datavorous/entrance-exam-dataset huggingface dataset:
A square loop of edge length $2 \mathrm{~m}$ carrying current of $2 \mathrm{~A}$ is placed with its edges parallel to the $x$ $y$ axis. A magnetic field is passing through the $x-y$ plane and expressed as $\vec{B}=B_0(1+4 x) \hat{k}$, where $B_0=5 \mathrm{~T}$. The net magnetic force experienced by the loop is _______ $\mathrm{N}$.

Options:

The correct answer is:
160

Solution:
$\begin{aligned} & \mathrm{B}(\mathrm{x}=0)=\mathrm{B}_0, \quad \mathrm{~B}(\mathrm{x}=2)=9 \mathrm{~B}_0 \\ & \text { Also, } \mathrm{F}=\mathrm{i} \ell \mathrm{B} \\ & \Rightarrow \mathrm{F}_1=\mathrm{i} \ell \mathrm{B}_0 \quad \& \quad \mathrm{~F}_2=9 \mathrm{i} \ell \mathrm{B}_0 \\ & \mathrm{~F}=\mathrm{F}_2-\mathrm{F}_1=8 \mathrm{i} \ell \mathrm{B}_0=8 \times 2 \times 2 \times 5 \\ & \mathrm{~F}=160 \mathrm{~N}\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (09 Apr Shift 1)
A square loop of side 1 m and resistance 1 Ω is placed in a magnetic field of 0.5 T. If the plane of loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is:

Options:
A 0.5 weber
B 1 weber
C zero weber
D 2 weber

The correct answer is:
0.5 weber

Solution:
As plane of loop is perpendicular to the direction of magnetic field, therefore direction of area vector will be parallel to magnetic field.

Magnetic flux is given by,

ϕ=B→·A→=0.512cos0°=0.5 Wb

Topic: Physics, Electromagnetic Induction, JEE Main
A square loop of side 2.0 cm is placed inside a long solenoid that has 50 turns per centimetre and carries a sinusoidally varying current of amplitude 2.5 A and angular frequency 700 rad s-1. The central axes of the loop and solenoid coincide. The amplitude of the emf induced in the loop is x×10-4 V. The value of x is ___________

( Take, π=227

Options:

The correct answer is:
44

Solution:
It is given that the current is varying sinusoidally. The current can be written as I=I0sinωt.

The magnetic field through the solenoid B=μ0nI

The flux through the square is ϕ=μ0nIA

The emf is

ε=μ0nA×d(I0sinωt)dt⇒ε=μ0nAI0ωcosωt

The amplitude of the emf is,

ε=μ0nAI0ω=4π×10-7×5010-2×4×10-4×2.5×700⇒ε=44×10-4 V

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (10 Apr Shift 2)
A square loop of side 2a, and carrying current I is kept in XZ plane with its centre at origin. A long wire carrying the same current I is placed parallel to the z-axis and passing through the point (0, b, 0),(b>> a). The magnitude of the torque on the loop about z-axis is given by.

Options:
A μ0I2a22πb
B μ0I2a32πb2
C 2μ0Ia22πb
D 2μ0Ia32πb2

The correct answer is:
2μ0Ia22πb

Solution:
Net force acting on wire AB & CD are zero

F1=ILB=I×2a×μ0I2πa2+b2=μ0I2aπa2+b2

τ=F1cosθ×2a=μ0I2aπa2+b2×ba2+b2×2a

τ=2μ0I2a2bπa2+b2≈2μ0I2a2πb, a≪b

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A square loop of side $15 \mathrm{~cm}$ being moved towards right at a constant speed of 2 $\mathrm{cm} / \mathrm{s}$ as shown in figure. The front edge enters the $50 \mathrm{~cm}$ wide magnetic field at $t=$ 0 . The value of induced emf in the loop at $t=10 \mathrm{~s}$ will be :

Options:
A $0.3 \mathrm{mV}$
B zero
C $4.5 \mathrm{mV}$
D $3 \mathrm{mV}$

The correct answer is:
zero

Solution:
At $\mathrm{t}=10 \mathrm{sec}$ complete loop is in magnetic field therefore no change in flux
$\mathrm{e}=\frac{\mathrm{d} \phi}{\mathrm{dt}}=0$
$\mathrm{e}=0$ for complete loop

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2024 (09 Apr Shift 2)
A square matrix A of order 3 satisfies A2=Î-2A, where Î is an identity matrix of order 3. If An=29A-12Î, then the value of n is equal to

Options:
A 3
B 4
C 5
D 6

The correct answer is:
5

Solution:
A3=A2A=Ι-2AA=A-2A2=A-2Ι-2A=5A-2Ι
A4=A3A=5A-2ΙA=5A2-2A=5Ι-2A-2A=5Ι-12A
A5=A4A=5Ι-12AA=5A-12A2=5A-12Ι-2A=29A-12Ι

Topic: Mathematics, Matrices, JEE Main
A square metal loop of side 10 cm and resistance 1 Î© is moved with a constant velocity partly inside a uniform magnetic field of 2 Wb m-2, directed into the paper, as shown in the figure. The loop is connected to a network of five resistors each of value 3 Î©. If a steady current of 1 mA flows in the loop, then the speed of the loop is -

Options:
A 0.5  cm s −1
B 1  cm s −1
C 2â cm s â1
D 4  cm s −1

The correct answer is:
2â cm s â1

Solution:
The network appears as a Balanced Wheatstone bridge. Therefore, no current will flow through the bridge resistance. Thus, the resistance will have no effect on the network. Set of two series resistors each values 3âÎ© is now parallel to each other. Therefore the equivalent resistance of the balanced network is,

â´ R= 1 ( 1 3+3 + 1 3+3 ) = 1 ( 1 6 + 1 6 ) = 1 ( 2 6 ) = 6 2 =3âÎ©
The balanced network is in series with the 1âÎ©âinternal resistance. Therefore, the equivalant resistance of the network is,
R'=3â+1=4âÎ©
The direction of motion of free electron in rod in magnetic field is towards right and it has a velocity v. Theâinducedâemfâis,
e=vBl=IR'
Therefore, the velocity is,

v= IR' Bl
v= ( 1Ã 10 â3 )( 4 ) ( 2 )( 10Ã 10 â2 ) =2Ã 10 â2 â m s â1
v=2â cm s â1

Topic: Physics, Electromagnetic Induction, JEE Main
A square, of each side 2, lies above the x-axis and has one vertex at the origin. If one of the sides passing through the origin makes an angle 30° with the positive direction of the x-axis , then the sum of the x-coordinates of the vertices of the square is:

Options:
A 23-2
B 3-2
C 23-1
D 3-1

The correct answer is:
23-2

Solution:
For A,

xcos30°=ysin30°=2

⇒x=232= 3 and y=1

For C,

xcos120°=ysin120°=2

x=-1, y=3

For B,

xcos75°=ysin75°=22

x= 3-1, y=3+1

Sum of the x-coordinates =0+3+3-1+-1

=23-2

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2017 (09 Apr Online)
A square of side a lies above the $x$-axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha\left(0 < \alpha < \frac{\pi}{4}\right)$ with the positive direction of $\mathrm{x}$-axis. The equation of its diagonal not passing through the origin is

Options:
A
$y(\cos \alpha+\sin \alpha)+x(\cos \alpha-\sin \alpha)=a$

B
$y(\cos \alpha-\sin \alpha)-x(\sin \alpha-\cos \alpha)=a$

C
$\mathrm{y}(\cos \alpha+\sin \alpha)+\mathrm{x}(\sin \alpha-\cos \alpha)=\mathrm{a}$

D
$y(\cos \alpha+\sin \alpha)+x(\sin \alpha+\cos \alpha)=a$

The correct answer is:
$y(\cos \alpha+\sin \alpha)+x(\cos \alpha-\sin \alpha)=a$

Solution:
Co-ordinates of $\mathrm{A}=(a \cos \alpha, a \sin \alpha)$
Equation of $\mathrm{OB}, \mathrm{y}=\tan \left(\frac{\pi}{4}+\alpha\right) \mathrm{x}$
$\mathrm{CA} \perp \mathrm{r}$ to OB $\quad \therefore$ slope of $\mathrm{CA}=-\cot \left(\frac{\pi}{4}+2\right)$
Equation of $\mathrm{CA} \mathrm{y}-\mathrm{a} \sin \alpha=-\cot \left(\frac{\pi}{4}+2\right)(\mathrm{x}-\mathrm{a} \cos \alpha)$
$\mathrm{y}(\sin \alpha+\cos \alpha)+\mathrm{x}(\cos \alpha-\sin \alpha)=\mathrm{a}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2003
A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in cm2) is equal to

Options:
A 800
B 675
C 1025
D 900

The correct answer is:
800

Solution:
Given that, the side of square is 30 cm and x cm squares are cut off.

