datavorous/entrance-exam-dataset huggingface dataset:
A single current carrying loop of wire carrying current I flowing in anticlockwise direction seen from +ve z direction and lying in xy plane is shown in figure. The plot of j^ component of magnetic field (By) at a distance a (less than radius of the coil) and on yz plane vs z coordinate looks like

Options:
A
B
C
D

The correct answer is:

Solution:
Magnetic field in plane of coil is By = 0

The direction of j^ component of magnetic field By is opposite of each other in –z and +z positions.

So, the variation of magnetic field is shown below.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A single letter is selected at random from the word "PROBABILITY". The probability that the selected letter is a vowel is

Options:
A $\frac{2}{11}$
B $\frac{3}{11}$
C $\frac{4}{11}$
D $0$

The correct answer is:
$\frac{4}{11}$

Solution:
Since there are one $A$, two 1 and one $O$, hence the required probability $=\frac{1+2+1}{11}=\frac{4}{11}$.

Topic: Mathematics, Probability, JEE Main
A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000 Å and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is:

Options:
A 9 mm
B 3 mm
C 4.5 mm
D 1.5 mm

The correct answer is:
9 mm

Solution:
a=0.1 mm=10-4 m
λ=6000×10-10=6×10-7 m
D=0.5 m
For 3rd dark
asin⁡θ=3λ
sin⁡θ=3λa=xD
x=3λDa
=3×6×10-7×0.510-4
=9 mm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2017 (09 Apr Online)
A single slit of width a is illuminated by a monochromatic light of wavelength 600 nm. The value of a for which first minimum appears at θ=30o on the screen will be :

Options:
A 1.2 μm
B 3 μm
C 1.8 μm
D 0.6 μm

The correct answer is:
1.2 μm

Solution:
For the first minima

a sinθ=λ

The given data is

θ=30°λ=600 nm

The value of a is

a=λsin30°=600×10-912=1.2 μm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2023 (15 Apr Shift 1)
A single slit of width b is illuminated by a coherent monochromatic light of wavelength λ. If the second and fourth minima in the diffraction pattern at a distance 1 cm from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum? (i.e. distance between first minimum on either side of the central maximum)

Options:
A 4.5 cm
B 1.5 cm
C 6.0 cm
D 3.0 cm

The correct answer is:
3.0 cm

Solution:
For secondary minima dsin⁡θ=nλ
sin⁡θ=nλd
For second minima
n=2
sin⁡θ=2λd=tan⁡θ1=x1D
For fourth minima n=4
sin⁡θ2=4λd=x2D
Distance between both minima is x2-x1=4λd-2λd=2λd= 6 -3 =3 cm
Width of central maxima,=2λd=3 cm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2017 (08 Apr Online)
A single turn current loop in the shape of a right angle triangle with sides 5 cm, 12 cm, 13 cm is carrying a current of 2 A. The loop is in a uniform magnetic field of magnitude 0.75 T whose direction is parallel to the current in the 13 cm side of the loop. The magnitude of the magnetic force on the 5 cm side will be x130 N. The value of x is _______.

Options:

The correct answer is:
9

Solution:
Here, B=0.75 T

The magnitude of magnetic force on the 5 cm side is FAB=i×lABBsinθ angle between AC & AB

=2×5100×0.75×1213

=9130 N

Hence, the value of x=9.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (24 Jan Shift 2)
A singly ionized magnesium atom A=24 ion is accelerated to kinetic energy 5keV, and is projected perpendicularly into a magnetic field $B$ of the magnitude 0.5 T. The radius of path formed will be _____ cm.

