datavorous/entrance-exam-dataset huggingface dataset:
Three capacitors C1=2 μF, C2=6 μF and C3=12 μF are connected as shown in the figure. Find the ratio of the charges on capacitors C1, C2 and C3 respectively.

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

The correct answer is:
1:2:2

Solution:
From the above circuit diagram, the equivalent capacitance of the combination of capacitors is,

Ceq=C1+C2C3C2C3,

Ceq=2+6×126+12=6 μF

Then, QC1QC2=C1VC2C3C2+C3V=2×(6+12)6×12=12

and QC2=QC3

So, the ratio of charges are, 1:2:2.

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (27 Aug Shift 2)
Three capacitors each of 4 μF are to be connected in such a way that the effective capacitance is 6 μF. This can be done by connecting them

Options:
A all in series
B all in parallel
C two in parallel and one in series
D two in series and one in parallel

The correct answer is:
two in series and one in parallel

Solution:
For top branch,
1Ceq=14+14
⇒Ceq=2 μF
For parallel combination, Cnet=2+4=6 μF

Topic: Physics, Capacitance, JEE Main, JEE Main 2016 (09 Apr Online)
Three capacitors each of capacitance $C$ and of breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be

Options:
A $\frac{\mathrm{C}}{3}, \frac{\mathrm{V}}{3}$
B $3 \mathrm{C}, \frac{\mathrm{V}}{3}$
C $\frac{\mathrm{C}}{3}, 3 \mathrm{~V}$
D $3 \mathrm{C}, 3 \mathrm{~V}$

The correct answer is:
$\frac{\mathrm{C}}{3}, 3 \mathrm{~V}$

Solution:
In series arrangement charge on each plate of each capacitor has same magnitude.
The potential difference is distributed inversely in the ratio of capacitors ie,
$$
\mathrm{V}=\mathrm{V}_1+\mathrm{V}_2+\mathrm{V}_3 \ldots
$$
Here, $V=3 V$
The equivalent capacitance $\mathrm{C}_s$ is given by
$$
\begin{aligned}
& \frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots \\
& C_s=\frac{\mathrm{C}}{3}
\end{aligned}
$$

Topic: Physics, Capacitance, JEE Main
Three capacitors each of capacitance $C$ and of breakdown voltage $\mathrm{V}$ are joined in series. The capacitance and breakdown voltage of the combination will be :

Options:
A $3 \mathrm{C}, 3 \mathrm{~V}$
B $\frac{\mathrm{C}}{3}, \frac{\mathrm{V}}{3}$
C $3 \mathrm{C}, \frac{\mathrm{V}}{3}$
D $\frac{\mathrm{C}}{3}, 3 \mathrm{~V}$

The correct answer is:
$\frac{\mathrm{C}}{3}, 3 \mathrm{~V}$

Solution:
In series, $\mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}}{3}$,
$$
\mathrm{V}_{\mathrm{eq}}=3 \mathrm{~V}
$$

Topic: Physics, Capacitance, NEET, NEET 2009 (Mains)
Three capacitors each of capacity $4 \mu \mathrm{F}$ are to be connected in such a way that the effective capacitance is $6 \mu \mathrm{F}$. This can be done by:

Options:
A connecting all of them in series
B connecting them in parallel
C connecting two in series and one in parallel
D connecting two in parallel and one in series

The correct answer is:
connecting two in series and one in parallel

Solution:
$$
\begin{aligned}
& \mathrm{C}_{\mathrm{AB}}=(2+4) \mu \mathrm{F} \\
& =6 \mu \mathrm{F}
\end{aligned}
$$

Topic: Physics, Capacitance, JEE Main
Three capacitors of capacitances and are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases will be

Options:
A $1: 15$
B $15: 1$
C $\quad 1: 1$
D $\quad 1: 3$

The correct answer is:
$1: 15$

Solution:
$\begin{aligned} & \frac{1}{C_s}=\frac{1}{3}+\frac{1}{9}+\frac{1}{18}=\frac{1}{2} \Rightarrow C_s=2 \mu \mathrm{F} \\ & C_p=3+9+18=30 \mu \mathrm{F} \Rightarrow \frac{C_s}{C_p}=\frac{2}{30}=\frac{1}{15}\end{aligned}$

