Tag: measurements and experimentation

Quizzes (17) Questions (185)

Questions Related to measurements and experimentation

Multiple choice physics measurements and experimentation measuring distance of celestial bodies unconventional units of measurements units of mass

 _______________displacement or difference in the apparent position of an object viewed along two different lines of sight, and is measured by the angle or semi-angle of inclination between those two lines.

  1. Echo

  2. Parallax

  3. Triangulation

  4. None of These
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Parallax displacement or difference in the apparent position of an object vied along two different lines of sight, and is measured by the angle or semi-angle of inclination between those two lines.

Multiple choice physics measurements and experimentation measuring distance of celestial bodies unconventional units of measurements units of mass

A uniform metre scale is balanced at $40\ cm$ mark, when weighs of $25\ gf$ and $10\ gf$ are suspended at $5\ cm$ mark and $75\ cm$ mark respectively. Calculate weight of metre scale.(in gf)

  1. 50.5

  2. 72.5

  3. 52.5

  4. 80
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let W be the weight of the scale acting at the 50 cm mark. Taking moments about the pivot (40 cm): Clockwise moments = 10 gf * (75 - 40) cm = 350 gf cm. Counter-clockwise moments = 25 gf * (40 - 5) cm + W * (50 - 40) cm = 875 + 10W. Equilibrium: 875 + 10W = 350. This implies W is negative, suggesting the scale is balanced differently or the pivot is on the other side. Re-evaluating: 25 gf at 5 cm (35 cm from pivot) and 10 gf at 75 cm (35 cm from pivot). 25*35 = 875, 10*35 = 350. The weight W must be at 50 cm (10 cm from pivot). 25*35 = 10*35 + W*10. 875 = 350 + 10W. 525 = 10W. W = 52.5 gf.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The period of oscillation of a simple pendulum is Given by $ T=2\pi \sqrt { \frac { \ell  }{ g }  }$  where  $\ell$ is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measured by a stop watch of least count 0.1 s. The percentage error in g is:-

  1. 0.1 %

  2. 1 %

  3. 0.2 %

  4. 0.8 %
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

T = 2*pi*sqrt(L/g) implies g = 4*pi^2*L/T^2. Relative error dg/g = dL/L + 2*dT/T. dL = 0.1 cm, L = 100 cm, dL/L = 0.001. dT = 0.1/100 = 0.001 s, T = 2 s, dT/T = 0.0005. dg/g = 0.001 + 2(0.0005) = 0.002 or 0.2%.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

In an experimental set up, the density of a small sphere is to be determined. The diameter of the small sphere is measured with the help of a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the sphere has a relative error of 2%, the relative percentage error in the density is

  1. $0.03$%
  2. $3.11$%
  3. $0.08$%
  4. $8.2$%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

First find the least count of the screw gauge, which is pitch/divisions = 0.5 mm / 50 = 0.01 mm. The measured diameter is main scale reading + circular scale reading * least count = 2.5 mm + 20 * 0.01 mm = 2.7 mm. The relative error in diameter is least count / measured diameter, and the relative error in density (mass / volume) combines mass error and three times the diameter error.