Physics 9-Ch7 (Properties of Matter)-Numerical Problems

 

Table of Contents

• A wooden block measuring 40 cm $\times10 cm \times 5$ cm has a mass 850 g. Find the density of 3 wood. (425 $kgm^{-3}$)
• How much would be the volume of ice formed by freezing
• Calculate the volume of the following objects
• The density of air
• A student presses her palm by her thumb with a force
• The head of a pin is a square of a side of 10 mm.
• A uniform rectangular block of wood
• A cube of a glass
• An object weights
• A solid block of wood of density
• The diameter of the piston of a hydraulic press
• A steel wire of cross-sectional area

A wooden block measuring 40 cm $\times 10��\times 5$ cm has a mass 850 g. Find the density of 3 wood. (425 $��{�}^{-3}$)

Difficulty: Medium

Solution: 

Volume of wooden block = V = 40 cm $\times 10��\times 5��$ = $2000�{�}^3$
                                                                       =$\frac{2000\times 1}{100}$ $\times \frac{1}{100}$ $\times \frac{1}{100}$  ${�}^3$
                                                                       = $0.002{�}^3$

                                         Mass = m = 850 g =$\frac{850}{1000}$  = 0.85 kg
                          Density of wood = $�$ = ?
                          Density =$\frac{����}{������}$ =$\frac{0.85}{0.02}$ = $425�{�}^{-3}$ 

How much would be the volume of ice formed by freezing 1 liter of water? (1.09 liter)

Difficulty: Easy

Solution:           

Volume of water = 1 litre
                           The volume of ice  =?
1 litre of water = 1 kg mass and density = $1000�{�}^{-3}$ 
Since the density of ice is 0.92 times of the liquid water,, therefore,

                             Volume of ice   =  $\frac{����}{�������}$  
                                                      =  $\frac{1000}{920}$ 
                     Volume of ice = 1.09 litre

Calculate the volume of the following objects:

 

(i) An iron sphere of mass 5 kg, the density of iron is 8200 $��{�}^{-3}$. 
                                                                                                            $(6.1\times 10^{-4}{�}^3)$
                                                                                                            
(ii) 200 g of lead shot having density 11300 $��{�}^{-3}$. 
                                                                                                         $(1.77\times 10^{-5}{�}^3)$
                                                                                                         
(iii) A gold bar of mass 0.2 kg. The density of gold is $19300��{\text{–}}^{-3}$.
                                                                                                         $(1.04\times 10^{-5}{�}^3)$

Difficulty: Hard

Solution:      

(i)            Mass of iron sphere = m = 5 kg

                     Density of iron = $�$ = $8200��{�}^{-3}$

                      Volume of iron sphere = V =?

                      Volume =$\frac{����}{�������}$ 

                 Volume =$\frac{5}{8200}$

                              =  0.00060975 = $6.0975\times 10^{-4}$ 

                      Volume = $6.1\times 10^{-4}{�}^3$

 

(ii)                 Mass of lead shot = m = 200 g =$\frac{200}{1000}$  kg = 0.2 kg

                     Density of lead = $�$ = $11300��{�}^{-3}$

                      Volume of lead shot = V =?

                      Volume =$\frac{����}{�������}$

               Volume =$\frac{0.2}{11300}$

                              = 0.000017699 = $1.76699\times 10^{-5}$ 

                      Volume = $1.77\times 10^{-5}{�}^3$

 

(iii)                Mass of gold bar = m = 0.2 kg

                     Density of gold = $�$ = $19300��{�}^{-3}$

                      Volume of gold bar = V =?

