datavorous/entrance-exam-dataset huggingface dataset:
A circular coil having N turns and radius r carries a current I. It is held in the XZ plane in a magnetic field Bi^ . The torque on the coil due to the magnetic field is:

Options:
A Bπr2IN
B Br2IπN
C Bπr2IN
D Zero

The correct answer is:
Bπr2IN

Solution:
Magnetic moment M→=NIπr2j^
Torque τ→=M→×B→
=NIπr2j^×(Bi^)
=BNI πr2 (-k^) N m
τ→=BNI πr2 N m

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (08 Apr Shift 1)
A circular coil of 1000 turns each with area 1 m2 is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of 0.07 T. The maximum voltage generation will be _____ V.

Options:

The correct answer is:
440

Solution:
Given, N=1000, A=1 m2 and B=0.07 T

The maximum emf generated is εmax=NBAω,

=NBA2πf

=1000×0.07×1×2π1=140π

Thus, maximum emf is 439.6 V≈440 V.

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A circular coil of area $0.01 \mathrm{~m}^2$ and 40 turns is rotated about its vertical diameter with an angular speed of 50 $\mathrm{rad} \mathrm{s}^{-1}$ in a uniform horizontal magnetic field $0.05 \mathrm{~T}$. If the average power loss due to joule heating is $25 \mathrm{~mW}$, then the closed loop resistance of the coil is

Options:
A $50 \Omega$
B $12.5 \Omega$
C $75 \Omega$
D $20 \Omega$

The correct answer is:
$20 \Omega$

Solution:
We have
$\begin{aligned} & \mathrm{e}=\frac{\mathrm{d}}{\mathrm{dt}}=\mathrm{NBA} \frac{\mathrm{d}}{\mathrm{dt}}(\sin \omega \mathrm{t}) \\ & =\mathrm{NBA} \omega \sin \omega \mathrm{t}\end{aligned}$
$\begin{aligned} & \mathrm{e}_{\max }=\mathrm{NBA} \omega \\ & \mathrm{e}_{\mathrm{rms}}=\frac{\mathrm{NBA} \omega}{\sqrt{2}}\end{aligned}$
$\begin{aligned} & P_{\mathrm{avg}}=\frac{\mathrm{e}_{\mathrm{rms}}^2}{\mathrm{R}} \Rightarrow \mathrm{R}=\frac{\mathrm{e}_{\mathrm{rms}}^2}{\mathrm{P}_{\mathrm{avg}}}=\frac{\mathrm{N}^2 \mathrm{~B}^2 \mathrm{~A}^2 \omega^2}{2 \times \mathrm{P}_{\mathrm{avg}}} \\ & =\frac{40^2 \times 0.05^2 \times 0.01^2 \times 50^2}{2 \times 25 \times 10^{-3}} \\ & =20 \Omega\end{aligned}$

Topic: Physics, Electromagnetic Induction, JEE Main
A circular coil of area $100 \mathrm{~cm}^2$ and 20 turns is kept in magnetic field of flux density $2 \mathrm{~Wb}-\mathrm{m}^2$. It rotates from a position where its plane makes an angle of $30^{\circ}$ with the field to a position perpendicular to the field in a time $0.2 \mathrm{~s}$. Find the magnitude of the emf induced in the coil due to its rotation.

Options:
A 2 V
B 3 V
C 1.5 V
D 1 V

The correct answer is:
1 V

Solution:
Given that, area of coil, $A=100 \mathrm{~cm}^2$ $=100 \times 10^{-4} \mathrm{~m}^2$
Number of turns, $N=20$
Magnetic field, $B=2 \mathrm{~Wb}-\mathrm{m}^2$
Now, induced emf, $e=-\frac{\Delta \phi}{\Delta t}=\frac{-\left(\phi_2-\phi_1\right)}{t}$
$$
\begin{aligned}
e & =-\frac{\left(N B A \cos 0^{\circ}-N B A \cos 60^{\circ}\right)}{t}=\frac{-N B A\left(1-\frac{1}{2}\right)}{t} \\
& =\frac{-B A N}{2 t}=-\frac{2 \times 20 \times 100 \times 10^{-4}}{2 \times 0.2}=-1 \mathrm{~V}
\end{aligned}
$$
$\therefore$ Magnitude of induced emf $=1 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, JEE Main
A circular coil of diameter $7 \mathrm{~cm}$ has 24 turns of wire carrying current of $0.75 \mathrm{~A}$. The magnetic moment of the coil is

