datavorous/entrance-exam-dataset huggingface dataset:
A thermocouple of negligible resistance produces an emf of $40 \mu \mathrm{V} /{ }^{\circ} \mathrm{C}$ in the linear range of temperature. A galvanometer of resistance $10 \Omega$ whose sensitivity is $1 \mu \mathrm{A} /$ div, is employed with the termocouple. The smallest value of temperature difference that can be detected by the system will be

Options:
A $0.5^{\circ} \mathrm{C}$
B $1^{\circ} \mathrm{C}$
C $0.1^{\circ} \mathrm{C}$
D $0.25^{\circ} \mathrm{C}$

The correct answer is:
$0.25^{\circ} \mathrm{C}$

Solution:
For minimum deflection of 1 division require current $=1 \mu \mathrm{A}$
voltage required $=i R=1 \times 10=10 \mu \mathrm{V}$
$$
\begin{aligned}
\therefore \quad 40 \mu \mathrm{V} & \cong 1^{\circ} \mathrm{C} \\
10 \mu \mathrm{V} & \cong \frac{1}{4}{ }^{\circ} \mathrm{C} \\
& =0.25^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A thermocouple of negligible resistance produces an emf of $40 \mu \mathrm{V} /{ }^{\circ} \mathrm{C}$ in the linear range of temperature. A galvanometer of resistance $10 \Omega$ whose sensitivity is $1 \mu \mathrm{A} /$ div, is employed with the termocouple. The smallest value of temperature difference that can be detected by the system will be

Options:
A $0.5^{\circ} \mathrm{C}$
B $1^{\circ} \mathrm{C}$
C $0.1^{\circ} \mathrm{C}$
D $0.25^{\circ} \mathrm{C}$

The correct answer is:
$0.25^{\circ} \mathrm{C}$

Solution:
For minimum deflection of 1 division require current $=1 \mu \mathrm{A}$
voltage required $=i R=1 \times 10=10 \mu \mathrm{V}$
$$
\begin{aligned}
\therefore \quad 40 \mu \mathrm{V} & \cong 1^{\circ} \mathrm{C} \\
10 \mu \mathrm{V} & \cong \frac{1}{4}{ }^{\circ} \mathrm{C} \\
& =0.25^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Current Electricity, NEET, NEET 2011 (Mains)
A thermodynamic cycle in the pressure \( (\mathrm{P}) \)-volume \( (\mathrm{V}) \) plane is given below: \( \mathrm{AB} \) and \( \mathrm{CD} \) are isothermal processes while \( \mathrm{BC} \) and \( \mathrm{DA} \) are adiabatic processes. The same cycle in the temperature (T)- entropy \( (S) \) plane is :

Options:
A
B
C
D

The correct answer is:

Solution:
AB is isothermal reversible expansion process i.e. ΔT=0 and S increases as there is increase in volume.

BC is adiabatic reversible expansion process qrev=0 i.e. temperature decreases and ΔS=0

CD is isothermal reversible compression process i.e. ΔT=0 and S decreases as there is decrease in volume.

DA is adiabatic reversible compression process qres=0 i.e. temperature decreases and ΔS=0

Topic: Chemistry, Thermodynamics (C), JEE Main
A thermodynamic process is shown in the figure. The pressures and volumes corresponding to some points in the figure are: $P_A=3 \times 10^4 \mathrm{~Pa}, P_B=8 \times 10^4 \mathrm{~Pa}$ and $V_A=2 \times 10^{-3} \mathrm{~m}^3, V_D=5 \times 10^{-3} \mathrm{~m}^3$ In process $A B, 600 \mathrm{~J}$ of heat is added to the system and in process $B C, 200 \mathrm{~J}$ of heat is added to the system. The change in internal energy of the system in process $A C$ would be

Options:
A $560 \mathrm{~J}$
B $800 \mathrm{~J}$
C $600 \mathrm{~J}$
D $640 \mathrm{~J}$

