datavorous/entrance-exam-dataset huggingface dataset:
A smooth semicircular wire track of radius R is fixed in a vertical plane. One end of a massless spring of natural length 3R 4 is attached to the lowest point O of the wire track. A small ring of mass m which can slide on the track is attached to the other end of the spring. The ring is held stationary at point P such that the spring makes an angle of 60Â° with the vertical. The spring constant K=mgR. Consider the instant when the ring is released. The normal reaction on the ring by the track is

Options:
A 3mg8
B mg
C mg4
D 3mg4

The correct answer is:
3mg8

Solution:
In ÎOCP, OC = CP = R.

â´ The triangle is isosceles

â´ â COP= â CPO=60Â° âââââ OCP=60Â°

â´ ÎOCP is an equilateral triangle

â OP = R

â´ Extension of string = Râââââ 3R 4 ââ=ââ R 4 ââ=ââx

The forces acting are shown in the figure (i)

The free-body diagram of the ring is shown in fig (ii)

Force in the tangential direction

=Fcos30Â°+mgcos30Â°=[ kx+mg ]cos30Â°

â´ââ F t ââ=ââ 5mg 8 â 3 â´ Ft = mat

âââââ 5mg 3 8 âââ=ââm a t ââââââ a t ââ=ââ 5 3 8 âg

Also when the ring is just released

N + F sin30Âº = mg sin 30Âº

=( mgâââââ mg 4 )âââÃââ 1 2 ââ=ââ 3mg 8

Topic: Physics, Laws of Motion, JEE Main
A smooth sphere is moving on a horizontal surface with a velocity vector 2i^+2j^ m s-1 immediately before it hits a vertical wall. The wall is parallel to the vector j ^ and the coefficient of restitution between the sphere and the wall is e=12. The velocity of the sphere, after it hits the wall, is

Options:
A i ^ - j ^
B - i ^ + 2 j ^
C - i &Hat; - j &Hat;
D 2 i &Hat; - j &Hat;

The correct answer is:
- i ^ + 2 j ^

Solution:
12=0-vx2
vx=-1
v→=-i^+2j^

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A smooth sphere of mass m is moving on a horizontal plane with a velocity 3i^+j^ . It collides with smooth a vertical wall which is parallel to the vector j^ . If coefficient of restitution e=12 then impulse that acts on the sphere is

Options:
A -92mi^
B -32i^+j^
C 32mj^
D 32mj^+12mi^

The correct answer is:
-92mi^

Solution:
u=3i^+j^
component of velocity parallel to the wall i.e. y- component remains unchanged.
Also component of velocity perpendicular to the wall becomes e times,
v⊥→=-eu⊥→
v⊥→=-123i^
Impulse J→=mv→-u→
J→=mv⊥→-u⊥→
J→=m-32i^-3i^
J→=-92mi^

Topic: Physics, Center of Mass Momentum and Collision, NEET
A smooth sphere of mass M moving with velocity u directly collides elastically with another sphere of mass m at rest. After the collision, their final velocities are V and v respectively. Then v is

Options:
A 2uMm
B 2umM
C 2u1+mM
D 2u1+Mm

The correct answer is:
2u1+mM

Solution:
By law of conservation of momentum,
Mu=MV+mv…(i)
e=|v1−v2||u1−u2|=v-Vu⇒e=1=v-Vu⇒u=v-V
⇒Mu=Mv−MV ....(ii)
From (i) and (ii), 2Mu=(M+m)v
⇒v=2uMM+m⇒v=2u1+mM

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A smooth sphere of radius $\mathrm{R}$ is made to translate line with a constant acceleration $\mathrm{a}=\mathrm{g}$. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of angle $\theta$ as it slides down.

