datavorous/entrance-exam-dataset huggingface dataset:
The conversion of a moving coil galvanometer into a voltmeter is done by

Options:
A introducing a resistance of large value in series.
B introducing a resistance of small value in parallel
C introducing a resistance of large value in parallel.
D introducing a resistance of small value in series.

The correct answer is:
introducing a resistance of large value in series.

Solution:
Galvanometer is a very sensitive instrument, therefore it can not measure high potential difference. In order to convert a Galvanometer into voltmeter, a very high resistance known as "series resistance" is connected in series with the galvanometer.

Topic: Physics, Magnetic Properties of Matter, JEE Main
The conversion of benzene diazonium chloride to bromobenzene can be accomplished by

Options:
A
Reimer-Tiemann reaction

B
Friedel-Crafts reaction

C
Gattermann reaction

D
Azo-coupling reaction

The correct answer is:
Gattermann reaction

Solution:
Diazonium salts react with copper powder and hydrogen bromide to form bromobenzene. This reaction is known as Gattermann reaction.

Topic: Chemistry, Amines, JEE Main, JEE Main 2012 (12 May Online)
The coordinate axes is rotated and shifted in such a way that the IVth quadrant direction of line 4x+3y-35=0 becomes the new positive x-axis direction and the Ist quadrant direction of line 3x-4y+5=0 becomes the new positive y-axis direction. If origin as per old coordinate system is O, then according to the new coordinate system, the coordinates of O are

Options:
A 1,7
B -1,7
C 1,-7
D 7,-1

The correct answer is:
1,-7

Solution:
Distance of 0,0 from lines 4x+3y-35=0 (new x -axis) and 3x-4y+5=0 (new y -axis) are 7 units and 1 unit respectively.
Since, according to the new coordinate system O lies in IVth quadrant, hence the new coordinates of O are 1,-7

Topic: Mathematics, Straight Lines, JEE Main
The co-ordinates of a moving particle at any time ' $t$ ' are given by $x=\alpha t^3$ and $y=\beta t^3$. The speed of the particle at time ' $t$ ' is given by

Options:
A
$3 t \sqrt{\alpha^2+\beta^2}$

B
$3 t^2 \sqrt{\alpha^2+\beta^2}$

C
$\mathrm{t}^2 \sqrt{\alpha^2+\beta^2}$

D
$\sqrt{\alpha^2+\beta^2}$

The correct answer is:
$3 t^2 \sqrt{\alpha^2+\beta^2}$

Solution:
$V_x=\frac{d x}{d t}=3 \alpha t^2, V_y=\frac{d y}{d t}=3 \beta t^2$
$\overrightarrow{\mathrm{v}}=\sqrt{\mathrm{V}_{\mathrm{x}^2}+\mathrm{V}_{\mathrm{y}^2}}=3 \mathrm{t}^2 \sqrt{\alpha^2+\beta^2}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2003
The coordinates of a moving particle at any time t are given by x = Î± tâ¡ 3 and yâ¡ = Î² tâ¡ 3. The speed of the particle at time t is given by

Options:
A 3 tâ¡ α 2 + β 2
B tâ¡ 2 α 2 + β 2
C α 2 + β 2
D 3 tâ¡ 2 Î± 2 + Î² 2

The correct answer is:
3 t⁡ 2 α 2 + β 2

Solution:
∵ x = α t⁡ 3
∴d⁡xd⁡t⁡=3αt⁡2⇒vx=3αt⁡2
Again y⁡ = β t⁡ 3
∴d⁡y⁡d⁡t⁡⇒vy⁡=3βt⁡2
Since, v2=vx2+vy2
v2=(3αt⁡2)2+(3βt⁡2)2=(3t⁡2)2(α2+β2)
Or v=3t⁡2α2+β2

Topic: Physics, Motion In Two Dimensions, JEE Main
The coordinates of a moving particle at time \(t\) are given by \(x=c t^2\) and \(y=b t^2\). The speed of the particle is given by