The correct answer is:
$560 \mathrm{~J}$

Solution:
By adjoining graph $W_{A B}=0$ and
$\begin{gathered}W_{B C}=8 \times 10^4[5-2] \times 10^{-3}=240 \mathrm{~J} \\ \therefore W_{A C}=W_{A B}+W_{B C}=0+240=240 \mathrm{~J}\end{gathered}$
Now, $\Delta Q_{A C}=\Delta Q_{A B}+\Delta Q_{B C}=600+200=800 \mathrm{~J}$
From FLOT $\Delta Q_{A C}=\Delta U_{A C}+\Delta W_{A C}$
$\Rightarrow 800=\Delta U_{A C}+240 \Rightarrow \Delta U_{A C}=560 \mathrm{~J}$

Topic: Physics, Thermodynamics, JEE Main
A thermodynamic system is taken from an original state D to an intermediate state E by the linear process shown in the figure. Its volume is then reduced to the original volume from E to F by an isobaric process. The total work done by the gas from D to E to F will be

Options:
A -450 J
B 450 J
C 900 J
D 1350 J

The correct answer is:
450 J

Solution:
Total work done by the gas from D to E to F=Area of ∆DEF

Work done during the process DE is WDE=12600+3005-2 J

=1350 J

Work done during the process EF is WEF=-300×3=-900 J

WDEF=WDE+WEF=450 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (29 Jul Shift 2)
A thermodynamic system is taken through cyclic process. The total work done in the process is :

Options:
A 200 J
B 300 J
C 100 J
D Zero

The correct answer is:
300 J

Solution:
The work done is given by the area of the PV graph of the loop.

W=12VE-VC(PD-PC)=2×3002 J

=300 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (15 Apr Shift 1)
A thermodynamic system is taken through the cycle $P Q R S P$ process. The net work done by the system is

Options:
A $20 \mathrm{~J}$
B $-20 J$
C $400 \mathrm{~J}$
D - $374 J$

The correct answer is:
$-20 J$

Solution:
Work done by the system \(=\) Area of shaded portion on \(\mathrm{P}-\mathrm{V}\) diagram
\(=(300-100) 10^{-} 6 \times(200-10) \times 10^3=20 \mathrm{~J}\)
and direction of process is anticlockwise so work done will be negative i.e.
\(\Delta \mathrm{W}=-20 \mathrm{~J}\)

Topic: Physics, Thermodynamics, JEE Main
A thermodynamic system undergoes a cyclic process \( A B C \) as shown in the diagram. The work
done by the system per cycle is

Options:
A \( -750 \mathrm{~J} \)
B \( 750 \mathrm{~J} \)
C \( 1250 \mathrm{~J} \)
D \( -1250 \mathrm{~J} \)

The correct answer is:
\( -750 \mathrm{~J} \)

Solution:
(A)
Work done \( =p \times \Delta v \)
Work done = Area of triangle ABC
\( \therefore W=\frac{1}{2}(10-5) \times(100-400) \)
\( =\frac{5}{2} \times(-300)=-750 \mathrm{~J} \)
The negative sign means the work is done by the system.

Topic: Physics, Thermodynamics, JEE Main
A thermo-dynamical system is changed from state $\left(P_1, V_1\right)$ to $\left(P_2, V_2\right)$ by two different process. The quantity which will remain same will be

Options:
A $\Delta Q$
B $\Delta W$
C $\Delta Q+\Delta W$
D $\Delta Q-\Delta W$

The correct answer is:
$\Delta Q-\Delta W$

Solution:
by float $\quad \Delta U=\Delta Q-\Delta W$ And $\quad \Delta U$ is a state function.

Topic: Physics, Thermodynamics, JEE Main
A thermodynamical system is changed from state $\left(\mathrm{P}_{1}, \mathrm{~V}_{1}\right)$ to $\left(\mathrm{P}_{2}, \mathrm{~V}_{2}\right)$ by two different process, the quantity which will remain same will be

Options:
A $\Delta \mathrm{Q}$
B $\Delta \mathrm{W}$
C $\Delta \mathrm{Q}+\Delta \mathrm{W}$
D $\Delta \mathrm{Q}-\Delta \mathrm{W}$

The correct answer is:
$\Delta \mathrm{Q}-\Delta \mathrm{W}$

Solution:
For all process $\Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W}$

does not change as it depends on initial final states.

Topic: Physics, Thermodynamics, JEE Main
A thermodynamical system is changed from state P1, V1 to P2, V2 by two different processes. The quantity which will remain the same will be

Options:
A Q
B W
C Q+W
D Q-W

The correct answer is:
Q-W

Solution:
For all processes, the change in internal energy ΔU=Q-W does not change. It depends only on initial and final states.

