Physical Quantities and Measurement - Physics For Class IX (Science Group) - Numericals And Text book Exercise

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Physics For Class IX (Science Group)
UNIT 1: Physical Quantities and Measurement
Numericals And Text Book Exercise

Numerical From Examples

Worked Example 1
Convert mass of Sun 2 000 000 000 000 000 000 000 000 000 000 kg into Scientific Notation.
Data:
Mass of Sun in standard form = 2 000 000 000 000 000 000 000 000 000 000 kg
Mass of Sun in scientific notation = ?

Solution:
Step 1:
Since, M_sun = 2 000 000 000 000 000 000 000 000 000 000 kg
It's obvious that in this value decimal lies at the end.
Step 2:
Converting into scientific notation
Move the decimal to left writing in terms of base of ten
M_sun = 2.00 x 10³⁰ kg. Ans.
(Note: power of exponent is taken as positive because we move decimal from right to left  but not numbers.)

Worked Example 2
Convert mass of an electron 9.11 x 10^-31 kg into standard form.
Data:
Mass of electron in standard form =
Mass of electron in scientific notation = ? 

Solution:
Step 1:
The decimal lies in the middle of the value.
Since, m_electron = 9.11 and n=10^-31 kg
Step 2:
Move the decimal 31 steps towards left (because exponent is negative integers.)
m_electron = 0.000 000 000 000 000 000 000 000 000 000 911 kg Ans.

Worked Example 3:
What is the mass of a solid iron wrecking ball of radius 18 cm. if the density of iron is 7.8 gm/cm³?
Data:
Radius of iron ball = 18 cm
Density of iron is 7.8 gm/cm³
Mass of a solid iron ball =?

Solution:
Step 1: write known physical quantities with units and point out the quantity to be found.

• Density of iron ball = ρ = 7.8 gm/cm³ = 7.8 x 1000 kg/ m³
• Radius of iron ball = r = 18 cm = 18 x 10^-2 m = 0.18 m
• Volume of the iron ball = V = (4/3) x π x r³ 
  V = (1.33) x 3.14 x (0.18 m)³ 
  V = 0.024m³

Step 2: write down the formula and rearrange if necessary

m = ρ x V

Step 3: put the values in formula and calculate
Since mass of iron ball is m = ρ x V = (7.8 x 10³) x (0.024)
m = 187.2 kg
Answer: The mass Of iron ball = 187.2 kg

Worked example 5
How many significant figures are there in the area of a cylinder whose diameter is 5 cm.

Solution:
Step 1: write known physical quantities and point out the unknown quantity
Data:

• Diameter of the cylinder is d = 5 cm = 5 x 10^-2 m = 0.05 m
• Radius of cylinder is r = d/2 = 2.5 x 10^-2 m = 0.025 m

Step 2: write down formula and rearrange if necessary
Formula:

• The area of the cylinder is A = π x r² = 3.14 x (0.025 m)² = 0.0019 mm²

Step 3: put value in formula and calculate
Solution:
Thus area of cylinder can be written as A = 1.9 mm²
Ans: Thus, there are two significant numbers in the value i.e. 1 and 9.

Section (B) Structured Questions

1.

Column A Action	Column B Branch
Cooking Bar B.Q	Thermodynamics
Turning the Bulb on	Electricity
Riding a bicycle	Mechanics
Looking for Giant Galaxies	Astrophysics
Producing a loud sound	Sound
Describing an atom	Atomic Physics
Obtaining energy from Earth	Geo Physics

2.

Physical Quantities	S. I. Units	Type
Ampere	A	Base
Volume	m3	Derived
Time	Sec.	Base
Temperature	K	Base
Force	N	Derived
Density	Kg per m3	Derived
Acceleration	m / s2	Derived