Options:
A \(2 \mathrm{t}(\mathrm{c}+\mathrm{b})\)
B \(2 t \sqrt{\left(c^2-b^2\right)}\)
C \(t \sqrt{\left(c^2+b^2\right)}\)
D \(2 t \sqrt{\left(c^2+b^2\right)}\)

The correct answer is:
\(2 t \sqrt{\left(c^2+b^2\right)}\)

Solution:
\(\begin{aligned}
& \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{ct} \text { and } \frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{bt} \\
& \therefore \mathrm{v}=\sqrt{\left\{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)+\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)^2\right\}} \\
& =\sqrt{\left\{(2 c t)^2+(2 b t)^2\right\}}=2 t \sqrt{\left(c^2+b^2\right)} \\
&
\end{aligned}\)

Topic: Mathematics, Differentiation, JEE Main
The co-ordinates of a particle moving in $x-y$ plane are given by : $x=2+4 \mathrm{t}, y=3 \mathrm{t}+8 \mathrm{t}^2$.
The motion of the particle is :

Options:
A uniformly accelerated having motion along a parabolic path.
B uniform motion along a straight line.
C uniformly accelerated having motion along a straight line.
D non-uniformly accelerated.

The correct answer is:
uniformly accelerated having motion along a parabolic path.

Solution:
$\begin{aligned}
& x=2+4 t \\
& \frac{d x}{d t}=v_x=4 \\
& \frac{d v_x}{d t}=a_x=0 \\
& y=3 t+8 t^2 \\
& \frac{d y}{d t}=v_y=3+16 t \\
& \frac{d v_y}{d t}=a_y=16
\end{aligned}$
the motion will be uniformly accelerated motion.
For path
$\begin{aligned}
& \mathrm{x}=2+4 \mathrm{t} \\
& \frac{(\mathrm{x}-2)}{4}=\mathrm{t}
\end{aligned}$
Put this value of $t$ is equation of $y$
$y=3\left(\frac{x-2}{4}\right)+8\left(\frac{x-2}{4}\right)^2$
this is a quadratic equation so path will be parabola.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2024 (04 Apr Shift 1)
The co-ordinates of a point $P$ are $(3,12,4)$ with respect to origin $O$, then the direction cosines of $O P$ are

Options:
A $3,12,4$
B $\frac{1}{4}, \frac{1}{3}, \frac{1}{2}$
C $\frac{3}{\sqrt{13}}, \frac{1}{\sqrt{13}}, \frac{2}{\sqrt{13}}$
D $\frac{3}{13}, \frac{12}{13}, \frac{4}{13}$

The correct answer is:
$\frac{3}{13}, \frac{12}{13}, \frac{4}{13}$

Solution:
Required direction cosines are
$\frac{3}{\sqrt{3^2+12^2+4^2}}, \frac{12}{\sqrt{3^2+12^2+4^2}}, \frac{4}{\sqrt{3^2+12^2+4^2}}$
i.e., $\frac{3}{13}, \frac{12}{13}, \frac{4}{13}$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The coordinates of a point P on the line x+y+13=0, nearest to the circle x2+y2+4x+6y-5=0, are

Options:
A -15,2
B -5,-6
C -6,-7
D -7,-6

The correct answer is:
-6,-7

Solution:
The distance of the centre -2,-3 from the line x+y+13=0 is -2-3+132=42 units which is greater than the radius 32
Therefore, the line is neither tangent nor secant to the circle

Hence, nearest point P on the line from the circle is the foot of the perpendicular from the centre on the line.
⇒ If P=h,k, then

h+21=k+31=--2-3+132
⇒h=-6 & k=-7

⇒P=-6,-7

Topic: Mathematics, Circle, JEE Main
The co-ordinates of a point which is equidistant from the points \( (0,0,0),(a, 0,0),(0, b, 0) \) and \( (0,0, c) \) are given by

