datavorous/entrance-exam-dataset huggingface dataset:
A hollow spherical shell of radius $r$ has a uniform charge density $\sigma$. It is kept in a cube of edge $3 r$ such that the centres of the cube and the shell coincide. Then the electric flux coming out of one face of a cube is ( $\varepsilon_0$ - permittivity of free space)

Options:
A $\frac{\pi \mathrm{r}^2 \sigma}{\varepsilon_0}$
B $\frac{5 \varepsilon_0}{2 \pi r^2 \sigma}$
C $\frac{\pi \mathrm{r}^2 \sigma}{6 \varepsilon_0}$
D $\frac{2 \pi r^2 \sigma}{3 \varepsilon_0}$

The correct answer is:
$\frac{2 \pi r^2 \sigma}{3 \varepsilon_0}$

Solution:
Charge on the hollow sphere, $q=\sigma \times 4 \pi r^2$
According to Gauss's law, The flux through a single face
of the cube, $\oint^{\prime}=\frac{1}{6} \frac{\mathrm{q}}{\varepsilon_0}=\frac{\sigma \times 4 \pi \mathrm{r}^2}{6 \varepsilon_0}=\frac{2 \pi \mathrm{r}^2 \sigma}{3 \varepsilon_0}$

Topic: Physics, Electrostatics, JEE Main
A hollow vertical cylinder of radius R and height h has a smooth internal surface. A small particle is held in contact with the inner side of the upper rim at a point P. It is given a horizontal speed v0 tangential to the rim. It leaves the lower rim at point Q, vertically below P. The number of revolutions made by the particle will be [Take acceleration due to gravity g]

Options:
A h2πR
B v0hh2g
C v02πR2hg
D v0πg2gh2R

The correct answer is:
v02πR2hg

Solution:
A hollow vertical
h = 1 2 gt 2
n 2 π R = V 0 t
n = V 0 2 π R 2 h g

Topic: Physics, Laws of Motion, JEE Main
A homogeneous ideal gaseous reaction AB2g⇌Ag+2 Bg is carried out in a 25 litre flask at 27°C. The initial amount of AB2 was 1 mole and the equilibrium pressure was 1.9 atm. The value of Kp is x×10-2. The value of x is

(Integer answer) R=0.08206 dm3 atm K-1 mol-1

Options:

The correct answer is:
73

Solution:
AB211-α0.535 ⇌Aα0.465+2B2α0.9

1.9×25=nT×0.08206×300

nT=1.93=1+2α

α=0.465

Kp=0.4651.93×190.931.93×1.920.5351.93×1.9

≃73×10-2 atm2

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2021 (26 Feb Shift 1)
A homogeneous sphere floats with half its volume immersed in mercury (density 13.6 gm cc-1 ) and the other half in oil (density 0.8 gm cc-1 ) in a vessel containing oil over mercury. What is the density of the material of the sphere in gm cc-1?

Options:
A 3.3
B 6.4
C 7.2
D 12.8

The correct answer is:
7.2

Solution:
For equilibrium, the total upward pull will be equal to the downward pull. If V is the volume of the sphere, we have
V⁡ 2 1 3 . 6 g + V⁡ 2 0 . 8 g = V⁡ ρ g
∴ ρ = 1 3 . 6 + 0 . 8 2 g cm - 3 = 7 . 2 g cm - 3

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and Tc respectively, then:

Options:
A Th=2Tc
B Th=Tc
C Th=0.5Tc
D Th=1.5Tc

The correct answer is:
Th=Tc

Solution:
Time period of any magnetic body place in an external magnetic field is:-
T=2πIMB
Thoop=2πmR2MB ...... (1) Ihoop=mR2
Tc=2πmR22M2B ...... (2) ISolid cylinder=mR22
From (1) and (2)
Th=Tc

Topic: Physics, Oscillations, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and Tc respectively, then:

