datavorous/entrance-exam-dataset huggingface dataset:
The carrier wave is given by $(C) t=2 \sin (8 \pi t)$ volt . The modulating signal is a square wave as shown. Find modulation index.

Options:
A 5
B 0.5
C 1
D 10

The correct answer is:
0.5

Solution:
$y(t)=\left[A_c+m(t)\right] c(t)$
where,
$A_c$ is the amplitude of carrier wore.
The modulated waveform $y(t)$ as shown in fig.
Modulation index, $\mu=\frac{M}{A}=0.5$

Topic: Physics, Communication System, JEE Main
The Cartesian equation of the plane whose vector equation is \( \bar{\gamma}=(1+\lambda-\mu) \hat{i}+(2-\lambda) \hat{j}+(3-2 \lambda+2 \mu) \hat{k} \) where \( \lambda, \mu \) are scalars is

Options:
A \( 2 x+y=5 \)
B \( 2 x-y=5 \)
C \( 2 x-z=5 \)
D \( 2 x+z=5 \)

The correct answer is:
\( 2 x+z=5 \)

Solution:
r→=i^+2j^+3k^+λi^- j^-2k^+μ-i^+2k^
x-1y-2z-31-1-2-102=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The cartesian product $A \times A$ has 9 elements among which are found $(-1,0)$ and $(0,1)$. Find the set $A$ and the remaining elements of $A \times A$.

Options:

None

Solution:
Clearly, set $A$ has 3 elements since $(-1,0) \in A \times \operatorname{A}$ and $(0,1)$
$\in A \times A$. Therefore, $(-1,0) \in A \times A$
$\Rightarrow-1,0 \in A$ and $(0,1) \in A \times A \Rightarrow 0,1 \in A$.
Therefore, $-1,0,1 \in A$.
Hence, $\quad A=\{-1,0,1\}$
Thus remaining elements of
$A \times A=(-1,-1),(-1,1),(0,-1),(0,0),(1,-1)$
$(1,0),(1,1)$

Topic: Mathematics, Relations and Functions (Class 11)
The catalyst and promoter respectively used in the Haber's process of industrial synthesis of ammonia are

Options:
A $\mathrm{Mo}, \mathrm{V}_2 \mathrm{O}_5$
B $\mathrm{V}_2 \mathrm{O}_5, \mathrm{Fe}$
C $\mathrm{Fe}, \mathrm{Mo}$
D $\mathrm{Mo}, \mathrm{Fe}$

The correct answer is:
$\mathrm{Fe}, \mathrm{Mo}$

Solution:
In the Haber's process for the synthesis of ammonia from $\mathrm{N}_2$ and $\mathrm{H}_2$, $\mathrm{Fe}$ is used as catalyst and Mo works as activator.
$\mathrm{N}_2+3 \mathrm{H}_2 \stackrel{\mathrm{Fe}, \mathrm{Mo}}{\rightleftharpoons} 2 \mathrm{NH}_3+22.4 \mathrm{kcal}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The catalytic activity of transition metals and their compounds is ascribed mainly to

Options:
A their magnetic behaviour
B their unfilled $d$-orbitals
C their ability to adopt variable oxidation states
D their chemical reactivity

The correct answer is:
their ability to adopt variable oxidation states

Solution:
Transition metals show variable oxidation states, due to which they can perform as catalyst.
$$
\begin{aligned}
& \text { e.g., } 2 \mathrm{SO}_2+\mathrm{O}_2 \underset{\text { (catalyst) }}{\stackrel{\mathrm{V}_2 \mathrm{O}_5}{\longrightarrow}} 2 \mathrm{SO}_3 \\
& \mathrm{SO}_2+\stackrel{+5}{\mathrm{~V}_2} \mathrm{O}_5 \longrightarrow \mathrm{SO}_3+\stackrel{+4}{\mathrm{~V}_2} \mathrm{O}_4 \\
& \stackrel{+4}{\mathrm{~V}_2} \mathrm{O}_4+\mathrm{O}_2 \longrightarrow \stackrel{+5}{\mathrm{~V}} \mathrm{~V}_5 \\
&
\end{aligned}
$$

