datavorous/entrance-exam-dataset huggingface dataset:
A circular hole of diameter $\mathrm{R}$ is cut from a disc of mass $M$ and radius $R$; the circumference of the cut passes through the centre of the disc. The moment of inertia of the remaining portion of the disc about an axis perpendicular to the disc and passing through its centre is

Options:
A
$\left(\frac{15}{32}\right) M R^2$

B
$\left(\frac{1}{8}\right) M R^2$

C
$\left(\frac{3}{8}\right) M R^2$

D
$\left(\frac{13}{32}\right) M R^2$

The correct answer is:
$\left(\frac{13}{32}\right) M R^2$

Solution:
M.I. of complete disc about its centre $O$.
$$
I_{\mathrm{Total}}=\frac{1}{2} M R^2
$$
Mass of circular hole (removed)
$$
=\frac{M}{4} \quad\left(\text { As } M=\pi R^2 t \therefore M \propto R^2\right)
$$
M.I. of removed hole about its own axis
$$
=\frac{1}{2}\left(\frac{M}{4}\right)\left(\frac{R}{2}\right)^2=\frac{1}{32} M R^2
$$
M.I. of removed hole about $O^{\prime}$
$$
\begin{aligned}
I_{\text {removed hole }} & =I_{\mathrm{cm}}+m x^2 \\
& =\frac{M R^2}{32}+\frac{M}{4}\left(\frac{R}{2}\right)^2 \\
& =\frac{M R^2}{32}+\frac{M R^2}{16}=\frac{3 M R^2}{32}
\end{aligned}
$$
M.I. of complete disc can also be written as
$I_{\text {Total }}=I_{\text {removed hole }}+I_{\text {remaining disc }}$
$I_{\text {Total }}=\frac{3 M R^2}{32}+I_{\text {remaining disc }}$
From eq. (i) and (ii),
$$
\begin{aligned}
& \frac{1}{2} M R^2=\frac{3 M R^2}{32}+I_{\text {remaining disc }} \\
& \Rightarrow I_{\text {remaining disc }} \\
& \quad=\frac{M R^2}{2}-\frac{3 M R^2}{32}=\left(\frac{13}{32}\right) M R^2
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2012 (07 May Online)
A circular hole of radius 1 cm is cut off from a disc of radius 6 cm. The centre of the hole is 3 cm from the centre of the disc. Then the distance of the centre of mass of the remaining disc from the centre of the disc is

Options:
A 335 cm
B 135 cm
C 310 cm
D none of these

The correct answer is:
335 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
=-3πσ35πσ=-335 cm

Topic: Physics, Center of Mass Momentum and Collision, NEET
A circular hole of radius $3 \mathrm{~cm}$ is cut out from a uniform circular disc of radius $6 \mathrm{~cm}$. The centre of the hole is at $3 \mathrm{~cm}$, from the centre of the original disc. The distance of centre of gravity of the resulting flat body from the centre of the original disc is

Options:
A 0.5 cm
B 1 cm
C 1.5 cm
D 0.75 cm

The correct answer is:
1 cm

Solution:
Let mass per unit area of the disc $=m$
Total mass of disc, $M=\pi R^2 m$
Mass of the scooped out portion of the disc,
$M^{\prime}=\pi r^2 m$
For small objects, centre of gravity and centre of mass are at same position.
We take the centre $O$ of the disc as the origin. The masses $M$ and $M^{\prime}$ can be supposed to be concentrated at the centres of the disc and scooped out portion respectively. The mass $M^{\prime}$ of the removed portion is taken negative.
The $x$-coordinate of the centre of mass of the remaining portion of the disc will be
$\begin{aligned}
x_{\mathrm{CM}} & =\frac{M x_1-M^{\prime} x_2}{M-M^{\prime}}=\frac{M \times 0-M^{\prime} \times 3}{\pi R^2 m-\pi r^2 m}=\frac{-3 M^{\prime}}{\pi m\left(R^2-r^2\right)} \\
& =\frac{-3 \pi r^2 m}{\pi m\left(R^2-r^2\right)}=-\frac{-3 r^2}{R^2-r^2}=\frac{-3 \times 3^2}{6^2-3^2}=-1 \mathrm{~cm} \\
\Rightarrow x_{\mathrm{CM}} & =-\mathrm{lcm}
\end{aligned}$
Hence, distance of centre of gravity of the resulting flat body from the centre of the original disc will be $1 \mathrm{~cm}$ left side.

