datavorous/entrance-exam-dataset huggingface dataset:
Two capillary tubes $P$ and $Q$ are dipped in water. The height of water level in capillary $P$ is $2 / 3$ to the height in $Q$ capillary. The ratio of their diameters is

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

The correct answer is:
$3: 2$

Solution:
$r \propto \frac{1}{h} \Rightarrow \frac{r_p}{r_Q}=\frac{h_Q}{h_p}=\frac{h}{\frac{2}{3} h}=\frac{3}{2}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then PX=1+PX=2 equals:

Options:
A 24169
B 52169
C 49169
D 25169

The correct answer is:
25169

Solution:
Using the concept of complementary events,
Px=1+Px=2=1-Px=0
=1-4852×4852
=1-144169=25169

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Two Carnot engines $\mathrm{A}$ and $\mathrm{B}$ are operated in series. Engine A receives heat from a reservoir at $600 \mathrm{~K}$ and rejects heat to a reservoir at temperature T. Engine B receives heat rejected by engine A and in turn rejects it to a reservoir at $100 \mathrm{~K}$. If the efficiencies of the two engines $\mathrm{A}$ and $\mathrm{B}$ are represented by $\eta_A$ and $\eta_B$ respectively, then what is the value of $\frac{\eta_{\mathrm{A}}}{\eta_{\mathrm{B}}}$

Options:
A
$\frac{12}{7}$

B
$\frac{12}{5}$

C
$\frac{5}{12}$

D
$\frac{7}{12}$

The correct answer is:
$\frac{7}{12}$

Solution:
Efficiency of engine $A, n_A=\frac{T_1-T_2}{T_1}$ and $n_B=\frac{T_2-T_3}{T_2} ; T_2=\frac{T_1+T_3}{2}=350 \mathrm{~K}$ or $\frac{n_A}{n_B}=\frac{\frac{600-350}{600}}{\frac{350-100}{350}}=\frac{7}{12}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Two carnot engines A and B are operated in series. The first one, A, receives heat at T1=600K and rejects to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and, in turn, rejects to a heat reservoir at T3=400K. Calculate the temperature T2 if the work outputs of the two engines are equal:

Options:
A 500 K
B 400 K
C 300 K
D 600 K

The correct answer is:
500 K

Solution:
W=Q1-Q2=Q2-Q3
⇒Q1Q2+Q3Q2=2
T1T2+T3T2=2
T2=T1+T32=500K

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with speed of 20 m s-1 with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source in car B?
(speed of sound in air =340 m s-1 )

Options:
A 2150 Hz
B 2300 Hz
C 2250 Hz
D 2060 Hz

The correct answer is:
2250 Hz

Solution:
From Doppler effect: f'=V∓ V0V±VSf0
2000=340-20340+20f0=320360f0
f0=98×2000=2250 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Two cars A and B are moving with a speed of 15 m s-1 and 5 m s-1 and acceleration of the cars are 1 m s-2 and 6 m s-2 respectively. Then the minimum separation between them is

Options:
A 5 m
B 10 m
C Zero
D None

The correct answer is:
10 m

Solution:
vAB=10 m s-1, aAB=-5 m s-2
sAB=10t-52t2

Minimum displacement of A w.r.t. B
sABmin is at t=105=2
Minimum separation =20-10=10 m

Topic: Physics, Motion In One Dimension, JEE Main
Two cars are approaching each other at an equal speed of 7.2 km hr-1. When they see each other, both blow horns having a frequency of 676 Hz. The beat frequency heard by each driver will be _______Hz. [Velocity of sound in air is 340 m s-1.

Options:

The correct answer is:
8

Solution:
Frequency of sound heard by car-1, which comes by reflection from car-2

f1=f0340+2340-2340+2340-2

=f03423382

Frequency of sound coming directly from car-2

f2=f0340+2340-2

∴ f1-f2=f0342338342338-1=8.09≃8 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Two cars are travelling towards each other at speed of $20 \mathrm{~m} \mathrm{~s}^{-1}$ each. When the cars are $300 \mathrm{~m}$ apart, both the drivers apply brakes and the cars retard at the rate of $2 \mathrm{~m} \mathrm{~s}^{-2}$. The distance between them when they come to rest is :

Options:
A $200 \mathrm{~m}$
B $100 \mathrm{~m}$
C $50 \mathrm{~m}$
D $25 \mathrm{~m}$

