datavorous/entrance-exam-dataset huggingface dataset:
Two stones are thrown up simultaneously from the edge of cliff with initial speeds v and 2v. The relative position of the second stone with respect to first varies with time till both the stones strike the ground as

Options:
A Linearly
B First linearly then parabolically
C Parabolically
D First parabolically then linearly

The correct answer is:
First linearly then parabolically

Solution:
Initially when both are in air, the relative acceleration between them is zero. Therefore, relative position verus-time graph is a straight line.
S1=vt-12gt2
S2=2vt-12gt2
S21=S2-S1
S21=vt
Once the first ball reaches the ground, its acceleration becomes zero. Therefore, the relative acceleration is now g and the graph is parabolic.
S2=2vt-12gt2

Topic: Physics, Motion In Two Dimensions, JEE Main
Two stones of mass $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ (such that $\left.\mathrm{m}_{1}>\mathrm{m}_{2}\right)$ are dropped $\Delta \mathrm{t}$ time apart from the same height towards the ground. At a later time $t$ the difference in their speed is $\Delta \mathrm{V}$ and their mutual separation is $\Delta \mathrm{S}$. While both stones are in flight

Options:
A $\Delta V$ decreases with time and $\Delta S$ increases with time
B Both $\Delta V$ and $\Delta S$ increase with time
C $\Delta \mathrm{V}$ remains constant with time and $\Delta \mathrm{S}$ decreases with time
D $\Delta V$ remains constant with time and $\Delta S$ increases with time

The correct answer is:
$\Delta V$ remains constant with time and $\Delta S$ increases with time

Solution:
$\Delta \mathrm{V}=$ Const $\& \Delta \mathrm{S}$ increases with time

Topic: Physics, Motion In Two Dimensions, KVPY, KVPY 2013 (SA)
Two straight lines having variable slopes m1 and m2 pass through the fixed points a,0 and -a,0 respectively. If m1m2=2, then the eccentricity of the locus of the point of intersection of the lines is

Options:
A 2
B 3
C 2
D 32

The correct answer is:
3

Solution:
Let, the lines are y=m1x-a and y=m2x+a
which intersects at the point h,k.
âk=m1h-a & k=m2h+a
âk2=m1m2h2-a2
âk22=h2-a2 (given, m1m2=2)
âx21-y22=a2 (Hyperbola)
Eccentricity =1+21=3

Topic: Mathematics, Hyperbola, JEE Main
Two straight parallel wires, both carrying 10 ampere in the same direction attract each other with a force of $1 \times 10^{-3} \mathrm{~N}$. If both currents are doubled, the force of attraction will be

Options:
A $1 \times 10^{-3} \mathrm{~N}$
B $2 \times 10^{-3} \mathrm{~N}$
C $4 \times 10^{-3} \mathrm{~N}$
D $0.25 \times 10^{-3} \mathrm{~N}$

The correct answer is:
$4 \times 10^{-3} \mathrm{~N}$

Solution:
\(\mathrm{F}=\frac{\mu_0}{4 \pi} \frac{2 i_1 i_2}{a} \times l=10^{-3} \mathrm{~N}\)
When current in both the wires is doubled, then
\(\mathrm{F}^{\prime}=\frac{\mu_0}{4 \pi} \frac{2\left(2 \mathrm{i}_1 \times 2 \mathrm{i}_2\right)}{\mathrm{a}} \times \mathrm{l}=4 \times 10^{-3} \mathrm{~N}\)

Topic: Physics, Magnetic Effects of Current, JEE Main
Two straight roads OA and OB intersect at O. A tower is situated within the angle formed by them and subtends angles of 45o and 30o at the points A and B where the roads are nearest to it. If OA=100 meters and OB=50 meters, then the height of the tower is

Options:
A 252 meters
B 50 meters
C 256 meters
D 253 meters

The correct answer is:
256 meters

Solution:
Let, PQ be the tower of height h.

