9th Class Physics Chapter 1 (Numerical Problems with Solutions)

 


CHAPTER NO.1

Physical Quantities and Measurement

(Numerical Problems)

1.1          Express the following quantities using prefixes.

(a) 5000 g        (b) 2000 000 W            (c) 52 x10-10 kg            (d) 225x10-8 s

Solution

      (a)    5000 g = 5x103 g = 5 kg       

      (b)   2000 000 W = 2x106 W = 2 MW

      (c)    52 x10-10 kg = 5.2x101x10-10 kg = 5.2 x10-9 kg = 5.2 x10-9 x103 g = 5.2 x10-6 g = 5.2 µg

      (d)    225x10-8 s = 2.25x102x10-8 s = 2.25x10-6 s = 2.25 µg

 

1.2          How do the prefixes micro, nano and pico relate to each other?

Solution

micro = 10-6

nano= 10-9

pico= 10-12

                    i.    Relation between micro and nano

  micro= 10-6 = 1x10-6 =103x10-3 x10-6 = 103x 10-9 = 103 nano.

           ii.    Relation between micro and pico

   micro= 10-6 = 1x10-6 =106x10-6 x10-6 = 106x 10-12 = 106 pico.

          iii.       Relation between nano and micro

    nano= 10-9 = 1x10-9 =10-3 x10-6 = 10-3 micro.

          iv.    Relation between nano and pico

    nano = 10-9 = 1x10-9 =103x10-3 x10-9 = 103x 10-12 = 103 pico.

          v.         Relation between pico and micro

   pico= 10-12 =10-6 x10-6 = 10-6 micro. 

          vi.      Relation between pico and nano

   pico= 10-12 = 10-3 x10-9 = 10-3 nano.

1.3          Your hair grows at the rate of 1 mm per day. Find their growth rate in nm s-1.

Data

Hair growth (mm/day) = 1

Hair growth (nms-1) =?

Solution

Hair growth = 1 mm/day = 1x10-3 m/day

 = 1x10-3 m/ (24x60x60 s)

 = 1x10-3 m/86400 s

 = 1.1574x10-8 ms-1

 = 11.574 x10-1x10-8 ms-1

= 11.574x10-9 ms-1

Hair growth in nms-1      = 11.574 nms-1  Ans

1.4          Rewrite the following in standard form.

(a) 1168x10-27              (b) 32x10-5                   (c) 725 x10-5 kg                        (d) 0.02 x10-8

      (a)    1168x10-27

Solution

1168x10-27 = 1.168x103 x10-27 = 1.168x10-27+3 = 1.168x10-24 

      (b)   32x10-5

Solution

32x10-5 = 3.2x101x10-5 = 32x10-5+1 = 3.2x10-4

      (c)    725 x10-5 kg

Solution

725 x10-5 kg = 7.25x102 x10-5 kg

        = 7.25 x10-5+2 kg

        = 7.25 x10-3 x103 g

        = 7.25 x10-3+3 g

        = 7.25 x100 g

        = 7.25 x 1

        = 7.25 g

      (d)   0.02 x10-8

Solution

0.02 x10-8 = 2x10-2 x10-8 = 2 x10-8-2 = 2 x10-10 = 0.02 x10-8 = 2 x10-8

 

Write the following quantities in standard form.
(a) 6400 km
(b) 380 000 km
Solution

1.5          (c) 300 000 000 ms-1
(d) seconds in a day 

      (a)  6400 km

Solution

6400 km = 6.4x103 km

      (b)  380 000 km

Solution

380 000 km = 3.8x105 km 

      (c)  300 000 000 ms-1 

Solution

300 000 000 ms-1  = 3x108 ms-1 

      (d)  seconds in a day

Solution

seconds in a day = 24x60x60 s = 86400 s = 8.64x104 s

1.6          On closing the jaws of a Vernier Callipers, zero of the vernier scale is on the right to its main scale such that 4th division of its vernier scale coincides with one of the main scale division. Find its zero error and zero correction.

Solution

Main scale reading = 0.0 cm

Vernier division coinciding with main scale = 4 div.

Vernier scale reading = (Vernier division coinciding with main scale) (Least Count of vernier Calliper)

                                      = 4 x 0.01 cm = 0.04 cm

Zero error = (Main Scale reading) + (Vernier Scale Reading)

 = 0.0 cm + 0.04 cm

         As zero line of vernier scale is on the right of main scale zero, therefore zero error is positive.

         Zero error = + 0.04 cm

        zero correction (Z.C) = - 0.04 cm 

1.7          A screw gauge has 50 divisions on its circular scale. The pitch of the screw gauge is 0.5 mm. What is its least count?

Solution

Divisions on Circular Scale = 50

Pitch of Screw Gauge = 0.5 mm

Least Count of Screw Gauge =?

 \[Least Count=\frac{Pitch of Screw gauge}{Number of divisions on circular scale}\] 

\[Least Count=\frac{0.5 mm}{50}\]

\[Least Count=0.01 mm\]  

1.8          Which of the following quantities have three significant figures?

(a) 3.0066 m       (b) 0.00309 kg             (c) 5.05x10-27 kg           (d) 301.0 s

(a)           3.0066 m

Solution

All the 5 digits are significant.

(b)          0.00309 m

Solution

The first two zeros are not significant. They are used to space the decimal point. The digits 3 and 9 are significant. The zero between the two significant figures 3 and 9 are significant. Thus, there are three significant figures. In scientific notation this number can be written

0.00309 m = 3.09x10-2 m

(c)           5.05x10-27 kg

Solution

Both the 5 are significant. The zero between the two significant figures 5 and 5 are significant. Thus, there are three significant figures

(d)          301.0 s

Solution

The final zero is significant since it comes after the decimal point. The zero between 3 and1 is also significant because it comes between the significant figures. Thus, the number of significant figures in this case is four. In scientific notation, it can be written as
301.0 s = 3.010x102 s

1.9          What are the significant figures in the following measurements?

(a) 1.009 m         (b) 0.00450 kg             (c) 1.66x10-27 kg           (d) 2001 s

(a)           1.009 m

Solution

The digits 1 and 9 are significant. The zeros between the two significant figures 1 and 9 are also significant.

    

(b)          0.00450 kg

Solution

The first two zeros are not significant. They are used to space the decimal point. The digits 4 and 5 are significant. The final or ending zero is also significant. Thus, there are three significant figures. In scientific notation this number can be written

0.00450 m = 4.50x10-3 m  

(c)           1.66x10-27 kg

Solution

All the three (03) digits are significant.      

(d)          2001 s

Solution

All the four digits are significant.

Solution

1.10      A chocolate wrapper is 6.7 cm long and 5.4 cm wide. Calculate its area up to reasonable number of significant figures.

Solution

Length of Chocolate wrapper = 6.7 cm

Width of chocolate wrapper = 5.4 cm

Area of chocolate wrapper =?

Area of chocolate wrapper = (Length of chocolate wrapper) (width of chocolate wrapper)

Area of chocolate wrapper = 6.7 cm x 5.4 cm

Area of chocolate wrapper = 36.18 cm2


 

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