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where you can see a collection of useful answers:
Q. How do we know atoms exist if we can't see them?
A. Even the
smallest speck which can be seen through an
ordinary microscope contains billions of atoms.
However, scientists use very powerful instruments
and techniques to study atoms. When X rays are
passed through a crystal, the atoms in the
crystal diffracts the X-rays in a certain way.
These diffracted rays produce patterns on
photographic film. In this way scientists are
able to find out how the atoms are arranged and
how far apart they are.Extremely powerful
Scanning Electron Microscopes allow scientists to
observe the positions of individual atoms.
Electricity
Q. Explain Ohm's Law and resistance?
A. Ohm's law gives the relationship between current and the
resistance to it. The unit of measuring
resistance is called Ohm.
This law states that the voltage equals the current multiplied by
the resistance, through which the current flows.
For example if a current of 5 amperes passes
through a resistance of 4 ohms, the voltage which
will be needed will be 20 volts ( 5 amperes X 4
Q. Is it possible, under special conditions, for
the energy in an electric circuit to be 100%
conserved?
This meaning that none of it is ever lost through
heat energy caused by friction through the
conduit or load, or any other type of
transformation of the energy that eventually
leaves the circuit.
When would this happen, if possible, or why not,
if impossible?
A. The electrical receptivity of a metal arises from the
interactions of the conduction electrons with
impurities, defects and the vibrating ions of the
crystal lattice.As the temperature is lowered,
the amplitudes of the lattice vibrations
diminish, so one would expect the receptivity
also to decrease gradually toward a small, but
finite, value determined by impurities and
defects.Many materials manifest this behavior.
However, in 1911, H Kammerlingh Ownes discovered
that as the temperature of a specimen of mercury
was reduced its resistance suddenly dropped to
extremely small values at 4.15 K. The metal had
made transition to a new superconducting state.
The resistivity of a superconductor is at least a
factor of 10-12 less than that for an
ordinary conductor. We can usually take it to be
An electric current induced in a
superconductor has been observed to flow for
several years without any applied potential
difference-provided the temperature is maintained
below the critical temperature.
Q. What are the factors involved in the functioning of a light bulb?
A. The light bulb glows due to
the conversion of electric energy into heat and light energy. As the
electric current passes through the filament of a light bulb,
the atoms of the filament try to prevent the flow of this
current. In other words the filament act as a resistance.This
leads to the release of heat energy, which in turn increases
the temperature of the filament. The filament becomes white
hot and starts to glow.If the light bulb did not have a
vacuum in side, the filament would fuse or burn due to this
Q. In the circuit
below... a) what is the potential difference between points A
and B in Fig1. When switch S is open? b) Which point A or B
is at the higher potential? c) What is the final potential of
point B when switch S is closed? d) How much charge flows
through switch S when it is closed?
A. i = 18/(6+3) = 2 A
potential at A = 18-(2x6) = 6V
At steady state, capacitor
will be fully charged
Let Ceff be the
effective capacitance of the 6uF and 3uF capacitors in
Ceff = 1/6 + 1/3 =
Ceff = 2 uF
Some charge will flow through
6 uF and 3 uF
Q=2x10-6 x 18 = 36
Potential at B = 18 - (36 x 10-6
)/(6 x 10-6 ) = 12V
Therefore B is at an higher
potential by 6V
once the switch is closed the
potential at B will become 6V
potential difference across
charge on 6 uF= (6 x 10-6
this is the charge which will
flow from B to A
A capacitor of 6uf is connected in parallel with the
combination of 4uF capacitor and a 3uF capacitor that are
in series. Find (a) the net capacitance of the entire
combination. and (b) The Potential Difference across the 3uF
capacitor when 20V is maintained across the 6uF capacitor.
A. Capacitors
3 uF and 4 uF are in series
= capacitance of a single capacitor which replaces the above
two capacitors in series. Then
= 1/4 + 1/3 =(4+3)/12=7/12
therefore Cs
= 12/7 (see Fig a above)
= capacitance of the capacitor which replaces the capacitors
in parallel on Fig b above
= (12/7+6) uF = 54/7 uF = net capacitance
20 V potential
difference is across 6 uF as well as 12/7 uF as they are in
Charge through
12/7 uF capacitor = capacitance x voltage
=12/7 uF x 20
= 240/7 uC
Charge on 4
uF and 3 uF capacitors will be the same as they are in
difference across 3 uF = (240 uC) / (7 x 3 uF) = 80/7 V =
What is the difference between analog and digital devices?
A. Electronic devices work
with two basic types of signals, analog and digital. Digital
signals represent all information with a limited number of
voltage signals. Each signal has a distinct value. Analog
signals vary continuously in voltage or current,
corresponding to input information. Many digital circuits can
process information much faster than analog circuits. Digital
circuits do the majority of processing.
soft drink from Australia is labelled 'low Joule Cola'. The
label says 100 ml yields1.7 kJ. The can contains 375 ml.
Sally drinks the cola and then offsets this input of food
energy by climbing stairs. How high would she have to climb
if Sally has a mass of 65kg?
A. The energy given by 375ml
of cola is 1.7? 375/100=6.375 kJ. In climbing the stairs the
energy will be used up to gain potential energy mgh where m
is the mass of the body, g is the acceleration due to gravity
and h is the height above the ground. Thus, mgh=6375 gives
750 kg car moving at 23m/s brakes to a stop. The brakes
contain about 15 kg of iron that absorbs the energy. What is
the increase in the temperature of the brakes?
A. To find the retardation
produced in the car due to the application of brakes, we use
the third equation of motion, v?-u?=2as where v is the
final velocity, u is the initial velocity, s is the distance
it covers before it comes to rest and a is the retardation.
We get a=529/2s. Now force applied on the car to stop is
F=ma=750 x529/2s.