Options:
A $\sqrt{R g(1+\sin \theta-\cos \theta)}$
B $\sqrt{2 R g(2+\sin \theta-\cos \theta)}$
C $\sqrt{2 R g(1+\sin \theta-\cos \theta)}$
D $\sqrt{2 R g(1-\sin \theta-\cos \theta)}$

The correct answer is:
$\sqrt{2 R g(1+\sin \theta-\cos \theta)}$

Solution:
A force $f$ acts on the sphere towards left, so the particle experience pseudo force $\mathrm{f}=$ ma towards right.
Initial kinetic energy of the particle is zero.
Let the speed of the particle at point $\mathrm{C}$ be $\mathrm{v}$.
From work-energy theorem, $\mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{f}}=\Delta \mathrm{K} . \mathrm{E}=\mathrm{K} . \mathrm{E}_{\mathrm{f}}$
where $\mathrm{W}_{\mathrm{g}}$ is work done by gravity and $\mathrm{W}_{\mathrm{f}}$ is work done by pseudo force.
From figure, we get $A B=R-R \cos \theta=R(1-\cos \theta)$
$\begin{aligned}& \text { Also } A C=R \sin \theta \\& \therefore \mathrm{mg}(\mathrm{AB})+\mathrm{f}(\mathrm{AC})=\frac{1}{2} \mathrm{mv} v^2 \\& \text { Or } \mathrm{mgR}(1-\cos \theta)+(\mathrm{ma})(\mathrm{R} \sin \theta)=\frac{1}{2} \mathrm{mv}^2 \\& \text { Or } \mathrm{mgR}(1+\sin \theta-\cos \theta)=\frac{1}{2} \mathrm{mv}^2(\because \mathrm{a}=\mathrm{g}) \\& \Rightarrow \mathrm{v}=\sqrt{2 g R(1+\sin \theta-\cos \theta)}\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
A smooth steel ball strikes a fixed smooth steel plate at an angle θ with the vertical. If the coefficient of restitution is e, the angle at which the re-bounce will take place is :

Options:
A θ
B tan-1â¡tanθe
C eÂ tanâ¡Î¸
D tan-1â¡etanÎ¸

The correct answer is:
tan-1â¡tanθe

Solution:
Let ball strikes the plate with speed v. Suppose v ′ is the speed after the collision and α is the angle of reflection. Parallel component of velocity of the ball remain constant as there is no force parallel to surface.
v ′ sinâ¡α=vsinâ¡θ ...(i)
Velocity of sepration = v ′ cosâ¡α
Velocity approach =vcosâ¡θ
∴e= v ′ cosâ¡αvcosâ¡θ
v ′ cosâ¡α=e vcosâ¡θ ...(ii)
from eq (i) and (ii)
tanâ¡α=tanâ¡θe
α=tan-1â¡tanâ¡θe

Topic: Physics, Center of Mass Momentum and Collision, NEET
A smooth tunnel is dug along the radius of the Earth that ends at the centre. A ball is released from the surface of Earth along the tunnel. If the coefficient of restitution for the collision between the ball and the surface of the tunnel is 0.2 then the distance travelled by the ball before the second collision at the centre is nRS. Find the value of n.

Options:

The correct answer is:
7

Solution:
a= -GMR3x
a= -Ï2x
vc=ÏA= GMR3ÃR
After collision velocity of the ball towards A,
vC=evC=0.2GMR=15GMR
Let now amplitude be A, then
Aâ²=vCÏ=15GMRGMR3=R5
Net distance =R+R5+R5=7R5

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A smooth vertical tube having two different cross-sections is open from both the ends but closed by two sliding pistons as shown in the diagram and tied with an inextensible string. One mole of an ideal gas at a pressure P is enclosed between the piston. The difference in cross-sectional areas of the two pistons is given ÎS. The masses of the pistons are m1 and m2 for larger and smaller one, respectively. Find the temperature by which the tube is raised so that the pistons will be displaced by a distance l. Take atmospheric pressure equal to P0 .

Options:
A T2-T1=PlÎSR
B T2-T1=PlΔS
C T2-T1=PΔSlR
D T2-T1=RPl∆S

The correct answer is:
T2-T1=PlΔSR

Solution:
If we consider both the pistons and the string together as a system and balance the forces, we obtain

m1+m2g=P-P0âS

âP=P0+m1+m2ÎSg

It is clear from the above equation that the pressure of the gas has to remains constant for the system to be in equilibrium. Hence, on increasing the temperature the volume of the gas increases isobarically.