Topic: Physics, Center of Mass Momentum and Collision, TS EAMCET, TS EAMCET 2020 (10 Sep Shift 1)
A circular hole of radius a2 is cut out of a circular disc of radius a as shown in figure. The centroid of the remaining circular portion with respect to point O will be:

Options:
A 23a
B 56a
C 16a
D 1011a

The correct answer is:
56a

Solution:
Let σ be the uniform mass density of disc then

xCOM=σπa2a-σπa24×3a2σπa2-σπa24

=a-3a81-14=5a6

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2021 (24 Feb Shift 2)
A circular hole of radius R4 is made in a thin uniform disc having mass and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is-

Options:
A 219MR2256
B 237MR2512
C 197MR2256
D 19MR2512

The correct answer is:
237MR2512

Solution:
ID=mr22
Moment of inertia of removed portion about the axis.
Iremoved=12 m16 r216+m16 9r216 (by parallel axis theorem)
=mr2+18mr2512
=19mr2512
Iremaining=mr22-19512 mr2
=237512 mr2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2017 (09 Apr Online)
A circular loop and a square loop are formed from two wires of same length and cross section. Same current is passed through them. Then, the ratio of their dipole moments is

Options:
A 4
B $\frac{2}{\pi}$
C 2
D $\frac{4}{\pi}$

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

Solution:
Dipole moments due to current carrying circular wire $M=i A$
Here,
$$
\begin{aligned}
\frac{M_1}{M_2} & =\frac{i \pi r^2}{i a^2}=\pi\left[\frac{\left(\frac{1}{2 \pi}\right)^2}{\left(\frac{1}{4}\right)^2}\right] \\
& =\pi \frac{16}{4 \pi^2}=\frac{4}{\pi}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with a bigger loop is:

Options:
A 3.3×10-11weber
B 6.6×10-9weber
C 9.1×10-11weber
D 6×10-11weber

The correct answer is:
9.1×10-11weber

Solution:
ϕ2=B1×A2=μ0IR122(R12+x2)32×πR22
= μ0(2)(20×10−2)22[(0.2)2+(0.15)2]32×π(0.3×10−2)
On solving
= 9.216 × 1 0 - 1 1
≈ 9.2 × 1 0 - 1 1 weber

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2013 (07 Apr)
A circular loop of radius R carrying current I is kept in XZ plane. A uniform and constant magnetic field B→=B0i^+2B0j^+3B0k^ exists in the region (B0- a positive constant) . Then the magnitude of the torque acting on the loop will be

Options:
A 2IπR2B0
B 5IπR2B0
C 10IπR2B0
D 6IπR2B0

The correct answer is:
10IπR2B0

Solution:
Magnetic dipole moment, M→=±IπR2j^
τ→=M→×B→=±IπR2j^×B0i^+2B0j^+3B0k^
τ→=±IπR2B0-k^+3i^

Topic: Physics, Magnetic Effects of Current, JEE Main
A circular loop of radius R is carrying current i A. The ratio of magnetic field at the centre of circular loop and at a distance R from the center of the loop on its axis is :

Options:
A 1:32
B 32:2
C 22:1
D 1:2

The correct answer is:
22:1

Solution:
The magnitude of magnetic field at the centre of loop is

B1 =Bcentre =μ∘i2R …1

The magnitude of magnetic field at a point at a distance x from the centre along axis of loop is

Baxis =μ∘iR22R2+x232

B2 =μ∘iR22232R3 …2 (since x=R)

Dividing equation 1 by equation 2, we get

B1B2=22:1

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (24 Jan Shift 1)
A circular loop of radius $r$ of conducting wire connected with a voltage source of zero internal resistance produces a magnetic field
$B$ at its centre. If instead, a circular loep of radius $2 r,$ made of same material, having the same cross-section is connected to the same voltage source, what will be the magnetic ficld at its centre?

