∮ B d s = 4pI/ c

Where
I denotes the enclosed current.

\ ∮ B d s = 0
and

**÷**B**÷****= 0 Ans. (A)**
77.
From the same relation as in (76), we can determine:

2pr

**÷**B**÷**= 4 pI/ c
Where
r is the distance from the center of the
tube. Then:

| B | =
2I/ r c Ans. (E)

78.
By Gauss’ Law: EA = Q/ e

_{o}
Then:
V A/ d = Q/ e

_{o}
And:
V = Q d / A e

_{o}
So
the capacitance is: C = Q/ V = Q/ [ Q d
/ A e

_{o}]
= A e

_{o}/ d Ans. (**)**__B__
79.A
potential difference is established between the plates – hence assume a
constant source of emf. Then from Kirchoff’s 2

^{nd}law:
e = iR + Q
[1 – exp (-t/ RC))/ C

The power is: P =

e i=
i

^{2}R + iQ[[1 – exp (-t/ RC))/ C
Where
i

^{2}R is the constant rate at which electrical energy is converted into heat energy ni the resistance.
i
= (e/ R) exp (-t/ RC)

So:

dW/dt
= (e) exp (-t/ RC) Q [1 – exp (-t/ RC)]/ RC ]

_{o}

^{¥}exp (-t/ RC) [-dt/RC + RC/2]

x ∫

_{o}^{¥}exp (-2t/ RC)(-2dt/ RC)}
Or:
W = ½ e Q

^{2 }= Q^{2}/ 2C
Ans. (

**)**__E__
80.
Let Q represent the total charge on one
plate. Then the total force on the other plate is obtained from Coulomb’s law:

F
= s Q/ 2 e

_{o }where s is the surface charge density, or
s = Q/ A

Then:
F = Q/ A
[Q/ 2 e

_{o }] = Q^{2}/2A e_{o }_{}

But: = V A
e

_{o}/ d = Q (from # 78)
So:
F = {V A e

_{o}/ d }^{2}/ 2A e_{o }
= e

_{o }A [V^{2}/ 2d^{2}] Ans. (__D)__
81.
By the method of images we can consider there is a point charge (-q) at x = -a,
y – z = 0.

Then:

F
= - q

^{2}/ (2 a)^{2 }= q^{2}/ 4 a^{2 }Ans. (__E)__V = ∫

_{¥}

^{a }(- q

^{2}dx)/ 4 x

^{2}

V
= - q

^{2}/ 4 a Ans. (__E)__
83. We can do a simple optical sketch for the
object in relation to mirror surface, i.e.

Sketching the object at 4” from the mirror axis (below),
and the focal point (focus) at 6” (since f = r/2 = 12”/ 2) we can easily see
the magnification for an ‘object’ about 1.33” tall is 3x since the y dimension
shown is 3x greater.

Also by
calculation, assume a 2" object:

f = r/ 2 = 12 in. / 2 = 6 in.

For an object with x = 2 “ :

x/ 2”
= y/ 6” so that M = y/ x = 6”/ 2” = 3x

Ans. (

__C)__
84. The optical path the light has to travel in
water

*must be equal to the path*it would travel in air.
So

n (H20)
d’ = n (air) d

where d, d’ are the respective paths

We know: n(H20) = 4/3 and n (air) = 1.0 therefore:

4d’ / 3 =
6 in. or d’ = 6
in./ (4/3) = 6 in (3/4) = 4.5 inches

Ans. (

__B)__
85. We use the lens maker’s formula:

n/ f =
(n’ – n) [1/ r1 = 1/r2]

__In air__:

1/ f1 = (1.50 – 1.00)(1/ r1 – 1/r2)

__In water__: 1.33/ f2 = (1.50 – 1.33) (1/r1 – 1/ r2)

Then:

1/f1 = 0.50 (1/ r1 – 1/ r2)

1.33/ f2 =
0.17 [1/r1 – 1/ r2]

f2/
(1.33 f1) = 0.50/ 0.17 so f2 = 3.91
f1 Ans. (

__E)__
86.
The operation of a laser depends on filling a metastable energy level higher
than the ground state, depleting the ground state. This inverts the population
levels of the two states. Ans. (

__B)__
87. Treat the system as a Carnot engine for
which the efficiency:

h =
W / Q

_{h}
Where W is the mechanical energy (work) produced and Q

_{h}is the heat removed from the high temperature reservoir. For the refrigerator, the coefficient of performance is the inverse of the efficiency so:
COP
= 1/ h = Q

_{h}/ W
We have Q

_{h}= 2 400 calories and COP = 6 then solving for W:
W = Q

_{h}/ COP = 2400 cal/ 6 = 400 cal
Ans. (

__C)__
88. If we compute the determinant for the matrix:

D =
0(0) - 0(0) + 3(0) = 0

__E)__(I.e. statement '

**E**' is NOT true)

89. The positron and electron completely
annihilate each other so by conservation of energy-momentum, spin, baryon and
lepton numbers two photons will be produced.

Ans. (

__C)__
90. According to

**Kepler’s 3**the period of a planet varies^{rd}law of planetary motion__directly__as the cube of its semi-major axis, a. That is:
T

^{2}~ a^{3}Since the semi-major axis is equal to the mean distance R from the Sun then T

^{2}varies

*inversely*as

*the third root*of the mean distance, or: R

^{1/3}

Ans. (

__E)__
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