Let, PA,PB be perpendiculars from P upon OA and OB respectively.
Then, ∠PAQ=45o and ∠PBQ=30o
OA=100,OB=50
PAh=cot⁡45o=1∴PA=h
PBh=cot⁡30o=3∴PB=3h
Now, OP2=PA2+OA2=PB2+OB2
⇒h2+1002=3h2+502⇒2h2=1002-502
⇒h=256 meters

Topic: Mathematics, Heights and Distances, JEE Main
Two straight roads OA and OB intersect at O. A tower is situated within the angle formed by them and subtends angles of 45o and 30o at the points A and B where the roads are nearest to it. If OA=400 meters and OB=300 meters, then the height of the tower is

Options:
A 2502 meters
B 500 meters
C 5014 meters
D 1007 meters

The correct answer is:
5014 meters

Solution:
Let, PQ be the tower of height h.

Let, PA,PB be the perpendiculars from P upon OA and OB respectively.
Then, ∠PAQ=45o and ∠PBQ=30o
OA=400, OB=300
⇒PAh=cot⁡45o=1∴ PA=h
⇒PBh=cot⁡30o=3∴ PB=3h
Now, OP2=PA2+OA2=PB2+OB2
⇒h2+4002=3h2+3002⇒2h2=4002-3002
⇒h=42-322×100=5014 meters

Topic: Mathematics, Heights and Distances, JEE Main
Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x:3 The value of x is

Options:

The correct answer is:
1

Solution:
KEmax=hν-ϕ

12mv2=hν-ϕ

v=2hν-ϕm

Given hν1=2ϕ

hν2=10ϕ

∴ v1v2=hν1-ϕhν2-ϕ

v1v2=2ϕ-ϕ10ϕ-ϕ=13

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Two streams of photons, possessing energies to five and ten times the work function of metal are incident on the metal surface successively. The ratio of the maximum velocities of the photoelectron emitted, in the two cases respectively, will be

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

The correct answer is:
2:3

Solution:
Using Einstein's photoelectric equation,

Maximum kinetic energy, KEmax=E-ϕ, where, E is energy of photon and ϕ is work function.

For two cases,

12mv12=5ϕ-ϕ=4ϕ and 12mv22=10ϕ-ϕ=9ϕ

Thus, the ratio v1v2=49=23=2:3

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Two substances At12=5 min and Bt12=15 min follow first order kinetics, are taken in such a way that initially A=4B. The time after which the concentration of both substances will be equal is:

Options:
A 5Â min.Â
B 10Â min.
C 15 min.
D 20Â min.

The correct answer is:
15 min.

Solution:

Topic: Chemistry, Chemical Kinetics, JEE Main
Two substances A t12=10min & B t12=20min follow I order kinetics in such a way that [A]i = 8 [B]i. Time when [B] = 2[A] in min is:

Options:
A 20
B 40
C 60
D 80

The correct answer is:
80

Solution:
For A:
8a→ 10 → 10 → 10 → 10 → 10 → 10 → 10 → 10 a32
For B:
a→ 20 → 20 → 20 → 20 a16

Topic: Chemistry, Chemical Kinetics, JEE Main
Two substances $\mathrm{R}$ and $\mathrm{S}$ decompose in solution independently, both following first order kinetics. The rate constant of $\mathrm{R}$ is twice that of $\mathrm{S}$. In an experiment, the solution initially contained 0.5 millimoles of $\mathrm{R}$ and 0.25 of $\mathrm{S}$. The molarities of $\mathrm{R}$ and $\mathrm{S}$ will be equal just at the end of time equal to

Options:
A twice the half life of $\mathrm{R}$
B twice the half life of $S$
C the half life of $\mathrm{S}$
D the half life of $\mathrm{R}$

The correct answer is:
twice the half life of $\mathrm{R}$

Solution:
Substance $\mathrm{R}$ Substance $\mathrm{S}$ $2 \mathrm{k} \quad \mathrm{k}$ rate constant $\mathrm{t}_{1 / 2}=\mathrm{n} \times \mathrm{t}_{1 / 2}$

where $\mathrm{n}=$ number of half life period

Amount of $\mathrm{R}$ left $=\frac{0.5}{(2)^{\mathrm{T} / \mathrm{t}_{/ 2}}}$

Amount of $\mathrm{S}$ left $=\frac{0.25}{(2)^{\mathrm{T} / 2 \mathrm{t} / 2}}$

Equating both $\frac{0.5}{0.25}=\frac{(2)^{\mathrm{T} / \mathrm{t}_{1 / 2}}}{(2)^{\mathrm{T} / 2 \mathrm{t}_{1 / 2}}}$

or $2=(2)^{\mathrm{T} / \mathrm{t}_{1 / 2}}$

$\therefore \mathrm{T}=2 \mathrm{t}_{1 / 2} \cdot 2 \mathrm{t}_{1 / 2}$ is half life of $\mathrm{S}$ and

twice the half -life of $\mathrm{R}$

Topic: Chemistry, Chemical Kinetics, JEE Main
Two system of rectangular axes have the same origin. If a plane cuts them at distances $\mathrm{a}, \mathrm{b}, \mathrm{c}$ and $\mathrm{a}^{\prime}, \mathrm{b}^{\prime}, \mathrm{c}^{\prime}$ from the origin then