Options:

The correct answer is:
10

Solution:
The radius of circle formed by a charged particle in a magnetic field perpendicular to the velocity is given by,

R=mVqB⇒R=2mKqB=2×24×1.6×10-27×5000×1.6×10-191.6×10-19×0.5=979×10-4 m≃10 cm

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (28 Jun Shift 1)
A sinusoidal carrier voltage is amplitude modulated. The resultant amplitude modulated wave has maximum and minimum amplitude of 120 V and 80 V respectively. The amplitude of each side band is :

Options:
A 15 V
B 10 V
C 20 V
D 5 V

The correct answer is:
10 V

Solution:
Given here,

Ac+Am=120 ...1

Ac-Am=80 ...2

From equation 1 and 2, we have

2Ac=200⇒Ac=100 V and Am=20 V

Modulation index μ=20100=15

Amplitude of each side band

=Acμ2

=100×110=10 volt

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (30 Jan Shift 1)
A sinusoidal voltage of peak value 250 V is applied to a series LCR circuit, in which R=8 Ω,L=24 mH and C=60 μF. The value of power dissipated at resonant condition is x kW. The value of x to the nearest integer is _______.

Options:

The correct answer is:
4

Solution:
At resonance power, P=Vrms2R

P=250/228=3906.25 W

≈4 kW

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (16 Mar Shift 1)
A sinusoidal voltage of peak value 283 V and angular frequency 320 s-1 is applied to a series LCR circuit. Given that R=5 Ω, L=25 mH and C=1000 μF. The total impedance and phase difference between the voltage across the source and the current will respectively be-

Options:
A 10 Ω and tan-1⁡53
B 7 Î© and 45o
C 7 Î© and tan-1â¡53
D 10 Î© and tan-1â¡83

The correct answer is:
7 Î© and 45o

Solution:
ε0=283 volt, ω=320 s-1
XL=320×25×10-3=8 Ω
XC=1ωC=1320×1000×10-6
XC=10000320=3.1 Ω
R=5Ω
Z= R2+XL-XC2= 50 ≈ 7Ω
tan⁡ϕ=XL-XCR=1
⟹ϕ=45o

Topic: Physics, Alternating Current, JEE Main, JEE Main 2017 (09 Apr Online)
A sinusoidal voltage $\mathrm{V}(\mathrm{t})=100 \sin (500 \mathrm{t})$ is applied across a pure inductance of $L=0.02 \mathrm{H}$. The current through the coil is:

Options:
A
$10 \cos (500 \mathrm{t})$

B
$-10 \cos (500 \mathrm{t})$

C
$10 \sin (500 t)$

D
$-10 \sin (500 t)$

The correct answer is:
$-10 \cos (500 \mathrm{t})$

Solution:
In a pure inductive circuit current always lags behind the emf by $\frac{\pi}{2}$.
If $\mathrm{v}(\mathrm{t})=\mathrm{v}_0 \sin \omega \mathrm{t}$ then $I=I_0 \sin \left(\omega t-\frac{\pi}{2}\right)$ Now, given $v(t)=100 \sin (500 t)$ and $\mathrm{I}_0=\frac{\mathrm{E}_0}{\omega \mathrm{L}}=\frac{100}{500 \times 0.02}[\because \mathrm{L}=0.02 \mathrm{H}]$ $\mathrm{I}_0=10 \sin \left(500 \mathrm{t}-\frac{\pi}{2}\right)$ $\mathrm{I}_0=-10 \cos (500 \mathrm{t})$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2014 (12 Apr Online)
A sinusoidal voltage Vt=210sin3000t volt is applied to a series LCR circuit in which L=10 mH,C=25 μF and R=100Ω. The phase difference Φ between the applied voltage and resultant current will be

Options:
A tan-10.17
B tan-19.46
C tan-10.30
D tan-113.33

The correct answer is:
tan-10.17

Solution:
Capacitive reactance is XC=1ωC=13000×25×10-6=403Ω

Inductive reactance is XL=ωL=3000×10×10-3=30Ω

Now, total reactance is X=XL-XC

=30-403=503Ω

tanϕ=XR=503100=16

tanϕ=16=0.17

Hence, the phase difference between the applied voltage and resultant current is tan-10.17.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (25 Jun Shift 2)
A sinusoidal wave yt=40sin10×106πt is amplitude modulated by another sinusoidal wave xt=20sin1000πt. The amplitude of minimum frequency component of modulated signal is

Options:
A 0.5
B 0.25
C 20
D 10

The correct answer is:
10

Solution:
Here, modulation index, μ=AmAc=2040=12

Modulated amplitude =μAc2

=12×402=10

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (26 Jun Shift 2)
A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?