Topic: Physics, Thermodynamics, NEET
A thermometer graduated according to a linear scale reads a value $x_{0}$ when in contact with boiling water, and $x_{0} / 3$ when in contact with ice. What is the temperature of an object in ${ }^{\circ} \mathrm{C}$, if this thermometer in the contact with the object reads $x_{0} / 2 ?$

Options:
A 25
B 60
C 40
D 35

The correct answer is:
25

Solution:
Let required temperature $=\mathrm{T}^{\circ} \mathrm{C}$

$\Rightarrow \mathrm{T}^{\circ} \mathrm{C}=\frac{\mathrm{x}_{0}}{2}-\frac{\mathrm{x}_{0}}{3}=\frac{\mathrm{x}_{0}}{6}$

$\&\left(\mathrm{x}_{0}-\frac{\mathrm{x}_{0}}{3}\right)=\left(100-0^{\circ} \mathrm{C}\right)$

$\Rightarrow \frac{2 \mathrm{x}_{0}}{3}=100 \Rightarrow \mathrm{x}_{0}=\frac{300}{2}$

$\Rightarrow \mathrm{T}^{\circ} \mathrm{C}=\frac{\mathrm{x}_{0}}{6}=\frac{150}{6}=25^{\circ} \mathrm{C}$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A thick walled hollow sphere has outer radius $\mathrm{R}$. It rolls down an inclined plain Without slipping and its speed at the bottom is v. If the inclined plane is frictionless and the sphere slides down without rolling, its speed at the bottom will be $5 \mathrm{v} / 4$. What is radius of gyration of the sphere.

Options:
A $\frac{R}{2}$
B $\frac{\mathrm{R}}{\sqrt{2}}$
C $\frac{3 R}{4}$
D $\frac{\sqrt{3} R}{4}$

The correct answer is:
$\frac{3 R}{4}$

Solution:
Initially, when the sphere is rolling down the include without any slipping, then let the angular velocity of the sphere at the bottom of the incline be w.
No slipping condition, $\mathrm{V}=\mathrm{wR}$
Furthermore, due to conservation of energy
PE lost $=$ KE gained
$\mathrm{mgh}=1 / 2 \mathrm{mv}^{2}+1 / 2{ }^* \mathrm{I}^* \mathrm{w}^{2}$ (Where $\mathrm{h}$ is the height of the incline and $\mathrm{I}$ is the moment of inertia of sphere around its center)
Also, $\mathrm{I}=\mathrm{m}^* \mathrm{Rg}^{2}$ where $\mathrm{Rg}$ is the radius of gyration and also $\mathrm{v}=\mathrm{wR}$ (No slipping condition)
Hence, $\mathrm{mgh}=1 / 2 \mathrm{mv}^{2}+1 / 2{ }^* \mathrm{~m} * \mathrm{Rg}{ }^{2}{ }^* \mathrm{v}^{2} / \mathrm{R}^{2}$ $2 \mathrm{gh}=\left[1+(\mathrm{Rg} / \mathrm{R})^{2}\right] * \mathrm{v}^{2}(1)$
Furthermore, under no rolling case
$\begin{aligned}& \mathrm{mgh}=1 / 2 \mathrm{~m}^*(5 \mathrm{v} / 4)^{2} \\& \mathrm{v}^{2}=32 / 25 \mathrm{gh}\end{aligned}$
Putting it in the equation 1
$2 \mathrm{gh}=\left[1+(\mathrm{Rg} / \mathrm{R})^{2}\right]^* 32 / 25 \mathrm{gh}$
Hence, $\mathrm{Rg}=3 / 4 \mathrm{R}$
So, the radius of gyration of the sphere around its center is $3 / 4 \mathrm{R}$

Topic: Physics, Rotational Motion, JEE Main
A thick-walled hollow sphere has outside radius $R_0$. It rolls down an incline without slipping and its speed at the bottom is $v_0$. Now the incline is waxed, so that it is practically frictionless and the sphere is observed to slide down (without any rolling). Its speed at the bottom is observed to be $5 v_0 / 4$. The radius of gyration of the hollow sphere about an axis through its centre is