Options:
A
$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{a^{\prime 2}}-\frac{1}{b^{\prime 2}}-\frac{1}{c^{\prime 2}}=0$

B
$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^{22}}+\frac{1}{b^{22}}+\frac{1}{c^{22}}=0$

C
$\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}-\frac{1}{\mathrm{c}^2}+\frac{1}{\mathrm{a}^{\prime 2}}+\frac{1}{\mathrm{~b}^{\prime 2}}-\frac{1}{\mathrm{c}^{\prime 2}}=0$

D
$\frac{1}{a^2}-\frac{1}{b^2}-\frac{1}{c^2}+\frac{1}{a^{\prime 2}}-\frac{1}{b^{\prime 2}}-\frac{1}{c^{\prime 2}}=0$

The correct answer is:
$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{a^{\prime 2}}-\frac{1}{b^{\prime 2}}-\frac{1}{c^{\prime 2}}=0$

Solution:
Eq. of planes be $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1 \& \frac{\mathrm{x}}{\mathrm{a}_1}+\frac{\mathrm{y}}{\mathrm{b}_1}+\frac{\mathrm{z}}{\mathrm{c}_1}=1(\perp \mathrm{r}$ distance on plane from origin is same. $)$
$\left|\frac{-1}{\sqrt{\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}+\frac{1}{\mathrm{c}^2}}}\right|=\left|\frac{-1}{\sqrt{\frac{1}{\mathrm{a}_1^2}+\frac{1}{\mathrm{~b}_1^2}+\frac{1}{\mathrm{c}_1^2}}}\right|$
$\therefore \Sigma \frac{1}{a^2}-\Sigma \frac{1}{a_1^2}=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2003
Two tall buildings are 40 m apart. With what speed must a ball be thrown horizontally from a window 145 m above the ground in one building, so that it will enter a window 22.5 m above from the ground in the other?

Options:
A 5 m s-1
B 8 m s-1
C 10 m s-1
D 16 m s-1

The correct answer is:
8 m s-1

Solution:
Given, ux=v and uy=0
Separating the motion in x and y directions
Using x=uxt+12at2
40=vt (i)
⟹t=40v (ii)
y = hnet=145-22.5

Using y=uyt+12at2
122.5=0+12×10t2
122.5=540v2
v=405122.5=8 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main
Two tangent galvanometers A and B have coils of radii 8 cm and 16 cm respectively and resistance 8 Ω each. They are connected in parallel with a cell of emf 4 V and negligible internal resistance. The deflections produced in the tangent galvanometers A and B are 30° and 60° respectively. If A has 2 turns, then B must have

Options:
A 18 turns
B 12 turns
C 6 turns
D 2 turns

The correct answer is:
12 turns

Solution:
Current in tangent galvanometer
I=2rBHμ0Ntanâ¡θ ..i

From Eq. (i), we get
rtanâ¡θN=μ0I2BH= same for both
∴ rAtanâ¡θANA=rBtanâ¡θBNB
⇒ 8×1 2 3=16 3 NB
∴ NB=12 turns

Topic: Physics, Experimental Physics, JEE Main
Two tangent galvanometers A and B have coils of radii 8 cm and 16 cm respectively and resistance 8 Î© each. They are connected in parallel with a cell of emf 4 V and negligible internal resistance. The deflections produced in the tangent galvanometers A and B are 30Â° and 60Â° respectively. If A has 2 turns, then the number of turns B must have is

Options:

The correct answer is:
12

Solution:
Current in tangent galvanometer
I=2rBHÎ¼0Ntanâ¡Î¸ ..i

From Eq. (i), we get
rtanâ¡Î¸N=Î¼0I2BH= same for both
â´ rAtanâ¡Î¸ANA=rBtanâ¡Î¸BNB
â 8Ã1 2 3=16 3 NB
â´ NB=12 turns