3. Convert the following values.
a) 230 cm = 2.3 m
Solution:
1 m = 100 cm
(Hint: Converting small unit into bigger so we divide the values)
230 cm = 230 /100 = 2.3 m Ans

b) 250 g = 0.25 kg
Solution:
1 kg = 1000 g
(Hint: Converting small unit into bigger so we divide the values)
250 g = 250 / 1000 = 0.25 kg Ans

c) 0.5 s = 500 ms
Solution:
1 sec = 1000 ms
(Hint: Converting bigger unit into small so we multiply the values)
0.5 sec = 0.5 x 1000 = 500 ms Ans

d) 0.8 m = 800 mm
Solution:
1 m = 1000 mm
(Hint: Converting bigger unit into small so we multiply the values)
0.8 m = 1000 x 0.8 = 800 mm Ans

e) 350ms = 0.35 s
Solution:
1 sec = 1000 ms
(Hint: Converting small unit into bigger so we divide the values)
350ms = 350 / 1000 = 0.35 s Ans

f) 1.2Kg = 1200 g
Solution:
1 kg = 1000 g
(Hint: Converting bigger unit into small so we multiply the values)
1.2 Kg = 1000 x 1.2 = 1200 g Ans

4. An engineer measures the width of an aluminum sheet using Vernier caliper as shown in fig 1.29
a)What is the measurement of the width of aluminum sheet
b)Which gives more precise measurement Vernier caliper, Screw Gauge or meter rule?

Ans: Solution:
a) Main scale reading = 3.9 mm
Vernier scale reading = 1 div
Leact count of V.S = 0.1 mm
V.S reading = 1 x 0.1 = 0.1 mm
Total reading = 3.9 + 0.1 = 4.0 mm Ans.

b) Screw Guage Is More Precise:
As compare to vernier calipers and meter rule, screw guage is more precise because the least count of vernier caliper is 0.1 mm, least count of meter rule is 1 mm and for screw guage, it's least count is 0.01 mm. Therefore, a screw guage cam measure more precisely than a vernier scale and meter rule.

5. A pendulum swings as shown if figure 1.30 from X to Y and back to X again.
i) What would be the most accurate way of measuring time for one oscillation? with the help of a Stop Watch.

a) Record time for 10 oscillations and multiply by 10
b) Record time for 10 oscillation and divide by 10 Ans.
c) Record time for one oscillation
d) Record time from X to Y and double it

ii) Suggest an instrument for measuring time period more accurately.
Ans: A digital stop watch can measure a time interval more accurately which is upto 0.01 second.

Prefixes

6. write the correct prefix of notion
a) 75000 m = 750 x 10² = 750 hm (hecto meter)
b) 2/1000 sec = 2 x 10^-3 = 2 msec (milli second)
c) 1/1000000 g = 1 x 10^-6 = 1 μg (micro gram)
d) 1000000000 m = 1 x 10⁹ = 1 Gm (giga meter)

Scientific Notation

7. Write values in standard and scientific notation
a) The radius of 1st orbit of Hydrogen atom is r = 0.53 A° = __________
b) 1 light year is 2628000000000m = ___________
c) Vacuum pressure 2.7 x 10^-4 torr = __________
Ans:

	Standard Notation	Scientific Notation
0.53 A°	53 pm	53 x 10-12
2628000000000m	2.682 Tm	2.682 x 1012
2.7 x 10-4 torr	0.00027 torr	2.7 x 10-4 torr

Density and Volume

8. A wooden piece is made in different shapes take length (l) = radius (r) = 2m Calculate its volume as a:
a) Sphere b) Cube c) Cylinder d) Pyramid
Data:

• Length = 2m
• Radius = 2m
• a) Volume of Sphere = ?
• b) Volume of Cube = ?
• c) Volume of Cylinder = ?
• d) Volume of Pyramid = ?

Solution:

9. Find the density of wood as sphere and cube if the mass of wood is 1kg. Is there any change in density due to shape?
Solution:

Ans: No there is no change in density due to shape, because change in shape does not change the mass or volume of an object.

10. A measuring cylinder (fig 1.31) is filled with 500 cc water. A stone of mass 20g is immersed in to the cylinder such that ,water level rises up to 800 cc. Which statement is correct?
a) The difference between the readings gives the density of stone.
b) The difference between the readings gives volume of the stone. ✓
c) The final reading gives the density of stone
d) The final reading gives the volume of stone

Significant Figures

11. Write significant numbers in the following values.
a) 980 has 2 Significant numbers.
b) 91.60 has 4 Significant numbers.
c) 10010.100 has 8 Significant numbers.
d) 0.0086 has 2 Significant numbers.