Chapter 1 Problems, Problem # 1.5 (Estimate the number of liters of gasoline used by all Pakistan’s each year. Given: Number cars in Pakistan = 500000 Average distance covered by the each car = 16000 Km Rate of Consumption of gasoline = 6 Km/liter.)

 Sindh Text Board Jamshoro 

Class XI / HSC-Ι丨Chapter # 1丨Problems

Problem # 1.5

Estimate the number of liters of gasoline used by all Pakistan’s each year. Given:

Number cars in Pakistan = 500000

Average distance covered by the each car = 16000 Km

Rate of Consumption of gasoline = 6 Km/liter.

 

Solution:

Total distance covered by an equivalent car = (500000) (16000)

= 8000000000 = 8 × 10⁹ km

Total liters of gasoline consumed =

Total distance covered/Rate of consumption

= 8000000000/6

=1.333 × 109 liter

Chapter 1 Problems, Problem # 1.4 ( Suppose the displacement of a particles is related to a time according to expression S=Ct3 what are the dimension of C?)

                                                                                  Sindh Text Board Jamshoro 

Class XI / HSC-Ι丨Chapter # 1丨Problems

Problem # 1.4

Suppose the displacement of a particles is related to a time according to expression  S=Ct³ what are the dimension of C?

Chapter 1 Problems, Problem # 1.3 (Show that the expression S = vit + ½ at2 is dimensionally correct, when S is a coordinate and has a unit of length, Vi is velocity ‘a’ is acceleration and ‘t’ is time.)

                                                                                      Sindh Text Board Jamshoro 

Class XI / HSC-Ι丨Chapter # 1丨Problems

Problem # 1.3

Show that the expression S = v_it + ½ at² is dimensionally correct, when S is a coordinate and has a unit of length, V_i is velocity ‘a’ is acceleration and ‘t’ is time.

Solution

Chapter 1 Problems, Problem # 1.2 (Find the area of rectangular plate having length 21.3 ± 0.2 cm and with 9.80 ± 0.10 cm.)

                                                                        Sindh Text Board Jamshoro 

Class XI / HSC-Ι丨Chapter # 1丨Problems

Problem # 1.2

Calculate

(a)  the circumference of circle of radius 3 cm

                and

(b) area of circle of radius 4.65 cm.

Solution

(a) Circumference of the circle.

Data:

Radius = r = 3.5 cm

Formula

Circumference of the circle = 2πr

C= 2 (3.147)( 3.5)

C= 21.994

 

After reducing

C = 22.0 cm

(b) Are of the circle:

Data:

Radius = r = 4.65 cm

Formula

Area of the circle = πr²

A = (3.147)( 4.65) ²

A = (3.147)( 21.6225)

A = 67.929

After reducing

A= 67.93 cm²

Chapter 1 Problem, Problem # 1.1 (Find the area of rectangular plate having length 21.3 ± 0.2 cm and with 9.80 ± 0.10 cm.)

 Sindh Text Board Jamshoro 

Class XI / HSC-Ι丨Chapter # 1丨Problems

Problem # 1.1

Find the area of rectangular plate having length 21.3 ± 0.2 cm and with 9.80  ± 0.10 cm.

Data:

Length =  L = 21.3 ± 0.2 cm

With = W =  9.80  ± 0.10 cm

Formula

A = L × W

Find Maximum Probable Area

Length =  L = 21.3 + 0.2 = 21.5 cm

With = W =  9.80 + 0.10 = 9.9 cm

A_max = L × W

A_max = 21.5 × 9.9

A_max = 212.83 cm²

After reducing to 2 Significant figures

A_max = 213 cm²

Find Minimum  Probable Area

Length =  L = 21.3 - 0.2 = 21.1 cm

With = W =  9.80 + 0.10 = 9.7 cm

A_min = L × W

Amin = 21.1 × 9.7

Amin = 204.67 cm²

After reducing to 2 Significant figures

A_max = 205 cm²

Find average of two values.

A_av = (213 + 205 ) / 2

A_av = 209 cm²

Difference b/w average value with maximum and minimum value:         

213-209 = + 4

205-209 = - 4

Hence the area will be:

A = 209 ± 4 cm²