Topic: Chemistry, d and f Block Elements, NEET, NEET 2012 (Mains)
The catalytic activity of transition metals and their compounds is ascribed mainly to

Options:
A Their ability of adopt variable oxidation states
B Their chemical reactivity
C Their magnetic behaviour
D Their unfilled d- orbitals

The correct answer is:
Their ability of adopt variable oxidation states

Solution:
The catalytic activity of transition metals and their compounds is ascribed mainly to their ability of adopt variable oxidation states for example
2SO2+O2→V2O52SO3
SO2+V2O5+5→SO3+V2O4+4
V2O4+4+O2→V2O5+5

Topic: Chemistry, d and f Block Elements, NEET
The catenation tendency of $\mathrm{C}, \mathrm{Si}$ and $\mathrm{Ge}$ is in the order $\mathrm{Ge} < \mathrm{Si} < \mathrm{C}$. The bond energies (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ ) of $\mathrm{C}-\mathrm{C}, \mathrm{Si}-\mathrm{Si}$ and $\mathrm{Ge}-\mathrm{Ge}$ bonds are respectively ;

Options:
A
$348,297,260$

B
$297,348,260$

C
$348,260,297$

D
$260,297,348$

The correct answer is:
$348,297,260$

Solution:
The linking of identical atoms with each other to form long chains is called catenation. However, this property decreases from carbon to lead. Decrease of this property is associated with $\mathrm{M}-\mathrm{M}$ bond energy which decreases from carbon to lead.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The catenation tendency of $\mathrm{C}, \mathrm{Si}$ and $\mathrm{Ge}$ is in the order $\mathrm{Ge} < \mathrm{Si} < \mathrm{C}$. The bond energies (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ ) of $\mathrm{C}-\mathrm{C}, \mathrm{Si}-\mathrm{Si}$ and $\mathrm{Ge}-\mathrm{Ge}$ bonds are respectively ;

Options:
A
$348,297,260$

B
$297,348,260$

C
$348,260,297$

D
$260,297,348$

The correct answer is:
$348,297,260$

Solution:
The linking of identical atoms with each other to form long chains is called catenation. However, this property decreases from carbon to lead. Decrease of this property is associated with $\mathrm{M}-\mathrm{M}$ bond energy which decreases from carbon to lead.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2013 (25 Apr Online)
The catenation tendency of $\mathrm{C}, \mathrm{Si}$ and $\mathrm{Ge}$ is in the order $\mathrm{Ge} < \mathrm{Si} < \mathrm{C}$. The bond energies (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of $\mathrm{C}-\mathrm{C}, \mathrm{Si}-\mathrm{Si}$ and $\mathrm{Ge}-\mathrm{Ge}$ bonds, respectively are

Options:
A $167,180,348$
B $180,167,348$
C $348,167,180$
D $348,180,167$

The correct answer is:
$167,180,348$

Solution:
Catenation tendency $\approx$ bond energy.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The cation that will not be precipitated by H2S in the presence of dil HCl is:

Options:
A Co2+
B As3+
C Pb2+
D Cu2+

The correct answer is:
Co2+

Solution:
Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt.)
Other are precipitated as sulphide in presence of dil HCl in group II.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The cation that will not be precipitated by H2S in the presence of dil HCl is:

Options:
A Co2+
B As3+
C Pb2+
D Cu2+

The correct answer is:
Co2+

Solution:
Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt.)
Other are precipitated as sulphide in presence of dil HCl in group II.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2015 (10 Apr Online)
The cause of periodicity of properties is

Options:
A Increasing atomic radius
B Increasing atomic weights
C Number of electrons in the valency orbit
D The re-occurrence of similar outer electronic configuration

