datavorous/entrance-exam-dataset huggingface dataset:
Two vertical parallel mirrors A and B are separated by 10 cm. A point object O is placed at a distance of 2 cm from mirror A. The distance of the second nearest image behind mirror A from the mirror A is _______ cm.

Options:

The correct answer is:
18

Solution:
As can be seen from the image above, object distance is 8 cm [For image by B]

In a plane mirror the image distance is equal to the object distance. Thus, the required distance is

8+8+2 cm=18 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Two vertical poles AB=15 m and CD=10 m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m ) above the line AC is :

Options:
A 20/3
B 5
C 10/3
D 6

The correct answer is:
6

Solution:
ΔAQP~ΔACD

⇒ xh=x+y10 ...1

∵ ΔCQP~ΔCAB

⇒yh=x+y15 ....2

Adding 1 & 2, we get

x+yh=x+y110+115

⇒ h=6.

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Two vertical poles AL and BM of height 4 m and 16 m respectively stand apart on a horizontal plane. If A,B be the feet of the poles and AM and BL intersect at P, then the height of P from the horizontal plane is equal to

Options:
A 3.2 m
B 2.5 m
C 4 m
D 8 m

The correct answer is:
3.2 m

Solution:
4x+y=zy ∵∆LAB~∆PNB
⇒y=zx+y4….i
Also, 16x+y=zx ∵∆MBA~∆PNA
⇒x=zx+y16….ii
Adding equations i & ii, we get,

x+y=x+yz4+z16⇒z=165

Topic: Mathematics, Heights and Distances, JEE Main
Two vertical poles AL and BM of heights 25m and 100m respectively stand apart on a horizontal plane. If A,B be the feet of the poles and AM and BL intersect at P, then the height of P from the horizontal plane is equal to

Options:
A 20m
B 18m
C 16m
D 15m

The correct answer is:
20m

Solution:
25x+y=zy ∵∆LAB~∆PNB
⇒y=zx+y25…i
Also, 100x+y=zx ∵∆MBA~∆PNA
⇒x=zx+y100…ii
Adding equation i & ii, we get

x+y=x+yz25+z100⇒z=20

Topic: Mathematics, Heights and Distances, JEE Main
Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

Options:
A 25
B 30
C 203
D 253

The correct answer is:
253

Solution:
tanθ=h75=753 h

⇒ h2=7523

h=253 m

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Two vertical poles of height 10 m and 40 m stand apart on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is

Options:
A 8
B 10
C 6
D 4

The correct answer is:
8

Solution:
Let A0,0, D0,10, Bx1,0, Cx1,40
Equation of AC⇒y=40x1x … 1
Equation BD⇒y-10=-10x1x … 2
From 1 and 2, we get,
x1⋅y40=-y-10⋅x110
y=-4y+40⇒y=8

Topic: Mathematics, Heights and Distances, JEE Main
Two vertical poles of height, 20 m and 80 m stand apart on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is:

Options:
A 16
B 12
C 18
D 15

The correct answer is:
16

Solution:
Equation of line OQ y=80ax

Equation of line AB xa+y20=1

Solving for Px0,y0,ay0/80a+y020=1

⇒y0180+120=1⇒y0=16

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Two vertices of a triangle $\mathrm{ABC}$ are $\mathrm{A}(3,-1)$ and $\mathrm{B}(-2,3)$, and its orthocentre is $\mathrm{P}(1,1)$. If the coordinates of the point $\mathrm{C}$ are $(\alpha, \beta)$ and the centre of the of the circle circumscribing the triangle $\mathrm{PAB}$ is $(\mathrm{h}, \mathrm{k})$, then the value of $(\alpha+\beta)+2(\mathrm{~h}+\mathrm{k})$ equals