Topic: Physics, Experimental Physics, JEE Main
Two tangent lines l1 and l2 are drawn from the point 2,0 to the parabola 2y2=-x. If the lines l1 and l2 are also tangent to the circle x-52+y2=r, then $17 r^2$ is equal to

Options:

The correct answer is:
9

Solution:
Given, from point 2,0 tangents are drawn on 2y2=-x, these tangents also touches the circle x-52+y2=r2, then value of 17r2 is:

Now equation of tangents when point is outside will be y=mx-18m, now 2,0 will lie here so m=±14, putting the value of m we get,

x-4y=2 and 4y+x=2

Using distance of centre from tangent equal to radius, we get

5-0-217=r or 5+0-217=r

∴ r2=917

⇒ 17r2=9

Topic: Mathematics, Parabola, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Two tangents are drawn from a point $(-2,-1)$ to the curve, $\mathrm{y}^2=4 \mathrm{x}$. If $\alpha$ is the angle between them, then $|\tan \alpha|$ is equal to:

Options:
A
$\frac{1}{3}$

B
$\frac{1}{\sqrt{3}}$

C
$\sqrt{3}$

D
3

The correct answer is:
3

Solution:
The locus of the point of intersection of tangents to the parabola $y^2=4 a x$ inclined at an angle $\alpha$ to each other is $\tan ^2 \alpha(x+a)^2=y^2-4 a x$
Given equation of Parabola
$$
y^2=4 x\{a=1\}
$$
Point of intersection $(-2,-1)$
$$
\begin{aligned}
&\tan ^2 \alpha(-2+1)^2=(-1)^2-4 \times 1 \times(-2) \\
&\Rightarrow \tan ^2 \alpha=9 \\
&\Rightarrow \tan \alpha=\pm 3 \\
&\Rightarrow|\tan \alpha|=3
\end{aligned}
$$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2014 (12 Apr Online)
Two tangents are drawn from a point (â4, 3) to the parabola y2=16x . If Î± is the angle between them, then the value of cos Î± is

Options:
A 0
B 12
C 32
D 12

The correct answer is:
0

Solution:
âµ (â 4, 3) lies on the directrix of y2=16x .
â´ Angle between tangents is 90Â° .
â´cos 90o=0

Topic: Mathematics, Parabola, JEE Main
Two tangents are drawn from a point P to the circle x2+y2-2x-4y+4=0, such that the angle between these tangents is tan-1125, where tan-1125∈(0,π). If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of ΔPAB and ΔCAB is :

Options:
A 11:4
B 9:4
C 3:1
D 2:1

The correct answer is:
9:4

Solution:
Let, the angle between the tangents PA and PB from the point P to the circle x2+y2-2x-4y+4=0 with centre C is θ.

Given θ=tan-1125 ⇒tanθ=125.

We know that the centre and radius of a circle x2+y2+2gx+2fy+c=0 are respectively -g, -f and g2+f2-c.

Thus, the centre and radius of the circle x2+y2-2x-4y+4=0 are respectively 1, 2 and 12+22-4=1 unit.

We know that the line joining the centre of a circle to the point of contact of the tangent is perpendicular to the tangent, hence ∠CAP=∠CBP=90° also CP bisects the angle ∠APB as ∆CAP≅∆CBP by SSS congruency.

Thus, we have ∠CPA=∠CPB=θ2

In ∆CAP, we have tanθ2=ACAP

⇒tanθ2=1AP

⇒AP=cotθ2

Now, area of ΔPAB=12PAPBsinθ

Since, tangents to a circle from an external point are equal, hence PA=PB

Area of ΔPAB=12PA2sinθ

=12cot2θ2sinθ

=12cos2θ2sin2θ2sinθ

Using cos2x=2cos2x-1=1-2sin2x, we get

Area of ΔPAB=121+cosθ1-cosθsinθ

We have tanθ=125 ⇒sinθ=1213 & cosθ=513

Area of ΔPAB=121+5131-5131213

=12×188×1213=2726 sq units.

And, area of ΔCAB=12ACBCsinπ-θ

=12sinθ=121213=613 sq units.

∴ area of ΔPAB area of ΔCAB=2726613=94.