Thus the work done in stopping
the car is W=F x d=750 x (529/2s) x s=198375J. This work done
is used up in increasing the temperature of the brakes. It
can be calculated by W = mc x T. The specific heat of iron is
448 J/kg deg C. Thus the increase in temperature of the
brakes is 29.5 deg C.
you have a 200kg lump of ice hanging from a roof at a height
of 5 metres above the ground, a) if the ice were to detach
and fall, what would its kinetic energy be at the
&instant& before it touched the ground? b) How many
grams of snow (at 0 degrees) would that amount of heat be
able to melt? ( the heat of fusion of snow is 334J/g)
A. As the lump falls, the
potential energy decreases and the kinetic energy increases.
Just before the instant when it touches the ground, the
energy is totally kinetic and is equal to the potential
energy at the top. Thus K.E. = mgh = 200 x 9.8 x 5 = 9800 J.
Let us assume that this entire
energy is converted to heat energy. We can use the formula Q
= mL where Q is the amount of heat evolved or absorbed when
change of state takes place, m is the mass of the substance
that undergoes change of state and L is the latent heat.
Thus, m = Q/L = g.
fluid is rotating at constant angular velocity w about the
central vertical axis of a cylindrical container. Show that
the variation of pressure in the radial direction is given by
dP/dr = p'w?r, where p' is the density of the fluid.
b) Take P=P& at the axis of rotation (r=0) and show that
the pressure P at any point r is P=P&+1/2(p'w?r?).
c) Show that the liquid surface
is, a vertical cross section of the surface is the curve
y=w?r?/(2g).
d) Show that the variation of pressure with depth is P=p'gh.
A. a) Centrifugal force on the rotating liquid F = mw?r.
Pressure P = F/area . Area = lateral area of the cylinder = 2
Πrl where r is the radius of the cylinder and l is the length
of the cylinder. Also, mass of the liquid in the container =
density x volume = p' x Πr?l. Combining these relations, we
get, P = p'r?w?/2. Differentiating, we get, dP/dr = p'rw?
b) If pressure at the axis is
P& then from a) pressure at any point r is P = P& +
c) Pressure due to rotation =
p'r?w?/2 and hydrostatic pressure = p'gy where y is the
depth of the liquid. Since the liquid surface is at
equilibrium, these two pressures must be equal. Equating, we
get, y = r?w?/2g. This is the equation of parabola.
d) Pressure at any point at a
depth h is the weight of the liquid above that depth divided
by the area. It is P = mg/A = mgh/Ah = mgh/V = p'gh.
is moving on a plane. What is the direction of the frictional
A. If a car is moving on a
plane, then since the bottom of the wheel is pushing backward
the force of friction is in the forward
direction.
In case of a freely rolling
wheel, the force of friction is backward whereas in a driven
wheel, the force of friction on the wheel exerted by the road
is in the forward direction. In a driven wheel, the torque
due to the axle is opposite to the torque due to the friction
and the vertical force N which acts at a point ahead of the
center. If the wheel does not slip, the maximum value of
friction is the maximum value of static friction.
kg crate is to be pulled a distance of 20m requiring 1210 J
of work being done. If the job is done by attaching a rope
and pulling with a force of 75 N, at what angle is the rope
A. The work done by a force F
in moving a body by a distance d is given by the dot product,
F.d=Fd Cos Θ where Θ is the angle between the force and
distance. Substituting, we get 1210=75 x 20CosΘ . Thus, Cos
Θ =121/150.
kg. horizontal beam is supported at each end. A 200 kg piano
rests a quarter of the way from the end. What is the vertical
force on each of the supports?
A. The total vertical force on the support which is closer to
the piano will be [200 x (3/4)] + 180/2 = 240kgwt and that on
the other support will be [200 x (1/4)] + 180/2 = 140kgwt.
force acting at 90 degrees and a 44N force acting at 60
degrees act concurrently on point P. What is the magnitude
and direction of a third force that produces equilibrium at
A. The forces have to be
resolved into components along the x axis and y axis. The
component along the x axis is 44Cos60 = 22N and the component
along the y axis is 33 + 44Sin60 = 71.1N. The resultant of
these two forces is (22? + 71.1?)1/2 = 74.4 N.
The direction of this resultant force will be = Tan-1(71.1/22)
= 72.8 degrees to the x axis. The force which will keep the
point P in equilibrium will be equal to 74.4 N in magnitude
and in a direction opposite to the above direction.
6'1& man swings a 48& golf club weighing 275 grams
and launches a 20 gram ball at a 45 degree angle. How many
miles per hour must the man swing the golf club for the ball
to travel 400 yards? How many more yards will each additional
5 mph of club speed generate?
A. 1mph = 0.447m/s and 1 yard
The ball is set into
projectile motion. The range of projectile is given by the
formula - R = u?Sin2q /g where u is the initial velocity of
the projectile, q is the angle with which the body is
projected and g is the acceleration due to gravity and R is
the range or the horizontal distance that the projectile
Now, R = 400 x 0.914 = 365.6m
and angle = 45 degrees.
Substituting, we get, u =
59.8m/s. This will be the initial velocity of the ball.
When the golf club comes and
strikes the ball, it will transfer some of its kinetic energy
to the stationary ball. If we assume that an elastic
collision takes place, then since the ball is 13.75 times
lighter than the golf club, it will transfer 13.75% of its
kinetic energy to the ball (this can be shown
mathematically).
Thus, (13.75/100)(1/2 m1 v? ) = 1/2 m2 u? where m1 is the
mass of the golf club and m2 is the mass of the golf ball and
v is velocity with which the club strikes the ball and u is
the initial velocity of the ball. Calculating we get, v =
43.1m/s. So this is the velocity with which the golf club
will have to be swung.