Options:
A
$3 R_0 / 2$

B
$3 R_0 / 4$

C
$9 R_0 / 16$

D
$3 R_0$

The correct answer is:
$3 R_0 / 4$

Solution:
When body rolls dawn on inclined plane with velocity $\mathrm{V}_0$ at bottom then body has both rotational and translational kinetic energy.
Therefore, by law of conservation of energy,
$$
\begin{aligned}
& P . E .=K . E_{\text {trans }}+K . E_{\text {rotational }} \\
& =\frac{1}{2} m V_0^2+\frac{1}{2} I \omega^2 \\
& =\frac{1}{2} m V_0^2+\frac{1}{2} m k^2 \frac{V_0^2}{R_0^2} \quad \ldots \text { (i) } \\
& {\left[\because \mathrm{I}=m k^2, \omega=\frac{V}{R_0}\right]}
\end{aligned}
$$
When body is sliding down then body has only translatory motion.
$$
\begin{aligned}
\therefore P . E . & =K . E_{\text {trans }} \\
& =\frac{1}{2} m\left(\frac{5}{4} v_0\right)^2
\end{aligned}
$$
Dividing (i) by (ii) we get
$$
\begin{aligned}
& \frac{P . E .}{P . E .}=\frac{\frac{1}{2} m v_0^2\left[1+\frac{K^2}{R_0^2}\right]}{\frac{1}{2} \times \frac{25}{16} \times m V_0^2} \\
& =\frac{25}{16}=1+\frac{K^2}{R_0^2} \Rightarrow \frac{K^2}{R_0^2}=\frac{9}{16} \\
& \text { or, } K=\frac{3}{4} R_0 .
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2012 (26 May Online)
A thin 1 m long rod has a radius of 5 mm. A force of 50π×103 N is applied at one end to determine its Young's modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01 mm, which of the following statements is false?

Options:
A The maximum value of Y that can be determined is 2×1014 N m-2
B âYY gets minimum contribution from the uncertainty in the length.
C âYY gets its maximum contribution from the uncertainty in strain.
D The figure of merit is the largest for the length of the rod.

The correct answer is:
The maximum value of Y that can be determined is 2×1014 N m-2

Solution:
Given, l=1 m, r=5×10-3 m and F=50π×103 N.

Least count in length is =0.01 mm=0.01×10-3 m
Young's modulus is given by Y=FA∆ll
For, ∆l→Min, Y→Max
⇒YΔll=FA
Y=50π×103π×5×10-32×lΔl
Y=50×10325×10-6×1Δl⇒Y=2×109Δl
Here Δl is unknown so we can not find the Y

The change in young’s modulus ratio is ∆YY=2∆rr+∆ll.

As seen from above, the minimum contribution to change in young’s modulus is given by the uncertainty in the length and the maximum contributing to the uncertainty in young’s modulus is given by uncertainty in radius.

Hence, correct option is A.

Topic: Physics, Mathematics in Physics, JEE Main
A thin 1 m long rod has a radius of 5 mm. A force of 50π×103 N is applied at one end to determine its Young's modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01 mm, which of the following statements is false?

Options:
A The maximum value of Y that can be determined is 2×1014 N m-2
B âYY gets minimum contribution from the uncertainty in the length.
C âYY gets its maximum contribution from the uncertainty in strain.
D The figure of merit is the largest for the length of the rod.

The correct answer is:
The maximum value of Y that can be determined is 2×1014 N m-2

Solution:
Given, l=1 m, r=5×10-3 m and F=50π×103 N.

Least count in length is =0.01 mm=0.01×10-3 m
Young's modulus is given by Y=FA∆ll
For, ∆l→Min, Y→Max
⇒YΔll=FA
Y=50π×103π×5×10-32×lΔl
Y=50×10325×10-6×1Δl⇒Y=2×109Δl
Here Δl is unknown so we can not find the Y

The change in young’s modulus ratio is ∆YY=2∆rr+∆ll.

As seen from above, the minimum contribution to change in young’s modulus is given by the uncertainty in the length and the maximum contributing to the uncertainty in young’s modulus is given by uncertainty in radius.

Hence, correct option is A.

