datavorous/entrance-exam-dataset huggingface dataset:
A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference $P$. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is

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

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

Solution:
From $V=\frac{P \pi r^4}{8 \eta l} \Rightarrow P=\frac{V 8 \eta l}{\pi^4}$
$\Rightarrow \frac{P_2}{P_1}=\frac{V_2}{V_1} \times \frac{l_2}{l_1} \times\left(\frac{r_1}{r_2}\right)^4=2 \times 2 \times\left(\frac{1}{2}\right)^4=\frac{1}{4}$
$\Rightarrow P_2=\frac{P_1}{4}=\frac{P}{4}$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A liquid is immiscible in water was steam distilled at 95.2oC at a pressure of 0.983 atm. What is the mass of the liquid present per gram of water in the distillate. Molar mass of the liquid is 134.3 g/mol and the vapour pressure of water is 0.84 atm. Also, Vapour pressure of pure liquid is 0.143 atm.

Options:
A 1 g
B 1.27 g
C 0.787 g
D 13.43 g

The correct answer is:
1.27 g

Solution:
p T = 0.983 atm
p water â = 0.84 atm
â´ p liquid â = 0.143 atm
Apply Dalton's law of vapour phase
p o water p o liquid = 0.84 0.143 = n H 2 O n Liquid
0.84 0.143 = 1 18 w e 134.3 = 134.3 18 Ã w e
w e = 134.3 18 Ã 0.143 0.84 = 1.27 g

Topic: Chemistry, Solutions, JEE Main
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(a) What is the initial effect of the change on vapour pressure?
(b) How do rates of evaporation and condensation change initially?
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

Options:

None

Solution:
(a) Initially, the vapour pressure will decrease.
(b) The rate of evaporation remains constant at constant temperature in a closed vessel. However, the rate of condensation will be low initially because there are fewer molecules per unit volume in the vapour phase and hence the number of collisions per unit time with the liquid surface decreases.
(c) When equilibrium is restored, rate of evaporation = rate of condensation. The final vapour pressure will be same as it was initially.

Topic: Chemistry, Equilibrium
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. What is the initial effect on vapour pressure?

Options:
A The vapour pressure increases.
B The vapour pressure decreases.
C The vapour pressure remains the same.
D The effect on vapour pressure cannot be determined.

The correct answer is:
The vapour pressure decreases.

Solution:
Correct Option is : (B)
The vapour pressure decreases.

Topic: Chemistry, Ionic Equilibrium, NEET
A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at its sides. If the radius of the vessel is 0.05 m and the speed of rotation is 2 rev s-1, the difference in the height of liquid at the centre of the vessel and its sides is

Options:
A 0.001 m
B 0.002 m
C 0.01 m
D 0.02 m

The correct answer is:
0.02 m

Solution:
h=r2Ï22 g
h=0.052 2Ã2Ï22Ãg
=0.02 m

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A liquid mixture boils without changing constituent is called

Options:
A Stable structure complex
B Binary liquid mixture
C Zeotropic liquid mixture
D Azeotropic liquid mixture

The correct answer is:
Azeotropic liquid mixture

Solution:
Azeotropic mixture is a mixture of two liquids which boils at on particular temperature like a pure liquid and distils over in the same composition.

Topic: Chemistry, Solutions, JEE Main
A liquid of bulk modulus $k$ is compressed by applying an external pressure such that its density increases by $0.01 \% .$ The pressure applied on the liquid is

Options:
A $\frac{k}{10000}$
B $\frac{k}{1000}$
C $1000 \mathrm{k}$
D $0.01 k$

The correct answer is:
$\frac{k}{10000}$

Solution:
The Bulk modulus $k$
$$
k=-\frac{p}{\Delta V / V}
$$
Where, $p=$ Pressure
$$
\begin{aligned}
\Delta V &=\text { Change in volume } \\
V &=\text { Volume of liquid }
\end{aligned}
$$
From (i)
$$
\frac{p}{k}=-\frac{\Delta V}{V}=\frac{\Delta p}{p}
$$
$\Rightarrow \quad p=\frac{k \Delta p}{p}$
$$
\begin{aligned}
\Delta p &=0 \cdot 01 \%=0 \cdot 01 / 100 \\
p &=\frac{k}{10000}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A liquid of density 0.85 g cm-3 flows through a calorimeter at the rate of 8.0 cm3 s-1. Heat is added by means of a 250 W electric heating coil and a temperature difference of 15Â° C is established in steady-state conditions between the inflow and the outflow points of the liquid. The specific heat for the liquid will be (1 kcal =4186 J )

