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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 }
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?
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 $s^{-1}. (11.57 nm s^{-1})$
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 $10^{-9} ms^{-1} = 1n ms^{-1})$.
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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) $7.25g (d) 2\times10^{-10}$
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) $0.64\times10^{4s}$
Solution:
(a) 64000 km
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)
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
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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 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)}
Solution:
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}
Solution:
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A chocolate wrapper is 6.7 cm long and 5.4 cm wide. Calculate its area up to a reasonable number of significant figures.
$(36 cm^{2})$
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 $cm^{2}$
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