The correct answer is:
The re-occurrence of similar outer electronic configuration

Solution:
The re-occurrence of similar outer electronic configuration

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The cell has an emf of $2 \mathrm{~V}$ and the internal resistance of this cell is $0.1 \Omega$, it is connected to resistance of $3.9 \Omega$, the voltage across the cell will be

Options:
A $1.95 \mathrm{~V}$
B $1.5 \mathrm{~V}$
C $2 \mathrm{~V}$
D $1.8 \mathrm{~V}$

The correct answer is:
$1.95 \mathrm{~V}$

Solution:
Key Idea When cell is giving current then the potential difference across its plates is less than its emf as potential drop across internal resistance occurs.
When a cell of $\operatorname{emf} E$ is connected to a resistance of $3.9 \Omega$, then the emf $E$ of the cell remains constant, while voltage $V$ goes on decreasing on taking more and more current from the cell.
$$
\therefore \quad V=E-i r
$$
where $r$ is internal resistance.
$$
\begin{array}{ll}
\text { Also current } & i=\frac{E}{R+r} \\
\therefore & V=E-\left(\frac{E}{R+r}\right) r
\end{array}
$$
Putting the numerical values, we have
$$
\begin{aligned}
E=2 V, r &=0.1 \Omega, R=3.9 \Omega \\
V &=2-\left(\frac{2}{3.9+0.1}\right) \times 0.1 \\
V &=1.95 \mathrm{volt}
\end{aligned}
$$

Topic: Physics, Wave Optics, MHT CET, MHT CET 2007
The cell junctions called tight, adhering and gap junctions are found in :

Options:
A Neural tissue
B Muscular tissue
C Connective tissue
D Epithelial tissue

The correct answer is:
Epithelial tissue

Solution:
No Solution Available

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2009 (Mains)
The cell membranes are mainly composed of:

Options:
A fats
B proteins
C phospholipids
D carbohydrates

The correct answer is:
phospholipids

Solution:
The fundamental building blocks of all cell membranes are phospholipids, which are amphipathic molecules, consisting of two hydrophobic fatty acid chains linked to a phosphatecontaining hydrophilic head group
Related Theory
Both prokaryotic and eukaryotic cells have a plasma membrane a phospholipid bilayer with embedded proteins, that separates the internal contents of the cell from its surrounding environment.

Topic: Chemistry, Biomolecules, NEET, NEET 2005
The cell potential for the following cell PtH2gH+aq||Cu2+0.01MCus is 0.576 V at 298 K. The pH of the solution is (Nearest integer)

(Given : ECu2+/Cuo=0.34 V and 2.303RTF=0.06 V)

Options:

The correct answer is:
5

Solution:
Anode ⇒H2g⟶2H+aq+2e-

Cathode ⇒Cu2+aq+2e-⟶Cus

Overall ⇒H2q+Cu2+aq⟶2H+aq+Cus¯

Ecell°=ECu2+Cu°-EH+H2°=0.34 V

Ecell=Ecell°-0.062logH+2Cu2+

0.576=0.34+0.03-logH+2+logCu2+

0.576=0.34+0.032pH+logCu2+

0.236=0.032pH-2

7.866=2pH-2

pH=4.93≈5

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2022 (24 Jun Shift 1)
The cell potential for ZnZn2+(aq)‖Snx+Sn is 0.801V at 298 K. The reaction quotient for the above reaction is 10-2. The number of electrons involved in the given electrochemical cell reaction is _____ .
(Given EZn2+∣Zn0=-0.763 V,ESnx+|Sn0=+0.008 V and 2.303RTF=0.06 V)

Options:

The correct answer is:
4

Solution:
E=E0-2.303RTnFlogQ

Here,

E=+0.801 VE0=ECathode∘-EAnode∘Both are SRP valuesE0=0.008--0.763

=+0.771 V

∴0.801=+0.771-0.06nlog10-2

⇒n=4

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The cell reaction $\mathrm{Cu}+2 \mathrm{Ag}^{+} \rightarrow \mathrm{Cu}^{+2}+\mathrm{Ag}$ is best represented by

Options:
A $C u_{(s)}\left|C u^{+2}{ }_{(a q)} \| A g^{+}{ }_{(a q)}\right| A g_{(s)}$
B $P t\left|C u^{+2} \| A g^{+}{ }_{(a q)}\right| A g_{(s)}$
C $\mathrm{Cu}^{+2}|\mathrm{Cu} \| \mathrm{Pt}| \mathrm{Ag}$
D None of the above representations

The correct answer is:
$C u_{(s)}\left|C u^{+2}{ }_{(a q)} \| A g^{+}{ }_{(a q)}\right| A g_{(s)}$

Solution:
The cell reaction \(\mathrm{Cu} \mid+2 \mathrm{Ag}^{+} \rightarrow \mathrm{Cu}^{+2}+\mathrm{Ag}\) is best represented by
\(C u_{(s)}\left|C u_{(a q)}^{+2} \| A g_{(a q)}^{+}\right| A g_{(s)}\)

Topic: Chemistry, Electrochemistry, JEE Main
The cell which lacks cytoskeleton is

Options:
A prokaryotic bacterial cells
B eukaryotic plant cell
C both (a) and (b)
D prokaryotic and eukaryotic animal cells

The correct answer is:
prokaryotic bacterial cells

Solution:
They are extremely minute, fibrous, and tubular structures that form the structural framework inside the cell. They occur only in eukaryotic cells. They maintain the shape of the cell, regulate orientation and distribution of cell transport and movement of cells.

Topic: Biology, Cell - The Unit of Life, NEET
The cell, Zn/Zn2+(1M)||Cu2+(1M)/Cu(Ecell0=1.10V) was allowed to be completely discharged at 298K. The relative concentration of Zn2+ to Cu2+Zn2+Cu2+ is 10x . The value of x is:
( Take2.303RTF=0.059 Round off your answer up to one decimal)

Options:

The correct answer is:
37

Solution:
Ecell=0 ; when cell is completely discharged
Ecell=Ecell0-0.0592log[Zn2+][Cu2+]
0=1.1-0.0592log[Zn2+][Cu2+]
log[Zn2+][Cu2+]=2×1.10.059=37.3
` Zn2+Cu2+=1037.3=10x
∴x=37.3

Topic: Chemistry, Electrochemistry, JEE Main
The cell, $\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{Cu}^{2+}(1 \mathrm{M})\right| \mathrm{Cu}\left(\mathrm{E}_{\text {cell }}^0=1.10 \mathrm{~V}\right)$, was allowed to be completely discharged at $\mathrm{298~K}$. The relative concentration of $\mathrm{Zn}^{2+}$ to $\mathrm{Cu}^{2+}\left[\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\right]$ is

Options:
A
antilog $(24.08)$

B
$37.3$

C
$10^{37.3}$

D
$9.65 \times 10^4$

The correct answer is:
$10^{37.3}$

Solution:
$\begin{aligned}
& \mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{0.0591}{\mathrm{n}} \log \mathrm{Q} \\
& \text { Where } \mathrm{Q}=\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}
\end{aligned}$
For complete discharge $\mathrm{E}_{\text {cell }}=0$
So $\mathrm{E}_{\text {cell }}^{\circ}=\frac{0.591}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}$
$\Rightarrow\left|\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\right|=10^{37.3}$
Hence, (C) is correct.