Options:
A 0.6 kcal kg-1K-1
B 0.3 kcal kg-1K-1
C 0.5 kcal kg-1K-1
D 0.4 kcal kg-1K-1

The correct answer is:
0.6 kcal kg-1K-1

Solution:
This is a problem on 'flow calorimeter' used to measure the specific heat of the liquid.
Amount of heat supplied to the water per second by the heating coil = Qs = 250 J = 2 5 0 4 1 8 6 kcal
The volume of liquid flowing out per second = 8.0 cm3 = 8 × 10-6 m3

Mass of this liquid = (0.85) × 1000 × 8 × 10-6 kg
The temperature rise of this mass of liquid = 15° C

Hence, = 2 5 0 4 1 8 6 = mst = 0.85 × 8 × 1 0 - 3 × s × 1 5
Hence, s=250×1034186×0.85×8×15=0.6 kcal kg-1K-1

Topic: Physics, Thermodynamics, JEE Main
A liquid of density 750 kg m-3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from A1=1.2×10-2 m2 to A2=A12. The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is _____ ×10-3 m3 s-1.

Options:

The correct answer is:
24

Solution:
From the equation of continuity, v2=2v1.

Applying the Bernoulli's Theorem,

P1+ρv122=P2+ρ4v122 ⇒P1-P2=3ρv122

⇒4500=3×750×v122 ⇒v1=2 m s-1

Now volume flow rate,Q=Av1=1.2×10-2×2=24×10-3 m3 s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2022 (28 Jun Shift 2)
A liquid of mass 250 g is kept warm in a vessel using an electric heater. The liquid is maintained at 57 °C when the power supplied by the heater is 30 W and surrounding temperature is 27 °C. As the heater is switched off, it took 10 s time for the temperature of the liquid to fall from 47 °C to 46.9 °C. The specific heat capacity of the liquid is

Options:
A 8000 J kg-1 K-1
B 9000 J kg-1 K-1
C 6000 J kg-1 K-1
D 12000 J kg-1 K-1

The correct answer is:
8000 J kg-1 K-1

Solution:
The expression for the rate of heat flow is,

dqdt=-kT-T0

Here T0=Surrounding temperature

From first condition-|

30=-k57-27

⇒ k=-1

Using second condition-

dqdt=-kT-T0

but dqdt=msdTdt

∴ msdTdt=-k T-T0

Substitute the given values in the above equation.

2501000s47-46.910=--1 47-27

⇒ s=8000 J kg-1 K-1

Topic: Physics, Kinetic Theory of Gases, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 2)
A liquid was mixed with ethanol and a drop of concentrated $\mathrm{H}_2 \mathrm{SO}_4$ was added. A compound with a fruity smell was formed. The liquid was:

Options:
A
$\mathrm{CH}_3 \mathrm{OH}$

B
$\mathrm{HCHO}$

C
$\mathrm{CH}_3 \mathrm{COCH}_3$

D
$\mathrm{CH}_3 \mathrm{COOH}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{COOH}$

Solution:
Esterification reaction is involved

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
A liquid was mixed with ethanol and a drop of concentrated $\mathrm{H}_2 \mathrm{SO}_4$ was added. A compound with a fruity smell was formed. The liquid was:

Options:
A
$\mathrm{CH}_3 \mathrm{OH}$

B
$\mathrm{HCHO}$

C
$\mathrm{CH}_3 \mathrm{COCH}_3$

D
$\mathrm{CH}_3 \mathrm{COOH}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{COOH}$

Solution:
Esterification reaction is involved

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2009
A list of species having the formula XZ4 is given below. XeF4,SF4,SiF4,BF4-,BrF4-,CuNH342+,FeCl42-,CoCl42- and PtCl42-
Defining shape on the basis of the location of X and Z atoms, the total number of species having an square planar shape is

Options:

The correct answer is:
4

Solution:
Square planar (sp3d2)
See-saw (sp3d)
Tetrahedral (sp3)
Tetrahedral (sp3)
Square planar (sp3d2)
Square planar dsp2
Tetrahedral sp3
Tetrahedral sp3
Square planar dsp2

