Units and Measurements

The side is $1\text{ cm} = 10^{-2}\text{ m}$.

Volume of cube $= a^3 = (10^{-2})^3\text{ m}^3 = 10^{-6}\text{ m}^3$.

So the blank is $1\times 10^{-6}\text{ m}^3$.

Total surface area of a cylinder is

$$2\pi r(r+h)$$

Here $r=2.0\text{ cm}$ and $h=10.0\text{ cm}$.

$$\text{Area}=2\pi(2.0)(2.0+10.0)=48\pi\text{ cm}^2$$

$$48\pi \approx 150.8\text{ cm}^2$$

Now convert to $\text{mm}^2$:

$$1\text{ cm}^2 = 100\text{ mm}^2$$

$$150.8\text{ cm}^2 = 150.8\times 100 = 1.508\times 10^4\text{ mm}^2$$

To significant figures, this is $1.5\times 10^4\text{ mm}^2$ if rounded to 2 s.f., but the commonly accepted exact evaluation from the given values is about $1.51\times 10^4\text{ mm}^2$.

Convert $18\text{ km h}^{-1}$ to m s$^{-1}$:

$$18\times \frac{1000}{3600} = 5\text{ m s}^{-1}$$

So in $1\text{ s}$, the vehicle covers

$$5\times 1 = 5\text{ m}$$

Relative density is the ratio of density of the substance to density of water. Since water has density $1\text{ g cm}^{-3}$,

$$\text{density of lead} = 11.3\times 1 = 11.3\text{ g cm}^{-3}$$

Convert to SI:

$$1\text{ g cm}^{-3} = 10^3\text{ kg m}^{-3}$$

So,

$$11.3\text{ g cm}^{-3} = 11.3\times 10^3 = 1.13\times 10^4\text{ kg m}^{-3}$$

Convert each unit:

- $1\text{ kg} = 10^3\text{ g}$
- $1\text{ m} = 10^2\text{ cm}$, so $1\text{ m}^2 = 10^4\text{ cm}^2$

Therefore,

$$1\text{ kg m}^2\text{ s}^{-2} = 10^3\times 10^4\text{ g cm}^2\text{ s}^{-2} = 10^7\text{ g cm}^2\text{ s}^{-2}$$

Use $1\text{ ly} = 9.46\times 10^{15}\text{ m}$ approximately.

So,

$$1\text{ m} = \frac{1}{9.46\times 10^{15}}\text{ ly} \approx 1.06\times 10^{-16}\text{ ly}$$

Convert $3.0\text{ m s}^{-2}$ to km h$^{-2}$.

Since

$$1\text{ m s}^{-2} = \frac{1\text{ m}}{1\text{ s}^2}$$

and

$$1\text{ m} = 10^{-3}\text{ km},\quad 1\text{ s} = \frac{1}{3600}\text{ h}$$

So

$$1\text{ m s}^{-2} = 10^{-3}\times 3600^2\text{ km h}^{-2}$$

$$=10^{-3}\times 12{,}960{,}000 = 12960\text{ km h}^{-2}$$

Therefore,

$$3.0\text{ m s}^{-2} = 3.0\times 12960 = 38880\text{ km h}^{-2}$$

Given

$$G = 6.67\times 10^{-11}\text{ N m}^2\text{ kg}^{-2}$$

Now,

$$1\text{ N} = 1\text{ kg m s}^{-2}$$

So

$$G = 6.67\times 10^{-11}\times (\text{kg m s}^{-2})\times \text{m}^2\text{ kg}^{-2}$$

$$=6.67\times 10^{-11}\text{ m}^3\text{ s}^{-2}\text{ kg}^{-1}$$

Convert to cgs:

$$1\text{ m}^3 = 10^6\text{ cm}^3,\quad 1\text{ kg}^{-1}=10^{-3}\text{ g}^{-1}$$

Thus,

$$G=6.67\times 10^{-11}\times 10^6\times 10^{-3}$$

$$=6.67\times 10^{-8}\text{ cm}^3\text{ s}^{-2}\text{ g}^{-1}$$

A calorie $=4.2\text{ J}$. Also,

$$1\text{ J} = 1\text{ kg m}^2\text{ s}^{-2}$$

If the new units of mass, length and time are

$$1\text{ mass unit}=\alpha\text{ kg},\quad 1\text{ length unit}=\beta\text{ m},\quad 1\text{ time unit}=\gamma\text{ s},$$

then in new units,

$$1\text{ kg} = \alpha^{-1}\text{ new mass unit}$$
$$1\text{ m} = \beta^{-1}\text{ new length unit}$$
$$1\text{ s} = \gamma^{-1}\text{ new time unit}$$

Hence,

$$1\text{ J} = 1\text{ kg m}^2\text{ s}^{-2} = \alpha^{-1}\beta^{-2}\gamma^2\text{ (new units)}$$

Therefore,

$$1\text{ calorie} = 4.2\text{ J} = 4.2\alpha^{-1}\beta^{-2}\gamma^2\text{ (new units)}$$

The statement means that a quantity can be called large or small only when compared with some standard. A length, mass, speed, or number has meaningfully large or small only relative to a chosen reference.

Reframed statements:

- (a) Atoms are very small compared with ordinary objects.
- (b) A jet plane moves with great speed compared with a car or train.
- (c) The mass of Jupiter is very large compared with the mass of the Earth.
- (d) The air inside this room contains a large number of molecules compared with the number in a small sample of gas.
- (e) A proton is much more massive than an electron.
- (f) The speed of sound is much smaller than the speed of light.

The light travel time is

$$8\text{ min }20\text{ s} = 8\times 60 + 20 = 500\text{ s}$$

Since the new unit is chosen so that the speed of light is unity, the distance covered in $500\text{ s}$ is 500 new units.

So the Sun-Earth distance is $500$ units, i.e. 500 light-second units.

The instrument that measures length to within a wavelength of light has the greatest precision, because a wavelength is much smaller than the least count of a vernier calipers or screw gauge.