Numerical Problems

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

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

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 $10^{-9} ms^{-1} = 1n ms^{-1})$.

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) $0.02 \times 10^{-8}  = 2\times 10^{-2}\times  10^{-8} = 2 \times  10^{-10}$

Solution:

(a) 64000 km

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

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

Solution:

Solution:

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 $cm^{2}$