# Introductory Physics for the Life Sciences, PHYS*1070 - Sample Exam 2

Note: Not all questions may be applicable in the current semester.

1. At fairly high velocities, the drag force (F) acting on a sphere of radius R moving through a fluid of density r at a velocity (V) is given by:

$F = (1/4)\pi R^2\rho v^2$

Which graph below currently illustrates the relationship between F and v? A, B, C, D or E?

D

(Material from Appendix I in the Text)

Since $F$ is proportional to $v^2$, a graph of $F$ vs $v^2$ gives a straight line through 0,0.

Therefore D is correct.

2. At very low velocities, the drag force acting on a sphere of radius (R) moving through a fluid with a velocity (V) is given by:
$F = 6\pi \eta Rv$
where $\eta$ is a property of the fluid called its "viscosity''. According to this equation, the dimensions of viscosity ( $\eta$ ) are:

(A) $M L^{-1} T^{-1}$
(B) $M L^2 T^{-2}$
(C) $M^{-1} L^2 T^{-1}$
(D) $M^2 L^{-1} T^2$
(E) $M^2 L^{-1} T^{-2}$

A

(Material from Appaendix I in the Text)
Since $\eta = F/(6\pi Rv) = \mathrm {(mass)(acceleration)}/(6\pi Rv)$

Dimensions of $\eta = (M L/T^2)/(L L/T) = M L^{-1} T^{-1}$

3. Mary and Jane decided to try making some wine. Each started with yeast culture containing an identical number of cells. The yeast thrived in Mary's wine and increased exponentially with a growth constant of 1.0 day-1. Unfortunately, Jane kept her wine at too low a temperature and her yeast population decayed exponentially with a decay constant of 2.0 day-1. How long will it be before Mary's wine contains four times as many yeast cells as Jane's?

(A) $\ln 3.0$ days
(B) $\ln 3.0$ days
(C) $(\ln 3)/4$ days
(D) $4 \ln 3.0$ days
(E) $(\ln 4)/3$ days

E

(Material from Appendix II in the Text)
For Mary: $N_m = N_0e^{+1.0t}$

For Joan: $N_j = N_0e^{-2.0t}$

Find $t$ for which $N_m = 4N_j$

or, $N_0e^{+1.0t} = 4 N_0e^{+2.0t}$

$e^{+1.0t}/e^{-2.0t} = 4$

take natural logs

$3.0t = \ln 4$

$t = (\ln 4)/3.0 \; \mathrm {days}$

4. A travelling sound wave has a frequency of 1000 Hz and a speed in air of 340 m s-1. Which equation below could describe this wave?

(A) $y = y_o \cos (1000\pi t)\sin(340\pi x)$
(B) $y = -2y_o \cos(2000\pi t)\sin(5.88\pi x)$
(C) $y = y_o\sin(2000\pi t + 5.88\pi x)$
(D) $y = y_o\sin(1000\pi t - 0.340\pi x)$
(E) $y = -y_o\sin(1000\pi t + 340\pi x)$

C

Answers A and B represent standing waves.
$\omega = 2\pi f = 2\pi \; 1000 = 2000\pi \; \mathrm {rad/s}$

Therefore the answer must be C

Check: $v = \lambda f, \lambda = v/f = 340/1000 = 0.34 \mathrm{m}$

$k = 2\pi /\lambda = 2\pi /0.34 = 5.88\pi \; \mathrm {rad/m}$

5. A wave on a string is described by the equation:

$y = 0.03\cos(150t)\sin(15x) \mathrm {(All\;quantities \; are \; S.I.)}$

Which one of the following statements is NOT correct?

(A) This represents a standing wave with a node at the origin (i.e., where $x=-0$).
(B) This wave could be produced by two travelling waves each of amplitude $0.015 \mathrm {\; m}$.
(C) Each particle oscillates with a period of $0.042 \mathrm {\; s}$.
(D) The speed of travelling waves on this string is $5.00\mathrm {\; m/s}$.
(E) The wavelength of this wave is $0.42 \mathrm {\; m}$.

D

A is correct-a standing wave; $y = 0$ at $x = 0$ for all values of t (a node)

B is correct since $2\times 0.015 \mathrm {\;m} = 0.03 \mathrm {\;m}$

C is correct: $\omega = 2\pi /T = 150$. Therefore $T = 2\pi /150 = 0.042 \mathrm {\;s}$

D is not correct: $k = 2\pi /\lambda ,\; \lambda = 2p /15 = 0.42 \mathrm {\;m}. v = f\lambda = (0.42)(1/.042) = 10 \mathrm {\;m/s}$

E is correct: See value of $\lambda$  in previous line.

6. The diagram shows a tube which is closed at one end and open at the other. If the tube is in air, which of the following sound frequencies will induce a standing wave or resonance in the tube? (See Equation Sheet for required data.) (Answers given to 2 significant figures.)

