datavorous/entrance-exam-dataset huggingface dataset:
Three isolated metal spheres A, B, C have radius R, 2R, 3R respectively, and the same charge, $Q . U_A, U_B$ and $U_C$ be the energy density just outside the surface of the spheres. The relation between $\mathrm{U}_{\mathrm{A},} \mathrm{U}_{\mathrm{B}}$ and $\mathrm{U}_{\mathrm{C}}$ is

Options:
A $\mathrm{U}_{\mathrm{A}}>\mathrm{U}_{\mathrm{B}}<\mathrm{U}_{\mathrm{C}}$
B $\mathrm{U}_{\mathrm{A}}>\mathrm{U}_{\mathrm{B}}>\mathrm{U}_{\mathrm{C}}$
C $\mathrm{U}_{\mathrm{A}}<\mathrm{U}_{\mathrm{B}}<\mathrm{U}_{\mathrm{C}}$
D $\mathrm{U}_{\mathrm{A}}<\mathrm{U}_{\mathrm{B}}>\mathrm{U}_{\mathrm{C}}$

The correct answer is:
$\mathrm{U}_{\mathrm{A}}>\mathrm{U}_{\mathrm{B}}>\mathrm{U}_{\mathrm{C}}$

Solution:
The correct option is (B).
Concept: Energy density in the free space is given by, $\mathrm{U}=\frac{1}{2} \varepsilon_0 \mathrm{E}^2$
The electric field for an isolated sphere is $E=\frac{Q}{2 \pi \varepsilon_0 r^2}$
Therefore, Energy density $U \propto \frac{1}{\mathrm{r}^4}$
And, $\mathrm{U}_{\mathrm{A}}>\mathrm{U}_{\mathrm{B}}>\mathrm{U}_{\mathrm{C}}$ is the correct trend.

Topic: Physics, Electrostatics, JEE Main
Three isomers A, B and C (mol. formula C8H11N ) give the following results:

A and C→DianotizationP+Q→(ii) acidationKMnO4+H+(i) HydrolydsR product of A+S product of C

R has lower boiling point than S

B→C6H5SO2Cl alkali-insoluble product

A, B and C, respectively are:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Three isothermal plots (P versus V) A, B and C are plotted at three temperature T1,T2 and T3 respectively

The correct order of temperature will be

Options:
A T1<T2<T3
B T1=T2=T3
C T1>T2>T3
D T1>T2<T3

The correct answer is:
T1<T2<T3

Solution:
At constant volume and definite mass
P∝T
Thus, the pressure and temperature both will increase in the sequence given below
T1<T2<T3

Topic: Chemistry, States of Matter, JEE Main
Three kids $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are pulling a rope as shown in the figure. What are the magnitudes of force with which $B$ and $C$ must pull in order to keep this system in equilibrium?

Options:
A $20 \sqrt{ } 3 \mathrm{~N}, 15 \mathrm{~N}$
B $20 \mathrm{~N}, 15 \mathrm{~N}$
C $10 \sqrt{ } 3 \mathrm{~N}, 10 \mathrm{~N}$
D $10 \mathrm{~N}, 10 \mathrm{~N}$

The correct answer is:
$10 \sqrt{ } 3 \mathrm{~N}, 10 \mathrm{~N}$

Solution:
Equilibrium of the system is achieved when the net force acting on the system is zero i.e. the sum of the horizontal component of forces must be zero and the sum vertical components of the force must be zero.
$\Rightarrow \Sigma F_x=0$ and $\Sigma F_y=0$
Therefore, for equilibrium $A \sin 30^{\circ}=B$ and $A \cos 30^{\circ}=C$
$\Rightarrow B=20 \cos 30^{\circ}=10 \sqrt{ } 3 \mathrm{~N}$
$\Rightarrow C=20 \sin 30^{\circ}=10 \mathrm{~N}$
Hence the correct option is (3).

Topic: Physics, Mathematics in Physics, JEE Main
Three large identical plates are kept parallel to each other. The outer two plates are maintained at temperatures T and 2T, respectively. The temperature of the middle plate in steady state will be close to

Options:
A 1.1 T
B 1.3 T
C 1.7 T
D 1.9 T

The correct answer is:
1.7 T

Solution:
Plates does not touch each other, so exchange of heat takes place in form of radiations.

In steady state, heat gained by middle plate = heat lost by it.

