Physics with Applications, PHYS*1130 - Sample Exam 2
- The equation for a certain standing wave on a string is:
\(\mathrm{y = 5\times 10^{-3}(\cos 2t)\sin\; 16x}\)
What is the approximate length of the shortest string that could be used? The amplitude and \(\mathrm{x}\) are in metres and \(\mathrm{t}\) is in seconds.
(A) \(\mathrm{0.5 \;m}\)
(B) \(\mathrm{0.6 \;m}\)
(C) \(\mathrm{0.4 \;m}\)
(D) \(\mathrm{0.3\; m}\)
(E) \(\mathrm{0.2 \;m}\)
Answer is E
\(\mathrm{Y = 5 \times 10^{-3}(\cos2t)(\sin16x)}\)
\(\mathrm{k = 2\pi /\lambda = 16}\)\(\mathrm{\lambda = \pi /8}\)
Shortest string will contain 1 loop i.e., length \(\mathrm{l = ½\lambda = \pi /16 = 0.2 \;m}\)
- Two travelling waves, \(\mathrm{y = 0.10 \sin(3t + 4\pi x) \; metres}\) and \(\mathrm{y = 0.20 \sin(3t - 5\pi x)\; metres}\) meet at \(\mathrm{t = 0.}\) What is the displacement at \(\mathrm{x = 4.5 \;m?}\)
(A) \(\mathrm{0.14 \;m}\)
(B) \(\mathrm{0 \;m}\)
(C) \(\mathrm{-0.20 \;m}\)
(D) \(\mathrm{-0.14\; m}\)
(E) \(\mathrm{0.20 \;m}\)
Answer is D
\(\mathrm{y_1 = 0.1\sin(3t + 4\pi x)}\)
at \(\mathrm{t = 0,\; y_1 = 0.1\sin 4\pi x}\)
at \(\mathrm{x = 4.5,\; y_1 = 0.1 \sin (4\pi )(4.5) = 0.1 \sin 9(2\pi ) = 0}\)
\(\mathrm{y_2 = 0.2 \sin(3t - 5\pi\;x)}\)
at \(\mathrm{t = 0,\; y_2 = 0.2 \sin (-5\pi \; x) = 0.2 \sin[-(5/2)2\pi \; x]}\)
at \(\mathrm{x = 4.5,\; y_2 = 0.2 \sin [-(5/2)(2\pi )(4.5)] = 0.2sin [-11.25(2\pi )]}\)
\(\mathrm{= 0.2 \sin[-0.25(2\pi )] = 0.2 \sin(-\pi /4) = 0.2 \sin(-45\pi ) = -0.14}\)
\(\mathrm{y = y_1 + y_2 = 0 - 0.14 = -0.14}\)
- The intensity level \(\mathrm{5.0 \;m}\) from a speaker is \(\mathrm{75\; db.}\) What acoustical power would pass through a rectangular opening \(\mathrm{(2.0 \;m \times 1.0\; m)}\) in a wall at this distance from the speaker? Assume that the speaker is a point source radiating in all directions.
(A) \(\mathrm{6.32 \times 10^{-5}\; W}\)
(B) \(\mathrm{7.98 \times 10^{-5} \;W}\)
(C) \(\mathrm{4.85 \times 10^{-4} \; W}\)
(D) \(\mathrm{1.36 \times 10^{-6}\; W}\)
(E) \(\mathrm{1.78 \times 10^{-6} \; W}\)
Answer is A
\(\mathrm{IL = 10 \log(I/10^{-12})}\)
\(\mathrm{75 = 10 \log(I/10^{-12})}\)
\(\mathrm{\log(I/10^{-12}) = 7.5}\)
\(\mathrm{I/10^{-12} = 3.16\times 10^7}\)
\(\mathrm{I = 3.16 \times 10^{-5} \;W/m^2}\)
\(\mathrm{P = IA = 3.16 \times 10^{-5} \times 2 = 6.32 \times 10^{-5} \; W}\)
- A whistle is emitting a frequency of \(\mathrm{1000\; Hz.}\) You approach it at a speed of \(\mathrm{4.00\; m\; s^{-1}.}\) What frequency will you hear?