Now, since the golf club is
swung in a circular movement, the taller the golfer and
longer the club, the more will be the radius of the circle.
From, v = ω r. So for a particular velocity v, the more the
value of r, the less will be the value of ω or the angular
velocity or the golfer will have to swing his arms more
For calculating the increase
in the range, for each 5mph increase in the club speed, we
will first have to calculate the increase in the initial
velocity of the ball using the energy relation and then find
the new range and the increase in range.
horse pulls a cart by a force F1 in the forward direction.
From Newton's third law of motion, the cart pulls the horse
by an equal force F2 in the backward direction. The sum of
these forces is thus zero. Why should then the cart
accelerate forward?
A. In the above argument we
have not considered all the forces acting on the horse. While
trying to pull a cart, the horse pushes the ground backward
with a certain force F at an angle. The ground offers an
equal reaction F in the opposite direction, on the feet of
the horse. The forward component of this reaction when it
exceeds F2 will accelerate the horse in the forward
direction. Thus the horse will be able to pull the cart in
the forward direction.
a force of 20KN acting on a tripod get distributed equally?
This structure consists of three sticks with the same
dimensions that are holding up a weight of 20KN in three
faced triangular position.
Yes, the force will get
distributed equally.
Friction does not depend upon the surface area of contact.
Then why should a tire be properly inflated?
A tire has a complex
structure. In case it is not filled properly, there will be
greater flexing of the tire walls. This will generate heat
which in turn will adversely affect the ability of the tire
to steer a vehicle properly. The steering of a vehicle is
achieved by the stretching of the rubber in the tire to road
contact patch. The heat transfer from the tire to a cold road
will be very small as compared to the amount of heat
generated due to the flexing of the tire. As a tire has a
tread, the adhesion of tire to a road can't be explained in
simple terms. This area, is in fact, a hi-tech area on which
the tire manufacturers are doing much research.
broom balances at its CG. If you cut the broom in half at the
CG and weigh each part of the broom, which end would weigh
A. Both parts of the broom
will have equal weights. If it were not so, there will be a
net torque on the broom when you try to balance it at its CG.
Q. Two people pull a
box resting on a frictionless surface. One person pulls with
a force of 10 N to the north. The other person pulls with a
force of 15 N to the west. Find the magnitude and direction
of the resultant using a graphical and algebraic approach.
A. Algebraic
The resultant
force will be found by squaring the two given forces, adding
them and then finding the square root.
force= square root of (100 + 225) = 18 N
Here 10 ? =
15 ? = 225
The graphical
method would be to plot the forces on a graph and then to
find the resultant as shown in the figure below.
What is the term moment of force measured in?of the
line of action of the force from the axis of rotation.
A. The turning effect of a
force about the axis of rotation is called moment of force or
torque due to the force. It is measured as the product of the
magnitude of the force and the perpendicular distance of the
line of action of the force from the axis of rotation.
Its unit is Nm in SI units. It
may be pointed out that no doubt, Nm is equivalent to joule
(the unit of work), but joule is not used as the unit of the
moment of force.
What is Universal law of gravitation. How do Kepler's laws
apply to 2-d analysis? What formula describes the motion of a
pendulum and SHM?
A. The Universal law of
gravitation states that every body in this universe attracts
every other body with a force which is directly proportional
to the product of their masses and is inversely proportional
to the square of the distance. Kepler's laws of planetary
motion are by their very definition, applicable to 2-d motion
of the planets around the sun. In 2-d analysis, they can be
applied either by resolving the motion along the x and y
directions or by using (r,?) coordinates.
The equation of SHM is F = -ky
where F is the restoring force and y is the displacement from
the mean position and k is a constant. The negative sign
shows that the restoring force is always directed opposite to
the displacement The equation for displacement in SHM is y =
a Sin(wt + ? ) where y is the displacement of the particle
undergoing SHM, a is the amplitude of vibration, w is the
angular frequency and ? is the initial phase. These
equations can describe the motion of a simple pendulum as it
executes SHM. The time period of simple pendulum is given by
T = 2? ? (l/g) where l is the length of the pendulum and g is
the acceleration due to gravity.
What is the force on a 1kg ball that is falling freely due to
the pull of gravity ?
A. F = mg where m is the mass
of the body and g is the acceleration due to gravity (9.8
m/s?). Substituting the value we get, F = 9.8 N.
Q. How does a rainbow form?
A. Rainbow is an arch of
brilliant colors that appears in the sky when the sun shines
during or shortly after a shower of rain. It forms in that
part of the sky opposite the sun. If the rain has been heavy,
the bow may spread all the way across the sky, and its two
ends seem to rest on the earth.
The reflection, refraction,
and diffraction of the sun's rays as they fall on drops of
rain cause this interesting natural phenomenon. These
processes produce all the colors of the color
spectrum--violet, blue, green, yellow, orange, and red.
However, the colors of a rainbow blend into each other so
that an observer rarely sees more than four or five clearly.
The width of each color band varies, and depends chiefly on
the size of the raindrops in which a rainbow forms. Larger
drops cause narrow bands.
there a critical angle for light going from glass to water?
What about water to glass?
A. Critical angle is the angle
of incidence for which the corresponding angle of refraction
is 90 degrees. Since the refracted ray is moving further away
from the normal, the light ray should be traveling from
denser medium to rarer medium. So there is a critical angle
for light going from glass to water, but not from water to
Q. Why is the sky blue?
A. Sky is the region of space
visible from the earth. The sky consists of the atmosphere,
which extends hundreds of kilometers above the earth. The
atmosphere is composed chiefly of nitrogen and oxygen. In
addition, the atmosphere contains tiny water droplets and ice
crystals in the form of clouds and precipitation. Smoke, dust
particles, and chemical pollutants may also fill the sky over
The colors of the sky result
from the scattering of sunlight by the gas molecules and dust
particles in the atmosphere. Sunlight consists of light waves
of varying wavelengths, each of which is seen as a different
color. The shortest light waves appear blue and the longest
red. The blue light waves are readily scattered by tiny
particles of matter in the atmosphere, but the red light
waves travel undisturbed unless they are struck by larger
particles.
When the sky is clear, the
waves of blue light are scattered much more than those of any
other color. As a result, the sky appears blue. When the sky
is full of dense clouds or smoke, the light waves of all
colors are scattered, causing the sky to turn gray. At
sunrise or sunset, sunlight must travel farther through the
atmosphere than when the sun is overhead. Light waves of most
colors are scattered. Undisturbed red light waves give the
sun and sky near the horizon a red or orange appearance.
Q. How are mass and inertia related?
A. A body continues to be in a state of
rest or uniform motion unless some external force is applied
to it. Galileo called this property of objects inertia. The
word inertia, which means unchanging, has been derived from
the Latin word inert. Inertia of a body may be defined as,
the property of a body to resist any change in its state of
rest or of uniform motion in a straight.
From our common
experience, we know that it is easier to move a small and
light object than a large and heavy one. Thus, a large/ heavy
body shows a greater resistance to change in its state of
rest or of uniform motion. Thus, a heavier body has more
inertia than a lighter body. Therefore we can say that the larger the
mass the larger is the
inertia, and the smaller the mass, the smaller is the
inertia. In other words we
can say that the mass of a body is a measure of its inertia.
ball is dropped from the top of a tower h meters high and
another ball is projected upwards simultaneously from the
bottom. They meet when the upper ball has covered 1/n of the
distance. Show that when they meet , the velocities are in
the ratio 2:n-2 and the velocity of the lower ball is
[ngh/2]1/2.
A. Upper ball: h/n=(1/2)gt?
or t=[(2h)/(ng)]1/2. And (v1)=gt=[(2gh)/n]1/2. For the lower
ball (h-h/n)=(u2)t +(1/2)gt?. Substitute for t and simplify
for (u2). (u2)=[ngh/2]1/2. By substituting in the
(v2)=(u2)+gt solve for (v2). Take the ratio (v1) : (v2).
10kg brick falls from a height of 2m. What is its momentum as
it reaches the ground?
A. We can use the third
equation of motion to find the velocity of the brick as it
reaches the ground, v?-u? = 2gh. Substituting, we get,
v=6.28 m/s. Thus, the momentum is p=mv=10×6.28=62.8 kg m/s.
A 110 kg football player running at 3 m/s collides head on
with a 55 kg referee by accident.
This collision gives an
impulse to the referee at the expense of the player.
Irrespective of how large the impulse is, how will the
magnitude of the change in the referee’s velocity during
the collision compare with that of the player? Explain
carefully.
, u1 , v1 , be the mass, initial
velocity and final velocity of the player and m2 ,
u2 , v2 , be the mass, initial velocity
and final velocity of the referee. Then
= 110 kg, u1 = 3 m/s , v1 ,= 0 m/s
,= 55 kg, u2 = 0 m/s, v2 = not known
u1 , + m2 u2 , =
m1 v1 , + m2 v2
110 x 3 + 0 =
0 + 55 x v2
referee's velocity = 6-0 = 6 m/s
player's velocity= 0 -3 = -3 m/s
referees velocity as compared to player's velocity = 6-(-3) =
ball rolling west at 3.0 m/sec has a mass of 1.0 kg. It
collides with a second ball whose mass is 2.0 kg and is
stationary. After the collision, the first ball moves, south.
Determine the momentum and speed of each ball after the
collision.
A. We assume that the
collision is elastic. We apply the conditions of conservation
of momentum and kinetic energy. After the collision the first
ball moves south. Thus, the second ball will move in the
north west direction at an angle Θ to the negative x axis so
that the y component of the momentum of the second ball will
equalize the final momentum of the second ball and the x
component will give the total final momentum equal to the
initial momentum. If m1 and m2 are the
masses of the two balls, u1 and u2 are
the initial velocities of the two balls (u2=0) and
v1 and v2are the final velocities of
the two balls. Writing the equations for conservation of
momentum along the x and y axes, we get,
m1u1 =m2v2Cos
Θ and m1v1 = m2v2Sin
Applying the condition for the
conservation of kinetic energy, we get
= 1/2m1v1? + 1/2m2v2?.
Now, m1=1kg, u1
= -3m/s, and m2 = 2kg. Substituting these values
and solving the above equations, we get, v1=v2=3
1/2 . Thus the momenta of the two balls are the product
of their mass and the speed .
ball rolling west at 3.0 m/sec has a mass of 1.0 kg. It
collides with a second ball whose mass is 2.0 kg and is
stationary. After the collision, the first ball moves, south.
Determine the momentum and speed of each ball after the
collision.
A. We assume that the
collision is elastic. We apply the conditions of conservation
of momentum and kinetic energy. After the collision the first
ball moves south. Thus, the second ball will move in the
north west direction at an angle ? to the negative x axis so
that the y component of the momentum of the second ball will
equalize the final momentum of the second ball and the x
component will give the total final momentum equal to the
initial momentum. If m1 and m2 are the
masses of the two balls, u1 and u2 are
the initial velocities of the two balls (u2=0) and
v1 and v2are the final velocities of
the two balls. Writing the equations for conservation of
momentum along the x and y axes, we get,
m1u1 =m2v2Cos
Θ and m1v1 = m2v2SinΘ
Applying the condition for the
conservation of kinetic energy, we get
= 1/2m1v1? + 1/2m2v2?.
Now, m1=1kg, u1
= -3m/s, and m2 = 2kg. Substituting these values
and solving the above equations, we get, v1=v2=3
1/2 . Thus the momenta of the two balls are the product
of their mass and the speed .
boat heading due north crosses a wide river with a speed of
10 m/s relative to the water. The river has uniform speed of
5 m/s due east. Find the velocity of the boat w.r.t. a
stationary ground observer.
A. We will have to use vectors
to solve this problem. The velocity of the boat w.r.t. water
is denoted by vbw and velocity of water w.r.t.
ground is denoted by vwg and velocity of boat
w.r.t. ground is denoted by vbg. Then we have, vbw
+ vwg = vbg. Since vbw is
along the north direction, it can be denoted along the
positive y-axis and vwg is along the east
direction, it can be represented along the positive x-axis,
the two vectors forming the perpendicular and base of the
right angled triangle. The resultant of the two vectors will
be given by the hypotenuse. Thus, vbg = (10 ?+
5?)1/2= 1251/2= 11.18 m/s.
The direction of the boat with
respect to the ground will be tan-110/2 = at an
angle of 63.4° to the x-axis.
car A is towing car B with a fixed towing connection at 90
MPH. Car B starts his engine and accelerates to 90 MPH. How
many MPH will cars A and B be travelling at that time? Why?
A. Once car B has accelerated
to 90 MPH, the cars A and B will travel at a speed of 90MPH
with no tension in the tow bar. As the speed of the car B
increases, the tension in the tow bar will decrease, reaching
a value 0 when the car B has attained the speed of 90MPH. The
speed of the system of cars A and B together will be given by
the speed of the center of mass which is equal to (m1v1 +
m2v2)/(m1 + m2). Assuming that the cars have the same mass,
their speed will be simply 90MPH, where v1=v2=90MPH.
car drives straight off the edge of a cliff that is 54m high.
The police at the scene of the accident note that the point
of impact is 130m from the base of the cliff. How fast was
the car travelling when it went over the cliff?
A. For motion in two
dimensions, we can resolve the motion into motion along
x-axis and y-axis and look at the two components
independently.
Looking at the motion in the
vertical direction, when the car just takes off from the
cliff, it has got only initial horizontal velocity and zero
initial vertical velocity. However, there is no acceleration
in the horizontal direction and the acceleration in the
vertical direction is acceleration due to gravity,
g(9.8m/s?). The distance covered in the horizontal direction
is 130m and that in the vertical direction is 54m.
Vertical motion: We apply the
equation of motion, h = ut + 1/2 gt?. Here u = 0 and h = 54.
Substituting we get, t = 3.3s.
Horizontal motion: Since there
is no acceleration in the horizontal direction, s = vt where
s is the distance covered and v is velocity in the vertical
direction. Substituting for time from above, we get v = s/t =
130/3.3 = 39.4m/s. So the car was travelling with a speed of
39.4m/s when it went over the cliff.
jet fighter is traveling horizontally at 111metres per second
at an altitude of 3.00x10squared metres, when the pilot
accidentally releases an outboard fuel tank.
(a) How much time elapses before the tank hits the ground?
(b) What is the velocity of the tank just before it hits the
A. We will follow the logic of
the previous question.
Vertical motion: initial
velocity u = 0 and vertical distance h = 300m. Applying the
equation of motion, h = ut + 1/2 gt?, we get t = 7.8s.
For finding the final velocity
in the vertical direction, we apply the equation of motion,
v? - u? = 2gh. Substituting, we get, vy =
Now, the velocity in the
horizontal direction will remain 111m/s as there is no
acceleration in the horizontal direction. Thus, vx
The velocity with which the
tank hits the ground, v = (vx? + vy?)1/2
= 134.9 m/s and the direction will be as follows: ? = tan-1vy/vx
where ? is the angle made with the x-axis.
object weighing 1,000 pounds is attached to a rope and fed
through a suspended pulley. The other end of the rope is
wrapped around a 12& drum which is fixed on a rotatable
shaft. The weight is allowed to fall (say 500 feet), spinning
the shaft. How fast will the object fall? How fast will it
spin the shaft (with only minor bearing resistance)? How much
horse power will it produce at the shaft and does this change
at different points of travel?
A. Let the speed of the object
be v when it descends through a height h. So is the speed of
the rope and hence of a particle at the rim of the drum. The
angular velocity of the wheel is v/r and its kinetic energy
at this instant is 1/2 I(v/r)? where r is the radius of the
drum and I is moment of inertia of the drum. If the drum is a
solid cylinder, then I = 1/2mr? where m is the mass of the
drum and r is the radius of the drum. Using the principle of
conservation of energy, the gravitational potential energy
lost by the object must be equal to the kinetic energy gained
by the object and the drum. Thus,
Mgh = 1/2Mv? + 1/2 I (v/r)?
where M is the mass of the object. Or v = [2Mgh/(M+I/r?)]1/2.
The angular velocity of the
spinning shaft will be ω = v/r.
The power produced at the
shaft will be given by P = ω T where T is the torque which
is rotating the drum and ω is the angular velocity of the
rotating drum. F will be given by rt where t is the tension
in the rope. Yes, the power produced will change at different
points of travel because as the object has fallen by
different heights, the velocity of the object and hence
velocity of the drum will be different. Thus, the angular
velocity will be different. The power can be expressed in
Horse Power by using the relation, 1HP = 746 watts.
1/p of the distance between two stations a train is uniformly
accelerated and for 1/q it is uniformly retarded. It started
from rest from the first station and comes to rest at the
second. Prove that the ratio of the greatest velocity to the
average velocity is 1+1/p+1/q.
A. Let total distance be x.
Then x/p=1/2(a1)(t1)? and v=(a1)(t1). Rearranging these we
get (t1)=2x/pv . Now the distance for which the train moves
with uniform speed is x(1-1/p-1/q). Thus t2=(x/v)(1-1/p-1/q).
For the x/q part of the distance, -v?= -2(a2)(x/q) and
0=v-(a2)(t3). Rearranging these we get (t3)=2x/vq. Now
average velocity is total distance/total time or
x/[(t1)+(t2)+(t3)]. Substitute for the values of time from
above and the highest velocity attained is v and then take
the ratio. The result is [1+1/p+1/q]
examples to demonstrate Newton's Laws of Motion.
A. First Law: Take a tumbler.
Place a piece of cardboard on it. On top of the cardboard
place a coin. Now flick the cardboard away with your finger.
The coin will not follow the cardboard but drop into the
tumbler. This is because there is no force acting on the coin
and it continues to stay in a state of rest.
Second Law: A cricketer moves
his hands back when taking a catch. This to increase the time
period over which the velocity of the ball reduces from a
high value to zero. Or in other words the acceleration is
reduced. Thus the force transferred to the cricketer's hand
is reduced.
Third Law: When a rocket is
launched, the downward action of the exhausted gases results
in the upward reaction on the rocket, causing it to rise.
do you solve two dimensional trajectory problems?
A. Trajectory problems may be
put into two categories. Horizontal projection and angular
projection. During its motion the object covers horizontal
distance due to horizontal velocity and vertical distance due
to vertical velocity. So each problem can be simplified into
two one dimensional problems by taking the components of
position, velocity and acceleration along the horizontal and
vertical directions. Let us consider the case of horizontal
projection. The object is projected with initial horizontal
velocity u. Since the velocity of the object in the
horizontal direction is constant so the acceleration ax
along horizontal direction is zero. If the initial position
of the object was ( x0, y0) the
position of the object at any time t along the horizontal
direction i.e. along the x-axis is given by x = x0
+ uxt + 1/2 axt2 , u and ax
= 0. If we put the origin of the co-ordinate system at the
initial position then the co-ordinates of the initial
position become (0,0). Our equation becomes x = 0 + ut + 1/2
(0)t 2 = ut. This means t = x / u.
Let us take the downward
direction as positive. Thus the acceleration in the vertical
direction is g (9.8 m / s 2). The position of the
object at any time t is given by y = y0 + uyt
+ 1/2 ayt 2. Here y 0 = 0 ,
uy = 0 and ay = g . The equation then
becomes y = 0 + (0) t + 1/2 g t 2 = 1/2 g t 2
= 1/2 g ( x / u) 2 = ( 1/2 g / u 2 ) x
2 = k x 2 where ( ( 1/2 g / u 2 )
= k, a constant. This is the equation of a parabola, which is
symmetric about the y-axis. Hence the path of the projectile
projected horizontally from a certain height from the ground
is a parabolic path. The above conditions can be substituted
in the equation of motion v = u + at , separately for motion
along the x and the y axis and the horizontal and the
vertical can be calculated at any instant. The resultant
velocity v is given by v = (v x 2 + v
y 2 )1/2 .
fast must a roller coaster travel around a vertical circular
track with radius 10m if it is not to fall from the track?
A. The minimum speed at the
lowest point of the circular track, so that it completes the
vertical circular track is (5gr)1/2 where r is the
radius and g is the acceleration due to gravity (9.8m/s?).
Substituting the values we get minimum speed=22.14 m/s.
long would it take for a ball to fall from a 70 feet tower?
A. This is a problem of free
fall. We use the equation of motion, h = ut + 1/2 gt? to
solve it. Here, h is the height by which the body falls, u is
the initial velocity, t is the time taken to fall by a
distance h and g is the acceleration due to gravity. Now, u =
0 and h = 70 feet. By substituting we can get the answer.
you fire a bullet from a level gun and drop a bullet at the
same time, both the bullets will hit the ground at the same
time. Explain.
A. Both the bullets have the
same acceleration in the vertical direction which is the
acceleration due to gravity. Also both the bullets have zero
initial velocity in the vertical direction. The bullet fired
from the gun has an initial velocity in the horizontal
direction but it will not affect the motion in the vertical
direction. Since, both the bullets have to cover the same
vertical distance, they will take the same time to do to it
and will hit the ground at the same time.
is the minimum speed the pilot of an aircraft should have so
as to successfully loop a vertical loop without falling at
the top of the loop? Also what is the thrust acting on the
A. The forces acting on the
aircraft are - the weight mg acting downwards and the force F
by the air upward. Let the aircraft be at any position of the
vertical loop such that the radius vector of the position of
the aircraft makes an angle θ with the vertical. The weight
mg can be resolved into two components- mgCos θ and mgSin θ
. The net force acting on the aircraft will be F - mgCos θ
which will provide the necessary centripetal force mv?/r
where r is the radius of the loop. Thus, F = mgCosθ +
mv?/r. Now, F will be minimum when the value of Cosθ is
minimum which will be -1 and it will be minimum at the
highest point of the loop.
F(min) = mv?/r - mg. The
aircraft will be able to complete the loop only if F(min)
& 0. If v' is the velocity at the highest point, then
mv'?/r ? mg or v' ? ? (gr) . This is the minimum value of
velocity at the highest point.
If we apply the principle of
conservation of energy, then total mechanical energy at the
lowest point = total mechanical energy at the highest point.
1/2mv? = 1/2mv'? + mg(2r).
If we substitute the value for v', we get, v ? ? (5gr). Thus,
for looping the loop the minimum speed the aircraft should
have at the lowest point should be ? (5gr). Also the maximum
thrust on the aircraft will be at the lowest point. F(max) =
mv?/r + mg.
a ball is moving on a smooth plane, it will have acceleration
only when there is an external force. Can it be said that the
above statement is wrong because when ball is on a smooth
inclined plane, it still has acceleration?
A. To understand this
question, let us first understand Newton's first law. It says
that if the vector sum of all the forces acting on a body is
zero then and only then the body remains unaccelerated (i.e.
remains at rest or moves with uniform velocity).