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Two tangents are drawn from the point P-1,1 to the circle x2+y2-2x-6y+6=0. If these tangents touch the circle at points A and B, and if D is a point on the circle such that length of the segments AB and AD are equal, then the area of the triangle ABD is equal to:

Options:
A 2
B 32+2
C 4
D 32-1

The correct answer is:
4

Solution:
Given, the equation of circle is x2+y2-2x-6y+6=0

So, the centre of the circle is 1,3 and the radius is 12+32-6=2

Assume, the coordinate of point B is -1,k and the coordinate of point A is h,1.

So, h+12=2⇒h=1 and k-12=2⇒k=3

So, the coordinate of B is -1,3 and the coordinate of A is 1,1.

From the diagram, area of ∆ABD=12×2×4

=4 sq. units

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Two thermally insulated vessels 1 and 2 are filled with air at temperatures $\left(T_1, T_2\right)$, volume $\left(V_1, V_2\right)$ and pressure $\left(P_1, P_2\right)$ respectively. If the valve joining two vessels is opened, the temperature inside the vessel at equilibrium will be

Options:
A
$\mathrm{T}_1+\mathrm{T}_2$

B
$\left(\mathrm{T}_1+\mathrm{T}_2\right) / 2$

C
$\frac{\mathrm{T}_1 \mathrm{~T}_2\left(\mathrm{P}_1 \mathrm{~V}_1+\mathrm{P}_2 \mathrm{~V}_2\right)}{\mathrm{P}_1 \mathrm{~V}_1 \mathrm{~T}_2+\mathrm{P}_2 \mathrm{~V}_2 \mathrm{~T}_1}$

D
$\frac{\mathrm{T}_1 \mathrm{~T}_2\left(\mathrm{P}_1 \mathrm{~V}_1+\mathrm{P}_2 \mathrm{~V}_2\right)}{\mathrm{P}_1 \mathrm{~V}_1 \mathrm{~T}_1+\mathrm{P}_2 \mathrm{~T}_2 \mathrm{~T}_2}$

The correct answer is:
$\frac{\mathrm{T}_1 \mathrm{~T}_2\left(\mathrm{P}_1 \mathrm{~V}_1+\mathrm{P}_2 \mathrm{~V}_2\right)}{\mathrm{P}_1 \mathrm{~V}_1 \mathrm{~T}_2+\mathrm{P}_2 \mathrm{~V}_2 \mathrm{~T}_1}$

Solution:
No Solution Available

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2004
Two thin dielectric slabs of dielectric constants K1 and K2K1<K2 are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field ‘ E ’ between the plates with distance ‘ d ’ as measured from plate P is correctly shown by:

Options:
A
B
C
D

The correct answer is:

Solution:
Here we have to calculate the graph between E and d in the given situation

We know that in dielectric E becomes E/k

where

E = Electric fjeld

k = Dielectric constants

Since K1<K2, The third option will be correct.

Topic: Physics, Capacitance, JEE Main
Two thin lenses of focal lengths $f_1$ and $f_2$ are in contact and coaxial. The combination is equivalent to a single lens of power

Options:
A $f_1+f_2$
B $\frac{f_1 f_2}{f_1+f_2}$
C $\frac{1}{2}\left(f_1+f_2\right)$
D $\frac{f_1+f_2}{f_1 f_2}$

The correct answer is:
$\frac{f_1+f_2}{f_1 f_2}$

Solution:
$P=\frac{1}{F}=\frac{f_1+f_2}{f_1 f_2}$

Topic: Physics, Ray Optics, JEE Main
Two thin lenses of focal lengths $f_1$ and $f_2$ are in contact and coaxial. The power of the combination is

Options:
A $\sqrt{\frac{f_1}{f_2}}$
B $\sqrt{\frac{f_2}{f_1}}$
C $\frac{f_1+f_2}{f_1 f_2}$
D $\frac{f_1+f_2}{f_1 f_2}$

The correct answer is:
$\frac{f_1+f_2}{f_1 f_2}$

Solution:
If two thin lenses of focal lengths $f_1, f_2$ are placed in contact coaxially, then equivalent focal length of combination is
$$
\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{0}{f_1 f_2}=\frac{1}{f_1}+\frac{1}{f_2}
$$
Power for the combination is
$$
P=\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{f_1+f_2}{f_1 f_2}
$$

