Physics 205A Quiz 5, Fall Semester 2009
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 9.12
At the surface of a oil storage tank the air pressure is 1.0 atm. (The density of this oil is 880 kg/m^3.) At what depth under oil in this tank is the oil pressure 1.1 atm? (1 atm = 1.013e+5 Pa.)
(A) 1.2 m.
(B) 12 m.
(C) 13 m.
(D) 25 m.
Correct answer: (A)
The pressure at location [2], a depth d in oil under the surface is greater than compared to the pressure at a location [1] at the surface of the oil, as given by:
P_2 = P_1 + rho_oil*g*d,
where P_2 and P_1 are pressures in units of Pa. With the atm to Pa conversion built in, the depth is:
d = (P_2 - P_1)*((1.013e+5 Pa)/(1 atm))/(rho_oil*g).
Response (B) is P_1/(rho_oil*g); response (C) is P_2/(rho_oil*g); response (D) is (P_2 + P_1)/(rho_oil*g).
Student responses
Sections 70854, 70855
(A) : 36 students
(B) : 11 students
(C) : 1 student
(D) : 0 students
Success level: 75%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.67
Student responses
Section 72177
(A) : 11 students
(B) : 1 student
(C) : 0 students
(D) : 0 students
Success level: 91%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.25
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