datavorous/entrance-exam-dataset huggingface dataset:
A silver cup is plated with silver by passing 965 C of electricity. The amount of Ag deposited is

Options:
A 107.89 g
B 9.89 g
C 1.0002 g
D 1.08 g

The correct answer is:
1.08 g

Solution:
Ag++ e- ⟶ Ag
∵ 96500 C are required to deposite Ag = 108 g
∴ 965 C are required to deposite Ag
= 10896500 × 965 = 1.08 g

Topic: Chemistry, Electrochemistry, JEE Main
A silver ingot weighing $2.1 \mathrm{~kg}$ is held by a string so as to be completely immersed in a liquid of relative density 0.8 . The relative density of silver is 10.5 . The tension in the string in $\mathrm{kg}-\mathrm{wt}$ is

Options:
A 1.6
B 1.94
C 3.1
D 5.25

The correct answer is:
1.94

Solution:
Apparent weight $=V(\rho-\sigma) g=\frac{M}{\rho}(\rho-\sigma) g$
$=M\left(1-\frac{\sigma}{\rho}\right) g=2.1\left(1-\frac{0.8}{10.5}\right) g=1.94 g_N$
$=1.94 \mathrm{Kg}$-wt

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of 200 nm wavelength electromagnetic waves. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is AÃ10Z (where 1<A<10). The value of Z is

Options:

The correct answer is:
7

Solution:
Photo emission will stop when potential on silver sphere becomes equal to the stopping potential.
Stopping potential
= hc λ - w
=6.2-4.7=1.5 eV
V = Kq r = 1 . 5 = K × n × e r
⇒ n = 1 . 5 × 1 0 - 2 1 . 6 × 1 0 - 1 9 × 9 × 1 0 9 = 1 . 0 5 × 1 0 7
⇒ z=7
Therefore answer is 7.

Topic: Physics, Dual Nature of Matter, JEE Main
A silver sphere of radius $1 \mathrm{~cm}$ and work function $4.7 \mathrm{eV}$ is suspended from an insulating thread in free-space. It is under continuous illumination of $200 \mathrm{~nm}$ wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is $A \times 10^Z$ (where $1 < A < 10$ ). The value of $Z$ is

Options:

The correct answer is:
7

Solution:
Photoemission will stop when potential on silver sphere becomes equal to the stopping potential.
$\therefore \quad \frac{h c}{\lambda}-W=e V_0$
Here, $\quad V_0=\frac{1}{4 \pi \varepsilon_0} \frac{n e}{r}$
$$
\begin{aligned}
& \therefore \quad\left(\frac{1240}{1200} \mathrm{eV}\right)-(4.7 \mathrm{eV}) \\
&=\frac{9 \times 10^9 \times n \times 1.6 \times 10^{-19}}{10^{-2}} \\
&(6.2-4.7)=\frac{9 \times 10^9 \times n \times 1.6 \times 10^{-19}}{10^{-2}} \\
& \text { or } \quad n= \frac{1.5 \times 10^{-2}}{9 \times 1.6 \times 10^{-10}}=1.04 \times 10^7
\end{aligned}
$$
$\therefore$ Answer is 7 .
Analysis of Question
(i) Question is moderately difficult.
(ii) In $\mathrm{eV}_0$ if we substitute the value of $e=1.6 \times 10^{-19} \mathrm{C}$, then answer is in J. If we leave it then, answer is in $e \mathrm{~V}$.

Topic: Physics, Dual Nature of Matter, JEE Main
A similarity between optical and geometrical isomerism is that

Options:
A
each forms equal number of isomers for a given compound

B
If in a compound one is present then so is the other

C
both are included in stereoisomerism

D
they have no similarity

The correct answer is:
both are included in stereoisomerism

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2002
A simple example of a coordinate covalent bond is exhibited by

Options:
A $\mathrm{C}_2 \mathrm{H}_2$
B $\mathrm{H}_2 \mathrm{SO}_4$
C $\mathrm{NH}_3$
D $\mathrm{HCl}$