Options:
A 5
B 81
C 15
D 51

The correct answer is:
5

Solution:
$\mathrm{M}_{\mathrm{AB}}=\frac{4}{-5} \Rightarrow \mathrm{M}_{\mathrm{DP}}=\frac{5}{4}$
Equation of $\mathrm{PC}$ is $\mathrm{y}-1=\frac{5}{4}(\mathrm{x}-1) \ldots(1)$
$\mathrm{M}_{\mathrm{AP}}=\frac{2}{-2}=-1 \Rightarrow \mathrm{M}_{\mathrm{BC}}=+1$
Equation of $\mathrm{BC}$ is $\mathrm{y}-3=(\mathrm{x}+2)\ldots(2)$
On solving (1) and (2)
$\begin{aligned}
& x+4=\frac{5}{4}(x-1) \Rightarrow 4 x+16=5 x-5 \Rightarrow \alpha=21 \\
& \Rightarrow \beta=y=x+5=26 \\
& \alpha+\beta=47
\end{aligned}$
Equation of $\perp$ bisector of AP
$y-0=(x-2)\ldots(3)$
Equation of $\perp$ bisector of $\mathrm{AB}$
$y-1=\frac{5}{4}\left(x-\frac{1}{2}\right)\ldots(4)$
$\begin{aligned} & \text { On solving (3) \& (4) } \\ & (\mathrm{x}-3) 4=5 \mathrm{x}-\frac{5}{2} \\ & \mathrm{x}=\frac{-19}{2}=\mathrm{h} \\ & \mathrm{y}=\frac{-23}{2}=\mathrm{k} \\ & \Rightarrow 2(\mathrm{~h}+\mathrm{k})=-42\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Two vertices of a triangle are 0,2 and 4,3. If its orthocenter is at the origin, then its third vertex lies in which quadrant?

Options:
A Fourth
B Second
C Third
D First

The correct answer is:
Second

Solution:
Let the third vertex be h,k.

Since H is orthocenter of ΔABC
Since the slope of BH is 34 hence the slope of AC will be -43

⇒ Equation of line AC is
4x+3y=6 ...1
Similarly, since AH is parallel to y-axis hence BC will be parallel to x-axis and
Equation of line BC is
y=3 ...2
On solving 1 and 2
h=-34, k=3
∴ The point Ch,k≡-34,3 lies in the 2nd quadrant.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Two very long, straight, and insulated wires are kept at 90o angle from each other in xy-plane as shown in figure.
These wires carry currents of equal magnitude I , whose direction are shown in the figure. The net magnetic field at point P will be:

Options:
A -μ0I2πd x^+y^
B +μ0Iπd z^
C μ0I2πd x^+y^
D Zero

The correct answer is:
Zero

Solution:
Magnetic field due to wire along x-axis:
B1=μ0I2πdz^
Magnetic field due to wire along z-axis:
B2=μ0I2πd(-z^)
B→net=B→1+B→2=0

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Two vessels A and B are of the same size and are at same temperature. A contains 1 g of hydrogen and B contains 1 g of oxygen. PA and PB are the pressures of the gases in A and B respectively, then PAPB is :

Options:
A 16
B 8
C 4
D 32

The correct answer is:
16

Solution:
For both the gases, it follows that

PAVAPBVB=nARTAnBRTB ...1

Given VA=VB and TA=TB.

Thus, from equation (1), it follows that

PAPB=nAnB=12132=16

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weights of A and B is

Options:
A 34
B 2
C 12
D 23

The correct answer is:
34

Solution:
PA= ρARTMA, PB= ρBRTMB

⇒ PAPB= ρAρB MBMA=2

⇒32MBMA=2
So, MAMB =34

Topic: Physics, Kinetic Theory of Gases, JEE Main
Two vibrating tuning forks produce waves given by \( y_{1}=4 \sin (500 \pi t) \) and \( y_{2}=2 \sin (506 \pi t) \). The number of beats
produced per minute is

Options:
A \( 360 \)
B \( 180 \)
C \( 60 \)
D \( 3 \)

The correct answer is:
\( 180 \)

Solution:
y1=4sin500πt, y2=2sin506πt
ω1=500π rad s-1, ω1=2πν1, ν1=250 Hz
ω2=506π rad s-1, ω2=2πν2, ν2=253 Hz
Number of beats per second =ν2-ν1=253-250=3 Hz
Number of beats per minute =3×60=180 Hz

Topic: Physics, Waves and Sound, JEE Main
Two voltameters one of copper and another of silver, are joined in parallel. When a total charge $q$ flows through the voltameters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are $z_1$ and $z_2$ respectively the charge which flows through the silver voltameter is

