# Introductory Physics for the Life Sciences, PHYS*1070 - Sample Exam 1

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

- A traveling sound wave has a speed in air of \(340 \; \mathrm m \; \mathrm s^{-1}\) and a wavelength of \(0.50 \; \mathrm m\). Which equation below could describe this wave?

(A) \(y = -y_0 \sin (1360\pi t + 4\pi x)\)

(B) \(y = 2y_0 \cos (340 \pi t) \sin (0.50 \pi x)\)

(C) \(y = y_0 \sin (680 t - 0.50 x)\)

(D) \(y = y_0 \cos (1360 \pi t) \sin (4 \pi x)\)

(E) \(y = y_0 \sin (340\pi t - 0.50 \pi x)\)A

The equation \(y = y_0 \sin (\omega t -kx)\) describes a travelling wave.

In this case: \(k = 2\pi/\lambda = 2\pi/(0.5\; \mathrm{m}) = 4\pi \; \mathrm{rad/m}\)\(v = f\lambda \; \mathrm {or} f = v/\lambda = (340 \; \mathrm{m/s}) / (0.5 \; \mathrm{m}) = 680 \; \mathrm {s^{-1}}= 680 \; \mathrm {Hz}\)

\(\omega = 2\pi f = 2 \pi (680 \; \mathrm{s^{-1}}) = 1360\pi \; \mathrm {rad/s}\)

Equation is: \(y = y_0 \sin(1360 \pi t - 4 \pi x)\)

- A standing wave is described by the equation:

\(y = -20.0 \cos (3.00\; t) \sin (8.00\; x)\)

when \(x\) and \(y\) are in cm and \(t\) is in seconds. At what time will the shape of this wave be perfectly flat i.e. the reflected wave will cancel the incident wave everywhere where the equation is valid.

(A) 0.370 s

(B) 0.524 s

(C) 0.631 s

(D) 0.823 s

(E) 0.967 sB

\(y = 0\) for all \(x\) whenever \(\cos(3.00t) = 0\)

i.e., when \(3.00t = \pi /2 or t = \pi /(3X2) = \pi /6 = 0.524 s.\)

- Below are four statements ((i) - (iv)) concerning the human ear:

(i) The tympanic membrane, oscicles and oval window form a piston-lever system which amplifies the pressure fluctuations in the sound wave.

(ii) The average human ear is most sensitive to sounds of frequency about \(3500 Hz\); at this frequency the average human can just hear sounds of intensity about \(1 \times 10^{-12} W/m^2\).

(iii) The basilar membrane contains the hair cells which change mechanical oscillations to electrical nerve signals.

(iv) The region of maximum oscillation of the basilar membrane depends on the sound frequency; this may be one way we recognize sounds of different pitch.

Which of the above statements is/are correct?

(A) All of the above statements are correct.

(B) Statements (i) and (iii) are correct; the rest are false.

(C) Statements (ii) and (iv) are correct; the rest are false.

(D) Statements (i), (ii) and (iv) are correct; (iii) is false.

(E) None of the statements is correct.A

See text, section 2-4

- Cloud-to-cloud lightning produces thunder which can be heard by observers on the ground below. One particularly brilliant flash produced a loud thunder crash for \(200 \; \mathrm {ms} (1 \; \mathrm {ms} = 10^{-3} \mathrm s)\). The sound of the thunder reached the ear of an observer \(4.0 \; \mathrm s\) after the flash was seen. The intensity level of the thunder measured by the observer was \(92 \; \mathrm {db}\). What acoustical energy (in J) was released when the thunder was produced? \((I_0 = 1.0 \times 10^{-12}\; \mathrm {Wm}^{-2})\)

(A) \(7.4 \times 10^3\)

(B) \(29\)

(C) \(1.6 \times 10^5\)

(D) \(6.7 \times 10^{-6}\)