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

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

Solution:
$$
B=\frac{\mu_{0}}{2} \cdot \frac{I}{r}
$$
$\because$
$$
I_{1}=\frac{V}{R_{1}}
$$
Case (I) $l_{1}=2 \pi r$ (length of the loop).
$$
R_{1}=\rho \frac{l_{1}}{A} \text { and } B_{1}=\frac{\mu_{0}}{2} \cdot \frac{I_{1}}{r}
$$
Case (II) $I_{2}=2 \pi \cdot 2 r=2 l_{1}$
$$
\begin{array}{l}
R_{2}=\rho \cdot \frac{l_{2}}{A}=\rho \cdot \frac{2 l_{1}}{A}=2 R_{1} \\
I_{2}=\frac{V}{R_{2}}=\frac{V}{2 R_{1}}=\frac{I_{1}}{2} \\
B_{2}=\frac{\mu_{0}}{2} \cdot \frac{I_{2}}{2 r}=\frac{\mu_{0}}{2} \cdot \frac{I_{1}}{2 \cdot 2 r}
\end{array}
$$
From Eqs. (i) and (ii),
$$
\frac{B_{2}}{B_{1}}=\frac{\frac{\mu_{0}}{2} \cdot \frac{I_{1}}{2 \cdot 2 r}}{\frac{\mu_{0}}{2} \cdot \frac{I_{1}}{r}}=\frac{1}{4}
$$
Magnetic field $B_{2}=\frac{B_{1}}{4}$

Topic: Physics, Magnetic Effects of Current, JEE Main
A circular metal plate of radius $R$ is rotating with a uniform angular velocity $\omega$ with its plane perpendicular to a uniform magnetic field $B$. Then the emf developed between the centre and the rim of the plate is

Options:
A $\pi \omega B R^2$
B $\omega B R^2$
C $\pi \omega B R^2 / 2$
D $\omega B R^2 / 2$

The correct answer is:
$\omega B R^2 / 2$

Solution:
We write $d \in=\operatorname{Blv}(\epsilon \mathrm{HF})$
Now l $=\mathrm{dx} \mathrm{v}=\omega \mathrm{x}$
$\int \mathrm{d} \in=\mathrm{B} \omega \int_0^r \mathrm{x} d x$
$\in=\frac{\mathrm{B} \omega \mathrm{r}^2}{2}$

Topic: Physics, Electromagnetic Induction, JEE Main
A circular plate $A$ of radius $1.5 \mathrm{r}$ is removed from one edge of a uniform circular plate $B$ of radius $2 r$. The distance of centre of mass of the remaining portion from the centre of the plate $B$ is

Options:
A $\frac{5 \mathrm{r}}{12}$
B $\frac{9 \mathrm{r}}{14}$
C $\frac{3 r}{4}$
D $\frac{7 r}{8}$

The correct answer is:
$\frac{9 \mathrm{r}}{14}$

Solution:
Radius of removed circular plate, $\mathrm{A}=1.5 \mathrm{r}$
Radius of circular plate, $\mathrm{B}=2 \mathrm{r}$
Let mass of circular plate $A, M_1=M$
So surface mass density, $\sigma=\frac{M}{\pi(2 r)^2}=\frac{M}{4 \pi r^2}$
Then mass of removed circular plate, A
$\begin{aligned} & \mathrm{M}_2=\sigma \mathrm{A} \\ & =\frac{\mathrm{M}}{4 \pi \mathrm{r}^2} \times \pi(1.5 \mathrm{r})^2 \\ & =\frac{\mathrm{M}}{4} \times 1.5 \times 1.5=0.5625 \mathrm{M}\end{aligned}$
The distance of centre of mass of the remaining portion from the centre of the plate $B$ is
$\begin{aligned} & X_{c m}=\frac{M_1 x_1-M_2 x_2}{M_1-M_2} \\ & =\frac{M \times 0-0.5625 M \times 0.5 g}{M-0.5625 M}=\frac{9 r}{14}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2023 (16 May Shift 1)
A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now the platform is given an angular velocity ω0. When the tortoise moves along a chord of the platform with a constant velocity (w.r.t. the platform), which of the following graphs shows properly the variation of angular velocity of the platform with the time t?