(A) \(\mathrm{1012 \;Hz}\)
(B) \(\mathrm{980\; Hz}\)
(C) \(\mathrm{1020\; Hz}\)
(D) \(\mathrm{1000\; Hz}\)
(E) \(\mathrm{1004\; Hz}\)
Answer is A
\(\mathrm{\Delta\; f/f = v/c}\)
\(\mathrm{\Delta \; f = 1000(4/340) = 11.8\; Hz = 12 \;Hz}\)\(\mathrm{f' = f + \Delta \; f = 1012\; Hz}\)
- A black body at \(\mathrm{800\; K}\) sits in a room at \(\mathrm{27^\circ C.}\) Its surface area is \(\mathrm{3.00\; m^2.}\) What net power does it radiate? (Stefan's constant is \(\mathrm{5.62 \times 10^{-8} WK^{-4}m^{-2}}\))
(A) \(\mathrm{3.65 \times 10^3\; W}\)
(B) \(\mathrm{4.98 \times 10^3\; W}\)
(C) \(\mathrm{2.63 \times 10^4\; W}\)
(D) \(\mathrm{6.83 \times 10^4\; W}\)
(E) \(\mathrm{3.95 \times 10^5\; W}\)
Answer is D
\(\mathrm{P = \sigma T^4A \\ T_0 = 273 + 27 = 300\; K}\)
\(\mathrm{\Delta P = \sigma (T^4 - T_0{^4})A = 5.62\times 10^{-8}(800^4 - 300^4)(3.00) = 6.83 \times 10^4\; W}\)
- A city is at \(\mathrm{50^\circ}\) south latitude. At noon on a certain day, the Earth's axis of rotation is tilted \(\mathrm{10^\circ}\) towards the Sun (i.e., the N pole is closer to the Sun than the S pole). What is the angle \(\theta\) you would use to calculate the average solar intensity at this city at this time?
(A) \(\mathrm{20^\circ}\)
(B) \(\mathrm{30^\circ}\)
(C) \(\mathrm{40^\circ}\)
(D) \(\mathrm{50^\circ}\)
(E) \(\mathrm{60^\circ}\)
Answer is E
\(\mathrm{\theta = 50 + 10 = 60^\circ}\)
- A beam of light enters side 1 of a triangular piece of plastic (a prism) of index of refraction 1.5 as shown. The three sides of the prism are of equal length. Which of the following statements is correct?
(A) The beam will emerge from side 1.
(B) The beam will emerge from side 2.
(C) The beam will emerge from side 3.
(D) The beam will not emerge.
(E) The problem cannot be solved because the wavelength of the light has not been given.
Answer is C
Beam emerges from side 3
- A glass hemisphere of radius 5 cm is placed over a spot on a table and viewed from above. Where does the spot appear to be?
(A) 5 cm below the top of the hemisphere.
(B) At the top surface of the hemisphere.
(C) 2.5 cm above its real position.
(D) 5 cm above the top of the hemisphere.
(E) At infinity.
Answer is A
\(\mathrm{n_1/p + n_2/q = (n_2 - n_1)/R \\ 1.5/5 + 1/q = (1 - 1.5)/(-5) = 1/10 \\ 1/9 = 1/10 - 3/10 = 2/10 \\ 9 = -5 \;cm}\)The spot appears to be \(\mathrm{5 \;cm}\) below the surface.
- Determine the wavelength of an \(\mathrm{n = 3}\) electron in a linear conjugated \(\mathrm{\pi -system}\) of length \(\mathrm{1.05 \;nm.}\)
(A) \(\mathrm{0.4 \;nm}\)
(B) \(\mathrm{0.7 \;nm}\)
(C) \(\mathrm{2.1 \;nm}\)
(D) \(\mathrm{2.8\;nm}\)
(E) \(\mathrm{4.5 \;nm}\)
Answer is B
\(\mathrm{\lambda = (2/3)(1.05) = 0.7\; nm}\)
- If the chromophore of question 9 contained 8 carbon atoms, what is the probability density of the \(\mathrm{n = 3}\) electron at the 2nd or the 7th carbon atom.
(A) \(\mathrm{2.4\; nm^{-1}}\)
(B) \(\mathrm{1.8 \;nm^{-1}}\)
(C) \(\mathrm{3.2 \;nm^{-1}}\)
(D) \(\mathrm{0.6\; nm^{-1}}\)
(E) \(\mathrm{4.8\; nm^{-1}}\)
Answer is B
\(\mathrm{P = (2/\ell )\sin^2(n\pi x/\ell)}\)
\(\mathrm{= (2/1.05)\sin^2[3\pi (6/7)\ell]/\ell,}\) at the 7th carbon
\(\mathrm{P = (2/1.05)\sin^2(3/2)(6/7)2\pi = (2/1.05)\sin^2(9/7)2\pi = (2/1.05)\sin^2(9/7)360^\circ}\)
\(\mathrm{= (2/1.05)\sin^2462.86^\circ = 1.905(0.9505) = 1.81\; nm^{-1}}\)
- A toaster is rated at \(\mathrm{700 \;watts}\) when plugged into a \(\mathrm{115\; volt \;(rms)}\) circuit. Determine the resistance of the toaster.