Topic: Physics, Ray Optics, JEE Main
Two thin lenses of focal lengths $f_1$ and $f_2$ are in contact and coaxial. The power of the combinations is

Options:
A $\frac{f_1+f_2}{f_1 f_2}$
B $\sqrt{\frac{f_1}{f_2}}$
C $\sqrt{\frac{f_2}{f_1}}$
D $\frac{f_1+f_2}{2}$

The correct answer is:
$\frac{f_1+f_2}{f_1 f_2}$

Solution:
$$
\begin{aligned}
& \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2} \\
\Rightarrow \quad P & =\frac{f_1+f_2}{f_1 f_2}
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main
Two thin lenses of focal lengths $f_1$ and $f_2$ are in contact and coaxial. The power of the combinations is

Options:
A $\frac{f_1+f_2}{f_1 f_2}$
B $\sqrt{\frac{f_1}{f_2}}$
C $\sqrt{\frac{f_2}{f_1}}$
D $\frac{f_1+f_2}{2}$

The correct answer is:
$\frac{f_1+f_2}{f_1 f_2}$

Solution:
$$
\begin{aligned}
& \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2} \\
\Rightarrow \quad P & =\frac{f_1+f_2}{f_1 f_2}
\end{aligned}
$$

Topic: Physics, Ray Optics, NEET, NEET 2008 (Mains)
Two thin, long parallel wires separated by a distance 'd' carry a current of 'i' $A$ in the same direction. They will

Options:
A
attract each other with a force of $\mu_0 i^2 /(2 \pi \mathrm{d})$

B
repel each other with a force of $\mu_0 i^2 /(2 \pi \mathrm{d})$

C
attract each other with a force of $\mu_0 \mathrm{i}^2\left(2 \pi \mathrm{d}^2\right)$

D
repel each other with a force of $\mu_0 \mathrm{i}^2\left/(2 \pi \mathrm{d}^2\right)$

The correct answer is:
attract each other with a force of $\mu_0 i^2 /(2 \pi \mathrm{d})$

Solution:
Using the definition of force per unit length due to two long parallel wires carrying currents.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2005
Two thin metallic spherical shells of radii $20 \mathrm{~cm}$ and $30 \mathrm{~cm}$, respectively are placed with their centres coinciding. A material of thermal conductivity $\alpha$ is filled in the space between the shells. The inner shell is maintained at $300 \mathrm{~K}$ and the outer shell at $310 \mathrm{~K}$. If the rate at which heat flows radially through the material is $40 \mathrm{~W}$, find the value of $\alpha$ (in units of $\mathrm{J} \mathrm{s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ ).

Options:
A $\frac{3}{\pi}$
B $\frac{4 \pi}{3}$
C $\frac{5}{3 \pi}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{5}{3 \pi}$

Solution:
Given, radius of two thin metallic spherical
shells,
$\begin{aligned} & r_1=20 \mathrm{~cm}=0.2 \mathrm{~m} \\ & r_2=30 \mathrm{~cm}=0.3 \mathrm{~m}\end{aligned}$
temperature of inner shell, $T_1=300 \mathrm{~K}$, temperature of outer shell, $T_2=310 \mathrm{~K}$, and rate of heat flow, $H=40 \mathrm{~W}$.
Radial rate of flow of heat through the shell in the steady state,
$\begin{aligned} H & =\frac{d Q}{d t}=\alpha A \frac{d T}{d r}=\alpha\left(4 \pi r^2\right) \frac{d T}{d r} \\ \frac{d r}{r^2} & =\frac{4 \pi \alpha}{H} \cdot d T\end{aligned}$
Integrate on both sides, we get
$\begin{array}{rlrl}\Rightarrow & \quad \int_{r_1}^{r_2} \frac{d r}{r^2} & =\frac{4 \pi \alpha}{H} \int_{T_1}^{T_2} d T \\ \frac{r_2-r_1}{r_1 r_2} & =\frac{4 \pi \alpha}{H}\left(T_2-T_1\right) \\ \therefore \quad & \alpha & =\frac{H\left(r_2-r_1\right)}{4 \pi r_1 r_2\left(T_2-T_1\right)}\end{array}$
Putting the given values, we get
$\alpha=\frac{40(0.3-0.2)}{4 \pi(0.3)(0.2)(310-300)}=\frac{4}{2.4 \pi}=\frac{5}{3 \pi}$
Hence, the value of $\alpha=\frac{5}{3 \pi} \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$.