Options:
A
$\frac{q}{1+\frac{z_1}{z_2}}$

B
$\frac{q}{1+\frac{z_2}{z_1}}$

C
$\mathrm{q} \frac{z_1}{z_2}$

D
$\mathrm{q} \frac{z_2}{z_1}$

The correct answer is:
$\frac{q}{1+\frac{z_2}{z_1}}$

Solution:
$q_1 Z_1=q_2 Z_2$
$q=q_1+q_2$
$\therefore q_2=\frac{q}{1+\frac{Z_2}{Z_1}}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2005
Two water pipes of diameter \(2 \mathrm{~cm}\) and \(4 \mathrm{~cm}\) are separately connected to a main supply line. The velocity of flow of water in the pipe of \(2 \mathrm{~cm}\) diameter is

Options:
A 4 times that in the other pipe
B \(\frac{1}{4}\) times that in the other pipe
C 2 times that in the other pipe
D \(\frac{1}{2}\) times that in the other pipe

The correct answer is:
4 times that in the other pipe

Solution:
Diameter of first pipe, \(d_1=2 \mathrm{~cm}\)
\(\therefore\) Radius, \(r_1=1 \mathrm{~cm}=10^{-2} \mathrm{~cm}\)
Diameter of second pipe, \(d_2=4 \mathrm{~cm}\)
\(\therefore \quad r_2=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}\)
If \(v_1\) and \(v_2\) are the velocities of water in first and second pipe respectively, then according to principle of continuity,
\(\begin{aligned}
& & A_1 v_1=A_2 v_2 \Rightarrow \pi r_1^2 v_1=\pi r_2^2 v_2 \\
\Rightarrow & v_2 & =\left(\frac{r_1}{r_2}\right)^2 v_1=\left(\frac{10^{-2}}{2 \times 10^{-2}}\right)^2 v_1 \\
\Rightarrow & & v_2=\frac{v_1}{4} \Rightarrow v_1=4 v_2
\end{aligned}\)

Topic: Physics, Mechanical Properties of Fluids, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 2)
Two water pipes of diameters $2 \mathrm{~cm}$ and $4 \mathrm{~cm}$ are connected with the main supply line. The velocity of flow of water in the pipe of $2 \mathrm{~cm}$ diameter is

Options:
A 4 times that in the other pipe
B $\frac{1}{4}$ times that in the other pipe
C 2 times that in the other pipe
D $\frac{1}{2}$ times that in the other pipe

The correct answer is:
4 times that in the other pipe

Solution:
$d_A=2 \mathrm{~cm}$ and $d_B=4 \mathrm{~cm} \quad \therefore r_A=1 \mathrm{~cm}$ and $r_B=2 \mathrm{~cm}$
From equation of continuity av = constant
$\therefore \frac{v_A}{v_B}=\frac{a_B}{a_A}=\frac{\pi\left(r_B\right)^2}{\pi\left(r_A\right)^2}=\left(\frac{2}{1}\right)^2 \Rightarrow v_A=4 v_B$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two water pipes of diameters 2 cm and 4 cm are connected with the main supply line. The velocity of flow of water in the pipe of 2 cm diameter is

Options:
A 4 times that in the other pipe
B 14 times that in the other pipe
C 2 times that in the other pipe
D 12 times that in the other pipe

The correct answer is:
4 times that in the other pipe

Solution:
Given, dA=2 cm and dB=4 cm
∴rA=1 cm and rB=2 cm
From the equation of continuity av= constant
∴VAVB=aBaA=π(rB)2π(rA)2=212
⇒VA=4VB

Topic: Physics, Mechanical Properties of Fluids, NEET
Two water pipes P and Q having diameter 2Ã10-2 m and 4Ã10-2 m respectively are joined in series with the main supply line of water. The velocity of water flowing in pipe P is

Options:
A 4 times that of Q
B 2 times that of Q
C 1/2 times that of Q
D 1/4 times that of Q

The correct answer is:
4 times that of Q

Solution:
Using theorem of continuity, we have
πDp2vp=πDQ2vQ;
vp=DQDP2vQ=4×10-22×10-22×vQ=4vQ

Topic: Physics, Mechanical Properties of Fluids, NEET
Two wavelengths $\lambda_1$ and $\lambda_2$ are used in Young's double slit experiment. $\lambda_1=450 \mathrm{~nm}$ and $\lambda_2=650 \mathrm{~nm}$. The minimum order of fringe produced by $\lambda_2$ which overlaps with the fringe produced by $\lambda_1$ is $n$. The value of $n$ is _____.