(A) \(\mathrm{11.3\; \Omega}\)
(B) \(\mathrm{0.16\; \Omega}\)
(C) \(\mathrm{18.9\; \Omega}\)
(D) \(\mathrm{2.80\; \Omega}\)
(E) \(\mathrm{5.11\; \Omega}\)
Answer is C
\(\mathrm{W = V^2/R \\ R = V^2/W = 115^2/700 = 18.9\; \Omega}\)
- The radioactive decay product of \(\mathrm{^{15}{_8}O}\) is (\(\mathrm{^{15}{_8}O}\) decays by beta-decay):
(A) \(\mathrm{^{15}{_9}F}\)
(B) \(\mathrm{^{16}{_8}O}\)
(C) \(\mathrm{^{14}{_9}F}\)
(D) \(\mathrm{^{14}{_7}N}\)
(E) \(\mathrm{^{15}{_7}N}\)
Answer is A
\(\mathrm{^{15}{_8}O --> ^{15}{_9}F + ^0{_{-1}}\beta}\)
- Determine the maximum kinetic energy of the -ray in the radioactive decay of \(\mathrm{^{31}{_{14}}Si.}\) The mass of the electron is 0.000550 amu.
(A) \(\mathrm{7.3 \times 10^{-11}\; J}\)
(B) \(\mathrm{2.4 \times 10^{-13} \; J}\)
(C) \(\mathrm{1.6 \times 10^{-13} \;J}\)
(D) \(\mathrm{4.9 \times 10^{-14} \;J}\)
(E) \(\mathrm{9.2 \times 10^{-12} \; J}\)
Atomic number Z Element symbol Mass number A Atomic mass 12 Mg 24 23.985045 13 Al 27 26.981541 14 Si 28 27.976928 14 Si 31 30.975364 15 P 31 30.973763 15 P 32 31.973908 16 S 32 31.972072 Answer is C
\(\mathrm{^{31}{_{14}}Si --> ^{31}{_{15}}P + ^0{_{-1}}\beta}\)
- m(amu) m(amu) \(\mathrm{^{31}{_{14}}Si}\) - 30.975364 1531P 30.973763 - 0-1$ 00.000550 - Total 30.974313 --> 30.974313 Diff. - 00.001051
Since \(\mathrm{1\; amu}\) represents \(\mathrm{931\; MeV,}\)\(\mathrm{E = 1.05\times 10^{-3} \times 931 = 0.98\; MeV = 0.98 \times 1.6\times 10^{-13} = 1.57\times 10^{-13}\; J}\)
- Suppose that 2 years after a spill of radioactive tritium \(\mathrm{( ^3{_1} T)}\) into Lake Ontario a count rate of 670 counts/hour is obtained from a 10 ml sample of lake water. After another 2 years, the count rate from a similar sample is 410 counts/hour. If all the non-radioactive clearance processes (river outflow, evaporation, absorption into lake mud, etc.) can be assumed to have a half-life \(\mathrm{T_x,}\) determine \(\mathrm{T_z.}\) (Assume that complete mixing of the tritium throughout the lake has occurred. The radioactive half life of tritium is 12.33 yr)
(A) 0.7 yrs
(B) 1.7 yrs
(C) 2.7 yrs
(D) 3.7 yrs
(E) 4.7 yrs
Answer is D
\(\mathrm{T_p = 12.33\; yr \\ \lambda_ p = 0.693/12.33 = 0.0562\; yr^{-1}}\)
\(\mathrm{410 = 670\; e^{-\lambda}{^{eff{^2}}}}\)
\(\mathrm{-0.4911 = -\lambda _{eff}(2)}\)
\(\mathrm{\lambda _{eff} = 0.2456\; yr^{-1}}\)
\(\mathrm{\lambda_{eff} = \lambda _x + \lambda _p}\)
\(\mathrm{\lambda_x = 0.2456 - 0.0562 = 0.1894\; yr^{-1}}\)
\(\mathrm{T_x = 0.693/0.1894 = 3.7\; yr}\)
- Which one of the following statements is FALSE?
(A) The light radiated by the Earth is primarily in the infrared region of the electromagnetic spectrum.
(B) The greenhouse effect is due to the fact that the atmosphere is clear to incoming radiation in the ultraviolet and visible, but opaque to outgoing radiation in the infrared.