If the gas temperature is increased from T1 to T2, the volume of gas increase from V to V+lÎS as l is the displacement of pistons, then from gas law we must have

PlÎS=RT2-T1

or T2-T1=PlÎSR

Topic: Physics, Thermodynamics, NEET
A sniper fires a rifle bullet into a gasoline tank making a hole 53.0âm below the surface of gasoline. The tank was sealed at 3.10âatm. The stored gasoline has a density of 660âkg m-3. The velocity with which gasoline begins to shoot out of the hole is

Options:
A 27.8âm s-1
B 41.0âm s-1
C 9.6âm s-1
D 19.7âm s-1

The correct answer is:
41.0âm s-1

Solution:
According to Bernoulli's theorem,
PB+hρg=PA+12ρvA2 As vA>>vB
3.10P+53×660×10=P+12×660vA2
⇒2.1×1.01×105+3.498×105=12×660×vA2
⇒5.619×105=12×660×vA2
∴ vA=2×5.619×105660=41 m s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A sniper fires a rifle bullet into a gasoline tank making a hole $53.0 \mathrm{~m}$ below the surface of gasoline. The tank was sealed at $3.10 \mathrm{dtm}$. The stored gasoline has a density of $660 \mathrm{kgm}^{-3}$. The velocity with which gasoline begins to shoot out of the hole is

Options:
A $27.8 \mathrm{~ms}^{-1}$
B $41.0 \mathrm{~ms}^{-1}$
C $9.6 \mathrm{~ms}^{-1}$
D $19.7 \mathrm{~ms}^{-1}$

The correct answer is:
$41.0 \mathrm{~ms}^{-1}$

Solution:
According to Bernoulli's theorem,
$P_B+h \rho g=P_A+\frac{1}{2} \rho v_A^2$
$\left(\mathrm{A} s v_A \gg v_B\right)$
$3.10 P+53 \times 660 \times 10=P+\frac{1}{2} \times 660 v_A^2$
$\Rightarrow 2.1 \times 1.01 \times 10^5+3.498 \times 10^5=\frac{1}{2} \times 660 \times v_A^2$
$\Rightarrow \quad 5.619 \times 10^5=\frac{1}{2} \times 660 \times v_A^2$
$\therefore v_A=\sqrt{\frac{2 \times 5.619 \times 10^5}{660}}=41 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A soap bubble A of radius 0.03 m and another bubble B of radius 0.04 m coalesce to form a combined bubble such that the radius of curvature of their common interface is r. Then, the value of r is

Options:
A 0.24 m
B 0.48 m
C 0.12 m
D none of these

The correct answer is:
0.12 m

Solution:
Let the radius of curvature of the common internal film surface of the double bubble formed by two bubbles A and B be r.
Excess of pressure as compared to the atmosphere inside A is
p1=4Tr1=4T0.03
Excess of pressure inside B is
p2=4Tr2=4T0.04
In the double bubble, the pressure difference between A and B on either side of the common surface is
4T0.03-4T0.04=4Tr
⟹ 10.03-10.04=1r
⟹ r=0.03×0.040.01=0.12 m

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A soap bubble has radius R and thickness d (<<R) as shown in the diagram. It collapses into a spherical drop. The ratio of excess pressure in the drop to the excess pressure inside the bubble is

Options:
A R3d13
B R6d13
C R24d13
D None of these

The correct answer is:
R24d13

Solution:
The volume of soap solution in a bubble =4πR2d
now we make a drop
4πR2d=43πr3 ⇒ r=3R2d1/3
Excess pressure initially =4SR
Excess pressure finally =2Sr=2S3R2d1/3
Ratio =R24d13

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A soap bubble having radius of 1 mm is blown from a detergent solution having a surface tension of 2.5×10-2 N m-1. The pressure inside the bubble equals at a point Z0 below the free surface of water in a container. Taking g=10 m s-2, density of water =103 kg m-3, the value of Z0 is

Options:
A 100 cm
B 10 cm
C 1 cm
D 0.5 cm

The correct answer is:
1 cm

Solution:
Pressure inside soap bubble is given by, ∆P=4TR
Pressure below water surface at depth Z0 is given by, ∆P=ρgZ0
According to question,
⇒4TR=ρgZ0
⇒Z0=4Tρg=4×2.5×10-2103×10×10-3
⇒Z0=1100 m
⇒Z0=1 cm

