Physics clicker question: pressure underwater

Physics 205A, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 9.18

Students were asked the following clicker questions (Classroom Performance System, einstruction.com) in the middle of their learning cycle:

P_atm = 101.3 kPa
Density of water rho_water = 1,000 kg/m^3

The pressure at point [2] is:
(A) 6.5 x 10^3 Pa.
(B) 1.1 x 10^5 Pa.
(C) 2.1 x 10^5 Pa.
(D) (I'm lost, and don't know how to answer this.)

Sections 70854, 70855
(A) : 1 student
(B) : 16 students
(C) : 20 students
(D) : 1 student

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 70854, 70855
(A) : 0 students
(B) : 7 students
(C) : 28 students
(D) : 0 students

Correct answer: (C)

The pressure at point [1] is P_1 = P_atm = 1.013 x 10^5 Pa. Due to the depth underwater of point [2], the pressure there is:

P_2 = P_1 + rho_water*g*d,

where g = 9.80 m/s^2, and d = 11.0 m. Response (B) is the rho_water*g*d term only, while response (A) is P_1 - rho_water*g*d.

Pre- to post- peer-interaction gains:
pre-interaction correct = 53%
post-interaction correct = 80%
Hake, or normalized gain = 58%