datavorous/entrance-exam-dataset huggingface dataset:
A parallel plate capacitor (without dielectric) is charged by a battery and kept connected to the battery. A dielectric slab of dielectric constant \( k \) is inserted between the plates, fully occupying the space between the plates. The energy density of electric field between the plates will

Options:
A increase \( k^{2} \) times.
B decrease \( k^{2} \) times.
C increase \( k \) times.
D decrease \( k \) times.

The correct answer is:
increase \( k \) times.

Solution:
As the battery remains connected, potential difference across the plate will remain the same.

Here, C,U,V and C',U',V' are capacitance, energy density of electric field and potential difference between the plates of the capacitor before insertion and after insertion of a slab.

As the slab is introduced,

C'=(k)(C)

U=12CV2

V'=V

U'=12C'V'2

U'=12(kC)(V)2

U'=k(U)

Energy density increases k times.

Topic: Physics, Capacitance, JEE Main
A parallel plate condenser has a capacitance $50 \mu \mathrm{F}$ in air and $110 \mu \mathrm{F}$ when immersed in an oil. The dielectric constant ' $k$ ' of the oil is

Options:
A 0.45
B 0.55
C 1.10
D 2.20

The correct answer is:
2.20

Solution:
$C_{\text {medium }}=K C_{\text {air }} \Rightarrow K=\frac{C_{\text {medum }}}{C_{\text {air }}}=\frac{110}{50}=2.20$

Topic: Physics, Capacitance, JEE Main
A parallel plate condenser is immersed in an oil of dielectric constant 2 . The field between the plates is

Options:
A Increased proportional to 2
B Decreased proportional to $\frac{1}{2}$
C Increased proportional to $\sqrt{2}$
D Decreased proportional to $\frac{1}{\sqrt{2}}$

The correct answer is:
Decreased proportional to $\frac{1}{2}$

Solution:
Let before immersed in oil the capacitance is \(\mathrm{C}\), potential is \(\mathrm{V}\). and field, \(\mathrm{E}=\mathrm{V} / \mathrm{d}\) where \(\mathrm{d}=\) separation of plates.
After immersed in oil the capacitance becomes \(\mathrm{C}^{\prime}=\mathrm{kC}=2 \mathrm{C}\) and potential becomes \(\mathrm{V}^{\prime}=\mathrm{V} / 2\) because \(\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}\).
Now filed, \(E^{\prime}=V^{\prime} / d=\frac{1}{2}(V / d)=\frac{E}{2}\)

Topic: Physics, Capacitance, JEE Main
A parallel plate condenser with a dielectric of dielectric constant $\mathrm{K}$ between the plates has a capacity $\mathrm{C}$ and is charged to a potential $\mathrm{~V ~volts}$. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

Options:
A
$1 / 2(\mathrm{~K}-1) \mathrm{CV}^2$

B
$\mathrm{CV}^2(\mathrm{~K}-1) / \mathrm{K}$

C
$(\mathrm{K}-1) \mathrm{CV}^2$

D
zero

The correct answer is:
zero

Solution:
No Solution Available

Topic: Physics, Capacitance, JEE Main, JEE Main 2007
A parallelogram is constructed on the vectors \( \overrightarrow{\mathbf{a}}=3 \vec{\alpha}-\vec{\beta}, \overrightarrow{\mathbf{b}}=\vec{\alpha}+3 \vec{\beta} \), if \( |\overrightarrow{\alpha \alpha}|=|\vec{\beta}|=2 \) and angle between \( \vec{\alpha} \) and \( \vec{\beta} \) is \( \frac{\pi}{3} \), then length of diagonal of the parallelogram is

Options:
A \( 4 \sqrt{7} \)
B \( 4 \sqrt{3} \)
C \( 4 \sqrt{17} \)
D None of these

The correct answer is:
\( 4 \sqrt{3} \)

Solution:
Let ABâ=aâ=3Â Î±â-Î²â,Â BCâ=bâ=Î±â+3Î²â
Diagonal ACâ=ABâ+BCâ=aâ+bâ
â¹ACâ=aâ+bâ
â¹ACâ=4Â Î±â+2Î²â
â¹ACâ2=16Î±â2+4Î²â2+16Î±â⋅Î²â
â¹ACâ2=64+16+16Î±âÎ²âcosÏ3
â¹ACâ2=80+16Ã4Ã12=112
â¹Â ACâ=47
Other diagonal is |BDâ=|aâ-bâ|
â¹BDâ2=|2Î±â-4Î²â|2
=4Î±â|2+16Î²â|2-16Â Î±âÎ²âcosâ¡Ï3
=64+16-16Ã4Ã12=48
â¹|BDâ=48=43