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A soap bubble is blown to a diameter of $7 \mathrm{~cm} .36960 \mathrm{erg}$ of work is done in blowing it further. If surface tension of soap solution is $40 \mathrm{dyne} / \mathrm{cm}$ then the new radius is______ $\mathrm{cm}$ Take $\left(\pi=\frac{22}{7}\right)$

Options:

The correct answer is:
7

Solution:
$\begin{aligned} & \omega=\Delta \mathrm{U}=\mathrm{S} \Delta \mathrm{A} \\ & 36960 \mathrm{erg}=\frac{40 \text { dyne }}{\mathrm{cm}} 8 \pi\left[(\mathrm{r})^2-\left(\frac{7}{2}\right)^2\right] \mathrm{cm}^2 \\ & \mathrm{r}=7 \mathrm{~cm}\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (04 Apr Shift 1)
A soap bubble of the radius 3 cm is formed inside another soap bubble of radius, 6 cm. The radius of an equivalent soap bubble that has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is____cm.

Options:

The correct answer is:
2

Solution:
The formula of excess pressure inside the soap bubble is, 4TR here T is the surface tension and R is the radius of the soap bubble,

⇒P0+4T6+4T3=P0+4TR

⇒16+13=1R⇒R=2 cm.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2021 (26 Aug Shift 1)
A soft ferromagnetic material is placed in an external magnetic field. The magnetic domains:

Options:
A have no relation with external magnetic field.
B decrease in size and changes orientation.
C increase in size but no change in orientation.
D may increase or decrease in size and change its orientation.

The correct answer is:
may increase or decrease in size and change its orientation.

Solution:
Soft ferromagnetic materials are materials which can be easily magnetised and demagnetised by external magnetic field. When external field is applied, the domains experiences a net torque hence change its orientation.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2021 (24 Feb Shift 2)
A solenoid has 2000 turns wound over a length of 0.30 m . The area of its cross section is 1.2Ã10-3 m2. Around its central section, a coil of 300 turns is wound. If an initial current of 2 A in the solenoid is reversed in 0.25 s , then the emf induced in the coil is equal to

Options:
A 8.6×10-2 V
B 4.8Ã10-2 V
C 1.2×10-2 V
D 6.0×10-2 V

The correct answer is:
4.8Ã10-2 V

Solution:
Emf= Mdi dt = μ 0 N 1 N 2 A l di dt
= 4ÏÃ10-7Ã2000Ã300Ã1.2 Ã10-340.3Ã0.25
=4.8 Ã10-2 V

Topic: Physics, Electromagnetic Induction, JEE Main
A solenoid of \( 1.5 \mathrm{~m} \) length and \( 4.0 \mathrm{~cm} \) diameter has \( 10 \) turns per \( \mathrm{cm} \). A current of \( 5 \) A is flowing through it. The magnetic field induction at axis inside the solenoid is

Options:
A \( 2 \pi \times 10^{-7} \) Gauss
B \( 2 \pi \times 10^{-5} \) Gauss
C \( 2 \pi \times 10^{-5} \mathrm{~T} \)
D \( 2 \pi \times 10^{-7} \mathrm{~T} \)

The correct answer is:
\( 2 \pi \times 10^{-5} \mathrm{~T} \)

Solution:
Magnetic field inside the solenoid and is given by,
B=μ0ni=4π×10-7100.15=2π×10-5 T

Topic: Physics, Magnetic Effects of Current, JEE Main
A solenoid of 1000 turns per metre has a core with relative permeability 500. Insulated windings of the solenoid carry an electric current of 5 A. The magnetic flux density produced by the solenoid is:

(Permeability of free space=4π×10-7 H m-1)

Options:
A π T
B 2×10-3π T
C π5 T
D 10-4π T

The correct answer is:
π T

Solution:
B=μnI=μ0μrnI

B=4π×10-7×500×1000×5

B=π Tesla

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (17 Mar Shift 1)
A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is:

Options:
A 2.4×103 A m-1
B 1.2×103 A m-1
C 1 A m-1
D 2.4×10-3 A m-1

The correct answer is:
1.2×103 A m-1

Solution:
Given,

Number of turns per metre n=12002

l= 2 m, d=0.2 m and I=2 A

Magnetic field at centre inside the solenoid is given by B=μ0nI, where, μ0 is the permeability constant.