(C) The main problem with excess \(\mathrm{CO_2}\) is that it increases atmospheric absorption in the near-ultraviolet region of the electromagnetic spectrum.
(D) Determination of the Earth's albedo (the fraction of sunlight reflected to that absorbed) can be made by examining the moon at first or third quarter.
(E) Some anthropogenic gases such as the chlorofluorocarbons can contribute to the greenhouse effect.
Answer is C
C is incorrect- \(\mathrm{CO_2}\) increases absorption in the infra red.
- A crude telescope is constructed of 2 thin lenses of focal length 50 cm and 15 cm (the eyepiece) in air. The lenses are of glass of index of refraction 1.6. The front surface of the objective is placed just under the surface of a remarkably clear lake. What now will be the new separation between the lenses if both object viewed and the final image are at infinity? (Assume both surfaces of the objective have the same magnitude of radius of curvature. The index of refraction of water is 1.33)
For object and image at infinity the separation = sum of focal lengthsIn air separation is \(\mathrm{50 + 15 = 65\; cm}\).
Objective lens:
\(\mathrm{1/f = 1/50 (n - 1)(1/R_1 - 1/R_2) = (1.6 - 1)(2/R) since |R_1|, = |R_2| = R \\ 1/50 = (0.6 \times 2)/R \\ R = 60\; cm}\)
One surface of the objective in water:
\(\mathrm{P = P_1 + P_2 = (n_2 - n_1)/R + (n_3 - n_2)/R \\ = (1.6 - 1)/60 + (1.33 - 1.6)/(-60) \\ = 0.01 + 0.0045 = 0.0145 \\ f = 1/P = 1/0.0145 = 69\; cm}\)Separation for object and image at infinity = sum of focal lengths = \(\mathrm{69 + 15 = 84\; cm}\)
- A spring is suspended vertically from the ceiling. When a mass of 2.0 kg is hung on the end of the spring, the mass lowers 25 cm before it reaches its new equilibrium position. The mass is then raised 20 cm above the equilibrium position and released. Find the velocity of the mass the second time it is 10 cm below the equilibrium position.
\(\mathrm{F = -kx \\ -2(9.8) = -k(0.25) \\ k = 78.4\; N/m \\ w = (k/m)^{1/2} = (78.4/2) = 6.26\; rad/s}\)
For a mass released from rest above the equilibrium position the equation is a cosine curve.
\(\mathrm{y = A \cos \omega t = 0.2\cos6.26t \\ -0.1 = 0.2 \cos6.26t \\ 6.26t = 120^\circ \; or\; -120^\circ}\)
For the second time it is at \(\mathrm{-10 \;cm,}\) the angle is \(\mathrm{-120^\circ}\) or \(\mathrm{240^\circ}\) (see the figure)
\(\mathrm{6.26t = 240^\circ = 4.189\; rad \\ t = 4.189/6.26 = 0.669\; s}\)
\(\mathrm{v = -A\omega \sin \omega t = -0.2(6.26)\sin6.26t}\)
At \(\mathrm{t = 0.669\; s}\)
\(\mathrm{v = -0.2(6.26)\sin(6.26)(0.669) = -0.2(6.26)\sin4.189 \;rad = -0.2(6.26)\sin240^\circ}\)
\(\mathrm{v = -1.08\; m/s}\)
- A manufacturer of plastics needs to make a light filter which will absorb 80% of the photons \(\mathrm{(\lambda = 488\; nm)}\) striking it. A measurement on a sample piece of the plastic \(\mathrm{1.0\; cm}\) thick indicated that \(\mathrm{30\%}\) of the photons striking it were absorbed.
(A) Determine the pathlength (thickness) required to absorb \(\mathrm{80\%}\) of the photons.
\(\mathrm{1 \;cm \;sample: \%T = 70 = 100(I/I_0) \\ I_0/I = 1/0.7 = 1.429 \\ A = \log(I_0/I) = 0.155 = \varepsilon cl = \varepsilon c(1) = \varepsilon c \\ For \;length = l' \\ \log(I_0/I') = \varepsilon cl' = 0.155\; l' \\ I_0/I' = 100.155\; l' \\ I'/I_0 = 0.2 = 10-0.155\; l', (i.e., \; 80\% \;T) \\ \log 0.2 = -0.699 = -0.155\; l' \\ l' = 0/699/0.155 = 4.5 \;cm}\)
(B) Determine absorbance of the plastic which absorbs \(\mathrm{80\%}\) of the photons.
\(\mathrm{A = \log\; I_0/I' = \log 1/0.2 = \log 5 = 0.699}\)