Options:
A \( \left(\frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right) \)
B \( \left(-\frac{a}{2},-\frac{b}{2}, \frac{c}{2}\right) \)
C \( \left(\frac{a}{2},-\frac{b}{2}, \frac{c}{2}\right) \)
D \( \left(-\frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right) \)

The correct answer is:
\( \left(\frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right) \)

Solution:
Let point be x, y, z then x2+y2+z2
=x-a2+y2+z2=x2+y-b2+z2=x2+y2+y-c2
Therefore x=a2, y=b2 and z=c2 .

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The co-ordinates of pole of line $l x+m y+n=0$ with respect to circles $x^2+y^2=1$, is

Options:
A $\left(\frac{1}{n}, \frac{m}{n}\right)$
B $\left(-\frac{1}{n},-\frac{m}{n}\right)$
C $\left(\frac{1}{n},-\frac{m}{n}\right)$
D $\left(-\frac{1}{n}, \frac{m}{n}\right)$

The correct answer is:
$\left(-\frac{1}{n},-\frac{m}{n}\right)$

Solution:
Let pole be $\left(x_1, y_1\right)$ then polar will be $x x_1+y y_1=1$ comparing with $l x+m y+n=0$
$\Rightarrow x_1=-\frac{l}{n}, y_1=-\frac{m}{n}$.

Topic: Mathematics, Circle, JEE Main
The co-ordinates of the extremities of the latus rectum of the parabola $5 y^2=4 x$ are

Options:
A $(1 / 5,2 / 5),(-1 / 5,2 / 5)$
B $(1 / 5,2 / 5),(1 / 5,-2 / 5)$
C $(1 / 5,4 / 5),(1 / 5,-4 / 5)$
D None of these

The correct answer is:
$(1 / 5,2 / 5),(1 / 5,-2 / 5)$

Solution:
$y^2=4 \cdot \frac{1}{5} x ; a=\frac{1}{5}$. So, Focus is $\left(\frac{1}{5}, 0\right)$ and co-ordinates of latus rectum are $y^2=\frac{4}{25} \Rightarrow y= \pm \frac{2}{5}$ or end points of latus rectum are $\left(\frac{1}{5}, \pm \frac{2}{5}\right)$

Topic: Mathematics, Parabola, JEE Main
The coordinates of the feet of the perpendiculars from the vertices of a triangle on the opposite sides are \( (20,25),(8,16) \) and \( (8,9) \). If the orthocenter of the triangle is \( (h, k) \), then \( \left(k^{3}+h-3380\right) \) is equal to

Options:

The correct answer is:
5

Solution:
Orthocenter of ∆ABC is same as the incenter of ∆DEF.

Now,

DE=122+92=15

FE=72=7

FD=20-82+25-92=122+162=20

Therefore,

h,k≡20×7+8×20+8×157+20+15,25×7+16×20+9×157+20+15

h,k≡10,15

Then,

k3+h-3380=153+10-3380=5

Topic: Mathematics, Straight Lines, JEE Main
The coordinates of the foci of the rectangular hyperbola $x y=c^2$ are

Options:
A $( \pm c, \pm c)$
B $( \pm c \sqrt{2}, \pm c \sqrt{2})$
C $\left( \pm \frac{c}{\sqrt{2}}, \pm \frac{c}{\sqrt{2}}\right)$
D None of these

The correct answer is:
$( \pm c \sqrt{2}, \pm c \sqrt{2})$

Solution:
$x y=c^2$ as $c^2=\frac{a^2}{2}$. Here, co-ordinates of focus are $\left(a e \cos 45^{\circ}, a e \sin 45^{\circ}\right) \equiv(c \sqrt{2}, c \sqrt{2}),\{$as $e=\sqrt{2}, a=c \sqrt{2}\}$
Similarly other focus is $(-c \sqrt{2},-c \sqrt{2})$
Note : Students should remember this question as a fact.