Options:

None

Solution:
The wire is stretched both end so frequency of stretched wire is
$$
v=\frac{n}{2 l} \sqrt{\frac{T}{m}}
$$
Mass of wire, $M=v p=A l \rho=\pi r^2 l \rho$
Mass per unit length
$$
m=\frac{\text { Mass of wire }}{\text { Length }}=\frac{\left(\pi \pi^2 l\right) \rho}{l}=\pi r^2 \rho
$$
So the ratio of $m_2$ and $m_1$ is
$$
\begin{aligned}
&\therefore \frac{m_2}{m_1}=\frac{\pi r_2^2 \rho}{\pi r_1^2 \rho} \\
&\frac{m_2}{m_1}=\left(\frac{r_2}{r_1}\right)^2 \Rightarrow \frac{r_2}{r_1}=\sqrt{\frac{m_2}{m_1}}
\end{aligned}
$$
We also know that
$$
v \propto \frac{1}{\sqrt{m}} \Rightarrow \frac{v_1}{v_2}=\frac{\sqrt{m_2}}{\sqrt{m_1}}
$$
So, $\frac{v_1}{v_2}=\frac{r_2}{r_1} \Rightarrow \frac{v_1}{v_2}=\frac{3}{1} \Rightarrow v_2=\frac{v_1}{3}$

Topic: Physics, Waves and Sound
A six faced die is biased such that 3×P(a prime number)=6×P(a composite number)=2×P1. Let X be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of X is

Options:
A 311
B 511
C 711
D 811

The correct answer is:
811

Solution:
Let Pa prime number2=Pa composite number1=P13=k

So, P(a prime number)=2k,

P (a composite number)=k,

and P1=3k

i.e. 6k+2k+3k=1

⇒k=111

P(success)=P1 or 4=3k+k=411

Number of trials, n=2

∴ required mean=np=2×411=811

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (27 Jul Shift 2)
A skew-symmetric matrix of order n has the maximum number of distinct elements equal to 73, then the order of the matrix is

Options:
A 7
B 8
C 9
D 10

The correct answer is:
9

Solution:
If the order of skew-symmetric matrix is n, then the maximum number of distinct elements is n2-n+1
n2-n+1=73
n2-n-72=0
n=9,-8 (rejected)

Topic: Mathematics, Matrices, JEE Main
A slab is subjected to two forces $\overrightarrow{\mathrm{F}_{1}}$ and $\overrightarrow{\mathrm{F}_{2}}$ of same

magnitude $F$ as shown in the figure. Force $\overrightarrow{\mathrm{F}_{2}}$ is in XY-

plane while force $\mathrm{F}_{1}$ acts along $z$ -axis at the point

$(2 \vec{i}+3 \vec{j})$. The moment of these forces about point $\mathrm{O}$ will be:

Options:
A $(3 \hat{i}-2 \hat{j}+3 \hat{k})$ F
B $(3 \hat{i}-2 \hat{j}-3 \hat{k}) \mathrm{F}$
C $(3 \hat{i}+2 \hat{j}-3 \hat{k}) \mathrm{F}$
D $(3 \hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{F}$

The correct answer is:
$(3 \hat{i}-2 \hat{j}+3 \hat{k})$ F

Solution:
Given, $\overrightarrow{\mathrm{F}}_{1}=\frac{\mathrm{F}}{2}(-\hat{\mathrm{i}})+\frac{\mathrm{F} \sqrt{3}}{2}(-\hat{\mathrm{j}})$

$\overrightarrow{\mathrm{r}}_{1}=0 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}$

