# Class 9 Physics Chapter 1 Numerical Problems Class 9 Physics Chapter 1 numerical problems are according to the new syllabus of the Punjab Board. Chapter 1 is related to Physical Quantities and Measurements in which you will study Physics, applications of Physics, and daily used measuring instruments.

## Class 9 Physics Chapter 1 Numerical Problems

The problems have been solved with easy-to-use methods using the following formulae:

micro = µ = 10-6

nano = n = 10-9

pico = p = 10-12

kilo = k = 103

mega = M = 106

mili = m = 10-3

one day = 24 x 60 x 60 seconds

Area of rectangle = Length x Width

### Numerical Problems Chapter 1 – Physical Quantities and Measurements

#### {(a)5kg   (b) 2 MW     (c) 5.2 µg (d) 2.25µs }

Solution:

(a)     5000 g

= 5 x 1000g

= 5 x 1kg                     (Since 1000g = 1kg)

= 5kg

(b)    2000,000 W

=2 x 1000000

= 2 x 106 W              (106 = 1 Mega)

= 2 MW

(c)    52 x 10-10 kg

= 5.2 x l0 x 10-10 kg

= 5.2 x 10-9 kg

= 5.2 x 10-9 x 1000g      (Since 1 kg = 1000 g)

= 5.2 x 10-9 x 10g

= 5.2 x 10-9+3

=5.2 x 10-6 g

= 5.2 µg               (10-6 = 1micro(µ))

(d)    225 x 10-8 s

= 225 x 102 x 10-8 s

= 2.25 x 10-6 s

= 2.25 µs               (10-6 = 1micro(µ))

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

Solution:

As we know

micro = µ = 10-6

nano = n = 10-9

pico = p = 10-12

The relation between micro, nano, and pico can be written as.

micro = 10-6

nano =  10-6 x 10-3 = 10-6  micro

pico  =  10-6 x 10-6 = 10-6   micro

#### 1.3 Your hair grows at the rate of 1 mm per day. Find their growth rate in nm s-1. (11.57 nm s-1)

Solution:

Growth rate Of hair in nm s-1  = Imm per day

Growth rate of hair in one day = 24 x 60 x 60 s

(Since 1 mm 10-3 m and one day = 24 x 60 x 60 s), hence

1 mm per day = I x 10-3 m x 1/24 x 60 x 60 s

= I x 10-3  m x 1/8400 m s-1

= I x 10-3  m x 0.00001157

= I x 10-3  m x 1157 x 10-8 ms-l

= 1157 x 10-2m x 10-9ms-1

=11.57 x 10-9 ms-1

1 mm per day =11.57  nms-1

(because 10-9 ms-1 = 1n ms-1).

#### {(a) 1.168×10-24  (b) 3.2×106 (c) 7.25g (d) 2 x10-10  }

Solution:

(a) 1168 x 10-27 = 1.168 x 10x 10-27 = 1.168 x 10-24

(b) 32 x 10= 3.2 x 101 x 10 = 3.2 x 106

(c) 725 x 10-5 kg=7.25X 102 x 10-5 kg = 7.25 x 10-3 kg

As (10-3 kg = 1g), therefore

7.25 x 10-3 kg = 7.25g

(d) 0.02 x 10-8  = 2 x 10-2 x  10-8  = 2 x  10-10

#### {(a) 6.4×103 km (b) 3.8 x 105 km (c) 3 x 108 ms-1   (d) 0.64×104s)

Solution:

(a) 64000 km

Multiplying and dividing by “103

=6400 m/1000 x 103km

=64 m/ 10 x 103km

=6.4×10km

(b) 38000 km

Multiplying and dividing by “105

=38000 / 10 x 105km

=380000 / 1050000 x 105km

= 3.8 x 10km

(c) 300000000 ms-1

Multiplying and dividing by “108

300000000 ms-1 / 100000000 x 108km

= 3 x 10km

(d) seconds in a day

As we know

1 day =24 hours

1 hour= 60 minutes

1 minute = 60 seconds so

1 day = 24 x 60 x 60 seconds

1 day = 86400 s

Multiplying and dividing by 104

=86400 / 10000 x 10s

=8.4 x10s

#### 1.6 On closing the jaws Of a Vernier Calipers, 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.                          (+0.04cm, -0.04 cm)

Solution:

Main scale reading = 0.0 cm.

Vernier division coinciding with main scale = 4th division

Vernier scale reading = 4 x 0.01 cm = 0.04 cm

Zero error = 0.0 cm + 0.04 cm = 0.04 cm

Zero correction (Z.C) = -0.04 cm

The zero error of the Vernier scale is 0.04cm and its zero correction is -0.04cm

(Vernier division coinciding with the main scale) = 4 div

Vernier scale reading = 4 x 0.01 cm

= 0.04 cm

Since zero of the Vernier scale is on the right side of the zero of the main scale, thus the instrument has measured more than the actual reading. It IS said to be positive zero error.

Zero correction is the negative of zero error. Thus

Zero error = +0.04 cm

and Zero correction = – 0.04 cm

#### 1.7 A screw gauge has 50 divisions on its circular sale. The pitch of the screw gauge is 0.5 mm. What is its least count?                                                                          (0.001 cm)

Solution:

Number Of division on the circular scale = 50

Pitch of screw gauge = 0.5 mm

Least count Of screw gauge L.C. = ?

Least count = Pitch / Number Of division on the circular scale

Least count = 0.5mm / 50

= 0.01 mm = 0.01 x 1/10 cm

Least count = 0.001 cm

#### {(b)  and (c)}

Solution:

(a) 3.0066m

Zeros between significant digits are significant. Therefore, there are 5 significant figures in 3.0066m.

(b) 0.00309kg

Zeros used for spacing the decimal point are not significant. Therefore, there are 3 significant figures in 0.00309kg.

(C) 5.05 x 10-27kg

Only the digits before the exponent are considered, thus there are 3 significant figures.

(d) 301.0s

Final zeros or zeros after the decimal are significant. Therefore, there are 4 significant figures.

Result:

Quantities (b) and (c) have three significant figures

#### {(a) 4 (b) 3 (c) 3 (d) 4}

Solution:

(a) 1.009m

Since zeros between two significant figures are significant, so there are 4 significant figures.

(b) 0.00450

Zeros used for spacing the decimal point are not significant. Hence, there are 3 significant figures.

(c) 1.66 x 10-27kg

Only the digits before the exponent are considered so there are 3 significant figures.

(d) 2001s

Since zeros between two significant figures are significant so there are 3 significant figures.

#### 1.10 A chocolate wrapper is 6.7 cm long and 5.4 cm wide. Calculate its area up to a reasonable number of significant figures.                                                                (36 cm2)

Solution:

Length of the chocolate wrapper I = 6.7 cm

Width of chocolate wrapper w = 5.4 cm

Area = A =?

Area = Length x Width

A = l x w

A=6.7cm x 5.4 cm =36.18 cm2 = 36cm2

Note:

The answer should be in two significant figures because in data the least significant figures are two therefore the answer is 36 cm2.

Class 9 Physics Full Book Numerical Problems with Solution

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