Options:
A Th=2Tc
B Th=Tc
C Th=0.5Tc
D Th=1.5Tc

The correct answer is:
Th=Tc

Solution:
Time period of any magnetic body place in an external magnetic field is:-
T=2πIMB
Thoop=2πmR2MB ...... (1) Ihoop=mR2
Tc=2πmR22M2B ...... (2) ISolid cylinder=mR22
From (1) and (2)
Th=Tc

Topic: Physics, Magnetic Properties of Matter, JEE Main
A hoop of mass \( M \) and radius \( R \) is hung from a support fixed in a wall. Its moment of inertia about the support is

Options:
A \( 2 \mathrm{MR}^{2} \)
B \( 3 \mathrm{MR}^{2} \)
C \( 4 \mathrm{MR}^{2} \)
D \( 6 \mathrm{MR}^{2} \)

The correct answer is:
\( 2 \mathrm{MR}^{2} \)

Solution:
A hoop is a circular ring ∴ I 0 = MR 2
By theorem of parallel axes,
I = I 0 + MR 2 = MR 2 + MR 2 = 2 MR 2

Topic: Physics, Rotational Motion, JEE Main
A hoop of radius r and mass m rotating with an angular velocity ω0 is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

Options:
A r Ï 0 2
B r ω 0
C r ω 0 4
D r ω 0 3

The correct answer is:
r ω 0 2

Solution:
By conservation of angular momentum at the point of contact

mr2ω0=mvr+mr2ω=mvr+mr2.v/r
⇒ r2ω0=2vr⇒v=rω0/2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2013 (07 Apr)
A horizontal disk is rotating with angular velocity Ï about a vertical axis passing through its centre. A ball is placed at the centre of groove and pushed slightly. The velocity of the ball when it comes out of the groove-

Options:
A 32ÏR
B ωR2
C ωR
D ωR2

The correct answer is:
32ωR

Solution:
Let us consider the motion of the ball with respect to disk
Net force along groove=mÏ2r sinâ¡Î¸
=mÏ2r xr
=mÏ2x
â´ ma=mÏ2x
â v dvdx=Ï2x

ââ«0vvdv=â«0R2Ï2xdx

âv22=Ï22R22
â v=ÏR2
vâBall,Ground=vâBall,Disk+ vâDisk,Ground
â´ vâBall,Ground
= ÏR2+ÏR22+2ÏRÏR2cosâ¡120Â°12
=32ÏR

Topic: Physics, Laws of Motion, JEE Main
A horizontal disk of moment of inertia $4.25 \mathrm{~kg}-\mathrm{m}^{2}$ with respect to its axis of symmetry is spinning counter clockwise at 15 revolutions per second about its axis, as viewed from above. A second disk of moment of inertia $1.80 \mathrm{~kg}-\mathrm{m}^{2}$ with respect to its axis of symmetry is spinning clockwise at 25 revolutions per second as viewed from above about the same axis and is dropped on top of the first disk. The two disks stick together and rotate as one about their axis of symmetry. The new angular velocity of the system as viewed from above is close to.

Options:
A 18 revolutions/second and clockwise
B 18 revolutions/second and counter clockwise
C 3 revolutions/second and clockwise
D 3 revolutions/second and counter clockwise

The correct answer is:
3 revolutions/second and clockwise

Solution:
Take anticlockwise positive
$$
\begin{array}{l}
I_{1} \omega_{1}+I_{2} \omega_{2}=I \omega \\
4.25 \times 15-1.80 \times 25=6.05 \times \omega
\end{array}
$$
$\omega=3$ revolutions/second and clockwise

Topic: Physics, Rotational Motion, JEE Main
A horizontal force F is applied at the top of an equilateral triangular block having mass m. The minimum coefficient of friction required to topple the block before translation will be

Options:
A 23
B 12
C 13
D None of these

The correct answer is:
13

Solution:
I=ma26+ma32⇒I=ma22
⇒F×a32=mga2⇒F=mg3
F=Fmax⇒mg3μmg⇒μ=13

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Topic: Physics, Laws of Motion, JEE Main
A horizontal force of $10 \mathrm{~N}$ is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2 . the weight of the block is