Topic: Mathematics, Hyperbola, JEE Main
The co-ordinates of the focus of the parabola described parametrically by $x=5 t^2+2, y=10 t+4$ are

Options:
A $(7,4)$
B $(3,4)$
C $(3,-4)$
D $(-7,4)$

The correct answer is:
$(7,4)$

Solution:
Hints: $\mathrm{x}=5 \mathrm{t}^2+2 ; \mathrm{y}=10 \mathrm{t}+4,\left(\frac{y-4}{10}\right)^2=\left(\frac{x-2}{5}\right)$ or, $(\mathrm{y}-4)^2=20(\mathrm{x}-2)$

Topic: Mathematics, Parabola, JEE Main
The coordinates of the focus of the parabola described parametrically by x=5t2+2, y=10t+4 are

Options:
A (7,4)
B (3,4)
C (3,-4)
D (-7,4)

The correct answer is:
(7,4)

Solution:
Given parametric curves are
x=5t2+ 2, y=10 t+4
or x-25=t2, y-410=t
⇒ x-25= y-4102
⇒ y-42=20x-2
⇒ Y2=20X, where Y=y-4, X=x-2
∴ Coodinate of focus is (5, 0)
ie, x-2=5, y-4=0
⇒ x=7, y=4
Hence, the required coordinate is (7, 4)

Topic: Mathematics, Parabola, JEE Main
The co-ordinates of the foot of the perpendicular drawn from the origin to a plane is $(2,4,-3)$. The equation of the plane is

Options:
A $2 x-4 y-3 z=29$
B $2 x-4 y+3 z=29$
C $2 x+4 y-3 z=29$
D None of these

The correct answer is:
$2 x+4 y-3 z=29$

Solution:
According to question, $2(x-2)+4(y-4)-3(z+3)=0$
$2 x+4 y-3 z-29=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The coordinates of the foot of the perpendicular from the point 1,-2, 1 on the plane containing the lines x+16=y-17=z-38 and x-13=y-25=z-37, is:

Options:
A 2,-4, 2
B 1, 1, 1
C 0, 0, 0
D -1, 2,-1

The correct answer is:
0, 0, 0

Solution:
Direction ratio's of the plane perpendicular to two given lines is, n→= n→1×n→2

=i^j^k^678357

=i^49-40-j^42-24+k^30-21

=9i^-18j^+9k^

Hence, dr's=9,-18,9

Now, plane passes through -1,1, 3 (A point on the first line),

Hence, the equation of plane containing lines is

1x+1-2y-1+z-3=0

⇒x-2y+z=0

∴ Foot of perpendicular of point 1,-2,1 is

x-11=y+2-2=z-11= -1+4+16=-1

x-1=-1 , y+2=2, z-1=-1

x=0, y=0, z=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2017 (08 Apr Online)
The coordinates of the foot perpendicular from the point $(1,0,0)$ to the line
$$
\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8} \text { are }
$$

Options:
A
$(2,-3,8)$

B
$(1,-1,-10)$

C
$(5,-8,-4)$

D
$(3,-4,-2)$

The correct answer is:
$(3,-4,-2)$

Solution:
Let the equation of $A B$ is
$$
\frac{x-1}{2}=\frac{y-(-1)}{-3}=\frac{z-(-10)}{8}=k
$$
Let $L$ be the foot of the perpendicular drawn form $P(1,0,0)$.
$\therefore \quad L=(2 k+1,-3 \mathrm{k}-1,8 k-10)$.
Now, direction ratio of $P L=(2 k,-3 k-1,8$ $-10)$ and direction ratio of $A B=(2,-3,8)$ Since, $P L$ is perpendicular to $A B$
$$
\therefore \quad 2(2 k)-3(-3 k-1)+8(8 k-10)=0
$$
Now, $k=\frac{2(1-1)+(-3)(0+1)+8(0+10)}{(2)^2+(-3)^2+(8)^2}$
$$
=\frac{0-3+80}{4+9+64}=\frac{77}{77}=1
$$
$\therefore \quad$ Required co-ordinate
$$
=L=(2+1,-3-1,8-10)=(3,-4,-2) \text {. }
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2012 (12 May Online)
The coordinates of the orthocenter of the triangle that has the coordinates of midpoints of its sides as 0,0, 1,2 and -6,3 is

