Download Midterm Exam - 2 Set B Solution

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Physics 122, Summer 2005
Midterm Examination Solution (Set B)
August 8th , 2005
Exam policy for students and proctors:
• Write your name, SB ID number and Lab section (A/B) on each blue book you use.
• Solve any four of the five problems and start a new problem solution on a new page
• You can use one A4 size formula sheet and use any non programmable calculator.
Constants: c = 3 × 108 m/s, h = 6.626 × 10−34 J s, e = 1.6 × 10−19 C, 1 eV = 1.6 × 10−19 J
1.) NEWS FLASH !!!! A star Alpha Lyrae the fourth brightest star in the galaxy, is going to go supernova (explode). You are on a planet Orpheus orbiting the star studying emission lines from the star for PHY122
class when you hear the NEWS. You keep calm with rest heart beat of 70 beats/minute, jump aboard your space
ship FIRE BALL XL5 and escape the planet with a speed of 0.95c
a.) While studying the emission lines, you measured the peak wavelength emitted by the star to be 290 nm.
In that case, find the surface temperature of the star Alpha Lyrae. (10 points)
b.) Find your heartbeat rate as measured by a still observer on the planet (Round to closest integer). (10 points)
SOLN: a.) Use Wien’s Law that gives a relation between peak wavelength and temperature
λp T = 2.9 × 10−3 m K
T =
2.9 × 10−3
= 10, 000 K
290 × 10−9
(1)
b.) Use time dilation for time for one beat
∆t = p
∆to
1−
(v/c)2
1(min)/70
1(min)
=p
=
2
21.85
1 − (0.95)
(2)
Thus heart beat is 1/∆t ∼ 22 beats/min (Are you alive)
2.) A convex lens with a power of 50 diopters is placed 6 cm from a lighted object.
a.) At what distance will an image be formed? (7 points)
b.) Draw the ray diagram for the same (Show incident and refracted rays). (7 points)
c.) State the nature of the image. (magnified/diminished), (inverted/upright), (real/virtual) (2+2+2 points)
SOLN: a.) Image distance is given using lens equation
1
1
1
=
+
f
do di
c.) Image is diminished M = −0.5, inverted, real
di =
do × f
6 × (102 /50)
=
= 3 cm
do − f
6−2
(3)
2
3.) An unpolarized light beam of intensity 200 watt/m2 & wavelength 400 nm is incident on two polarizers.
The transmission axis of the first polarizer makes an angle of 45◦ with the horizontal. The transmission axis of the
second polarizer makes an angle of 90◦ with the horizontal.
a.) What is the intensity of the light after passing through the two polarizers? (10 points)
b.) The emergent light enters a photocell and is incident on a sodium surface whose work function is W = 1.1 eV,
determine the stopping potential. (10 points)
SOLN: If Io = 200 W/m2 is the intensity of the incident light, then the 1st polarizer reduces its intensity to
Io /2 = 100 W/m2
Angle between the transmission axis of 1st and 2nd polarizer is 45◦ . Hence final intensity is
If = (Io /2) cos2 θ = 100 cos2 45 = 50 watt/m2
b.) Stopping potential Vs can be obtained from
eVs = Ef − W =
6.626 × 10−34 × 3 × 108
hc
−W =
− 1.1 × 1.6 × 10−19 = 3.2095 × 10−19
λ
400 × 10−9
This gives Vs = 2 V
4.) A hydrogen bomb has a rest mass of 20 kg and about 0.8 % of this mass is converted to energy.
a.) If all of this energy were emitted as 400 nm photons, how many photons are produced? (10 points)
b.) If 20% of the emitted photons are absorbed by 2 kg rock at rest in 1 sec, then find its acceleration. (10 points)
SOLN: a.) The number of photons can be obtained as
N=
Etotal
0.008moc2
0.008 ∗ 20 ∗ (3 × 108 )2 × 400 × 10−9
=
==
= 2.897 × 1034 photons
Ephoton
hc/λ
6.626 × 10−34 × 3 × 108
b.) Use KE imparted to rock KE = mc2 − mo c2 on the rock is
0.1N hc
= mc2 − mo c2 = (γ − 1)mo c2
λ
0.2N h
= 0.01599
(γ − 1) =
mo cλ
(4)
Gives a = v/t = 5.30 × 107 m/s2
If you did this I gave you correct..however there is a difference in both. b.) Use Force F = ma = dp/dt on the rock is
dp
0.2 ∗ N h/λ
=
= 4.79 × 106 m/s2
mdt
m
5.) A solenoid with 100 turns is placed in a uniform magnetic field of strength 0.3 T. Assume that the face of the
coil is in the plane of the paper towards you and the magnetic field is pointing inside the plane of the paper. The
area of the loop begins shrinking at a constant rate dA/dt = 0.006 m2 /s.
a=
a.) What is the magnitude of the emf (voltage) induced in the loop while it is shrinking?(10 points)
b.) If the self inductance of the coil is 300 mH find (i) the rate at which induced current changes in the loop and
(ii) the direction (clockwise/anticlockwise) of the induced current. (7+3s points)
SOLN:
a.) The emf induced in the loop is
V =N ×B
dA
= 100 × 0.3 × 0.006 = 0.18 V
dt
b.) Rate of change of current in the circuit
V =L
The current is clockwise
dI
dt
dI
= 0.18/300 × 10−3 = 0.6A/s
dt