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
A litre of dry air at STP allowed to expand to a volume of 3 L under adiabatic conditions. If Î³=1.40, the work done is (31.4=4.6555)

Options:
A 48 J

B 60.7 J

C 90.5 J

D 100.8 J

The correct answer is:
90.5 J

Solution:
Here, V1=1 L=10-3 m3, V2=3L=3Ã10-3m3
p1=1 atm =1.013Ã105Nm-2, Î³=1.40, W=?
As changes are adiabatic,
p1V1Î³=p2V2Î³
p1p2=V2V1Î³=31.4=4.6555
â´ p2=p14.6555=1.013Ã1054.6555
=0.217Ã105 Nm-2
Work done, W=p1V1-p2V2Î³-1
W=1.013Ã105Ã10-3-0.217Ã105Ã3Ã10-31.4-1
W=90.5 J

Topic: Physics, Thermodynamics, JEE Main
A litre of dry air at STP expands adiabatically to a volume of 3 litres. If γ=1.40, the work done by air is: 31.4=4.6555 [Take air to be an ideal gas]

Options:
A 60.7J
B 90.5J
C 100.8J
D 48J

The correct answer is:
90.5J

Solution:
P1=1atm,T1=273K
P1V1Î³=P2V2Î³
P2=P1V1V2Î³
=1atm131.4
Now work done =P1V1-P2V2Î³-1=90.5 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (07 Jan Shift 1)
A little charged bead is inside the hollow frictionless sphere manufactured from the insulating material. Sphere has a diameter of 50 cm. The mass of the bead is 90 mg, its charge is 0.5ââÎ¼C. What minimum charge must carry an object at the bottom of the sphere to keep hold the charged bead at the vertex of the sphere in stable equilibrium?

Options:
A 4.9×10-8  C
B 9.8×10-8  C
C 19.6×10-8  C
D 30.2×10-8  C

The correct answer is:
9.8×10-8  C

Solution:
Let give small displacement to q . The equilibrium will be stable if net force after giving displacement will be in direction of position of equilibrium

Fesin⁡α≥ mg sin⁡2α
then net force will be directed towards vertex i.e. initial equilibrium position
kqQd2  sin⁡α≥mg sin⁡2α
d=diameter of sphere and  α is very small
kqQd2×α ≥ mg × 2α
Q ≥ 2 mgd2Kq=9.8× 10-8 C

Topic: Physics, Electrostatics, JEE Main
A lizard, at an initial distance of $21 \mathrm{~cm}$ behind an insect, moves from rest with an acceleration of $2 \mathrm{~cm} / \mathrm{s}^2$ and pursues the insect which is crawling uniformly along a straight line at a speed of $20 \mathrm{~cm} / \mathrm{s}$. Then the lizard will catch the insect after

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

B
$1 \mathrm{~s}$

C
$21 \mathrm{~s}$

D
$24 \mathrm{~s}$

The correct answer is:
$21 \mathrm{~s}$

Solution:
$$
\begin{aligned}
& \frac{1}{2} 2 t^2=21+20 t \\
& \Rightarrow t=21 .
\end{aligned}
$$

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2005
A load of 31.4 kg is suspended from a wire of radius 10-3 m and density 9 Ã 103 kg m-3. Calculate the change in temperature of the wire if 75% of the stored elastic potential energy is converted into heat. The Young's modulus and heat capacity of the material of the wire is 9.8 Ã 1010 N m-2 and 490 J kg-1 K-1 respectively.

Options:
A 4.33 × 10-2 K

B 8.33 × 10-3 K

C 2.44 × 10-5 K

D 6.22 × 10-2 K

The correct answer is:
8.33 × 10-3 K

Solution:
As work done in stretching an elastic body per unit volume is given by
W V = 1 2 stress Ã strain = 1 2 stress 2 Y as Y = stress strain ........ (1)
So W = 1 2 Mg A 2 V Y as stress = F A = Mg A
Now according to given problem heat produced H = (75/100) W = 3 4 .W
â ms Î Î¸ = 3 4 Ã 1 2 Mg Ï r 2 2 V Y as H = ms Î Î¸ using equation (1)
or Ï V s Î Î¸ = 3 4 Ã 1 2 Mg Ï r 2 2 V Y as m = Ï V
â´ Î Î¸ = 3 8 3 1 . 4 Ã 9 . 8 Ï Ã 1 0 - 6 2 1 9 . 8 Ã 1 0 1 0 Ã 1 9 Ã 1 0 3 Ã 4 9 0
i.e., Î Î¸ = 1 1 2 0 K = 8 . 3 3 Ã 1 0 - 3 K