The required diagram is,

Now the dimensions of the cuboid formed will be

lx=30-2x, bx=30-2x and hx=x.

The Volume of the cuboid will be Vx=30-2x2x

Now to get Maximum value,

⇒dVxdx=0

⇒230-2x-2x+30-2x21=0

⇒30-2x-4x+30-2x=0

On simplifying we get,

⇒x=15 cm, 5 cm

But x cannot be 15 cm as the volume becomes zero.

Hence x=5 cm.

Now to find the surface area of the cuboid,

Surface area will be =30-2x×x×4+30-2x2

=30-2×5×5×4+30-2×52

=800 cm2.

Therefore, the required surface area will be 800 cm2.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (10 Apr Shift 1)
A square planar complex is formed by hybridisation of which atomic orbitals ?

Options:
A
$s, p_x, p_y, d_{y z}$

B
$\mathbf{s}, \mathbf{p}_x, \mathbf{p}_y, \mathrm{~d}_{x^2-y^2}$

C
$s, p_x, p_y, d_{z^2}$

D
$\mathbf{s}, \mathbf{p}_y, \mathrm{p}_z, \mathrm{~d}_{\mathrm{xy}}$

The correct answer is:
$\mathbf{s}, \mathbf{p}_x, \mathbf{p}_y, \mathrm{~d}_{x^2-y^2}$

Solution:
No Solution Available

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2002
A square plate is hinged as shown in figure and it is held stationary by means of a light thread as shown in figure. Then find out the force exerted by the hinge.

Options:
A 0
B $\mathrm{mg}$
C $\mathrm{mg}-\mathrm{T}$
D $\mathrm{mg}+\mathrm{T}$

The correct answer is:
0

Solution:
Let $\mathrm{N}$ be the force exerted by the hinge.
From FBD of square plate we can see that tension $T$ and weight mg have same line of action.
Under equilibrium condition,
$\begin{aligned}
& \tau_{\text {hinge }}=0 \\
& \Rightarrow \mathrm{T}=\mathrm{mg} \ldots
\end{aligned}$
Also,
$\begin{aligned}
& F_{\text {net }}=O \\
& \Rightarrow N+T=m g \\
& \Rightarrow N=O \text { Using (1) }
\end{aligned}$
So, the force exerted by the hinge, i.eN = 0 .

Topic: Physics, Rotational Motion, JEE Main
A square plate of $0.1 \mathrm{~m}$ side moves parallel to a second plate with a velocity of $0.1 \mathrm{~m} / \mathrm{s}$, both plates being immersed in water. If the viscous force is $0.002 \mathrm{~N}$ and the coefficient of viscosity is 0.01 poise, distance between the plates in $m$ is

Options:
A 0.1
B 0.05
C 0.005
D 0.0005

The correct answer is:
0.0005

Solution:
$A=(0.1)^2=0.01 \mathrm{~m}^2, \eta=0.01$ Poise $=0.001$ decapoise (M.K.S. unit), $d v=0.1 \mathrm{~m} / \mathrm{s}$ and. $E=0.002 \mathrm{~N}$
$F=\eta A \frac{d v}{d x} \quad \therefore d x=\frac{\eta A d v}{F}=\frac{0.001 \times 0.01 \times 0.1}{0.002}=0.0005 \mathrm{~m}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A square shaped coil of area 70 cm2 having 600 turns rotates in a magnetic field of 0.4 Wb m–2, about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field. If the coil completes 500 revolution in a minute, the instantaneous emf when the plane of the coil is inclined at 60° with the field, will be ______ V.

