Removed Charge Problem (Kvpy '10)
Let's discuss a problem based on Removed Charge from Kishore Vaigyanik Protsahan Yojana, KVPY, 2010. First, try the problem yourself, then read the solution here.
The Problem:
(12) positive charges of magnitude (q) are placed on a circle of radius (R) in a manner that they are equally spaced. A charge (Q) is placed at the center. If one of the charges (q) is removed, then the force on (Q) is
(A) zero
(B)$$ \frac{qQ}{4\pi\epsilon_0R^2}$$ away from the position of the removed charge
(C) $$ \frac{11qQ}{4\pi\epsilon_0R^2}$$ away from the position of the removed charge
(D)
$$ \frac{qQ}{4\pi\epsilon_0R^2}$$ towards from the position of the removed charge
Discussion:
If one of the charges is removed, then the net force on Q is $$ \frac{qQ}{4\pi\epsilon_0R^2}$$ towards the position of removed charge
Maximum Height of Water in A Tank With A Hole
Let's discuss a problem based on the maximum height of water in a tank with a hole. Try the problem yourself first and then read the solution.
The Problem: Maximum Height of Water
A large tank is filled with water of (70 cm^3/s). A hole of cross-section (0.25cm^2) is punched at the bottom of the tank. Find the maximum height to which the tank can be filled.
Solution:
For the water level to remain stationary volume efflux= rate of filling = (x)
The velocity $$ v=\sqrt{2gh}$$ where (g) is the acceleration due to gravity.
Hence,$$ vA=\sqrt{2gh}A$$ $$ =x$$ $$=70cm^3//s$$
The maximum height $$ h=\frac{x^2}{2gA^2}$$ $$=\frac{70^2}{2\times 980\times(0.25)^2}$$ $$=40cm$$
Efflux Velocity of Fluid through a Small Orifice in a Tube
Let's discuss a problem based on efflux velocity of fluid through a small orifice in a tube. Try it yourself first, then read the solution.
The Problem:
A horizontal tube of length (L), open at (A) and closed at (B), is filled with an ideal fluid. The end (B) has a small orifice. The tube is set in rotation in the horizontal plane with angular velocity (\omega) about a vertical axis passing through (A). Show that the efflux velocity of the fluid is given by $$ v=\omega l\sqrt{\frac{2L}{l}-1}$$ where (l) is the length of the fluid.
Solution:
Consider a mass element (dm) of the fluid at a distance (x) from the vertical axis. The centrifugal force on (dm) is
$$ Df=dm\omega^2x$$ $$=dm\frac{dv}{dt}$$ $$=dm \frac{dv}{dx}v$$
$$ vdv=\omega^2 xdx$$
$$ \int vdv=\omega^2 \int xdx$$ $$
\frac{v^2}{2}=\frac{\omega^2}{2} \int_{L}^{L-l}
$$
So,
$$ v=\omega l\sqrt{\frac{2L}{l}-1}$$
Work Done By A Force Acting on The Piston
Let's discuss a problem and find out the work done by a force acting on the piston. First, try the problem and then read the solution.
The Problem:
A cylinder filled with water of volume (V) is fitted with a piston and placed horizontally. There is a hole of cross-sectional area (s) at the other end of the cylinder, (s) being much smaller than the cross-sectional area of the piston. Show that the work to be done by a constant force acting on the piston to squeeze all water from the cylinder in time (t) is given by $$ W=\frac{\rho V^3}{2s^2t^2}$$ where (\rho) is the density of water.
Solution:
The volume of water flowing out per second $$ Q=sv$$
where (v) is the speed of sound and (s) is the cross-sectional area.
Volume flowing out $$ V=Qt=svt$$ $$ \frac{ \rho v^2}{2}=P$$$$= \frac{F}{A}$$ $$=\frac{FL}{AL}$$ $$ =\frac{W}{V}$$
where (L) is the length of the cylinder and (W) is the work done.
$$ W=\frac{1}{2}\frac{\rho V^3}{s^2t^2}$$
More Problem: https://cheenta.com/velocity-of-efflux-at-the-bottom-of-a-tank/
Losing Seconds - Solve Problem and Learn
Let's discuss the problem based on losing seconds where we find the depth below the earth's surface. Try yourself and then read the solution.
The Problem: Losing Seconds
Taking the earth's radius as (6400 Km) and assuming that the value of (g) inside the earth is proportional to the distance from the earth's centre, at what depth below the earth's surface would a pendulum which beats seconds at the earth's surface lose (5) min in a day?
Discussion:
Let us assume (g) is the acceleration due to gravity at depth (d) below the surface of the earth and (g) is the acceleration due to gravity on the surface.