Options:
A $6.9 \times 10^{-2} a m p-m^2$
B $2.3 \times 10^{-2} \mathrm{amp}-\mathrm{m}^2$
C $10^{-2} a m p-m^2$
D $10^{-3} \mathrm{amp}-\mathrm{m}^2$

The correct answer is:
$6.9 \times 10^{-2} a m p-m^2$

Solution:
Given: \(d=7 \mathrm{~cm}, \mathrm{~N}=24, \mathrm{i}=0.75 \mathrm{~A}\)
The magnetic moment is given by, \(\mathrm{M}=\mathrm{NiA}\)
\(\begin{aligned}
& M=24 \times 0.75 \times 3.14 \times\left(3.5 \times 10^{-2}\right)^2 \\
& M=6.9 \times 10^{-2} \mathrm{Am}^2
\end{aligned}\)
Hence, option (A) is correct.

Topic: Physics, Magnetic Effects of Current, JEE Main
A circular coil of mean radius of $7 \mathrm{~cm}$ and having 4000 turns is rotated at the rate of 1800 revolutions per minute in the earth's magnetic field ( $B=$ 0.5 gauss), the maximum e.m.f. induced in coil will be

Options:
A $1.158 \mathrm{~V}$
B $0.58 \mathrm{~V}$
C $0.29 \mathrm{~V}$
D $5.8 \mathrm{~V}$

The correct answer is:
$0.29 \mathrm{~V}$

Solution:
$e_0=\omega N B A=(2 \pi v) N B\left(\pi r^2\right)=2 \times \pi^2 v N B r^2$
$=2 \times(3.14)^2 \times \frac{1800}{60} \times 4000 \times 0.5 \times 10^{-4} \times\left(7 \times 10^{-2}\right)^2$
$=0.58 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, JEE Main
A circular coil of radius $4 \mathrm{~cm}$ and of 20 turns carries a current of 3 amperes. It is placed in a magnetic field of intensity of 0.5 weber $/ \mathrm{m}^2$. The magnetic dipole moment of the coil is

Options:
A 0.15 ampere- $m^2$
B 0.3 ampere $-m^2$
C 0.45 ampere-m
D 0.6 ampere-m

The correct answer is:
0.3 ampere $-m^2$

Solution:
0.3 ampere \(-m^2\)
\(M=N L A=20 \times \frac{22}{7}\left(4 \times 10^{-2}\right)^2 3=0.3 \mathrm{~A}-\mathrm{m}^2\)

Topic: Physics, Magnetic Effects of Current, JEE Main
A circular coil of radius $4 \mathrm{~cm}$ has 50 turns. In this coil a current of $2 A$ is flowing. It is placed in a magnetic field of 0.1 weber $/ \mathrm{m}^2$. The amount of work done in rotating it through $180^{\circ}$ from its equilibrium position will be

Options:
A $0.1 \mathrm{~J}$
B $0.2 \mathrm{~J}$
C $0.4 \mathrm{~J}$
D $0.8 \mathrm{~J}$

The correct answer is:
$0.1 \mathrm{~J}$

Solution:
Magnetic moment of current carrying loop \(=\mathrm{M}=\mathrm{iAN}=\mathrm{i}\left(\pi \mathrm{r}^2\right) \mathrm{N}\)
\(\begin{aligned}
& \mathrm{W}=\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right)=\mathrm{MB}\left(\cos 0^{\circ}-\cos 180^{\circ}\right)=2 \mathrm{MB} \\
& =0.1 \mathrm{J}
\end{aligned}\)
\(\therefore\) Option A is correct.

Topic: Physics, Magnetic Effects of Current, JEE Main
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s-1 in a uniform horizontal magnetic field of 3.0×10-2 T. The maximum emf induced in the coil will be _______×10-2 volt (rounded off to the nearest integer).

