Université d\'Ottawa University of Ottawa Faculté des sciences Faculty of Scienc

Question

Université d'Ottawa University of Ottawa Faculté des sciences Faculty of Science Physique Physics PHY 1122 February 28, 2012 Dr. Z. M. Stadnik Mid-Term Examination Page 1 of 6 pages The answers should be entered carefully on a computer readable sheet using an HB pencil. When the exam time is over, you hand over only the computer sheet and keep this questionnaire for yourself. 1. A U-shaped tube open on both sides contains water. Each arm has a 30-cm high column of water in it. A 10-cm column of oil (density 800 kg/m is added to one arm. How high is the column of water in the arm that has no oil in it, when the system is in equilibrium? A) 28 cm B) 30 cm C) 32 cm D) 34 cm E) 36 cm 2. 3.00 g has a -4.00 uc charge on it. If the mass is released, in vacuum near the earth's surface, what electric field is necessary to create a zero net force on the mass? A) 7.35 N/C pointing up B) 7.35 x 103 N/C pointing up c) 7.35 x 103 N/C pointing down D) 7.35 x 106 N/C pointing up E) 7.35 x 106 N/C pointing down

Explanation / Answer

Given

   water in u shaped open tube with 30 cm high column

and oil , added to one arm to a hight of 10 cm

then

using Bernouli's equation

   P0+(l+h1)rho_w*g = P0+rho_o*g*l

solving for h1

       h1 = l(rho_o/rho_w - 1)

substituting the values

       h1 = 10(800/1000 -1) cm
       h1 = 2 cm

so the hight of water column in the arm that has no oil is 30-2 = 28 cm

answer is option A--------------->>>Answer

2.

   F = mg

   F = Eq

       given mass is m = 3.00 g = 0.003 kg.charge q = -4*10^-6 C

   F = m*g = 0.003*9.8 = 0.0294 N which is downward

and the electric force is F = E*q ==> E = F/q = 0.0294 /4*10^-6 N/C

                   E = 7350 N/C
the electric force should be in the upward dierection so the electric field also

so the answer is 7.35*10^3 N/C poointing up

option B ---------------->>>> Anwswer

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