⇒ Heat radiations from plate A + Heat radiations from plate B= Heat radiations emitted from both surfaces

⇒AεσT4+Aεσ2T4=2AεσT14

⇒T4+24T4=2T14⇒172T4=T14

⇒T14=17214·T or T1=1.7 T

Topic: Physics, Thermal Properties of Matter, KVPY, KVPY 2019 (SB/SX)
Three lines $3 x-y=2,5 x+a y=3$ and $2 x+y=3$ are concurrent, then $a=$

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

The correct answer is:
-2

Solution:
$\begin{aligned} & \left|\begin{array}{ccc}3 & -1 & -2 \\ 5 & a & -3 \\ 2 & 1 & -3\end{array}\right|=0 \\ & \Rightarrow 3(-3 a+3)+1(-15+6)-2(5-2 a)=0 \Rightarrow a=-2\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main
Three lines are drawn from the origin $O$ with direction cosines proportional to (1,-1,1) (2,-3,0) and $(1,0,3) .$ The three lines are

Options:
A not coplanar
B coplanar
C perpendicular to each other
D coincident

The correct answer is:
coplanar

Solution:
$\quad$ $\Delta=\left|\begin{array}{ccc}1 & -1 & 1 \\ 2 & -3 & 0 \\ 1 & 0 & 3\end{array}\right|$
$=1(-9-0)-(-1) \cdot(6-0)+1(0-(-3))$
$=-9+6+3=0$
since, $\Delta=0$
$\therefore$ Lines are coplanar.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Three liquids with masses $m_1, m_2, m_3$ are thoroughly mixed. If their specific heats are $c_1, c_2, c_3$ and their temperatures $T_1, T_2, T_3$ respectively, then the temperature of the mixture is

Options:
A $\frac{c_1 T_1+c_2 T_2+c_3 T_3}{m_1 c_1+m_2 c_2+m_3 c_3}$
B $\frac{m_1 c_1 T_1+m_2 c_2 T_2+m_3 c_3 T_3}{m_1 c_1+m_2 c_2+m_3 c_3}$
C $\frac{m_1 c_1 T_1+m_2 c_2 T_2+m_3 c_3 T_3}{m_1 T_1+m_2 T_2+m_3 T_3}$
D $\frac{m_1 T_1+m_2 T_2+m_3 T_3}{c_1 T_1+c_2 T_2+c_3 T_3}$

The correct answer is:
$\frac{m_1 c_1 T_1+m_2 c_2 T_2+m_3 c_3 T_3}{m_1 c_1+m_2 c_2+m_3 c_3}$

Solution:
Let the final temperature be $T^{\circ} \mathrm{C}$.
Total heat supplied by the three liquids in coming down to $0^{\circ} \mathrm{C}=m_1 c_1 T_1+m_2 c_2 T_2+m_3 c_3 T_3$ ...(i)
Total heat used by three liquids in raising temperature from $0^{\circ} \mathrm{C}$ to $T^{\circ} \mathrm{C}$
$=m_1 c_1 T+m_2 c_2 T+m_3 c_3 T$ ...(ii)
By equating (i) and (ii) we get
$\begin{aligned} & \left(m_1 c_1+m_2 c_2+m_3 c_3\right) T \\ = & m_1 c_1 T_1+m_2 c_2 T_2+m_3 c_3 T_3 \\ \Rightarrow & T=\frac{m_1 c_1 T_1+m_2 c_2 T_2+m_3 c_3 T_3}{m_1 c_1+m_2 c_2+m_3 c_3}\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main
Three long, straight, parallel wires carrying different currents are arranged as shown in the diagram. In the given arrangement, let the net force per unit length on the wire ' $C$ ' be $\vec{F}$. If the wire ' $B$ ' is removed without disturbing the other two wires, then the force per unit length on wire ' $\mathrm{A}$ ' is

Options:
A $-\overrightarrow{\mathrm{F}}$
B $3 \overrightarrow{\mathrm{F}}$
C $2 \overrightarrow{\mathrm{F}}$
D $-3 \overrightarrow{\mathrm{F}}$

The correct answer is:
$-3 \overrightarrow{\mathrm{F}}$

Solution:
Force on one wire. due to the other current carrying wive is $\quad F=\frac{\mu_0^2 i_1 i_2 \times e}{4 \pi x}$
Given $\left(\frac{F}{l}\right)_c=\vec{F}$
$\begin{aligned} & =+\frac{\mu_0}{4 d} 2(2 i)(i)+\left(-\frac{\mu_0}{4 \pi(2 d)} 2(2 i)(3 i)\right) \\ & =\frac{\mu_0 i^2}{4 \pi d}(-6+4) \Rightarrow \vec{F}=-\frac{2 \mu_0 i^2}{4 \pi d}\end{aligned}$
Now we need to calculate force per unit length on wire $A$ if $B$ is removed
$\begin{aligned} \Rightarrow\left(\frac{F}{l}\right)_A & =-\frac{\mu_0}{4 \pi} \frac{2(2 i)(3 i)}{2 d} \\ \left(\frac{F}{l}\right)_A & =3\left(\frac{2 \mu_0 i^2}{4 \pi}\right) \\ \left(\frac{F}{e}\right)_A & =-3 \vec{F} \end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main
Three machines \( E_{1}, E_{2} \) and \( E_{3} \) in a certain factory produce \( 50 \%, 25 \% \) and \( 25 \% \), respectively of the total daily output of
electric tubes. It is known that \( 4 \% \) of the tubes produced one each of machines \( E_{1} \) and \( E_{2} \) are defective and that \( 5 \% \) of
those produced on \( E_{3} \) are defective. If one tube is picked up at random from a day's production, the probability that it is
defective, is

Options:
A \( 0.025 \)
B \( 0.125 \)
C \( 0.325 \)
D \( 0.0425 \)

The correct answer is:
\( 0.0425 \)

Solution:
Let

E→Event that the tube is defective.

B1→Event when E1 produces the tube.

B2→Event when E2 produces the tube.

B3→Event when E3 produces the tube.

PB1=50100=12 and PEB1=4100=125.

PB2=25100=14 and PEB2=4100=125.

PB3=25100=14 and PEB3=5100=120.

Now,

PE=PEB1PB1+EB3PB+EB3PB3.

⇒ PE=12×125+14×125+14×120

⇒ PE=0.0425

Topic: Mathematics, Probability, JEE Main
Three masses are placed on the $x$-axis : $300 \mathrm{~g}$ at origin, $500 \mathrm{~g}$ at $x=40 \mathrm{~cm}$ and $400 \mathrm{~g}$ at $x=70 \mathrm{~cm}$. The distance of the centre of mass from the origin is

Options:
A $40 \mathrm{~cm}$
B $45 \mathrm{~cm}$
C $50 \mathrm{~cm}$
D $30 \mathrm{~cm}$

The correct answer is:
$40 \mathrm{~cm}$

Solution:
$$
\text { } \begin{aligned}
x_{\mathrm{CM}} & =\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3} \\
& =\frac{300(0)+500(40)+400(70)}{300+500+400} \\
& =\frac{20000+28000}{1200} \\
& =\frac{48000}{1200} \\
& =40 \mathrm{~cm}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Three masses are placed on the $x$-axis : $300 \mathrm{~g}$ at origin, $500 \mathrm{~g}$ at $x=40 \mathrm{~cm}$ and $400 \mathrm{~g}$ at $x=70 \mathrm{~cm}$. The distance of the centre of mass from the origin is

Options:
A $40 \mathrm{~cm}$
B $45 \mathrm{~cm}$
C $50 \mathrm{~cm}$
D $30 \mathrm{~cm}$

The correct answer is:
$40 \mathrm{~cm}$

Solution:
$$
\text { } \begin{aligned}
x_{\mathrm{CM}} & =\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3} \\
& =\frac{300(0)+500(40)+400(70)}{300+500+400} \\
& =\frac{20000+28000}{1200} \\
& =\frac{48000}{1200} \\
& =40 \mathrm{~cm}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2012 (Mains)
Three masses $\mathrm{m}, 2 \mathrm{~m}$ and $3 \mathrm{~m}$ are moving in $\mathrm{x}-\mathrm{y}$ plane with speed $3 \mathrm{u}, 2 \mathrm{u}$ and $\mathrm{u}$ respectively as shown in figure. The three masses collide at the same point at $\mathrm{P}$ and stick together. The velocity of resulting mass will be:

Options:
A
$\frac{\mathrm{u}}{12}(\hat{\mathrm{i}}+\sqrt{3} \hat{\mathrm{j}})$

B
$\frac{\mathrm{u}}{12}(\hat{\mathrm{i}}-\sqrt{3} \hat{\mathrm{j}})$

C
$\frac{\mathrm{u}}{12}(-\hat{\mathrm{i}}+\sqrt{3} \hat{\mathrm{j}})$

D
$\frac{\mathrm{u}}{12}(-\hat{\mathrm{i}}-\sqrt{3} \hat{\mathrm{j}})$

The correct answer is:
$\frac{\mathrm{u}}{12}(-\hat{\mathrm{i}}-\sqrt{3} \hat{\mathrm{j}})$

Solution:
From the law of conservation of momentum we know that,
$$
\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2+\ldots .=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2+\ldots .
$$
Given $\mathrm{m}_1=\mathrm{m}, \mathrm{m}_2=2 \mathrm{~m}$ and $\mathrm{m}_3=3 \mathrm{~m}$ and $\mathrm{u}_1=3 \mathrm{u}, \mathrm{u}_2=2 \mathrm{u}$ and $\mathrm{u}_3=\mathrm{u}$
Let the velocity when they stick $=\overrightarrow{\mathrm{v}}$
$$
\text { Then, according to question, }
$$
$$
\begin{aligned}
&m \times 3 u(\hat{i})+2 m \times 2 u\left(-\hat{i} \cos 60^{\circ}-\hat{j} \sin 60^{\circ}\right) \\
&+3 \mathrm{~m} \times \mathrm{u}\left(-\hat{\mathrm{i}} \cos 60^{\circ}+\hat{\mathrm{j}} \sin 60^{\circ}\right) \\
&=(\mathrm{m}+2 \mathrm{~m}+3 \mathrm{~m}) \overrightarrow{\mathrm{v}} \\
&\Rightarrow 3 m u \hat{\mathrm{i}}-4 \mathrm{mu} \frac{\hat{\mathrm{i}}}{2}-4 \mathrm{mu}\left(\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)-3 \mathrm{mu} \frac{\hat{\mathrm{i}}}{2} \\
&+3 m u\left(\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)=6 \mathrm{~m} \mathrm{v} \\
&\Rightarrow m u \hat{i}-\frac{3}{2} m u \hat{i}-\frac{\sqrt{3}}{2} m \hat{j}=6 \mathrm{~m} \mathrm{v} \\
&\Rightarrow-\frac{1}{2} m u \hat{i}-\frac{\sqrt{3}}{2} m u \hat{j}=6 \mathrm{~m} \mathrm{v} \\
&\Rightarrow \overrightarrow{\mathrm{v}}=\frac{\mathrm{u}}{12}(-\hat{\mathrm{i}}-\sqrt{3} \hat{\mathrm{j}}) \\
&
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Three masses $\mathrm{m}, 2 \mathrm{~m}$ and $3 \mathrm{~m}$ are moving in $\mathrm{x}-\mathrm{y}$ plane with speed $3 \mathrm{u}, 2 \mathrm{u}$ and $\mathrm{u}$ respectively as shown in figure. The three masses collide at the same point at $\mathrm{P}$ and stick together. The velocity of resulting mass will be:

Options:
A
$\frac{\mathrm{u}}{12}(\hat{\mathrm{i}}+\sqrt{3} \hat{\mathrm{j}})$

B
$\frac{\mathrm{u}}{12}(\hat{\mathrm{i}}-\sqrt{3} \hat{\mathrm{j}})$

C
$\frac{\mathrm{u}}{12}(-\hat{\mathrm{i}}+\sqrt{3} \hat{\mathrm{j}})$

D
$\frac{\mathrm{u}}{12}(-\hat{\mathrm{i}}-\sqrt{3} \hat{\mathrm{j}})$

The correct answer is:
$\frac{\mathrm{u}}{12}(-\hat{\mathrm{i}}-\sqrt{3} \hat{\mathrm{j}})$