Options:

The correct answer is:
60

Solution:
Emf=BωNAsinωt

Emfmax=BωNA=BωNπR2

=3×10-2×50×20×π×64104≈60×10-2 V

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A circular coil of radius 10 cm is placed in a uniform magnetic field of 3.0×10-5 T with its plane perpendicular to the field initially. It is rotated at constant angular speed about an axis along the diameter of coil and perpendicular to magnetic field so that it undergoes half of rotation in 0.2 s. The maximum value of EMF induced (in μV) in the coil will be close to the integer....

Options:

The correct answer is:
15

Solution:
Flux as a function of time ϕ=B→·A→=ABcos(ωt)

Emf induced,

e=-dϕdt=ABωsin(ωt)

Max. value of Emf = ABω

=πR2Bω

=3.14×0.1×0.1×3×10-5×π0.2=15

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2020 (02 Sep Shift 1)
A circular conducting coil of radius 1 m is being heated by the change of magnetic field B→ passing perpendicular to the plane in which the coil is laid. The resistance of the coil is 2 μΩ. The magnetic field is slowly switched off such that its magnitude changes in time as

B=4π×10-3 T1-t100

The energy dissipated by the coil before the magnetic field is switched off completely is E=______mJ.

Options:

The correct answer is:
80

Solution:
ϕ=B→·S→

ϕ=4π×10-31-t100·πR2

ϕ=4×10-3×121-t100

ε=-dϕdt

ε=-ddt4×10-31-t100

ε=4×10-31100=4×10-5 V

When B=0

1-t100=0

t=100 sec

Heat =ε2Rt

Heat =4×10-522×10-6×100 J

Heat =16×10-10×1002×10-6 J

Heat =0.08 J

Heat =80 mJ

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2021 (25 Jul Shift 1)
A circular disc $D_{1}$ of mass $M$ and radius $R$ has two identical $\operatorname{discs} D_{2}$ and $D_{3}$ of the same mass $M$ and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO', passing through the centre of $D_{1}$, as shown in the figure, will be

Options:
A $\mathrm{MR}^{2}$
B $3 \mathrm{MR}^{2}$
C $\frac{4}{5} \mathrm{MR}^{2}$
D $\frac{2}{3} \mathrm{MR}^{2}$

The correct answer is:
$3 \mathrm{MR}^{2}$

Solution:
Moment of inertia of disc $D_{1}$ about $O O^{\prime}=I_{1}=\frac{M R^{2}}{2}$

M.O.I of $\mathrm{D}_{2}$ about $\mathrm{OO}^{\prime}$ $=I_{2}=\frac{1}{2}\left(\frac{M R^{2}}{2}\right)+M R^{2}=\frac{M R^{2}}{4}+M R^{2}$

M.O.I of $\mathrm{D}_{3}$ about $\mathrm{OO}^{\prime}$ $=I_{3}=\frac{1}{2}\left(\frac{M R^{2}}{2}\right)+M R^{2}=\frac{M R^{2}}{4}+M R^{2}$

so, resultant M.O.I about $\mathrm{OO}^{\prime}$ is $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}$

$$

\begin{aligned}

\Rightarrow & I=\frac{M R^{2}}{2}+2\left(\frac{M R^{2}}{4}+M R^{2}\right) \\

&=\frac{M R^{2}}{2}+\frac{M R^{2}}{2}+2 M R^{2}=3 M R^{2}

\end{aligned}

$$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A circular disc is to be made by using iron and aluminium, so that it aquires momentum of inertia about its geometrical axis. It is possible with

Options:
A iron and aluminium layers in alternate order
B aluminium at interior and iron surrounding it
C iron at interior and aluminium surrounding it
D Either (a) or (c)

The correct answer is:
aluminium at interior and iron surrounding it

Solution:
By doing so the distribution of mass can be made away from the axis of rotation.

Topic: Physics, Rotational Motion, JEE Main
A circular disc of mass M and radius R is rotating about its axis with angular speed ω1. If another stationary disc having radius R2 and same mass M is dropped co-axially on to the rotating disc. Gradually both discs attain constant angular speed ω2. The energy lost in the process is p% of the initial energy. Value of p is ______

Options:

The correct answer is:
20

Solution:
I1ω1+I2ω2=I1+I2×ωf

MR22×ω+0MR22+MR281625

ωf=45ω

Final K.E., Kf=12MR22+MR281625ω2

Kf=MR2ω25 ; Ki=12MR22ω2=MR2ω24

Percentage loss in kinetic energy % loss=MR2ω24−MR2ω25MR2ω24×100=20%

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (04 Sep Shift 1)
A circular disc of radius $0.2 \mathrm{~m}$ is placed in a uniform magnetic field of induction $\frac{1}{\pi}\left(\frac{W b}{m^2}\right)$ in such a way that its axis makes an angle of $60^{\circ}$ with $\overrightarrow{\mathbf{B}}$. The magnetic flux linked with the disc is

Options:
A $0.02 \mathrm{~Wb}$
B $0.06 \mathrm{~Wb}$
C $0.08 \mathrm{~Wb}$
D $0.01 \mathrm{~Wb}$

The correct answer is:
$0.02 \mathrm{~Wb}$

Solution:
The magnetic flux $\phi$ passing surface of area $A$ placed in a u field $B$ is given by
$\phi=B A \cos \theta$
where $\theta$ is the angle between
$\begin{aligned}
& \text {Here, } \theta=60^{\circ}, B=\frac{1}{\pi} \mathrm{Wb} / \mathrm{m}^2, A=\pi(0.2)^2 \\
& \text {Therefore, } \phi=\frac{1}{\pi} \times \pi(0.2)^2 \times \cos 60^{\circ} \\
& =(0.2)^2 \times \frac{1}{2} \\
& =0.02 \mathrm{~Wb} \\
&
\end{aligned}$

Topic: Physics, Electromagnetic Induction, JEE Main
A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction linked with the disc is

Options:
A $0.01 \omega b$
B $0.02 \omega b$
C $0.06 \omega b$
D $0.08 \omega b$

The correct answer is:
$0.02 \omega b$

Solution:
$$
\begin{aligned}
& \text { Magnetic flux }=B A \cos \theta=\frac{1}{\pi} \times \pi(0.2)^2 \times \cos 60^{\circ} \\
& =0.04 \times \frac{1}{2}=0.02 \mathrm{wb}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, NEET, NEET 2008 (Mains)
A circular disc of radius b has a hole of radius a at its centre(see figure). If the mass per unit area of the disc varies as σ0r then, the radius of gyration of the disc about its axis passing through the center is

Options:
A a+b3
B a2+b2+ab3
C a+b2
D a2+b2+ab2

The correct answer is:
a2+b2+ab3

Solution:
Solution: (B)
dm=σ   2πrdr
=σ0r2 πrdr
m=2πσ0∫abdr=2π σ0(b-a)
I=2π σ0∫abr2dr=23πσ0b3-a3
Radius of gyration =Im =23π σ0b3-a32π σ0b-a
=b3-a33(b-a)=(b-a)(b2+a2+ab)3  (b-a)
=b2+a2+ab3.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A circular disc of radius $\mathrm{R}$ and thickness $\frac{R}{6}$ has moment inertia I about an axis passing through its centre perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter is

Options:
A $I$
B $\frac{2 I}{8}$
C $\frac{I}{5}$
D $\frac{I}{10}$

The correct answer is:
$\frac{I}{5}$

Solution:
According to problem disc is melted and recasted into a solid sphere so their volume

$$

\begin{array}{c}

\text { will be same. } \\

V_{\text {Disc }}=V_{\text {Sphere }} \Rightarrow \pi R_{\text {Disc }}^{2} t=\frac{4}{3} \pi R_{\text {Sphere }}^{3} \\

\Rightarrow \pi R_{\text {Disc }}^{2}\left(\frac{R_{\text {Disc }}}{6}\right)=\frac{4}{3} \pi R_{\text {Sphere }}^{3}\left[t=\frac{R_{\text {Disc }}}{6}, \text { given }\right] \\

\Rightarrow R_{\text {Disc }}^{3}=8 R_{\text {Sphere }}^{3} \Rightarrow R_{\text {Sphere }}=\frac{R_{\text {Disc }}}{2}

\end{array}

$$

Moment of inertia of disc

$$

\begin{array}{l}

I_{\text {Disc }}=\frac{1}{2} M R_{\text {Disc }}^{2}=I \text { (given) } \\

\therefore \text { M }\left(R_{\text {Disc }}\right)^{2}=2 I

\end{array}

$$

Moment of inertia of sphere

$$

\begin{array}{l}

I_{\text {Sphere }}=\frac{1}{2} M R_{\text {Sphere }}^{2} \\

=\frac{2}{5} M\left(\frac{R_{\text {Disc }}}{2}\right)^{2}=\frac{M}{10}\left(R_{\text {Disc }}\right)^{2}=\frac{2 I}{10}=\frac{I}{5}