Options:
A $2 \mathrm{~N}$
B $20 \mathrm{~N}$
C $50 \mathrm{~N}$
D $100 \mathrm{~N}$

The correct answer is:
$2 \mathrm{~N}$

Solution:
$F=\frac{W}{\mu} \quad \therefore W=\mu F=0.2 \times 10=2 N$

Topic: Physics, Laws of Motion, JEE Main
A horizontal force of $10 \mathrm{~N}$ is necessary to just hold a block stationary against a wall. The co-efficient of friction between the block and the wall is $0.2$. The weight of the block is

Options:
A
$20 \mathrm{~N}$

B
$50 \mathrm{~N}$

C
$100 \mathrm{~N}$

D
$2 \mathrm{~N}$

The correct answer is:
$2 \mathrm{~N}$

Solution:
$\mathrm{f}=\mathrm{mg} \Rightarrow \mu \mathrm{N}=\mathrm{W} \Rightarrow \mu .10=\mathrm{W}$
$\Rightarrow 0.2 \times 10=\mathrm{W}$
$\therefore \mathrm{W}=2 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2003
A horizontal metallic rod of mass m and length l is supported by two vertical identical springs of spring constant, K each and natural length l0. A current i is flowing in the rod in the direction shown. If the rod is in equilibrium then the length of each spring in this state is:

Options:
A l0+ilB-mgK
B l0+ilB-mg2K
C l0+mg-ilB2K
D l0+mg-ilBK

The correct answer is:
l0+ilB-mg2K

Solution:
The force on the rod due to the magnetic field and gravity is
ilB-mg (upwards)
Hence the extension in the springs is ilB-mg2k
(Note that effective spring constant is 2k)
Therefore the length of the spring is l0+ilB-mg2k

Topic: Physics, Magnetic Effects of Current, JEE Main
A horizontal overhead power line is at a height of $4 \mathrm{~m}$ from the ground and carries a current of $100 \mathrm{~A}$ from east to west. The magnetic field directly below it on the ground is $\left(\mu_0=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1}\right)$

Options:
A
$2.5 \times 10^{-7} \mathrm{~T}$ southward

B
$5 \times 10^{-6} \mathrm{~T}$ northward

C
$5 \times 10^{-6}$ T southward

D
$2.5 \times 10^{-7}$ northward

The correct answer is:
$5 \times 10^{-6}$ T southward

Solution:
$$
B=\frac{\mu_0}{2 \pi} \frac{i}{R}=\frac{4 \pi \times 10^{-7}}{2 \pi} \times \frac{100}{4}=5 \times 10^{-6} \mathrm{~T} \text { southward }
$$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2008
A horizontal plane supports a fixed vertical cylinder of radius R and a particle is attached to the cylinder by a horizontal thread AB as shown in figure. The horizontal velocity v0 is imparted to the particle normal to the thread, then during subsequent motion which of the following statements is true?

Options:
A angular momentum of particle about O remains constant
B angular momentum about B remains constant
C momentum and kinetic energy both remain constant
D kinetic energy alone remains constant

The correct answer is:
kinetic energy alone remains constant

Solution:
From the diagram, we can see that net torque about O and B is not zero. Therefore angular momentum changes about these points.
As the tension is always perpendicular to velocity therefore work done is zero and speed remains constant and hence the kinetic energy.
But the direction of velocity changes continuously and hence momentum changes.

Topic: Physics, Rotational Motion, NEET
A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time a viscous fluid of mass $m$ is dropped at the centre and is allowed to spread out and finally fall. The angular velocity during this period

Options:
A decreases continuously
B decreases initially and increases again
C remains unaltered
D increases continuously.