(A) 60 Hz
(B) 140 Hz
(C) 280 Hz
(D) 12 Hz
(E) 340 Hz

B

For a tube closed at one end, the fundamental frequency is $f_1 = v/4\mathrm {L} = (340\; \mathrm {m/s})/(4\times 0.60\mathrm {\;m}) = 142 \mathrm {\;Hz}.$

Can have only odd harmonics: e.g. $3\times142 = 425 \mathrm {\;Hz}, 5\times142 = 710 \mathrm {\; Hz}$ etc

7. When John is shouting, the intensity level 1.0 m from his mouth is 80.0 dB. What is the intensity level due to this sound, 20.0 m from John? (Assume no loss of acoustic power in the air; ($I_o = 1.0 \times 10^{-12} W m^{-2}.$) (Answers given to 2 significant figures.)

(A) 80 dB
(B) 63 dB
(C) 54 dB
(D) 4.0 dB
(E) 0.20 dB

C

At $1.0 \mathrm {\; m}: \mathrm L = 80.0 = 10 \; \log \mathrm {(I_1/I_0)}$

$\mathrm {I_1/I_0} = 10^8$

$\mathrm {I_1} = \mathrm {10^8I_0} (\mathrm {when \; I_0 = 10^{-12} W/m^2})$

Inverse square law: $\mathrm {I = P/4\pi r^2}$

$\mathrm {I_1/I_2 = (r_2/r_1)^2}$

$\mathrm {I_2 = (I0\times10^8)(1.0/20.0)^2 = I0\times 10^8\times 20^{-2}}$

$\mathrm {IL_2 = 10 \log(I_2/I_0) = 10 \log [(I0\times 10^8\times 20^{-2})/I_0] = 80 - 26 = 54 \;dB}$

8. Referring to John's shouting in Question 7, the acoustical power being produced by John is:

(A) $\mathrm {4\pi \times 10^{-4} W}$
(B) $\mathrm{4\pi \; W}$
(C) $\mathrm {1.0 \times 10^{-4} W}$
(D) $\mathrm {320\pi \; W}$
(E) $\mathrm {\pi \;W}$

A

From question 7 $\mathrm {I \;at \;1 \;m = I0\times10^8 = 1\times 10^{-12}\times 10^8 = 1\times 10^{-4} W/m^2}$
Also $\mathrm {I = P/4\pi \; r^2}$

$\mathrm {P = I(4\pi \; r^2) = (1\times 10^{-4} W/m^2)(4\pi )(1.0 m)^2 = 4\pi \times 10^{-4} W}$

9. Which statement concerning the human ear is NOT correct?

(A) On average, the human ear is most sensitive to sounds of frequency about 3500 Hz.
(B) The tympanic membrane, oscicles and oval window form a piston-lever system which amplifies the pressure fluctuations in the sound wave,
(C) In adults, the air in the auditory canal has a natural resonance frequency of about 500 Hz.
(D) The basilar membrane contains the hair cells which change mechanical oscillations to electrical nerve signals.
(E) The region of maximum amplitude oscillation of the basilar membrane depends on the sound frequency; this may be one way we recognize sounds of different pitch.

C

The resonance of the auditory canal is at about 3500 Hz not 500 Hz.

10. The diagram illustrates a camera with a 50.0 mm focal length lens, taking a picture of a person. The image on the film is 34 mm high and the film plane is 51.0 mm behind the lens. Which row below correctly gives the lens-person distance and the person's height?

Answer Lens-Person Distance in m Height of Person in m
3.5 1.7
B 2.6 1.7
C 3.5 1.4
D 2.6 1.4
E 3.5 2.0

B

$\mathrm {1/p + 1/q = 1/f}$

$\mathrm {1/p + 1/0.051 = 1/0.050}$

$\mathrm {p = 2.55\; m = 2.6 \; m}$

$\mathrm {m = y'/y = -q/p}$

$\mathrm {-0.034\; m/y = -(+0.051 \; m)/+2.55\; m}$

$\mathrm {y = 1.7\; m}$

11. John is myopic and he wears corrective eyeglasses with a power of -2.25 diopters for his left eye. The near point of his corrected vision for his left eye is 25 cm. Where are the near and far points of John's uncorrected left eye?

Answer Near Point (m) Far Point (m)
A 0.25 infinity
B infinity  0.25
C 0.16 0.25
D 0.16  0.44
E 0.25 0.44

D

Far point: For an object at infinity, the corrective lens must produce a virtual image at the uncorrected far point.

