datavorous/entrance-exam-dataset huggingface dataset:
The best reagent to convert pent -3- en-2-ol into pent -3-en-2-one is

Options:
A
Acidic permanganate

B
Acidic dichromate

C
Chromic anhydride in glacial acetic acid

D
Pyridinium chloro – chromate

The correct answer is:
Chromic anhydride in glacial acetic acid

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2005
The B-H curve for a ferromagnet is shown in the figure. The ferromagnet is placed inside a long solenoid with 1000 turns/cm.. The current that should be passed in the solenoid to demagnetize the ferromagnet completely is:

Options:
A
$2 \mathrm{~mA}$

B
$1 \mathrm{~mA}$

C
$40 \mu \mathrm{A}$

D
$20 \mu \mathrm{A}$

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

Solution:
Given Number of turns, $\mathrm{n}=1000$ turns $/ \mathrm{cm}=1000 \times 100$ turns $/ \mathrm{m}$ Coercivity of ferromagnet, $\mathrm{H}=100 \mathrm{~A} / \mathrm{m}$ Current to demagnetise the ferromagnet, $\mathrm{I}=?$
Using, $\mathrm{H}=\mathrm{nI}$
or, $\quad 100=10^5 \times \mathrm{I}$
$\therefore \quad \mathrm{I}=\frac{100}{10^5}=1 \mathrm{~mA}$

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The binding energy of a satellite of mass m in an orbit of radius r is (R = radius of the earth, g = acceleration due to gravity)

Options:
A mgR2r
B mgR22r
C mgR24r
D 2mgR2r

The correct answer is:
mgR22r

Solution:
The energy required to remove the satellite from its orbit around the earth to infinity is called binding energy of the satellite. It is equal to negative of total mechanical energy of a satellite in its orbit.
Thus,
The binding energy of the Earth-satellite system,
E = - GMm 2 r
using, g = GM R 2
â GM = gR 2
â´ BE=gmR22r

Topic: Physics, Gravitation, JEE Main
The binding energy of an electron in the ground state of He is equal to 24.6 eV. What is the amount of energy (in eV) required to remove both the electrons?

Options:

The correct answer is:
79

Solution:
Helium atom has 2 electrons. When one electron is removed, the remaining atom is hydrogen like atom, whose energy in first orbit is
E1=-2213.6 eV=-54.4 eV
Therefore, to remove the second electron from the atom, the additional energy of 54.4 eV is required. Hence, total energy required to remove both the electrons =24.6+54.4=79.0 eV

Topic: Physics, Atomic Physics, JEE Main
The binding energy per nucleon for a $6 \mathrm{C}^{12}$ the nucleus is
$\begin{array}{ll} \text{nuclear mass of } 6 \mathrm{C}^{12} & =12 . 00000 \mathrm{~amu} \\ \text{Mass of hydrogen nucleus } & =1 . 007825 \mathrm{~amu} \\ \text{Mass of neutron } & =1 . 008665 \mathrm{~amu} \end{array}$

Options:
A $2.675 \mathrm{MeV}$
B $7.675 \mathrm{MeV}$
C $0 \mathrm{MeV}$
D $3.675 \mathrm{MeV}$

The correct answer is:
$7.675 \mathrm{MeV}$

Solution:
Given,
nuclear mass of ${ }_\epsilon C^{12}=12.00000$ ane $=M_{\varepsilon^{12}}$
Mass of hydrogen nucleus $M_H=1.007825$ amu
Mass of neutron $m_r=1.008665$ amu.
Now binding energy of $\varepsilon^{12}$ nucleus is.
$B \cdot E=\left(z m_H+N m_N\right) c^2-\left(m_c c^{12}\right) \cdot c^2$
where $z=$ proton number $=6$
$r=$ neutron number $=12-6=6$
we know 1 amu $=931.478 \mathrm{mev} / \mathrm{c}^2$
so,
$\begin{aligned} B . E & =\left[z m_H-N m_N-m_{C_1}\right] c^2 \\ & =[6 \times 1.007825+6 \times 1.008665-12.00] \text { amce. } c^2 \\ & =0.00894 \times 931.478 \frac{\mathrm{meV}}{\mathrm{c}^2} \times c^2 \\ B . E & =92.16 \mathrm{meV}\end{aligned}$
Now Binding energy per nucleon is
$\frac{B . E}{12}=\frac{92.16}{12}=7.68$ mev