Torque due to $\mathrm{F}_{1}$ force

$$

\vec{\tau}_{\mathrm{F}_{1}}=\overrightarrow{\mathrm{r}}_{1} \times \overrightarrow{\mathrm{F}}_{1}=6 \hat{\mathrm{j}} \times\left(\frac{\mathrm{F}}{2}(-\hat{\mathrm{i}})+\frac{\mathrm{F} \sqrt{3}}{2}(-\hat{\mathrm{j}})\right)=3 \mathrm{~F}(\hat{\mathrm{k}})

$$

Torque due to $\mathrm{F}_{2}$ force

$$

\begin{aligned}

\vec{\tau}_{\mathrm{F}_{2}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \times & \mathrm{F} \hat{\mathrm{k}}=3 \mathrm{~F} \hat{\mathrm{i}}+2 \mathrm{~F}(-\hat{\mathrm{j}}) \\

\vec{\tau}_{\mathrm{net}}=\vec{\tau}_{\mathrm{F}_{1}}+\vec{\tau}_{\mathrm{F}_{2}} &=3 \mathrm{Fi}+2 \mathrm{~F}(-\hat{\mathrm{j}})+3 \mathrm{~F}(\hat{\mathrm{k}}) \\

&=(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \mathrm{F}

\end{aligned}

$$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (11 Jan Shift 1)
A slab of copper of thickness \( \frac{d}{2} \) is introduced between the plates of a parallel plate capacitor, where \( d \) is the separation between its two plates. If the capacitance of the capacitor without copper slab is \( C \) and with copper slab is \( C^{\prime} \), then \( \frac{C^{\prime}}{C} \) is

Options:
A \( \sqrt{2} \)
B \( 2 \)
C \( 1 \)
D \( \frac{1}{\sqrt{2}} \)

The correct answer is:
\( 2 \)

Solution:
We know that the capacitance of parallel plate capacitor is given by, C=Aε0d.

Figure shows the parallel plate capacitor with and without copper slab.

And, the capacitance of parallel plate capacitor with metallic plate of thickness t is given by, C'=Aε0d-t.

Here, thickness, t=d2. So,

C'=Aε0d-d2=2Aε0d.

Now, dividing C' by C, we get,

C'C = 21 or 2.

Topic: Physics, Capacitance, JEE Main
A slab of dielectric constant K has the same crosssectional area as the plates of a parallel plate capacitor and thickness 34d, where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :
(Given C0= capacitance of capacitor with air as medium between plates.)

Options:
A 4KC03+K
B 3KC03+K
C 3+K4KC0
D K4+K

The correct answer is:
4KC03+K

Solution:
From the above diagram, we can write,

x+y+3d4=d

x+y=d4

Initially, when there is no dielectric inserted between the plates the capacitance will be, C0=ϵ0A d.

When the dielectric slab is inserted, we can assume there are 3 capacitors that are connected in series.

C1=ϵ0Ax, C2=4Kϵ0A3d and C3=ϵ0Ay

The equivalent capacitance,

1C=1C1+1C2+1C3⇒1C=1ϵ0Ax+3d4K+y⇒1C=1ϵ0Ad4+3d4K⇒C=4ϵ0AKd3+K⇒C=4KC03+K

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A slab of stone of area of $0.36 \mathrm{~m}^2$ and thickness $0.1 \mathrm{~m}$ is exposed on the lower surface to steam at $100^{\circ} \mathrm{C}$. A block of ice at $0^{\circ} \mathrm{C}$ rests on the upper surface of the slab. In one hour $4.8 \mathrm{~kg}$ of ice is melted. The thermal conductivity of slab is
(Given latent heat of fusion of ice $=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$ )

Options:
A $1.24 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$
B $1.29 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$
C $2.05 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$
D $1.02 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$

The correct answer is:
$1.24 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$

Solution:
$$
\begin{aligned}
\frac{\delta Q}{\delta t} & =\frac{K A}{L}\left(T_1-T_2\right) \\
Q & =\frac{K A}{L}\left(T_1-T_2\right) t \\
Q & =m L_f \\
\frac{K A}{L}\left(T_1-T_2\right) t & =m L_f \\
K & =\frac{m L_f L}{A\left(T_1-T_2\right) t} \\
K & =\frac{4.8 \times 3.36 \times 10^5 \times 0.1}{0.36 \times 100 \times 3600}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{4.8 \times 3.36}{0.36 \times 36} \\
& =1.24 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
A slab of stone of area of $0.36 \mathrm{~m}^2$ and thickness $0.1 \mathrm{~m}$ is exposed on the lower surface to steam at $100^{\circ} \mathrm{C}$. A block of ice at $0^{\circ} \mathrm{C}$ rests on the upper surface of the slab. In one hour $4.8 \mathrm{~kg}$ of ice is melted. The thermal conductivity of slab is
(Given latent heat of fusion of ice $=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$ )