$\mathrm {i.e.,\;1/p + 1/q = P}$

$\mathrm {1/inf + 1/q = -2.25 \;diopters}$

$\mathrm {q = -0.44 \;m}$

Therefore the uncorrected far point is at $\mathrm {0.44 \;m}$

Near Point: For an object at $\mathrm {0.25 \;m}$, the lens produces a virtual image at the uncorrected near point.

$\mathrm {i.e.,\; 1/p + 1/q = P}$

$\mathrm {1/0.25 + 1/q = -2.25 \;diopters}$

$\mathrm {q = -0.16\; m}$

Therefore the uncorrected near point is at $\mathrm {0.16\; m}$

12. The 2 bright wing lights of a Boeing 747 are 30 m apart. The pilot of another plane suddenly sees the two lights, which previously appeared as one, resolve into two separate lights. Assuming the pilot has a pupil diameter of 2.0 mm and the wavelength of the light is 550 nm, the distance from the pilot to the Boeing 747 is: (assume visual resolution is limited only by diffraction in the eye.)

(A) 100 m
(B) 1.2 km
(C) 20 km
(D) 90 km
(E) 250 km

D

$\mathrm {\alpha = (1.22\lambda )/na = 30/D,\; for \;a \;small \; angle.}$

$\mathrm {\alpha = (1.22\times 550\times10^{-9} m)/(1.00\times2.0\times10^{-3} m) = 30/D}$

$\mathrm {D =89\times10^3 m = 90\; km}$

13. Which line below correctly describes the electric field at point x, midway between the two ions. The ions are in water with a dielectric constant of 80.

(A) $\mathrm {13.5 \times 10^8 \;N/C\; toward \;the \; Cl^- \;ion}$
(B) $\mathrm {13.5 \times 10^8 \; N/C \; toward \; the\; Ca^{++} \; ion}$
(C) $\mathrm {4.5 \times 10^8 \; N/C \; toward \; the \; Cl^- \; ion}$
(D) $\mathrm {4.5 \times 10^8 \; N/C \; toward\; the \; Ca^{++} \;ion}$
(E) $\mathrm {9.0 \times 10^8 \;N/C \;toward\; the\; Cl^- \;ion}$

A

The field $E_1$ at $x$ due to the $Ca^{++}$ ion acts toward the $Cl^-$ ion (or away from the $+ve \;Ca^{++}$)

$\mathrm {E_1 = (kQ)/(\kappa \;r^2) = (9\times10^{-9})(2\times1.6\times10^{-19})/(80\times2\times10^{-9}) = 9.00\times10^8 \;N/C}$

The field $E_2$ at $x$ due to the $Cl^-$ ion acts toward it (or in the same direction as $E_1$)

$\mathrm {E_2 = (kQ)/(\kappa r^2) = (9\times10^{-9})(1.6\times10^{-19})/(80\times 2\times 10^{-9}) = 4.50\times 10^8 \; N/C}$

The total field is $\mathrm {13.5\times10^8\; N/C}$ toward the $Cl^-$ ion.

14. In the electrical circuit below, what must be the value of gx (in S) in order that I = 20 A?

(A) 0.17
(B) 2.1
(C) 3.3
(D) 4.0
(E) 5.9

D

The total equivalent conductance $\mathrm {G_T}$ is given by

$\mathrm {I = G_TV}$

$\mathrm {G_T = I/V = 20A/10V = 2 \;S}$

The circuit simplifies as shown in the figure:

$\mathrm {1/G_T = 1/G_x + 1/G_{12}}$

$\mathrm {1/2=1/G_x + 1/4}$

$\mathrm {G_x = 4 \;S}$

15. A light source is emitting 4.2 watts of light energy at a wavelength of 550 nm. How many photons are emitted by this source in one hour?

(A) $5.5 \times 10^{-3}$
(B) $4.2 \times 10^4$
(C) $5.5 \times 10^{11}$
(D) $1.7 \times 10^{14}$
(E) $4.2 \times 10^{22}$

E

$\mathrm{Energy \;of \;1\; photon= \\ E_{ph} = hc/\lambda = (6.63\times 10^{-34} Js)(3.00\times 10^8 m/s)/550\times 10^{-9} m = 3.62\times 10^{-19} J}$

$\mathrm {Power = 42 \;W = E/T = (N E_{ph})/t = N 3.62\times10^{-19} J/3600 s}$

$\mathrm {N = 4.2\times10^{22} \;photons}$

16. The graph of the probability density of a p -electron as a function of x in a linear, conjugated carbon chain is shown below:

Which one of the lines below is correct?