The correct answer is:
$\mathrm{H}_2 \mathrm{SO}_4$

Solution:
$\mathrm{H}_2 \mathrm{SO}_4$ has co-ordinate covalent bond.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
A simple harmonic motion is represented by $\alpha \frac{d^2 x}{d t^2}+\beta x=0$. Its period is

Options:
A $\frac{2 \pi \beta}{\alpha}$
B $2 \pi \sqrt{\frac{\alpha}{\beta}}$
C $2 \pi \sqrt{\frac{\beta}{\alpha}}$
D $\frac{2 \pi \alpha}{\beta}$

The correct answer is:
$2 \pi \sqrt{\frac{\alpha}{\beta}}$

Solution:
The standard equation of SHM is: $\frac{d^2 x}{d t^2}=-\omega^2 x$
On comparing with equation: $\alpha \frac{d^2 x}{d t^2}+\beta x=0$
The angular frequency is given by: $\omega=\frac{\beta}{\alpha}$.
Therefore, the time period of SHM is:
$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\alpha}{\beta}}$

Topic: Physics, Oscillations, JEE Main
A simple harmonic motion is represented by $F(t)=10 \sin (20 t+0.5)$. The amplitude of the S.H.M. is

Options:
A $a=30$
B $a=20$
C $a=10$
D $a=5$

The correct answer is:
$a=10$

Solution:
Given,
$F(t)=10 \sin (20 t+0.5)$
Equation of $\mathrm{SHM}$
$y=a \sin (\omega t+\varphi)$
on comparing,
$a=$ Amplitude $=10$

Topic: Physics, Oscillations, JEE Main
A simple harmonic motion is represented by:
y=5sin3πt+3cos3πt cm
The amplitude and time period of the motion are:

Options:
A 5 cm, 23 s
B 10 cm, 23 s
C 5 cm, 32 s
D 10 cm, 32 s

The correct answer is:
10 cm, 23 s

Solution:
y=5sin3πt+53cos3πt
A2=52+532=102
A=10 cm
ω=3π
T=2πω=23 s

Topic: Physics, Oscillations, JEE Main, JEE Main 2019 (12 Jan Shift 2)
A simple harmonic oscillator has an amplitude A and time period 6π second. Assuming the oscillation starts from its mean position, the time required by it to travel from x=A to x=32A will be πx s, where x=_______.

Options:

The correct answer is:
2

Solution:
From phasor diagram particle has to move from P to Q in a circle of radius equal to amplitude of SHM.

cosϕ=3 A2 A=32

⇒ϕ=π6

Now, π6=ωt

⇒π6=2πTt

⇒π6=2π6πt

⇒t=π2

So, x=2

Topic: Physics, Oscillations, JEE Main, JEE Main 2024 (29 Jan Shift 2)
A simple harmonic oscillator of angular frequency 2 rad s-1 is acted upon by an external force F=sin⁡t N. If the oscillator is at rest in its equilibrium position at t=0, its position at later times is proportional to:

Options:
A sin⁡t+12cos⁡2t
B cost-12sin⁡2t
C sin⁡t-12sin⁡2t
D sin⁡t+12sin⁡2t

The correct answer is:
sin⁡t-12sin⁡2t

Solution:
It is given that oscillator at rest at t=0 i.e. at t=0, v=0. So, in option, we can check by putting v=dxdt=0.
(1) If x∝sin⁡t+12cos⁡2t
⟹v∝cos⁡t+12×2 (-sin⁡2t)
⟹ at t=0, v∝1-0≠0
(2) If x ∝cos⁡t-12sin⁡t
⟹ v∝-sin⁡t-12cos⁡t
⟹at t=0, v∝-12≠0
(3) If x∝sin⁡t-12 sin 2t
then υ∝cos⁡t-12×2cos⁡2t
⟹at t=0, v∝1-1=0
(4) If x∝sin⁡t+12sin⁡2t
⟹v∝cos⁡t+12×2cos⁡2t
⟹at t=0, v∝1+1
⟹ v∝2≠0
∴ in option 3 v=0 at t=0

Topic: Physics, Oscillations, JEE Main, JEE Main 2015 (10 Apr Online)
A simple harmonic wave of amplitude 8 unit travels along positive x-axis. At a given instant of time, for a particle at a distance of 10 cm from the origin, the displacement is + 6 unit and for a particle at a distance of 25 cm from the origin, the displacement is + 4 unit. What is the wavelength (in metre) of the wave?