Options:
A 0,0
B -4,5
C -5,5
D -4,4

The correct answer is:
-5,5

Solution:
Line joining the midpoints of two sides is ∥ to the third side and half of it.
∠O=90o (as AO⊥OB)
Hence P will be the orthocentre
AOBP forms a rectangle.

∴ P=A+B-O (using the concept that diagonals bisect each other)
Px,y= x=-6+1-0=-5y=2+3-0=5

Topic: Mathematics, Straight Lines, JEE Main
The co-ordinates of the point $P$ are $(x, y, z)$ and the direction cosines of the line $O P$ when $O$ is the origin, are $1, m, n$. If $O P=r$, then

Options:
A $I=x, m=y, n=z$
B $I=x r, m=y r, n=z r$
C $x=I r, y=m r, z=n r$
D None of these

The correct answer is:
$x=I r, y=m r, z=n r$

Solution:
$x=l r, y=m r, z=n r$
Explanation for correct option:
Finding the values:
Given that $P \rightarrow(x, y, z)$ and $O \rightarrow(0,0,0)$
New direction cosines are $l, m, n$.
We know that, direction cosines are written as
$\begin{aligned}
& l=\frac{z}{\sqrt{z^4+y^2+z^5}} \rightarrow(i) \\
& m=\frac{y}{\sqrt{z^2+y^2+z^2}} \rightarrow(i i) \\
& n=\frac{z}{\sqrt{x^2+y^4+z^2}} \rightarrow(\dot{i} i i)
\end{aligned}$
Also, $O P=r$
We know that distance between $O$ and $P$ is calculated as
$\begin{aligned}
r & =\sqrt{(x-0)^2+(y-0)^2+(z-0)^2} \\
& =\sqrt{x^2+y^2+z^2}
\end{aligned}$
Substituting the value of $r$ in equations $(i),(i i)$ and (iii)
Therefore
$\begin{aligned}
& l=\frac{x}{r} \Rightarrow x=l r \\
& m-\frac{y}{r} \Rightarrow y-m r \\
& n=\frac{z}{r} \Rightarrow z=n r
\end{aligned}$
Hence, option (C) is the correct answer

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The coordinates of the points on the line 3x+2y=5 which are equidistant from the lines 4x+3y-7=0 and 2y-5=0 can be

Options:
A -114,7328
B 114,-7328
C 116,-7732
D -116,7732

The correct answer is:
-114,7328

Solution:
We know that the bisectors of the angles between two lines contain all those points which are equidistant from the lines.

So, the required point lies on the bisectors of the angles between the given lines.
The equations of the bisectors are given by
4x+3y-742+32=±2y-502+22
⇒4x+3y-75=±2y-52
⇒8x-4y+11=0 and 8x+16y-39=0
Let, h,k be the coordinates of the required point. Then,
8h-4k+11=0 …(i)
or, 8h+16k-39=0 …(ii)
It is given that h,k lies on 3x+2y=5. Therefore,
3h+2k=5 …(iii)
Solving (iii) with (i) and (ii) respectively, we obtain the coordinates of the required points as
-114,7328 and 116,7732

Topic: Mathematics, Straight Lines, JEE Main
The coordination complex CoOH262+ has one unpaired electron, which of the following statements are true?

(i) The complex is octahedral.

(ii) The complex is an outer orbital complex.

(iii) The complex is diamagnetic.