(E) \(0.20\)A

The distance \(R\) from source to observer is given by \(v = R/t\) or

\(R = vt (340 m/s)(4.0s) = 1360 m\)Since \(IL = 92 db = 10 \log(I/I_0)\)

then \(I/I_0 = 10^{9.2}\) and \(I = 10^{9.2}(10^{-12}) = 1.58\times 10^{-3} W/m^2\)

\(I = P/4\pi R^2\) and \(P = E/t\)

Therefore \(E = Pt = I(4\pi R^2)t\)

\(E = (1.58\times10^{-3} W/m^2)(4\pi 1360^2m^2)(0.200s) = 7.4\times 10^3 J\)

- In the acoustics experiment an audio oscillator (A) drove a small speaker (S) located at the mouth of a tube containing a moveable piston P. In one experiment, "loud sound" was heard when the piston was at the four locations indicated by the four vertical arrows. The distances (5.2 cm and 36.4 cm)were measured from the speaker end of the tube to the first and fourth piston position, as indicated.

The frequency of the sound was: (answers are to 3 significant figures)

(A) 980 Hz

(B) 1210 Hz

(C) 1420 Hz

(D) 1640 Hz

(E) 3120 HzD

For a "loud sound" or resonance, the piston must be at the position of a node.

Therefore the distance the piston is moved between 2 adjacent resonances is \(\lambda /2\).\((36.4 \; \mathrm {cm} - 5.2 \; \mathrm {cm}) = 3X(\lambda /2)\)

\(\lambda = 20.8 \; \mathrm {cm}\)

\(v = f\lambda \; \mathrm {or} f = (340 \mathrm {m/s})/0.208 \mathrm m = 1640\; \mathrm {Hz}\)

- Which ray diagram below, for a single, spherical refracting surface, illustrates the concepts of both a
**virtual object and virtual image**. The solid straight lines represent light rays.

(A)

(B)

(C)

(D)

(E)A: both O and I are real

B: O is real, I virtual

C: O is virtual I real

**D: both O and I are virtual - Correct!**

E: O is real, I virtual

- A goldfish (F) is floating in a spherical bowl of radius 20 cm. The bowl is filled with water (n = 1.33) and the fish is 6.0 cm to the right of the center (C) of the bowl, as shown. Which row below describes the fish as seen by a viewer (eye at X) 1.0 m to the right of the bowl? (
)**Answers to 2 significant figures.**

Answer Location of Image Size of image relative to actual fish A 7.3 cm to right of centre 14% larger B 7.3 cm to right of centre 21% larger C 6.2 cm to right of centre 11% larger D 6.2 cm to left of centre 12% larger E 13 cm to left of centre same size B

Rays go from fish to eye.\(n_1 = 1.33 \) (water), \(n_2 = 1.00\) (air)

\(p = (20 - 6) = 14 \; \mathrm {cm}\) (from surface to fish)

\(n_1/p + n_2/q = (n_2 - n_1)/r\)

\(1.33/14 + 1.00/q = (1.00 - 1.33)/-20\)

\(q = -12.7 \; \mathrm {cm}\)

Therefore the image is virtual and \(12.7 \; \mathrm {cm}\) to the left of the surface or \(7.3 \; \mathrm {cm}\) to the right of centre.

\(m = y'/y = -(n_1q)/(n_2p) = -1.33(-12.7)/(1.00 \times 14) = +1.21\)

Image is erect (+) and 21% larger.

- A person has his eyesight corrected to the standard 25 cm nearpoint with a lens of +2.6 diopters Where is his nearpoint without the correcting eyeglasses?

(A) 0.18 m

(B) 0.27 m

(C) 0.45 m

(D) 0.63 m

(E) 0.71 mE

The "nearpoint without glasses" is at the image position of the glasses for an object at 25 cm.\(1/p + 1/q = P\)

\(1/0.25 + 1/q = +2.6\)

\(q = -0.71\; \mathrm {m}\)

Therefore the nearpoint without glasses is 0.71 m in front of the eye..