(Take π=227)

Options:

The correct answer is:
44

Solution:
Given here, N=600, A=70×10-4 m2, and B=0.4 T

Angular speed of coil is ω=2πn60=500×2π60=100π6 rad s-1

Instantaneous emf in coil is given by ε=-Ndϕdt, where, magnetic flux ϕ=BAcosωt and ωt=30° is angle between A and B.

So, ε=-NdBAcosωtdt=-N-BAωsinωt=NBAωsinωt

Putting the values, we get

ε=600×70×10-4×0.4×100π6×12

=44 V

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (01 Feb Shift 2)
A square shaped hole of side l= a2 is carved out at a distance d=a2 from the centre 'O' of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from O is -ax, value of X (to the nearest integer) is :

Options:

The correct answer is:
23

Solution:
Xa=A·X-A·X1A-A1=πa2×0-a24×a2πa2-a24
=-a3/8π-14a2=-a2(4π-1)=-a8π-2=-a23
x=23

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (02 Sep Shift 2)
A stair-case of length $l$ rests against a vertical wall and a floor of a room. Let $\mathrm{P}$ be a point on the stair-case, nearer to its end on the wall, that divides its length in the ratio $1: 2$. If the staircase begins to slide on the floor, then the locus of $\mathrm{P}$ is:

Options:
A
an ellipse of eccentricity $\frac{1}{2}$

B
an ellipse of eccentricity $\frac{\sqrt{3}}{2}$

C
a circle of radius $\frac{1}{2}$

D
a circle of radius $\frac{\sqrt{3}}{2} l$

The correct answer is:
an ellipse of eccentricity $\frac{\sqrt{3}}{2}$

Solution:
Let point $\mathrm{A}(a, 0)$ is on $x$-axis and $\mathrm{B}$ $(0, b)$ is on $y$-axis.
Let $\mathrm{P}(h, k)$ divides $\mathrm{AB}$ in the ratio $1: 2$. So, by section formula
$$
\begin{aligned}
&h=\frac{2(0)+1(a)}{1+2}=\frac{a}{3} \\
&k=\frac{2(b)+1(0)}{3}=\frac{2 b}{3} \\
&\Rightarrow a=3 h \text { and } b=\frac{3 k}{2}
\end{aligned}
$$
Now, $a^2+b^2=l^2$
$$
\begin{aligned}
&\Rightarrow 9 h^2+\frac{9 k^2}{4}=l^2 \\
&\Rightarrow \frac{h^2}{\left(\frac{l}{3}\right)^2}+\frac{k^2}{\left(\frac{2 l}{3}\right)^2}=1
\end{aligned}
$$
Now $\mathrm{e}=\sqrt{1-\left(\frac{l^2}{9} \times \frac{9}{4 l^2}\right)}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$
Thus, required locus of $\mathrm{P}$ is an ellipse with eccentricity $\frac{\sqrt{3}}{2}$.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2014 (11 Apr Online)
A standard cell of 1.08 V is balanced by the potential difference across 90 cm of a 1 meter, long wire supplied by a cell of emf 2V through a series resistor of resistance 2 Î©. If the internal resistance of cell in the primary circuit is zero then the resistance per unit length (in Î© m-1) of potentiometer wire is

Options:

The correct answer is:
3

Solution:
E=X.l (X=Potential gradient)
E =EoL×RwRw+r+Rh×l ∵ r = 0, Rh = 2 Ω
∴ 1.08 = 21 m×RwRw+2 × 0.90 m
1.082×0.90= RwRw+2
Hence : 0.6 =RwRw+2
0.6 Rw + 1.2 = Rw⇒1.2 = 0.4 Rw ⇒3 = Rw
∴ Resistance per unit length is 3 Ω m-1

Topic: Physics, Experimental Physics, JEE Main
A standard filament lamp consumes $100 \mathrm{~W}$ when connected to $200 \mathrm{~V}$ a.c. mains supply. The peak current through the bulb will be :

Options:
A $0.707 \mathrm{~A}$
B $1 \mathrm{~A}$
C $1.414 \mathrm{~A}$
D $2 \mathrm{~A}$