Options:
A
B
C
D

The correct answer is:

Solution:
As there is no external torque, angular momentum will remain constant. When the tortoise moves from A to C, figure, moment of inertia of the platform and tortoise decreases. Therefore, angular velocity of the system increases. When the tortoise moves from C to B, moment of inertia increases. Therefore, angular velocity decreases
If, M=mass of platform
R=radius of platform
m=mass of tortoise moving along the chord AB
a=perpendicular distance of O from AB
Initial angular momentum, I1=mR2+MR22
At any time t, let the tortoise reach D moving with velocity v
â´ AD=vt
AC=R2-a2
As DC=AC-AD=(R2-a2-vt)
â´ OD=r=a2+R2-a2-vt2
Angular momentum at time t
I2=mr2+MR22
As angular momentum is conserved
â´ I1Ï0=I2Ït
This shows that variation of Ï(t) with time is nonlinear. Choice

Topic: Physics, Rotational Motion, JEE Main
A circular platform is mounted on a frictionless vertical axle. Its radius $R=2 \mathrm{~m}$ and its moment of inertia about the axle is $200 \mathrm{~kg} \mathrm{~m}^2$. It is initially at rest. A $50 \mathrm{~kg}$ man stands on the edge of the platform and begins to walk along the edge at the speed of $1 \mathrm{~ms}^{-1}$ relative to the ground. Time taken by the man to complete one revolution is

Options:
A $\pi \sec$
B $\frac{3 \pi}{2} \sec$
C $2 \pi \mathrm{sec}$
D $\frac{\pi}{2} \sec$

The correct answer is:
$2 \pi \mathrm{sec}$

Solution:
From conservation of angular momentum
$$
\begin{aligned}
I \omega & =m v r \\
200 \times \omega & =50 \times 2 \times 1 \\
\omega & =\frac{1}{2} \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
$$
\begin{aligned}
\quad v & =r \omega=1 \mathrm{~m} / \mathrm{s} \\
\therefore \quad T & =\frac{2 \pi r}{1-(-1)}=\frac{2 \pi r}{2}=\pi r=2 \pi \mathrm{s}
\end{aligned}
$$

Topic: Physics, Rotational Motion, NEET, NEET 2012 (Mains)
A circular portion of radius $R_2$ has been removed from one edge of a circular disc of radius $R_1$. The correct expression for the centre of mass for the remaining portion of the disc is

Options:
A $-\frac{R_2^2}{R_1+R_2}$
B $-\frac{R_2^2}{R_1-R_2}$
C $\frac{R_2^2}{R_1+R_2}$
D $-\frac{R_1^2}{R_1+R_2}$

The correct answer is:
$-\frac{R_2^2}{R_1+R_2}$

Solution:
Given situation is shown below.
Here we treat removed mass as negative mass. As solid is symmetrical about $X$ - axis, its CM lies along $X$ - axis.
Now
$x_{\mathrm{CM}}=\frac{m_1 x_1-m_2 x_2}{m_1-m_2}$
$=\frac{d \times \pi R_1^2 \times 0-d \times \pi R_2^2\left(R_1-R_2\right)}{d \times \pi R_1^2-d \times \pi R_2^2}$
Here, $d=$ Area wise mass density
$\Rightarrow x_{\mathrm{CM}}=\frac{-R_2^2\left(R_1-R_2\right)}{R_1^2-R_2^2}$
$=\frac{-R_2^2\left(R_1-R_2\right)}{\left(R_1+R_2\right)\left(R_1-R_2\right)}=\frac{-R_2^2}{R_1+R_2^{\prime}}$

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 2)
A circular racing car track has a radius of curvature of \( 500 \mathrm{~m} \). The maximum speed of the car is \( 180 \mathrm{~km} \mathrm{hr}^{-1} \). The angle of banking \( \theta \) is \( \left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right) \)

Options:
A \( \theta=\tan ^{-1}(0) \)
B \( \theta=\tan ^{-1}(0.5) \)
C \( \theta=\tan ^{-1}(0.3) \)
D \( \theta=\tan ^{-1}(0.1) \)

The correct answer is:
\( \theta=\tan ^{-1}(0.5) \)

Solution:
Angle of banking θ is given by

tanθ=v2rg ...1

where v=speed of the vehicle

r= radius of the road

∴ Given, v=180 km h-1=180×518 m s-1=50 m s-1

and r=500 m.

Putting in equation 1 we get

tanθ=502500×10=0.5

⇒θ=tan-10.5.