The correct answer is:
decreases initially and increases again

Solution:
According to law of conservation of momentum,
$$
I \omega=\text { constant }
$$
When viscous fluid of mass $m$ is dropped and start spreading out then its moment of inertia increases and angular velocity decreases. But when it start falling then its moment of inertia again starts decreasing and angular velocity increases.

Topic: Physics, Rotational Motion, AIIMS, AIIMS 2005
A horizontal platform, with an object placed on it, is executing SHM in the vertical direction. The amplitude of oscillation is 2.5 cm. Let the least period of these oscillations, so that the object is not detached, be equal to Ï2a. Then, the value of a is (Given: g=10 m s-2)

Options:

The correct answer is:
5

Solution:
The object would not leave the platform when
amax<g
∴ w2A<g
∴ w<102.5×10-2=10425=1005=20
∴ 2πT<20 ⇒T>π10

Topic: Physics, Oscillations, JEE Main
A horizontal ray of light is incident on the right-angled prism with prism angle $6^{\circ}$. If the refractive index of the material of the prism is 1.5 , then the angle of emergence will be

Options:
A $4^{\circ}$
B $6^{\circ}$
C $9^{\circ}$
D $10^{\circ}$

The correct answer is:
$9^{\circ}$

Solution:
Deviation produced by a thin prism
$\delta=(\mu-1) A=\left(\frac{3}{2}-1\right) 6^{\circ}=3^{\circ}$
$\begin{aligned} \text { Angle of emergence }=e & =6^{\circ}+3^{\circ} \\ & =9^{\circ}\end{aligned}$

Topic: Physics, Ray Optics, JEE Main
A horizontal rod of mass 0.01 kg and length 10 cm is placed on a frictionless plane inclined at an angle 60Â° with the horizontal and with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field is applied vertically downwards. If the current through the rod is 1.73 A, then the value of magnetic field induction B for which the rod remains stationary on the inclined plane is

Options:
A 1 T
B 3 T
C 2.5 T
D 4 T

The correct answer is:
1 T

Solution:
Here two forces are acting on the rod simultaneously. From the hypothetical free body diagram.
mgsin⁡60=Bilcos⁡60o
B=mgiltan⁡60o
=0.01×101.73×0.1×3
B=1 T

Topic: Physics, Laws of Motion, NEET
A horizontal rod of mass $0.01 \mathrm{~kg}$ and length 10 $\mathrm{cm}$ is placed on a frictionless plane inclined at an angle $60^{\circ}$ with the horizontal and with the length of rod parallel to the edge of the inclined plane. A uniform magnetic field is applied 'Vertically downwards. If the current through the rod is $1.73 \mathrm{~A}$, then the value of magnetic field induction $B$ for which the rod remains stationary on the inclined plane is

Options:
A $1 \mathrm{~T}$
B $3 \mathrm{~T}$
C $2.5 \mathrm{~T}$
D $4 \mathrm{~T}$

The correct answer is:
$1 \mathrm{~T}$

Solution:
Here two forces acting on the rod simultaneously.

From FBD, mg sin $60=$ Bil $\cos 60^{\circ}$
$\begin{array}{c}
\mathrm{B}=\frac{\mathrm{mg}}{\mathrm{il}} \tan 60^{\circ} \\
=\frac{0.01 \times 10}{173 \times 0.1} \times \sqrt{3}=1 \mathrm{~T}
\end{array}$

Topic: Physics, Magnetic Effects of Current, VITEEE, VITEEE 2015
A horizontal rod of mass m=3kÏ2 kg and length L is pivoted at one end. The rod at the other end is supported by a spring of force constant k. The rod is displaced by a small angle Î¸ from its horizontal equilibrium position and released. The time period (in second) of the subsequent simple harmonic motion is

Options:

The correct answer is:
2

Solution:
E=12mL23Ï2+12kx-x02+mgx2
dEdt=12mL232vLaL+kx-x0v+mg2v=0
=m3a+kx-kx0+mgv=0
At equilibrium,
kx0=mg2
â´ma3=-kx
a=-3kmx
T=2ÏÏ=2Ïm3k
=2Ï3k3kÏ2=2 s