Let the corresponding time periods be (T_0) and (T).
$$ g=g_0(1-\frac{d}{R}) $$
where (g) and (g_0) are the acceleration due to gravity at depth respectively, and (R) is the radius of the earth
$$ T=T_0\sqrt{\frac{g_0}{g}}=T_0(1-d/R)^{-1/2}$$ $$=T_0(1+\frac{d}{2R})
$$
Time registered for the whole body will be proportional to the time period. Thus,
$$ \frac{T}{T_0}=\frac{t}{t_0}=1+d/2R$$
$$ = \frac{86400}{86400-300}=1+\frac{d}{2R}$$
Substituting
(R=6400Km), we find (d=44.6Km)
Time Period of a Rolling Cylinder
In this post, we have discussed a problem based on the time period of a rolling cylinder. Try the problem yourself first, then read the solution.
The Problem: Time Period of a Rolling Cylinder
A solid uniform cylinder of radius (r) rolls without sliding along the inside the surface of a hollow cylinder of radius (R), performing small oscillations. Determine time period.
Solution:
Translational kinetic energy + rotational kinetic energy + potential energy= constant
$$ \frac{1}{2}mv^2+{\frac{1}{2}I\omega^2+mg(R-r)(1-cos\theta)}=C$$
Now $$ I=1/2mr^2$$
$$ 3/4m(dx/dt)^2+mg(R-r)\theta^2/2=C$$
Differentiating with respect to time,
$$ \frac{3}{2}m(\frac{d{^2}x}{dt{^2}})^+mg(R-r)\theta\frac{d\theta}{dt} $$
Now, $$ x=(R-r)\theta$$
$$ \frac{3}{2} d^2x/dt^2(R-r)d\theta/dt+gxd\theta/dt=0$$
Cancelling (\frac{d\theta}{dt}) throughout
$$ \frac{d^2x}{dt^2}+\frac{2}{3}\frac{gx}{R-r}=0$$
this is the equation for SHM, with
$$ \omega^2=\frac{2}{3}\frac{g}{R-r}
$$
$$ T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{3(R-r)}{2g}}$$
Some useful links:
Magnetic Field at Focus of Parabola
Let's solve the problem based on Magnetic Field at Focus of Parabola and learn how to solve it. First, try it yourself, then check your solution.
The Problem:
An infinite wire carrying current (I) is bent in the form of a parabola. Find the magnetic field at the focus of the parabola. Take the distance of the focus from the apex as (a).
Solution:
From Biot-Savart law, the magnetic field ar (S) is given by $$ \vec{B}= \frac{\mu_0}{4 \pi} \int\frac{I\vec{dl}\times\vec{r}}{r^3}$$
From the figure, we note that
$$ |\vec{dl}\times \vec{r}|$$=area of the parallelogram by (\vec{dl}) and (\vec{r}) $$ = 2\times1/2\times r.rd\theta$$$$=r^2d\theta$$
Hence, $$ \vec{B}=\frac{\mu_0I}{4 \pi}\int_{0}^{2\pi}\frac{d\theta}{r}$$ Using (r(1-cos\theta)=2a) as the equation to the parabola, we get $$ \vec{B}=\mu_0I/4a
$$
Some Useful Links:
Orbit of Planet (KVPY '10)
Let's discuss a problem where we will find the change in the orbit of the planet. Give it a try first, then read the solution.
The Problem:
A planet of mass (m) is moving around a star of mass (M) and radius (R) in a circular orbit of radius (r). The star abruptly shrinks to half its radius without any loss of mass. What change will there in the orbit of the planet?
Discussion:
A planet of mass (m) is moving around a star of mass (M) and radius (R) in a circular orbit of radius (r). The star abruptly shrinks to half its radius without any loss of mass.
As we know, the formula for centripetal force is $$ F=\frac{mv^2}{r}$$
Since the radius of the star is decreasing without any change in mass will not have any effect on force exerted by star on planet which is the required centripetal force.
Some Useful Links
Angular Velocity of Ice Skater
An ice skater spins at (4\pi )rad/s with her arms extended. If her moment of inertia with arms folded is (80\%) of that with her arms extended, what is her angular velocity when she folded her arms?
Discussion:
Conservation of angular momentum gives $$ I_1\omega_1=I_2 \omega_2$$
$$ (I_1) (4\pi)=0.8 I_1 \omega_2
$$
Hence, (\omega_2=5\pi)
Let's discuss a problem on angular momentum and prove that the tension in the thread is inversely proportional to the cube of the distance from the hole.
The Problem:
A small mass m tied to a piece of thread moves over a smooth horizontal plane/ The other end of the thread is drawn through a hole with constant velocity. Show that the tension in the thread is inversely proportional to the cube of the distance from the hole.
Discussion:
Angular momentum of the mass is assumed to be constant. The particle velocities (v) and (r) are perpendicular. The angular momentum $$ J=mvr=constant$$
Hence,
$$ v\propto 1/r$$
The tension T arises from centripetal force $$ T=\frac{mv^2}{r}
$$ Hence
$$ T\propto \frac{1}{r^2}\frac{1}{r}$$ or
$$ T\propto 1/r^3$$
Some useful links: