Nov 4, 2010

Interpherence, Young's Double-slit Formula

Q. Interpherence, Young's Double-slit Formula

Question.
A police cruiser has an unusual radar speed trap set up. It has two transmitting antennae at the edge of a main road that runs north and south. One antenna is 2.0 m[west] of the other, and they are both fed from a common transmitter with a frequency of 3.0 x10^9 Hz. These antennae can be considered as point sources of continuous radio waves. The trap is set for cars travelling south. A student drives his car along an east-west road that crosses the main road at a level intersection 100 m[north] of the radar trap. The student's car has a "radar detector", so he hears a series of beeps as he drives west through the intersection. If the time interval between successive quiet spots is 1/5 s, as he crosses the main road, what is his speed?


Answer.
This is the problem of Young's double-slit interference experiment.

d = 2.0 m
D = 100 m
λ = c/f = 3x10^8 / 3 x 10^9 = 10^(-1) m
distance between successive destructive interference,
x = λD/d = 10^(-1) * 100/2 = 5 m
=> speed of the car 
= 5/(1/5) m/s
= 25 m/s
= 25 * 3.6 km/h
= 90 km/h.


Refraction of light

Topic :Light

Refraction of light

Question.
What is the final y-position of the ray?














Answer.
For the green slab, the angle of refraction, r, is given by
sinr / sin35° = 1/1.3
=> r = 26.1813°
=> y-position at interface
= 5 tan(26.1813°) cm
= 2.4583 cm




For the blue slab, the angle of refraction, r', is given by         
sinr' / sinr = 1.3/1.8
=> r' = (1.3)/1.8) * sin(26.1813°)
= 18.5815°
=> final y-position 
= 2.4583 cm + 5 tan(18.5815°) cm
= (2.4583 + 1.6809) cm
= 4.1392 cm.

Total internal reflection, refraction of light, prism

Topic :Light

Total internal reflection, refraction of light, prism.

Question.
Refer to the figure as under. Monochromatic light is incident normally on the face OY of a prism made of glass whose index of refraction is such that the critical angle is 42 degrees.
Redraw this figure showing the path of light through the prism, any reflections as well as refractions, until some of the light emerges from the prism.



Answer.

As the incident ray is normal to the hypotenuse, it passes undeviated and hits the vertical surface of the prism at 30 degrees with the vertical or 60 degrees with the norm.al at 8 cm below O.
As the critical angle is 42 degrees, the ray wiull undergo total internal reflection as its incident angle is 60 degrees which is more than the critical angle and hit the bottom surface at some point with the incident angle of 30 degrees and come out with refraction.



If r = angle of refraction                       
sin i / sin r = 1/sinC
=> sin r = sin i / sinC = sin30 / sin42
=> sinr = 0.7472
=> r = 48.35 degrees.

Heat:

Topic : Heat / Thermodynamics



Heat:



Question.
An insulated beaker with negligible mass contains a mass of 0.345 kg of water at a temperature of 60.4 deg. Celsius.
How much ice at a temperature of -11.2deg. Celsius must be dropped in the water so that the final temperature of the system will be 20.2deg. Celsius?
Take the specific heat for water to be 4190 J/(kg*K), the specific heat for ice to be 2100 J/(kg* K), and the heat of fusion for water to be 334 kJ/kg.



 Answer. 


Let the mass of ice required = x kg. 
Now, heat lost by 0.345 kg water when its temperature falls from 60.4 to 20.2 degrees
= 0.345 *1000*( 60.4 - 20.2 ) = 13869 cal.

Heat gained by x kg of ice when its temperature rises from - 11.2 to 0 degree
= x*1000*0.5*11.2 = 5600x cal.

Heat gained by x kg of ice at 0 degree to melt to water at 0 degree
= x*1000*80 = 80000x cal

Heat gained by x kg of water when its temperature rises from 0 to 20.2 degrees
= x*1000*1*20.2 = 20200x cal.