Topic: Physics, Nuclear Physics, JEE Main
The binding energy per nucleon in deuterium and helium nuclei are $1.1 \mathrm{MeV}$ and $7.0 \mathrm{MeV}$, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is

Options:
A $23.6 \mathrm{MeV}$
B $2.2 \mathrm{MeV}$
C $28.0 \mathrm{MeV}$
D $30.2 \mathrm{MeV}$

The correct answer is:
$23.6 \mathrm{MeV}$

Solution:
Mass of ${ }_1 \mathrm{H}^2=2.01478 \mathrm{amu}$
Mass of ${ }_2 \mathrm{He}^4=4.00388 \mathrm{amu}$
Mass of two deuterium $=2 \times 2.01478$
$=4.02956 \mathrm{amu}$
Energy equivalent to ${ }_2 \mathrm{H}^2$
$\begin{aligned}
& =4.02956 \times 1.2 \\
& =4.4 \mathrm{MeV}
\end{aligned}$
Energy equivalent to ${ }_2 \mathrm{He}^4$
$=4.00388 \times 7=28 \mathrm{MeV}$
$\text {Energy released }=(28-4.4)$
$=23.6 \mathrm{MeV}$

Topic: Physics, Nuclear Physics, NEET, NEET 2010 (Mains)
The binding energy per nucleon is practically independent of the atomic number for nuclei of mass number in the range. $\qquad$

Options:
A $10 \lt \mathrm{A} \lt 50$
B $A \lt 30$
C $30 \lt A \lt 170$
D $A\gt170$

The correct answer is:
$30 \lt A \lt 170$

Solution:
The binding energy per nucleon is practically independent of the atomic number for nuclei of mass number in the range $30 \lt A \lt 170$.

Topic: Physics, Nuclear Physics, JEE Main
The binding energy per nucleon of  16O is 7.97 MeV and that of  17O is 7.75 MeV. What is the energy in MeV required to remove a neutron from  17O ?

Options:

The correct answer is:
4.23

Solution:
BE per nucleon O 16=7.97 MeV
BE per nucleon O 17=7.75 MeV
O17→n10+O16
The energy required to remove a neutron
=17×7.75-16×7.97
=4.23 MeV

Topic: Physics, Nuclear Physics, JEE Main
The binding energy per nucleon of deuterium and helium nuclei are 1.1 MeV and 7.0 MeV respectively. When two deuterium nuclei fuse together to form a helium nucleus, the energy released in the fusion is

Options:
A 2.2 MeV
B 23.6 MeV
C 28.0 MeV
D 30.2 MeV

The correct answer is:
23.6 MeV

Solution:
The fusion reaction is as given below:
1H2+1H2â2He4+Q
The binding energy of the reacting nuclei
=1.1 Ã2+1.1 Ã2=4.4 MeV
The binding energy of the product nuclei
=7.0 Ã4=28.0 MeV
Hence, the energy released in the fusion reaction,
Q=28.0-4.4=23.6 MeV .

Topic: Physics, Nuclear Physics, JEE Main
The binding energy per nucleon of deuteron $\left({ }_1^2 \mathrm{H}\right)$ and helium nucleus $\left({ }_2^4 \mathrm{He}\right)$ is $1.1 \mathrm{MeV}$ and $7 \mathrm{MeV}$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

Options:
A
$13.9 \mathrm{MeV}$

B
$26.9 \mathrm{MeV}$

C
$23.6 \mathrm{MeV}$

D
$19.2 \mathrm{MeV}$

The correct answer is:
$23.6 \mathrm{MeV}$

Solution:
Energy released $=$ total binding energy of product $-$ total binding energy of reactants $\Rightarrow 28-(2 \times 2.2)=28-4.4=236 \mathrm{MeV}$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2004
The binding energy per nucleon of deuteron (H)12and helium nucleus (He)24is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

Options:
A 13.9 MeV
B 26.9 MeV
C 23.6 MeV
D 19.2 MeV

The correct answer is:
23.6 MeV

Solution:
As given
1H2 + 1H2â 2He4 +energy
The binding energy per nucleon of a deuteron( 1H2)
=1.1 MeV
â´ Total binding energy of one deuteron nucleus
= 2 Ã 1.1=2.2 MeV
â´ The binding energy per nucleon of Helium( 2He4)
=7 MeV
â´ Total binding energy
= 4 Ã 7 =28 MeV
Hence, energy released in the above process
=28-2 Ã2.2
= 28 â 4.4 = 23.6 MeV