Solution:
From the law of conservation of momentum we know that,
$$
\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2+\ldots .=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2+\ldots .
$$
Given $\mathrm{m}_1=\mathrm{m}, \mathrm{m}_2=2 \mathrm{~m}$ and $\mathrm{m}_3=3 \mathrm{~m}$ and $\mathrm{u}_1=3 \mathrm{u}, \mathrm{u}_2=2 \mathrm{u}$ and $\mathrm{u}_3=\mathrm{u}$
Let the velocity when they stick $=\overrightarrow{\mathrm{v}}$
$$
\text { Then, according to question, }
$$
$$
\begin{aligned}
&m \times 3 u(\hat{i})+2 m \times 2 u\left(-\hat{i} \cos 60^{\circ}-\hat{j} \sin 60^{\circ}\right) \\
&+3 \mathrm{~m} \times \mathrm{u}\left(-\hat{\mathrm{i}} \cos 60^{\circ}+\hat{\mathrm{j}} \sin 60^{\circ}\right) \\
&=(\mathrm{m}+2 \mathrm{~m}+3 \mathrm{~m}) \overrightarrow{\mathrm{v}} \\
&\Rightarrow 3 m u \hat{\mathrm{i}}-4 \mathrm{mu} \frac{\hat{\mathrm{i}}}{2}-4 \mathrm{mu}\left(\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)-3 \mathrm{mu} \frac{\hat{\mathrm{i}}}{2} \\
&+3 m u\left(\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)=6 \mathrm{~m} \mathrm{v} \\
&\Rightarrow m u \hat{i}-\frac{3}{2} m u \hat{i}-\frac{\sqrt{3}}{2} m \hat{j}=6 \mathrm{~m} \mathrm{v} \\
&\Rightarrow-\frac{1}{2} m u \hat{i}-\frac{\sqrt{3}}{2} m u \hat{j}=6 \mathrm{~m} \mathrm{v} \\
&\Rightarrow \overrightarrow{\mathrm{v}}=\frac{\mathrm{u}}{12}(-\hat{\mathrm{i}}-\sqrt{3} \hat{\mathrm{j}}) \\
&
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2014 (12 Apr Online)
Three moles of AgCl get precipitated when one mole of an octahedral co-ordination compound with empirical formula CrCl3.3NH3.3H2O reacts with excess of silver nitrate. The number of chloride ions satisfying the secondary valency of the metal ion is ______ .

Options:

The correct answer is:
0

Solution:
Mole of AgCl precipitated is equal the mole of Cl- present in ionization sphere.

Since none of Cl- is present in the co ordination sphere. Therefore answer is zero.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Three moles of an ideal gas are compressed isothermally from \(60 \mathrm{~L}\) to \(20 \mathrm{~L}\) using constant pressure of \(5 \mathrm{~atm}\). Heat exchange \(\mathrm{Q}\) for the compression is - ______ Lit. atm.

Options:

The correct answer is:
200

Solution:
As isothermal $\Delta \mathrm{U}=0$
and process is irreversible
$\begin{aligned}
& \mathrm{Q}=-\mathrm{W}=-\left[-\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right)\right] \\
& \mathrm{Q}=5(20-60)=-200 \mathrm{~atm}-\mathrm{L}
\end{aligned}$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2024 (04 Apr Shift 2)
Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be

Options:
A infinite
B $3 \mathrm{~J}$
C $9 \mathrm{~J}$
D zero

The correct answer is:
zero

Solution:
$W=\mathrm{p}_{\text {ext }} \cdot \Delta \mathrm{V}$
For vacuum, $\quad \mathrm{P}_{\text {ext }}=0$
$\therefore \quad \mathrm{W}=0 \cdot \Delta \mathrm{V}=0$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2010 (Mains)
Three moles of an ideal gas undergo a cyclic process shown in figure. The work done by the gas during the process is [Take ln(2)=0.693]

Options:
A 2.36 RT0
B 0.58 RT0
C 1.16 RT0
D –0.58 RT0

The correct answer is:
1.16 RT0

Solution:
If V0 be volume of the gas at C, P0V0=3RT0
W=3R(2T0)ln2âP0V0
=6RT0ln2â3RT0
=3RT0[2Ã0.693â1]
=3RT0[1.386â1]
=1.16RT0

Topic: Physics, Thermodynamics, JEE Main
Three moles of B2H6 are completely reacted with methanol. The number of moles of boron containing product formed is

Options:

The correct answer is:
6

Solution:
B2H6+6MeOH→2B(OMe)3+6H2
Diborane reacts with methanol to give hydrogen and trimethoxyborate ester.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Three moles of \( \mathrm{B}_{2} \mathrm{H}_{6} \) are completely reacted with methanol. The number of moles of boron containing product formed is

Options:

The correct answer is:
6

Solution:
B2H6+6CH3OH⟶2BOCH33+6H2
For 3 moles of B2H6 mole of B containing product formed = 6

