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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.

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 Ohms)

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
zero.

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.

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 heat.

A. i = 18/(6+3) = 2 A

potential at A = 18-(2x6) = 6V

At steady state, capacitor will be fully charged

Let C_{eff} be the
effective capacitance of the 6µF and 3µF capacitors in
series

C_{eff} = 1/6 + 1/3 =
1/2

C_{eff} = 2 µF

Some charge will flow through 6 µF and 3 µF

Q=2x10^{-6} x 18 = 36
x 10^{-6}

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 6µF=6V

charge on 6 µF= (6 x 10^{-6}
x 6) µC

this is the charge which will flow from B to A

A. Capacitors 3 µF and 4 µF are in series

let C_{s}
= capacitance of a single capacitor which replaces the above
two capacitors in series. Then

1/C_{s}
= 1/4 + 1/3 =(4+3)/12=7/12

therefore C_{s}
= 12/7 (see Fig a above)

let C_{p}
= capacitance of the capacitor which replaces the capacitors
in parallel on Fig b above

C_{p}
= (12/7+6) µF = 54/7 µF = net capacitance

20 V potential difference is across 6 µF as well as 12/7 µF as they are in parallel

Charge through 12/7 µF capacitor = capacitance x voltage

=12/7 µF x 20 = 240/7 µC

Charge on 4 µF and 3 µF capacitors will be the same as they are in series.

Potential difference across 3 µF = (240 µC) / (7 x 3 µF) = 80/7 V = 11.4 V

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.

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 h=10m.

A. To find the retardation produced in the car due to the application of brakes, we use the third equation of motion, v²-u²=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.

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 =9800/334 g.

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'w²r²).

c) Show that the liquid surface is of parabolic form; that is, a vertical cross section of the surface is the curve y=w²r²/(2g).

d) Show that the variation of pressure with depth is P=p'gh.

b) If pressure at the axis is P" then from a) pressure at any point r is P = P" + p'r²w²/2.

c) Pressure due to rotation = p'r²w²/2 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 = r²w²/2g. 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.

A. If a car is moving on a plane, then since the bottom of the wheel is pushing backward against the road; 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.

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.

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.

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 (22² + 71.1²)^{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.

A. 1mph = 0.447m/s and 1 yard = 0.914m.

The ball is set into projectile motion. The range of projectile is given by the formula - R = u²Sin2q /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 covers.

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 v² ) = 1/2 m2 u² 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 slowly.

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.

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.

Yes, the force will get distributed equally.

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.

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.

A. Algebraic approach

The resultant force will be found by squaring the two given forces, adding them and then finding the square root.

Resultant force= square root of (100 + 225) = 18 N

Here 10 ² = 100 and

15 ² = 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.

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.

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 + Ø ) where y is the displacement of the particle undergoing SHM, a is the amplitude of vibration, w is the angular frequency and Ø 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.

A. F = mg where m is the mass of the body and g is the acceleration due to gravity (9.8 m/s²). Substituting the value we get, F = 9.8 N.

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.

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 glass.

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 cities.

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.

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.

A. Upper ball: h/n=(1/2)gt² or t=[(2h)/(ng)]1/2. And (v1)=gt=[(2gh)/n]1/2. For the lower ball (h-h/n)=(u2)t +(1/2)gt². 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).

A. We can use the third equation of motion to find the velocity of the brick as it reaches the ground, v²-u² = 2gh. Substituting, we get, v=6.28 m/s. Thus, the momentum is p=mv=10×6.28=62.8 kg m/s.

A. Let m_{1}
, u_{1} , v_{1} , be the mass, initial
velocity and final velocity of the player and m_{2} ,
u_{2} , v_{2} , be the mass, initial velocity
and final velocity of the referee. Then

m_{1}
= 110 kg, u_{1} = 3 m/s , v_{1} ,= 0 m/s

m_{2}
,= 55 kg, u_{2} = 0 m/s, v_{2} = not known

m_{1}
u_{1} , _{+ }m_{2} u_{2} , =
m_{1} v_{1} , _{+ }m_{2} v_{2}

110 x 3 + 0 =
0 + 55 x v_{2}

v_{2}
= 6 m/s

change in referee's velocity = 6-0 = 6 m/s

change in player's velocity= 0 -3 = -3 m/s

change in referees velocity as compared to player's velocity = 6-(-3) = 9 m/s

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 m_{1} and m_{2} are the
masses of the two balls, u_{1} and u_{2} are
the initial velocities of the two balls (u_{2}=0) and
v_{1} and v_{2}are the final velocities of
the two balls. Writing the equations for conservation of
momentum along the x and y axes, we get,

m_{1}u_{1} =m_{2}v_{2}Cos
Θ and m_{1}v_{1} = m_{2}v_{2}Sin
Θ

Applying the condition for the conservation of kinetic energy, we get

1/2m_{1}u_{1}²
= 1/2m_{1}v_{1}² + 1/2m_{2}v_{2}².

Now, m_{1}=1kg, u_{1}
= -3m/s, and m_{2} = 2kg. Substituting these values
and solving the above equations, we get, v_{1}=v_{2}=3^{
1/2 }. Thus the momenta of the two balls are the product
of their mass and the speed .

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 m_{1} and m_{2} are the
masses of the two balls, u_{1} and u_{2} are
the initial velocities of the two balls (u_{2}=0) and
v_{1} and v_{2}are the final velocities of
the two balls. Writing the equations for conservation of
momentum along the x and y axes, we get,

m_{1}u_{1} =m_{2}v_{2}Cos
Θ and m_{1}v_{1} = m_{2}v_{2}SinΘ

Applying the condition for the conservation of kinetic energy, we get

1/2m_{1}u_{1}²
= 1/2m_{1}v_{1}² + 1/2m_{2}v_{2}².

Now, m_{1}=1kg, u_{1}
= -3m/s, and m_{2} = 2kg. Substituting these values
and solving the above equations, we get, v_{1}=v_{2}=3^{
1/2 }. Thus the momenta of the two balls are the product
of their mass and the speed .

A. We will have to use vectors
to solve this problem. The velocity of the boat w.r.t. water
is denoted by v_{bw} and velocity of water w.r.t.
ground is denoted by v_{wg} and velocity of boat
w.r.t. ground is denoted by v_{bg}. Then we have, v_{bw}
+ v_{wg} = v_{bg}. Since v_{bw} is
along the north direction, it can be denoted along the
positive y-axis and v_{wg} 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, v_{bg} = (10 ²+
5²)^{1/2}= 125^{1/2}= 11.18 m/s.

The direction of the boat with
respect to the ground will be tan^{-1}10/2 = at an
angle of 63.4° to the x-axis.

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.

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/s²). 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 gt². 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.

(a) How much time elapses before the tank hits the ground?

(b) What is the velocity of the tank just before it hits the ground?

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 gt², we get t = 7.8s.

For finding the final velocity
in the vertical direction, we apply the equation of motion,
v² - u² = 2gh. Substituting, we get, v_{y} =
76.7m/s.

Now, the velocity in the
horizontal direction will remain 111m/s as there is no
acceleration in the horizontal direction. Thus, v_{x}
= 111m/s.

The velocity with which the
tank hits the ground, v = (v_{x}² + v_{y}²)^{1/2}
= 134.9 m/s and the direction will be as follows: ? = tan^{-1}v_{y}/v_{x}
where ? is the angle made with the x-axis.

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)² 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 = 1/2mr² 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 = 1/2Mv² + 1/2 I (v/r)²
where M is the mass of the object. Or v = [2Mgh/(M+I/r²)]^{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.

A. Let total distance be x. Then x/p=1/2(a1)(t1)² 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, -v²= -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]

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.

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 a_{x}
along horizontal direction is zero. If the initial position
of the object was ( x_{0}, y_{0}) the
position of the object at any time t along the horizontal
direction i.e. along the x-axis is given by x = x_{0}
+ u_{x}t + 1/2 a_{x}t^{2} , u and a_{x}
= 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 = y_{0} + u_{y}t
+ 1/2 a_{y}t^{ 2}. Here y_{ 0 }= 0 ,
u_{y} = 0 and a_{y} = 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} .

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/s²).
Substituting the values we get minimum speed=22.14 m/s.

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.

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 mv²/r where r is the radius of the loop. Thus, F = mgCosθ + mv²/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) = mv²/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'²/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.

1/2mv² = 1/2mv'² + 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) = mv²/r + mg.

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.

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.

(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 formula

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 real image; 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 = -12cm

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) = -3

Thus size of the image is 3×6 = 18cm.

(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 u.

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 7.84 cm.

(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.

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 .

A. The focal length of a convex or a concave lens can be calculated by using the lens formula,

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 = 20.8 cm.

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.

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.

A. The intensity level L( in bel ) of a sound of intensity I is represented by

L = log I/I_{0} where
I_{0} 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/I_{0}
. Let I' be the intensity of sound when x number of machines
are used to produce decibel rating of 95dB. Then, 9.5 = log
I' /I_{0} .

Taking the antilog of the
above equations, I/I_{0} = 10^{6.5} and I'/I_{0}
= 10^{9.5}. Taking the ratio of these two relations,
we get, I/I' = 10³ 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.

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.

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.