Topic: Physics, Laws of Motion, JEE Main
A circular ring of mass $10 \mathrm{~kg}$ rolls along a horizontal floor. The center of mass of the ring has a speed $1.5 \mathrm{~m} / \mathrm{s}$. The work required to stop the ring is

Options:
A $10 \mathrm{~J}$
B $-6 J$
C $14.5 \mathrm{~J}$
D $-22.5 \mathrm{~J}$

The correct answer is:
$-22.5 \mathrm{~J}$

Solution:
Given,
mass of a circular ring, $m=10 \mathrm{~kg}$
speed of centre of mass of the ring
i.e, linear speed of ring, $u=1.5 \mathrm{~m} / \mathrm{s}$
Total initial kinetic energy of rolling ring,
$$
K_i=K_{\text {rotational }}+K_{\text {linear }}=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2
$$
Here, $I=m R^2$ and $\omega=\frac{v}{R}$
$$
\begin{aligned}
& =\frac{1}{2} m R^2\left(\frac{v}{R}\right)^2+\frac{1}{2} m v^2=\frac{1}{2} m v^2+\frac{1}{2} m v^2=m v^2 \\
& =10 \times(1.5)^2=22.5 \mathrm{~J}
\end{aligned}
$$
According to work-energy theorem, work required to stop the ring.
$\omega=$ change in kinetic energy
$$
=K_f-K_i=0-22.5=-22.5 \mathrm{~J}
$$

Topic: Physics, Rotational Motion, JEE Main
A circular road of $1000 \mathrm{~m}$ radius has banking angle $45^{\circ}$, the maximum safe speed of a car having 2000 $\mathrm{kg}$. mass will be, if the coefficient of friction between tyre and road is 0.5 .

Options:
A $86 / \mathrm{m} / \mathrm{sec}$
B $99 \mathrm{~m} / \mathrm{sec}$
C $124 \mathrm{~m} / \mathrm{sec}$
D $172 \mathrm{~m} / \mathrm{sec}$

The correct answer is:
$172 \mathrm{~m} / \mathrm{sec}$

Solution:
$172 \mathrm{~m} / \mathrm{sec}$
Given that $r=1000 \mathrm{~m} ; \theta=45^{\circ}$; and $\mu=0.5$;
Maximum safe velocity:
$\begin{aligned} & v=\sqrt{\frac{r g(\tan \theta+\mu)}{1-\mu \tan \theta}} \\ & v=\sqrt{\frac{1000 \times 9.8\left(\tan 45^{\circ}+0.5\right)}{1-0.5 \times \tan 45^{\circ}}}\end{aligned}$
$v=172 \mathrm{~m} / \mathrm{s}$
So the correct answer is option 4.

Topic: Physics, Laws of Motion, JEE Main
A circular wire loop of radius $1 \mathrm{~cm}$ carries a total charge $1 \times 10^{-6} \mathrm{C}$ distributed uniformly over its length. If $0.01 \%$ of its length (circumference) is cut-off, then the electric field at the centre of the loop due to the remaining wire is
$$
\left(\text { Take, } \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \text { SI unit }\right)
$$

Options:
A $3 \times 10^3 \mathrm{~N} / \mathrm{C}$
B $6 \times 10^3 \mathrm{~N} / \mathrm{C}$
C $9 \times 10^3 \mathrm{~N} / \mathrm{C}$
D $1.2 \times 10^2 \mathrm{~N} / \mathrm{C}$

The correct answer is:
$9 \times 10^3 \mathrm{~N} / \mathrm{C}$

Solution:
Radius of the loop, $r=1 \mathrm{~cm}=0.01 \mathrm{~m}$
Total charge on the loop, $Q=1 \times 10^{-6} \mathrm{C}$
$$
\begin{aligned}
& \therefore \text { Charge per unit length, } \lambda=\frac{Q}{2 \pi r}=\frac{10^{-6}}{2 \pi \times 0.01} \\
& =\frac{10^{-4}}{2 \pi} \mathrm{C} / \mathrm{m} \\
& l^{\prime}=0.01 \% \text { of the length }=\frac{0.01}{100} \times 2 \pi r \\
& \Rightarrow \quad l^{\prime}=\frac{2 \pi \times 0.01 \times 0.01}{100} \\
& =2 \pi \times 10^{-6} \mathrm{~m}
\end{aligned}
$$
For the whole circular loop, the electric field at the centre is zero due to symmetry. But when a certain portion of the loop is cut off, then there will exist an electric field at the centre due to that cutting part.
Which is equal to the electric field due to remaining part of wire as per charge conservation.
$\therefore$ Charge of the part $l^{\prime}, Q^{\prime}=l^{\prime} \lambda=\frac{2 \pi \times 10^{-6} \times 10^{-4}}{2 \pi}$
$$
=10^{-10} \mathrm{C}
$$
$\therefore$ Electric field at centre,
$$
\begin{aligned}
E & =\frac{Q^{\prime}}{4 \pi \varepsilon_0 r^2} \\
E & =\frac{10^{-10} \times 9 \times 10^9}{(0.01)^2} \\
& =9 \times 10^3 \mathrm{~N} / \mathrm{C}
\end{aligned}
$$