- The diagrams represent "images" (diffraction patterns) on the retina corresponding to 2 point objects of different angular separation. Recall that each image or diffraction pattern consists primarily of a bright central disk surrounded by a minimum (dark ring) which is surrounded by a weak. maximum (a weak bright ring). In these diagrams the dark areas correspond to bright areas in the patterns. Which diagram represents the 'Rayleigh Criterion' for resolution of the two images?

(A) (Diffraction patterns well separated)

(B) (Weak bright rings coincide)

(C) (Minima or dark rings coincide)

(D) (Minima of one pattern coincides with center of the other)

(E) (Central disks overlap 80% by area)D

See the Rayleigh Criterion and Fig 3.20 on p 34 of the text.

- The butadiene molecule is illustrated below. \((C-C \; \mathrm {bond\; length} = 0.15 nm; 1 nm = 10-9m)\)

What is the probability of finding an \(n = 2 \pi\)-electron in a region of length 0.0030 nm, centered half-way between carbon 1 and carbon 2? You may assume that the probability density is constant in this 0.0030 nm interval.

(A) 0.010

(B) 0.020

(C) 0.030

(D) 0.200

(E) 1.00A

\(x = (1/6)\ell \) at the required point

Therefore \(P_x = (2/\ell) sin^2 (n\pi x /\ell )\)\(P_x = (2/3\times0.15) sin^2 [2\pi (1/6)\ell / \ell ]\)

\(= (2/0.45) sin^2 (\pi /3) = 3.33 nm^{-1}\)

\(P = P_x dx = (3.33 nm^{-1})(0.0030 nm) = 0.010\)

- The diagram below shows the \(\pi\)-electron energy levels of a linear molecule with 4 electrons in a possible arrangement. A photon of wavelength 568 nm is incident on the molecule. Which diagram A to D below illustrates the new electron arrangement if the photon is absorbed (assume that it is absorbed if absorption is possible). If it is not possible for this photon to be absorbed by this molecule, select answer E (\(1 \; \mathrm {nm} = 10^{-9} \mathrm m\)).

(A)

(B)

(C)

(D)

(E) This photon cannot be absorbed by this molecule.D

Photon energy = \(E = hc/\lambda\)

\(= 6.63 \times 10^{34} Js)(3.00\times10^8 m/s)/568\times10^{-9}m = 3.50\times10^{-19} J\)This photon energy is equal to the energy difference between levels 2 and 3, i.e. \((6.3 - 2.8) \times 10^{-19}J\). and no other pair of levels.

Moreover the transition of the electron from the 2nd to the 3rd level is permitted by the Pauli Principle. Assuming that absorption does occur the answer is D

- A
**circular**carbon-chain molecule contains 10 carbon atoms; the C-C bond length is \(0.15 \mathrm {nm}\). What photon energy is required to raise this molecule from its electronic ground state to its first excited state. (\(1 \; \mathrm {nm} = 10^{-9} \mathrm m\))

(A) \(2.1\times10^{-19} \mathrm J\)

(B) \(3.0 \times10^{-19} \mathrm J\)

(C) \(3.7\times10^{-19} \mathrm J\)

(D) \(4.5\times10^{-19} \mathrm J\)

(E) \(5.4\times10^{-19} \mathrm J\)E

10 C atoms, therefore 10 \(\pi \; - \; \mathrm {electrons}\).

Ground state; levels 0, 1, 2 are fullTherefore the electron goes from n = 2 to n = 3

\(E_\mathrm {photon} = E_3 -E_2 = (3^2 - 2^2)h^2/2m\ell ^2 = [5(6.63\times 10^{-34} Js)]/[2(9.1\times10^{-31} kg)(10\times 0.15\times 10^{-9} m)] = 5.37\times 10^{-19} J\) - A molecule can absorb radiation of wavelengths: \(4.0 \times 10-3\mathrm m\), \(4.0 \times 10^{-6} \mathrm m\) and \(4.0 \times 10^{-7} \mathrm m\). Choose the two that correspond to an electronic transition and a pure rotational transition and determine the ratio of the photon energy for the electronic transition to that for the pure rotational transition. The ratio is:

(A) \(10^4:1\)

(B) \(10^3:1\)

(C) \(1^2: 1\)

(D) \(20:1\)

(E) \(5:1\)A

Electronic transitions: \(4.0\times10-7 m\) (shortest wavelength)

Pure rotation: \(4\times 10^{-3} m\) (longest wavelength)\(E_e/E_r = \lambda _r/\lambda _ e = 4.0\times10^{-3}/4.0\times10^{-7} = 10^4/1\)

- The diagram, illustrates molecular energy levels as discussed in this course. Below are 5 statements ((i) - (v)) concerning this diagram:

(i) Arrow 1 represents absorption of light.