Topic: Chemistry, Electrochemistry, JEE Main
The cell, $\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{Cu}^{2+}(1 \mathrm{M})\right| \mathrm{Cu}\left(\mathrm{E}_{\text {cell }}^0=1.10 \mathrm{~V}\right)$, was allowed to be completely discharged at $\mathrm{298~K}$. The relative concentration of $\mathrm{Zn}^{2+}$ to $\mathrm{Cu}^{2+}\left[\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\right]$ is

Options:
A
antilog $(24.08)$

B
$37.3$

C
$10^{37.3}$

D
$9.65 \times 10^4$

The correct answer is:
$10^{37.3}$

Solution:
$\begin{aligned}
& \mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{0.0591}{\mathrm{n}} \log \mathrm{Q} \\
& \text { Where } \mathrm{Q}=\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}
\end{aligned}$
For complete discharge $\mathrm{E}_{\text {cell }}=0$
So $\mathrm{E}_{\text {cell }}^{\circ}=\frac{0.591}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}$
$\Rightarrow\left|\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\right|=10^{37.3}$
Hence, (C) is correct.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2007
The cells lining the blood vessels belong to the category of

Options:
A smooth muscle tissue
B squamous epithelium
C columnar epithelium
D connective tissue

The correct answer is:
squamous epithelium

Solution:
The squamous epithelium is made up of a single thin layer of flattened cells with irregular boundaries. They are found in the walls of blood vessels and air sacs of lungs and are involved in a functions like forming a diffusion boundary.

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2011 (Mains)
The ‘cells of Rauber’ are

Options:
A secretory cells of endometrium in uterus
B inner cell mass of blastocoel
C outer cells of trophoblast in contact with
uterine wall
D cells of trophoblast, in contact with inner cell mass of blastocyst.

The correct answer is:
cells of trophoblast, in contact with inner cell mass of blastocyst.

Solution:
The blastomeres in the blastocyst are arranged into an outer layer called trophoblast and an inner group of cells attached to trophoblast called the inner cell mass. Inner cell mass looks like a small knob at one pole which gives rise to the embryo and is called the embryonal knob. The trophoblast does not take part in the formation of the embryo proper. It remains external to the embryo and gives rise to the extraembryonic membranes, namely, chorion and amnion, for the protection and nourishment of the embryo. The trophoblast cells in contact with the embryonal knob are known as cells of Rauber.

Topic: Biology, Human Reproduction, AIIMS, AIIMS 2012
The cells of the main structural tissue of the body are present in particular spaces called

Options:
A ampulla
B lacunae
C lamellae
D canaliculi

The correct answer is:
lacunae

Solution:
Bones have a hard and non-pliable ground substance rich in calcium salts and collagen fibres which give bone its strength. It is the main tissue that provides the structural frame to the body. Bones support and protect softer tissues and organs. The bone cells (osteocytes) are present in the spaces called lacunae. These lacunae also have extensions known as canaliculi. Lacunae are present in the matrix of cartilage too but are without canaliculi.

Topic: Biology, Structural Organisation in Animals, NEET
The cells of the quiescent centre are characterised by:

Options:
A having dense cytoplasm and prominent nuclei
B having light cytoplasm and small nuclei
C dividing regularly to add to the corpus
D dividing regularly to add to tunica

The correct answer is:
having light cytoplasm and small nuclei

Solution:
According to Quiescent Centre Theory, root apical meristem is present as the tip of main root and has a biconvex quiescent centre in its centre which is characterized by presence of inactive cells with less number of cell organelles, light cytoplasm, small nuclei and low synthesis of DNA, RNA and protein caused by low frequency of cell division, mitosis.
Related Theory
Behind the root cap is the quiescent centre. a region of inactive cells. They function to replace the meristematic cells of the root cap meristem. The quiescent centre is also important in organizing the patterns of primary growth in the root.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2003
The cells of the quiescent centre are in

Options:
A G1 phase
B G2 phase
C G0 phase
D S-phase

The correct answer is:
G0 phase

Solution:
Some cells in the adult animals do not appear to exhibit division (e.g., heart cells) and many other cells divide only occasionally, as needed to replace cells that have been lost because of injury or cell death. These cells that do not divide further exit G1 phase to enter an inactive stage called quiescent stage (G0 ) of the cell cycle. Cells in this stage remain metabolically active but no longer proliferate unless called on to do so depending on the requirement of the organism.