Topic: Physics, Oscillations, JEE Main
A horizontal smooth square platform ABCD is moving towards the right with a uniform speed v. At what angle Î¸ must a particle slide from A, on the platform, with speed u so that it strikes the point B? (Assume u to be uniform and u > v)

Options:
A sin-1uv
B cos-1vu
C cos-1uv
D sin-1vu

The correct answer is:
cos-1vu

Solution:
⇒ u cos θ t = v t
⇒ u cos θ = v
cos θ = v u
θ = cos -1 v u

Topic: Physics, Motion In Two Dimensions, JEE Main
A horizontal steel railroad track has a length of \( 100 \mathrm{~m} \) when the temperature is \( 25^{\circ} \mathrm{C} \). The track is constrained from expanding or bending. The stress on the track on a hot summer day, when the temperature is \( 40^{\circ} \mathrm{C} \), is (Note: The linear coefficient of thermal expansion for steel is \( 1.1 \times 10^{-5} \mathrm{CC}^{-1} \) anc the Young's modulus of steel is \( 2 \times 10^{11} \mathrm{~Pa} \) )

Options:
A \( 6.6 \times 10^{7} \mathrm{~Pa} \)
B \( 8.8 \times 10^{7} \mathrm{~Pa} \)
C \( 3.3 \times 10^{7} \mathrm{~Pa} \)
D \( 5.5 \times 10^{7} \mathrm{~Pa} \)

The correct answer is:
\( 3.3 \times 10^{7} \mathrm{~Pa} \)

Solution:
The stress is,
FA=Y×α×∆T. where, Y is Young's modulus, α is coefficient of thermal expansion & ∆T is change in temperature.

∴Stress=2×1011×1.1×10-5×40-25∴Stress=3.3×107 Pa

Topic: Physics, Mechanical Properties of Solids, JEE Main
A horizontal straight wire 5 m long extending from east to west falling freely at right angle to horizontal component of earth's magnetic field 0.60×10-4 Wb m-2. The instantaneous value of emf induced in the wire when its velocity is 10 m s-1 is _______×10-3 V.

Options:

The correct answer is:
3

Solution:
Horizontal component of earth's magnetic field,

BH=0.60×10-4 Wb m-2.

Induced emf e=BHvℓ

=0.60×10-4×10×5

=3×10-3 V

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2024 (29 Jan Shift 2)
A horizontal turn table in the form of a disc of radius $\mathrm{r}$ carries a gun at $\mathrm{G}$ and rotates with angular velocity $\omega_0$ about a vertical axis passing through the centre $O$. The increase in angular velocity of the system if the gun fires a bullet of mass $\mathrm{m}$ with a tangential velocity $\mathrm{v}$ with respect to the gun is (moment of inertia of gun+table about $\mathrm{O}$ is $\mathrm{I}_{\mathrm{o}}$ )

Options:
A $\frac{\mathrm{mvr}}{\mathrm{I}_{\mathrm{o}}+\mathrm{mr}^2}$
B $\frac{2 m v r}{I_o}$
C $\frac{v}{2 r}$
D $\frac{\mathrm{mvr}}{2 \mathrm{I}_{\mathrm{o}}}$

The correct answer is:
$\frac{\mathrm{mvr}}{\mathrm{I}_{\mathrm{o}}+\mathrm{mr}^2}$

