
TP-TQ=T1-T28R5 3R5
=38×120=45

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Temperature difference of 120oC is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB and length 3L2, is connected across AB (See figure). In steady state, temperature difference between P and Q will be close to:

Options:
A 45oC
B 35oC
C 60oC
D 75oC

The correct answer is:
45oC

Solution:
T1-T2=120oC
We define thermal resistance
R=LKA
Given circuit can be reduced to
Equivalent resistance Req=8R5
Thermal current i=T1-T28R5
TA-TB=T1-T28R5 3R5
=38Ã120=45

Topic: Physics, Thermal Properties of Matter, NEET
Temperature of an ideal gas is 300 K. The final temperature of the gas when its volume changes from V to 2V in the process p=Î±V (here Î± is a positive constant) is

Options:
A 900 K
B 1200 K
C 300 K
D 500 K

The correct answer is:
1200 K

Solution:
The given relation is pâV
Therefore, pâV
When V changes from V to 2V, pressure p is also doubled
For an ideal gas, pVT=constant
â´ TâpV. Hence, T becomes 2Ã2=4 times
T=4Ã300 K=1200 K

Topic: Physics, Thermodynamics, JEE Main
Temperature of cold junction in a thermocouple is $10^{\circ} \mathrm{C}$ and neutral temperature is $270^{\circ} \mathrm{C}$, then the temperature of inversion is

Options:
A $540^{\circ} \mathrm{C}$
B $530^{\circ} \mathrm{C}$
C $280^{\circ} \mathrm{C}$
D $260^{\circ} \mathrm{C}$

The correct answer is:
$530^{\circ} \mathrm{C}$

Solution:
Temperature of inversion is
$\begin{aligned} & T_i=2 T_n-T_0 \\ & \therefore \quad T_i=2 \times 270-10=530^{\circ} \mathrm{C} \\ & \end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
Ten chairs are numbered as 1 to 10 . Three women and two men wish to occupy one chair each. First the women choose the chairs marked 1 to 6 , then the men choose the chairs from the remaining. The number of possible ways is

Options:
A ${ }^6 P_3 \times{ }^4 P_2$
B ${ }^6 C_3 \times{ }^4 P_2$
C ${ }^6 P_3 \times{ }^4 C_2$
D ${ }^6 C_3 \times{ }^4 C_2$

The correct answer is:
${ }^6 P_3 \times{ }^4 P_2$

Solution:
We have, 3 women and 2 men and first
women choose the chairs marked 1 to 6 .
$\because$ Total number of chair $=6$
So, the number of arrangement $={ }^6 P_3$
Now, 2 men choose from the remaining 4 chairs.
So, the number of arrangement $={ }^4 p_2$ ways
$\therefore$ Total number of arrangements $={ }^6 p_3 \times{ }^4 p_2$

Topic: Mathematics, Permutation Combination, KCET, KCET 2023
Ten charges are placed on the circumference of a circle of radius R with constant angular separation between successive charges. Alternate charges1,3,5,7,9 have charge (+q) each, while 2,4,6,8,10 have charge (–q) each. The potential V and the electric field E at the centre of the circle are respectively : (TakeV=0 at infinity)

Options:
A V=10q4πϵ0R;E=0
B V=0,E=10q4πϵ0R2
C V=0; E=0
D V=10q4πϵ0R;E=10q4πϵ0R2

The correct answer is:
V=0; E=0

Solution:
vnet=5kqR+5k(-q)R
vnet=0, Qnet=0
Enet=0, by symmetry

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (05 Sep Shift 2)
Ten charges are placed on the circumference of a circle of radius \( R \) with constant angular separation between successive charges. Alternate charges \( 1,3,5,7,9 \) have charge \( (+q) \) each, while \( 2,4,6,8,10 \) have charge \( (-q) \) each. The potential V and the electric field \( \mathrm{E} \) at the centre of the circle are respectively : (Take \( V=0 \) at infinity)

Options:
A \( V=\frac{10 q}{4 \pi \epsilon_{0} R} ; E=0 \)
B \( \mathrm{V}=0, \mathrm{E}=\frac{10 \mathrm{q}}{4 \pi \epsilon_{0} \mathrm{R}^{2}} \)
C \( V=0 ; E=0 \)
D \( V=\frac{10 q}{4 \pi \epsilon_{0} R} ; E=\frac{10 q}{4 \pi \epsilon_{0} R^{2}} \)

The correct answer is:
\( V=0 ; E=0 \)

Solution:
Vnet=5kqR+5k(-q)R
⇒Vnet=0, Qnet=0
Enet=0, by symmetry

Topic: Physics, Electrostatics, JEE Main
Ten moles of an ideal gas are filled in a closed vessel. The vessel has cylinder and piston type arrangement and pressure of the gas remains constant at 0.821 atm. Which of the following graph represents correct variation of log V vs log T?
(V = Volume in litre and T = temperature in Kelvin)

Options:
A
B
C
D

The correct answer is:

Solution:
log⁡V∝log⁡T at constant pressure
PV=nRT
0.821V=10×0.082×T
VT= constant = 1
Slope of the line will be 1
i.e., angle will be 45° .

Topic: Chemistry, States of Matter, JEE Main
Ten points lie in a plane so that no three of them are collinear. The number of lines passing through exactly two of these points and dividing the plane into two regions each containing four of the remaining points is

Options:
A $1$
B $5$
C $10$
D dependent on the configuration of points

The correct answer is:
$5$

Solution:

Topic: Mathematics, Permutation Combination, JEE Main