Options:

The correct answer is:
0.25

Solution:
Y=Asinâ¡2ÏÎ»vt-x
YA=sinâ¡2ÏtT-xÎ»
In first case, Y1A=sinâ¡2ÏtT-x1Î»
Here, Y1=+ 6,A=8,x1=10 cm
68=sinâ¡2ÏtT-10Î» .... (i)
Similarly in the second case,
48=sinâ¡2ÏtT-25Î» ... (ii)
From equation (i),
2Ïâ tT-10Î»=sin-1â¡68=0.85 rad
â tT-10Î»=0.14 .... (iii)
Similarly from equation (ii),
2ÏtT-25Î»=sin-1â¡48=Ï6 rad
â tT=25Î»+â0.08 .... (iv)
Subtracting equation (iv) from equation (iii), we get
15Î»=0.06
â Î»=0.25 m

Topic: Physics, Oscillations, JEE Main
A simple harmonic wave of amplitude 8 unit travels along the positive x-axis. At given instant of time, for a particle at a distance of 10 cm from the origin, the displacement is +6 unit and for a particle at a distance of 25 cm from the origin, the displacement is +4 unit. Calculate the wavelength.

Options:
A 200 cm

B 230 cm

C 210 cm

D 250 cm

The correct answer is:
250 cm

Solution:
Y=Asin⁡2πλvt-x
YA=sin⁡2πtT-xλ
In first case, Y1A=sin⁡2πtT-x1λ
Here, Y1=+ 6,A=8,x1=10 cm
68=sin⁡2πtT-10λ .... (i)
Similarly in the second case,
48=sin⁡2πtT-25λ ... (ii)
From equation (i),
2π​ tT-10λ=sin-1⁡68=0.85 rad
⇒ tT-10λ=0.14 .... (iii)
Similarly from equation (ii),
2πtT-25λ=sin-1⁡48=π6 rad
⇒ tT=25λ+​0.08 .... (iv)
Subtracting equation (iv) from equation (iii), we get
15λ=0.06
⇒ λ=250 cm

Topic: Physics, Oscillations, JEE Main
A simple microscope is used to see the object first in blue light and then in red
light. Due to the change from blue to red light, what is the effect on its magnifying
power?

Options:
A Magnifying power increases.
B Magnifying power decreases.
C Magnifying power is independent of colour of light.
D Magnifying power remains constant.

The correct answer is:
Magnifying power decreases.

Solution:
The magnifying power of the simple microscope is given by
$\text { M.P. }=1+\frac{D}{f}$
The focal length of red light is greater than that of blue light. Hence the magnifying power decreases.

Topic: Physics, Ray Optics, MHT CET, MHT CET 2020 (13 Oct Shift 2)
A simple pendulum consisting of a mass M attached to a string of length L is released from rest at an angle Î±. A pin is located at a distance l below the pivot point. When the pendulum swings down, the string hits the pin as shown in the figure. If the particle doesn't undergo a complete vertical circle, then the maximum angle Î¸ <Ï2 through which the string swings after hitting the pin is

Options:
A cos-1â¡Lcosâ¡Î±+l L+l
B cos-1â¡Lcosâ¡Î±-l L-l
C cos-1⁡Lcos⁡α+l L-l
D cos-1⁡Lcos⁡α-l L+l

The correct answer is:
cos-1â¡Lcosâ¡Î±-l L-l

Solution:
At the bottom most point, square of speed of bob,
v2=2gL1 â cos Î±
It will rise further to a height,
h=v22g= L1-cosâ¡Î±
or L-l 1-cosâ¡Î¸=L1-cosâ¡Î±
â´ Î¸=cos-1â¡Lcosâ¡Î±-lL-l

Topic: Physics, Laws of Motion, JEE Main
A simple pendulum doing small oscillations at a place $\mathrm{R}$ height above earth surface has time period of $T_1=4 \mathrm{~s}$. $T_2$ would be it's time period if it is brought to a point which is at a height $2 \mathrm{R}$ from earth surface. Choose the correct relation $[R=$ radius of earth $]$ :