Options:

The correct answer is:
9

Solution:
$\begin{aligned} & \mathrm{n}_2 \lambda_2=\mathrm{n}_1 \lambda_1 \\ & \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{\lambda_1}{\lambda_2}=\frac{450}{650}=\frac{9}{13} \\ & \mathrm{n}_2=9\end{aligned}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Two wavelengths of light \( \lambda_{1} \) and \( \lambda_{2} \) are sent through Young's double slit apparatus simultaneously. If \( 3^{r d} \) order bright fringe of \( \lambda_{1} \) coincide with the fourth order bright fringe for \( \lambda_{2} \), then \( \frac{\lambda_{1}}{\lambda_{2}}= \)

Options:
A \( \frac{1}{2} \)
B 1
C \( \frac{1}{\frac{1}{2}} \)
D \( \frac{4}{3} \)

The correct answer is:
\( \frac{4}{3} \)

Solution:
The expression of the distance of nth order maximum from the central maximum in Young's double slit experiment is given as, yn=nλDd, here λ is the wavelength, n=1,2,3..., D is the distance between the slits and screen, d is the distance between slits.

The position of 3rd order bright fringe, y1=3λ1Dd.

The position of 4th order bright fringe, y2=4λ1Dd.

Given third order bright fringe coincides with the fourth order bright fringe.

Thus, y1=y2.

3λ1Dd=4λ2Dd

Hence, the ratio λ1λ2=43.

Topic: Physics, Wave Optics, JEE Main
Two wavelengths of light, \( \lambda_{1}=4500 \mathrm{~A} \) and \( \lambda_{2}=6000 \mathrm{~A} \) are sent through a Young's double-slit apparatus simultaneously.
Then:

Options:
A no interference pattern will be formed.
B the third order bright fringe of \( \lambda_{1} \) will coincide with the fourth order bright fringe of \( \lambda_{2} \).
C the third order bright fringe of \( \lambda_{2} \) will coincide with fourth order bright fringe of \( \lambda_{1} \).
D the fringes of wavelength \( \lambda_{1} \) will be wider than the fringes of wavelength \( \lambda_{2} \).

The correct answer is:
the third order bright fringe of \( \lambda_{2} \) will coincide with fourth order bright fringe of \( \lambda_{1} \).

Solution:
Let nth bright fringe of λ1 overlap with mth bright fringe of λ2.

Distance of nth bright for λ1=Distance of mth bright for λ2

⇒nβ1=mβ2, where β is fringe width.

⇒nλ1Dd=mλ2Dd

⇒nλ1=mλ2

⇒nm=λ2λ1

⇒nm=60004500

⇒nm=43

Therefore, 4th bright fringe of λ2 coincides with 3rd bright fringe of λ1.

Topic: Physics, Wave Optics, JEE Main
Two waves are described by the equations:
y1=Acosâ¡0.5 Ïx-100Ït
And y2=Acosâ¡0.46 Ïx-92Ït
Here x and y are in m and t is in s.
Find the number of times y1+y2 becomes zero per second, at x = 0.

Options:
A 46
B 48
C 192
D 100

The correct answer is:
100

Solution:
At x = 0 ,y=y1+y2=2Acos⁡96 πtcos⁡4πt
Frequency of cos⁡96πt=48 Hz and that of cos⁡(4πt) function is 2 Hz
In one-second cos function becomes zero at 96 times and second at 4 times. Hence net y will become zero 100 times in 1 second.

