GRE Physics Problem Solutions (4) - Part B

We now see the remaining solutions for GRE problems, (4):

 

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

 

Factoring:

(1 - l) [ ((1 - l)² – 1] = 0

 

Or:

(1 - l) (l² –  2l) = 0

 

Yielding eigenvalues: l= 0, l = 2

 

Then:

 

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

 

So (B) is the answer.

 

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(R² I – RR)

 

D I =  I _o -  I_G   =    M(R² 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/m²

 

(Remember the conversion factor: 1G = 10 ^-4  Wb/m²   )

 

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 r² )  dr  =

 

1/8 p  ∫ ₀  ^b   (q/ r²)² ( 4p r² )  dr    +

 

1/8 p  ∫ _a  ^b   (q/ r²)² ( 4p r² )  dr    +

1/8 p  ∫ _a  ^¥   (q/ r²)² ( 4p r² )  dr   

 

And:

 

W2 =   1/8 p  ∫ ₀  ^b   (q/ r²)² ( 4p r² )  dr    +

1/8 p  ∫ _b  ^a  (kq/ r²)² (q/ r²) ( 4p r² )  dr    +

 

1/8 p  ∫ _a  ^¥   (q/ r²)² ( 4p r² )  dr   

 

Work done =   W2 – W1  =

1/8 p  ∫ _b  ^a  (kq/ r²) (q/ r²) ( 4p r² )  dr    

 

-  1/8 p  ∫ _b  ^a   (q/ r²)² ( 4p r² )  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 change in entropy of the gas, given

dS =  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