Class 11 Physics

# Units & Measurement

## NCERT Solution

### Part 4

Question 20: The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?

Answer: 1 light year = 9.46 × 1015 m

So, 4.29 light year = 4.29 × 9.46 × 1015 m

= 4.058 × 1016 m

1 parsec = 3 × 1016 m

So, 4.29 light years =(4.058×10^(16))/(3×10^(16))=1.3 parsec

Distance between two locations of the Earth six months apart in its orbit = diameter of Earthâ€™s orbit = 2r

Parallax angle subtended by 1 parsec distance at this basis = 2”

So, parallax angle subtended by Alpha Centauri at the given basis = 1.3 ÷ 2 = 0.65”

Question 21: Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

Answer: Some examples are as follows:

• Launching a spaceship to be sent to International Space Station
• Using laser to operate a patient
• Measuring a single fragment of RNA

Question 22: Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):

(a) the total mass of rain-bearing clouds over India during the Monsoon

Answer: Average rainfall in the country is recorded in terms of centimeter or meter during the monsoon. Mass of average rainfall is calculated. After that, this quantity is multiplied by total area of the country to get an estimate of total mass of rain-bearing clouds over India during monsoon.

(b) the mass of an elephant

Answer: Let us take a huge boat that can accommodate an elephant. Depth of empty boat in water is measured. Area of boat is multiplied by depth to get volume of water displaced by boat. After that, elephant is made to get on the boat. Now, depth of water is measured. This will help in finding volume of water displaced by boat and elephant together. Difference between this figure and previous figure will help in estimating the mass of the elephant.

(c) the wind speed during a storm

Answer: For this, a gas-filled balloon is kept at a known height h. In case of no wind, the balloon remains at position A. When the wind starts blowing, the balloon shifts to position B in 1 second.

AB = d = hθ

Value of d gives the wind speed.

Answer: Area of cross section of a strand of hair is measured with the help of suitable instrument. Then, area of head (covered with hairs) is calculated. Area of head is divided with area of cross section of a single strand of hair to get the number of strands on head.

(e) the number of air molecules in your classroom.

Answer: We know that 1 mole of an ideal gas has the volume of 22.4 liter

Volume of room is calculated by measuring length, breadth and height of the room.

This is divided by 22.4 liter to find number of moles of air in room. Then, number of moles is multiplied by 6.023 × 1023 to find the number of air molecules in the room.

Question 23: The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: mass of the Sun = 2.0 × 1030 kg, radius of the Sun = 7.0 × 108 m.

Answer: Volume of Sun = 4/3 πr3

=4/3×(22)/7×7^3×10^(24)

=4/3×22×49×10^(24)

Density = (2 × 1030) ÷ (4/3 × 22 × 49 × 1024)

= 1.4 × 103 kg m-3

Question 24: When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72" of arc. Calculate the diameter of Jupiter.

Answer: Angular diameter = 35.72” = 35.72 × 4.85 × 10-6 rad

Diameter of Jupiter D = θ × d

= 1.73 × 10-4 × 824.7 × 109 m

= 1.43 × 108 m