Topic: Physics, Waves and Sound, JEE Main
Two waves are given by $y_1=a \sin (\rho t-k x)$ and $y_2=a \cos (\omega t-k x)$ The phase difference between the two waves is

Options:
A $\frac{\pi}{4}$
B $\pi$
C $\frac{\pi}{8}$
D $\frac{\pi}{2}$

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

Solution:
$y_1-a \sin (u t-k x)$
and $y_2-a \cos (\omega t-k x)-a \sin \left(\omega t-k x+\frac{\pi}{2}\right)$
Hence phase difference between these two is $\frac{\pi}{2}$.

Topic: Physics, Waves and Sound, JEE Main
Two waves are simultaneously passing through a string and their equations are : y1=A1sink(x-vt),y2=A2sinkx-vt+x0. Given amplitudes A1=12 mm and A2=5 mm, x0=3.5 cm and wave number k=6.28 cm-1. The amplitude of resulting wave will be _____ mm.

Options:

The correct answer is:
7

Solution:
Δϕ=kx0=2πλx0=2π×3.5 λ=2πk=1

=2π×3510=7π

AR=122+52+2×12×5×cos(7π)=(12-5)2=7 mm

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Two waves of intensity ratio 1 : 9 cross each other at a point. The resultant intensities at the point, when (a) Waves are incoherent is I1 b Waves are coherent is I2 and differ in phase by 60o. If I1I2=10x, then x= _________.

Options:

The correct answer is:
13

Solution:
For incoherent wave, I1=IA+IB⇒I1=I0+9I0

⇒I1=10I0

For coherent wave I2=IA+IB+2IAIBcos60o

⇒I2=I0+9I0+29I02×12=13I0

⇒I1I2=1013

Therefore, x=13.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Two waves of wavelengths $99 \mathrm{~cm}$ and $100 \mathrm{~cm}$ both travelling with velocity $396 \mathrm{~m} / \mathrm{s}$ are made to interfere. The number of beats produced by them per second is

Options:
A 1
B 2
C 4
D 8

The correct answer is:
4

Solution:
Velocity of wave $\mathrm{v}=\mathrm{n} \lambda$ where $n=$ frequency of wave $\Rightarrow n=\frac{v}{\lambda}$

$\mathrm{n}_{2}=\frac{\mathrm{v}_{2}}{\lambda_{2}}=\frac{396}{100 \times 10^{-2}}=396 \mathrm{~Hz}$

no. of beats $=\mathrm{n}_{1}-\mathrm{n}_{2}=4$

Topic: Physics, Waves and Sound, JEE Main
Two waves represented by the following equations are travelling in the same medium
$y_1=5 \sin 2 \pi(75 t-0.25 x), y_2=10 \sin 2 \pi(150 t-0.50 x)$

Options:
A $1: 2$
B $1: 4$
C $1: 8$
D $1: 16$

The correct answer is:
$1: 4$

Solution:
$\frac{I_1}{I_2}=\frac{a_1^2}{a_2^2} \Rightarrow \frac{I_1}{I_2}=\frac{25}{100}=\frac{1}{4}$

Topic: Physics, Waves and Sound, JEE Main
Two waves
$y_1=A_1 \sin \left(\omega t-\beta_1\right), y_2=A_2 \sin \left(\omega t-\beta_2\right)$
Superimpose to form a resultant wave whose amplitude is

Options:
A $\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \left(\beta_1-\beta_2\right)}$
B $\sqrt{A_1^2+A_2^2+2 A_1 A_2 \sin \left(\beta_1-\beta_2\right)}$
C $A_1+A_2$
D $\left|A_1+A_2\right|$

The correct answer is:
$\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \left(\beta_1-\beta_2\right)}$

Solution:
Phase difference between the two waves is
$\phi=\left(\omega t-\beta_2\right)=\left(\omega t-\beta_1\right)=\left(\beta_1-\beta_2\right)$
$\therefore$ Resultant amplitude $A=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \left(\beta_1-\beta_2\right)}$

Topic: Physics, Waves and Sound, JEE Main
Two waves $y=0.25 \sin 316 t$ and $y=0.25 \sin 310 t$ are travelling in same direction. The number of beats produced per second will be