Topic: Physics, Capacitance, JEE Main
Three capacitors of capacity $C_1, C_2 C_3$ are connected in series. Their total capacity will be

Options:
A $C_1+C_2+C_3$
B $1 /\left(C_1+C_2+C_3\right)$
C $\left(C_1^{-1}+C_2^{-1}+C_3^{-1}\right)^{-1}$
D None of these

The correct answer is:
$\left(C_1^{-1}+C_2^{-1}+C_3^{-1}\right)^{-1}$

Solution:
$\frac{1}{C_R}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \Rightarrow C_R=\left(C_1^{-1}+C_2^{-1}+C_3^{-1}\right)^{-1}$

Topic: Physics, Capacitance, JEE Main
Three charged particles A, B and C with charges -4q,2q and -2q are present on the circumference of a circle of radius d. The charged particles A,C and centre O of the circle formed an equilateral triangle as shown in the figure. The electric field at the point O is

Options:
A 3qπε0d2
B 23qπε0d2
C 3q4πε0d2
D 33q4πε0d2

The correct answer is:
3qπε0d2

Solution:
Enet=4kqd2×2cos30°=q3πε0d2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Three charged particles A, B and C with charges -4q,2q and -2q are present on the circumference of a circle of radius d. The charged particles A,C and centre O of the circle formed an equilateral triangle as shown in the figure. The electric field at the point O is

Options:
A 3qπε0d2
B 23qπε0d2
C 3q4πε0d2
D 33q4πε0d2

The correct answer is:
3qπε0d2

Solution:
Enet=4kqd2×2cos30°=q3πε0d2

Topic: Physics, Electrostatics, JEE Main
Three charges, each $+q$, are placed at the corners of an isosceles triangle $A B C$ of sides $B C$ and $A C, 2 a . D$ and $E$ are the mid points of $B C$ and $C A$. The work done in taking a charge $Q$ from $D$ to $E$ is

Options:
A $\frac{e q Q}{8 \pi \varepsilon_0 a}$
B $\frac{q Q}{4 \pi \varepsilon_0 a}$
C zero
D $\frac{3 q Q}{4 \pi \varepsilon_0 a}$

The correct answer is:
zero

Solution:
$$
\text { Here, } A C=B C
$$
$$
\begin{aligned}
& V_D=V_E=V \\
& W=Q\left[V_E-V_D\right] \\
& W=Q[V-V] \\
& W=0
\end{aligned}
$$

Topic: Physics, Electrostatics, NEET, NEET 2011 (Mains)
Three charges $-q_1,+q_2$ and $-q_3$ are placed as shown in the figure. The $x$-component of the force on $-q_1$ is proportional to

Options:
A
$\frac{\mathrm{q}_2}{\mathrm{~b}^2}-\frac{\mathrm{q}_3}{\mathrm{a}^2} \cos \theta$

B
$\frac{\mathrm{q}_2}{\mathrm{~b}^2}+\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta$

C
$\frac{\mathrm{q}_2}{\mathrm{~b}^2}+\frac{\mathrm{q}_3}{\mathrm{a}^2} \cos \theta$

D
$\frac{\mathrm{q}_2}{\mathrm{~b}^2}-\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta$

The correct answer is:
$\frac{\mathrm{q}_2}{\mathrm{~b}^2}+\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta$

Solution:
The force body diagram
$F_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_3}{a^2}$ $\mathrm{F}_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1 \mathrm{q}_3}{\mathrm{~b}^2}$
$\mathrm{F}_{\mathrm{X}}=\mathrm{F}_1 \sin \theta+\mathrm{F}_2=\frac{\mathrm{q}_1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta+\frac{\mathrm{q}_1}{\mathrm{~b}_2}\right] \Rightarrow \mathrm{F}_{\mathrm{X}} \propto\left(\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta+\frac{\mathrm{q}_2}{\mathrm{~b}^2}\right)$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2003
Three charges $-q_1,+q_2$ and $-q_3$ are placed as shown in the figure. The $x$-component of the force on $-q_1$ is proportional to