\end{array}

$$

Topic: Physics, Rotational Motion, JEE Main
A circular disc of radius \( R \) is free to rotate about an axis passing through its centre. An external tangential force \( F \) is applied on the disc along its edge. If the angular velocity of disc is increased from \( 0 \) to \( \mathrm{w} \) in a time \( \mathrm{t} \) then the work done by \( \mathrm{F} \) during same time \( t \) is

Options:
A RFwt
B 2RFwt
C \( \frac{\text { RFwt }}{2} \)
D \( \mathrm{RF}(\mathrm{wt})^{2} \)

The correct answer is:
\( \frac{\text { RFwt }}{2} \)

Solution:
Work done W=12RFωt

Topic: Physics, Rotational Motion, JEE Main
A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ such that the circumferences of the discs touch. The centre of mass of the new disc is at a distance $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is

Options:
A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{1}{4}$
D $\frac{1}{6}$

The correct answer is:
$\frac{1}{3}$

Solution:
Let centre $O_1$ of disc be the origin. Due to symmetry the centre of mass of remaining part will lie at $x$-axis. Let $\mathrm{CM}$ of new disc is at point $\mathrm{O}_2 \quad$ where, $\mathrm{O}_1 \mathrm{O}_2=\alpha \mathrm{R}$ (given).
here mass of cut off portion,
$m=\frac{\pi(R)^2 \cdot M}{\pi(2 R)^2}=\frac{M}{4}$
and position of its centre of mass, $O_1 O_1=R$ Hence, for remaining part (new disc)
$\begin{aligned}
x_{\mathrm{CM}}=\alpha R & =\frac{M \times 0-\left(\frac{M}{4}\right) \times R}{M-\frac{M}{4}} \\
& =-\frac{R}{3} \Rightarrow \alpha=\frac{1}{3}
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2011
A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ such that the circumferences of the discs touch. The centre of mass of the new disc is at a distance $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is

Options:
A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{1}{4}$
D $\frac{1}{6}$

The correct answer is:
$\frac{1}{3}$

Solution:
Let centre $O_1$ of disc be the origin. Due to symmetry the centre of mass of remaining part will lie at $x$-axis. Let $\mathrm{CM}$ of new disc is at point $\mathrm{O}_2 \quad$ where, $\mathrm{O}_1 \mathrm{O}_2=\alpha \mathrm{R}$ (given).
here mass of cut off portion,
$m=\frac{\pi(R)^2 \cdot M}{\pi(2 R)^2}=\frac{M}{4}$
and position of its centre of mass, $O_1 O_1=R$ Hence, for remaining part (new disc)
$\begin{aligned}
x_{\mathrm{CM}}=\alpha R & =\frac{M \times 0-\left(\frac{M}{4}\right) \times R}{M-\frac{M}{4}} \\
& =-\frac{R}{3} \Rightarrow \alpha=\frac{1}{3}
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ such that the circumferences of the discs touch. The centre of mass of the new disc is at a distance $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is

Options:
A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{1}{4}$
D $\frac{1}{6}$

The correct answer is:
$\frac{1}{3}$

Solution:
Let centre $O_1$ of disc be the origin. Due to symmetry the centre of mass of remaining part will lie at $x$-axis. Let $\mathrm{CM}$ of new disc is at point $\mathrm{O}_2 \quad$ where, $\mathrm{O}_1 \mathrm{O}_2=\alpha \mathrm{R}$ (given).
here mass of cut off portion,
$m=\frac{\pi(R)^2 \cdot M}{\pi(2 R)^2}=\frac{M}{4}$
and position of its centre of mass, $O_1 O_1=R$ Hence, for remaining part (new disc)
$\begin{aligned}
x_{\mathrm{CM}}=\alpha R & =\frac{M \times 0-\left(\frac{M}{4}\right) \times R}{M-\frac{M}{4}} \\
& =-\frac{R}{3} \Rightarrow \alpha=\frac{1}{3}
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, TS EAMCET, TS EAMCET 2011
A circular disc of radius $\mathrm{R}$ is removed from a bigger circular disc of radius $\mathrm{2R}$ such that the circumferences of the discs coincide. The centre of mass of the new disc is $\mathrm{\alpha / R}$ from the centre of the bigger disc. The value of $\mathrm{\alpha}$ is