The correct answer is:
$0.707 \mathrm{~A}$

Solution:
Power,
$\begin{aligned}
\mathrm{P} & =\mathrm{I}_{\mathrm{rms}} \mathrm{V}_{\mathrm{rms}} \\
\mathrm{I}_{\mathrm{rms}} & =\frac{\mathrm{P}}{\mathrm{V}_{\mathrm{rms}}}=\frac{100}{200}=\frac{1}{2} \\
\mathrm{I}_0 & =\sqrt{2} \mathrm{I}_{\mathrm{rms}} \\
\mathrm{I}_0 & =\sqrt{2} \times \frac{1}{2} \\
\mathrm{I}_0 & =\frac{1}{\sqrt{2}}=0.707 \mathrm{~A}
\end{aligned}$

Topic: Physics, Alternating Current, JEE Main
A standard filament lamp consumes $100 \mathrm{~W}$ when connected to $200 \mathrm{~V}$ a.c. mains supply. The peak current through the bulb will be :

Options:
A $0.707 \mathrm{~A}$
B $1 \mathrm{~A}$
C $1.414 \mathrm{~A}$
D $2 \mathrm{~A}$

The correct answer is:
$0.707 \mathrm{~A}$

Solution:
Power,
$\begin{aligned}
\mathrm{P} & =\mathrm{I}_{\mathrm{rms}} \mathrm{V}_{\mathrm{rms}} \\
\mathrm{I}_{\mathrm{rms}} & =\frac{\mathrm{P}}{\mathrm{V}_{\mathrm{rms}}}=\frac{100}{200}=\frac{1}{2} \\
\mathrm{I}_0 & =\sqrt{2} \mathrm{I}_{\mathrm{rms}} \\
\mathrm{I}_0 & =\sqrt{2} \times \frac{1}{2} \\
\mathrm{I}_0 & =\frac{1}{\sqrt{2}}=0.707 \mathrm{~A}
\end{aligned}$

Topic: Physics, Alternating Current, NEET, NEET 2022 (Phase 2)
A standing wave having 3 nodes and 2 antinodes is formed between two atoms having a distance 1.21 Aâ between them. The wavelength of the standing wave is

Options:
A 1.21 A∘

B 1.42 A∘
C 6.05 A∘

D 3.63 A∘

The correct answer is:
1.21 A∘

Solution:
The given standing wave is shown in the figure
As the length of loop or segment is λ2
So the length of 2 segments is 2λ2
∴2λ2=1.21 A∘
⟹λ=1.21 A∘

Topic: Physics, Waves and Sound, JEE Main
A standing wave having 3 nodes and 2 antinodes is formed between two atoms having a distance $1.21 Å$ between them. The wavelength of the standing wave is

Options:
A $1.21 Å$
B $2.42 Å$
C $6.05 Å$
D $3.63 Å$

The correct answer is:
$1.21 Å$

Solution:
$\lambda=1.21 Å$

Topic: Physics, Waves and Sound, JEE Main
A standing wave is formed by the superposition of two waves travelling in opposite directions. The transverse displacement is given by, yx, t=0.5sin⁡5π4xcos⁡200πt. What is the speed of the travelling wave moving in the positive x direction? (x and t are in meter and second, respectively)

Options:
A 120 m s-1
B 90 m s-1
C 160 m s-1
D 180 m s-1

The correct answer is:
160 m s-1

Solution:
y=2asinkxcosωt
⇒k=5π4 m-1, ω=200π rad s-1,
v=ωk=200π5π4=160 m s-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2017 (09 Apr Online)
A star $2.5$ times the mass of the sun and collapsed to a size of the $12 \mathrm{~km}$ rotates with a speed of $1.5$ revolution per second. Will an object placed on its equator remain struck to its surface due to gravity? (Mass of the sun $=2 \times 10^{30} \mathrm{~kg}$ )

Options:

None

Solution:
The object will remain struck to the surface of star due to gravity, if the acceleration due to gravity is more than the centrifugal acceleration due to its rotation.
Acceleration due to gravity $=g=\frac{G M}{R^2}$
$$
=\frac{6.67 \times 10^{-11} \times 2.5 \times 2 \times 10^{30}}{(12000)^2}=2.3 \times 10^{12} \mathrm{~m} / \mathrm{s}^2
$$
Centrifugal acceleration $=\mathrm{r} \omega^2=r(2 \pi)^2$
$=12000 \times(2 \times 3.14 \times 1.5)^2=1.1 \times 10^6 \mathrm{~m} / \mathrm{s}^2$
Here $g>r \omega^2$
$\therefore \quad$ The objected will remain struck with the surface.