Topic: Biology, Cell Cycle and Cell Division, NEET
The center of mass of a system of particles does not depend upon,

Options:
A Position of particles
B Relative distance between particles
C Masses of particles
D Force acting on particle.

The correct answer is:
Force acting on particle.

Solution:
The location of the centre of mass of a variety of bodies does not rely on the internal or external force acting on the system. Rather, the movement of the centre of mass like acceleration relies on the system's net external force.
The variables on which the centre of mass depends on the

The mass of a rigid body.

Position of mass from the axis.

Distribution of mass.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The centre of a wheel rolling on a plane surface moves with a speed v0. A particle on the rim of the wheel at the same level as the centre will be moving at a speed xv0. Then the value of x is .

Options:

The correct answer is:
2

Solution:
For no slipping, v0=ωR
Now, vA=vB=v02+(ωR)2
=2v0

⇒x=2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (22 Jul Shift 1)
The centre of mass of a solid hemisphere of radius 8 cm is x cm from the centre of the flat surface. Then value of x is

Options:

The correct answer is:
3

Solution:
Centre of Mass of a solid hemisphere from centre of the flat surface is given by,hcm=3R8

It is given that R= 8 cm

Therefore,

x=hcm=3R8=3×88x=3 cm

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The centre of mass of a system of two particles divides. The distance between them is

Options:
A In inverse ratio of square of masses of particles
B In direct ratio of square of masses of particles
C In inverse ratio of masses of particles
D In direct ratio of masses of particles.

The correct answer is:
In inverse ratio of masses of particles

Solution:
Let the centre of mass of the system lies at origin.
$\begin{aligned} & \text { Using } X_{C M}=\frac{M_1 X_1+M_2 X_2}{M_1+M_2} \\ & \text { Oro }=\frac{M_1\left(-X_1\right)+M_2 X_2}{M_1+M_2} \\ & \Rightarrow M_1 X_1+M_2 X_2 \\ & \Rightarrow \frac{X_1}{X_2}=\frac{M_2}{M_1}\end{aligned}$
So, centre of mass divides the distance between the two particles in the inverse ratio of masses of particles.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The centre of mass of a system particles does not depend on :

Options:
A masses of the particles
B Internal forces on the particle
C Position of the particles
D Relative distance between the particles

The correct answer is:
Internal forces on the particle

Solution:
Explanation:
- Masses of the particles (1): The center of mass calculation takes into account the masses of all particles in the system.
- Position of the particles (3): The calculation also depends on the positions of these particles.
- Relative distance between the particles (4): Since the center of mass is derived from the positions of the particles, the relative distances indirectly influence the calculation.
- Internal forces on the particle (2): Internal forces do not affect the center of mass. They may affect the motion of particles relative to each other but do not change the center of mass of the system as a whole.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The centre of mass of three particles of masses $1 \mathrm{~kg}, 2 \mathrm{~kg}$ and $3 \mathrm{~kg}$ is at $(2,2,2)$. The position of the fourth mass of $4 \mathrm{~kg}$ to be placed in the system as that the new centre of mass is at $(0,0,0)$ is.

