# Introductory Physics for the Life Sciences, PHYS*1070 - Sample Exam 3

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

- A travelling wave is given by the expression \(\mathrm {y = -15\sin(12t+6x)}\). What are the correct values for the amplitude \(\mathrm {A}\), wavelength \(\lambda\) , angular frequency \(\omega\) , and period \(\mathrm {T}\)?

A) \(\mathrm {amplitude = 15, wavelength = \pi/3, \omega = 12, T= \pi /6}\)

B) \(\mathrm {amplitude = -15, wavelength = 6, \omega = 12, T = 1/12}\)

C) \(\mathrm {amplitude = 15, wavelength = \pi /6, \omega = 6, T = \pi /6}\)

D) \(\mathrm {amplitude = -15, wavelength = \pi /3, \omega = 6, T = 1/6}\)

E) \(\mathrm {amplitude = 15, wavelength = \pi /3, \omega = 12, T = 3/\pi}\)

A

Compare with the general equation

\(\mathrm {y = y_0 \sin(\pi t - kx)}\)

\(\mathrm {Amplitude\; y_0 = 15}\)

\(\mathrm {\omega = 12\;rad/s = 2\pi/T}\)

\(\mathrm {Therefore \; T = 2\pi/12 = \pi/6 \;s}\)

\(\mathrm {k = 6 \; rad/m = 2\pi/\lambda}\)

Therefore \(\mathrm {\lambda = 2\pi /6 = \pi /3 m}\)

- A porpoise sends an echo locating pulse (60 kHz) as it tracks the path of a shark. The power associated with the pulse is \(\mathrm {3.0 \times 10^{-5} \;watts.}\) The intensity of the pulse at the position of the shark is \(\mathrm {1.5 \times 10^{-5} watts/m^2.}\) What is the displacement amplitude of the water molecules adjacent to the shark?

A) \(\mathrm {1.2 \times 10^{-11} m}\)

B) \(\mathrm {1.7 \times 10^{-10} m}\)

C) \(\mathrm {2.5 \times 10^{-11} m}\)

D) \(\mathrm {3.8 \times 10^{-10} m}\)

E) \(\mathrm {5.2 \times 10^{-11} m}\)

A

\(\mathrm {I = 2\pi ^2 \rho vf^2y_0\;^2}\)

\(\mathrm {Therefore\; y_0 = [I/(2\pi^2\rho vf^2)]^{1/2}}\)\(\mathrm {Density\; of\; water = 1000\; kg/m^3}\)

\(\mathrm {Speed\; of\; sound\; in \;water = 1550\; m/s}\)

\(\mathrm {y_0 = [(1.5\times 10^{-5} W/m^2)/(2\pi^2\times 1000\; kg/m^3\times 1550 \;m/s(60\times 10^3 \;Hz)^2)]^{1/2} \\ = 1.2\times 10^{-11} m}\)

- A beat frequency of 17 Hz is observed when the fundamental mode of two hollow tubes is excited. The tubes differ in length but have similar construction with an open end and one closed end. The shorter tube length is 80 cm. What is the length of the longer tube?

A) 85 cm

B) 90 cm

C) 95 cm

D) 100 cm

E) 105 cm

C

For a tube closed at one end, in its lowest frequency mode, the closed end is a node and the open end is a loop with no other nodes in between. Therefore the length L of the tube is 1/4 wavelength.

\(\mathrm {\lambda = 4L}\)For the shorter tube: \(\mathrm {\lambda_ 1 = 4\times0.8\; m = 3.2 \;m}\) and its frequency:

\(\mathrm {f1 = v/\lambda_ 1 = 340\; m/s/3.2 \;m = 106 \;Hz}\)

For the longer tube the fundamental frequency is \(\mathrm {106 - 17 = 89\; Hz}\)

\(\mathrm {\lambda_ 2 = v/f_2 = 340/89 = 4L_2}\)

Therefore \(\mathrm {L_2 = 0.95\;m}\)

- The sound intensity of a train whistle is \(\mathrm {1.0 \times 10^{-3} \;watts/m^2}\) at a distance of 100 m. If the whistle sounds for 2 seconds and the frequency of the sound is 500 Hz, what is the power and energy emitted by the train's whistle?

