53) We have:
I = K x

(1….0…..0)

(0….1…..1)

(0….1…
..1)

We
write out the determinant with eigenvalue l:

(1
- l….0…..0)

(0….1
- l ..1)

(0….1…
..1 - l)

Leading
to the characteristic equation:

(1
- l)

^{3}– (1 - l) = 0
(1
- l) [ ((1 - l)

^{2}– 1] = 0
Or:

(1
- l) (l

^{2}– 2l) = 0
Yielding
eigenvalues: l=
0, l = 2

Then:

T
= Kl, so T1 = 0, T2 = K, and T3 =2K

So
(

**) is the answer.**__B__
54)We will
have to take: (

**I**–**T1**)__C__
So
that:

K
[(1….0…..0)

[(0….1…..1)

[(0….1… ..1)

**-**

K
(1….0…..0)](x)

(0….1….. 0)] (y)

(0….0… ..1)] (z)

=

(0….0…..0) (x)

(0….0…..1) (y)

(0….1… ..0) (z) = 0

So
that: 0
=

(0)

(z)

(y)

Therefore
the ans. is (

**A**) or**i****(**Since the x-coordinate in the column vector is absent

**)**

55)
By the analog of the

*parallel axis theorem*:
I

_{o }= I_{G}- M(**R**^{2}**I**– RR)**D**I = I

_{o }- I

_{G}= M(

**R**

^{2}

**I**– RR)

RR
= r

_{o}^{2 }x
(0….0………0)

(0….1/2…..1/2)

(0….1/2…
..1/2)

**D**I = M r

_{o}

^{2 }x

[(1….0…..0)

[(0….1…..0)

[(0….0… ..1)

**-**

(0….0……..0)]

(0….1/2.. 1/2)]

(0….1/2… ..1/2)]

=
M r

_{o}^{2 }x
(1….0………..0)

(0….1/2…..-1/2)

(0….-1/2…
..1/2)

56)
For a

*circular loop*: B = m_{o}I / 2r at the center of the loop, where m_{o}is the**magnetic permeability of free space**.
We
have: r = 0.05 m, B = (0.70) (10

^{-4}) Wb/m^{2}
(Remember
the conversion factor: 1G = 10

^{-4}Wb/m^{2 })
Then,
solving for I: I = 2 Br/ m

_{o}_{}

where m

_{o}= 4p x 10^{-7}H/m
I= 2 (0.070)
10

^{-4}A/ (4p x 10^{-7}H/m) = 5.6 A
57)
The work done in bringing the charge is equal
to the change in the electrostatic energy of the system. Let W1 and W2
be the electrostatic energy of the system without and with the presence of the
dielectric, respectively.

Then:

W1 = 1/8 p ∫

_{0}^{¥}^{ }**D****×****E**( 4p r^{2}) dr =
1/8 p ∫

_{0}^{b }(q/ r^{2})^{2}**( 4p r**^{2}) dr +
1/8
p ∫

_{a}^{b }(q/ r^{2})^{2}**( 4p r**^{2}) dr +
1/8
p ∫

_{a}^{¥}^{ }(q/ r^{2})^{2}**( 4p r**^{2}) dr
And:

W2
= 1/8 p ∫

_{0}^{b }(q/ r^{2})^{2}**( 4p r**^{2}) dr +
1/8
p ∫

_{b}^{a}^{ }(kq/ r^{2})^{2}**(q/ r**^{2}) ( 4p r^{2}) dr +
1/8
p ∫

_{a}^{¥}^{ }(q/ r^{2})^{2}**( 4p r**^{2}) dr
Work
done = W2 – W1 =

1/8
p ∫

_{b}^{a}^{ }(kq/ r^{2})**(q/ r**^{2}) ( 4p r^{2}) dr
- 1/8 p ∫

_{b}^{a }(q/ r^{2})^{2}**( 4p r**^{2}) dr
=
(k – 1) q

^{2 }[1/b - 1/a]
58) We have an ideal
monatomic gas so:

U
(internal energy) = 3/2 [n RT/2]

The
change in internal energy is:

DU = 3 nR [ 373 K
- 273K] / 2

=
3 (8.32 J/K) (100K) /2= 1250 J

59) The

dS = dQ/ T , is:

**change in entropy**of the gas, givendS = dQ/ T , is:

D S = ∫

_{273}^{373 }nR dT/ T
=
5/2 nR ln (373 – 273) = 6.56 J/K

60) The f electron has ℓ =3 so that the

*total angular momentum quantum number*possibilities are:
j = ℓ + ½,
ℓ - ½

Then: j = 7/2,
5/2

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