University Science for the Life Sciences – Course Summary and Study Material

,ACCESS Test Bank for University Physics for Life Sciences 1st Edition
Randall




PHYSICS FOR THE LIFE SCIENCES
1
QUESTIONS
Q1.1 REASON: We know from the reading that diffusion distances scale with the square root of time. Specifically,
Equation 1.3 tells us xrms  2Dt . So, if the time were quadrupled, we would expect the diffusion distance to

double from d to 2d .

ASSESS: This is not specific to one dimension. Equation 1.5 for three dimensions reads rrms  6Dt . It has a
different numerical constant, but the dependence on time is the same.
1
Q1.2 REASON: The diffusion constant is given by Equation 1.2, D  vd . The fact that Xenon atoms are heavier
2
won’t affect the distance between collisions very much, but the fact that they are slower means the diffusion
constant will be smaller for Xenon.
ASSESS: It is intuitive that a heavy, slow-moving thing will spread more slowly than a light, fast-moving
thing.
Q1.3 REASON: We know that higher temperature indicates faster molecular motion on a microscopic scale. So, we
expect diffusion to be faster in hot water than in cold water. This means the diffusion constant will be larger
for hot water.
ASSESS: This can be seen easily by putting drops of food coloring or other dyes into glasses of water with
different temperatures.
Q1.4 REASON: One reason might be that the elephant trunk snake is larger. The surface area of skin available for
gas exchange scales with ℓ  r (treating the snake as cylindrical). The volume available for lungs scales with

ℓ  r 2 . So as snake size increases, the surface area increases linearly with r , whereas the volume increases
quadratically in r , meaning V  r 2 . Thus, as snake size increases, volume increases faster than surface area,
and using the volume for lungs becomes more efficient than using the skin for gas exchange.
ASSESS: This is not the only possible reason. A larger snake may also have a thicker skin, such that diffusion
of gases through the skin would take longer for the larger snake.




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, ACCESS Test Bank for University Physics for Life Sciences 1st Edition
Randall
1-2 Chapter 1

Q1.5 REASON: We can use the formula for the volume of a sphere, and check our answer by considering scaling.
4 1 1 1 1 1 
For a sphere, V   r 3   d 3 . So if V A   d 3A and V   d 3   2d   8  d 3 , then clearly
3

3 6 6 B B A  A 
6 6 6
 

VB  8VA.
ASSESS: We know that the volume of a sphere depends on the diameter cubed. So doubling the diameter

should result in the volume increasing by a factor of 23  8.

Q1.6 REASON: We can express this proportionality as T  c L , where c is some unknown quantity that does not

depend on length. If we have two such pendula, we can write TA  c LA and TB  c LB , such that
2
T c L T 
B  B
 LB   B  LA .
TA c LA  TA 
Inserting numbers, we have
2
2 s
LB    1m  4 m .
 1s 
ASSESS: If the period depended linearly on the length, doubling the period would require doubling the
length. We know the period doesn’t depend that strongly on the length; it depends on the square root of the
length. So, it makes sense that we would need to do more than double the length to double the period.

Q1.7 REASON: This is a diffusion problem. We will use the expression xrms  2Dt , where xrms is the root mean
squared average distance that the gas diffuses. The decay time is
24 h   3600 s  1.73 106 s
20 days 
1 day 1 h
We can use this in the diffusion equation above, and we find

xrms  2Dt   2 m

So, the correct answer is C.
ASSESS: This is consistent with the problem statement in that it means radon from deep inside the earth will
not reach the atmosphere. It is important to remember that this is the length that the gas will diffuse on
average; some radon particles will diffuse farther than that. If we are concerned with trace amounts of radon,
we may need to consider a depth of more than 2 m.

Q1.8 REASON: Another way of writing d  m3/8 is d  cm3/8 , where c is an unknown constant. Then we can

write dE  cmE3/8 and dH  cmH3/8 , from which it follows that
3/8
3/8 3/8  5000 kg 
cm3/8 m   dE  dH  mE 
  0.020 m 
d
E
 E
  E  

 0.1 m
dH cm3/8
H  mH   mH   70 kg 
This is the same as 10 cm. The correct answer is C.
ASSESS: It is certainly reasonable that an elephant’s aorta would be larger than a human’s.



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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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, ACCESS Test Bank for University Physics for Life Sciences 1st Edition
Randall
Physics for the Life Sciences 1-3

Q1.9 REASON: The line goes up two orders of magnitude every time it goes over one order of magnitude. On the
plot, this corresponds to a slope of 2. So, the correct answer is B.
ASSESS: As described in the chapter, the slope of a line on a log-log is equivalent to the exponent in a scaling
law; such that y  x2 .

Q1.10 REASON: The given scaling law can be written as f  m1/2 or f  cm1/2 , where c is an unknown

constant. Call the 200 g mass A, and the other mass B. Then fA  cmA1/2 and fB  cmB1/2 , such that
1/ 2
fB cm
     200 g 
B mA
fB fA 2.0 cycles/s 1.4 cycles/s
fA cmA1/ 2 mB 400 g

So, the correct answer is B.
ASSESS: It makes sense that a heavier mass would oscillate more slowly if the spring stiffness is unchanged.


PROBLEMS
P1.1 PREPARE: In a random walk in one dimension, the particle moves to the left or right with equal probability.
SOLVE: (a) After one step, half the particles (500) will move to x  d and half to x  d . In the second
step, half the particles on x  d will move further to x  2d and half (250) will move back to x  0 . It is
similar on the left, such that another 250 move back to x  0 . Thus, we expect a total of 500 random walkers
at x  0 after 2 steps.
(b) There cannot be any particles at x  d after two steps. They have to move after each time step. Any
particles that moved to x  d in the first time step, now must move to either x  2d or back to x  0 in the
second step.
(c) As described in part (a), 500 will initially move to x  d , and in the second step 250 of those will move

on to x  2d .
ASSESS: The presence of more random walkers in the center than on the edges is consistent with what we
know of diffusion.

P1.2 PREPARE: We will use Equation 1.1: xrms  (x2 )avg  n d for part (a). For part (b), we remember that

diffusion occurs equivalently in all directions.

SOLVE: (a) Since xrms  (x2 )avg  n d , we can write (x2 )avg  nd 2 , or equivalently


 
2
d  (x2 )avg / n   6
100  10 m / 10 6  108 m  10 nm
 
(b) The average position along the x -axis is zero, because diffusion to the left is equivalent to diffusion to
the right.

ASSESS: The answer to part (b) is why we often speak of xrms   . The average position is zero.

But average of the square of the position is not.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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