Topic: Physics, Mechanical Properties of Solids, JEE Main
A load of mass m falls from a height h on to the scale pan hung from a spring, as shown in the figure. If the spring constant is k, mass of the scale pan is zero and the mass m does not bounce, relative to the pan, then the amplitude of vibration is

Options:
A mgk
B mgk 1+2hkmg
C mgk+mgkÂ 1+2hkmg
D mgkÂ 1+2hkmg-mgk

The correct answer is:
mgk 1+2hkmg

Solution:
According to conservation principle,
mgh=12 kx02-mgx0
where x0 is maximum elongation in spring (when particle is in its lowest extreme position)
or 12kx02-mgx0-mgh=0
or x02-2mgk x0-2mgk h=0
â´ x0=2mgk Â± 2mgk2+ 4 Ã 2mgkh2
âµ Amplitude = elongation in spring for lowest extreme position - elongation in spring for equilibrium position =x0-x1
=mgk 1+2hkmg

Topic: Physics, Oscillations, JEE Main
A load of mass M kg is suspended from a steel wire of length 2m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is:

Options:
A 4.0 mm
B Zero
C 5.0 mm
D 3.0 mm

The correct answer is:
3.0 mm

Solution:
Î´=Ï0vgÃLAy
Î´â²=Ï0-ÏLvgÃLAyâÎ´^â²Î´=Ï0-ÏLÏ0=8-28
Î´â²=3 mm

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2019 (12 Jan Shift 2)
A loaded spring gun of mass M fires a shot of mass m with a velocity Ï at an angle of elevation Î¸. The gun is initially at rest on a horizontal frictionless surface. After firing, the centre of mass of the gun-shot system

Options:
A moves with a velocity Ï m M
B moves with a velocity Ï m M cos Î¸ in the horizontal direction
C remains at rest
D moves with a velocity Ï M - m M + m in the horizontal direction

The correct answer is:
remains at rest

Solution:
If we consider a system consists of a gun plus mass m then no external force is present on the system. So, the acceleration of the centre of mass of the system will be zero. Since initially the system is in rest. So, the centre of mass will always remain at rest.

Topic: Physics, Center of Mass Momentum and Collision, NEET
A log of wood of mass $120 \mathrm{Kg}$ floats in water. The weight that can be put on the raft to make it just sink, should be (density of wood $=600 \mathrm{Kg} / \mathrm{m}^5$ )

Options:
A $80 \mathrm{Kg}$
B $50 \mathrm{Kg}$
C $60 \mathrm{Kg}$
D $30 \mathrm{Kg}$

The correct answer is:
$80 \mathrm{Kg}$

Solution:
Volume of log of wood $V=\frac{\text { mass }}{\text { density }}=\frac{120}{600}=0.2 \mathrm{~m}^3$
Let $x$ weight that can be put on the log of wood.
So weight of the body $=(120+x) \times 10 \mathrm{~N}$
Weight of displaced liquid $=V \sigma g=0.2 \times 10^3 \times 10 \mathrm{~N}$
The body will just sink in liquid if the weight of the body will be equal to the weight of displaced liquid.
$\therefore(120+x) \times 10=0.2 \times 10^3 \times 10$
$\Rightarrow 120+x=200 \quad \therefore x=80 \mathrm{~kg}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A logistic growth curve depicting a population that is limited by a definite carrying capacity is shaped like the letter

Options:
A J
B L
C M
D S

The correct answer is:
S

Solution:
Logistic growth curve show sigmoid shape or S-shape curve. This growth curve is shown by the population which grows in a habitat with limited resources. The population growth shows a lag phase initially, followed by phases of increases and decrease and finally the population density reaches the carrying capacity where the population struggle to keep itself maintained, and ultimately, the population seems to be in balance or equilibrium with the limited resources.