A) \(\mathrm {126\;W, 63\; J}\)

B) \(\mathrm {126 \;W, 252 \;J}\)

C) \(\mathrm {8.0 \times 10^{-9} W, 4 \times 10^{-2} J}\)

D) \(\mathrm {8.0 \times 10^{-9} W, 1.6 \times 10^{-8} J}\)

E) \(\mathrm {1.0 \times 10^{-2} W, 20 \times 10^{-2} J}\)

B

\(\mathrm {I = P/4\pi\; r^2}\)

\(\mathrm {P = 4\pi\; r^2I = (1.0\times 10^{-3} W/m^2)4\pi (100m)^2 = 126 \;W}\)

\(\mathrm {P = \Delta E/\Delta t}\)

\(\mathrm {\Delta E = P\Delta t = (126\; W)(2\; s) = 252\; J}\)

- The intensity level of a sound at a distance of 25 m from the source is 80 db. What is the intensity level at a distance of 50 m?

A) 40 db

B) 20 db

C) 7 db

D) 77 db

E) 74 db

E

\(\mathrm {At\; 25\; m, L = 80dB = 10 \log(I/I_0)}\)

\(\mathrm {I/I_0 = 10^8}\)

\(\mathrm {I = 10^8\; I_0= I_1}\)

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

\(\mathrm {I_2 = I_1(r_1/r_2)^2 = (I_010^8)(25/50)^2 = I_010^8/4}\)

\(\mathrm {Therefore\; at\; 50\;m, L = 10 \log[(1/4)I_010^8/I_0] \\ = 10\; \log 10^8 - 10\; \log 4 = 80 - 6 = 74 dB}\)

- Which of the following diagrams correctly describes the refraction occurring as a light ray passes from air (n = 1) to glass (n = 1.50) to air again? A, B, C, D or E?

B

For top surface (surface A)

\(\mathrm {n_1\sin \theta_1 = n_2\sin\theta_ 2}\)

\(\mathrm {1 \sin 30 = 1.5 \;sin \theta_ 2}\)

\(\mathrm{\theta_ 2 = 19.5^{\circ} = 20^{\circ}}\)

For bottom surface (surface B)

\(\mathrm {\theta_{1B} = \theta_ 2= 20^{\circ}}\)

\(\mathrm {1.5 \sin 20^{\circ} = 1 \;sin \;\theta_{2B}}\)

\(\mathrm {\theta _{2B} = 30^{\circ} = \theta_ 1}\)

- The end of a long glass rod 8 cm in diameter has a convex hemispherical surface 4 cm in radius. The rod is placed in a liquid and an object is placed 60 cm from the end of the rod and on its axis. This object is imaged at a point 100 cm inside the rod. What is the refractive index of the liquid if the refractive index of the rod is 1.5?

A) 1.15

B) 1.25

C) 1.35

D) 1.45

E) 1.55

C

\(\mathrm{n_1/p + n_2/q = (n_2 - n_1)/R}\)

\(\mathrm {n_1/0.60 + 1.5/1.00 = (1.5 - n_1)/+0.04}\)

\(\mathrm{n_1 = 1.35}\)

- An object 5 cm high is placed 10 cm in front of a thin lens with a 15 cm focal length. What is the image position, height, and orientation?