Options:
A $(-3,-3,-3)$
B $(-3,3,-3)$
C $(2,3,-3)$
D $(2,-2,3)$

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

Solution:
$m_1=1 \mathrm{~kg}, m_2=2 \mathrm{~kg}, m_3=3 \mathrm{~kg}$
Position of centre of mass $(2,2,2)$
$m_4=4 \mathrm{~kg}$
New position of centre of mass $(0,0,0)$.
For initial position,
$\begin{aligned}
& \qquad X_{\mathrm{CM}}=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3} \\
& \qquad 2=\frac{m_1 \times x_1+m_2 x_2+m_3 x_3}{1+2+3} \\
& m_1 x_1+m_2 x_2+m_3 x_3=12 \\
\end{aligned}$
Similarly,
$\begin{aligned}
& m_1 y_1+m_2 y_2+m_3 y_3=12 \\
& m_1 z_1+m_2 z_2+m_3 z_3=12
\end{aligned}$
and
$m_1 z_1+m_2 z_2+m_3 z_3=12$
For new position,
$\begin{gathered}
X_{\mathrm{CM}}^{\prime}=\frac{m_1 x_1+m_2 x_2+m_3 x_3+m_4 x_4}{m_1+m_2+m_3+m_4} \\
0=\frac{12+4+x_4}{1+2+3+4} \\
4 x_4=-12 \\
x_4=-3
\end{gathered}$
Similarly, $y_4=-3$
$z_4=-3$
$\therefore$ Position of fourth mass $(-3,-3,-3)$.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The centre of the circle passing through $(0,0)$ and $(1,0)$ and touching the circle $x^2+y^2=9$ is

Options:
A
$\left(\frac{1}{2}, \frac{1}{2}\right)$

B
$\left(\frac{1}{2},-\sqrt{2}\right)$

C
$\left(\frac{3}{2}, \frac{1}{2}\right)$

D
$\left(\frac{1}{2}, \frac{3}{2}\right)$

The correct answer is:
$\left(\frac{1}{2},-\sqrt{2}\right)$

Solution:
No Solution Available

Topic: Mathematics, Circle, JEE Main, JEE Main 2002
The centre of the circle passing through the point 0, 1 and touching the parabola y=x2 at the point 2, 4 is :

Options:
A -5310, 165
B 65, 5310
C 310, 165
D -165, 5310

The correct answer is:
-165, 5310

Solution:
y=x2

Tangent at 2,4 is 12y+4=2x, using Cartesian form.

⇒4x-y-4=0

Equation of circle

x-22+y-42+λ4x-y-4=0

It passes through 0,1

∴ 4+9+λ0-1-4=0

13=5λ⇒λ=135

∴Circle is x2-4x+4+y2-8y+16+1354x-y-4=0

⇒x2+y2+525-4x-8+135y+20-525=0

⇒x2+y2+325x-535y+485=0

∴ Centre is -165,5310

Topic: Mathematics, Parabola, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The centre of the circle passing through the points of intersection of the curves 2x+3y+43x+2y-1=0 and xy=0 is

Options:
A ( 56,-512)
B -56,-512
C 512,-56
D -512,56

The correct answer is:
-56,-512

Solution:
Family of curves passing through points of intersection of curves 2x+3y+43x+2y-1=0 and xy=0 is
2x+3y+43x+2y-1+Î»xy=0
â 2x2+y2+13xy+10x+5y-4+Î»xy=0 .... (1)
If equation 1 represents a circle then coefficient of xy is zero âÎ»=-13
â equation of circle is x2+y2+53x+56y-23=0
â center of the circle is -56,-512

Topic: Mathematics, Circle, JEE Main
The centre of the ellipse $\frac{(x+y-3)^2}{9}+\frac{(x-y+1)^2}{16}=1$, is

Options:
A $(-1,2)$
B $(1,-2)$
C $(-1,-2)$
D $(1,2)$

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

Solution:
We have,
$$
\frac{(x+y-3)^2}{9}+\frac{(x-y+1)^2}{16}=1
$$
To determine the centre of ellipse, put $x+y-3=0$
$$
\Rightarrow \quad x+y=3
$$
and $\quad x-y+1=0$
$$
\Rightarrow \quad x-y=-1
$$
On adding Eqs. (i) and (ii), we get
$$
x=1 \text { and } y=2
$$
Hence, the centre of the ellipse is $(1,2)$.

Topic: Mathematics, Ellipse, JEE Main