Options:
A $2 \mathrm{~T}_1=\mathrm{T}_2$
B $2 \mathrm{~T}_1=3 \mathrm{~T}_2$
C $\mathrm{T}_1=\mathrm{T}_2$
D $3 \mathrm{~T}_1=2 \mathrm{~T}_2$

The correct answer is:
$3 \mathrm{~T}_1=2 \mathrm{~T}_2$

Solution:
$\begin{aligned} & \mathrm{T}_1=2 \pi \sqrt{\frac{\ell}{\mathrm{GM}}(2 \mathrm{R})^2} \\ & \mathrm{~T}_2=2 \pi \sqrt{\frac{\ell}{\mathrm{GM}}(3 \mathrm{R})^2} \\ & \therefore \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{2}{3}\end{aligned}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (05 Apr Shift 1)
A simple pendulum has a period $T$ inside a lift, when it is stationary. The lift is accelerated upwards with constant acceleration $a$. The period

Options:
A decreases
B increases
C remains same
D becomes infinite

The correct answer is:
decreases

Solution:
The time period of a pendulum is given by
$T=\frac{1}{2 \pi} \sqrt{\frac{l}{g}}$
When a lift containing a simple pendulum is accelerated upwards with acceleration $a$, then its time period becomes
$T^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{l}{(g+a)}}$
Clearly, $\quad T^{\prime} \angle T$
Hence, its time period decreases.

Topic: Physics, Motion In One Dimension, COMEDK, COMEDK 2016
A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stopwatch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is:

Options:
A 5.40%
B 3.40%
C 4.40%
D 2.40%

The correct answer is:
4.40%

Solution:
Time period of the simple pendulum, T=2πlg ⇒g=4π2lT2

∆gg=2∆TT+∆LL⇒∆gg=2150+0.125.0

⇒∆gg=4.4%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2020 (08 Jan Shift 2)
A simple pendulum is placed inside a lift, which is moving with a uniform acceleration. If the time periods of the pendulum while the lift is moving upwards and downwards are in the ratio $1: 2$, then the acceleration of the lift is (Acceleration due to gravity, $g=10 \mathrm{~ms}^{-2}$ )

Options:
A $6 \mathrm{~ms}^{-2}$
B $0 \mathrm{~ms}^{-2}$
C $3 \mathrm{~ms}^{-2}$
D $2 \mathrm{~ms}^{-2}$

The correct answer is:
$6 \mathrm{~ms}^{-2}$

Solution:
Given,
time period of the pendulum while the lift is moving upwards and downward are in the ratio,
$$
T_1: T_2=1: 2
$$
Acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$
We know that,
If the lift is moving upward, then total time-period,
when the lift is moving downwards, then the total time period,
By dividing Eq. (i) to (ii), we get
$$
\begin{array}{rlrl}
\therefore & \frac{T_1}{T_2} & =\sqrt{\frac{g-a}{g+a}} \\
& \text { Now, } & \frac{1}{2} & =\sqrt{\frac{g-a}{g+a}}
\end{array}
$$
Square on the both sides, we get
$$
\begin{aligned}
& \text { or } & \left(\frac{1}{2}\right)^2 & =\frac{g-a}{g+a} \text { or } \frac{g-a}{g+a}=\frac{1}{4} \\
& \text { or } & 4 g-4 a & =g+a \text { or } 3 g=5 a \\
& \text { or } & a & =\frac{3 g}{5} \\
\Rightarrow & & a & =\frac{30}{5}=6 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
So, the acceleration of the lift is $6 \mathrm{~m} / \mathrm{s}^2$.

Topic: Physics, Oscillations, JEE Main
A simple pendulum is taken to 64 km above the earth's surface. It's time period will :

Options:
A Increase by 1%
B Decrease by 1%
C Increase by 2%
D Decrease by 2%

The correct answer is:
Increase by 1%

Solution:
T=2πlg=2πlR+h2GM
T∝R+h
or T'T=R+hR=1+hR=1+1100
∴ T'T-1=1100
∴ T'-TT×100=1
∴ : Time period will increase by 1%

Topic: Physics, Oscillations, JEE Main
A simple pendulum is vibrating in an evacuated chamber. It will oscillate with

Options:
A constant amplitude
B decreasing amplitude
C increasing amplitude
D amplitude, the variation of which cannot be predicted

The correct answer is:
constant amplitude

Solution:
Since, the amplitude of simple pendulum depends on the net force acting on bob. As, in the vacuum, no external air dragging force acts on bob other than gravity.
That's why amplitude remains constant in vacuum.