A) q = 6.0 cm, height = 3 cm, orientation = erect

B) q = 30 cm, height = 15 cm, orientation = inverted

C) q = -30 cm, height = 3 cm, orientation = erect

D) q = 30 cm, height = 15 cm, orientation = inverted

E) q = -30 cm, height = 15 cm, orientation = erect

E

\(\mathrm {1/p + 1/q = 1/f}\)

\(\mathrm {1/+10 + 1/q = 1/+15}\)

\(\mathrm {q = -30 \;cm\; (minus\; sign\; means \;virtual)}\)

\(\mathrm {m = y'/y = -q/p = -(-30)/(+10) = +3\;(plus \;sign\; means\; erect)}\)

\(\mathrm {y' = 3y = 3\times 5 = 15 \;cm \;high}\)

- An electron \(\mathrm {(m = 9.11\times 10^{-31} kg)}\) has a kinetic energy of \(\mathrm {6.0 \times 10^{-21} J}\). What is the energy of a photon which has the same wavelength as that of the electron?

A) \(\mathrm {9.9 \times 10^{-18} J}\)

B) \(\mathrm{4.2 \times 10^{-18} J}\)

C) \(\mathrm {1.0 \times 10^{-17} J}\)

D) \(\mathrm {3.1 \times 10^{-17} J}\)

E) \(\mathrm {1.0 \times 10^{-16} J}\)

D

For the electron

\(\mathrm {K = (1/2)mv^2 = p^2/2m (since\; p = mv)}\)\(\mathrm {Therefore\; p = (2mK)^{1/2}}\)

\(\mathrm{\lambda = h/p = h/(2mK)^{1/2} \\ = (6.63\times10^{-34}Js)/[(2)(9.11\times 10^{-31} kg)(6.0\times 10^{21} J)]}\)

\(\mathrm {= 6.34\times10^{-9} m}\)

For the photon\(\mathrm {E = hc/\lambda = (6.63\times 10^{-34}Js)(3.0\times 10^8 m/s)/6.34\times 10^{-9} m \\ = 3.14\times 10^{-17} J}\)

- The percent transmittance of a 1.0 cm. length of a solution is 85%. What is the percent transmittance of a 3.0 cm length?

A) 85%

B) 61%

C) 55%

D) 41%

E) 45%

B

\(\mathrm {\%T = 100I/I_0 = 85}\)

\(\mathrm {I/I_0 = 0.85}\)

\(\mathrm {I_0/I = 1/0.85}\)

\(\mathrm {A1(for\; 1 \;cm) = \log(I_0/I) = \log(1/0.85) = 0.0706}\)

\(\mathrm {Since\; A = \varepsilon \;cl \;and\; e \;and \;c \;are \;constant \\ \;then\; A_2(for\; 3 \;cm) = 3.0\times0.0706}\)\(\mathrm {= 0.2117 = \log(I_0/I_2)}\)

\(\mathrm{I_0/I_2 = antilog\; 0.2117 = 1.628}\)

\(\mathrm {\%T = (I_2/I_0)100 = 100/1.628 = 61.4\%}\)

- A standing wave is given by \(\mathrm {y = 4\cos(5t)\sin(6x).}\) What is the standing wave pattern at \(\mathrm {t = T/8}\)?

IMAGE

A)

B)

C)

D)

E)

B

\(\mathrm {\omega = 2\pi /T = 5 \;rad/s}\)

\(\mathrm {T = 2p /5 s}\)

\(\mathrm {at \;t = T/8, y = 4 \cos(5\times 2\pi /5\times8)\sin 6x \\ = 4 \cos(\pi /4)\sin6x = 2.8 \sin6x}\)

\(\mathrm {Also\; k = 2\pi /\lambda = 6\; rad/m}\)

\(\mathrm {\lambda = 2\pi /6 = p /3 \;m}\)

\(\mathrm {At\; this\; instant\; (t = T/8), y = 0 (ie\; a \;node)\; \\ if \; \sin 6x = 0 \;or\; x = 0 (and \;every\; (1/2)\lambda \; from\; there).}\)

\(\mathrm {At \;x = +\pi /12, y = 2.8\sin6\pi /12 = 2.8 \sin(p\pi /2) = 2.8}\)

- A certain beam of light has a wavelength of \(\mathrm {500 \;nm; 6 \times 10^{18}}\) photons of this light are incident on an absorbing solution every second and \(\mathrm {7 \times 10^{17}}\) photons are transmitted. The concentration of the solution is \(\mathrm {3 \times 10^{-4}\; moles\; dm^{-3}}\) and the absorbing path length is \(\mathrm {2 \;cm}\). What is the extinction coefficient of the absorbing species?