Options:
A (i) and (iii) only
B (i), (ii) and (iii)
C (i) and (ii) only
D (ii) and (iii) only

The correct answer is:
(i) and (ii) only

Solution:
As in the given complex, 6 ligands are attached so this complex is octahedral complex.

Topic: Chemistry, Coordination Compounds, JEE Main
The coordination environment of \(\mathrm{Ca}^{2+}\) ion in its complex with \(\mathrm{EDTA}^{4-}\) is :

Options:
A tetrahedral
B trigonal prismatic
C octahedral
D square planar

The correct answer is:
octahedral

Solution:
$\begin{aligned} & \text { EDTA }^{4-} \rightarrow \text { Hexadentate ligand } \\ & {[\mathrm{Ca}(\text { EDTA })]^{2-}}\end{aligned}$
So Coordination environment is octahedral

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The coordination geometry around the manganese in decacarbonyldimanganese0

Options:
A Octahedral
B Trigonal bipyramidal
C Square pyramidal
D Square planar

The correct answer is:
Octahedral

Solution:
Around the manganese atom, there are five CO ligands and one Mn occupying octahedron vertices. For octahedral geometry, the more appropriate hybridization is d2sp3 that produces six hybrid orbitals pointing towards CO and Mn.

The octahedral shape of molecules is the shape of molecules where six atoms or ligands or groups of atoms are arranged in a systematic way around a central dogma or atom. The Octahedral Shape of Molecules molecules has eight faces.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (30 Jan Shift 2)
The coordination number and the oxidation state of the element ' $\mathrm{E}$ ' in the complex $\left[\mathrm{E}(\mathrm{en})_2\left(\mathrm{C}_2 \mathrm{O}_4\right)\right] \mathrm{NO}_2$ (where (en) is ethylene diamine) are, respectively,

Options:
A
6 and 2

B
4 and 2

C
4 and 3

D
6 and 3

The correct answer is:
6 and 3

Solution:
$$
\text { Coordination no. }=6 \text { and Oxidation no. }=3
$$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2008
The coordination number of a metal in coordination compounds is

Options:
A Same as primary valency
B Sum of primary and secondary valencies
C Same as secondary valency
D None of these

The correct answer is:
Same as secondary valency

Solution:
The secondary valency is equal to the coordination number the secondary valency are non ionizable valencies. These are satisfied by neutral molecules or negative ions.

Topic: Chemistry, Coordination Compounds, JEE Main
The coordination number of an atom in a body-centered cubic structure is______

[Assume that the lattice is made up of atoms.]

Options:

The correct answer is:
8

Solution:
All the corner atoms around body centre atom are nearest to that atom, so co-ordination number in this structure is eight.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The coordination number of central metal atom in a complex is determined by

Options:
A
the number of ligands around a metal ion bonded by sigma bonds

B
the number of only anionic ligands bonded to the metal ion

C
the number of ligands around a metal ion bonded by sigma and pi- bonds both

D
the number of ligands around a metal ion bonded by pi-bonds

The correct answer is:
the number of ligands around a metal ion bonded by sigma bonds

Solution:
No Solution Available

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2004
The coordination number of Th in $\mathrm{K}_{4}\left[\mathrm{Th}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]$ is:

$\left(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}=\right.$ oxalato $)$

Options:
A 14
B 6
C 8
D 10

The correct answer is:
10

Solution:
$\mathrm{K}_{4}\left[\mathrm{Th}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]$ $\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$ (oxalato) : bidentate ligand $\mathrm{H}_{2} \mathrm{O}$ (aqua): monodentate $\therefore$ Co-ordination no. of Th $=2 \times 4+2=10$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The coordination numbers of Co and Al in CoClen2Cl and K3AlC2O43, respectively are:
(en = ethane-1, 2-diamine)