                      Volume =$\frac{����}{�������}$ 

               Volume =$\frac{0.2}{19300}$ 

                              = 0.000010362 = $1.0362\times 10^{-5}$ 

                    Volume = $1.04\times 10^{-5}{�}^3$

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The density of air is $1.3��{�}^{-3}$. Find the mass of air in a room measuring $8�\times 5�\times 4�$. (208 kg)

Difficulty: Easy

Solution:       

Density of air = \rho = $1.3��{�}^{-3}$
                       Volume of room = v = $8�\times 5�\times 4$ m = $160{�}^3$
                       Mass of air = m =?
                       Mass of air = $������������\times ������������$
                       Mass of air = $1.3\times 160$
                       Mass of air = 208 kg

A student presses her palm by her thumb with a force of 75 N. How much would the pressure under her thumb have a contact area of $1.5�{�}^2$? 
$(5\times 105�{�}^{-2})$

Difficulty: Easy

Solution:        

Force = F = 75 N
                       Contact Area A = $1.5�{�}^2$ = $1.5\times \frac{1}{100}\times \frac{1}{100}{�}^2$ = $1.5\times 10^{-4}{�}^2$
                        Pressure under the thumb = P =?
                        P =$\frac{�}{�}$ 
                        P  =$\frac{75}{(1.5\times 10^{(}-4))}$ = $\frac{75}{1.5\times 10^{-4}}$= $5\times 10^5�{�}^{-2}$

The head of a pin is a square of a side of 10 mm. Find the pressure on it due to a force of 20 N. $(2\times 10^5�{�}^{-2})$

Difficulty: Medium

Solution:       

Force = F = 20 N
                      Area of head of a pin A = $0��\times 10��$=  $����1010��\times \frac{10}{10}$cm 
                         = $\frac{1}{100}�\times \frac{1}{100}$ m
                         = $10^{-4}{�}^2$

                        Pressure under the thumb = P =?
                        P =$\frac{�}{�}$ 
                        P  =$\frac{20}{(1\times 10^{-4})}$ =  $2\times 10^5�{�}^{-2}$

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A uniform rectangular block of wood $20��\times 7.5��\times 7.5��$ and of mass 1000g stands on a horizontal surface with its longest edge vertical. 

Find:

(i) The pressure exerted by the block on the surface
(ii) Density of the wood. 

$1778�{�}^{-2},889��{�}^{-3}$

Difficulty: Hard

Solution:       

Length of the smallest side of the block = 7.5 cm

                      Mass of the block    m = 1000g = 1kg

 

The pressure exerted by the block           P =?

The density of wood                            $�$ =?

 

Calculations:

 

(i) since the smallest edge of the block is rested on the horizontal surface. Therefore, area of the block will be:

                   Area = A = $7.5��\times 7.5��$ = $56.25�{�}^2$

                                   = $56.25\times \frac{1}{100}\times \frac{1}{100}{�}^2$ =$56.25\times 10^{-4}{�}^2$

 

                        Pressure under the thumb = P =?

                        P = $\frac{�}{�}=\frac{��}{�}$ 

                        P  = $\frac{1\times 10}{56.25\times 10^{-4}}$ = $0.1778\times 10^4=1778�{�}^{-2}$

 

(ii)                 Volume = V = $20��\times 7.5��\times 7.5$ cm $=1125�{�}^3$

                                   = $1125\times \frac{1}{100}�\times \frac{1}{100}$ m $\times \frac{1}{100}�$

= $1125\times 10^6{�}^3$

                       Or     V = $1.125\times 10^{-3}{�}^3$

                        

                        Density = $\frac{����}{������}$

                   Density $=\frac{1}{(1.125\times 10{(}^{-3})}(1)=0.8888\times 10^3=888.8��{�}^{-3}$

                   Density = $889��{�}^{-3}$

A cube of a glass of 5 cm side and mass $30^6$ g, has a cavity inside it. If the density of glass is $2.55�{�}^{-3}$. Find the volume of the cavity. 
 $5�{�}^3$

Difficulty: Medium

Solution:    

Size of the cube = 7.5 cm
                      Mass of the cube = m = 306 g
                      Density of glass $=�$ = $2.55��{�}^{-3}$
                      The volume of the cavity = V =?

                       Volume of the whole cube = $5��\times 5$ $��\times 5��=125�{�}^3$
                       Volume of the glass $=\frac{����}{�������}$ 
                       Volume $=\frac{306}{2.55}=120�{�}^3$
                       Volume of the cavity $=125�{�}^3\text{–}120�{�}^3$ $=5�{�}^3$   

 

An object weights 18 N in air. Its weight is found to be 11.4 N when immersed in water. Calculate its density. Can you guess the material of the object? ( $2727��{�}^{-3}$, Aluminium )

Difficulty: Medium

Solution:     

weight of object in air =${�}_1$ = 18 N
                     Weight of object immersed in water $={�}_2=11.4$ N
                      Density of glass =$�=1000��{�}^{-3}$

    The density of the object = D =?
    Nature of the material =? 