Magnetic intensity is related to magnetic field by

H=Bμ0=μ0nIμ0=nI=12002(2)

⇒H=1.2×103 A m-1

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (25 Jan Shift 1)
A solenoid of inductance L and resistance R has a soft iron core. It is connected to a battery of E.M.F E as shown in the figure and allowed to attain its steady state.

At an instant, the iron core is suddenly pulled out and because of it, inductance decreases to L3.
The work done by an external agent in pulling out the iron core is

Options:
A LE2R2
B 2LE2R2
C LE22R2
D LE24R2

The correct answer is:
LE2R2

Solution:
Flux does not change suddenly.

∴ L1I1=L2I2
∴  W=U2−U1=12×L3×3ER2−12×L×ER2
=3E2L2R2−E2L2R2=E2LR2

Topic: Physics, Electromagnetic Induction, JEE Main
A solid AB has NaCl type structure with edge length 580.4 pm. The radius of A+ is 100 pm. What is the radius of B- in pm?

Options:
A 190.2
B 540.13
C 525
D 78.12

The correct answer is:
190.2

Solution:
NaCl has fcc structure and thus,
rc+ra=a2
100+ra=580.42=290.2
100+ra=290.2
So ra=290.2-100=190.2 pm

Topic: Chemistry, Solid State, JEE Main
A solid ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at the speed $60 \mathrm{~cm}^{-1}$ on the string, when the car is at rest. When the car accelerates on a horizontal road, then speed of the pulse is $66 \mathrm{~cm}^{-1}$. The acceleration of the car is nearly $\left(g=10 \mathrm{~ms}^{-2}\right)$

Options:
A $4.3 \mathrm{~ms}^{-2}$
B $2.9 \mathrm{~ms}^{-2}$
C $6.8 \mathrm{~ms}^{-2}$
D $5.5 \mathrm{~ms}^{-2}$

The correct answer is:
$6.8 \mathrm{~ms}^{-2}$

Solution:
When car is at rest, tension in string is $T=m g$.
$$
\begin{aligned}
T & =M g=m v_1^2 \\
V_1=\sqrt{\frac{T}{r}} & =\sqrt{\frac{M g}{\mu}} &..(i)
\end{aligned}
$$
When car is accelerating tension,
$$
\begin{aligned}
T & =\sqrt{M\left(a^2+g^2\right)^{1 / 2}} \\
\therefore \quad v_2 & =\sqrt{\frac{M\left(a^2+g^2\right)^{1 / 2}}{\mu}}
\end{aligned}
$$
Dividing Eq (i) by Eq. (ii), we get
$$
\begin{gathered}
\frac{v_2}{v_1}=\frac{\sqrt{M\left(a^2+g^2\right)^{1 / 2}}}{\sqrt{M g}}=\frac{66}{60} \\
\frac{\left(a^2+g^2\right)^{1 / 2}}{g}=\frac{121}{100}
\end{gathered}
$$
Squaring, we get
$$
\begin{array}{rlrl}
& \Rightarrow & a^2+g^2 & =\left(\frac{121}{100}\right)^2 \\
\Rightarrow & a^2 & =146.41-100 \\
\Rightarrow & a^2 & =46.41
\end{array}
$$
So, $a=6.8 \mathrm{~ms}^{-2}$

Topic: Physics, Waves and Sound, JEE Main
A solid ball of radius R has a charge density ρ given by ρ=ρo1-rR for 0≤r≤R. The electric ﬁeld outside the ball is