Topic: Mathematics, Vector Algebra, JEE Main
A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k1, k2, k3 and k4 as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by

Options:
A k=k1+k2+k3+3k4
B k=23k1+k2+k3+2k4
C 2k=3k1+k2+k3+1k4
D k=2k43k1k1+k4+k2k2+k4+k3k3+k4

The correct answer is:
k=2k43k1k1+k4+k2k2+k4+k3k3+k4

Solution:
Here the capacitance, \(C=\frac{A k \epsilon_0}{d}\)
\(C_{K_1}, C_{K_2}, C_{K_3}\) are in parallel combination and they are connected in series with \(\mathrm{C}_{\mathrm{K}_4}\)

Here, \(\mathrm{C}_{\mathrm{K}_1}=\frac{(\mathrm{A} / 3) \mathrm{K}_1 \epsilon_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{~K}_1}{3} \mathrm{C}\),
\(\mathrm{C}_{\mathrm{k}_2}=\frac{(\mathrm{A} / 3) \mathrm{k}_2 \mathrm{\epsilon}_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_2}{3} \mathrm{C}\)
\(C_{K_3}=\frac{(\mathrm{A} / 3) \mathrm{k}_3 \epsilon_0}{\mathrm{~d} / 2} \frac{2 \mathrm{k}_3 \mathrm{C}}{3} \mathrm{C}\)
\(\mathrm{C}_{\mathrm{K}_4}=\frac{(\mathrm{A}) \mathrm{k}_4 \epsilon_0}{\mathrm{~d} / 2}=2 \mathrm{k}_4 \mathrm{C}\)
Now the equivalent capacitance for the combination of four capacitors is
\(\begin{aligned}
& \frac{1}{C_{c q}}=\frac{1}{\left(C_{k_1}+C_{k_2}+C_{k_3}\right)}+\frac{1}{C_{k_4}} \\
& C_{e q}=k C
\end{aligned}\)
\(\frac{1}{k C}=\frac{3}{2 C}\left[\frac{1}{k_1+k_2+k_3}\right]+\frac{1}{2 k_4 C}\)
or \(\frac{1}{k}=\frac{3}{2\left(k_1+k_2+k_3\right)}+\frac{1}{2\left(k_4\right)}\)
or \(k=\frac{2\left(k_1+k_2+k_3\right)}{3}+2\left(k_4\right)\)

Topic: Physics, Capacitance, JEE Main
A parallel-plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K1 and K2 of same area A2 and thickness d2 are inserted in the space between the plates. The capacitance of the capacitor will be given by :

Options:
A ε0 A d12+K1 K2 K1+K2
B ε0 A d12+2 K1+K2K1 K2
C ε0 A d12+K1+K2 K1 K2
D ε0 A d12+K1 K22 K1+K2

The correct answer is:
ε0 A d12+K1 K2 K1+K2

Solution:
Ceq=C1+C2C3C2+C3

Ceq=Aε02 d+K1 Aε0 d×K2 Aε0 dK1 Aε0 d+K2 Aε0 d

Ceq=Aε02 d+Aε0 dK1 K2 K1+K2=Aε0 d12+K1 K2 K1+K2

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A paramagnetic material has 1028 atoms m-3. Its magnetic susceptibility at temperature 350 K is 2.8Ãâ10-4. Its susceptibility at 300 K is

Options:
A 3.726x10-4
B 2.672x10-4
C 3.267x10-4
D 3.672x10-4

The correct answer is:
3.267x10-4

Solution:
From Curie's law, the magnetic susceptibility is inversely proportional to the temperature, i.e.,

χ∝1T

Hence, the ratio of magnetic susceptibility at different temperature T1=350 K and T2=300 K.