Topic: Physics, Oscillations, AP EAMCET, AP EAMCET 2021 (24 Aug Shift 1)
A simple pendulum made of a bob of mass m and a metallic wire of a negligible mass has a time period of 2 s at T=0°C. If the temperature of the wire is increased, and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope S. If the coefficient of linear expansion of metal is α, then the value of S is

Options:
A α2
B 2Î±
C Î±
D 1Î±

The correct answer is:
α

Solution:
△TT=12△ll=12α∆θ

⇒∆T=12Tα∆θ

∆T∆θ=T2α

But,

T=2s

∆T∆θ=α

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2016 (09 Apr Online)
A simple pendulum of length $1 \mathrm{~m}$ and having a bob of mass $100 \mathrm{~g}$ is suspended in a car, moving on a circular track of radius $100 \mathrm{~m}$ with uniform speed $10 \mathrm{~m} / \mathrm{s}$. If the pendulum makes small oscillation in a radial direction about its equilibrium position, then its time period can be given by $\mathrm{T}=2 \pi / \alpha^{1 / 4}$. The value of $\alpha$ is [Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ]
[Take g = 10 m/s2]

Options:
A 11
B 110
C 101
D 1100

The correct answer is:
101

Solution:
The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car. Acceleration due to gravity $=\mathrm{g}$ Centripetal acceleration $=\mathrm{v}^2 / \mathrm{R}$ Effective acceleration will be
$$
g^{\prime}=\sqrt{g^2+\frac{v^4}{R^2}}
$$
Time period, $T=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}^{\prime}}}$
$$
\begin{aligned}
& \Rightarrow \quad \frac{2 \pi}{2 / 4}=2 \pi \sqrt{\frac{1}{\sqrt{g^2+\frac{\mathrm{v}^4}{\mathrm{R}^2}}}} \\
& \Rightarrow \quad \alpha=\mathrm{g}^2+\frac{\mathrm{v}^4}{\mathrm{R}^2}=(10)^2+\frac{(10)^4}{(100)^2}=101
\end{aligned}
$$

Topic: Physics, Oscillations, JEE Main
A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass 10-2 kg moving with a speed of 2×102 m s-1. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use g=10 m s-2)

Options:
A 0.30 m
B 0.20 m
C 0.35 m
D 0.40 m

The correct answer is:
0.20 m

Solution:
Linear momentum will be conserved, therefore

mu=M+mV

⇒10−2×2×102≅1×V

⇒V≅2 m s-1

Again from energy conservation, we get height achieved, h=V22g=0.2m.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2024 (01 Feb Shift 1)
A simple pendulum of length $1 \mathrm{~m}$ is oscillating with an angular frequency $10 \mathrm{rad} / \mathrm{s}$. The support of the pendulum starts oscillating up and down with a small angular frequency of $1 \mathrm{rad} / \mathrm{s}$ and an amplitude of $10^{-2} \mathrm{~m}$. The relative change in the angular frequency of the pendulum is best given by :

Options:
A $10^{-3} \mathrm{rad} / \mathrm{s}$
B $1 \mathrm{rad} / \mathrm{s}$
C $10^{-1} \mathrm{rad} / \mathrm{s}$
D $10^{-5} \mathrm{rad} / \mathrm{s}$

The correct answer is:
$10^{-3} \mathrm{rad} / \mathrm{s}$

Solution:
Angular frequency of pendulum $\omega=\sqrt{\frac{g}{\ell}}$

$\therefore$ relative change in angular frequency

$\frac{\Delta \omega}{\omega}=\frac{1}{2} \frac{\Delta \mathrm{g}}{\mathrm{g}}[$ as length remains constant $]$