Total heat gained by ice 
= ( 5600 + 80000 + 20200 ) x cal.
= 105800 x cal.

Heat gained by ice = heat lost by water
=> 105800x = 13869
=> x = 0.131 kg.

Heating using electric heater

Topic : Heat / Thermodynamics


Heating using electric heater.

Question.
Mr Smith has a 3kW. water heater containing 120kg. of water initially as 20°C, and a clock radio which consumes 5W. No other appliances are using any electricity. There is 41p. of credit remaining on his electricity meter.
The alarm is set to sound after 8 hours, but if the meter runs out sooner than that then Mr Smith will oversleep.
What is the hottest temperature that his thermostat could be set to without running the electricity meter out of credit?
Specific Heat Capacity of water = 4200 J/(kg.K), Price of electricity = 8p/kWh.
Assume the water heater is thermally insulated, and neglect its own heat capacity.

Answer.

Let T = the maximum temperature to which the thermostat is set, then the temperature of water should not reach this value in 8 hours.
In 41 p, amount of electricity available 
= 41/8 
= 5.125 kWh
Of this, amount of electricity consumed by clock 
= 5 x 8 
= 40 Wh
= 0.040 kWh
Balance available for consumption by water heater
= 5.125 - 0.040 kWh
= 5.085 kWh 
= (5.085) * (3600000) J
 This should equal the energy for heating 120 kg of water
=> 120*(T - 20)*(4200) = (5.085) * ( 3600000)
=> T - 20 = (5.085) * (3600) / [(12) * (42)]
=> T - 20 = 36.32
=> T = 56.32° C.

Thermodynamics, Carnot Engine


Topic : Heat / Thermodynamics

Q. Thermodynamics, Carnot Engine.

Question
A reversible heat engine receives heat from two thermal reserviors at 870K and 580K,and rejects 50KW of heat to a sink at 290K.If the engine output is 85KW,make calculations for the energy efficiency and heat supplied by each reservior.


Answer  
Let Q1 = heat supplied by the first reservoir
and Q2 = heat received by the 2nd reservoir
=> Q1 + Q2 = work done + heat rejected 
=> Q1 + Q2 = 85 + 50 = 135 kW ... (1)

Efficiency of the engine for heat received from the first source
= (870 - 290) / (870) = 2/3
Efficiency of the engine for heat received from the 2nd source
= (580 - 290) / (580) = 1/2

=> (2/3)Q1 + (1/2)Q2 = 85
i.e., 4Q1 + 3Q2 = 510 ... (2)

Solving eqns. (1) and (2),
Q1 = 105 kW and
Q2 = 30 kW

Efficiency 
= W / (Q1 + Q2) x 100 
= 85 / (105 + 30) x 100
= 62.96 %.

Application of Kirchhoff's laws

Q. Application of Kirchhoff's laws

Question.
In the circuit below,  R1 and R2 are in parallel and R3 is in series with this combination. There is no immediate place to calculate a voltage drop by Ohm's Law because none of the resistors has both a known current and resistance. Neither can the total resistance be obtained because R2 is unknown. So how can to find the current in R3 by applying Kirchhoff's current Law to node A?

Answer.
Let current in R1 = i and in R3 = i'
Applying Kirchoff's first law to node A, i + .001 = i' 
Applying Kirchhoff's second law to the outermost loop, 
76 = 56000i + 56000 ( i + 0.001)
=> i = 20 /112000 = 0.00018 A = 0.18 mA
=> i' = 1.18 mA


Work done by a variable force

Q. Work done by a variable force

Question.
When a particle is located a distance x meters from the origin, a force of cos(πx/3) newton acts on it. How much work is done in moving the particle from x = 1 to x = 2 ? Interpret your answer by considering the work done from x = 1 to x = 1.5 and from x = 1.5 to x = 2.
Answer. 