χ2χ1=T1T2

χ2=χ1×T1T2=2.8×10-4×350300

χ2=3.267×10-4.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2019 (12 Jan Shift 2)
A paramagnetic material shows a net magnetization of 8 A m-1 when placed in an external magnetic field of intensity 0.6 T at a temperature of 4 K . When the same material is placed in an external magnetic field of intensity 0.2 T at a temperature of 16 K , the magnetization will be

Options:
A 323 A m-1
B 23 A m-1
C 6 A m-1
D 2.4 A m-1

The correct answer is:
23 A m-1

Solution:
According to Curie's law,

Magnetization, I∝Magnetic Induction(B)Absolute temperature(T)

⇒I2I1=B2T1B1T2

⇒I28=0.2×40.6×16

⇒I2=23 A m-1

Topic: Physics, Magnetic Properties of Matter, JEE Main
A paramagnetic sample shows a net magnetisation of 6 A/m when it is placed in an external magnetic field of 0.4 T at a temperature of 4 K. When the sample is placed in an external magnetic field of 0.3 T at a temperature of 24 K, then the magnetisation will be :

Options:
A 1 A/m
B 4 A/m
C 2.25 A/m
D 0.75 A/m

The correct answer is:
0.75 A/m

Solution:
According to Curie's Law,
χ∝1T
Therefore, χ1T1=χ2T2
Also, χ=MB
So,
M1T1B1=M2T2B2
On substituting the values,
6×40.4=M2×240.3

Therefore, M2=0.75 A m-1

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2020 (04 Sep Shift 2)
A paramagnetic substance in the form of a cube with sides $1 \mathrm{~cm}$ has a magnetic dipole moment of $20 \times 10^{-6} \mathrm{~J} / \mathrm{T}$ when a magnetic intensity of $60 \times 10^{3} \mathrm{~A} / \mathrm{m}$ is applied. Its magnetic susceptibility is:

Options:
A $3.3 \times 10^{-2}$
B $4.3 \times 10^{-2}$
C $2.3 \times 10^{-2}$
D $3.3 \times 10^{-4}$

The correct answer is:
$3.3 \times 10^{-4}$

Solution:
Magnetic susceptibility,

$\chi=\frac{\mathrm{I}}{\mathrm{H}}$

where, $\mathrm{I}=\frac{\text { Magnetic moment }}{\text { Volume }}=\frac{20 \times 10^{-6}}{10^{-6}}=20 \mathrm{~N} / \mathrm{m}^{2}$

Now , $\chi=\frac{20}{60 \times 10^{3}}=\frac{1}{3} \times 10^{-3}=3.3 \times 10^{-4}$

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A paramagnetic substance, in the form of a cube with sides 1 cm, has a magnetic dipole moment of 20Ã10-6 J T-1, when a magnetic intensity of 60Ã103 A m-1 is applied. Its magnetic susceptibility is

Options:
A 3.3×10-4
B 2.3×10-2
C 4.3×10-2
D 3.3×10-2

The correct answer is:
3.3×10-4

Solution:
M= Magnetic moment per unit volume =20Ã10-61Ã10-23 J T-1 m-3
=20 J T-1 m-3
H=60Ã103 A m-1
Ï=MH
=2060Ã103
=3.3Ã10-4

Topic: Physics, Magnetic Properties of Matter, JEE Main
A part of a complete circuit is shown in the figure. At some instant, the value of current I is 1A and it is decreasing at a rate of 102As-1. The value of the potential difference Vp-VQ, (in volts) at that instant is-

Options:

The correct answer is:
33

Solution:
Vp-Ldldt-30+RI=VQ

VP-Va=+50×10-3×102+30-1×2

= + 5 + 30 – 2 = 33

Topic: Physics, Alternating Current, JEE Main, JEE Main 2020 (06 Sep Shift 1)
A particle 'A' has a de-Broglie wavelength \( 1 Å \). If the particle B has mass \( 25 \% \) of that of A and velocity \( 75 \% \) of that of A, calculate the approximate de-Broglie wavelength of B.

Options:
A 1. \( 6 Å \)
B 5. \( 3 Å \)
C 3. \( 5 Å \)
D \( 5.9 Å \)

The correct answer is:
5. \( 3 Å \)

Solution:
λA=hmAvA=1 Å . ...........(i)
mB=0.25mA,vB=0.75vA
λB=h0.25mA×0.75vA ............(ii)
Dividing (i) by (ii)
λAλB=0.25×0.751
Or λB=λA0.1875=10.1875=5.33 A∘.