Options:
A $1.24 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$
B $1.29 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$
C $2.05 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$
D $1.02 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$

The correct answer is:
$1.24 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$

Solution:
$$
\begin{aligned}
\frac{\delta Q}{\delta t} & =\frac{K A}{L}\left(T_1-T_2\right) \\
Q & =\frac{K A}{L}\left(T_1-T_2\right) t \\
Q & =m L_f \\
\frac{K A}{L}\left(T_1-T_2\right) t & =m L_f \\
K & =\frac{m L_f L}{A\left(T_1-T_2\right) t} \\
K & =\frac{4.8 \times 3.36 \times 10^5 \times 0.1}{0.36 \times 100 \times 3600}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{4.8 \times 3.36}{0.36 \times 36} \\
& =1.24 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2012 (Mains)
A slender homogeneous rod of length 2L floats partly immersed in water, being supported by a string fastened to one of its ends, as shown in the figure. The specific gravity of the rod is 0.36. The length of the rod that extends out of the water is KL10. Find the value of K.

Options:

The correct answer is:
16

Solution:
Balancing torque about end A
MgLcosÎ¸=Bl+2L-l2 cosÎ¸
MgL=Ï.(2L-l)Ag2L+l2
0.36Ã2L2=4L2-l22
1.44L2=4L2-l2
l2=4L2-1.44L2
l2=2.56L2
l=1610L
â´ K=16

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is:

Options:
A 2g3lcos⁡θ
B 3g2lsin⁡θ
C 2g3lsin⁡θ
D 3g2lcos⁡θ

The correct answer is:
3g2lsin⁡θ

Solution:
Moment of a force about O is lα=mgl2sinθ
⇒α=mgl2ml23sinθ
⇒α=3g2lsin⁡θ

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2017 (02 Apr)
A slit of size $0.15 \mathrm{~cm}$ is placed at $2.1 \mathrm{~m}$ from a screen. On illuminated it by a light of wavelength $5 \times 10^{-5}$ $\mathrm{cm}$. The width of central maxima will be

Options:
A $70 \mathrm{~mm}$
B $0.14 \mathrm{~mm}$
C $1.4 \mathrm{~mm}$
D $0.14 \mathrm{~cm}$

The correct answer is:
$1.4 \mathrm{~mm}$

Solution:
Width of central maxima $=\frac{2 \lambda D}{d}$
$=\frac{2 \times 2.1 \times 5 \times 10^{-7}}{0.15 \times 10^{-2}}=1.4 \times 10^{-3} \mathrm{~m}=1.4 \mathrm{~mm}$

Topic: Physics, Wave Optics, JEE Main
A slit of width $a$ is illuminated by white light. For red light $(\lambda=6500 A)$, the first minima is obtained at $\theta=30^{\circ}$. Then the value of $a$ will be

Options:
A $3250 Å$
B $6.5 \times 10^{-4} \mathrm{~mm}$
C 1.24 microns
D $2.6 \times 10^{-4} \mathrm{~cm}$

The correct answer is:
1.24 microns

Solution:
For first minima ${ }^{\theta=\frac{\lambda}{a}}$ or $\quad a=\frac{\lambda}{\theta}$
$\therefore a=\frac{6500 \times 10^{-8} \times 6}{\pi} \quad\left(\right.$ As $30^{\circ}=\frac{\pi}{6}$ radian $)$
$=1.24 \times 10^{-4} \mathrm{~cm}=1.24$ microns