Topic: Biology, Organisms and Populations, NEET
A long conducting wire carrying a current $I$ is bent at $120^{\circ}$ (see figure). The magnetic field $B$ at a point $P$ on the right bisector of bending angle at a distance $d$ from the bend is $\left(\mu_{0}\right.$ is the permeability of free space)

Options:
A $\frac{3 \mu_{0} I}{2 \pi d}$
B $\frac{\mu_{0} I}{2 \pi d}$
C $\frac{\mu_{0} I}{\sqrt{3} n d}$
D $\frac{\sqrt{3} \mu_{0} I}{2 \pi d}$

The correct answer is:
$\frac{\sqrt{3} \mu_{0} I}{2 \pi d}$

Solution:
We know that

$$
\begin{aligned}
B_{\text {net }} &=2\left[\frac{\mu_{0} i}{4 \pi r}\left(\sin \theta_{1}+\sin \theta_{2}\right)\right] \\
&=2\left[\frac{\mu_{0}}{4 \pi} \times \frac{i}{\frac{d \sqrt{3}}{2}} \times\left(\sin 90^{\circ}+\sin 30^{\circ}\right)\right] \\
&=2\left[\frac{\mu_{0}}{4 \pi} \times \frac{2 i}{d \sqrt{3}} \times\left(1+\frac{1}{2}\right)\right] \\
&=2\left[\frac{\mu_{0}}{4 \pi} \times \frac{2 i}{d \sqrt{3}} \times \frac{3}{2}\right]=\frac{\sqrt{3} \mu_{0} i}{2 \pi d}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
A long conducting wire having a current I flowing through it, is bent into a circular coil of N turns. Then it is bent into a circular coil of n turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is:

Options:
A N:n
B n2 :N2
C N2:n2
D n:N

The correct answer is:
N2:n2

Solution:
As in both cases, length of the wire will remain the same, therefore

l=2πRN=2πrn⇒Rr=nN

Now magnetic field due to single circular coil is given by, B=μ0I2r.

Therefore, required ratio of total magnetic field wil be,

B1B2=Nμ0I2Rnμ0I2r=N2n2

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (31 Jan Shift 2)
A long cylindrical conductor of radius a and length l is made of a material whose resistivity depends only on the distance r from the axis of the conductor as Ï=Î±r2, where Î± is a constant. The resistance of the conductor across the two ends will be

Options:
A R=4αlπa3
B R=2αlπa4
C R=4απla3
D R=2lπαa3

The correct answer is:
R=2αlπa4

Solution:
We consider a long co-axial hollow cylinder of radius r and thickness dr. The cross-sectional area of considered element is dA = 2âπrdr
∴ Electric resistance between ends of conductor of considered element is
dR = ρ l dA = ρ l 2 π rdr = α l⁡ 2 π r 3 dr [as resistivity is ρ= α r 2 ]
The system may be assumed as a parallel combination of a number of such types of elementary resistance.
∴ 1 R = ∫ 1 dR = 2 π α l⁡ ∫ 0 a r 3 dr or 1 R = 2 π α l⁡ a 4 4
∴R=2αlπa4

Topic: Physics, Current Electricity, JEE Main
A long cylindrical pipe of radius $20 \mathrm{~cm}$ is closed atits upper end and has an airtight piston of negligible mass as shown. When a $50 \mathrm{Kg}$ mass is attached to the other end of the pistion, it moves down. If the air in the enclosure is cooled from temperature $T$ to $T-\Delta T$, the piston moves back to its original position. Then $\Delta T / T$ is close to (Assuming air to be an ideal gas, $g=10 \mathrm{~m} / \mathrm{s}^{2}$, atmospheric pressure is $10^{5}$ Pascal),

Options:
A $0.01$
B $0.02$
C $0.04$
D $0.09$

The correct answer is:
$0.04$

Solution:
Initially pressure $=P_{0}$ When $50 \mathrm{~kg}$ mass is suspended
$\begin{array}{l}
\text { pressure }=P_{0}-\frac{m g}{A} \\
\text { (temp }=\text { constant }) \\
P_{0} V_{1}=\left(V_{0}-\frac{m g}{A}\right) V_{f} \\
\left(P_{0}-\frac{m g}{A}\right) V_{f}=n R T \\
\left(P_{0}-\frac{m g}{A}\right) V_{i}=n R(T-\Delta T) \\
=\frac{V_{f}}{V_{i}}=\frac{T}{T-\Delta T} \\
\frac{T-\Delta T}{T}=\frac{V_{i}}{V_{f}} \\
=\frac{\Delta T}{T}=\left(P_{0}-\frac{m g}{A}\right) \frac{1}{P_{0}} \\
\end{array}$
$\begin{aligned} \frac{\Delta \mathrm{T}}{\mathrm{T}} &=\frac{\mathrm{mg}}{\mathrm{AP}_{0}} \\ &=\frac{50 \times 10}{3.14 \times(0.2)^{2} \times 10^{5}} \\ &=\frac{500}{12.56 \times 10^{3}} \\ &=\frac{5}{125.6}=\frac{1}{25}=0.04 \end{aligned}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A long cylindrical shell carries positive surface charge σ in the upper half and negative surface charge –σ in the lower half. The electric field lines around the cylinder will look like figure given in:
(figures are schematic and not drawn to scale)