Topic: Physics, Nuclear Physics, JEE Main
The binomial coefficients which are in decreasing order are

Options:
A ${ }^{15} C_5,{ }^{15} C_6,{ }^{15} C_7$
B ${ }^{15} C_{10},{ }^{15} C_9,{ }^{15} C_8$
C ${ }^{15} C_6,{ }^{15} C_7,{ }^{15} C_8$
D ${ }^{15} C_7,{ }^{15} C_6,{ }^{15} C_5$

The correct answer is:
${ }^{15} C_7,{ }^{15} C_6,{ }^{15} C_5$

Solution:
The series of binomial coefficient is ${ }^{15} C_0,{ }^{15} C_1,{ }^{15} C_2 \ldots{ }^{15} C_7,{ }^{15} C_8,{ }^{15} C_9 . .{ }^{15} C_{14},{ }^{15} C_{15}$
Form the above discussion, we can say that decreasing series is ${ }^{15} C_7,{ }^{15} C_6,{ }^{15} C_5$ therefore option (4) is correct.

Topic: Mathematics, Binomial Theorem, JEE Main
The binomial distribution has:

Options:
A only one parameter
B two parameters
C three parameters
D four parameters

The correct answer is:
two parameters

Solution:
Binomial distribution has two parameters $\mathrm{n}$ and $\mathrm{p}$, where $\mathrm{n}$ is number of trials and $\mathrm{p}$ is probability of success.

Topic: Mathematics, Probability, JEE Main
The biquadratic equation, two of whose roots are $1+i, 1-\sqrt{2}$, is

Options:
A $x^4-4 x^3+5 x^2-2 x-2=0$
B $x^4+4 x^3-5 x^2+2 x+2=0$
C $x^4+4 x^3-5 x^2+2 x-2=0$
D $x^4+4 x^3+5 x^2-2 x+2=0$

The correct answer is:
$x^4-4 x^3+5 x^2-2 x-2=0$

Solution:
When $1+i, 1-i$ are the roots, then
Sum of the roots $=1+i+1-i=2$
and product of the roots $=1+1=2$
The equation is $x^2-2 x+2=0$
When $1-\sqrt{2}, 1+\sqrt{2}$ are the roots, then
Sum of the roots $=1-\sqrt{2}+1+\sqrt{2}=2$
Product of the roots $=1-2=-1$
The equation is $x^2-2 x-1=0$
The bi-quadratic equation is
$\begin{array}{r}\left(x^2-2 x+2\right)\left(x^2-2 x-1\right)=0 \\ \left(x^4-(2+2) x^3=(-1+4+2) x^2\right. \\ =(2-4) x-2=0 \\ \Rightarrow \quad x^4-4 x^3+5 x^2-2 x-2=0\end{array}$

Topic: Mathematics, Quadratic Equation, JEE Main
The block of mass m0=Î¼L is attached to a uniform string of mass M=Î¼L and length L as shown in the figure. If a wave pulse is produced near the block, then the time it takes to reach the ceiling is

Options:
A 2-1Lg
B 22+1Lg
C 22-1Lg
D 2+1Lg

The correct answer is:
22-1Lg

Solution:
At x position tension in string
T=mog+μxg
V=Tμ=L+xg
∫otdxL+xg=∫otdt
T=22-1Lg

Topic: Physics, Waves and Sound, JEE Main
The block of mass 'm' initially at x = 0 is acted upon by a horizontal force F = a - bx as shown in the figure. The coefficient of friction between the surfaces of contact is μ . The net work done on the block is zero if the block travels a distance of ________.

Options:
A a - μ mg / 2 b
B a - μ mg / b
C 2 a - μ mg / b
D a - 2 μ mg / b

The correct answer is:
2 a - μ mg / b

Solution:
f = a - bx
f n ef = a - bx - f s
For block to move
at x = 0
a - b x > f s max = μ mg
a > μ mg , So we are assuming motion starts at x = 0
w = ∫ a - bx - μ mg dx
o = ax - bx 2 2 - μ mgx
⇒ x a - bx 2 - μ mg = 0
x = 0
a - bx 2 - μ mg = 0
bx 2 = a - μ mg
x = 2 b a - μ mg

Topic: Physics, Work Power Energy, JEE Main
The block of mass m is placed on a rough horizontal floor and it is pulled by an ideal string by a constant force F as shown. As the block moves towards right on the floor, then the frictional force on the block
âââ

Options:
A remains constant
B increases
C decreases
D can not be calculated

The correct answer is:
decreases

Solution:
As the block moves towards right normal reaction on it decreases therefore frictional force on it decreases.