Topic: Chemistry, Structure of Atom, JEE Main
A particle A is projected from the ground with an initial velocity of 10 m s-1 at an angle 60Â° with horizontal. From what height (in m) directly above the point of projection of the particle A, should another particle B be projected horizontally with velocity 5 m s-1 so that both the particles collide right before they strike the ground? [g=10 m s-2]

Options:

The correct answer is:
15

Solution:
Both the particles will meet at C, if their time of flight is the same. The time of flight of A is
T=2usinÎ¸g=2Ã10Ãsin 6 0 â 10=3s
For vertical downward motion of particle B from B to C
We have
h=12gT2=12 Ã10 Ã32=15 m

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle A of mass m and charge q is accelerated by a potential difference of 50 V. Another particle B of mass 4m and charge q is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths Î»AÎ»B is close to:

Options:
A 0.07
B 10.00
C 4.47
D 14.14

The correct answer is:
14.14

Solution:
de-Broglie wavelength λ=hρ
=h2meV
λAλB=4m×2500m×50
=2×52
=14.14

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (12 Jan Shift 1)
A particle A of mass m and charge q is accelerated by a potential difference of 50 V. Another particle B of mass 4m and charge q is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths Î»AÎ»B is close to:

Options:
A 0.07
B 10.00
C 4.47
D 14.14

The correct answer is:
14.14

Solution:
de-Broglie wavelength λ=hρ
=h2meV
λAλB=4m×2500m×50
=2×52
=14.14

Topic: Physics, Dual Nature of Matter, JEE Main
A particle A of mass m and initial velocity v collides with a particle B of mass m2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is:

Options:
A λAλB=12
B λAλB=13
C λAλB=2
D λAλB=23

The correct answer is:
λAλB=2

Solution:
mV=m V A + m 2 V B
V= V A + V B 2    −−−−−−−−−−−−  (i)
V= V B − V A     −−−−−−−−−−−−  (ii)
Solving (i) and (ii)
2V= 3 V B 2
∴ V B = 4V 3
V A = V B −V= V 3
P A = mV 3
P B = m 2 . 4V 3
λ A λ B = h× P B P A ×h = 2mV 3 mV 3 =2

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2017 (02 Apr)
A particle A of mass m and initial velocity v collides with a particle B of mass m2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is:

Options:
A λAλB=12
B λAλB=13
C λAλB=2
D λAλB=23

The correct answer is:
λAλB=2

Solution:
mV=m V A + m 2 V B
V= V A + V B 2    −−−−−−−−−−−−  (i)
V= V B − V A     −−−−−−−−−−−−  (ii)
Solving (i) and (ii)
2V= 3 V B 2
∴ V B = 4V 3
V A = V B −V= V 3
P A = mV 3
P B = m 2 . 4V 3
λ A λ B = h× P B P A ×h = 2mV 3 mV 3 =2

Topic: Physics, Dual Nature of Matter, JEE Main
A particle A of mass m and initial velocity v collides with a particle B of mass m2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is:

Options:
A λAλB=12
B λAλB=13
C λAλB=2
D λAλB=23

The correct answer is:
λAλB=2

Solution:
mV=m V A + m 2 V B
V= V A + V B 2 âââââââââââââââââ(i)
V= V B â V A ââââââââââââââââââ(ii)
Solving (i) and (ii)
2V= 3 V B 2
â´ V B = 4V 3
V A = V B âV= V 3
P A = mV 3
P B = m 2 . 4V 3
Î» A Î» B = hÃ P B P A Ãh = 2mV 3 mV 3 =2

Topic: Physics, Dual Nature of Matter, JEE Main
A particle A suffers an oblique elastic collision with a particle B which is initially at rest. If their masses are same, then after the collision

Options:
A they will move in opposite directions
B A continues to move in the original direction while B remains at rest
C they will move in mutually perpendicular directions
D A comes to rest and B starts moving in the direction of the original motion of A

The correct answer is:
they will move in mutually perpendicular directions

Solution:
as collision is elastic
12mu2=12mv12+12v22
u2=v12+v22 .........(1)
Momentum is conserved
muâ=mvâ1+mvâ2
uâ=vâ1+vâ2
u2=v12+v22+2v1.v2cosâ¡Î¸ .......(2)
Solving (1) and (2)
2v1.v2cosâ¡Î¸=0
cosâ¡Î¸=0
Î¸=90o
i.e. v1 & v2 are mutually perpendicular.

