9th Class Physics Notes “Physical Quantities & Measurement Numericals”

Q1: Write the no in prefix.

Ans:

a. Given data: Mass = m = 10^-21 kg

Solution: Mass = m = 10^-21 kg = 10^-18 x 10^-3 x 10^3 gm [1kg = 10^3] = 10^-18 gm Mass = 1 atto gm [1 atto = 10^-18]

Result: => 1 atto gm

b. Distance = 3.00 x 10^-18 meter Distance = 3.00 Exa – meter [10^-18 = Exa]

Result: Distance = 3.00 E.m

c. Distance between Earth and sun = 149.6 million – km

Solution: Distance = 149.6 million – km = 149.6 x 10^6 km = 149.6 x 10^6 x 10^3 m = 149.6 x 10^9 m = 149.6 Gega – meter : 1 million = 10^6 : Killo = 10^3 : 10^9 = Gega

Result: 149.6 million-km = 149.6 Gega-meter

Q2: An angstrom (Symbol “A”) is …..

Ans:

Solution:

a. As 1 Angstrom = 10^-10 m 1 Angstrom = 10 x 10^-11 m 1 Angstrom = 10^-1 nano-meter = 1 A°= 1/10 Nano – meter

Result: [1A°= 0.1 n.m]

b. As 1 Angstrom = 10^-10 m

Q3: The speed of light is ………….

Ans:

Solution:

a. Speed of light = c = 2.99792458 x 10^8 m/s Round off to 4 digits after the decimal. c = 2.9979 x 10^8 m/sec

b. Round off to 3 digits after decimal. c = 2.998 x 10^8 m/s Now we round off “8” and as a final result is c = 3.00 x 10^8 m/sec

Q4: Express the following in terms of power of 10?

a. 7 Nanometer 7 Nanometer = 7 x 10^-9 m [Nano = 10^-9]

b. 96 Mega watt 96 Mega watt = 96 x 10^6 watt = 9.6 x 10^7 watt [Mega = 10^6]

c. 2 Gigabite 2 x 10^9 bite

d. 43 picofarad 43 picofarad = 43 x 10^-12 farad 43 picofarad = 43.0 x 10^-12 farad 43 picofarad = 4.3 x 10^-11 farad

e. 2 millimeter 2 millimeter = 2 x 10^-3 m 2 mm = 2 x 10^-3 m

Q5: Write …… in standard notation?

Ans:

Solution:

a. mass = 0.000,000,000,005 kg mass = 5 x 10^-12 kg

b. Diameter = 1,390,000,000m Diameter = 1,390,000,000.0 x 10^8 m

Result: Diameter = 1.39 x 10^9 m