Options:
A
$1 / 3$

B
$1 / 2$

C
$1 / 6$

D
$1 / 4$

The correct answer is:
$1 / 3$

Solution:
In this question distance of centre of mass of new disc is $\alpha R$ not $\frac{\alpha}{R}$.
$\begin{aligned}
& -\frac{3 \mathrm{M}}{4} \alpha \mathrm{R}+\frac{\mathrm{M}}{4} \mathrm{R}=0 \\
& \Rightarrow \alpha=\frac{1}{3}
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2007
A circular disc of radius $R$ is removed from one end of a bigger circular disc of radius $2 R$. The centre of mass of the new disc is at a distance $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is

Options:
A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{1}{4}$
D $\frac{1}{6}$

The correct answer is:
$\frac{1}{3}$

Solution:
Let, mass of entire disc $=M$
Mass per unit area $=\frac{M}{\pi(2 R)^2}=\frac{M}{4 \pi R^2}$
Mass of removed disc of radius $R$,
$$
M_1=\frac{M}{4 \pi R^2} \times R^2=\frac{\mathrm{M}}{4}
$$
Mass of remaining disc,
$$
\Rightarrow \quad M_2=M-\frac{M}{4}=\frac{3 M}{4}
$$
Center of mass of removed disc is $C_1$ and centre of mass of remaining new disc is $C_2$.
And centre of mass of combination of $M_1$ and $M_2$ will be at $C(0,0)$.
$$
\begin{aligned}
\Rightarrow \quad & \frac{M_1 x_1-M_2 x_2}{M_1-M_2}=0 \\
\Rightarrow \quad & M_1 x_1=M_2 x_2 \\
& \frac{M}{4} \cdot R=\frac{3 M}{4}(\alpha R) \quad \Rightarrow \alpha=\frac{1}{3}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2018 (24 Apr Shift 1)
A circular disc reaches from top to bottom of an inclined plane of length L. When it slips down the plane, it takes time t1. When it rolls down the plane, it takes time t2. The value of t2t1 is 3x. The value of x will be

Options:

The correct answer is:
2

Solution:
If disk slips on inclined plane, then its acceleration, a1=gsinθ
L=12a1t12
⇒t1=2 La1 …i

If disk rolls on inclined plane, its acceleration,
a2=gsinθ1+ImR2
a2=gsinθ1+mR22 mR2
a2=23gsinθ
Now L=12a2·t22
⇒t2=2 La2 …ii

Now, t2t1=a1a2=32
⇒x=2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (20 Jul Shift 1)
A circular disc $X$ of radius $R$ is made from an iron plate of thickness $t$, and another disc $Y$ of radius $4 \mathrm{R}$ is made from an iron plate of thickness $\frac{\mathrm{t}}{4}$. Then the relation between the moment of inertia $\mathrm{I}_{\mathrm{X}}$ and $\mathrm{I}_{\mathrm{Y}}$ is

Options:
A
$\mathrm{I}_{\mathrm{Y}}=32 \mathrm{I}_{\mathrm{x}}$

B
$\mathrm{I}_{\mathrm{Y}}=16 \mathrm{I}_{\mathrm{x}}$

C
$\mathrm{I}_{\mathrm{Y}}=\mathrm{I}_{\mathrm{X}}$

D
$\mathrm{I}_{\mathrm{Y}}=64 \mathrm{I}_{\mathrm{X}}$

The correct answer is:
$\mathrm{I}_{\mathrm{Y}}=64 \mathrm{I}_{\mathrm{X}}$

Solution:
$$
\mathrm{I}=\frac{1}{2} \mathrm{mR} \quad \text { or } \quad \mathrm{M} \propto \mathrm{t} \propto \mathrm{R}^2
$$
For $\operatorname{disc} X, I_x=\frac{1}{2}(m)(R)^2=\frac{1}{2}\left(\pi r^2 t\right) .(R)^2$
For disc $Y, I_y=\frac{1}{2}\left[\pi(4 R)^2 \cdot t / 4\right][4 R]^2$
$$
\Rightarrow \frac{I_x}{I_y}=\frac{1}{(4)^3} \Rightarrow I_y=64 I_x
$$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2003
A circular disc $\mathrm{x}$ of radius $R$ is made from an iron plate of thickness $t$, and another disc $Y$ of radius $4 R$ is made from an iron plate of thickness $t / 4$. Then the relation between the moment of inertia $I_x$ and $I_Y$ is