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2016 (10 Apr Online)
A thin bar of length $\mathrm{L}$ has a mass per unit length $\lambda$, that increases linearly with distance from one end. If its total mass is $M$ and its mass per unit length at the lighter end is $\lambda_{\mathrm{O}}$, then the distance of the centre of mass from the lighter end is:

Options:
A
$\frac{\mathrm{L}}{2}-\frac{\lambda_0 \mathrm{~L}^2}{4 \mathrm{M}}$

B
$\frac{\mathrm{L}}{3}+\frac{\lambda_0 \mathrm{~L}^2}{8 \mathrm{M}}$

C
$\frac{\mathrm{L}}{3}+\frac{\lambda_0 \mathrm{~L}^2}{4 \mathrm{M}}$

D
$\frac{2 \mathrm{~L}}{3}-\frac{\lambda_0 \mathrm{~L}^2}{6 \mathrm{M}}$

The correct answer is:
$\frac{\mathrm{L}}{3}+\frac{\lambda_0 \mathrm{~L}^2}{4 \mathrm{M}}$

Solution:
No Solution Available

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2014 (11 Apr Online)
A thin bar of length L has a mass per unit length Î», that increases linearly with distance from one end. If its total mass is M and its mass per unit length at the lighter end is Î» 0 , then the distance of the centre of mass from the lighter end is

Options:
A L 3 + λ 0 L 2 8 M
B L 3 + λ 0 L 2 4 M
C L 2 - λ 0 L 2 4 M
D 2 L 3 - Î» 0 L 2 6 M

The correct answer is:
2 L 3 - λ 0 L 2 6 M

Solution:
Î» â x
Î» = Kx + Î» 0
dm = Î» dx
â´ X cm = â« 0 L dm Â· x â« 0 L dm = â« 0 L Kx + Î» 0 dx.x â« 0 L Kx + Î» 0 dx
= K Â· x 3 3 0 L + Î» 0 x 2 2 0 L K x 2 2 0 L + Î» 0 L
= KL 3 3 + Î» 0 L 2 2 KL 2 2 + Î» 0 L
= KL 3 + Î» 0 2 L 2 KL 2 2 + Î» 0 L
Now, M = â« 0 L dm
= KL 2 2 + Î» 0 L
2 M = KL 2 + 2 Î» 0 L
K = 2 M - Î» 0 L L 2
Putting the value of K.
X cm = 2 M - Î» 0 L 3 L + Î» 0 2 L 2 M
= 2 L 2 M - Î» 0 L 3 LM + Î» 0 L 2 2 M
= 2 ML 3 M - 2 Î» 0 L 2 3 M + Î» 0 L 2 2 M
= 2 L 3 - Î» 0 L 2 6 M
Hence, 2 L 3 - Î» 0 L 2 6 M

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A thin bi-convex lens made out of a material of refractive index Î¼1 is placed in a medium of refractive index Î¼2. A paraxial beam of light, parallel to the principal axis of the lens, is shown in the figure. Based on the ray diagram, we can conclude that

Options:
A μ1>μ2
B μ1≤μ2
C μ1=μ2
D μ1<μ2

The correct answer is:
μ1<μ2

Solution:
A bi-convex lens acts as a diverging lens when it is placed in a denser medium.

Topic: Physics, Ray Optics, JEE Main
A thin biconvex lens of refractive index 32 is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is then filled with water of refractive index 43. It is found that when a point object is placed 15 cm above the lens on its principal axis, the object coincides with its own image. On repeating with another liquid, the object and the image again coincide at a distance 25 cm from the lens. Calculate the refractive index of the liquid.

Options:

The correct answer is:
1.6

Solution:
Let R be the radius of curvature of both the surfaces of the equiconvex lens. In the first case
Let f1 be the focal length of planoconcave lens of index μ1. The focal length of the combined lens system will be given by
1 F = 1 f 1 + 1 f 2
= μ 1 - 1 1 R - 1 - R + μ 2 - 1 1 - R - 1 ∞
= 3 2 - 1 2 R + 4 3 - 1 - 1 R
=1R-13R=23R or F=3R2
Now, image coincides with the object when ray of light retraces its path or it falls normally on the plane mirror. This is possible only when object is at focus of the lens system.
Hence, F=15 cm ( Distance of object = 15 cm).
or 3R2=15 cm or R=10 cm
In the second case, let μ be the refractive index of the liquid-filled between lens and mirror and let F' be the focal length of new lens system. Then,