Options:
A
$\frac{\mathrm{q}_2}{\mathrm{~b}^2}-\frac{\mathrm{q}_3}{\mathrm{a}^2} \cos \theta$

B
$\frac{\mathrm{q}_2}{\mathrm{~b}^2}+\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta$

C
$\frac{\mathrm{q}_2}{\mathrm{~b}^2}+\frac{\mathrm{q}_3}{\mathrm{a}^2} \cos \theta$

D
$\frac{\mathrm{q}_2}{\mathrm{~b}^2}-\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta$

The correct answer is:
$\frac{\mathrm{q}_2}{\mathrm{~b}^2}+\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta$

Solution:
The force body diagram
$F_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_3}{a^2}$ $\mathrm{F}_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1 \mathrm{q}_3}{\mathrm{~b}^2}$
$\mathrm{F}_{\mathrm{X}}=\mathrm{F}_1 \sin \theta+\mathrm{F}_2=\frac{\mathrm{q}_1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta+\frac{\mathrm{q}_1}{\mathrm{~b}_2}\right] \Rightarrow \mathrm{F}_{\mathrm{X}} \propto\left(\frac{\mathrm{q}_3}{\mathrm{a}^2} \sin \theta+\frac{\mathrm{q}_2}{\mathrm{~b}^2}\right)$

Topic: Physics, Electrostatics, JEE Main
Three charges $Q_1+q$ and $+q$ are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if $Q$ is equal to

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

The correct answer is:
$\frac{-2 q}{2+\sqrt{2}}$

Solution:
Net electrostatic energy
$U=\frac{k Q q}{a}+\frac{k q^2}{a}+\frac{k Q q}{a \sqrt{2}}=0$
$\Rightarrow \frac{k q}{a}\left(Q+q+\frac{Q}{\sqrt{2}}\right)=0$
$\Rightarrow Q=-\frac{2 q}{2+\sqrt{2}}$

Topic: Physics, Electrostatics, JEE Main
Three charges +Q, q,+Q are placed respectively, at distance, 0,d/2 and d from the origin, on the x -axis. If the net force experienced by +Q, placed at x=0, is zero, then value of q is:

Options:
A +Q/2
B +Q/4
C -Q/4
D -Q/2

The correct answer is:
-Q/4

Solution:
For net force on charge placed at origin is given as
⇒ KQ.q d22+KQ2d2=0, where K is a constant.

Solving the above equation, we get, q=-Q4

Hence, the charge q=-Q4 such that net force on the charge +Q placed at origin is zero.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Three charges $Q,+y$ and $+q$ are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of $\mathrm{Q}$ is

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

The correct answer is:
$\frac{-\sqrt{2} q}{\sqrt{2}+1}$

Solution:
Net electrostatic energy for the system

$\mathrm{U}=\mathrm{K}\left[\frac{\mathrm{q}^{2}}{\mathrm{a}}+\frac{\mathrm{Qq}}{\mathrm{a}}+\frac{\mathrm{Qq}}{\mathrm{a} \sqrt{2}}\right]=0$

$\begin{aligned} \Rightarrow q &=-Q\left[1+\frac{1}{\sqrt{2}}\right] \\ & \Rightarrow Q=\frac{-q \sqrt{2}}{\sqrt{2}+1} \end{aligned}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Three children, each accompanied by a guardian, seek admission in a school. The principal want to interview all the 6 persons one after the other subject to the condition that no child is interviewed before its guardian. In how many ways can this be done -

Options:
A 60
B 90
C 120
D 180

The correct answer is:
90

Solution:
$\frac{6 !}{(2 !)^{3}}=90$

Topic: Mathematics, Permutation Combination, JEE Main
Three circles each of radius 1 , touch one another externally and they lie between two parallel lines. The minimum possible distance between the lines is

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

The correct answer is:
$2+\sqrt{3}$

Solution:
$\sin 60^{\circ}=\frac{A D}{2}$
$A D=2 \sin 60^{\circ}=\frac{2 \sqrt{3}}{2}=\sqrt{3}$
$d=1+A D+1$
$d=2+\sqrt{3}$

Topic: Mathematics, Circle, JEE Main
Three circles of radii a, b, c, a<b<c touch each other externally. If they have x- axis as a common tangent, then:

Options:
A 1a=1b+1c
B a, b, c are in A.P.
C 1b=1a+1c
D a, b, c are in A.P.