The correct answer is:
$100 \mathrm{~m}$

Solution:
$\begin{aligned} & \left|\overrightarrow{\mathrm{u}}_{\mathrm{BA}}\right|=40 \mathrm{~m} / \mathrm{s} \\ & \left|\overrightarrow{\mathrm{a}}_{\mathrm{BA}}\right|=4 \mathrm{~m} / \mathrm{s} \\ & \text { Apply }\left(\mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as}\right)_{\text {relative }} \\ & \mathrm{O}=(40)^2+2(-4)(\mathrm{S}) \\ & \mathrm{S}=200 \mathrm{~m} \\ & \text { Remaining distance }=300-200=100 \mathrm{~m}\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Two cars of masses $m_1$ and $m_2$ are moving in circles of radii $r_1$ and $r_2$, respectively. Their speeds are such that they make complete circles in the same time $t$. The ratio of their centripetal acceleration is

Options:
A
$m_1 r_1: m_2 r_2$

B
$m_1: m_2$

C
$r_1: r_2$

D
$1: 1$

The correct answer is:
$r_1: r_2$

Solution:
$a \propto r$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2012 (Offline)
Two cars of masses ' $\mathrm{m}_{1}$ ', and ' $\mathrm{m}_{2}$ ' are moving in the circles of radii ' $\mathrm{r}_{1}$ ' and ' $\mathrm{r}_{2}$ '
respectively. Their angular speeds ' $\omega_{1}$ ' and ' $\omega_{2}$ ' are such that they both complete one revolution in the same time ' $\mathrm{t}^{\prime}$. The ratio of linear speed of ' $\mathrm{m}_{1}$ ' to the linear
speed of ' $\mathrm{m}_{2}$ ' is

Options:
A $\mathrm{r}_{1}: \mathrm{r}_{2}$
B $\mathrm{~T}_{1}^{2}: \mathrm{T}_{2}^{2}$
C $\omega_{1}^{2}: \omega_{2}^{2}$
D $\mathrm{~m}_{1}: \mathrm{m}_{2}$

The correct answer is:
$\mathrm{r}_{1}: \mathrm{r}_{2}$

Solution:
They complete one revolution in the same time.
$$
\begin{array}{l}
\therefore \omega_{1}=\omega_{2} \\
\therefore \frac{V_{1}}{r_{1}}=\frac{V_{2}}{r_{2}} \\
\therefore \frac{V_{1}}{V_{2}}=\frac{r_{1}}{r_{2}}
\end{array}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main
Two cars P and Q start from a point at the same time in a straight line and their positions are represented by XPt=at+bt2 and XQt=ft-t2. At what time do the cars have the same velocity?

Options:
A a-f1+b
B a+f2b-1
C a+f21+b
D f-a21+b

The correct answer is:
f-a21+b

Solution:
vP=dxPdt=a+2bt
vQ=dxQdt=f-2t
vP=vQ
⇒a+2bt=f-2t
2t+2bt=f-a
⇒t=f−a2(b+1)

Topic: Physics, Motion In One Dimension, JEE Main
Two cars X and Y are approaching each other with velocities 36 km h-1 and 72 km h-1 respectively. The frequency of a whistle sound as emitted by a passenger in car X, heard by the passenger in car Y is 1320 Hz. If the velocity of sound in air is 340 ms-1, the actual frequency of the whistle sound produced is ________ Hz.

Options:

The correct answer is:
1210

Solution:
f'=f0V+V0 V-VS

f0=1320340-10340+20=1320×3336=1210 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Two cells are connected between points A and B as shown. Cell 1 has emf of 12 V and internal resistance of 3 Ω. Cell 2 has emf of 6 V and internal resistance of 6 Ω. An external resistor R of 4 Ω is connected across A and B. The current flowing through R will be ______ A.

Options:

The correct answer is:
1

Solution:
Equivalent internal resistance of two cells in parallel is req=r1r2r1+r2=3×63+6=2 Ω

Equivalent emf of given cells

Eeq=E1r1-E2r2req=123-662=6 V

Given: R=4 Ω

The circuit can be redrawn as shown below.

Current through resistor R is I=62+4=1 A.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Two cells are connected in opposition as shown. Cell E1 is of 8 V emf and 2 Ω internal resistance; the cell E2 is of 2 V emf and 4 Ω internal resistance. The terminal potential difference of cell E2 is ______ V.