1F'=μ1-11R-1-R+μ-11-R-1∞
or, 1F'=32-12R-μ-1R
or, 1F'=1R-μ-1R=2-μR
F'=R2-μ=102-μ R=10 cm
Now, the image coincides with object when it is placed at 25 cm distance.
Hence, F ′ = 2 5
or, 102-μ=25 or 50-25μ=10 or 25μ=40
∴ μ=4025=1·6 or μ=1·6

Topic: Physics, Ray Optics, JEE Main
A thin biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is then filled with water of refractive index 4/3. It is found that when a point object is placed 15 cm above the lens on its principle axis, the object coincides with its own image. On repeating with another liquid, the object and the image again coincide at a distance 25 cm from the lens. Calculate the refractive index of the liquid.

Options:
A 1 · 6
B 3.2
C 0.8
D 2

The correct answer is:
1 · 6

Solution:
Let R be the radius of curvature of both the surfaces of the equi-convex lens. In the first case:
Let f1 be the focal length of planoconcave lens of index μ 1 . The focal length of the combined lens system will be given by:
1 F = 1 f 1 + 1 f 2
= μ 1 - 1 1 R - 1 - R + μ 2 - 1 1 - R - 1 ∞
= 3 2 - 1 2 R + 4 3 - 1 - 1 R
= 1 R - 1 3 R = 2 3 R of F = 3 R 2
Now, image coincides with the object when ray of light retraces its path or it falls normally on the plane mirror. This is possible only when object is at focus of the lens system.
Hence, F 15 cm ( Distance of object = 15 cm).
or 3 R 2 = 1 5 cm or R = 1 0 cm
In the second case, let μ be the refractive index of the liquid filled between lens and mirror and let F' be the focal length of new lens system. Then,

1 F ′ = μ 1 - 1 1 R - 1 R - + μ - 1 1 - R - 1 ∞
or, 1 F ′ = 3 2 - 1 2 R - μ - 1 R
or, 1 R - μ - 1 R = 2 - μ R
F ′ = R 2 - μ = 1 0 2 - μ R = 1 0 cm
Now, the image coincides with object when it is placed at 25 cm distance.
Hence, F ′ = 2 5
or, 1 0 2 - μ = 2 5 or 5 0 - 2 5 μ = 1 0 or 2 5 μ = 4 0
∴ μ = 4 0 2 5 = 1 · 6 or μ = 1 · 6

Topic: Physics, Ray Optics, JEE Main
A thin charged rod is bent into the shape of a small circle of radius $\mathrm{R}$ the charge per unit length of the rod being $\lambda$. The circle is rotated about its axis with a time period $T$ and it is found that the magnetic field at a distance ' $d$ ' away $(d>>R)$ from the center and on the axis, varies as $\frac{\mathrm{R}^{\mathrm{m}}}{\mathrm{d}^{\mathrm{n}}} .$ The values of $\mathrm{m}$ and $\mathrm{n}$ respectively are

Options:
A $m=2, n=2$
B $m=2, n=3$
C $m=3, n=2$
D $m=3, n=3$

The correct answer is:
$m=3, n=3$

Solution:
$\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{2 \pi \mathrm{R}^{2} \mathrm{I}}{\left(\mathrm{R}^{2}+\mathrm{d}^{2}\right)^{3 / 2}} \quad \mathrm{I}=\frac{\mathrm{q}}{\mathrm{T}}$
$=\frac{2 \pi \mathrm{R} \lambda}{\mathrm{T}}$
$=\frac{\mu_{0}}{4 \pi} \frac{2 \pi \mathrm{R}^{2} \mathrm{I}}{\mathrm{d}^{3}}(\mathrm{~d}>>\mathrm{R})$
$\frac{\mu_{0}}{2} \frac{\mathrm{R}^{2}}{\mathrm{~d}^{3}} \times \frac{2 \pi \mathrm{R} \lambda}{\mathrm{T}}$
$\mathrm{B} \propto \frac{\mathrm{R}^{3}}{\mathrm{~d}^{3}}$
$\therefore \mathrm{m}=\mathrm{n}=3$