Options:
A $I_Y=64 I_X$
B $I_Y=32 I_X \mid$
C $I_Y=16 I_X$
D $I_Y=I_X$

The correct answer is:
$I_Y=64 I_X$

Solution:
Given,
The radius of the disc $=R$
The thickness of the disc $=t$
The radius of the second disc $=4 R$
The thickness of the second $\operatorname{disc}=\frac{t}{4}$
Now,
Moment of Inertia of $\operatorname{disc}^{\prime} X^{\prime}, I_x=\frac{M_x R_x^2}{2}$
Where $M_x=\rho\left(\pi R_x^2\right) t_x=\rho \pi R^2 t$
Moment of Inertia of $\operatorname{disc}^{\prime} Y^{\prime}, I_y=\frac{M_y R_y^2}{2}$
Where $M_y=\rho \pi(4 R)^2 \frac{t}{4}=4 \rho \pi R^2 t$
Ratio of the moment of inertia of both the discs
$\therefore \frac{I_x}{I_y}=\frac{\left(\rho \pi R^2 t\right) R^2}{2} \times \frac{2}{\left(4 \rho \pi R^2 t\right) 16 R^2}$
After solving
$\Rightarrow \frac{\mathrm{I}_x}{I_y}=\frac{1}{64} \Rightarrow I_y=64 I_x$
Hence, the moment of inertia of $Y$ is 64 times the moment of inertia of $X$

Topic: Physics, Rotational Motion, JEE Main
A circular freeway entrance and exit are commonly banked to control a moving car at $14 \mathrm{~m} / \mathrm{s}$. To design similar ramp for $28 \mathrm{~m} / \mathrm{s}$ one should

Options:
A increase the radius by factor 2
B increase the radius by factor 4
C decrease the radius by factor 4
D decrease the radius by factor 2

The correct answer is:
increase the radius by factor 4

Solution:
Given, $v_1=14 \mathrm{~m} / \mathrm{s}$ and $v_2=28 \mathrm{~m} / \mathrm{s}$
As we know that, $\tan \theta=\frac{v^2}{r g} \Rightarrow r=\frac{v^2}{g \tan \theta}$
In the first case, $r_1=\frac{v_1^2}{g \tan \theta}=\frac{(14)^2}{g \tan \theta}$ $\ldots$ (i)
In the second case,
$r_2=\frac{(28)^2}{g \tan \theta}$ [for similar ramp remains same]
$r_2=\frac{(14 \times 2)^2}{g \tan \theta}=4 \frac{(14)^2}{g \tan \theta} \Rightarrow r_2=4 r_1 \quad \text { [using Eq. (i)] }$
Hence, radius should be increased by factor 4 .

Topic: Physics, Laws of Motion, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 2)
A circular hole is cut from a disc of radius 6 cm in such a way that the radius of the hole is 1 cm and the centre of the hole is at a distance of 3 cm from the centre of the disc. The distance of the centre of mass of the remaining part from the centre of the original disc is

Options:
A 3/35 cm
B 1/35 cm
C 3/10 cm
D 7/35 cm

The correct answer is:
3/35 cm

Solution:
For the calculation of the position of centre of mass, cut off mass is taken as negative. The mass of disc is
m 1 = π r 1 2 σ
= π 6 2 σ = 36 π σ
where σ = surface mass density. The mass of cut out portion is
m 2 = π 1 2 σ = π σ
x cm = m 1 x 1 - m 2 x 2 m1-m2
Taking origin at the centre of disc,
x1 = 0, x2 = 3 cm
x cm = 36 π σ × 0 - π σ × 3 3 6 π σ - π σ
= - 3 π σ 3 5 π σ = - 3 3 5 cm

Topic: Physics, Center of Mass Momentum and Collision, NEET