Work done 
= integral (x1 to x2) cos (πx/3) dx 
1)
= (3/π) sin (πx/3) [ x = 1 to x = 2 ]
= (3/π) [sin (2π/3) - sin (π/3)]
= (3/π) (1/2 - 1/2) = 0
2)
= (3/π) sin (πx/3) [ x = 1 to x = 1.5 ]
= (3/π) [sin(π/2) - sin(π/3)]
= (3/π) (1 - 1/2)
= 3/(2π)
3)
= (3/π) sin (πx/3) [ x = 1.5 to x = 2.0 ]
= (3/π) [sin(2π/3) - sin(π/2)]
= (3/π) (1/2 - 1)
= - 3/(2π)

Interpretation:
This is the case of a variable force. The direction of force changes at x = 1.5. So work done is positive from x = 1 to x = 1.5. Same amount of negative work is done as the force is negative for x = 1.5 to x = 2. Hence, total work done for x = 1 to x = 2 is zero.

Electromagnetism

Q. Electromagnetism

Question.
A particle having mass m and charge q is released from the origin in a region in which magnetic field and electric field are given by B vector = - Bo*j vector and E vector = Eo* k vector. Find the speed of the particle as a function of its Z - coordinate.


Answer.

Lorentz force: F = (Fx, Fy, Fz) = q(E + V x B)
Here E = (0, 0, Eo), V = (Vx, Vy, Vz) and B = (0, -Bo, 0) ==>
V x B = (Vz*Bo, 0, -Vx*Bo)
m*(dV/dt) = q(Vz*Bo, 0, Eo -Vx*Bo)
(m/qBo)*(dV/dt) = (Vz, 0, Eo/Bo -Vx) ==>

(m/qBo)*(dVx/dt) = Vz ..... (1)
(m/qBo)*(dVz/dt) = Eo/Bo -Vx ..... (2)

Differentiate (1) wrt t ==> (m/qBo)*(d²Vx/dt²) = dVz/dt ..... (3)
Substitute dVz/dt from (3) in (2) ==>
(m/qBo)² * (d²Vx/dt²) = Eo/Bo -Vx
(m/qBo)² * (d²Vx/dt²) + Vx = Eo/Bo ..... (4)
The solution of the differential equation (4) is
Vx = (Eo/Bo)*[1-cos(qBo*t/m)] ..... (5)
From (1) ==>
Vz = (Eo/Bo)*sin(qBo*t/m) ..... (6)

From (5) and (6) ==>
X = (Eo/Bo)*[t - (m/qBo)sin(qBo*t/m)] ..... (7)
Z = (mEo/qBo²)*[1 - cos(qBo*t/m)] ..... (8) 

From (7) and (8) ==>
The particle is moving along the cycloid in the xz coordinate plane.

V² = Vx² + Vz² = (Eo/Bo)² * [2 - 2cos(qBo*t/m)]
V² = (Eo/Bo)² * 2[1 - cos(qBo*t/m)] ..... (9)
From (8) and (9) ==> V² = 2qEo*Z/m
V = √(2qEo*Z/m)

Laws of Motion

Q. Laws of Motion

Question.
Two blocks are in contact and pushing on one another. The blocks are on a frictionless surface. The mass of the blocks are 3.9kg and 9.9kg. The block with the mass of 3.9kg is being pushed into the 9.9kg block with a force of 69N. The 9.9kg block Is being pushed into the 3.3kg block with a force of 11N. What is the magnitude of the force exerted on the 3.9kg block by the 9.9kg block.

Answer.
Let R = force of action-reaction exerted by each block on the other.
Net force on block of mass 3.9 kg 
= 69 - R N 
and its acceleration will be 
= (69 - R) / 3.9 m/s^2

Net force on block of mass 9.9 kg 
= R - 11 N 
and its acceleration will be 
= (R - 11) / 9.9 m/s^2.

Both blocks will move with the same acceleration.
=> (69 - R) / 3.9 = (R - 11) / 9.9
=> R = 52.6 N.