Options:

The correct answer is:
6

Solution:
Current in circuit, I=8-22+4=66=1 A

Applying Kirchhoff's law from C to B,

VC-2-4×1=VB

⇒VC-VB=6 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Two cells, e.m.f. of each is $E$ and internal resistance $r$ are connected in parallel between the resistance $R$. The maximum energy given to the resistor will be, only when

Options:
A $R=r / 2$
B $R=r$
C $R=2 r$
D $R=0$

The correct answer is:
$R=r / 2$

Solution:
For maximum energy, we have External resistance of the circuit = Equivalent internal resistance of the circuit i.e. $R=r / 2$

Topic: Physics, Current Electricity, JEE Main
Two cells having internal resistance 0.2 Î© and 0.4 Î© are connected in parallel with the same polarity. The voltage across the battery terminals is 1.5 V. If the emf of the first cell is 1.2 V, then what is the emf of the second cell (in volt)?

Options:

The correct answer is:
2.1

Solution:
Given that, the resultant voltage across the battery terminal=1.5 V
Clearly, 1.2 Volt Battery is Charging and another battery is discharging given,
1.2+0.2I=1.5
âI=32 A
E-0.4I=1.5
âE=1.5+0.4Ã32
âE=2.1 V

Topic: Physics, Current Electricity, JEE Main
Two cells, having the same e.m.f. are connected in series through an external resistance $R$. Cells have internal resistances $r_1$ and $r_2\left(r_1>r_2\right)$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is:

Options:
A $r_1+r_2$
B $r_1-r_2$
C $\frac{r_1+r_2}{2}$
D $\frac{r_1-r_2}{2}$

The correct answer is:
$r_1-r_2$

Solution:
According to the question
$E-I r_1=0 \text { and } \mathrm{I}=\frac{E+E}{r_1+r_2+R}$
$\begin{array}{l}
\therefore \frac{E}{r_1} =\frac{2 E}{r_1+r_2+R} \\
\Rightarrow r_1+r_2+R =2 r_1-r_2 \\
\Rightarrow R =r_1-r_2
\end{array}$

Topic: Physics, Current Electricity, JEE Main
Two cells, having the same E.M.F. are connected in series through an external resistance R. Cells have internal resistances r1 and r2 r1>r2 respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is

Options:
A r1+r22
B r1-r22
C r1+r2
D r1-r2

The correct answer is:
r1-r2

Solution:
Current in the circuit
=E+Er1+r2+R=2Er1+r2+R
Potential difference across first cell = E-ir1
=E-2E×r1r1+r2+R
Now, E-2Er1r1+r2+R=0
⇒E=2Er1r1+r2+R⇒2r1=r1+r2+R
R=r1-r2

Topic: Physics, Current Electricity, JEE Main
Two cells of emf 2E and E with internal resistance r1and r2 respectively are connected in series to an external resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is

Options:
A r1+r2
B r12-r2
C r12+r2
D r1-r2

The correct answer is:
r12-r2

Solution:
i=3ER+r1+r2

T.P.D.=2E-ir1=0

2E=ir1

⇒2E=3E×r1R+r1+r2

2R+2r1+2r2=3r1

R=r12-r2

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Two cells of E.M.F. E1 and E2 E2 > E1 are connected in series in the secondary circuit of a potentiometer experiment for determination of E.M.F. The balancing length is found to be 825 cm. Now when the terminals of cell E1 are reversed, then the balancing length is found to be 225 cm. The ratio of E1 and E2 is

Options:
A 2Â :Â 3
B 4 : 7
C 7Â :Â 4
D none of these

The correct answer is:
4 : 7

Solution:
E2 + E1 = x825
E2 â E1 = x225
Dividing above equations, we get
E2+E1E2-E1=113
Appling componendo and dividendo, we get
E1E2=47

Topic: Physics, Current Electricity, JEE Main
Two cells of the same EMF E but different internal resistances r1 and r2 are connected in series with an external resistance R as shown in the figure. The terminal potential difference across, the second cell is found to be zero. The external resistance R must then be:

Options:
A r1r2
B r1-r2
C r2-r1
D r1+r2

The correct answer is:
r2-r1

Solution:
In this question we have to find the Terminal potential difference of second cell, we are given the internal resistances of both the cells and an external resistance. Applying the following expression:-

V = E-ir2

⇒0=E-ir2

Applying the expression of current

0=E-2Er1+r2+R·r2

⇒R=r2-r1

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (25 Jun Shift 2)
Two chambers containing $m_1 \mathrm{~g}$ and $m_2 \mathrm{~g}$ of a gas at pressures $P_1$ and $P_2$ respectively are put in communication with each other, temperature remaining constant. The common pressure reached will be