Topic: Physics, Center of Mass Momentum and Collision, NEET
A particle acted on by constant forces $4 \hat{i}+\hat{j}-3 \hat{k}$ and $3 \hat{i}+\hat{j}-\hat{k}$ to the point $5 \hat{i}+4 \hat{j}-\hat{k}$. The total work done by the forces is

Options:
A
50 units

B
20 units

C
30 units

D
40 units

The correct answer is:
40 units

Solution:
$\overrightarrow{\mathrm{F}}+\overrightarrow{\mathrm{F}}_1+\overrightarrow{\mathrm{F}}_2=7 \mathrm{i}+2 \mathrm{j}-4 \mathrm{k}$
$\overrightarrow{\mathrm{d}}=\mathrm{P} \cdot \mathrm{V}$ of $\overrightarrow{\mathrm{B}}-\mathrm{P} \cdot \mathrm{V}$ of $\overrightarrow{\mathrm{A}}=4 \mathrm{i}+2 \mathrm{j}-2 \mathrm{k}$
$\mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{d}}=28+4+8=40$ unit

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2003
A particle aimed at a target, projected with an angle 15Â° with the horizontal is short of the target by 10 m. If projected with an angle of 45Â° is away from the target by 15 m, then the angle of projection to hit the target is

Options:
A 12sin-1⁡110
B 12sin-1⁡310
C 12sin-1⁡910
D 12sin-1⁡710

The correct answer is:
12sin-1⁡710

Solution:
R1=R-10
R2=R+15
R-10R+15=sin⁡2θ1sin⁡2θ2=sin⁡30osin⁡90o=12
2R-20=R+15
R=35
Rmax=u2g
R=u2sin⁡2θg
50=u2g
35=50g×sin⁡2θg
u2=50 g
sin⁡2θ=3550=710
2θ=sin-1⁡710
θ=12sin-1⁡710

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle at a distance of $1 \mathrm{~m}$ from the origin starts moving such that $\mathrm{dr} / \mathrm{d} \theta=\mathrm{r}$, where $(\mathrm{r}, \theta)$ are polar coordinates. Then the angle between resultant velocity and tangential velocity component is

Options:
A 30 degrees
B 45 degrees
C 60 degrees
D Dependent on where the particle is

The correct answer is:
45 degrees

Solution:
$\overrightarrow{\mathrm{AB}}=$ Direction of resultant velocity $\overrightarrow{\mathrm{AD}}=$ Direction of tangential velocity
$\begin{array}{c}
\forall \tan \alpha=\frac{\mathrm{dr}}{\operatorname{rd} \theta}=\frac{\mathrm{r}}{\mathrm{r}} \\
\tan \alpha=1 \\
\alpha=45^{\circ}
\end{array}$

Topic: Physics, Rotational Motion, JEE Main
A particle at the end of a spring executes simple harmonic motion with a period $t_1$, while the corresponding period for another spring is $t_2$. If the period of oscillation with the two springs in series is $\mathrm{t}$, then

Options:
A
$\mathrm{T}=\mathrm{t}_1+\mathrm{t}_2$

B
$\propto \mathrm{x}^2$

C
$\mathrm{T}^{-1}=\mathrm{t}_1^{-1}+\mathrm{t}_2^{-1}$

D
$\mathrm{T}^{-2}=\mathrm{t}_1^2+\mathrm{t}_2^2$

The correct answer is:
$\propto \mathrm{x}^2$

Solution:
$t_1^2+t_2^2=T^2$

Topic: Physics, Oscillations, JEE Main, JEE Main 2004
A particle at the end of a spring executes simple harmonic motion with a period t1, while the corresponding period for another spring is t2. If the period of oscillation with the two springs in series is T, then -

Options:
A T=t1+t2
B T2=t12+t22
C T-1=t1-1+t2-1
D T-2=t1-2+t2-2

The correct answer is:
T2=t12+t22

Solution:
t1=2πmk1 ….i, t2=2πmk2 ….ii
when springs are in series then
T=2πmk1k2k1+k2=2πmk1+k2k1k2
squaring and adding (i) and (ii) we get
t12+t22=4π2mk1+4π2mk2
=4π2m k1+k2k1k2 or t12+t22=T2

Topic: Physics, Oscillations, JEE Main
A particle covers a distance from $A$ to $B$ over a period of time; the distance versus time plot is the shown below. Then which of the following is true for the motion of the particle?

Options:
A Both average speed and instantaneous speed are always zero.
B Average speed is always non-zero but instantaneous speed can be zero.
C Instantaneous speed is always non-zero but average speed can be zero.
D Both average speed and instantaneous speed are always non-zero.