Topic: Physics, Thermal Properties of Matter, JEE Main
Two thin metallic spherical shells of radii $20 \mathrm{~cm}$ and $30 \mathrm{~cm}$, respectively are placed with their centres coinciding. A material of thermal conductivity $\alpha$ is filled in the space between the shells. The inner shell is maintained at $300 \mathrm{~K}$ and the outer shell at $310 \mathrm{~K}$. If the rate at which heat flows radially through the material is $40 \mathrm{~W}$, find the value of $\alpha$ (in units of $\mathrm{J} \mathrm{s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ ).

Options:
A $\frac{3}{\pi}$
B $\frac{4 \pi}{3}$
C $\frac{5}{3 \pi}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{5}{3 \pi}$

Solution:
Given, radius of two thin metallic spherical
shells,
$\begin{aligned} & r_1=20 \mathrm{~cm}=0.2 \mathrm{~m} \\ & r_2=30 \mathrm{~cm}=0.3 \mathrm{~m}\end{aligned}$
temperature of inner shell, $T_1=300 \mathrm{~K}$, temperature of outer shell, $T_2=310 \mathrm{~K}$, and rate of heat flow, $H=40 \mathrm{~W}$.
Radial rate of flow of heat through the shell in the steady state,
$\begin{aligned} H & =\frac{d Q}{d t}=\alpha A \frac{d T}{d r}=\alpha\left(4 \pi r^2\right) \frac{d T}{d r} \\ \frac{d r}{r^2} & =\frac{4 \pi \alpha}{H} \cdot d T\end{aligned}$
Integrate on both sides, we get
$\begin{array}{rlrl}\Rightarrow & \quad \int_{r_1}^{r_2} \frac{d r}{r^2} & =\frac{4 \pi \alpha}{H} \int_{T_1}^{T_2} d T \\ \frac{r_2-r_1}{r_1 r_2} & =\frac{4 \pi \alpha}{H}\left(T_2-T_1\right) \\ \therefore \quad & \alpha & =\frac{H\left(r_2-r_1\right)}{4 \pi r_1 r_2\left(T_2-T_1\right)}\end{array}$
Putting the given values, we get
$\alpha=\frac{40(0.3-0.2)}{4 \pi(0.3)(0.2)(310-300)}=\frac{4}{2.4 \pi}=\frac{5}{3 \pi}$
Hence, the value of $\alpha=\frac{5}{3 \pi} \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$.

Topic: Physics, Thermal Properties of Matter, TS EAMCET, TS EAMCET 2019 (06 May Shift 1)
Two thin rings each of radius R are placed at a distance d apart. The charges on the rings are +q and âq. The potential difference between their centres will be -

Options:
A qR4πε0d2
B q2πε01R-1R2+d2
C zero
D q4πε01R-1R2+d2

The correct answer is:
q2πε01R-1R2+d2

Solution:
VA=kqR-kqR2+d2
VB=-kqR-+ +k2R2+d2-
â´VA-VB=2kqR-2kqR2+d2
â´VA-VB=q2ÏÎµ01R-1R2+d2

Topic: Physics, Electrostatics, JEE Main
Two thin spherical shells made of metal are at a large distance apart. One of radius 10 cm carries a charge of +0.5 Î¼C and the other of radius 20 cm carries a charge of +0.7 Î¼C. The charge on each, when they are connected by a suitable conducting wire is respectively

Options:
A 0.4 and 0.8 Î¼C
B +0.425 and +0.85Â Î¼C
C +0.5 and +0.7Â Î¼C
D +0.6 and +0.6Â Î¼C

The correct answer is:
0.4 and 0.8 Î¼C

Solution:
Let us assume that the initial charges on the spheres were Q1 and Q2 respectively and the final charges on the spheres are q1 and q2. Finally the potential of both the spheres is the same.

kq1R1=kq2R2

âq2=q1R2R1

q1+q2=Q1+Q2

q1=R1R1+R2Q1+Q2

âq1=1010+200.5+0.7=0.4 Î¼C

q2=R2R1+R2Q1+Q2

âq2=2010+200.5+0.7=0.8 Î¼C

Topic: Physics, Electrostatics, JEE Main