Topic: Physics, Gravitation
A star initially has 1040 deuterons. It produces energy via the processes 1 H 2 + 1 H 2 â¶ 1 H 3 + p & 1 H 2 + 1 H 3 â¶ 2 He 4 + n . If the average power radiated by the star is 1016 W, the deuteron supply of the star is exhausted in a time of the order of

(mass of H21=2.014 amu, mass of He42=4.001 amu, mp=1.007 amu, mn=1.008 amu)

Options:
A 106 s

B 108 s

C 1012 s

D 1016 s

The correct answer is:
1012 s

Solution:
The given reactions are:
1H2 + 1H2 → 1H3 + p
1H2 + 1H3 → 1He4 + n
31H2→ 2He4 +n + p

Mass defect
Δm = (3 × 2. 014 - 4. 001 - 1. 007 - 1. 008) amu
= 0.026 amu
Energy released = 0. 026 × 931 MeV
= 0. 026 × 931 × 1. 6 × 10-12J = 3. 87 × 10-13 J
This is the energy produced by the consumption of three deuteron atoms.
Therefore Total energy released by 1040 deuterons
= 1 0 4 0 3 × 3. 87 × 10-12 J = 1. 29 × 1028 J
The average power radiated is P=1016 W or 1016 J s. Therefore, the total time to exhaust all deuterons of the star, time taken will be
t=1.29×10281016=1.29×1012 ≈ 1012

Topic: Physics, Nuclear Physics, JEE Main
A stationary balloon is observed from three points A, B and C on the plane ground and is found that its angle of elevation from each point is 60o. If ∠ABC=30o and AC=5 meters, then the height of the balloon from the ground is

Options:
A 53 meters
B 543 meters
C 53 meters
D 543 meters

The correct answer is:
53 meters

Solution:
Let, PQ be vertical with height h meters and at its end Q there is a balloon and P is on the ground
⇒PA=PB=PC=hcot⁡60o=h3
∴ P is the circumcentre of ΔABC and its circumradius R=h3
In ΔABC,
2R=5sin⁡30o⇒2h3=5sin⁡30o
∴h=53 meters.

Topic: Mathematics, Heights and Distances, JEE Main
A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other, each with same kinetic energy E0. The energy of the explosion will be K times of E0, then the value of K is

Options:

The correct answer is:
6

Solution:
P 1 â =m v 0 i ^
P 2 â =m v 0 j ^
P3â=mv0j^
P4â=mvâ
0=P1â+P2â+P3â+P4â
Vâ=-v0(i^+j^+k^)
v=v03
Total energy =3Ã12mv02+12mv2=3E0+3E0
=6E0

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of θ, where θ is the angle by which it has rotated, is given as kθ2. If its moment of inertia is I then the angular acceleration of the disc is:

Options:
A 2kIθ
B k2Iθ
C k4Iθ
D kIθ

The correct answer is:
2kIθ

Solution:
Kinetic Energy =kθ2
12Iω2=kθ2
ω2=2kθ2I
Differentiate both side w.r.t. θ .
2ωdωdθ=4kθI
ωdωdθ=2kθI
α=2kθI

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A stationary hydrogen atom in the first excited state emits a photon. If the mass of the hydrogen atom is m and its ionization energy is E, then the recoil velocity acquired by the atom is [speed of light = c]

Options:
A 3E2m+c2-c
B 3E4m+c2-c
C 3E4mc
D Emc

The correct answer is:
3E2m+c2-c

Solution:
mv=hλ, here v is recoil speed of H-atom and λ is the wavelength of the photon.