Answer value of a Wavelength of electron probability that electron is between $x = \frac {1}{2}$ and $x = 3 \frac {1}{4}$ Graph of wave function
A $1/\ell$ $\ell/4$ 4
B $2/\ell$ $\ell$ 3/4
C $3/\ell$ $\ell/2$ 1/4
D $1/\ell$ $\ell$ 1/4
E $1/\ell$ $\ell/4$ 4

C

$\mathrm {P_x = \phi ^2 = (2/\ell)\sin^2(4\pi \; x /\ell), for\; an\; n = 4 \;electron}$

$\mathrm {a = peak\; value \;of\; P_x = 2/\ell \\ (i.e., the\; value \;of\; P_x \;at \;those\; x \;for\;which\; \sin^2() = 1)}$

$\mathrm {The \;wavelength = the\; length \;of \;2 \;humps = \ell /2\; for \;an \;n = 4 \;electron.}$

Probability = area relative to the whole area under the curve which is 1

The area for the space from $\mathrm { x = \ell /2\; to\; x = (3/4)\ell = \ell/4}$

Wave function or  $\mathrm {\phi = P_x\;^{1/2}}$ is given in the figure.

17. The speed of a $\mathrm {\beta-particle}$ just after $\mathrm {\beta -decay}$ is determined to be one half the speed of light. The corresponding wavelength of this $\mathrm {\beta -particle}$ would be:

(A) $\mathrm {1.7 \times 10^{-8} m}$
(B) $\mathrm {4.9 \times 10^{-12} m}$
(C) $\mathrm {7.3 \times 10^{-7} m}$
(D) $\mathrm {9.4 \times 10^{-14} m}$
(E) $\mathrm {6.2 \times 10^{-5} m}$

B

For particles, $\mathrm {\lambda = h/mv = 6.63\times10^{-34} Js/(9.1\times10^{-31} kg)(1/2\times3.00\times 10^8 m/s)}$

(a $\beta$ particle is an electron)

$\mathrm {\lambda = 4.9\times10^{-12} m}$

18. What is the wavelength of the photon emitted during the transition shown in the diagram? Assume a linear conjugated p -system with carbon-carbon bond lengths of 0.15 nm.

(A) 670 nm
(B) 570 nm
(C) 470 nm
(D) 370 nm
(E) 270 nm

E

$\mathrm {E_{ph} = hc/\lambda = (4^2 -3^2)h^2/8ml^2}$

$\mathrm {\lambda = 8ml^2c/(4^2 - 3^2)h}$

There are 6 carbon atoms so there are 5 bonds making up l

$\mathrm {\lambda = \\ [8(9.1\times10^{-31})(5\times0.15\times10^{-9})^2(3.00\times10^8)]/[7(6.63\times10^{-34})] \\ = 2.65\times10^{-7} m \\ = 265 \;nm}$

19. If $\mathrm {E_r}$, $\mathrm {E_v}$ and $\mathrm {E_e}$ are, respectively, the energies required to bring about rotational, vibrational and rotational transitions in molecules, then:

(A) $\mathrm {\lambda _e > \lambda _ v \; and\; \lambda _ v =\lambda _r}$
(B) $\mathrm {\lambda _ v < \lambda _e < \lambda _ r}$
(C) $\mathrm {\lambda _e < \lambda _ v < \lambda_r}$
(D) $\mathrm {\lambda _v = \lambda_e \;and \; \lambda_ r <\lambda _v}$
(E) $\mathrm {\lambda_ r < \lambda_v < \lambda_ e}$

C

Since $\mathrm {E_e > E_v > E_r and E = hc/\lambda}$

$\mathrm {\lambda_ e < \lambda_ v <\lambda_ r}$

(See the Text p 48 - 53, Sections 4-4 to 4-5)

20. Which of the following radiations are listed in increasing order of relative biological effectiveness or qualifying factor? (i.e., increase from left to right)

(A) $\mathrm {\gamma -rays, \alpha -rays, \beta -rays}$
(B) $\mathrm {\gamma -rays, \beta -rays, \alpha -rays}$
(C) $\mathrm {\beta -rays, \gamma -rays, \alpha -rays}$
(D) $\mathrm {\alpha -rays, \beta -rays,\gamma -rays}$
(E) $\mathrm {\alpha -rays, \gamma -rays, \beta -rays}$

B

See discussion in the Text pp 77, 78.

21. $\mathrm {Thorium\;^{ 232}\;_{90}Th}$ undergoes $\mathrm {\alpha -decay}$ to produce a daughter nucleus which in turn undergoes $\mathrm {\beta \; decay}$ yielding $\mathrm {^A\;_zA_c}$. Which of the following properly gives A and z?

A 226 90
B 227 88
C 228 89
D 228 87
E 227 86

HINT an  $\mathrm {\alpha -particle \; is\; ^4\;_2\alpha \; and \; a \; \beta -particle\; is \;^0\;_{-1}\beta}$

The $\mathrm {\alpha -decay}$ causes A to decrease by 4 and Z by 2.
($\mathrm {\beta\; decay}$ leaves A unchanged and Z increases by 1.)