Topic: Physics, Laws of Motion, NEET
The block of mass $M$ moving on the frictionless horizontal surface collides with a spring of spring constant $\mathrm{K}$ and compresses it by length $\mathrm{L}$. The maximum momentum of the block after collision is

Options:
A
$\sqrt{\mathrm{MK}} \mathrm{L}$

B
$\frac{\mathrm{KL}^2}{2 \mathrm{M}}$

C
zero

D
$\frac{M L^2}{\mathrm{~K}}$

The correct answer is:
$\sqrt{\mathrm{MK}} \mathrm{L}$

Solution:
$\frac{1}{2} \mathrm{KL}^2=\frac{\mathrm{P}^2}{2 \mathrm{~m}} \quad \therefore \mathrm{P}=\sqrt{\mathrm{MK}} \mathrm{L}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2005
The block shown in the diagram has a mass of 2 Î¼g and a charge of 2 Ã 10â9 C. If the block is given an initial velocity of 2 m s-1 in the x-direction at t= 0 and an electric field of 2 N C-1 is switched on in the x-direction, then the range ( R ) of the block on the ground will be

Options:
A 4 m
B 1 m
C 2 m
D 3 m

The correct answer is:
3 m

Solution:
At first, we will find time taken by block to hit ground
using, S y = u y t+ 1 2 a y t 2
â5=0+ 1 2 ( âg )Ã t 2
ât= 10 g =1 s
Now, in this time, block will move in x-direction.
S x = u x .t+ 1 2 a x . t 2
Here,Sx=R,ux=2,
ax=2Em=2Ã10â9Ã22Ã10â9=2
âR=2Ã1+ 1 2 Ã2Ã 1 2
=2+1=3 m

Topic: Physics, Electrostatics, JEE Main
The blocks A and B shown in the figure have masses MA=5 kg and MB=4 kg. the system is released from rest. The speed of B after A has travelled a distance 1 m along the incline is g=10 m s-2

Options:
A 152
B 158
C 56
D 52

The correct answer is:
56

Solution:
If A moves down the incline by 1 m, B shall move up by 12 m. If the speed of B is v then the speed of A will be 2v.
From the conservation of energy: Gain in K.E. = loss in P.E.
12MA2v2+12MBv2=MAg×35-MBg×12
Solving we get, v=12g3=56

Topic: Physics, Work Power Energy, JEE Main
The blood calcium level is lowered by the deficiency of

Options:
A parathormone
B thyroxine
C calcitonin
D Both (a) and (c)

The correct answer is:
parathormone

Solution:
The chief cells of the parathyroids secrete parathormone. Its deficiency causes the lowering of blood calcium level. This increases the excitability of nerves and muscles causing cramps and convulsions. This caused parathyroid tetany characterised by sustained contractions of the muscles of larynx, face, hands and feet.
Calcitonin is secreted when calcium level is high in blood. It has an opposite action to that of parathyroid hormone and lowers the calcium level by suppressing release of calcium ions from the bones.
Thyroxine is secreted from the thyroid gland. It regulates the metabolic rate of the body and thus, maintain basal metobolic rate, stimulate protein synthesis and therefore, promote growth of the body tissues.

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2008 (Screening)
The blood calcium level is lowered by the deficiency of

Options:
A Calcitonin
B Parathormone
C Thyroxine
D Both Calcitonin and Parathormone

The correct answer is:
Parathormone

Solution:
No Solution Available

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2008 (Mains)
The bob of a pendulum is released from a horizontal position A as shown in the figure. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point B, given that it dissipated 5% of its initial energy against air resistance?