Heat

Q. Heat

Question.
An insulated beaker with negligible mass contains a mass of 0.345 kg of water at a temperature of 60.4 deg. Celsius.
How much ice at a temperature of -11.2deg. Celsius must be dropped in the water so that the final temperature of the system will be 20.2deg. Celsius?
Take the specific heat for water to be 4190 J/(kg*K), the specific heat for ice to be 2100 J/(kg* K), and the heat of fusion for water to be 334 kJ/kg.

 
Answer. 
Let the mass of ice required = x kg. 
Now, heat lost by 0.345 kg water when its temperature falls from 60.4 to 20.2 degrees
= 0.345 *1000*( 60.4 - 20.2 ) = 13869 cal.

Heat gained by x kg of ice when its temperature rises from - 11.2 to 0 degree
= x*1000*0.5*11.2 = 5600x cal.

Heat gained by x kg of ice at 0 degree to melt to water at 0 degree
= x*1000*80 = 80000x cal

Heat gained by x kg of water when its temperature rises from 0 to 20.2 degrees
= x*1000*1*20.2 = 20200x cal.

Total heat gained by ice 
= ( 5600 + 80000 + 20200 ) x cal.
= 105800 x cal.

Heat gained by ice = heat lost by water
=> 105800x = 13869
=> x = 0.131 kg.

Electrostatics

Q. Electrostatics

Question.
A charge 4.59 x 10-6 C is held fixed at origin. A second charge 3.40 x 10-6 C is released from rest at the position (1.42 m, 0.746 m). If the mass of the second charge is 6.35 g, what is its speed when it moves infinitely far from the origin?


Answer.
Distance between the charges,
r = sqrt ( 1.42^2 + 0.746^2 )
= 1.60 m

Potential energy of the second charge w.r.t. the first charge, 
kqq' / r gets fully converted into kinetic energy, 
(1/2) mv^2 at an infinite distance.
=> kqq'/r = (1/2) mv^2
=> v = sqrt (2kqq' / ( r*m )
= √[2 (9*10^9) (4.59*10^-6) (3.40*10^-6) / (1.6) (0.00635)
= 3.72 m/s

Collision, Projectile Motion

Q. Collision, Projectile Motion.


Question.
Mass 1 hits mass 2 and causes the two masses to
fall on the floor as shown in the figure. The surfaces
are NOT frictionless.
a.
Using conservation of energy with consideration of the dissipated energy due to friction, calculate the velocity and momentum just before collision.
b.
Calculate the momentum of mass1 and mass2 after collision.




Answer.


Let u and v be the velocity of mass 1 before and after collision respectively.
Let v' be the velocity of mass 2 after collison. Initially, mass 2 is at rest.
First we calculate v and v' from the projectile motion of the two masses.
Time taken by both the masses to reach the ground
= √(0.7/4.9) s.
= 0.378 s.
In this time mass 1 covers a distance of 0.2 m with velocity v and mass 2 covers a distance of 1.1 m with velocity v'
So, v = 0.2 / 0.378 = 0.53 m/s
and v' = 1.1 / 0.378 = 2.91 m/s
 Now, by momentum balance,
25u + 0 = (25)*(0.53) + (15)*(2.91)
=> u = (13.25 + 43.65) / 25 = 2.276 m/s
 Dissipated energy due to friction
= P.E. at top - K.E. at the bottom of curved surface
= (0.025)*(9.8)*(0.3) - (1/2)*(0.025)*(2.276^2) joule
= (0.025)*[ (9.8)*(0.3) - (0.5)*(2.276^2) ]
= (0.025)*[ 2.94 - 2.59 ]
= 0.00875 joule
a)
Velocity of mass 1 before collision = 2.276 m/s
Momentum of mass 1 before collision
= (0.025)*(2.276) = 0.0569 (kg)(m)/s
Velocity and momentum of mass 2 before collision = zero.
b)
Momentum of mass 1 after collision
= (0.025)*(0.53) =0.0133 (kg)(m)/s
Momentum of mass 2 after collision
= (0.015)*(2.91) = 0.04365 (kg)(m)/s