The correct answer is:
Average speed is always non-zero but instantaneous speed can be zero.

Solution:
The distance versus time plot for movement of a particle from $A$ and $B$ is shown in the diagram.
From graph, it is a slope of distance-time graph for position $A$ to $C$ and $D$ to $B$ is always positive, therefore both the average speed and instantaneous speed are not always zero, therefore, option (1) is not correct.
$$
\text { Average velocity }=\frac{\text { Total displacement from A to B }}{\text { Total time interval }}
$$
Since, total displacement depends on straight distance between initial and final points, hence, displacement between $A$ to $B$ always non zero, therefore the average speed is always non zero. Since, no distance is travelled between point $C$ to $D$ is zero. Hence, instantaneous speed at every instant between $C$ to $D$ is zero.
Hence, option (2) is correct and option (3) and (4) are incorrect.

Topic: Physics, Motion In One Dimension, JEE Main
A particle covers half of its total distance with speed $v_1$ and the rest half distance with speed $v_2$. Its average speed during the complete journey is

Options:
A $\frac{y_1 y_2}{y_1+y_2}$
B $\frac{2 y_1 y_2}{y_1+y_2}$
C $\frac{2 y_1^2 y_2^2}{y_1^2+y_2^2}$
D $\frac{y_1+y_2}{2}$

The correct answer is:
$\frac{2 y_1 y_2}{y_1+y_2}$

Solution:
Velocity $y=\frac{s}{t} \Rightarrow s=v t$ The average speed of particle
$$
\begin{aligned}
& y_{\mathrm{av}}=\frac{s+s}{\frac{s}{y_1}+\frac{s}{y_2}} \\
& y_{\mathrm{av}}=\frac{2 y_1 y_2}{y_1+y_2}
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, NEET, NEET 2011 (Mains)
A particle describes a horizontal circle in a conical funnel, whose inner surface is smooth, with a speed of 0.5 m s-1. What is the height (in cm) of the plane of the circle from the vertex of the funnel?

Options:

The correct answer is:
2.5

Solution:
Nsinâ¡Î¸=mg
Ncosâ¡Î¸=mv2/r
Dividing above equations
â tanâ¡Î¸=rg/v2
â rh=rgv2
â h=v2g=2.5 cm

Topic: Physics, Laws of Motion, JEE Main
A particle executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude

Options:
A $T / 2$
B $T / 4$
C $T / 8$
D $T / 12$

The correct answer is:
$T / 12$

Solution:
$y=A \sin \omega t=\frac{A \sin 2 \pi}{T} t \Rightarrow \frac{A}{2}=A \sin \frac{2 \pi t}{T} \Rightarrow t=\frac{T}{12}$

Topic: Physics, Oscillations, JEE Main
A particle executes SHM of amplitude A. The distance from the mean position when it's kinetic energy becomes equal to its potential energy is

Options:
A $\sqrt{2 \mathrm{~A}}$
B $2 \mathrm{~A}$
C $\frac{1}{\sqrt{2}} \mathrm{~A}$
D $\frac{1}{2} \mathrm{~A}$

The correct answer is:
$\frac{1}{\sqrt{2}} \mathrm{~A}$

Solution:
Let the distance from the mean position is $\mathrm{X}$.
Given $\mathrm{KE}=\mathrm{PE}$
So, $\frac{1}{2} M \omega^2\left(A^2-x^2\right)=\frac{1}{2} M \omega^2 x^2$
$A^2-x^2=x^2 \Rightarrow A^2=2 \times 2$
$\therefore \mathrm{x}= \pm \frac{\mathrm{A}}{\sqrt{2}}$

Topic: Physics, Oscillations, JEE Main
A particle executes SHM of amplitude A. The distance from the mean position when its kinetic energy becomes equal to its potential energy is:

Options:
A 12A
B 2A
C 2A
D 12A

The correct answer is:
12A

Solution:
The potential energy is given by

U=mω2x22

The kinetic energy is given by K=mω2(A2-x2)2.

Using the condition that K=U,

12mω2A2-x2=12mω2x2

⇒x=A2

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (13 Apr Shift 2)
A particle executes S.H.M., the graph of velocity as a function of displacement is :

Options:
A a circle
B a helix
C a parabola
D an ellipse

The correct answer is:
an ellipse

Solution:
v2=ω2A2-x2

v2ω2+x2=A2
v2ωA2+x2A2=1

This is an equation of an ellipse.