Options:
A ρ0R3ε0r2
B 4ρ0R33ε0r2
C 3ρ0R34ε0r2
D Ï0R312Îµ0r2

The correct answer is:
ρ0R312ε0r2

Solution:
ρ=ρ01-rR
dq=ρdv
qin=∫dq=ρdv

⇒qin=∫0Rρ01-rR4πr2dr

=4πρ0∫0R1-rRr2dr

=4πρ0∫0Rr2dr-r3Rdr

=4πρ0r330R-r44R0R

=4πρ0 R33-R44R

=4πρ0R312

⇒qin=πρ0R33
E×4πr2=πρ0R33ε0
⇒E=ρ0R312ε0r2

Topic: Physics, Electrostatics, JEE Main
A solid ball rolls down a parabolic path ABC from a height h as shown in the figure. The portion AB of the path is rough while BC is smooth. How high will the ball climb in BC?

Options:
A H⁡ = 5 7 h⁡
B H&ApplyFunction; = 5 2 h&ApplyFunction;
C H&ApplyFunction; = 7 5 h&ApplyFunction;
D H&ApplyFunction; = 3 7 h&ApplyFunction;

The correct answer is:
H⁡ = 5 7 h⁡

Solution:
Using the conservation of mechanical energy between A and B
At B, total kinetic energy = mgh
Here, m = mass of the ball
The ratio of rotational to translational kinetic energy would be,
K⁡ R⁡ K⁡ T⁡ = 2 5
∴ K⁡R⁡=27mgh⁡andK⁡T⁡=57mgh
In portion BC, friction is absent. Therefore, rotational kinetic energy will remain constant and translational kinetic energy will convert into potential energy.

Hence, if H be the height to which ball climbs in BC, then
mgH = KT
or mgH=57mgh⁡orH⁡=57h⁡

Topic: Physics, Rotational Motion, NEET
A solid body of constant heat capacity 1 J ℃-1 is being heated by keeping it in contact with reservoirs in two ways: (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies the same amount of heat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies the same amount of heat. In both, cases the body is brought from initial temperature 100 K to final temperature 200 K . Entropy change of the body in the two cases respectively is: Note: This question was awarded as a bonus since temperatures were given in centigrade instead of in Kelvin. Proper corrections are made in the question to avoid it.

Options:
A 2ln2, 8ln2
B ln2, 4ln2
C ln2, ln2
D ln2, 2ln2

The correct answer is:
ln2, ln2

Solution:
Change in entropy,

dS=dQTΔQ= heat supplied=CΔT      {C=J °C-1}dQ=CdtdS=CdTTIntergrating both sides∫SiSfdS=C∫dTT  ⇒Sf−Si=ΔS=C.lnT|100200⇒ΔS=C[ln200−ln100]=Cln2and  C=J °C-1⇒  ΔS=ln2Entrpoy change in same for both cases as C is constant, andtemperature change (i.e. from 100 to 200) is same.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2015 (04 Apr)
A solid body of constant heat capacity $1 \mathrm{~J} /{ }^{\circ} \mathrm{C}$ is being heated by keeping it in contact with reservoirs in two ways:
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat.
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat.
In both the cases body is brought from initial temperature $100^{\circ} \mathrm{C}$ to final temperature $200^{\circ} \mathrm{C}$. Entropy change of the body in the two cases respectively is:

Options:
A $\ln 2,2 \ln 2$
B $2 \ln 2,8 \ln 2$
C $\ln 2,4 \ln 2$
D $\ln 2, \ln 2$

The correct answer is:
$\ln 2, \ln 2$

Solution:
The entropy change of the body in the two cases is same as entropy is a state function.

Topic: Physics, Thermodynamics, JEE Main
A solid body rotates about a stationary axis so that its angular velocity depends on the rotational angle Ï as Ï=Ï0-kÏ where Ï0 and k are positive constants. At the moment t=0, Ï=0, find the time dependence of rotation angle

Options:
A kÏ0Â e-kt
B Ï0kÂ e-kt
C Ï0k 1-e-kt
D kÏ0Â e-kt-1

The correct answer is:
Ï0k 1-e-kt

Solution:
∵ ω=ω0-kϕ or dϕdt=ω0-kϕ
Or ∫0ϕdϕω0-kϕ= ∫0tdt
Or -1k lnω0-kϕ0ϕ=t
Or lnω0-kϕω0= -kt
∴ ϕ=ω0k 1-e-kt

Topic: Physics, Laws of Motion, JEE Main
A solid body rotates an angle Î¸ about a stationary axis according to the law Î¸=6t-2t3 . What is the mean value of angular velocity (in rad s-1) over the time interval between t=0 and the time till the body comes to rest?