Options:
A 6
B 3
C $3 / \pi$
D $3 \pi$

The correct answer is:
$3 / \pi$

Solution:
$n_1=\frac{316}{2 \pi}$ and $n_2=\frac{310}{2 \pi}$ Number of beats heard per second $=n_1-n_2=\frac{316}{2 \pi}-\frac{310}{2 \pi}=\frac{3}{\pi}$

Topic: Physics, Waves and Sound, JEE Main
Two weights of the mass $m_{1}$ and $m_{2}\left(>m_{1}\right)$ are joined by an inextensible string of negligible mass passing over a fixed frictionless pulley. The magnitude of the acceleration of the loads is

Options:
A $g$
B $\frac{m_{2}-m_{1}}{m_{2}} g$
C $\frac{m_{1}}{m_{2}+m_{1}} g$
D $\frac{m_{2}-m_{1}}{m_{2}+m_{1}} g$

The correct answer is:
$\frac{m_{2}-m_{1}}{m_{2}+m_{1}} g$

Solution:
According to the question, we can draw the following diagram

Here, $T$ is the tension in the string.
In equilibrium condition.
For mass $m_{1}, \quad T-m_{1} g=m_{1} a$....(i)
For mass $m_{2}, \quad m_{2} g-T=m_{2} a$....(ii)
After adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
m_{2} g-m_{1} g &=m_{1} a+m_{2} a \\
\left(m_{2}-m_{1}\right) g &=\left(m_{1}+m_{2}\right) a \\
a &=\frac{\left(m_{2}-m_{1}\right)}{\left(m_{1}+m_{2}\right)} g
\end{aligned}
$$

Topic: Physics, Laws of Motion, JEE Main
Two wires $A$ and $B$ are made up of the same material and have the same mass. Wire $A$ has radius of $2.0 \mathrm{~mm}$ and wire $B$ has radius of $4.0 \mathrm{~mm}$. The resistance of wire $B$ is $2 \Omega$. The resistance of wire $A$ is _____$\Omega$.

Options:

The correct answer is:
32

Solution:
$\begin{aligned} & \because \mathrm{R}=\frac{\rho \ell}{\mathrm{A}}=\frac{\rho \mathrm{V}}{\mathrm{A}^2} \\ & \therefore \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\mathrm{A}_{\mathrm{B}}^2}{\mathrm{~A}_{\mathrm{A}}^2}=\frac{\mathrm{r}_{\mathrm{B}}^4}{\mathrm{r}_{\mathrm{A}}^4} \\ & \Rightarrow \frac{\mathrm{R}_{\mathrm{A}}}{2}=\left[\frac{4 \times 10^{-3}}{2 \times 10^{-3}}\right]^4 \\ & \Rightarrow \mathrm{R}_{\mathrm{A}}=32 \Omega .\end{aligned}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Two wires $A$ and $B$ are of lengths $40 \mathrm{~cm}$ and $30 \mathrm{~cm}$. $A$ is bent into a circle of radius $r$ and $B$ into an arc of radius $r$. A current $i_1$ is passed through $A$ and $i_2$ through $B$. To have the same magnetic inductions at the centre, the ratio of $i_1: i_2$ is

Options:
A $3: 4$
B $3: 5$
C $2: 3$
D $4: 3$

The correct answer is:
$3: 4$

Solution:
For wire $A$,
$B_1=\frac{\mu_0 \dot{l}_1}{2 r}$
where
$r=\frac{40}{2 \pi}$
For wire $B$,
$\begin{aligned} \text { Circumference } & =\text { length } \\ n \pi r & =30\end{aligned}$
or $\quad n \pi=\frac{30}{r}=\frac{30}{40 / 2 \pi}=\frac{3}{2} \pi$
$\theta=n \pi=\frac{3}{2} \pi$
$\therefore \quad B_2=\frac{\mu_0}{4 \pi}\left(\frac{i_2}{r}\right) \theta$
But $\quad B_1=B_2$
$\begin{array}{rlrl} & \text { or } & \frac{\mu_0 i_1}{2 r} & =\frac{\mu_0}{4 \pi}\left(\frac{i_2}{r}\right) \theta \\ \text { or } & \frac{i_1}{i_2} & =\frac{3}{4}\end{array}$

