Express the following quantities using prefixes.

(a) 5000 g                        (b) 2000000 W
(c) $52\times1Tt0kg$         (d) 225$\times10^{-8s}$

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

Difficulty: Medium

Solution:

(a) 5000 g
= $5 \times 1000g$
= $5 \times 1kg (Since 1000g = 1kg)$
= 5kg

(b) 2000,000 w
=$2 \times 1000000$
= $2 \times 10^{6}W (10^{6} = 1 Mega)$
= 2 MW

(c) $52 \times 10^{-10} kg$
= $5.2 \times l0\times 10^{-10}kg$
= $5.2 \times 10^{-9} kg$
= $5.2 \times10^{-9} \times 1000g$      (Since 1 kg = 1000 g)
= $5.2\times 10^{-9} \times 10^{3} g$
= $5.2\times 10^{-9} \times 10^{-6} g$
= $5.2 µg (10^{-6} = 1micro(µ))$

(d) $225 \times 10^{-8} s$
= $225\times10^{2} \times 10^{-8} s$
= $2.25 \times 10^{-6} s$
= $2.25 µs (10^{-6} = 1micro(µ))$

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

Difficulty: Easy

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.

$micro = 10^{-6}$
$nano = 10^{-6} \times 10^{-3} = 10^{-6}$  micro
pico  =  $10^{-6} \times10^{-6} = 10^{-6}$   micro

Your hair grows at the rate of 1 mm per day. Find their growth rate
in nm $s^{-1}. (11.57 nm s^{-1})$

Difficulty: Easy

Solution:

Growth rate Of hair in nm $s^{-1}$ = Imm per day

Growth rate of hair in one day = $24\times 60 \times$ 60 s

(Since 1 mm $10^{-3} m and one day = 24 \times 60 \times 60 s)$, hence

1 mm per day = $I\times 10^{-3} m \times \frac{1}{24} \times 60 \times$ 60 s
= $I \times 10^{-3} m \times \frac{1}{8400} m s^{-1}$
= $I \times10^{-3} m\times 0.00001157$
= $I \times 10^{-3} m \times1157 \times 10^{-8} ms^{-I}$
= $1157 \times 10^{-2}m \times 10^{-9}ms^{-1}$

=$11.57 \times10^{-9} ms^{-1}$
1 mm per day =$11.57 nms^{-1}$
(because $10^{-9} ms^{-1} = 1n ms^{-1})$.

Rewrite the following In Standard form. (Scientific notation)

(a) $1168\times 10^{-27}$                (b) $32\times 10^{5}$
(C) $725 \times 10 kg^{-5}$            (d) $0.02\times 10^{-8}$

{(a) $1.168\times10^{-24}$   (b) $3.2\times10^{6}$    (c) $7.25g (d) 2\times10^{-10}$

Difficulty: Easy

Solution:

(a) $1168 \times10^{-27} = 1.168 \times 10^{3} \times 10^{-27} = 1.168 \times 10^{-24}$
(b) $32 \times 10^{5} = 3.2 \times 10^{1} \times 10^{5} = 3.2 \times10^{6}$
(C) $725 \times 10^{-5} kg=7.25\times 10^{2} \times 10^{-5} kg = 7.25 \times 10^{-3}$ kg
As $(10^{-3} kg = 1g)$, therefore
$7.25 \times 10^{-3} kg = 7.25g$
(d) $0.02 \times 10^{-8} = 2\times 10^{-2}\times 10^{-8} = 2 \times 10^{-10}$

Write the following quantities in standard form.

(a) 6400 km                        (b) 38000 km
(c) $300000000 ms^{-1}$        (d) seconds in a day
{(a) $6.4\times10^{3} km$ (b) $3.8\times 10^{5}$ km (c) $3 \times10^{8} ms^{-1}$  (d) $0.64\times10^{4s}$

Difficulty: Medium

Solution:

(a) 64000 km

Multiplying and dividing by “$10^{3}$”
=6400 $\frac{m}{1000}\times 10^{3}$km
=$64 \frac{m}{10} \times 10^{3}$km
=$6.4\times10^{3}$ km

(b) 38000 km

Multiplying and dividing by “$10^{5}”$
=$\frac{38000}{10^{5}} \times10^{5}km$
=$\frac{380000}{380000}\times 10^{5}km$
= $3.8 \times 10^{5}$ km

(c) 300000000 $ms^{-1}$

Multiplying and dividing by “$10^{8}$”
$\frac{300000000 ms^{-1}}{100000000}\times10^{8}km$
= $3 \times 10^{8}$ km

(d) seconds in a day

As we know
1 day =24 hours
1 hour= 60 minutes
1 minute = 60 seconds so
1 day = $24 \times 60 \times 60$ seconds
1 day = 86400 s
Multiplying and dividing by $10^{4}$
=$\frac{86400}{10000}\times10^{4} s$
=$8.4 \times10^{4} s$

On closing the jaws Of a Vernier Calipers, zero of the Vernier scale is on the right to its main scale such that the 4th division of its Vernier scale coincides with one of the main scale divisions. Find its zero error and zero correction.(+0.04cm, -0.04 cm)

Difficulty: Easy

Solution:

Main scale reading = 0.0 cm.

Vernier division coinciding with main scale = 4th division

Vernier scale reading = $4\times$ 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 main scale) = 4 div
Vernier scale reading = $4 \times 0.01$ 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

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)

Difficulty: Easy

Solution:

Number Of divisions 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 = $\frac{0.5mm}{50}$
= 0.01 mm = $0.01\times \frac{1}{10}$cm
Least count = 0.001 cm

Which of the following quantities have three figures?

(a) 3.0066 m                       (b) 0.00309 kg
(c) 5.05\times10^{-27}kg      (d) 301.0 s

{(b)  and (c)}

Difficulty: Easy

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) $5.05 \times 10^{-27}$kg
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

What are the significant figures in the following measurements?

(a) 1.009 m                                          (b) 0.00450 kg
(c) $1.66 \times 10^{-27}$kg              (d) 2001 s

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

Difficulty: Easy

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) $1.66 \times 10^{-27}$kg
Only the digits before the exponent are considered so there are 3 significant figures.

(d) $2001^{s}$
Since zeros between two significant figures are significant so there are 3 significant figures.

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

$(36 cm^{2})$

Difficulty: Easy

Solution:

Length of the chocolate wrapper I = 6.7 cm

Width of chocolate wrapper w = 5.4 cm

Area = A = ?
Area = $Length \times Width$
A = $l \times w$
A=$6.7cm \times5.4$ cm = $36.18 cm^{2} = 36cm^{2}$

Note:
The answer should be in two significant figures because in data the least significant figures are two therefore answer is 36 $cm^{2}$