Let us consider a body moving
on a smooth horizontal surface with a uniform velocity. The
forces on the body are the gravitational force exerted by the
earth and the contact force or the normal reaction exerted by
the plane surface on the body. These forces are equal and
opposite and they cancel out. Thus the net force acting on
the body is zero and the body remains unaccelerated. Now, let
us consider the body moving on a smooth inclined surface.
Gravitational force acting on the body can be resolved into
two components -mgCosθ perpendicular to the inclined surface
and mgSinθ parallel to the inclined surface where m is the
mass of the body, g is the acceleration due to gravity and θ
is the angle of inclination of the incline with the
horizontal. The contact force which is exerted by the surface
on the body will be equal and opposite to mgCosθ . If we
take a resultant of all the forces acting on the body then
mgSinθ is the resultant force on the body acting down the
inclined plane. Thus the body will have acceleration.
(in which position) exactly a prism is used for
refraction/dispersion experiments?
A. The prism should be placed
in such a manner that if we view the prism from the top, we
see one of the triangular faces of the prism. Then one angle
of the prism can be taken as the refracting angle. Thus the
refracting edge will become vertical. The incident ray falls
on one of the faces which contains one of the sides making
the refracting angle. The refracted ray emerges from the face
containing the other edge of the refracting angle.
Locate the image of an object placed 10.0cm from a concave
spherical mirror of radius 15.0cm.
(2)A mouse 6.0cm tall stands 12cm in front of a concave
mirror whose radius of curvature is 18cm. Determine the size
and location of the mouse's image.
(3)Locate the image of an object placed 8.0cm in front of a
concave mirror whose radius of curvature is 20.0cm.
(4)A mouse 7.5cm tall stands at a point 4.0cm in front of a
concave mirror whose focal point is 6.8cm. Describe the
mouse's image in the mirror.
(5)A butterfly rests on a pin 8.6cm in front of a convex
mirror of focal length 22.5cm. Where is its image?
(6)The convex rear view mirror on a car door has a focal
length of 185cm. The image of an oncoming car is at 3.87m.
What is the actual position of the car?
(7)An ant 5.00mm long is placed 2.0cm from a converging lens
whose focal length is 3.85cm. How large and at what distance
from the lens will the ant appear to be?
(8)A student
looks at a mouse 8.05cm tall standing 21.8cm from a
converging lens of focal length 12.5cm. Describe the image of
the mouse as seen through the lens.
(9)In focusing a slide photograph on a screen, the convex
lens of a projector has to be moved to a distance of 23.2cm
from the slide. The focal length of the lens is 22.4cm. At
what distance is the screen from the lens? How large will a
figure 2.8mm tall on the slide appear to be on the screen?
(10)What is the focal length of a lens that produces a real
image three times as large as the object if the distance
between image and object is 1.0m?
A. We have to know the lens
and the mirror formulae and the sign convention to solve the
above problems.
1/v + 1/u = 1/f : mirror
1/v - 1/u = 1/f : lens formula
m = -v/u for a spherical
mirror and m = v/u for a lens and it is positive for a convex
lens when the image formed is virtual and is negative for a
it is always positive for a concave lens and f =
R/2 where v is the image distance, u is the object distance,
f is the focal length, R is the radius of curvature and m is
the magnification.
Sign convention is that all
distances measured in the same direction as the incident
light are taken as positive and in the opposite direction are
taken as negative. In case of mirrors, distances are measured
from the pole and in case of lenses, the distances are
measured from the optical center. Thus, the focal length of a
concave mirror will be negative and that of a convex mirror
will be positive. Also, the focal length of a convex lens
will be positive and that of a concave lens will be negative.
(1) f = -15/2cm and u = -10cm.
Substituting in the mirror formula, we get, v = -30cm. Thus
the image is real and inverted and at a distance of 30cm from
the mirror and on the same side as the object.
(2) R = -36cm, f = -9cm and u
Substituting in the mirror
formula, we get v = -36cm. Thus, image is on the same side as
the object at a distance of 36cm from the mirror.
Magnification m = -(-36/-12) =
Thus size of the image is 3×6
(3) Same method as (1)
(4) Same method as (2)
(5) As per the sign convention
for a convex mirror, f = 22.5cm, and u = -8.6cm. Substituting
in the mirror formula, 1/v = 1/22.5 + 1/8.6, we can find v.
(6) f = 185cm and v = 3.87cm
as the image formed in a convex mirror is always virtual thus
v will be positive. Substitute in the mirror formula to find
1/u = 1/185 - 1/3.87
(7) u = -2cm, f = 3.85cm.
Applying the lens formula, 1/v = 1/f + 1/u = 1/3.85 - 1/2
Thus, v = -4.17cm or the image
is virtual. Magnification m = 4.17/2 = 2.1 and the size of
the image is 5×2.1 = 10.5mm
(8) Same as (7)
(9) u = - 23.2 cm, f = 22.4
cm. Substituting in the lens formula, v = 6.49 m. The
magnification will be
m = - 649 / 23.2 = - 28. Thus,
the size of the image of an object of size 2.8 mm will be
(10) If the image is real,
then it is a convex lens. We have v = 1 - u. Now, v / u = (1
- u) / u = 3. Solving we get u = 1/4 m and v = 3/4 m. For a
real image u is negative and v is positive. Substituting in
the lens formula, we get f = 18.75 cm.
block of glass has a critical angle of 39 deg . What is the
index of refraction of the glass?
A. According to Snell's law,
Sin i/Sin r = refractive index of second medium w.r.t. to the
first medium where incident ray is coming from the first
medium and is getting refracted in the second medium. Also, i
is the angle of incidence and r is the angle of refraction.
In the problem, light is
passing from the medium glass to the medium air at the angle
of incidence being equal to the critical angle. In such a
case, the angle of refraction is equal to 90 deg .