Options:
A $\frac{P_1 P_2\left(m_1+m_2\right)}{\left(P_2 m_1+P_1 m_2\right)}$
B $\frac{P_1 P_2 m_1}{P_2 m_1+P_1 m_2}$
C $\frac{m_1 m_2\left(P_1+P_2\right)}{P_2 m_1+P_1 m_2}$
D $\frac{m_1 m_2 P_2}{P_2 m_1+P_1 m_2}$

The correct answer is:
$\frac{P_1 P_2\left(m_1+m_2\right)}{\left(P_2 m_1+P_1 m_2\right)}$

Solution:
According to Boyle'slaw, $P V=k$ (a constant)
or $P \frac{m}{\rho}=k$ or $\rho=\frac{P m}{k}$
$\left(\because V=\frac{m}{\rho}\right)$
$\rho=\frac{P}{K}$
$\left(\because \frac{k}{m}=K\right.$ a constant $)$
So, $\rho_1=\frac{P_1}{K}$ and $V_1=\frac{m_1}{\rho_1}=\frac{m_1}{P_1 / K}=\frac{K m_1}{P_1}$
Similarly, $V_2=\frac{K m_2}{P_2}$
$\therefore \quad$ Total volume $=V_1+V_2=K\left(\frac{m_1}{P_1}+\frac{m_2}{P_2}\right)$
Let $P$ be the common pressure and $\rho$ be the common density of mixture. Then
$\begin{aligned} & \rho=\frac{m_1+m_2}{V_1+V_2}=\frac{m_1+m_2}{K\left(\frac{m_1}{P_1}+\frac{m_2}{P_2}\right)} \\ \therefore \quad P & =K \rho=\frac{m_1+m_2}{\frac{m_1}{P_1}+\frac{m_2}{P_2}}=\frac{P_1 P_2\left(m_1+m_2\right)}{\left(m_1 P_2+m_2 P_1\right)}\end{aligned}$

Topic: Physics, Kinetic Theory of Gases, AIIMS, AIIMS 2017
Two charge metallic spheres are joined by a very thin metal wire. If the radius of the larger sphere is four times that of the smaller sphere, the electric field near the larger sphere is

Options:
A twice that near the smaller sphere
B quarter of that near the smaller sphere
C same as that near the smaller sphere
D half of that near smaller sphere

The correct answer is:
quarter of that near the smaller sphere

Solution:
Since the spheres are joined by a metal wire, their potentials will be same.
Let the radius of the smaller sphere be $r_1$ and the radius of the larger sphere be $r_2=4 r_1$ the $\mathrm{v}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1}{\mathrm{r}_1}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_2}{\mathrm{r}_2}$
$$
\therefore \frac{\mathrm{q}_2}{\mathrm{q}_1}=\frac{\mathrm{r}_2}{\mathrm{r}_1}
$$
The electric field near the surface of spheres are given by
$$
\begin{aligned}
& \mathrm{E}_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1}{\mathrm{r}_1^2} \text { and } \mathrm{E}_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_2}{\mathrm{r}_2^2} \\
& \therefore \frac{\mathrm{E}_2}{\mathrm{E}_1}=\frac{\mathrm{q}_2}{\mathrm{q}_1} \cdot \frac{\mathrm{r}_1^2}{\mathrm{r}_2^2}=\frac{\mathrm{r}_2}{\mathrm{r}_1} \cdot \frac{\mathrm{r}_1^2}{\mathrm{r}_2^2}=\frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{1}{4}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main
Two charged conducting spheres of radii $a$ and $b$ are connected to each other by a conducting wire. The ratio of charges of the two spheres respectively is:

Options:
A $\frac{a}{b}~v$
B $\sqrt{a b}$
C $\frac{b}{a}$
D $a b$

The correct answer is:
$\frac{a}{b}~v$

Solution:
Potential at surface will be same
$\begin{aligned} & \frac{\mathrm{Kq}_1}{\mathrm{a}}=\frac{\mathrm{Kq}_2}{\mathrm{~b}} \\ & \frac{\mathrm{q}_1}{\mathrm{q}_2}=\frac{\mathrm{a}}{\mathrm{b}}\end{aligned}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of radii of their circular paths is 6:5 and their respective masses ratio is 9:4. Then, the ratio of their charges will be

Options:
A 8:5
B 5:4
C 5:3
D 8:7

The correct answer is:
5:4

Solution:
Radius of charged particle moving in circular path R=2mKqB, here, B is magnetic field and K is kinetic energy.