Topic: Physics, Wave Optics, JEE Main
A small asteroid is orbiting around the sun in a circular orbit of radius $\mathrm{r}_{0}$ with speed $\mathrm{V}_{0}$. A rocket is launched from the asteroid with speed $\mathrm{V}=\alpha \mathrm{V}_{0}$, where $\mathrm{V}$ is the speed relative to the sun. The highest value of $\alpha$ for which the rocket will remain bound to the solar system is (ignoring gravity due to the asteroid and effects of other planets) -

Options:
A $\sqrt{2}$
B 2
C $\sqrt{3}$
D 1

The correct answer is:
1

Solution:
$\begin{array}{l}
\text { B.E }=\frac{G m M}{2 R} \\
\text { B.E }=\frac{1}{2} m v_{0}^{2} \\
\text { so, } \alpha=1
\end{array}$

Topic: Physics, Gravitation, KVPY, KVPY 2012 (SB/SX)
A small ball of mass 2 Ã 10-3 kg having a charge of 1 Î¼C is suspended by a string of length 0.8 m. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity which should be imparted to the lower ball so that it can make a complete revolution.

Options:
A 6.2 m s-1

B 9.8 m s-1

C 11.6 m s-1

D 5.86 m s-1

The correct answer is:
5.86 m s-1

Solution:
If the ball has to just complete the circle then the tension must vanish at the topmost point, i.e.,T2 = 0
From Newton's second law,
T2+mg-q24πε0l2=mv2l …i
At the topmost point T2 = 0,
mg-q24πε0l2=mv2l …ii
From energy conservation,
Energy at lowest point = Energy at topmost point
12mv2=12mv2+mg2l …iii
v2=u2-4gl …iv
From Eqn. (ii), v2=gl-q24πε0ml …v
From Eqns. (iv) and (v),
u=5gl-q24πε0ml
= 2 7 5 8 1 / 2
=5.86 m s-1

Topic: Physics, Laws of Motion, JEE Main
A small ball of mass $m$ and density $\rho$ is dropped in a viscous liquid of density $\rho_0$. After sometime, the ball falls with constant velocity. The viscous force on the ball is :

Options:
A $m g\left(1-\rho \rho_0\right)$
B $m g\left(1+\frac{\rho}{\rho_0}\right)$
C $m g\left(\frac{\rho_0}{\rho}-1\right)$
D $m g\left(1-\frac{\rho_0}{\rho}\right)_{\nabla}$

The correct answer is:
$m g\left(1-\frac{\rho_0}{\rho}\right)_{\nabla}$

Solution:
$\mathrm{mg}-\mathrm{F}_{\mathrm{B}}-\mathrm{F}_{\mathrm{v}}=\mathrm{ma}$
$\mathrm{a}=0$ for constant velocity
$\begin{aligned}
& m g-F_B=F_v \\
& F_v=m g-v \rho_0 g=m g-\frac{m}{\rho} \rho_0 g=m g\left(1-\frac{\rho_0}{\rho}\right)
\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (06 Apr Shift 1)
A small ball of mass m and radius r=R10 rolls without slipping along the track shown in the figure. The radius of circular part of the track is R. If the ball starts from rest at a height of 8R above the bottom, the normal force on the ball at the point P is

Options:
A 1003mg
B 1009mg
C 2009mg
D 507mg

The correct answer is:
1009mg

Solution:
Using Work-Energy Theorem

7mgR=12mv2+1225mr2×ω2

since the sphere is rolling without slipping v=rω

∴7mgR=12× 75mr2× v2r2

⇒v2=10gR

at point P, we can consider sphere as a point object doing circular motion in a circle of the radius (R-r) with velocity v,

Hence at point P

N=mv2R-r=10mgRR-R10=10mgR9R10 ( given r=R10 )

or N=1009mg

Topic: Physics, Rotational Motion, JEE Main
A small ball of mass m is thrown upward with velocity u from the ground. The ball experiences a resistive force mkv2 where v is it speed. The maximum height attained by the ball is :

