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ETEA SAMPLE PHYSICS QUESTIONS

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ETEA stands for Educational Testing and Evaluation Agency, it is an educational entity established in 1998 by KPK government. In order to get admission in the public sector Medical Colleges of KPK, the students have to undergo KPK ETEA entrance test, which is a requirement. Hundreds and thousands of students take this test each year. The test is based on the courses of FSc and A levels including the subjects: Physics, Chemistry, Mathematics and English. ETEA is a highly competitive and challenging test based on the student’s previous knowledge of FSc or A levels.

Pakprep.com is best online entry test preparation website equipped with more than 15000 solved MCQ’s along with explanation prepared by highly qualified instructors from all over Pakistan.  We also facilitate our students with past papers, MOCK test and smart analytics. Which gives our students clear and better picture of their preparation. We have more than 500 successful students from last year. To prepare yourself for any medical or engineering entry test please CLICK HERE.

PHYSICS QUESTIONS

TOPIC

Heat & Thermodynamics

QUESTION

At triple point of water, the pressure of gas is 2680 Pa. By changing T the pressure changed to 4870 Pa. The value of T is?

OPTIONS

  1. 496 K
  2. 438 K
  3. 0
  4. 438 F

CORRECT OPTION

496 K

EXPLANATION

p.V = n.R.T

Since volume V and mass hence number of moles is constant:

p/T = n.R/V = constant

=> p1/T1 = p2/T2

=> T2 = T1.p2/p1

T1 is the triple point temperature of water, i.e

T1 = 0.010C = 273.16 K

Hence:

T2 = 273.16 K * 4870Pa/2680Pa = 496.38K

TOPIC

Fluid Dynamics

QUESTION

Water hose with internal radius of 0.01m has water flowing at the rate of 10 kg per 20s. What is the water speed at the outlet? Density of water is 1000 kg/m3.

OPTIONS

  1. 1.6m/s
  2. 2.8m/s
  3. 3.2m/s
  4. 3.8m/s

CORRECT OPTION

1.6m/s

EXPLANATION

From continuity equation A*V = flow rate of water = volume/s;

Volume/s = mass/s * density = 0.5/1000;

A*V = 0.5/1000; A = 3.14 * (0.01)2;

V = 0.5/ (3.14*(0.01)2*1000) = 1.6m/s

TOPIC

Modern Physics

QUESTION

If energy of photon is 5 MeV. Kinetic energy of particles after pair production is?

OPTIONS

  1. 4 MeV
  2. 5 MeV
  3. 6 MeV
  4. Not enough information given

CORRECT OPTION

4 MeV

EXPLANATION

Energy of photon = Energy required for pair production + KE of particles. Now energy required for pair production is 1.02 MeV so KE is approximately equal to 5-1 = 4 MeV.

SAMPLE MATHEMATICS QUESTIONS

SAMPLE CHEMISTRY QUESTIONS

SAMPLE PHYSICS QUESTIONS 


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