Topic: Physics, Electrostatics, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 2)
A circular wire loop of radius, 10 cm carries a total charge of, 10-5C distributed uniformly over its length. A small length of, 3.14×10-6 m of wire is cut off. The magnitude of the electric field at the centre due to the remaining wire is,
(Assume 14πϵ0=9×109 SI units)

Options:
A 30 N C-1
B 40 N C-1
C 35 N C-1
D 45 N C-1

The correct answer is:
45 N C-1

Solution:
Let's divide the loop in two parts. 1. The small element of length, dl, which is to be cut off, 2. The Remaining part of the loop,
Initially, by symmetry, the net Electric field at the center is zero.
∴ E Remaining Wire â +Edl element â=0,
⇒E remaining wire=−Edl element .
The charge per unit length of the wire is, λ=Q2πa, the charge on the dl length of the wire, dQ=λdl=Qdl2πa.
ââââââThe small element can be considered as a point charge, So its Electric Field at the center,
Edl=kdQa2=kQdl2πa3=9×109×10-5×π×10-62×π×10×10-23=45 N C-1

Topic: Physics, Capacitance, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 1)
A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then b+3g is equal to

Options:

The correct answer is:
17

Solution:
Given, b boys and g girls

Also given total ways of selecting 3 boys and 2 girls which will be C3b×C2g=168

⇒ bb-1b-26×gg-12=168

⇒ bb-1b-2gg-1=7×6×4×3×2×2

⇒ bb-1b-2gg-1=8×7×6×3×2

Now on comparing both side we get, b=8 and g=3 ⇒b+3g=8+9=17

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A clock has $75 \mathrm{~cm}, 60 \mathrm{~cm}$ long second hand and minute hand respectively. In 30 minutes duration the tip of second hand will travel $x$ distance more than the tip of minute hand. The value of $x$ in meter is nearly (Take $\pi=3.14$ ) :

Options:
A 140.5
B 118.9
C 139.4
D 220.0

The correct answer is:
139.4

Solution:
$\begin{aligned} \mathrm{x}_{\min }= & \pi \times \mathrm{r}_{\min } \\ = & \pi \times \frac{60}{100} \mathrm{~m} . \\ \mathrm{x}_{\text {second }} & =30 \times 2 \pi \times \mathrm{r}_{\text {second }} \\ & =30 \times 2 \pi \times \frac{75}{100} \\ \mathrm{x}= & \mathrm{x}_{\text {second }}-\mathrm{x}_{\min } \\ = & 139.4 \mathrm{~m}\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (08 Apr Shift 1)
A clock has a continuously moving second's hand of 0.1m length. The average acceleration of the tip of the hand (in units of ms-2 ]) is of the order of :

Options:
A 10-3
B 10-4
C 10-2
D 10-1

The correct answer is:
10-3

Solution:
The instantaneous acceleration of tip is given by

ac=ω2R=2π602×0.1=1900≈10-3

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2020 (06 Sep Shift 1)
A clock pendulum having coefficient of linear expansion, α=9×10-7/Co has a period of 0.5s at 20oC. If the clock is used in a climate where the temperature is 30oC, how much time does the clock lose in each oscillation? g=constant

Options:
A 2.5×10-7s
B 5×10-7s
C 1.125×10-6s
D 2.25×10-6s

The correct answer is:
2.25×10-6s

Solution:
∵dTT=12∙α∝dT
dT=12129×10-710
dT=2.25×10-6

Topic: Physics, Thermal Properties of Matter, JEE Main
A clock pendulum made of invar has a period of $0.5 \mathrm{~s}$, at $20^{\circ} \mathrm{C}$. If the clock is used in a climate where the temperature averages to $30^{\circ} \mathrm{C}$, how much time does the clock lose in each oscillation? (For invar, $\alpha=9 \times 10^{-7} /{ }^{\circ} \mathrm{C}$, $g=$ constant $)$