Options:

The correct answer is:
4

Solution:
Given, Î¸=6t-2t3 ...(i)
â´dÎ¸dt=Ï=6-6t2
As, Ï=0 (given)
6-6t2=0
t2=Â±1
As, t=-1s is not possible, hence t=1s.
Substituting t=1s in Equation (i), we get
Î¸1=6-2=4 rad
and Î¸0=6Ã0-2Ã02t=0=0
â´ Mean value of Ï=Î¸1-Î¸0t1-t0=4-01-0=4 rad s-1

Topic: Physics, Laws of Motion, JEE Main
A solid circular disc of mass 50 kg rolls along a horizontal floor so that its center of mass has a speed of 0.4 m s-1. The absolute value of work done on the disc to stop it is ______ J.

Options:

The correct answer is:
6

Solution:
Using work-energy theorem, we get

W=∆KE

⇒W=0-12mv2+12Iω2

⇒W=0-12mv21+K2R2

=-12×50×0.421+12=-6 J

Absolute work =+6 J

Therefore, required value of W=6 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2024 (31 Jan Shift 1)
A solid compound X on heating gives CO2 gas and a residue. The residue mixed with water forms Y. On passing an excess of CO2 through Y in water, a clear solution Z is obtained. On boiling Z, compound X is reformed. The compound X is

Options:
A CaHCO32
B CaCO3
C Na2CO3
D K2CO3

The correct answer is:
CaCO3

Solution:
The given compound X must be CaCO3 . It can be explained by following reaction
CaCO3→ΔCaO+CO2↑
CaO+H2O→CaOH2Y
CaOH2+2CO2+H2O→CaHCO32Z
CaHCO32→ΔCaCO3X+CO2↑+H2O

Topic: Chemistry, s Block Elements, JEE Main
A solid compound $X Y$ has $\mathrm{NaCl}$ structure. If the radius of the cation is $100 \mathrm{pm}$, the radius of the anion ( $Y^{-}$) will be

Options:
A $275.1 \mathrm{pm}$
B $322.5 \mathrm{pm}$
C $241.5 \mathrm{pm}$
D $165.7 \mathrm{pm}$

The correct answer is:
$241.5 \mathrm{pm}$

Solution:
Radius ratio of $\mathrm{NaCl}$ like crystal $=\frac{r^{+}}{r^{-}}=0.414$ or $r^{-}=\frac{100}{0.414}=241.5 \mathrm{pm}$

Topic: Chemistry, Solid State, NEET, NEET 2011 (Mains)
A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –4Q, the new potential difference between the same two surfaces is:

Options:
A –2V
B 2 V
C V
D 4 V

The correct answer is:
V

Solution:
First condition:
Both conducting spheres are shown.
Vin-Vout=KQr1-KQr2
⇒V=KQ1r1-1r2
Second condition:
Shell is now given charge -4Q .
Vin-Vout=KQr1-4KQr2-KQr2-4KQr2
=KQr1-KQr2
=KQ1r1-1r2=V

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (08 Apr Shift 1)
A solid cone of height H, radius R suspended by a string is just submerged in a liquid of density ρ such that tension in string is zero as shown in the figure. The magnitude of the net force applied by the liquid on the curved surface of the cone is equal to (atmospheric pressure is P0)

Options:
A πR232P0+ρgH
B πR332P0+ρgH
C πR233P0+2ρgH
D πR333P0+2ρgH

The correct answer is:
πR233P0+2ρgH

Solution:
There will be only vertically downward force on the curved surface of the cone applied by liquid as the horizontal force will be cancelled.
Let this force is equal to F
Buoyancy force = P0+ÏgHÏR2-F
â 13ÏR2HÏg=P0+ÏgHÏR2-F
â F=ÏR233P0+2ÏgH

Topic: Physics, Mechanical Properties of Fluids, JEE Main