                                D = $\frac{({�}_1)}{({�}_1-{�}_2)\times �}$  
                                D = $\frac{(18)}{(18-11.4)}\times 1000$    
                              =$\frac{18}{(6.6)}\times 1000$ $=2.727\times 10^3$ = $2727��{�}^{-3}$ 

The density of aluminium is $2700��{�}^{-3}$, and the above-calculated value of density is 2727 $��{�}^{-3}$ nearest to the density of aluminium, so the material of the object is aluminium.

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A solid block of wood of density $0.6��{�}^{-3}$ weighs 3.06 N in air. Determine (a) volume of the block (b) the volume of the block immersed when placed freely in a liquid of density 0.9 $��{�}^{-3}$ ? $(510�{�}_3,340�{�}_3)$

Difficulty: Easy

Solution:    

(a)  Density of wood = D = $0.6��{�}^{-3}$

             Weight of the wooden block = w= 3.06 N

             Since w = mg       or          m $=\frac{�}{�}=\frac{3.06}{10}=0.306��=306$ g

             Density of liquid                  D = $0.9��{�}^{-3}$

(i) The volume of the block               V =?

 

(ii) The volume of the block immersed in a liquid   V =?

                   Density $=\frac{����}{������}$ 

                   Volume $=\frac{����}{�������}$

                   V =$\frac{306}{(0.6)}=510�{�}_3$

 

(b) Volume =$\frac{����}{�������}$ 

                      V =$\frac{306}{(0.9)}=340�{�}_3$

The diameter of the piston of a hydraulic press is 30 cm. How much force is required to lift a car weighing 20 000 N on its piston if the diameter of the piston of the pump is 3 cm? (200 N)

Difficulty: Hard

Solution: 

Diameter = D = 30 cm
                   Radius of the piston = R $=\frac{�}{2}$ = $\frac{(30��)}{2}$= 15 cm $=\frac{15}{100}�=0.15�$
                   Area of the piston = A = $�{�}^2=2\times 3.14\times (0.15{)}^2$
                                                      A = $0.1413{�}_2$

                  Weight of the car               w $={�}_2$= 20000 N
                   Diameter of the piston      d = 3 cm
                   Radius of the piston = R =$\frac{�}{2}=\frac{(3��)}{2}$ = $1.5��$ =$\frac{(1.5)}{100}$ m $=0.015�$
                   Area of the piston = A = $2�{�}^2=2\times 3.14\times (0.015{)}^2$

                                                       A $=1.413\times 10^{-3}{�}_2$
                   Force $={�}_1$ = ?
                                                             $\frac{({�}_1)}{�}$   $=\frac{({�}_2)}{�}$ 
${�}_1={�}_2\times \frac{(�)}{�}$ 
                                                                ${�}_1=\frac{200000�\times (1.413\times 10^{(}-3))}{0.1413}$ 
                                                     = $200000�\times 0.01$
                                                 ${�}_1=200�$

A steel wire of cross-sectional area $2\times 10^{-5}{�}^2$is stretched through 2 mm by a force of 4000 N. Find the Young's modulus of the wire. The length of the wire is 2 m.  $(2\times 10^{11}�{�}^{-2})$

Difficulty: Easy

Solution:      

Cross-sectional area = A = $2\times 10^{-5}{�}^2$
                            Extension =$��=2��=2\times$ $\frac{1}{1000}$  m = 0.002 m
                            Force = F = 4000 N

                           Length of the wire = L = 1m
                           Y =$\frac{��}{(�\mathrm{△}�)}$ 
                           Y = $\frac{(4000\times 2)}{(2\times 10{(}^3)\times 0.002)}$ = $\frac{8000}{0.004\times 10^{-5}}$
                           Y =$\frac{800}{0.004\times 10^{-5}}$ 
                           Y = $2,000,000\times 10^{-5}$ $=2\times 10^{11}�{�}^{-2}$