So, q=2mKRB

Now, ratio of the charges is q1q2=m1 m2×R2R1=94×56=54

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (25 Jul Shift 1)
Two charged thin infinite plane sheets of uniform charge density σ+ and σ-, where σ+>σ−, intersect at the right angle. Which of the following best represents the electric field lines for the system:

Options:
A
B
C
D

The correct answer is:

Solution:
The electric field intensity due to each uniformly charged infinite plane is uniform. The electric field intensity at points A, B, C and D as shown in figure. Hence, the electric lines of forces are as given in below figure.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Two charges \( +3.2 \times 10^{-19} \mathrm{C} \) and \( -3.2 \times 10^{-19} \mathrm{C} \) are placed at \( 2.4 \mathrm{~A} \) apart from an electric dipole. It is placed in a uniform electric field of intensity \( 4 \times 10^{5} \mathrm{Vm}^{-1} \). The electric dipole moment is

Options:
A \( 15.36 \times 10^{-29} \mathrm{Cm} \)
B \( 15.36 \times 10^{-19} \mathrm{Cm} \)
C \( 7.68 \times 10^{-29} \mathrm{Cm} \)
D \( 7.68 \times 10^{-19} \mathrm{Cm} \)

The correct answer is:
\( 7.68 \times 10^{-29} \mathrm{Cm} \)

Solution:
Electric dipole moment of a dipole is given by,

p=qd (where, q is the charge and d is the distance between the charges)

Given that q=±3.2×10-19 C and d=2.4 A∘=2.4×10-10 m

On substituting the given values in the above formula, we get,

p=3.2×10-192.4 ×10-10

p=7.68×10-29 C m

Topic: Physics, Electrostatics, JEE Main
Two charges, each equal to q , are kept at x = - a and x = a on the x-axis. A particle of mass m and charge q0=-q2 is placed at the origin. If charge q 0 is given a small displacement (y << a) along the y-axis, the net force acting on the particle is proportional to :

Options:
A 1 y
B - 1 y
C -y
D y

The correct answer is:
-y

Solution:
⇒ F net = 2Fcos θ
Fnet=2kqq2y2+a22yy2+a2
Fnet=2kqq2yy2+a23/2≈kq2a3y⇒F∝-y

Net force F is opposite the direction of displacement y. Hence, negative sign is used.

Topic: Physics, Oscillations, JEE Main, JEE Main 2013 (07 Apr)
Two charges of -4 μC and +4 μC are placed at the points A(1, 0, 4) m and B(2, -1, 5) m located in an electric field E→=0.20i^ V cm-1. The magnitude of the torque acting on the dipole is 8α×10-5 N m, where α=_________.

Options:

The correct answer is:
2

Solution:
Given the electric field is

E→=0.20 i^ Vcm=20i^ Vm

The formula for calculating the torque is

τ→=p→×E→ ...1

where,

p→=ql→=4×2i^+-1j^+5k^-1i^+0j^+4k^=4i^-j^+k^ μC m

Thus, from equation (1), it follows that

τ→=(4i^-4j^+4k^)×10-6×(20i^) N m=(8k^+8j^)×10-5 N m⇒τ→=82+82×10-5 N m=82×10-5 N m

Hence, α=2.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Two charges of 5Q and -2Q are situated at the points (3a, 0) and (-5a, 0) respectively. The electric flux through a sphere of radius 4a having centre at origin is:

Options:
A 2Qε0
B 5Qε0
C 7Qε0
D 3Qε0

The correct answer is:
5Qε0

Solution:
As can be seen from the figure above, only the 5Q charge is inside the spherical region.

The formula to calculate the flux through the surface is given by

ϕ=qencε0 ...1

From equation (1), the flux through the given surface can be calculated as follows:

ϕ=5Qε0

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Two charges of each magnitude 0.01 C and separated by a distance of 0.4 mm constitute an electric dipole. If the dipole is placed in an uniform electric field E→ of 10 dyne·C-1 making 30∘ angle with E→, the magnitude of torque acting on dipole is:

Options:
A 4.0×10-10 N m
B 1.0×10-8 N m
C 1.5×10-9 N m
D 2.0×10-10 N m

The correct answer is:
2.0×10-10 N m

Solution:
The formula to calculate the dipole moment is given by

p⇀=qd⇀ ...1

Substitute the values of the known parameters into equation (1) to calculate the dipole moment.

p⇀=0.01 C×0.4 mm×1 m103 mm= 4.0×10-6 C·m

The formula to calculate the torque on the dipole in the electric field is given by

|τ→|=|p→×E→| ...2

Substitute the values of the known parameters into equation (2) to calculate the required torque.

