## Thursday, February 27, 2014

### Solutions to GRE Sample Physics Problems (6)

76.  By Ampere’s relation:

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

Since E is const. within the capacitor then V= Ed

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.

Therefore, the instantaneous power output to the capacitor, is:

dW/ dt =  (iQ/C) [1 – exp (-t/ RC)]

Solve for the current:

i =  (e/ R) exp (-t/ RC)

So:

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

Therefore, on integrating:

W =  (e Q/ 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 [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)

82.  The potential energy is:

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]

Divide the top eqn. by the bottom eqn.:

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

So the matrix has no inverse. Ans.  (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 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)