A) \(\mathrm {982\; dm^3 \;mole^{-1} \;cm^{-1}}\)

B) \(\mathrm {1144\; dm^3 \;mole^{-1}\; cm^{-1}}\)

C) \(\mathrm {1555\; dm^3 \;mole^{-1} \;cm^{-1}}\)

D) \(\mathrm {2486\; dm^3 \;mole^{-1}\; cm^{-1}}\)

E) \(\mathrm {9423 \;dm^3\; mole^{-1}\; cm6{-1}}\)

C

\(\mathrm {A = \log(I_0/I) = \log(6\times10^{18}/7\times10^{17})=0.933 = \varepsilon\; cl}\)

\(\mathrm {\varepsilon = A/cl = 0.933/(3\times 10^{-4} moles/dm^3)(2 \;cm) = 1555 \;dm^3mole^{-1}cm^{-1}}\) - The graph of the probability distribution of a \(\mathrm{ \pi}\) electron on a covalently bonded molecule is given below.

Which one of the set of facts below is correct?

Answer Value of \(\mathrm{x}\) Area under the curve Value of Principle Quantum number, \(\mathrm{n}\) Graph of wave function A \(\mathrm{1/\ell }\) 3 1 B \(\mathrm{2/\ell}\) 1 1 C \(\mathrm{1/\ell}\) 3 3 D \(\mathrm{2/\ell}\) 1 3 E \(\mathrm{1/\ell}\) 1 3 D

Since \(\mathrm{P_x = (2/\ell) \sin^2(n\pi \; x/\ell),}\) with \(\mathrm{n = 3,}\) the peak value labelled \(\mathrm{"x"}\) is equal to \(\mathrm{2/\ell:}\) i.e.where the \(\mathrm{\sin = 1}\). These points are at \(\mathrm{(1/6)\ell,}\) \(\mathrm{(3/6)\ell = (1/2)\ell}\) and \(\mathrm{(5/6)\ell}\)

The area under the curve = total probability of finding the electron somewhere on the molecule = 1.

For \(\mathrm{n = 3}\) the wave function has 3 loops.

**Answer is D** - Two equal charges \(\mathrm {+Q}\) are separated by a distance d. Which of the following statements is
**NOT**true?

A) The electric field half way between the two charges is zero and the force on a charge \(\mathrm{+q}\) at that position is also zero.

B) The electric field, at a point on the connecting line but nearer to the left hand charge, is directed away from it; the force on a charge \(\mathrm{+q}\) at that place is in the same direction.

C) The electric field at a point on the connecting line but nearer to the right hand charge is directed to the left; the force on a charge \(\mathrm{-q}\) at that place is directed to the right.

D) A point has been found on the connecting line where the force on a charge \(\mathrm{+q}\) is zero; this point is therefore half way between the two charges.

E) A point is found on the connecting line where the force on a charge \(\mathrm{+q}\) is zero; the force on a charge -q will be zero at a different point.

E

From symmetry the vector sum of the fields at the mid-point = 0

Therefore A is true

B is also true as shown below

C is also true as shown in the third figure

The mid-point is the only point on the connecting line where \(\mathrm {E_{total} = 0}\)

Therefore D is also true

There is only one point on the connecting line where E (and the force) is 0

**Therefore E is false** - The circuit consists of four resistors with conductances given by:

\(\mathrm {G_1 = 2/3\; S}\)

\(\mathrm {G_2 = 2/3 \;S}\)

\(\mathrm {G_3 = 1\; S}\)

\(\mathrm {G_4 = 2 \;S}\)

The voltage of the battery is 2.0 volts. Which set of values below is correct?