(ii) Arrows 2 represent energy going into heat.

(iii) Arrow 3 represents fluorescence.

(iv) Arrow 4 represents a 'spin flip'.

(v) Arrow 5 represents phosphorescence

Which of the above statements is/are correct?

(A) All are correct.

(B) (i),(ii) and (v) are correct; the rest are false.

(C) (i), (iii) and (v) are correct; the rest are false.

(D) (iii),(iv) and (v) are correct; the rest are false.

(E) None is correct.A

All statements are correct. See Text, p 52, Fig 4-24.

- The graph shows the variation of extinction coefficient \(\varepsilon (\mathrm {dm}^3/\mathrm {mole \; cm})\) with wavelength \(\lambda (\mathrm{nm})\) for a sample of water from a local pond. The absorbance of this water at \(\lambda = 400 \mathrm\; {nm}\) as measured in a \(1.00 \mathrm\; {cm}\) cuvette is \(0.030\). What will be the \(\% \;\mathrm T\) at \(600 \mathrm \; {nm}\) for a \(50.0\mathrm {-cm-long}\) sample of this water?

(A) 0.13 %

(B) 0.56%

(C) 1.3 %

(D) 2.8%

(E) 6.8 %B

\(\mathrm {At} \;400 \; \mathrm {nm}, \mathrm A = 0.030 = \epsilon \; \mathrm {c\ell} = 2000 (\mathrm c) 1\)

\(c = 0.030/2000 = 1.5\times 10^{-5} \; \mathrm {and \; c \; is\; constant.}\)\(\mathrm{At} \; 600 \;nm \; \mathrm{and}\; \ell = 50.0 \;cm; A = \epsilon \;c\ell = 3000(1.5\times10^{-5})(50.0\;cm) = 2.25\)

\(A = 2.25 = \log(I_0/I)\)

\(I_0/I = \mathrm {antilog} \; 2.25 = 178\)

\(\%T = (I/I_0)100 = 100/178 = 0.56\%\)

- The graph below shows a computer simulation of the count rate for absorption of ,\(\gamma-\mathrm {ray}\) photons by lead, as a function of the thickness of lead. Note that the best fitting straight line passes through the points \((0, 3600)\) and \((0.5, 1950)\). From this graph, what is the linear attenuation coefficient of lead?

(A) \(5.0 \; \mathrm {cm}^{-1}\)

(B) \(4.3 \; \mathrm {cm}^{-1}\)

(C) \(3.5 \; \mathrm {cm}^{-1}\)

(D) \(2.7 \; \mathrm {cm}^{-1}\)

(E) \(1.2 \; \mathrm {cm}^{-1}\)\(C = C_0e^{-\mu x}\)

\(ln C = ln C_0 - ^{\mu}x\)\(\mu = |\mathrm {slope}| = |\delta lnC/\delta x| = |(ln 3600 - ln 1950)/(0 - 0.5)\; \mathrm {cm}| = 1.2 \; \mathrm {cm}^{-1}\)

- Ingested \(^{131} \mathrm I\) has been used medically to treat diseased thyroid tissue. One day a radiation detector placed near a patient's thyroid reads \(10^3\) counts/s. Forty-eight hours later it reads \(560\; \mathrm {counts/s}\). What is the biological halflife of \(^{131} \mathrm I\) in the thyroid? (The physical halflife of \(^{131} \mathrm I\) is \(8.1\) days.)

