∮ 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 2nd law:
e = iR + Q
[1 – exp (-t/ RC))/ C
The power is: P =
e i=
i2R + iQ[[1 – exp (-t/ RC))/ C
Where
i2R 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 ]
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 [V2 / 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
(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 3rd law of planetary motion the period of a
planet varies directly as the cube of its semi-major axis, a. That is:
T2 ~
a3
Since the semi-major axis is equal to the mean distance R from the Sun then T2 varies inversely as the third root of the mean distance, or: R1/3
Ans. (E)
No comments:
Post a Comment