Options:
A
B
C
D

The correct answer is:

Solution:
Using the properties of a dipole and laws of electric field line where two lines should not intersect, the graph would be,

Topic: Physics, Electrostatics, JEE Main, JEE Main 2015 (04 Apr)
A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of vessel is 5 cm and its rotational speed is 2 rotations per second, then the difference in the heights between the center and the sides, in cm, will be:

Options:
A 0.4
B 2.0
C 1.2
D 0.1

The correct answer is:
2.0

Solution:
When cylinder is rotated, water will rise at the edge and will dip at the centre. The difference in pressure at A and O is PA-P0=Ïgh. Due to height difference.
pressure difference P1-P0=ÏâacmR Due to rotation.
=ÏâÏ2.R2R
â´Ïgh=ÏÏ2R22âh= Ï2âR22g=2ÏÃ22Ã0.0522Ã10
=0.02 m
=2 cm

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (12 Jan Shift 2)
A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral.

Options:
A
a potential difference appears between the two cylinders when a charge density is given to the inner cylinder

B
a potential difference appears between the two cylinders when a charge density is given to the outer cylinder

C
no potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders

D
no potential difference appears between the two cylinders when same charge density is given to both the cylinders

The correct answer is:
a potential difference appears between the two cylinders when a charge density is given to the inner cylinder

Solution:
There will be an electric field between two cylinders (using Gauss theorem. This electric field will produce a potential difference.
$\therefore$ Answer is (a).

Topic: Physics, Electrostatics, JEE Main
A long horizontal rod has a bead which can slide along its length, and initially it is at a distance L from the end A of the rod. The rod is set in angular motion about A with constant angular acceleration Î± in a gravity free space. If the coefficient of friction between the rod and the bead is Î¼ , then the time after which the bead starts slipping is

Options:
A Î¼ Î±
B μ α
C 1 μ α
D infinitesimal

The correct answer is:
μ α

Solution:
N=mαL

When the bead starts slipping
fmax=μN=mω2L

μmαL=mαt2L
⇒t=μα

Topic: Physics, Laws of Motion, JEE Main
A long horizontal rod has a bead which can slide along its length and is initially placed at a distance \( L \) from one end \( A \) of the rod. The rod is set in angular motion about \( A \) with a constant angular acceleration \( a \). If the coefficient of friction between the rod and bead is \( \mu \), and gravity is neglected, then the time after which the bead starts slipping is

Options:
A \( \sqrt{\frac{\mu}{\alpha}} \)
B \( \frac{\mu}{\sqrt{\alpha}} \)
C \( \frac{1}{\sqrt{\mu \alpha}} \)
D Infinitesimal

The correct answer is:
\( \sqrt{\frac{\mu}{\alpha}} \)

Solution:
Tangential force $\left(F_{t}\right)$ of the bead will be given by the normal reaction $(\mathrm{N})$, while centripetal force $\left(F_{c}\right)$ is provided by friction $\left(f_{r}\right)$ The bead starts sliding when the centripetal force is just equal to the limiting friction,
Therefore,
$F_{t}=m a=m \alpha L=N$
Therefore, limiting value of friction
Angular velocity at time $t$ is $\omega=\alpha t(f r)_{\max }=\mu N=\mu m \alpha L \quad \ldots$ (i)
Centripetal force at time $t$ will be $F e=m L \omega 2=m L \alpha 2 t 2 \ldots(i i)$
Equating equations $(i)$ and $(i i)$, we get,
$t=\sqrt{\frac{\mu}{\alpha}}$
For $t>F_{c}>(f r)_{\max }$,
i. e., the bead starts sliding.
In the figure, $F_{t}$ is perpendicular to the paper inwards.

Topic: Physics, Motion In Two Dimensions, JEE Main