$\Delta g=2 \mathrm{~A} \omega_{s}^{2}\left[\omega_{s}=\right.$ angular frequency of support and, $\mathrm{A}=$

amplitude]

$\frac{\Delta \omega}{\omega}=\frac{1}{2} \times \frac{2 \mathrm{~A} \omega_{\mathrm{S}}^{2}}{\mathrm{~g}}$

$\Delta \omega=\frac{1}{2} \times \frac{2 \times 1^{2} \times 10^{-2}}{10}=10^{-3} \mathrm{rad} / \mathrm{sec}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E , as shown in figure. Its bob has mass m and charge q . The time period of the pendulum is given by:

Options:
A 2πLg2+qEm2
B 2πLg+qEm
C 2πLg-qEm
D 2πLg2-q2E2m2

The correct answer is:
2πLg2+qEm2

Solution:
Two forces will be acting on the mass m, gravity force and force by the electric field.

Thus, the effective force = qE2+mg2
From force equation, geff=g2+qEm2
Time period of the pendulum is T=2πLgeff
=2π Lg2+qEm2

Topic: Physics, Electrostatics, JEE Main
A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E , as shown in figure. Its bob has mass m and charge q . The time period of the pendulum is given by:

Options:
A 2πLg2+qEm2
B 2πLg+qEm
C 2πLg-qEm
D 2πLg2-q2E2m2

The correct answer is:
2πLg2+qEm2

Solution:
Two forces will be acting on the mass m, gravity force and force by the electric field.

Thus, the effective force = qE2+mg2
From force equation, geff=g2+qEm2
Time period of the pendulum is T=2πLgeff
=2π Lg2+qEm2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (10 Apr Shift 2)
A simple pendulum of mass ' m ', length ' l ' and charge '+q' suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be

Options:
A tan-1qmg×C1 V2-V1C1+C2(d-t)
B tan-1qmg×C2 V2-V1C1+C2(d-t)
C tan-1qmg×C2 V1+V2C1+C2(d-t)
D tan-1qmg×C1 V1+V2C1+C2(d-t)

The correct answer is:
tan-1qmg×C2 V1+V2C1+C2(d-t)

Solution:
Let E be electric field in air

Tsinθ=qE

Tcosθ=mg

tanθ=qEmg

Q=C1C2C1+C2V1+V2

E=QAÃÂµo=C1C2C1+C2V1+V2AÃÂµo

C1=ÃÂµ0 A d-t⇒E=C2 V1+V2C1+C2(d-t)

Now θ=tan-1q·Emg

θ=tan-1qmg×C2 V1+V2C1+C2(d-t)

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (27 Jul Shift 2)
A simple pendulum oscillating in air has period T . The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is 116th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is:

Options:
A 2T110
B 2T114
C 4T115
D 4T114

The correct answer is:
4T115

Solution:
Let the density of bob is σ & density of liquid is σ16 .
Now, Net downward force on bob = Weight - Buoyancy
=σVg-σ16Vg =1516σVg
Net effective gravitational acceleration = 1516σVgσV= 1516g
Time period T=2πlgeff=2πl1516g=2π1615lg=1615∙2πlg=415T
=4T115

Topic: Physics, Oscillations, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A simple pendulum performs simple harmonic motion about $\mathrm{x}=0$ with an amplitude a and time period T. The speed of the pendulum at $\mathrm{x}=\mathrm{a} / 2$ will be :

Options:
A $\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}$
B $\frac{\pi \mathrm{a} \sqrt{3}}{2 \mathrm{~T}}$
C $\frac{\pi a}{T}$
D $\frac{3 \pi^2 a}{T}$

The correct answer is:
$\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}$

Solution:
$\quad \mathrm{v}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2}$
$$
=\frac{2 \pi}{\mathrm{T}} \sqrt{\mathrm{a}^2-\frac{\mathrm{a}^2}{4}}=\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}
$$

Topic: Physics, Oscillations, NEET, NEET 2009 (Mains)
A simple pendulum performs simple harmonic motion about $\mathrm{x}=0$ with an amplitude a and time period T. The speed of the pendulum at $\mathrm{x}=\mathrm{a} / 2$ will be :

