## Wednesday, February 19, 2014

### GRE Physics Problem Solutions (5)

61) The net torque and net force must be zero if the cylinder is to remain stationary.

Then (Rem: ‘N’ denotes normal force):

For the horizontal forces: NA – F B = 0

For the vertical forces: P + W – NB – F A = 0

For the torque:  P d – r (F A     +   F B )  = 0

Factoring in the coefficient of static friction, m  :

F A    =  m NA        and  F B =   m NB       ,    NA   =   m NB

NB    +  m NA     = P  + W

And:

m r (N A     +   N B )  =  Pd

Or:   m r (m NB        +   N B )  =  Pd

So:   NB  (1 +    m2)  = P + W

m r  NB     (1 +    m)  = Pd

NB  =   Pd/  m r   (1 +    m)

Therefore:

P d/  m r   (1 +    m)    =   (P + W) (1 +    m2)

Solve for d:

d=    (P + W) m r (1 +    m)  /   P (1 +    m2)

Now, substitute in the values given:  P = 2W and m  = 1/3

d=    (2W + W) r/3  (1 + 1/3) /   2W( 1 + 1/9)

d=  [4Wr/ 3] /  20W/ 9  =   3r/ 5  or Ans. (E)

62) We use:  NB  =   Pd/  m r   (1 +    m)

Substituting the values given, or computed:

NB  =   [(3W)(3r/5)] /   [r/3(1 + 1/3)] =  [6Wr/5] /  [r/3 + r/9]

NB  =     6Wr/5 /  {4r/ 9}  =   (6Wr/ 5)   x   (9/ 4r )   =  27W/10 = 2.7W

Now,    NA   =   m NB       =   (1/3) NB

NA   =  (2.7W)/ 3  = 0.9W

But, we need:  F A    =  m NA        =   (0.9W/ 3) =  0.3W

Ans. (A)

63) From the result of the previous problem, we have: NB  =      2.7W

Ans. (B)

64) Also from the result of problem 62, we seek NA   =  0.9W

Ans.  (C)

65) We need:  F B =  NA =  0.9W

Ans. (C)

66) The intensity of the sound is proportional to the square of the power, and the power is proportional to the amplitude and the  frequency. Since the amplitude is constant, the ratio of the intensities is equal to the ratio of the  squares of the frequency, or:

I1/ I2  =   (f1/ f2)2

Then the ratio of the (higher) intensity 1200 cps wave to the 400 cps wave is:

I2/ I1 =   (f2 / f1)2  =   (1200 / 400)2   =   (3)2

I2/ I1 =     9:1

Ans. (E)

67) Let y be the vertical height and x the horizontal displacement, then:

y = ½ g t2    and   x  =   vo t

y =   ½ (32 ft/sec2)  t2  =   4ft  or:  t =   [2 (4ft) / (32 ft/sec2) ] ½

=   (1/4) ½ =  ½ s

But: x  =   vo t  =   20 ft.

So:  vo    =   20 ft./  (½ s) = 40 ft. /s

Ans.  C

68) By definition, the Young modulus  Y = stress/ strain.

Therefore:  Y =   FL/ A D L

The force constant (k) is equal to:  F/ D L  =  YA/ L

Ans.  (A)

69) By Ohm’s Law: 1)  V/R = I = 0.6 amp

2)  V/ (R + 4) =   0.5 amp

From (1):   V = 0.6R

From (2):  V = 0.5R  + 2

Subst:   V = 0.6R in (2):

0.6R = 0.5R  + 2

0.1R  = 2   and R =   2/ 0.1 = 20 ohms

Then: V =  0.6(20 ) = 12 V

The  emf   =   12 V

Ans.  C

70) At full scale we need a voltage drop of 120 V across the resistor and galvanometer while allowing a current of 0.01 amp flowing through the galvanometer. Therefore, the resistor and galvanometer must be in series, so:

0.01  A [ R + 10 W ] = 120 V

R + 10 W  = 120 V/ 0.01 A  = 12, 000 W

R =   12, 000 W   - 10 W  =   11,990 W

Ans. (E)

71) From the information and the Heisenberg Uncertainty Principle:

DP x (D x)  > h/ 2p  = 1.05 x 10-34 J-s  = 1.05 x 10-27 erg-s   (cgs units)

The uncertainty in the x-component is 0.5 angstrom where 1 A = 1.0 x 10-8 cm

Then the uncertainty in the x-component is: Dx =  0.5A = 5.0 x 10-9 cm

The uncertainty in the x-component of the  momentum of the electron is:

DP x»    (h/ 2p)  / D x  =    1.05 x 10-27 erg-s/  5.0 x 10-9  cm

Or, 2 x 10-19 g cm-s

72) This uncertainty is impossible to determine from the information given.

73) In classical theory the probability that the particle would be in the region between x and x + L/3 is 0.33.  Since no other information is given then it is equally probable that it’s in any place in the region – so one third of the time will be spent over any given line segment, i.e. which is 0ne-third the total length.

74) For a quantum mechanical interpretation, define the wave function (See e.g. http://brane-space.blogspot.com/2013/05/solutions-to-quantum-box-problems.html:

y = A sin 2kx

sin 2kL = 0  and k (wave number ) =   n p/L

y = A sin (n px/L)

P(x) =  y 2  = A2 sin 2 (n px/L)

For normalization:

1 =  =     ò L 0   P(x)   dx  =   A2  ò L 0    sin 2 (n px/L) dx

A2  (L/ n p) [ u/2 – sin 2u/ 4] 0 n p           = 1

A2  (L/ n p) (n p/2) = 1

Or:  A2  =  2/L

Then the probability it will lie between 0 and L/3 is:

òL/3 0    sin2 (npx/L) dx

=  2/ n p [n p/ 6  -  ¼ (sin 2 p/ 3 )]  =

2/ p [p/ 6  -  ¼ (sin 2 p/ 3 )]      = 0.19

(Since n = 1 is the lowest energy state)

75) Since n = 2 is the second lowest state the probability the particle is between 0 and L/3 is:

(2/L) L/ 2 p  [p/ 3  -  ¼ (sin 4 p/ 3 )]  =  0.40