τ⇀= 4.0×10-6 C·m×10 dyne·C-1× 1 N105 dyne×sin30°= 2×10-10 N m

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Two charges q and 3q are separated by a distance ‘r’ in air. At a distance x from charge q, the resultant electric field is zero. The value of x is :

Options:
A (1+3)r
B r3(1+3)
C r(1+3)
D r(1+3)

The correct answer is:
r(1+3)

Solution:
Given: E→netP=0

⇒kqx2=k×3qr-x2

⇒r-x2=3x2

⇒r-x=3x

⇒x=r3+1

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Two charges \( +q \) and \( -q \) are kept apart. Then at any point on the right bisector of line joining the two charges:

Options:
A The electric field strength is zero
B The electric potential is zero
C Both electric potential and electric field strength are zero.
D Both electric potential and electric field strength are non-zero.

The correct answer is:
The electric potential is zero

Solution:
At equatorial point,due to dipole electric field is
Ee=14πε0pr3
But electric potential due to dipole at a point on the equiorial line is
(directed from +q to -q) and Ve=0.

V=Q4πε0r⇒Ve=+q4πε0-q4πε0=0

Topic: Physics, Electrostatics, JEE Main
Two chemically non-reactive liquids are placed in a U-shaped tube as shown in the figure. The height of any liquid above a common reference line is

Options:
A directly proportional to their densities
B inversely proportional to their densities
C directly proportional to square of their densities
D equal

The correct answer is:
inversely proportional to their densities

Solution:
The pressure at the common reference must be the same, for both the liquids. Since both tubes are open to the atmosphere, we must have

h1ρ1g=h2ρ2g
⇒h1ρ1=h2ρ2
⇒hρ=constant
⇒h∝1ρ

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two chief functions of leaves are:

Options:
A Photosynthesis and respiration
B Photosynthesis and transpiration
C Transpiration and respiration
D Respiration and digestion

The correct answer is:
Photosynthesis and transpiration

Solution:
Photosynthesis and transpiration both are equally essential processes because photosynthesis helps to maintain the equilibrium position of O2 and CO2 in the atmosphere and transpiration maintains the atmospheric temperature.

Topic: Biology, Morphology of Flowering Plants, NEET
Two circles have a direct common tangent with length 36 cm. The shortest distance between these circles is 14 cm. If the radius of the longer circle is 4 times the radius of the smaller circle, then the diameter of the smaller circle (in cms) is

Options:
A 12
B 10
C 14
D 18

The correct answer is:
10

Solution:
From the figure,

9r2+362=5r+142
⇒4 r 2 +35r−25⋅11=0
⇒r-54r+55=0
Hence, r=5 cms⇒2r=10 cms

Topic: Mathematics, Circle, JEE Main
Two circles in the first quadrant of radii r1 and r2 touch the coordinate axes. Each of them cuts off an intercept of 2 units with the line x+y=2. Then r12+r22-r1r2 is equal to ____.

Options:

The correct answer is:
7

Solution:
Assuming a circle of radius r which touches both the axis is given by, x-r2+y-r2=r2

Now given line x+y=2 makes the intercept 2 the circle, so plotting the diagram we get,

Now from diagram,

AC=BC=1, OC=r2-1

Now using the formula of perpendicular distance of point from the line we get,

OC=2r-22=r2-1

⇒2r-12=r2-1

⇒r2-4r+3=0

⇒r1=1, r2=3

Hence, r12+r22-r1r2=1+9-3=7

Topic: Mathematics, Circle, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Two circles with centers at A and B touch each other externally at T. Let BD is the tangent at D and TC is a common tangent. If AT has length 3 units and BT has length 2 units, then the length (in units) of CB is

Options:
A 43
B 52
C 53
D 74

The correct answer is:
52

Solution:
From the diagram,

In ∆ABD, sin⁡B=35, cos⁡B=45
In ∆BTC, cos⁡B=TBBC=45
∴BC=52 units

Topic: Mathematics, Circle, JEE Main