Options:
A 3 and 3
B 5 and 3
C 5 and 6
D 6 and 6

The correct answer is:
5 and 6

Solution:
Compound CoClen2+
Since Cl- is monodentate ligand, en → bidentate ligand
Co-ordination number =1+2×2=5
Compound CoC2O433-
OX2-→ bidentate ligand co-ordination number = 3 × 2 = 6

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The copolymer formed by addition polymerization of styrene and acrylonitrile in the presence of peroxide is:

Options:
A

B

C

D

The correct answer is:

Solution:
No Solution Available

Topic: Chemistry, Polymers, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The copolymer formed by addition polymerization of styrene and acrylonitrile in the presence of peroxide is

Options:
A
B
C
D

The correct answer is:

Solution:
Polymerization takes place through head-head combination.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2018 (15 Apr)
The copolymer formed by addition polymerization of styrene and acrylonitrile in the presence of peroxide is:

Options:
A
B \( \left[\begin{array}{cc}\mathrm{CH}_{2}-\underset{\mid}{\mathrm{CH}}-\mathrm{CH}_{2}-\mathrm{CH} \mathrm{C}_{6} \mathrm{H}_{5} & \mathrm{CN}\end{array}\right]_{\mathrm{n}} \)
C
D

The correct answer is:

Solution:
Styrene is an organic compound with the chemical formula C6H5CH=CH2, and acrylonitrile is an organic compound with the formula CH2=CH-CN.

So, styrene and acrylonitrile will polymerize in the presence of peroxide to form a polymer. The reaction proceeds via free radical chain mechanism.

nC6H5CH=CH2+nCH2=CH-CN→H2O2-CH|C6H5-CH2-CH2-CH|CN-n

Topic: Chemistry, Polymers, JEE Main
The core of a transformer is laminated because:

Options:
A ratio of voltage in primary and secondary may be increased
B energy losses due to eddy currents may be minimised
C the weight of the transformer may be reduced
D rusting of the core may be prevented

The correct answer is:
energy losses due to eddy currents may be minimised

Solution:
The core of a transformers is laminated to minimise the energy losses due to eddy currents.

Topic: Physics, Alternating Current, JEE Main
The core of an electromagnet is made of:

Options:
A Soft iron
B Magnesium
C Copper
D Steel

The correct answer is:
Soft iron

Solution:
Soft iron defined as the iron that can easily be magnetized and demagnetized with a small quantity of hysteresis loss and has a low carbon content.

Topic: Physics, Magnetic Properties of Matter, JEE Main
The core of any transformer is laminated so as to

Options:
A
reduce the energy loss due to eddy currents

B
make it light weight

C
make it robust and strong

D
increase the secondary voltage

The correct answer is:
reduce the energy loss due to eddy currents

Solution:
No Solution Available

Topic: Physics, Alternating Current, JEE Main, JEE Main 2003
The correct answer to above question is based on

Options:
A Acceleration due to gravity in vacuum is same irrespective of size and mass of the body
B Acceleration due to gravity in vacuum depends on the mass of the body
C There is no acceleration due to gravity in vacuum
D In vacuum there is resistance offered to the motion of the body and this resistance depends on the mass of the body

The correct answer is:
There is no acceleration due to gravity in vacuum

Solution:
Acceleration due to gravity in earth inside a vacuum chamber is always same considering two different masses of different object subjected to a free fall. The mass actually cease to not exist and the net force cease to be the same.
Its like a feather and a baseball will fall at same rate in vacuum chamber given same height is provided.
But outside the earth far from any celestial body, acceleration due to gravity is 0 practically. And if we start moving in space, we will move at a constant velocity. But if we revolve around something, there will be acceleration, a simple acceleration due to change in direction. And it's because of the centrifugal-centripetal force.
Always remember acceleration, acceleration due to gravity, and velocity all are different. And have different effects. So when u mention acceleration due to gravity, it means always in earth or any heavy celestial body.

Topic: Physics, Gravitation, JEE Main