Topic: Physics, Magnetic Effects of Current, JEE Main
Two wires $A$ and $B$ are of the same material. Their lengths are in the ratio of $1: 2$ and the diameter are in the ratio $2: 1$. If they are pulled by the same force, then increase in length will be in the ratio of

Options:
A $2: 1$
B $1: 4$
C $1: 8$
D $8: 1$

The correct answer is:
$1: 8$

Solution:
We know that Young's modulus
$$
\mathrm{Y}=\frac{\mathrm{F}}{\pi \mathrm{r}^2} \times \frac{\mathrm{L}}{\ell}
$$
Since Y, F are same for both the wires, we have,
$$
\begin{aligned}
& \frac{1}{\mathrm{r}_1^2} \frac{\mathrm{L}_1}{\ell_1}=\frac{1}{\mathrm{r}_2^2} \frac{\mathrm{L}_2}{\ell_2} \quad \text { or, } \frac{\ell_1}{\ell_2}=\frac{\mathrm{r}_2^2 \times \mathrm{L}_1}{\mathrm{r}_1^2 \times \mathrm{L}_2}= \\
& \frac{\left(\mathrm{D}_2 / 2\right)^2 \times \mathrm{L}_1}{\left(\mathrm{D}_1 / 2\right)^2 \times \mathrm{L}_2} \\
& \text { or, } \frac{\ell_1}{\ell_2}=\frac{\mathrm{D}_2^2 \times \mathrm{L}_1}{\mathrm{D}_1^2 \times \mathrm{L}_2}=\frac{\mathrm{D}_2^2}{\left(2 \mathrm{D}_2\right)^2} \times \frac{\mathrm{L}_2}{2 \mathrm{~L}_2}=\frac{1}{8} \\
& \text { So, } \ell_1: \ell_2=1: 8
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two wires $A$ and $B$ of same length and of the same material have the respective radii ${ }^{r_1}$ and ${ }^{r_2}$. Their one end is fixed with a rigid support, and at the other end equal twisting couple is applied. Then the ratio of the angle of twist at the end of $A$ and the angle of twist at the end of $B$ will be

Options:
A $\frac{r_1^2}{r_2^2}$
B $\frac{r_2^2}{r_1^2}$
C $\frac{r_2^4}{r_1^4}$
D $\frac{r_1^4}{r_2^4}$

The correct answer is:
$\frac{r_2^4}{r_1^4}$

Solution:
Twisting couple $C=\frac{\pi m r^4 \theta}{2 l}$
If material and length of the wires $A$ and $B$ are equal and equal twisting couple are applied then
$\theta \propto \frac{1}{r^4} \quad \therefore$

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two wires A & B are carrying currents I1 and I2 as shown in the figure. The separation between them is d . A third wire C carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are:

Options:
A x=±I1dI1-I2
B x=I1I1+I2d and x=I2I1-I2d
C x=I2I1+I2d and x=I2I1-I2d
D x=I1I1-I2d and x=I2I1+I2d

The correct answer is:
x=±I1dI1-I2

Solution:
As current in A and B are opposite in direction, net force on C can never be zero if it is kept between
A and B .
∴ μ0I1I2πx=μ0I I22πx±d
I1x±d=I2 x
I2-I1x=± I1 d
x=±I1 dI2-I1

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Two wires are fixed in a sonometer. Their tensions are in the ratio 8:1. The lengths are in the ratio 36:35. The diameters are in the ratio 4:1. Densities of the materials are in the ratio 1:2. If the higher frequency in the setting is 360 Hz, the beat frequency when the two wires are sounded together, is

Options:
A 8
B 5
C 10
D 6

The correct answer is:
10

Solution:
Frequency of sonometer wire is given by
v=12lTm=12lTÏr2p
v1=12l1T1Ïr12Ï1
v2=12l2T2Ïr22Ï2
â´ v1v2=l2l1T1T2Ãr22r12ÃÏ2Ï1
v1v2=353681Ã116Ã21
âµv1<v2 and v2=360 Hz
Therefore,
v=360Ã3536
v1=350 Hz
So, number of beats produced =v1-v2
=360-350=10

Topic: Physics, Waves and Sound, JEE Main