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Three numbers a, b and c are in between 2 and 18 such that 2, a, b are in arithmetic progression and b, c, 18 are in geometric progression. If a+b+c=25, then the value of c-a is

Options:
A 4
B 3
C 7
D 0

The correct answer is:
7

Solution:
a+b+c=25; 2a=2+b; c2=18b
⇒2a+2b+2c=50
⇒2+b+2b+2c=50
⇒3b+2c=48⇒c26+2c=48
⇒c2+12c-48×6=0⇒c2+12c-24×12=0
⇒c+24c-12=0
⇒c=12,-24⇒c=12 (between 2 and 18 )
⇒b=c218=14418=8
⇒a=b+22=5
⇒a=5, b=8 and c=12
⇒c-a=7

Topic: Mathematics, Sequences and Series, JEE Main
Three numbers a, b and c are in geometric progression. If 4a, 5b and 4c are in arithmetic progression and a+b+c=70, then the value of c-a is equal to

Options:
A 10
B 20
C 30
D 40

The correct answer is:
30

Solution:
a+b+c=70
⇒a+ar+ar2=70
Also, 10b=4a+4c
⇒10ar=4a+4ar2
⇒2r2-5r+2=0⇒r=2,12
For r=2, we have,
a1+r+r2=70⇒a=701+r+r2=701+2+4=10
⇒a=10, b=20, c=40⇒c-a=30
For r=12, we have,
a=701+r+r2=701+12+14=70×44+2+1=40
⇒b=20, c=10
⇒c-a=10-40=30
So, in both the cases c-a=30

Topic: Mathematics, Sequences and Series, JEE Main
Three numbers are chosen at random without replacement from $\{1,2,3, \ldots . .8\}$. The probability that their minimum is $3$ , given that their maximum is $6$ , is

Options:
A
$\frac{3}{8}$

B
$\frac{1}{5}$

C
$\frac{1}{4}$

D
$\frac{2}{5}$

The correct answer is:
$\frac{1}{5}$

Solution:
Let $A$ be the event that maximum is $6$.
$B$ be event that minimum is $3$
$P(A)=\frac{{ }^5 C_2}{{ }^8 C_3}$ (the numbers $ < 6$ are $5$)
$P(B)=\frac{{ }^5 C_2}{{ }^8 C_3}$ (the numbers $>3$ are $5$)
$P(A \cap B)=\frac{{ }^2 C_1}{{ }^8 C_3}$
Required probability is $\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{{ }^2 \mathrm{C}_1}{{ }^5 \mathrm{C}_2}=\frac{2}{10}=\frac{1}{5}$.

Topic: Mathematics, Probability, JEE Main, JEE Main 2012 (Offline)
Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3r2, then r2-d is equal to :

Options:
A 7-3
B 7+33
C 7-73
D 7+3

The correct answer is:
7+3

Solution:
Three numbers are in an increasing geometric progression with common ratio r.

Let first term is a

⇒ar,a,ar ∈ G.P

Given that if middle term is doubled

⇒ar,2a,ar ∈ A.P

If a, b, c ∈ A. P ⇒ 2b =a + c
4a=ar+ar

4=r+1r

r2+1=4r

r2-4r+1=0

Ifax2 + bx +c = 0 ⇒ x = -b ± b2-4ac2ac

⇒r=4±122=2+3,2-3

But it is an increasing G. P

∴ r = 2 + 3

Given that Fourth term of G.P ⇒ t4 = ar2 = 3r2

⇒ a = 3.

Common difference of an A.P = d = t2 - t1

d=2a-ar=a2-1r=32-12+3

=3(2-2+3)=33
Hence r2-d=(2+3)2-(33)

= 4 + 3 + 43 - 33

=7+3

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (31 Aug Shift 1)
Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. These have masses m, 2m and m, respectively. The object A moves towards B with a speed 9 m s-1 and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m s-1) of the object C.

Options:
A 4
B 7
C 10
D 12

The correct answer is:
4

Solution:
After elastic collision,
Ï'A=m-2mm+2m9+22mm+2m0=-3 m s-1
Now from conservation of linear momentum after all collisions are complete,
m+9 m s-1=m-3 m s-1+3mÏC
or ÏC=4 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Three of the following pairs of the human skeletal parts are correctly matched with their respective inclusive skeletal category and one pair is not matched. Identify the non-matching pair.