Thus, Sin 39 /Sin 90 = refractive index of air w.r.t glass =
1/r where r is the refractive index of glass w.r.t. air. We
get, r = 1/Sin 39 .
do we determine the radii of curvature for a converging glass
lens(n=1.52) in order for it to have a focal length of f=
20cm. Assuming R1=R2.
A. The focal length of a
convex or a concave lens can be calculated by using the lens
1/f = ( n - 1 )( 1/R1 - 1/R2 )
where f is the focal length of the lens, n is the refractive
index of the material of the glass lens and R1, R2 are the
radii of curvature of the curved surfaces of the lens. For a
convex lens, R1 is positive and R2 is negative. For a concave
lens, R1 is negative and R2 is positive.
Here, f = 20, n = 1.52, R1 = R
and R2 = - R. Substituting into the lens makers formula, R =
the transmission axes of two Polaroid films are perpendicular
to each other, what is the percentage of incident light which
will pass the two films?
A. When incident light falls
on the first Polaroid film, it allows only those vibrations
to pass through which are parallel to its own transmission
axis. The emerging light has vibrations confined to one plane
only which is perpendicular to the transmission axis of the
second Polaroid film. Thus no light will pass through the
second Polaroid film.
would it be impossible to obtain interference fringes in a
double slit experiment if the separation of the slits is less
than the wavelength of light used?
The above diagram shows the double slit
experiment. The condition for maxima in this is dSinθ =mw
where d is the slit separation, θ is the angle S'SO and w is
the wavelength of the monochromatic source of light and m is
an integer. If d is less than w , then since m is an integer,
Sinθ becomes greater than 1 which is not possible. Thus it
is impossible to obtain interference fringes in a double slit
experiment if the separation of the slits is less than the
wavelength of the light used. Moreover, if d is less than w ,
then S and S' stop being two coherent sources but rather
behave like a single source.
noisy machine in a factory produces a decibel rating of 65dB.
How many identical machines could you add to the factory
without exceeding the 95-dB limit?
A. The intensity level L( in
bel ) of a sound of intensity I is represented by
L = log I/I0 where
I0 is the zero level of intensity or threshold of
hearing. Also, 1 decibel = 1/10 bel.
Let the intensity of sound
produced by one machine be I. Then, 6.5 = log I/I0
. Let I' be the intensity of sound when x number of machines
are used to produce decibel rating of 95dB. Then, 9.5 = log
Taking the antilog of the
above equations, I/I0 = 106.5 and I'/I0
= 109.5. Taking the ratio of these two relations,
we get, I/I' = 10? or I' = 1000 I or the intensity of sound
produced could be 1000 times that produced by one machine.
This means that 999 more machines can be used.
Explain the terms the fundamental mode of vibration and
resonant air columns.
A. Organ pipes are musical
instruments which are used for producing musical sounds by
blowing air into the pipe. Longitudinal stationary waves are
formed on account of superimposition of incident and
deflected longitudinal waves. The fundamental mode of
vibration is a simplest mode of vibration and the frequency
produced in this mode is the lowest frequency that can be
produced and is called the fundamental frequency that can be
produced. For example in a closed organ pipe ( closed at one
end ) in its fundamental mode of vibration the open end acts
as an antinode. This is because the air can move freely
there. The closed end acts as a node because air cannot move
too and fro there. If L is the length of the air column then
since the distance between an antinode and node is a quarter
wavelength, L = w /4 or w = 4L. Thus the fundamental
frequency is v / 4L, where v is the velocity of the wave.
Resonant air
column It is an apparatus
which consists of a long cylindrical tube filled with water
having its lower end joined by a rubber tubing to a moveable
reservoir of water. The cylindrical water tube is fixed along
a meter rod and the level of water in it can be lowered or
raised with the help of a reservoir.
A tuning fork of known
frequency is gently struck against a rubber pad and is held
horizontally at the mouth of the water tube. At the same time
the level of water in the tube is lowered till a loud sound
is heard. This increase in intensity of sound is due to
resonance between the tuning fork and the air column inside
the water tube. This happens when the compressions and
rarefactions sent down by the vibrating tuning fork reinforce
each other by reflection from the water surface and at the
mouth of the water tube. The first resonance position is the
fundamental mode of vibration. Lower water levels will give
us the next harmonics.
radar systems used by police to detect speeder are sensitive
to the Doppler shift of a pulse of radio waves. Discuss how
this device sensitive to the Doppler measures the speed of a
A. The Doppler effect is used
in radar to provide information regarding the speed of moving
targets by measuring the frequency shift between the emitted
and the reflected radiation. A transmitter produces pulsed
radio frequency radiation. It is fed to a movable aerial from
which it is transmitted as a beam. When the beam strikes the
moving vehicle a part of the energy of the radiation is
reflected back to the aerial. Signals received by the aerial
are passed to the receiver, where they are amplified and
detected. There will be a shift in frequency of the reflected
wave and emitted wave due to the Doppler effect. The apparent
frequency of the reflected wave is given by
F = f ( 1 - v/c ) where v is
the speed at which the source and the observer are moving
apart and c is the speed of electromagnetic radiation, f is
the real frequency or the frequency of the emitted signal
The output of the detector is
usually displayed on a cathode ray tube. The apparent
frequency is measured and thus the speed of the vehicle is
calculated.
A heterodyne device may also
be used in which beats are produced by superimposing the
emitted radio wave over the reflected (from the vehicle)
radio wave. In the heterodyne wave meter, a variable
frequency local oscillator is adjusted to give predetermined
beat frequency with the incoming reflected wave, enabling the
frequency of the reflected wave which has had Doppler shift
to be determined. Thus the speed of the vehicle can be
determined.
(C) Tutor 4 Physics}