9th Class Physics Notes “Chapter#1 (Assignment)”
Chapter#1 (Assignment)
Assignment No1:
The mass of the earth is 5,980,000,000,000,000,000,000,000kg. Write the number in standard notion.
Given Data: Mass = 5,980,000,000,000,000,000,000,000kg
Required date: Scientific notation.
Solution: Mass = 598 x 10^22 kg Mass = 5.98 x 10^22+2 kg Mass = 598 x 10^24 kg
Result: Mass of the earth in scientific Notation is 5.98 x 10^24 kg
Assignment No 2:
Calculate the numbers of seconds in a week? Express the number in power of 10 notation?
Solution: 1 day = 24 hours 1 hour = 3600 sec 1 day = 24 x 3600 sec 1 day = 86400 sec
Now No of second in 7 days. 7 days = 7 x 86400 seconds = 604800 seconds = 604800 x 10^0 sec = 6.048 x 10^5 sec
Result: One week = 6.048 x 10^5 sec.
Assignment No 3:
Adult housefly is having a mass only about 0.000,0214kg. Express this number in standard form.
Solution:
Mass of housefly = 0.000,0214 kg Mass of housefly = 0.000,0214 x 10^0 kg Mass of housefly = 2.14 x 10^-5 kg
Result: Mass of housefly = 2.14 x 10^-5 kg
Assignment No 4:
The smallest bird is the bee humming bird. Males measure only 0.057m. Convert this no in standard form and write this no in millimeters?
Solution:
Size of bee humming bird = 0.057 m Size of bee = 0.057 x 10^0 m Size of bee = 57 x 10^-3 m Size of bee = 57 x 10^-3 x 10^3 mm
Result: Size of bee = 57 mm
Assignment No 5:
Calculate the distance from Peshawar to Lahore in millimeters?
Solution:
Distance between Peshawar and Lahore is 489km. So we have the distance = d = 489 km. Distance = d = 489 km d = 489 x 10^3 m d = 489.0 x 10^3 m d = 489.0 x 10^3 x 10^3 mm d = 4.89 x 10^2 x 10^6 mm
Result: d = 4.89 x 10^8 mm
Assignment No 6:
Which of following is the accurate device for measuring length?
a) A vernier caliper with main scale of 1 mm marking and 50 divisions on the sliding scale. b) A screw gauge of pitch 1 mm and 25 divisions on the circular scale?
Solution:
a. Least count of vernier caliper is given by, L.C = 1/50 mm L.C = 0.020mm
Result: L.C = 0.02mm
b. Least count of screw gauge is given by L.C = 1/25mm
Result: L.C = 0.04mm
Conclusion:
As accurate measurement is related with the least count, smaller the L.C more accurate is the measurement and vice versa. So, in this case the least count of vernier caliper is smaller so it is more suitable for accurate measurement.
Assignment No 7:
A beaker contains 200ml of water. What is volume of water in cm³ and m³?
Solution:
Volume of water = 200 ml V = 200/1000 liter V = 1 liter = 1000 cm³ V = 200 x 10^-3 x 10^3 = 200 cm³
Volume in m³? V = 200/100x100x100 = 200 x 10^-6 m³