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(b)

Solution:
Clavicle and glenoid cavity are found in pectoral girdle. Each half of pectoral girdle consists of a clavicle and a scapula. Glenoid cavity is a depression below the acromion (the expanded process of scapula). The glenoid cavity articulates with the head of the humenes to form the shoulder joint.

Topic: Biology, Locomotion and Movement, NEET, NEET 2011 (Mains)
Three of the following statements about enzymes are correct and one is wrong. Which one is wrong?

Options:
A Enzymes require optimum $\mathrm{pH}$ for maximal activity
B Enzymes are denatured at high temperature but in certain exceptional organisms they are effective even at temperatures $80^{\circ}-90^{\circ} \mathrm{C}$
C Enzymes are highly specific
D Most enzymes are proteins but some are lipids

The correct answer is:
Most enzymes are proteins but some are lipids

Solution:
Enzymes show $\mathrm{pH}$ sensitivity and require an optimum $\mathrm{pH}$ for maximal activity.
Enzymes are thermolabile (heat sensitive) and denatured at high temperatures. However, extremophiles (thermophilic bacteria) are exceptions.
Most of the enzymes are highly specific in their action.
Most enzymes are proteins but exceptionally some are RNA enzymes, eg, ribozymes.

Topic: Biology, Biomolecules (B), NEET, NEET 2010 (Mains)
Three organic compounds A, B and C were allowed to run in thin layer chromatography using hexane and gave the following result (see figure). The Rf value of the most polar compound is _____ ×10–2

Options:

The correct answer is:
25

Solution:
In thin layer chromatography using hexane, the least polar compound will rise to maximum height and most polar compound will rise to minimum height.

The retardation factor of each spot can be determined by dividing the distance that a particular compound has travelled by the distance between solvent front and initial spotting site. This parameter is dependent on the thin layer chromatographic plate and the solvent.

∴ Rf value for most polar compound

=28=0.25

=25×10-2

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area A3 and the separation between the plates is d, 2d and 3d respectively. The equivalent capacity of combination is (Îµ 0 = absolute permittivity of free space)

Options:
A 7 ε 0 A 18d
B 11 ε 0 A 18d
C 13 ε 0 A 18d
D 17 ε 0 A 18d

The correct answer is:
11 ε 0 A 18d

Solution:
C=QV and C=Îµ0Ad
As given C1=Îµ0 . A3d; C2=Îµ0 A6d; C3=Îµ0 A9d
â´Ceq=C1+C2+C3for parallel combination of capacitors
â´ Ceq=Îµ0Ad13+16+19=11Îµ0A18d

Topic: Physics, Electrostatics, JEE Main
Three parallel straight lines $L_1, L_2$ and $L_3$ lie on the the same plane. Consider 5 points on $L_1, 7$ points on $L_2$ and 9 points on $L_3$. Then the maximum possible number of triangles formed with vertices at these points, is

Options:
A 1330
B 1200
C 1201
D 129

The correct answer is:
1201

Solution:
Number of triangles if one vertex at each line is
$$
={ }^5 C_1 \times{ }^7 C_1 \times{ }^9 C_1=315
$$
Number of triangles if two vertices at $L_1$ and remaining at either $L_2$ or $L_3$ is $={ }^5 C_2 \times{ }^{16} C_1=160$
Number of triangles if two vertices at $L_2$ and remaining at either $L_3$ or $L_1$ is $={ }^7 C_2 \times{ }^{14} C_1=294$ and number of triangles if two vertices at $L_3$ and remaining at either $L_1$ or $L_2$ is :
$$
{ }^9 C_2 \times{ }^{12} C_1=432
$$
So, total maximum possible number of triangles $=1201$

Topic: Mathematics, Permutation Combination, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 2)
Three particles (A, B, C) each of mass m are connected by three massless rods of length l. All three particles lie on a smooth horizontal plane. A particle of mass m moving along one of the rods with velocity V0 strikes on a particle and stops (as shown in the diagram). Tension in any of the rod just after the collision is

Options:
A mV0236 l
B mV0228 l
C mV0224 l
D mV0212 l

The correct answer is:
mV0236 l

Solution:
Applying conservation of angular momentum with respect to centroid.
Initial angular momentum = final angular momentum
r=l2tan30o=l23 ; mV0l23=Icmω
mV0l23=3ml32ω; ω=V023l
2Tcos30ο=mω2l3;T=mV0236l

Topic: Physics, Center of Mass Momentum and Collision, JEE Main