Topic: Physics, Magnetic Effects of Current, JEE Main
A thin circular disc of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity $\omega$. If another disc of same dimensions but of mass $\mathrm{M} / 2$ is placed gently on the first disc co-axially, then the new angular velocity of the system is :

Options:
A $\frac{3}{2} \omega$
B $\frac{5}{4} \omega$
C $\frac{2}{3} \omega$
D $\frac{4}{5} \omega$

The correct answer is:
$\frac{2}{3} \omega$

Solution:
$\begin{aligned} & \mathrm{I}_1 \omega=\mathrm{I}_2 \omega_2 \\ & \frac{\mathrm{MR}^2}{2} \omega=\frac{3}{2}\left(\frac{\mathrm{MR}^2}{2}\right) \omega_2 \\ & \omega_2=\frac{2}{3} \omega\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2024 (08 Apr Shift 2)
A thin circular disk is in the xy plane as shown in the figure. The ratio of its moment of inertia about z and z' axes will be:

Options:
A 1 :4
B 1:5
C 1:3
D 1:2

The correct answer is:
1:3

Solution:
IZ= mR22
(by parallel axis theorem)
Iz′=32mR2
IZIz′=13

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2018 (16 Apr Online)
A thin circular loop of radius $R$ rotates about its vertical diameter with an angular frequency $\omega$, Show that a small bead on the wire loop remains at its lowermost point for $\omega$ $\leq \sqrt{g} / R$. What is the angle made by the radius vector joining the center to the bead with the vertical downward direction for $\omega=\sqrt{2} g / R$ ? Neglect friction.

Options:

None

Solution:
$\theta$ - angle between radius vector joining the bead to the center and vertical downward direction.
$N$ - normal reaction.
$$
m g=N \cos \theta \text {...(i) }
$$
and $m r \omega^2=\mathrm{N} \sin \theta$
or, $m(R \sin \theta) \omega^2=N \sin \theta$
$\Rightarrow R \omega^2=$ N...(ii)
From (i) and (ii),
$m g=m R \omega^2 \cos \theta$
$\Rightarrow \cos \theta=g / R \omega^2$
Since $|\cos \theta| \leq 1$, bead will emain its lower most point. $\Rightarrow g / R \omega^2 \leq 1$ or, $\omega \leq \sqrt{ }(g / R)$
If, $\omega=\sqrt{(} 2 g / R), \cos \theta=g / R(2 g / R)=\frac{1}{2}$
i.e. $\theta=60^{\circ}$

Topic: Physics, Laws of Motion
A thin circular metal disc of the radius 500 mm is set rotating about a central axis normal to its plane. Upon raising in temperature gradually, the radius increases to 507.5 mm. The percentage change in the rotational kinetic energy will be

Options:
A 1.5%
B -1.5%
C 3%
D -3%

The correct answer is:
-3%

Solution:
Given dRR×100=507.5-500.0500.0×100=1.5%
Now, K=12Iω2=12IIω2=L22I
or, K=L22MR22=L2MR2
Taking log and differentiating, dK K = - 2 dR R
or, the percentage change in K = dK × 1 0 0 K = - 2 dR R × 1 0 0 = - 3 %

Topic: Physics, Rotational Motion, JEE Main
A thin circular plate of mass M and radius R has its density varying as ρ(r)=ρ0r with ρ0 as constant and r is the distance from its centre. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I=aMR2. The value of the coefficient a is:

Options:
A 35
B 12
C 85
D 32

The correct answer is:
85

Solution:
dm=2πrdr×ρ=2πρ0r2dr
M=2πρ0∫0Rr2dr=23πρ0R3
I=ρdmr2=2πρ0∫r4dr=25πρ0R5
Icm=35MR2
I=Icm+MR2=85MR2

Topic: Physics, Rotational Motion, JEE Main
A thin circular plate of mass M and radius R has its density varying as ρ(r)=ρ0r with ρ0 as constant and r is the distance from its centre. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I=aMR2. The value of the coefficient a is:

Options:
A 35
B 12
C 85
D 32

The correct answer is:
85

Solution:
dm=2πrdr×ρ=2πρ0r2dr
M=2πρ0∫0Rr2dr=23πρ0R3
ICM=ρdmr2=2πρ0∫r4dr=25πρ0R5
Icm=35MR2
I=Icm+MR2=85MR2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (08 Apr Shift 1)