Solution:
Let the final angular velocity of the "gun + table "system be w.
Given: velocity of bullet w.r.t gun is v.
Thus velocity of bullet w.r.t ground, $\mathrm{v}^{\prime}=\mathrm{v}-\mathrm{rw}$
Applying conservation of angular momemtum: $\mathrm{L}_i=\mathrm{L}_{\mathrm{f}}$
$\begin{aligned}& \left(\mathrm{mr}^2+\mathrm{I}_{\mathrm{o}}\right) \mathrm{w}_{\mathrm{o}}=\mathrm{I}_{\mathrm{o}} \mathrm{w}-\mathrm{mv} \mathrm{r} \\& \left(\mathrm{mr}^2+\mathrm{I}_{\mathrm{o}}\right) \mathrm{w}_{\mathrm{o}}=\mathrm{I}_{\mathrm{o}} \mathrm{w}-\mathrm{m}(\mathrm{v}-\mathrm{rw}) \mathrm{r} \\& \Rightarrow \mathrm{w}=\frac{\left(\mathrm{I}_{\mathrm{o}}+\mathrm{mr}^2\right) \mathrm{w}_{\mathrm{o}}+\mathrm{mvr}}{\mathrm{I}_{\mathrm{o}}+\mathrm{mr}^2}
\end{aligned}$
Increase in angular velocity, $\Delta \mathrm{w}=\mathrm{w}-\mathrm{w}_{\mathrm{o}}=\frac{\left(\mathrm{I}_{\mathrm{o}}+\mathrm{mr}^2\right) \mathrm{w}_{\mathrm{o}}+\mathrm{mvr}}{\mathrm{I}_{\mathrm{o}}+\mathrm{mr}^2}-\mathrm{w}_{\mathrm{o}}$
$\Rightarrow \Delta \mathrm{w}=\frac{\mathrm{mvr}}{\mathrm{I}_{\mathrm{o}}+\mathrm{mr}^2}$

Topic: Physics, Rotational Motion, JEE Main
'A horizontal uniform glass tube of $100 \mathrm{~cm}$, length sealed at both ends contains $10 \mathrm{~cm}$ mercury column in the middle. The temperature and pressure of air on either side of mercury column are respectively $31^{\circ} \mathrm{C}$ and $76 \mathrm{~cm}$ of mercury. If the air column at one end is kept at $0^{\circ} \mathrm{C}$ and the other end at $273^{\circ} \mathrm{C}$, then the pressure of air which is at $0^{\circ} \mathrm{C}$ is : (in $\mathrm{cm}$ of $\mathrm{Hg}$ )

Options:
A 76
B 88.2
C 102.4
D 122

The correct answer is:
102.4

Solution:
On keeping the temperature of the ends of tube at $0^{\circ} \mathrm{C}$ and $273^{\circ} \mathrm{C}$
Applying ideal gas equation
$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}=\frac{p_3 V_3}{T_3}$
$\frac{76 \times 45}{(273+31)}=\frac{p_2 \times l}{(273+0)}=\frac{p_3(90-l)}{273+273}$
$\frac{76 \times 45}{304}=\frac{p_2 \times l}{273}=\frac{p_3(90-l)}{546}$
$\begin{array}{lll}&\text { I } &&\text { II } && \text { III }\end{array}$
From Part II and III,
$\frac{p_2 \times l}{273}=\frac{p_3(90-l)}{546}$
(Mercury column is at rest, so pressure difference,
$p_2-p_3=0 \Rightarrow p_2=p_3$
$\therefore \quad \frac{p_2 \times l}{273}=\frac{p_2(90-l)}{546}$
$2 l=90-l$
$\Rightarrow \quad l=30 \mathrm{~cm}$
From I and II,
$\frac{76 \times 45}{304}=\frac{p_2 \times 30}{273}$
$\Rightarrow \quad p_2=\frac{76 \times 45 \times 273}{30 \times 304}$
$p_2=102.4$

Topic: Physics, Thermodynamics, JEE Main
A horizontal uniform glass tube of 100 cm length, sealed at both ends, contains 10 cm mercury column in the middle. The temperature and pressure of air on either side of mercury column are respectively 31 C ∘ and 76 cm of mercury. If the air column at one end is kept at 0 C ∘ and the other end at 273 C ∘, the pressure of air which is at 0 C ∘ is (in cm of Hg). Assure no heat flow through mercury

