Unit and Dimension Solved Example

University:
Concordia University Wisconsin
Course:
PHYS 1514 | General Physics I
Academic year:

2023
Views:

447

Pages:

1
Author:

Brent Cantu

UNIT AND DIMENSION SOLVED EXAMPLE Ex.1 Calculate the angle of (a) 1º (degree) (b) 1' (minute of arc or arcmin) and (c) 1" (second of arc or arc second) in radians. Use 360º = 2 rad, 10 = 60' and 1' = 60" Ans. (a) We have 360º = 2 rad 1º = ( /180) rad = 1.745 × 10–2 rad (b) 1º = 60' = 1.745 × 10–2 rad 1' = 2.908×10–4 rad ; 2.91 × 10–4 rad (c) 1' = 60" = 2.908 × 10–4 rad 1" = 4.847 × 10–4 rad ; 4.85 × 10–6 rad Ex.2 A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in front of the tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC up to B, a distance of 100 m, and looks at O and C again. Since O is very distant, the direction BO is practically the same as AO; but he finds the line of sight of C shifted from the original line of sight by an angle  = 40º ( is known as ‘parallax’) estimate the distance of the tower C from his original position A. Ans. We have, parallax angle  = 400 From Fig. 2.3, AB = AC tan  AC = AB/tan = 100 m/tan 40º = 100 m/0.8391 = 119 m Ex.3 The moon is observed from two diametrically opposite points A and B on Earth. The angle  subtended at the moon by the two directions of observation is 1º 54' . Given the diameter of the Earth to be about 1.276 × 107 m, compute the distance of the moon from the Earth. Ans.We have ? = 1° 54' = 114' = (114 × 60)" × (4.85 ×10-6 ) rad = 3.32 × 10–2 rad, since 1" = 4.85 ×10–6 rad. Also b = AB =1.276 × 107m Hence from Eq. (2.1), we have the earth-moon distance, D = b / = 1.276  107 3.32  10 2 = 3.84 ×108m Ex.4 The Sun’s angular diameter is measured to be 1920' '. The distance D ofthe Sun from the Earth is 1.496 × 1011 m. What is the diameter of the Sun ? Ans. Sun's angular diameter  = 1920" = 1920 × 4.85 × 10–6 rad = 9.31 × 10–3 rad Sun's diameter d =D = (9.31 × 10–3) × (1.496 × 1011) m = 1.39 ×109 m Ex.5 If the size of a nucleus (in the range of 10–15 to 10–14 m) i s scaled up to the tip of a sharp pin, what roughly is the size of an atom ? Assume tip of the pin to be in the range 10–5 m to 10–4 m. Ans. The size of a nucleus is in the range of 10–15 m and 10–14 m. The tip of a sharp pin is taken to be in the range of 10–5 m and 10–4 m. Thus we are scaling up by a factor of 1010. An atom roughly of size 10–10 m will be scaled up to a size of 1 m. Thus a nucleus in an atom is as small in size as the tip of a sharp pin placed at the centre of a sphere of radius about a metre long. Ex.6 The temperatures of two bodies measured by a thermometer are t1 = 20 ºC ± 0.5 ºC and t2 = 50 ºC ± 0.5 ºC. Calculate the temperature difference and the error theirin. Ans. t' = t2– t1 = (50 ºC±0.5 ºC) – (20ºC ± 0.5 ºC) t' = 30 ºC ± 1 ºC Ex.7 The resistance R = V/I where V = (100 ± 5)V and I = (10 ± 0.2)A. Find the percentage error in R. Ans. The psercentage error in V is 5% and in I it is 2%. The total error in R would therefore be 5% + 2% = 7%. Ex.8 5.74 g of a substance occupies 1.2 cm3. Express its density bykeeping the significant figures in view. Ans.There are 3 significant figures in the measured mass whereas there are only 2 significant figures in the measured volume. Hence the density should be expressed to only 2 significant figures. Density = 5.74 g cm–3 = 4.8 g cm–3 1.2