Options:
A $2.25 \times 10^{-6} \mathrm{~s}$
B $2.5 \times 10^{-7} \mathrm{~s}$
C $5 \times 10^{-7} \mathrm{~s}$
D $1.125 \times 10^{-6} \mathrm{~s}$

The correct answer is:
$5 \times 10^{-7} \mathrm{~s}$

Solution:
Time period of oscillation,
$\begin{aligned}
T & =2 \pi \sqrt{\frac{l}{g}} \\
\Rightarrow \quad \frac{d T}{T} & =\frac{1}{2} \frac{d l}{l} \\
\text { As, } \quad \frac{d l}{l} & =\alpha d t \\
\Rightarrow \quad \frac{d T}{T} & =\frac{1}{2} \alpha d t \\
& =\frac{1}{2} \times 9 \times 10^{-7} \times(30-20) \\
& =4.5 \times 10^{-6} \\
\therefore \quad \text { Loss in time }= & 4.5 \times 10^{-6} \times 0.5 \\
= & 2.25 \times 10^{-6} \mathrm{~s}
\end{aligned}$

Topic: Physics, Oscillations, JEE Main
A clock that keeps correct time at 20 Â°C, is subjected to 40 Â°C. If the coefficient of linear expansion of the pendulum is 12Ã10-6 Â°C-1, the time gained or lost, per day, is

Options:
A 10.3 s day-1
B 20.6Â sÂ day-1
C 5Â sÂ day-1
D 20minâ¡day-1

The correct answer is:
10.3 s day-1

Solution:
Time period, T=2Ïlg
âTT=12 âll=12Î±âÎ¸
= 1 2 Ã12Ã 10 â6 Ã( 40â20 )=12Ã 10 â5
âT=TÃ12Ã10-5
= 24Ã60Ã60Ã12Ã10-5
=10.3  s   day −1

Topic: Physics, Thermal Properties of Matter, JEE Main
A clock with iron pendulum keeps correct time at $15^{\circ} \mathrm{C}$. If the room temperature is $20^{\circ} \mathrm{C}$, the error in second per day will be nearly (coefficient of linear expansion of iron is $\alpha=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}$ )

Options:
A $3.1 \mathrm{~s}$
B $1.3 \mathrm{~s}$
C $6.2 \mathrm{~s}$
D $2.6 \mathrm{~s}$

The correct answer is:
$2.6 \mathrm{~s}$

Solution:
Time period of the pendulum is, $T=2 \pi \sqrt{\frac{l}{g}}$
$\therefore \Delta T=\frac{2 \pi}{\sqrt{g}} \times \frac{\Delta l}{2 \sqrt{l}}$
We know, $\frac{\Delta l}{l}=\alpha \Delta \theta$ the change in length due to thermal expansion.
$\therefore \frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta l}{l}=\frac{1}{2} \alpha \Delta \theta$
Given, $\quad \alpha=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}, \quad g=9.8 \mathrm{~m} / \mathrm{s}, \quad T=86,400 \mathrm{~s} \quad$ and $\Delta T=\alpha \Delta \theta T \approx 2.6 \mathrm{sec}$

Topic: Physics, Thermal Properties of Matter, JEE Main
A closed and an open organ pipe have same lengths. If the ratio of frequencies of their seventh overtones is $\left(\frac{a-1}{a}\right)$ then the value of $\mathrm{a}$ is ________

Options:

The correct answer is:
16

Solution:
For closed organ pipe
$\mathrm{f}_{\mathrm{c}}=(2 \mathrm{n}+1) \frac{\mathrm{v}}{4 \ell}=\frac{15 \mathrm{v}}{4 \ell}$
For open organ pipe
$\begin{aligned} & \mathrm{f}_{\mathrm{o}}=(\mathrm{n}+1) \frac{\mathrm{v}}{2 \ell}=\frac{8 \mathrm{v}}{2 \ell} \\ & \frac{\mathrm{f}_{\mathrm{c}}}{\mathrm{f}_{\mathrm{o}}}=\frac{15}{16}=\frac{\mathrm{a}-1}{\mathrm{a}} \\ & \Rightarrow \mathrm{a}=16\end{aligned}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2024 (08 Apr Shift 1)