Options:
A 76
B 88.2
C 102.4
D 12.2

The correct answer is:
102.4

Solution:
On keeping the temperature of the ends of tube at 0 C â and 273 C â .
Applying ideal gas equation
p1V1T1=p2V2T2=p3V3T3
76Ã45(273+31)=p2Ãl(273+0)=p3(90-l)273+273
76Ã45304=p2Ãl273=p3(90-l)546
p2Ãl273=p3(90-l)546
(Mercury column is at rest, so pressure
difference p2-p3=0 âp2=p3)
â´ p2Ãl273=p2(90-l)546
â 2l=90-l âl=30 cm
76Ã45304=p2Ã30273
â p2=76Ã45Ã27330Ã304
p2=102.4

Topic: Physics, Thermal Properties of Matter, JEE Main
A hormone responsible for normal sleep-wake cycle is

Options:
A
epinephrine

B
gastrin

C
melatonin

D
insulin

The correct answer is:
melatonin

Solution:
Melatonin hormone plays a very important role in maintaining the normal rhythms of sleep-wake cycle.

Topic: Biology, Chemical Coordination and Integration
A hormone responsible for normal sleep-wake cycle is

Options:
A epinephrine.
B gastrin.
C melatonin.
D insulin.

The correct answer is:
melatonin.

Solution:
The pineal gland is present in the forebrain. It secretes a hormone known as melatonin. This hormone is responsible for sleep-wake cycle. Other functions are maintaining the temperature of body, pigmentation, metabolism, menstrual cycle as well as controlling the defense mechanism of the body. Melatonin increases in darkness that causes initiation of sleep.

Topic: Biology, Chemical Coordination and Integration, NEET
A horse rider covers half the distance with 5 m s-1 speed. The remaining part of the distance was travelled with speed 10 m s-1 for half the time and with speed 15 m s-1 for other half of the time. The mean speed of the rider averaged over the whole time of motion is x7 m s-1. The value of x is ______.

Options:

The correct answer is:
50

Solution:
Let total distance travelled by horse rider be 2x.

Now, the time taken to cover first half distance is tAB=x5

Second half distance x between BC is covered in two parts, x=d1+d2

here d1 is the distance travelled with 10 m s-1 and d2 is the distance travelled with 15 m s-1 in equal time intervals t2 each.

Then, d1=10t2, d2=15t2

Second half distance x=t2(10+15)=25t2

Now, the mean speed of rider is v=2xx5+2x25=2×255+2=507 m s-1

Thus, the value of x=50.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (30 Jan Shift 1)
A hot body, obeying Newton's law of cooling is cooling down from its peak value $80^{\circ} \mathrm{C}$ to an ambient temperature of $30^{\circ} \mathrm{C}$. It takes 5 minutes in cooling down from $80^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. How much time will it take to cool down from $62^{\circ} \mathrm{C}$ to $32^{\circ} \mathrm{C}$ ? (Given $\operatorname{In} 2=0.693$, In $5=1.609$ )

Options:
A
$3.75$ minutes

B
$8.6$ minutes

C
$9.6$ minutes

D
$6.5$ minutes

The correct answer is:
$8.6$ minutes

Solution:
From Newton's law of cooling,
$$
t=\frac{1}{k} \log _e\left(\frac{\theta_2-\theta_0}{\theta_1-\theta_0}\right)
$$
From question and above equation,
$$
5=\frac{1}{k} \log _e \frac{(40-30)}{(80-30)}
$$
And, $t=\frac{1}{k} \log _e \frac{(32-30)}{(62-30)}$
Dividing equation (2) by (1),
$$
\frac{t}{5}=\frac{\frac{1}{k} \log _e \frac{(32-30)}{(62-30)}}{\frac{1}{k} \log _e \frac{(40-30)}{(80-30)}}
$$
On solving we get, time taken to cool down from $62^{\circ} \mathrm{C}$ to $32^{\circ} \mathrm{C}$, $t=8.6$ minutes.

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2014 (11 Apr Online)