Options:
A 12ktan-1ku2g
B 1kln1+ku22g
C 1ktan-1ku22g
D 12kln1+ku2g

The correct answer is:
12kln1+ku2g

Solution:
Fne =ma

-mg-mkv2=mvdvds

vdvds=-g-kv2

-∫v00vdvg+kv2=∫0hds=hmax

hmax=12King+kv02g

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2020 (04 Sep Shift 2)
A small ball of mass $\mathrm{m}$ starts at a point $\mathrm{A}$ with speed $\mathrm{v}_{\mathrm{o}}$ and moves along a frictionless track $\mathrm{AB}$ as shown. The track $\mathrm{BC}$ has coefficient of friction $\mu$. The ball comes to stop at $\mathrm{C}$ after travelling a distance $L$ which is:

Options:
A
$\frac{2 \mathrm{~h}}{\mu}+\frac{\mathrm{v}_{\mathrm{o}}^2}{2 \mu \mathrm{g}}$

B
$\frac{\mathrm{h}}{\mu}+\frac{\mathrm{v}_0^2}{2 \mu \mathrm{g}}$

C
$\frac{\mathrm{h}}{2 \mu}+\frac{\mathrm{v}_{\mathrm{o}}^2}{\mu \mathrm{g}}$

D
$\frac{\mathrm{h}}{2 \mu}+\frac{\mathrm{v}_{\mathrm{o}}^2}{2 \mu \mathrm{g}}$

The correct answer is:
$\frac{\mathrm{h}}{\mu}+\frac{\mathrm{v}_0^2}{2 \mu \mathrm{g}}$

Solution:
Initial speed at point A, $u=v_0$
Speed at point $\mathrm{B}, v=$ ?
$$
\begin{aligned}
&v^2-u^2=2 g h \\
&v^2=v_0^2+2 g h
\end{aligned}
$$
Let ball travels distance ' $S$ ' before coming to rest
$$
\begin{aligned}
&S=\frac{v^2}{2 \mu g}=\frac{v_0^2+2 g h}{2 \mu g} \\
&=\frac{v_0^2}{2 \mu g}+\frac{2 g h}{2 \mu g}=\frac{h}{\mu}+\frac{v_0^2}{2 \mu g}
\end{aligned}
$$

Topic: Physics, Work Power Energy, JEE Main
A small ball of mass $\mathrm{m}$ starts at a point $\mathrm{A}$ with speed $\mathrm{v}_{\mathrm{o}}$ and moves along a frictionless track $\mathrm{AB}$ as shown. The track $\mathrm{BC}$ has coefficient of friction $\mu$. The ball comes to stop at $\mathrm{C}$ after travelling a distance $L$ which is:

Options:
A
$\frac{2 \mathrm{~h}}{\mu}+\frac{\mathrm{v}_{\mathrm{o}}^2}{2 \mu \mathrm{g}}$

B
$\frac{\mathrm{h}}{\mu}+\frac{\mathrm{v}_0^2}{2 \mu \mathrm{g}}$

C
$\frac{\mathrm{h}}{2 \mu}+\frac{\mathrm{v}_{\mathrm{o}}^2}{\mu \mathrm{g}}$

D
$\frac{\mathrm{h}}{2 \mu}+\frac{\mathrm{v}_{\mathrm{o}}^2}{2 \mu \mathrm{g}}$

The correct answer is:
$\frac{\mathrm{h}}{\mu}+\frac{\mathrm{v}_0^2}{2 \mu \mathrm{g}}$

Solution:
Initial speed at point A, $u=v_0$
Speed at point $\mathrm{B}, v=$ ?
$$
\begin{aligned}
&v^2-u^2=2 g h \\
&v^2=v_0^2+2 g h
\end{aligned}
$$
Let ball travels distance ' $S$ ' before coming to rest
$$
\begin{aligned}
&S=\frac{v^2}{2 \mu g}=\frac{v_0^2+2 g h}{2 \mu g} \\
&=\frac{v_0^2}{2 \mu g}+\frac{2 g h}{2 \mu g}=\frac{h}{\mu}+\frac{v_0^2}{2 \mu g}
\end{aligned}
$$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2014 (11 Apr Online)