(A) 1.7 days

(B) 2.6 days

(C) 3.4 days

(D) 6.8 days

(E) 9.3 daysC

\(C = C_0 e^{-\lambda _{eff}{^t}} , \mathrm {or}\; \mathrm {ln} \; C/C_0 = -\lambda _{eff}{^t}\)

\(ln (560/1000) = -\lambda _{eff}(2.0 \; \mathrm {days})\)\(\lambda _{eff} = 0.290 \; \mathrm {d}^{-1}\)

\(T_{eff} = ln \; 2/0.290\; \mathrm d = 2.39 \; \mathrm {d}\)

\(1/2.39 = 1/8.1 + 1/T_b\)

\(T_b = 3.4 \; \mathrm {d}\)

- A zoologist conducting experiments on marine larvae, uses an aquarium containing artificial sea water (salt solution). He needs to know the electrical conductance of the salt solution. He places two metal electrodes in the aquarium as shown and connects them through an ammeter \((\mathrm A)\) to a \(12.0\) volt battery. The symbol \(\mathrm G_s\) represents the conductance of the solution. The symbol \(\mathrm {Gc = 1.0 \; S}\) represents the conductance of the rest of the circuit. If the ammeter \((\mathrm A)\) reads \(0.50 \mathrm A\), what is the value of the solution conductance \(\mathrm{Gs}\)?

(A) 0.013 S

(B) 0.027 S

(C) 0.036 S

(D) 0.043 S

(E) 0.068 SD

I = GeqV

where 1/Geq = 1/Gs + 1/Gc (conductors in series)

0.50 A = Geq (12.0 V)

Geq = 0.0417 S

1/0.0417 = 1/Gs + 1/1.0

Gs = 0.43 S

- Due to nearby charges, the point \(\mathrm P \) has the electrical properties shown, i.e. the electric field is \(3000\; \mathrm {N/C}\) to the right and the electrical potential is \(-200 \; \mathrm V \)(relative to \(\mathrm V = 0\) at infinity).

If an electron \((e = 1.60 \times 10^{-19} \mathrm C)\) is located at point \(\mathrm P\), which one, if any, of the statements A to D is**NOT**correct. If all statements A to D are correct, select answer E.

(A) The electrical force on the electron is \(4.8 \times 10^{-16} \mathrm N\) to the left.

(B) The electrical potential energy of the electron is \(+3.2 \times 10^{ -17} \mathrm J \)(relative to zero at infinity).

(C) The work done against the electrical field to move the electron from infinity to point \(\mathrm P\) was \(+3.2 \times 10^{ -17} \mathrm J \).

(D) If the electron at \(\mathrm P\) is released and the electric field accelerates it back to infinity without collisions (ie in a vacuum) the electron's kinetic energy at infinity will be \(+3.2 \times 10^{ -17} \mathrm J \).

(E) If all of statements A to D are correct, select E.E

Force on an electron at \(P = qE = (1.6\times 10^{-19} C)(3000 N/C) = 4.8\times 10^{-16} N\) to the left.

(opposite direction to \(E\) since electron is negative)

\(P_e\) of electron at \(P = qV = (-1.6\times 10^{-19})(-200 V) = +3.2\times 10^{-17} J\)

This also equals the electron's \(KE\) at infinity if released from rest at \(P\).

It is also the work done against \(E\) to bring the electron from infinity to \(P\).

**All statements are correct**

- The diagram shows a battery (terminal voltage \(\mathrm V\)) connected by copper wires (of very large conductance) to a tube containing NaCI solution of conductivity \(\sigma\). The tube has a length \(\mathrm L\) and uniform cross-sectional area \(\mathrm A\). Suppose V, A, L and s are varied (one at a time) and I is measured for each arrangement. Graphs of \(\mathrm I\) vs \(\mathrm {V, A, L}\) and \(\sigma\) are plotted. Which graph (A to D) below is
**NOT**correct. If all are correct, select answer E.

(A)

(B)

(C)

(D)

(E) All of the above graphs are correct

E

\(I = GV = (\sigma A/L)V\)

**All graphs are correct.**