The correct answer is:
1a=1b+1c

Solution:
We have,

LM=PN=a+b2-b-a2=2ab
NQ=MS=a+c2-c-a2=2ac
PQ=ST=b+c2-c-b2=2bc
∵ PQ=PN+NQ
⇒2bc=2ab+2ac
⇒1a=1b+1c

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Three coins are tossed. If one of them shows tail, then the probability that all three coins show tail, is

Options:
A $\frac{1}{7}$
B $\frac{1}{8}$
C $\frac{2}{7}$
D $\frac{1}{6}$

The correct answer is:
$\frac{1}{7}$

Solution:
Let $E$ be the event in which all three coins shows tail and $F$ be the event in which a coin shows tail.
$\therefore F=\{H H T, H T H, T H H, H T T, T H T, T T H, T T T\}$
and $E=\{T T T\}$.
Required probability
$=P(E / F)=\frac{P(E \cap F)}{P(E)}=\frac{1}{7}$

Topic: Mathematics, Probability, JEE Main
Three complexes,
CoClNH352+I, CoNH35H2O3+(II) and CoNH363+III
absorb light in the visible region. The correct order of the wavelength of light absorbed by them is:

Options:
A II>I>III
B III>II>I
C I>II>III
D III>I>(II)

The correct answer is:
I>II>III

Solution:
On increasing strength of ligand, wavelength of absorbed radiation decreases
Strength of ligands according to spectrochemical series is: NH3>H2O>Cl-
∴ Order of wavelength absorbed will be Cl->H2O>NH3
∴ Correct order of wavelength absorbed is
I>II>III

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +σ, -σ and +σ respectively. The potential of shell B is:

Options:
A σε0b2-c2c+a
B σε0a2-b2a+c
C σε0a2-b2b+c
D σε0b2-c2b+a

The correct answer is:
σε0a2-b2b+c

Solution:
Net potential at B is due to the superposition of potential due to all shells,
∨B=∨BI+∨BII+∨BIII
=K4πa2σb+K4πb2-σb+K4πc2σc
Required potential of b
=σa2ε0b-σb2ε0b+σc2ε0c
=σε0a2-b2b+c

Topic: Physics, Electrostatics, JEE Main, JEE Main 2018 (08 Apr)
Three concentric spherical metallic shells X, Y and Z of radius a, b and c respectively a<b<c have surface charge densities σ, – σ and σ, respectively. The shells X and Z are at same potential. If the radii of X & Y are 2 cm and 3 cm, respectively. The radius of shell Z is _____cm.

Options:

The correct answer is:
5

Solution:
The charges on the spheres are

qX=σ4πa2qY=-σ4πb2qZ=σ4πc2

It is given that VX=VZ

So,

qX4πε0a+qY4πε0b+qZ4πε0c=qX4πε0c+qY4πε0c+qZ4πε0c⇒σ4πa2a-σ4πb2b+σ4πc2c=σ4π(a2-b2+c2)c⇒c(a-b+c)=(a2-b2+c2)⇒c=a+b=2 cm+3 cm=5 cm

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Three concentric spherical shells have radii a, b, and $\mathrm{c}(\mathrm{a} < \mathrm{b} < \mathrm{c})$ and have surface charge densities $\sigma,-\sigma$ and $\sigma$ respectively. If $V_A, V_B$ and $V_C$ denote the potentials of the three shells, then, for $c=a+b$, we have :

Options:
A $V_C=V_B=V_A$
B $\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{A}} \neq \mathrm{V}_{\mathrm{B}}$
C $V_C=V_B \neq V_A$
D $\mathrm{V}_{\mathrm{C}} \neq \mathrm{V}_{\mathrm{B}} \neq \mathrm{V}_{\mathrm{A}}$

The correct answer is:
$\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{A}} \neq \mathrm{V}_{\mathrm{B}}$