Options:
A $\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}$
B $\frac{\pi \mathrm{a} \sqrt{3}}{2 \mathrm{~T}}$
C $\frac{\pi a}{T}$
D $\frac{3 \pi^2 a}{T}$

The correct answer is:
$\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}$

Solution:
$\quad \mathrm{v}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2}$
$$
=\frac{2 \pi}{\mathrm{T}} \sqrt{\mathrm{a}^2-\frac{\mathrm{a}^2}{4}}=\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}
$$

Topic: Physics, Oscillations, JEE Main
A simple pendulum, suspended from the ceiling of a stationary van, has time period $T$. If the van starts moving with a uniform velocity the period of the pendulum will be

Options:
A Less than $T$
B Equal to $2 T$
C Greater than $T$
D Unchanged

The correct answer is:
Unchanged

Solution:
Effective value of ' $g$ ' remains unchanged

Topic: Physics, Oscillations, JEE Main
A simple pendulum with a bob of mass m and a conducting wire of length L, swings under gravity with an angular amplitude Î¸. If the horizontal component of the earth's magnetic field perpendicular to the plane of motion of the pendulum is B, then the maximum emf induced across the pendulum is

Options:
A 2 BL sinâ¡Î¸2gL1/2
B BL sinâ¡θ22gL1/2
C 2BL sinâ¡θ2gL3/2
D BL sinâ¡θ2gL2

The correct answer is:
2 BL sinâ¡Î¸2gL1/2

Solution:
h=L-L cos ⁡θ
⇒ h=L(1-cos⁡θ) ...(i)
using energy conservation ΔKE=ΔPE
⇒12mv2=mgh=mgL(1−cosθ)
v=2g L2 sin2⁡θ2
⇒ v=2gL sin⁡θ2
Thus, maximum potential difference
Vmax=BvL
=B×2gLsinθ2L
=2BL sinθ2gL1/2

Topic: Physics, Electromagnetic Induction, JEE Main
A simple pendulum with length 100 cm and bob of mass 250 g is executing S.H.M of amplitude 10 cm. The maximum tension in the string is found to be x40 N. The value of x is _____.

Options:

The correct answer is:
99

Solution:
From the given figure we can see that

sinθ0=Al=10100=110

From conservation of energy mv22=mgl(1-cosθ0)

Maximum tension occurs at mean position

T=mg+mv2l⇒T=mg+2mg(1-cosθ0)

Substituting the values,

T=mg(1+2(1-1-sin2θ0)⇒T=mg(3-21-1100)⇒T=0.25×9.8(3-2(1-1200))⇒T=0.25×9.8×1.01=2.4745 N

It is given that

T=x40⇒x=40×2.4745=98.98≈99

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (06 Apr Shift 2)
A simple pendulwm of length $L$ carries a bob of mass $m$. When the bob is at its lowest position, it is given that the minimum horizontal speed necessary for it to move in a vertical circle about the point of suspension. When the string is horizontal, the net force on the bob is

Options:
A $\sqrt{10} m g$
B $\sqrt{5} m g$
C $4 \mathrm{mg}$
D $1 \mathrm{mg}$

The correct answer is:
$\sqrt{10} m g$

Solution:
According to the question, the bob is moving in a vertical circle, so the velocity at lowest point $v=\sqrt{5 r g}$ Now, applying conservation of mechanical energy, we get
$$
\frac{1}{2} m \times 5 r g=\frac{1}{2} m v^2+m g r
$$
or,
$$
\frac{5}{2} r g=\frac{v^2}{2}+r g \text { or, } \quad \frac{3}{2} r g=\frac{v^2}{2}
$$
or,
$$
v^2=3 r g
$$
Now, Horizontal force on the bob, $F x=F_c=\frac{m v^2}{r}$
$$
=\frac{m \times 3 r g}{r}=3 m g
$$
[from Eq. (i)]
Vertical force on the bob, $F_y=m g$
$$
\begin{aligned}
\therefore \quad F_{\text {net }} & =\sqrt{F_x^2+F_y^2} \\
& =\sqrt{(3 m g)^2+(m g)^2}=\sqrt{10} m g
\end{aligned}
$$

Topic: Physics, Work Power Energy, JEE Main