When the electron jumps down from the first excited state to the ground state the energy released is

∆E2→1=E1-122=3E4

3E4=12mv2+hcλ

3E4=12mv2+mvc

m2v2+mcv-3E4=0

⇒v=3E2m+c2-c

Topic: Physics, Atomic Physics, JEE Main
A stationary \( \mathrm{He}^{+} \)ion emitted a photon corresponding to first line of lyman series. The electron liberated from stationary H- atom after this elastic collision will have certain de Broglie wave length \( (\lambda) . \lambda \) will be expressed as \( (0.26 \times \mathrm{x}) Å \) for photo electron Take \( \sqrt{5.51}=2.34 \). The value of \( \mathrm{x} \) will be

Options:

The correct answer is:
9

Solution:
Photon emitted from He + = - 13.6 2 2 1 2 2 - 1
= 40.8 eV
Liberated electron energy = 40.8 - 13.6
âµ IE of H atom = 13.6 eV
i.e. as if accelerated by electric potential of 27.2 volts .
λ in Å for photo electron = 1 5 0 27.2 = 5.51 = 2.34
2.34 = 0.26 × x ∴ x = 9

Topic: Chemistry, Structure of Atom, JEE Main
A stationary object is released from a point P a distance 3R from the centre of the moon which has radius R and mass M. Which one of the following expressions gives the speed of the object on hitting the moon?

Options:
A 2GM3R12
B 4GM3R12
C 2GMR12
D GMR12

The correct answer is:
4GM3R12

Solution:
According to the law of conservation of energy,
-GmM3R+mv022=-GmMR+mv122
Where m is the mass of the object, v0 and v1 are the initial and final velocities of object, respectively.
The object is stationary i.e., v0=0.
â -GM3R=-GMR+v122
â v122=2GM3R
â v1=2GM3R=4GM3R12

Topic: Physics, Gravitation, JEE Main
A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec . The oscillation frequency of each tuning fork is v0=1400 Hz and the velocity of sound in air is 350 m/s . The speed of each tuning fork is close to:

Options:
A 12m/s
B 1m/s
C 14m/s
D 18m/s

The correct answer is:
14m/s

Solution:
f0CC-V-f0CC+V=2
V=14m/s

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2020 (07 Jan Shift 2)
A stationary particle breaks into two parts of masses $m_A$ and $m_B$ which move with velocities $v_A$ and $v_B$ respectively. The ratio of their kinetic energies $\left(K_B: K_A\right)$ is :

Options:
A $v_B: v_A$
B $m_B: m_A$
C $m_B v_B: m_A v_A$
D $1: 1$

The correct answer is:
$v_B: v_A$

Solution:
Initial momentum is zero.
$\begin{aligned} & \text { Hence }\left|\mathrm{P}_{\mathrm{A}}\right|=\left|\mathrm{P}_{\mathrm{B}}\right| \\ & \Rightarrow \mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{B}}=\mathrm{m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}} \\ & \frac{(\mathrm{KE})_{\mathrm{A}}}{(\mathrm{KE})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}^2}{\frac{1}{2} \mathrm{~m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}}^2}=\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}} \\ & \frac{(\mathrm{KE})_{\mathrm{B}}}{(\mathrm{KE})_{\mathrm{A}}}=\frac{\mathrm{v}_{\mathrm{B}}}{\mathrm{v}_{\mathrm{A}}}\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2024 (08 Apr Shift 1)
A stationary point source of sound emits sound uniformly in all directions in a non-absorbing medium. Two points $P$ and $Q$ are at a distance of $4 \mathrm{~m}$ and $9 \mathrm{~m}$ respectively from the source. The ratio of amplitudes of the waves at $P$ and $Q$ is

Options:
A $\frac{3}{2}$
B $\frac{4}{9}$
C $\frac{2}{3}$
D $\frac{9}{4}$

The correct answer is:
$\frac{9}{4}$

Solution:
For an isotropic point source of power P, intensity I at a distance $\mathrm{r}$ from it is
$$
\mathrm{I}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}}
$$
Since power P remains the same,
$$
\therefore \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2}=\left(\frac{9}{4}\right)^{2}
$$
$\because \quad I \propto \mathrm{A}^{2}$
where $A$ is the amplitude of a wave $\therefore \quad \frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\sqrt{\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}}=\frac{9}{4}$

Topic: Physics, Waves and Sound, KCET, KCET 2009