Solution:
$$
\begin{aligned}
\mathrm{q}_{\mathrm{A}} & =4 \pi \mathrm{a}^2 \sigma, \\
\mathrm{q}_{\mathrm{B}} & =-4 \pi \mathrm{b}^2 \sigma, \\
\mathrm{q}_{\mathrm{C}} & =4 \pi \mathrm{c}^2 \sigma, \mathrm{c}=\mathbf{a}+\mathrm{b} \\
\mathrm{V}_{\mathrm{A}} & =\frac{1}{4 \pi \in_0}\left(\frac{\mathrm{q}_{\mathrm{A}}}{\mathrm{a}}+\frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{b}}+\frac{\mathrm{q}_{\mathrm{C}}}{\mathrm{c}}\right) \\
& =\frac{2 \sigma \mathrm{a}}{\epsilon_0} \\
\mathrm{~V}_{\mathrm{B}} & =\frac{1}{4 \pi \epsilon_0}\left(\frac{\mathrm{q}_{\mathrm{A}}}{\mathrm{a}}+\frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{b}}+\frac{\mathrm{q}_{\mathrm{C}}}{\mathrm{c}}\right)
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{\sigma}{\epsilon_0}\left(\frac{\mathbf{a}^2}{\mathbf{b}}-\mathbf{b}+\mathrm{c}\right) \\
& =\frac{\sigma}{\epsilon_0}\left(\mathbf{a}+\frac{\mathbf{a}^2}{\mathbf{b}}\right) \\
\mathrm{V}_{\mathrm{C}} & =\frac{1}{4 \pi \in_0}\left(\frac{\mathrm{q}_{\mathrm{A}}}{\mathbf{a}}+\frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{b}}+\frac{\mathrm{q}_{\mathrm{C}}}{\mathrm{c}}\right) \\
& =\frac{\sigma}{\epsilon_0}\left(\frac{\mathrm{a}^2-\mathbf{b}}{\mathrm{c}}+\mathrm{c}\right)=\frac{2 \sigma \mathbf{a}}{\epsilon_0}
\end{aligned}
$$
So, $V_C=V_A \neq V_B$

Topic: Physics, Electrostatics, NEET, NEET 2009 (Mains)
Three concentric spherical shells have radii a, b and c (a<b<c) and have surface charge densities Ï , - Ï and Ï respectively. If VA, VB and VC denote the potentials of the three shells, then, for c=a+b, we have

Options:
A VC=VB≠VA

B VC≠VB≠VA
C VC=VB=VA
D VA=VC>VB

The correct answer is:
VA=VC>VB

Solution:
V A = 1 4 Ï Îµ 0 q A a + q B b + q C c
= 4 Ï 4 Ï Îµ 0 a 2 Ï a - b 2 Ï b + c 2 Ï c
VA=1Îµ0a2Ïa-b2Ïb+c2ÏcâVA=1Îµ0aÏ-bÏ+cÏ=2aÏÎµ0
VB=1Îµ0a2Ïb-b2Ïb+c2ÏcVB=1Îµ0a2Ïb-bÏ+cÏVB=aÏÎµ0ab-1
VC=1Îµ0a2Ïc-b2Ïc+c2ÏcVC=ÏÎµ0a2-b2a+b+cVC=ÏÎµ0a-b+cVC=2ÏaÎµ0

VA=VC>VB

Topic: Physics, Electrostatics, JEE Main
Three concurrent co-planar forces $1 \mathrm{~N}, 2 \mathrm{~N}$ and $3 \mathrm{~N}$ acting along different directions on a body

Options:
A can keep the body in equilibrium if $2 \mathrm{~N}$ and $3 \mathrm{~N}$ act at right angle
B can keep the body in equilibrium if $1 \mathrm{~N}$ and $2 \mathrm{~N}$ act at right angle
C cannot keep the body in equilibrium
D can keep the body in equilibrium in $1 \mathrm{~N}$ and $3 \mathrm{~N}$ act at an acute angle

The correct answer is:
cannot keep the body in equilibrium

Solution:
If we keep $1 \mathrm{~N}$ and $2 \mathrm{~N}$ forces act in same direction then these are balanced by $3 \mathrm{~N}$ force, but this is against statement of question. Hence, options (c) is correct.