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (26 Feb Shift 2)
A particle executes S.H.M. with a period of 6 second and amplitude of $3 \mathrm{~cm}$. Its maximum speed in $\mathrm{cm} / \mathrm{sec}$ is

Options:
A $\pi / 2$
B $\pi$
C $2 \pi$
D $3 \pi$

The correct answer is:
$\pi$

Solution:
$v_{\max }=a \omega=a \frac{2 \pi}{T}=3 \times \frac{2 \pi}{6}=\pi \mathrm{cm} / \mathrm{s}$

Topic: Physics, Oscillations, JEE Main
A particle executes S.H.M. with amplitude A and time period T. The displacement of the particle when its speed is half of maximum speed is xA2. The value of x is

Options:

The correct answer is:
3

Solution:
V=ωA2-x2 Vmax=Aω

Aω2=ωA2-x2

A24=A2-x2

x2=3 A24

x=32 A

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (26 Feb Shift 2)
A particle executes simple harmonic motion along a straight line with an amplitude $A$. The potential energy is maximum when the displacement is

Options:
A $\pm A$
B Zero
C $+\frac{A}{2}$
D $+\frac{A}{\sqrt{2}}$

The correct answer is:
$\pm A$

Solution:
P.E. $=\frac{1}{2} m \omega^2 x^2$
It is clear P.E. will be maximum when $x$ will be maximum i.e., at $x= \pm A$

Topic: Physics, Oscillations, JEE Main
A particle executes simple harmonic motion and it is located at x=a, b and c at time t0, 2t0 and 3t0 respectively. The frequency of the oscillation is:

Options:
A 12πt0cos-1⁡a+c2b
B 12πt0cos-1⁡a+2b3c
C 12πt0cos-1⁡a+b2c
D 12πt0cos-1⁡2a+3cb

The correct answer is:
12πt0cos-1⁡a+c2b

Solution:
a=Asin⁡ωt0...(1)
b=Asin⁡2ωt0...(2)
c=Asin⁡3ωt0...(3)

adding equation (1) and (3)
a+c=Asin⁡ωt0+sin⁡3ωt0=2Asin⁡2ωt0cos⁡ωt0...(4)
∵  sinC+sinD=  2sinC+D2cosC−D2

from equation (2) and (4)

a+cb=2cosωt0
ω= 1t0cos-1⁡a+c2b ⇒f=12πt0cos-1⁡a+c2b

Topic: Physics, Oscillations, JEE Main
A particle executes simple harmonic motion and it is located at x=a, b and c at time t0, 2t0 and 3t0 respectively. The frequency of the oscillation is:

Options:
A 12πt0cos-1⁡a+c2b
B 12πt0cos-1⁡a+2b3c
C 12πt0cos-1⁡a+b2c
D 12πt0cos-1⁡2a+3cb

The correct answer is:
12πt0cos-1⁡a+c2b

Solution:
a=Asin⁡ωt0...(1)
b=Asin⁡2ωt0...(2)
c=Asin⁡3ωt0...(3)

adding equation (1) and (3)
a+c=Asin⁡ωt0+sin⁡3ωt0=2Asin⁡2ωt0cos⁡ωt0...(4)
∵  sinC+sinD=  2sinC+D2cosC−D2

from equation (2) and (4)

a+cb=2cosωt0
ω= 1t0cos-1⁡a+c2b ⇒f=12πt0cos-1⁡a+c2b

Topic: Physics, Oscillations, JEE Main, JEE Main 2018 (16 Apr Online)
A particle executes simple harmonic motion and it is located at x=a, b and c at time t0, 2t0 and 3t0 respectively. The frequency of the oscillation is:

Options:
A 12πt0cos-1⁡a+c2b
B 12πt0cos-1⁡a+2b3c
C 12πt0cos-1⁡a+b2c
D 12πt0cos-1⁡2a+3cb

The correct answer is:
12πt0cos-1⁡a+c2b

Solution:
a=Asinâ¡Ït0
b=Asinâ¡2Ït0
c=Asinâ¡3Ït0
a+c=Asinâ¡Ït0+sinâ¡3Ït0=2Asinâ¡2Ït0cosâ¡Ït0
âµââsinC+sinD
=ââ2sin C+D 2 cos CâD 2
Ï= 1t0cos-1â¡a+c2b âf=12Ït0cos-1â¡a+c2b

Topic: Physics, Oscillations, JEE Main