Options:
A 5 m s-1

B 5.5 m s-1
C 5.3 m s-1
D 4.4 m s-1

The correct answer is:
5.3 m s-1

Solution:
At the point A the energy of the pendulum is entirely potential energy. At point B, the energy of the pendulum is entirely kinetic energy. It means that as the bob pendulum lowers from A to B, the potential energy is converted into kinetic energy. Thus, at B, KE = PE. But 5% of the potential energy is dissipated against air resistance.
KE at B=95% of PE at A .... (i)
∴ From equation (i), 12mv2=95100mgh
v2=2×95100gh=2×95100×9.8×1.5
v=19×9.8×1.510=27.93=5.285 m s-1
v=5.3 m s-1

Topic: Physics, Work Power Energy, JEE Main
The bob of a pendulum of mass m and length L is displaced 90Â° from the vertical and gently released. In order that the string may not break upon passing through the lowest point, its minimum strength must be

Options:
A mg
B 2 mg
C 3 mg
D 4 mg

The correct answer is:
3 mg

Solution:
If v is velocity of the bob on reaching the lowest point, then 12mv2=mgL
To avoid breaking, strength of the string
TL=mv2L+mg=2mg LL+mg=3 mg

Topic: Physics, Laws of Motion, JEE Main
The bob of a pendulum of mass $m$, suspended by an inextensible string of length $L$ as shown in the figure carries a small charge $a$. An infinite horizontal plane conductor with uniform surface charge density $\sigma$ is placed below it. What will be the time period of the pendulum for small amplitude oscillations?

Options:
A $2 \pi \sqrt{\frac{\mathrm{L}}{\left(\mathrm{g}-\frac{\mathrm{mq}}{\varepsilon_{0} \sigma}\right)}}$
B $\sqrt{\frac{\mathrm{L}}{\left(\mathrm{g}-\frac{\mathrm{mq} \sigma}{\varepsilon_{0}}\right)}}$
C $\frac{1}{2 \pi} \sqrt{\frac{\mathrm{L}}{\left(\mathrm{g}-\frac{\mathrm{q} \sigma}{\varepsilon_{0} \mathrm{~m}}\right)}}$
D $2 \pi \sqrt{\frac{\mathrm{L}}{\left(\mathrm{g}-\frac{\mathrm{q} \sigma}{\varepsilon_{0} \mathrm{~m}}\right)}}$

The correct answer is:
$2 \pi \sqrt{\frac{\mathrm{L}}{\left(\mathrm{g}-\frac{\mathrm{q} \sigma}{\varepsilon_{0} \mathrm{~m}}\right)}}$

Solution:
$\therefore$ Apparent weight of the bob, $w^{\prime}=m g-q E$
$$
m y^{\prime}=m g-q E
$$
$\therefore$
$g^{\prime}=g-\frac{q E}{m}$
Time period of a pendulum. $T=2 \pi \sqrt{\frac{L}{g_{e f f}}}$ $=2 \pi \sqrt{\frac{L}{g-\frac{q E}{m}}}=2 \pi \sqrt{\frac{L}{\left(g-\frac{q \sigma}{\varepsilon_{0} m}\right)}}$

Topic: Physics, Oscillations, JEE Main
The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10 m. If it dissipates 10% of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is: [Use, g=10 m s-2 ]

Options:
A 65 m s-1
B 56 m s-1
C 55 m s-1
D 25 m s-1

The correct answer is:
65 m s-1

Solution:
Given the length of the pendulum is ℓ=10 m.

The initial potential energy of the pendulum at the horizontal position is given by

U=mgℓ ...1

The final kinetic energy of the pendulum at the lowermost position is given by

K=12mv2 ...2

So, from equations (1) and (2), it follows that

910mgℓ=12mv2⇒910×10×10=12v2⇒v2=180⇒v=180=65 m s-1

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2024 (29 Jan Shift 2)
The bob of a simple pendulum executes simple harmonic motion in water with a period $t$, while the period of oscillation of the bob is $t_0$ in air. Neglecting frictional force of water and given that the density of the bob is $\left(\frac{4}{3}\right) \times 1000 \mathrm{~kg} / \mathrm{m}^3$. What relationship between $t$ and $t_0$ is true?

Options:
A
$t=t_0$

B
$t=t_0 / 2$

C
$t=2 t_0$

D
$\mathrm{t}=4 \mathrm{t}_0$

The correct answer is:
$t=2 t_0$

Solution:
$\frac{\mathrm{T}}{\mathrm{T}_0}=\sqrt{\frac{1}{\left(1-\frac{\rho^{\prime}}{\rho}\right)}}=\sqrt{\frac{1}{1-\frac{1}{3}}}$
$\Rightarrow \frac{\mathrm{T}}{\mathrm{T}_0}=2$
or, $\mathrm{T}=2 \mathrm{~T}_0$

Topic: Physics, Oscillations, JEE Main, JEE Main 2004
The bob of a simple pendulum has mass 2 g and a charge of 5.0 μC. It is at rest in a uniform horizontal electric field of intensity 2000 V/m At equilibrium, the angle that the pendulum makes with the vertical is:
take g=10 m/s2