Topic: Physics, Laws of Motion, JEE Main
Three conducting rods $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{BD}$ made of same material and same cross section are arranged as shown in the figure. Temperatures at the points A, D and C are maintained at $20^{\circ} \mathrm{C}, 90^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ respectively. The ratio of lengths of $\mathrm{BD}$ to $\mathrm{BC}$ when there is no heat flow in $A B$, is

Options:
A $\frac{2}{9}$
B $\frac{7}{2}$
C $\frac{2}{7}$
D $\frac{9}{2}$

The correct answer is:
$\frac{7}{2}$

Solution:
No solution. Refer to answer key.

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2017 (24 Apr Shift 1)
Three containers C1,C2 and C3 have water at different temperatures. The table below shows the final temperature T when different amounts of water (given in liters) are taken from each container and mixed (assume no loss of heat during the process)

C1
C2
C3
T

1l
2l
--
60oC

--
1l
2l
30oC

2l
--
1l
60oC

1l
1l
1l
θ

The value of θ (in oC to the nearest integer) is___________

Options:

The correct answer is:
50

Solution:
Let θ1, θ2 and θ3 be the temperatures of water in three containers.

1θ1+2θ2+0θ3=1+260
θ1+2θ2=180 …. (1)
0×θ1+1×θ2+2×θ3=1+230
θ2+2θ3=90 …… (2)
2×θ1+0×θ2+1×θ3=2+160
2θ1+θ3=180 …. (3)
and θ1+θ2+θ3=1+1+1θ
from (1)+(2)+(3)
3θ1+3θ2+3θ3=450⟹θ1+θ2+θ3=150
from (4) equation 150=3θ⟹θ=50 °C

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Three containers C1,C2 and C3 have water at different temperatures. The table below shows the final temperature T when different amounts of water (given in liters) are taken from each container and mixed (assume no loss of heat during the process)

C1	C2	C3	T

1l	2l	--	60oC

--	1l	2l	30oC

2l	--	1l	60oC

1l	1l	1l	θ

The value of θ (in oC to the nearest integer) is___________

Options:

The correct answer is:
50

Solution:
Let θ1, θ2 and θ3 be the temperatures of water in three containers.

1θ1+2θ2+0θ3=1+260
θ1+2θ2=180 …. (1)
0×θ1+1×θ2+2×θ3=1+230
θ2+2θ3=90 …… (2)
2×θ1+0×θ2+1×θ3=2+160
2θ1+θ3=180 …. (3)
and θ1+θ2+θ3=1+1+1θ
from (1)+(2)+(3)
3θ1+3θ2+3θ3=450⟹θ1+θ2+θ3=150
from (4) equation 150=3θ⟹θ=50 °C

Topic: Physics, Thermal Properties of Matter, JEE Main
Three critics review a book. For the three critics the odds in favor of the book are $2: 5$, $3: 4$ and $4: 3$ respectively. The probability that the majority is in favor of the book, is given by

Options:
A $\frac{183}{343}$
B $\frac{160}{343}$
C $\frac{209}{343}$
D $\frac{134}{343}$

The correct answer is:
$\frac{134}{343}$

Solution:
The probability that the first critic favors the book is $\mathrm{P}(\mathrm{A})=\frac{2}{2+5}=\frac{2}{7}$
$\therefore \quad \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\frac{2}{7}=\frac{5}{7}$
The probability that the second critic favors the book is $\mathrm{P}(\mathrm{B})=\frac{3}{3+4}=\frac{3}{7}$
$\therefore \quad \mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\frac{3}{7}=\frac{4}{7}$
The probability that the third critic favors the book is $\mathrm{P}(\mathrm{C})=\frac{4}{4+3}=\frac{4}{7}$
$\therefore \quad \mathrm{P}\left(\mathrm{C}^{\prime}\right)=1-\frac{4}{7}=\frac{3}{7}$
$\therefore \quad$ Majority will be in favor of the book if at least two critics favor the book.
Hence, the probability is
$\begin{aligned}
& \mathrm{P}\left(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}^{\prime}\right)+\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime} \cap \mathrm{C}\right) \\
& \quad+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B} \cap \mathrm{C}\right)+\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}) \\
& =\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}\left(\mathrm{C}^{\prime}\right)+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right) \cdot \mathrm{P}(\mathrm{C}) \\
& \quad+\mathrm{P}\left(\mathrm{A}^{\prime}\right) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{C}) \\
& =\frac{2}{7} \times \frac{3}{7} \times \frac{3}{7}+\frac{2}{7} \times \frac{4}{7} \times \frac{4}{7}+\frac{5}{7} \times \frac{3}{7} \times \frac{4}{7}+\frac{2}{7} \times \frac{3}{7} \times \frac{4}{7} \\
& =\frac{18}{343}+\frac{32}{343}+\frac{60}{343}+\frac{24}{343}=\frac{134}{343}
\end{aligned}$

