GRE Sample Problems Solutions (Conclusion)

91. Ans (E):

 

Since  R + O + Y  + G + B  + V = W,

 

R + BG = W, and B + Y = W then

 

O + Y + G + B + V= BG

 

92.  (A)

Since : R + O + G + B + V + Y = W,

 

B + Y  = W

 

Therefore, R + O + G + B + V = B

 

93.The shortest wavelength  x-ray will correspond to an x-ray of the highest energy, i.e. an x-ray which has been the only one emitted by the electron as it is stopped – and thus will have all the electron’s energy. Then, by Planck's law for quantization of energy:

 

E _max = hc/  l  = 2 Mev = 2 x 10 ⁶ eV (1.6 x 10^-12  erg/ eV)   


=  3.2 x 10 ^-6  erg

 

c = 3 x 10 ¹⁰  cm /s   and  h (Planck constant)  = 


 6.6261 × 10⁻²⁷ erg- s

 

Therefore:   l   = hc /  E _max   =  

 

 (6.6261 × 10⁻²⁷ erg- s) (3 x 10 ¹⁰  cm/s )/ 3.2 x 10^{- 6}  erg

l =   6.2 x 10 ^-11   cm

 

But 1 Å  = 10 ^-10 m = 10 ^-8  cm 

 

so  l =   (6.2 x 10 ^-11   cm)  / 10 ^-8  cm / Å  = 


6.2 x 10 ^{- 3}   Å » 0.01 Å

Ans. C

 

94 The de Broglie wavelength is defined from: l = h/ p

 

Where p is the momentum of the particle.

 

We may also write this as: p = Ö (2mE)

 

Where E = 3 kT/2

 

So that p = Ö (3mkT)

 

Then the de Broglie wavelength is:

l = h / Ö (3mkT)

 

Now, the neutron mass n = 1.66 x 10 ^-24 kg

 

The temperature T = 300 K

The Boltzmann constant is: k = 1.38 x 10 ^-23 J/K

 

And the Planck constant (MKS) is: h = 6.6261 × 10⁻³⁴ J- s

 

Solving for l:

 

l =  (6.6261 × 10⁻³⁴ J- s)/ Ö  [3 (1.67 x 10 ^-27 kg) 300 K (1.38 x 10 ^-23 J/K)]

 

=  1.79  x 10 ^-12  m

But: 1 Å  = 10 ^-10 m   so: 

 

l =   (1.79 x 10 ^-12  m)  / 10 ^-10  m / Å   =  1.79 Å

Ans. (E)

 

95. Just prior to the collision we will have:

T_A = 2 mgL (1 – cos  q) 

 

Also:  T_B= 2 mgL  after the collision (since cos  q = 0)

 

If the ball is to rise to a horizontal position the condition to be met is:

 

v_B – v_A = e(v_A – v_B) where e = 0.5

 

and:    m (v_A + v_B) = m (v’_A + v’ _B)

 

v_B – v_A =  0.5 v_A

 

v’_A + v’ _B  =   v_A

 

v’ _B  =    0.75 v_A

Ö (gL) = 0.75  Ö 2 gL (1 – cos  q) 

 

Square both sides:

 

gL  =  1.124 [ gL (1 – cos  q)  ]

 

Divide through by gL:

1 =   1.124 (1 – cos  q)   or:

1/ 1.124 = 1 – cos  q  

 

cos  q     =   1 - 1/ 1.124  =  1 – 0.89 = 0.11 

 

Or: cos  q     =   1/9  (In fractional format)

Ans. (E)

 

96  Ans. (E)

 

Since the coefficient of restitution is not 1 the collision is not elastic. Hence, part of the kinetic energy must be dissipated as heat.

 

97. We have:

 

v’_B – v’ _A =   v_A

 

v’_B +   v’ _A =   v_A

 

Then:  2 v’_B  =  2 v_A

 

Thus, the ball will be at rest after an elastic collision.

 

98. From electrodynamics:

 

E(r, t) = (e ₁ E₁ +  e ₂ E₂)  exp [ (ik r - iwt)

Is a plane-parallel wave with the polarization vector being:

[(E²₁ +  E²₂)² ] ^-1/2 (e ₁ E₁ +  e ₂ E₂) 

So Ans. (A)

99. Ans. (C)

If E₂ = i E₁   then E₂   is out of phase with E₁  by 90 degrees. So to the observer the electric vector will appear to rotate counter clockwise  since whenever E₁  is a maximum E₂  is zero and vice versa. When both have the same magnitude E₁  is leading E₂  by 90 degrees.  In other words, it is left circularly –polarized light.

100. Reinforcement of the reflected wave will occur when the light reflected back through the film is in phase with the reflected wave. This will occur when the optical path length is equal to one half wavelength. Hence:

2 n d = l/ 2

Solving:  d =   l/ 4 n 

I.e. in terms of the wavelength (l)    ¼  n

Ans. (B)