Options:
A tan-10.2
B tan-12.0
C tan-10.5
D tan-15.0

The correct answer is:
tan-10.5

Solution:
tan⁡θ=QEmg=5×10-6×20002×10-3×10=12
⇒θ=tan-1⁡0.5

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

Options:
A first increases then decrease back to the original value
B first decreases then increase back to the original value
C remains unchanged
D increases towards a saturation value

The correct answer is:
first increases then decrease back to the original value

Solution:
and T1>T2
Hence, time period first increases and then decreases to the original value.
The given system is like a simple pendulum, whose effective length between the point of suspension and the center of gravity of the hanging body is changing.
When water slowly flows out, the sphere, the C.G. of system is lowered & hence L increase

Topic: Physics, Mathematics in Physics, NEET
The bob of a simple pendulum is a spherical hollow ball, filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

Options:
A First increases and then decreases to the original value
B First decreases and then increases to the original value
C Remains unchanged
D Increases towards a saturation value

The correct answer is:
First increases and then decreases to the original value

Solution:
As T1>T2
Hence, time period first increases and then decreases to the original value.

Topic: Physics, Oscillations, NEET
The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would:

Options:
A remain unchanged
B increase towards a saturation value
C first increase and then decrease to the original value.
D first decrease and then increase to the original value.

The correct answer is:
first increase and then decrease to the original value.

Solution:
The time period of the simple pendulum is, T=2πlg Here, l is the effective length of the pendulum up to centre of mass.

When the hole at bottom of the hollow sphere (bob) filled with water is unplugged, the water will drain from the bob due to which the centre of mass will shift downward, therefore, the length of the pendulum will slightly increase, which will lead to increase in time period.

After the water completely drained out the centre of mass again shift to its original position that is up to the centre of the hollow sphere (bob) due to which the length of the pendulum will decrease and will reach to its initial state which causes the time period to reach its original value.

Topic: Physics, Oscillations, AP EAMCET, AP EAMCET 2021 (20 Aug Shift 1)
The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

Options:
A first increases then decrease back to the original value
B first decreases then increase back to the original value
C remains unchanged
D increases towards a saturation value

The correct answer is:
first increases then decrease back to the original value

Solution:
and T1>T2
Hence, time period first increases and then decreases to the original value.
The given system is like a simple pendulum, whose effective length between the point of suspension and the center of gravity of the hanging body is changing.
When water slowly flows out, the sphere, the C.G. of system is lowered & hence L increase

Topic: Physics, Mathematics in Physics, JEE Main
The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

Options:
A First increase and then decrease to the original value
B First decrease and then increase to the original value
C Remain unchanged
D Increase towards a saturation value

The correct answer is:
First increase and then decrease to the original value

Solution:
and T1>T2
The given system is like a simple pendulum whose effective length is between the point of suspension and the center of gravity of the hanging body.
When water slowly flows out, the sphere, the centre of gravity. of system is changing as shown in the figures.
Hence, time period first increases and then decreases to the original value.

Topic: Physics, Oscillations, NEET
The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillation bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

Options:
A
first increase and then decrease to the original value.

B
first decreased then increase to the original value.

C
remain unchanged.

D
increase towards a saturation value.

The correct answer is:
first increase and then decrease to the original value.

Solution:
First CM goes down and then comes to its initial position

Topic: Physics, Oscillations, JEE Main, JEE Main 2005
The bob of simple pendulum is hanging vertically down from a fixed identical bob by means of a sting of length $l$. If both bobs are charged with a charge $q$ each, time period of the pendulum is : (ignore the radii of the bobs)

Options:
A $2 \pi \sqrt{\frac{l}{g+\left(\frac{q^2}{l^2 m}\right)}}$
B $2 \pi \sqrt{\frac{l}{g-\left(\frac{q^2}{l^2 m}\right)}}$
C $2 \pi \sqrt{\frac{l}{g}}$
D $2 \pi \sqrt{\frac{l}{g-\left(\frac{q^2}{l}\right)}}$

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

Solution:
There will be a electrostatic repulsion between two charged bobs, but it does not affect the motion of pendulum. Thus, time period of pendulum remains same i.e., $T=2 \pi \sqrt{\frac{l}{g}}$.

Topic: Physics, Oscillations, AP EAMCET, AP EAMCET 2006