Topic: Mathematics, Probability, JEE Main
Three crops that contribute maximum to global food grain production are

Options:
A wheat, rice and maize
B wheat, maize and sorghum
C rice, maize and sorghum
D wheat, rice and barley

The correct answer is:
wheat, rice and maize

Solution:
Wheat, rice and maize belongs to family Poaceae or Graminae. The main fruit type of these crops is caryopsis in which fruit wall is fused with seed coat. These crops are cultivated in all over the world and are contributed maximum to global food grain production.

Topic: Biology, Strategies for Enhancement in Food Production, JIPMER, JIPMER 2007
Three crops that contribute maximum to global food grain production are:

Options:
A wheat, rice and maize
B wheat, rice and barley
C wheat, maize and sorghum
D rice, maize and sorghum.

The correct answer is:
wheat, rice and maize

Solution:
Wheat, rice and maize belongs to family Poaceae or Graminae. The main fruit type of these crops is caryopsis in which fruit wall is fused with seed coat. These crops are cultivated in all over the world and are contributed maximum to global food grain production.

Topic: Biology, Strategies for Enhancement in Food Production, NEET, NEET 2005
Three dice are rolled. If the probability of getting different numbers on the three dice is pq, where p and q are co-prime, then q-p is equal to

Options:
A 2
B 1
C 3
D 4

The correct answer is:
4

Solution:
Given that three dice are thrown.

The total number of outcomes when three dice are thrown together is 63=6×6×6.

The number of outcomes such that all the outcomes are different is =6×5×4.

For ex: If the outcome in dice 1 is "6" then the number of outcomes for dice 2 should be 5 (1,2,3,4,5) which excludes the outcome "6".

Hence, the required probability is =6×5×46×6×6

=2036=59=pq

⇒q-p=9-5=4

Therefore, the required answer is 4.

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Three different types of dielectric slabs have been arranged between the plates of a parallel plate capacitor, as shown in the figure. The equivalent capacitance of the system between the points P and Q will be

Options:
A K ε 0 A d
B 2K ε 0 A d
C 3K Îµ 0 A d
D 3K ε 0 A 2d

The correct answer is:
3K ε 0 A d

Solution:
C1=(2K)ε0A2d, C2=(3K)ε0Ad, C3=(6K)ε0Ad

Ceq=C1+C2C3C2+C3=3Kε0Ad

Topic: Physics, Electrostatics, JEE Main
Three discs, A, B and C, having radii 2 m, 4 m and 6 m, respectively, are coated with carbon black on their outer surfaces. The wavelengths corresponding to maximum intensity are 300 nm, 400 nm and 500 nm, respectively. The power radiated by them are Q A , Q B and Q C respectively. The correct statement, among the following, is

Options:
A Q A is maximum
B Q B is maximum
C Q C is maximum
D Q A = Q B = Q C

The correct answer is:
Q B is maximum

Solution:
Q ∝ AT 4 and λ m T = Constant
Hence ,
Q ∝ A λ m 4 or Q ∝ r 2 λ m 4
QA : QB : QC=2234 : 4244 : 6254
= 4 8 1 : 1 1 6 : 3 6 6 2 5 = 0.05 : 0.0625 : 0.0576
i.e. , Q B is maximum .

Topic: Physics, Thermal Properties of Matter, JEE Main