61)
The net torque and net force must be zero if the cylinder is to remain
stationary.

Then
(Rem: ‘N’ denotes normal force):

For
the

*horizontal forces*: N**– F**_{A}**= 0**_{B}
For
the

*vertical forces*: P + W – N**– F**_{B}**= 0**_{A}
For
the

*torque*: P d – r (F**+ F**_{A}**) = 0**_{B}
Factoring
in the coefficient of static friction, m :

F

**= m N**_{A}**and F**_{A }**= m N**_{B}**N**_{B , }**= m N**_{A }_{B }

_{}
N

**+ m N**_{B}**= P + W**_{A }_{}
And:

m r (N

**+ N**_{A}**) = Pd**_{B}
Or: m r (m N

**+ N**_{B }**) = Pd**_{B}
So: N

**(1 + m**_{B}^{2}) = P + W
m r N

**(1 + m) = Pd**_{B }
N

**= Pd/ m r**_{B}**(1 + m)**_{ }
Therefore:

P d/ m r

**(1 + m) = (P + W) (1 + m**_{ }^{2})
Solve
for d:

d= (P + W) m r (1 +
m)

**/**P (1 + m^{2})
Now,
substitute in the values given: P = 2W and
m = 1/3

d= (2W + W) r/3 (1 + 1/3) /
2W( 1 + 1/9)

d= [4Wr/ 3] /
20W/ 9 = 3r/ 5
or Ans.

__(E)__
62)
We use: N

**= Pd/ m r**_{B}**(1 + m)**_{ }
Substituting
the values given, or computed:

N

**= [(3W)(3r/5)] / [r/3(1 + 1/3)] = [6Wr/5] / [r/3 + r/9]**_{B}
N

**= 6Wr/5 / {4r/ 9} = (6Wr/ 5) x (9/ 4r ) = 27W/10 = 2.7W**_{B}
Now, N

**= m N**_{A }**= (1/3) N**_{B }_{B }
N

**= (2.7W)/ 3 = 0.9W**_{A }
But,
we need: F

**= m N**_{A}**= (0.9W/ 3) = 0.3W**_{A }
Ans.

__(__**A**)

**= 2.7W**

_{B}
Ans.

__(B)__
64)
Also from the result of problem 62, we seek N

**= 0.9W**_{A }
Ans.

__(C)__
65)
We need: F

**= N**_{B}**= 0.9W**_{A}
Ans.
(

__C)__
66)
The intensity of the sound is

*proportional to the square of the power*, and the power is proportional to the amplitude and the frequency. Since the amplitude is constant, the ratio of the intensities is equal to the ratio of the squares of the frequency, or:
I1/
I2 =
(f1/ f2)

^{2}^{}

Then
the ratio of the (higher) intensity 1200 cps wave to the 400 cps wave is:

I2/
I1 = (f2 / f1)

^{2 }= (1200 / 400)^{2 }= (3)^{2 }^{}

I2/
I1 = 9:1

Ans.
(

__E)__
67)
Let y be the

*vertical height*and x*the horizontal displacement*, then:
y
= ½ g t

^{2}and x = v_{o}t
y
= ½ (32 ft/sec

^{2}) t^{2}= 4ft or: t = [2 (4ft) / (32 ft/sec^{2}) ]^{½}
= (1/4)

^{½}= ½ s
But:
x =
v

_{o}t = 20 ft.
So: v

_{o}= 20 ft./ (½ s) = 40 ft. /s
Ans.

__C__
68)
By definition, the Young modulus Y =

*stress/ strain*.
Therefore: Y =
FL/ A D
L

The
force constant (k) is equal to: F/ D L = YA/
L

Ans.

__(A)__
69)
By

*Ohm’s Law*: 1) V/R = I = 0.6 amp
With
the added resistor:

2) V/ (R + 4) =
0.5 amp

From
(1): V = 0.6R

From
(2): V = 0.5R + 2

Subst: V = 0.6R in (2):

0.6R
= 0.5R + 2

0.1R = 2 and
R = 2/ 0.1 = 20 ohms

Then:
V = 0.6(20 ) = 12 V

The emf = 12
V

Ans.

__C__
70)
At full scale we need a voltage drop of 120 V across the resistor and
galvanometer while allowing a current of 0.01 amp flowing through the
galvanometer. Therefore, the resistor and galvanometer must be in series, so:

0.01
A
[ R + 10 W ] = 120 V

R + 10 W =
120 V/ 0.01 A = 12, 000 W

R
= 12, 000 W - 10 W
= 11,990 W

Ans.
(

**E)**
71)
From the information and the

*Heisenberg Uncertainty Principle*:
DP x (D x)

__>__h/ 2p = 1.05 x 10^{-34}J-s = 1.05 x 10^{-27}erg-s (cgs units)
The
uncertainty in the x-component is 0.5 angstrom where 1 A = 1.0 x 10

^{-8}cm
Then
the uncertainty in the x-component is: Dx =
0.5A = 5.0 x 10

^{-9}cm
The
uncertainty in the x-component of the
momentum of the electron is:

DP x» (h/ 2p) / D x = 1.05 x 10

^{-27}erg-s/ 5.0 x 10^{-9}cm
Or,
2 x 10

^{-19}g cm-s
72)
This uncertainty is impossible to determine from the information given.

73)
In classical theory the probability that the particle would be in the region between
x and x + L/3 is 0.33. Since no other
information is given then it is equally probable that it’s in any place in the
region – so one third of the time will be spent over any given line segment,
i.e. which is 0ne-third the total length.

74)
For a quantum mechanical interpretation, define the wave function (See e.g. http://brane-space.blogspot.com/2013/05/solutions-to-quantum-box-problems.html:

y
= A sin 2kx

sin
2kL = 0 and k (wave number ) = n p/L

y
= A sin (n px/L)

P(x)
= y

^{ 2 }= A^{2}sin^{2}(n px/L)
For
normalization:

1
= =

**ò**^{L }_{0}_{ }P(x)^{ }dx = A^{2}**ò**^{L }_{0}_{ }sin^{2}(n px/L) dx
A

^{2}(L/ n p) [ u/2 – sin 2u/ 4]_{ 0}^{n }^{p}^{ }^{ }= 1
A

^{2}(L/ n p) (n p/2) = 1
Or: A

^{2}= 2/L
Then the probability it will lie between 0 and L/3 is:

**ò**

^{L/3 }

_{0}_{ }sin

^{2}(npx/L) dx

= 2/ n p [n p/ 6 - ¼
(sin 2 p/
3 )] =

2/ p [p/ 6 - ¼ (sin 2 p/ 3 )] = 0.19

2/ p [p/ 6 - ¼ (sin 2 p/ 3 )] = 0.19

(Since
n = 1 is the lowest energy state)

75)
Since n = 2 is the second lowest state the probability the particle is between
0 and L/3 is:

(2/L) L/ 2 p [p/ 3
- ¼ (sin 4 p/ 3 )] = 0.40

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