Answer \(\mathrm {I_1(amp)}\) \(\mathrm {I_2(amp)}\) \(\mathrm {I_3(amp)}\) \(\mathrm {I_4(amp)}\) \(\mathrm {I_5(amp)}\) A 2 2/3 2/3 2/3 2 B 2 2/3 2/3 2/3 1 C 2 1/3 1/3 1/3 2 D 4 2/3 2/3 1/3 4 E 4 4/3 4/3 4/3 4 E

\(\mathrm {The \;equivalent\; for \;G_3 \;and \;G_4 \;is \;(for \;series)}\)

\(\mathrm {1/G_{34} = 1/G_3 + 1/G_4}\)

\(\mathrm {G_{34} = (2/3) \;S}\)

\(\mathrm {Since\; G_1, G_2 \;and\; G_{34}\; are\; parallel}\)

\(\mathrm {G_{1234} = 2/3 + 2/3 + 2/3 = 2 \;S}\)

\(\mathrm {I_1 = (G_{1234})(V) = (2 \;S)(2 \;V) = 4 \;amp \;(also = I_5)}\)

\(\mathrm {I_2 = G_1V = (2/3 \;S)(2 \;V) = 4/3 \;amp}\)

\(\mathrm {I_3 \;is\; the\; same \;as\; I_2}\)

\(\mathrm {I_4 = G_{34} V = (2/3)(2) = 4/3 \;amp}\)

\(\mathrm {Check\; I_2 + I_3 + + I_4 = 4/3 + 4/3 + 4/3 = 4 \;amp = I_1}\)

- The element Thorium decays radioactively to the element Protactinium. The decay is represented by the equation:

\(\mathrm {90^{234_{Th}}\rightarrow{91^{234_{Pa}}}+ \;\Rule{10mm}{5mm}{0mm} + \Delta + \gamma}\)

The particle missing in the box is:

A) an alpha particle

B) a beta particle

C) a proton

D) a neutron

E) a neutrino

B

Since the mass number 234 is unchanged and the atomic number 90 increases by 1 to 91, a \(\mathrm {\beta -particle}\) has been emitted.

- A particular radio-isotope has a physical half life of 4 d. The isotope, when administered to a human, is eliminated in the urine. Several hours after injection of the isotope, the count rate from a 50 cc sample of urine is 3000 counts per minute and 2 hours later it is lower by 20% in a similar sample of 50 cc. What is the biological half life of the isotope? .

A) 0.06 d

B) 0.19 d

C) 0.28 d

D) 0.35 d

E) 1.02 d

C

\(\mathrm {N = N_0 e^{-\lambda _{eff} t}}\)

\(\mathrm {at\; t = 2\; hours, N = 0.80 \;N_0}\)\(\mathrm {0.80 = e^{-\lambda _{eff} 2}}\)

\(\mathrm {\lambda_{ eff} = ln0.80/(-2\; hr) = 0.1116\; h^{-1}}\)

\(\mathrm {Also; \lambda_ p = ln2/Tp = ln2/(4\times24 hr) = 7.22\times 10^{-3} h^{-1}}\)

\(\mathrm {\lambda_{ eff} = \lambda_p + \lambda_b}\)\(\mathrm {0.1116 = 7.22\times 10^{-3} + \lambda _b}\)

\(\mathrm {\lambda _b = 0.1046 \;h^{-1}}\)

\(\mathrm {T_b = ln2/\lambda _ b = ln2/(0.1046\; h^{-1}) = 6.64 \;h = 0.28 \;d}\)

- The count rate of a small radioactive source at a distance of 20 cm is 300 counts per minute. What will it be at a distance of 30 cm?

A) 927

B) 672

C) 453

D) 133

E) 87

D

\(\mathrm {C_1/C_2 = (r_2/r_1)^2}\)

\(\mathrm {300/C_2 = (30 cm/20 cm)^2}\)

\(\mathrm {C_2 = 133 \;c/min}\)