Answer

(a) 1 cm $=\frac{1}{100} ~m$

Volume of the cube $=1 ~cm^{3}$

But, $1 ~cm^{3}=1 ~cm \times 1 ~cm \times 1 ~cm=(\frac{1}{100}) ~m \times(\frac{1}{100}) ~m \times(\frac{1}{100}) ~m$

$\therefore 1 ~cm^{3}=10^{-6} ~m^{3}$

Hence, the volume of a cube of side $1 ~cm$ is equal to $10^{-6} ~m^{3}$.

(b) The total surface area of a cylinder of radius $r$ and height $h$ is

$S=2 \pi r(r+h)$.

Given that,

$r=2 ~cm=2 \times 1 ~cm=2 \times 10 ~mm=20 ~mm$

$h=10 ~cm=10 \times 10 ~mm =100 ~mm$

$\therefore S=2 \times 3.14 \times 20 \times(20+100)=15072=1.5 \times 10^{4} ~mm^{2}$

(c) Using the conversion,

$1 ~km / h=\frac{5}{18} ~m / s$

$18 ~km / h=18 \times \frac{5}{18}=5 ~m / s$

Therefore, distance can be obtained using the relation:

Distance $=$ Speed $\times$ Time $=5 \times 1=5 ~m$

Hence, the vehicle covers $5 ~m$ in $1 ~s$.

(d) Relative density of a substance is given by the relation,

Relative density $=\frac{\text{ Density of substance }}{\text{ Density of water }}$

Density of water $=1 ~g / ~cm^{3}$

Density of lead $=$ Relative density of lead $\times$ Density of water

$ =11.3 \times 1=11.3 g / ~cm^{3} $

Again, $1 g=\frac{1}{1000} kg$

$1 ~cm^{3}=10^{-6} ~m^{3}$

$1 g / ~cm^{3}=\frac{10^{-3}}{10^{-6}} kg / ~m^{3}=10^{3} kg / ~m^{3}$

$\therefore 11.3 g / ~cm^{3}=11.3 \times 10^{3} kg / ~m^{3}$

Answer

(a) $1 kg=10^{3} ~g$

$1 m^{2}=10^{4} cm^{2}$

$1 ~g m^{2} s^{-2}=1 ~kg \times 1 m^{2} \times 1 ~s^{-2}$

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

(b) Light year is the total distance travelled by light in one year.

$1 ~ly=$ Speed of light $\times$ One year

$=(3 \times 10^{8} ~m / s) \times(365 \times 24 \times 60 \times 60 s)$

$=9.46 \times 10^{15} ~m$

$\therefore 1 ~m=\frac{1}{9.46 \times 10^{15}}=1.057 \times 10^{-16} ~ly$

(c) $1 ~m=10^{-3} ~km$

Again, $1 ~s=\frac{1}{3600} ~h$

$1 ~s^{-1}=3600 ~h^{-1}$

$1 ~s^{-2}=(3600)^{2} ~h^{-2}$

$\therefore 3 ~m s^{-2}=(3 \times 10^{-3} ~km) \times((3600)^{2} h^{-2})=3.88 \times 10^{-4} ~km h^{-2}$

(d) $1 ~N=1 ~kg m s^{-2}$

$1 ~kg=10^{-3} ~g^{-1}$

$1 ~m^{3}=10^{6} ~cm^{3}$

$\therefore 6.67 \times 10^{-11} ~N m^{2} kg^{-2}=6.67 \times 10^{-11} \times(1 ~kg m s^{-2})(1 m^{2})(1 s^{-2})$

$ \begin{aligned} & =6.67 \times 10^{-11} \times(1 kg \times 1 m^{3} \times 1 s^{-2}) \\ & =6.67 \times 10^{-11} \times(10^{-3} g^{-1}) \times(10^{6} cm^{3}) \times(1 s^{-2}) \\ & =6.67 \times 10^{-8} cm^{3} s^{-2} g^{-1} \end{aligned} $

Answer

Given that,

1 calorie $=4.2(1 kg)(1 m^{2})(1 s^{-2})$

New unit of mass $=\alpha kg$

Hence, in terms of the new unit, $1 kg=\frac{1}{\alpha}=\alpha^{-1}$

In terms of the new unit of length,

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

And, in terms of the new unit of time,

$ \begin{aligned} & 1 s=\frac{1}{\gamma}=\gamma^{-1} \\ & 1 s^{2}=\gamma^{-2} \\ & 1 s^{-2}=\gamma^{2} \\ & \therefore 1 \text{ calorie }=4.2(1 \alpha^{-1})(1 \beta^{-2})(1 \gamma^{2})=4.2 \alpha^{-1} \beta^{-2} \gamma^{2} \end{aligned} $

Answer

The given statement is true because a dimensionless quantity may be large or small in comparision to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.

(a) An atom is a very small object in comparison to a soccer ball.

(b) A jet plane moves with a speed greater than that of a bicycle.

(c) Mass of Jupiter is very large as compared to the mass of a cricket ball.

(d) The air inside this room contains a large number of molecules as compared to that present in a geometry box.

(e) A proton is more massive than an electron.

(f) Speed of sound is less than the speed of light.

Answer

Distance between the Sun and the Earth:

$=$ Speed of light $\times$ Time taken by light to cover the distance

Given that in the new unit, speed of light $=1$ unit

Time taken, $t=8 \min 20 s=500 s$

$\therefore$ Distance between the Sun and the Earth $=1 \times 500=500$ units

Answer

(a) A device with minimum count is the most suitable to measure length.

Least count of vernier callipers

$=1$ standard division $(SD)-1$ vernier division (VD)

$=1-\frac{9}{10}=\frac{1}{10}=0.01 cm$

(b) Least count of screw gauge $= \frac{\text{Pitch}}{\text{Number of divisions}}$

$=\frac{1}{1000}=0.001 cm$

(c) Least count of an optical device $=$ Wavelength of light $\sim 10^{-5} cm$

$=0.00001 cm$

Hence, it can be inferred that an optical instrument is the most suitable device to measure length.

Answer

Magnification of the microscope $=100$

Average width of the hair in the field of view of the microscope $=3.5 ~mm$

$\therefore$ Actual thickness of the hair is $\frac{3.5}{100}=0.035 ~mm$.

Answer

Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a metre scale. The diameter of the thread is given by the relation,

Diameter $=\frac{\text{ Length of thread }}{\text{ Number of turns }}$

It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only.

A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.

Answer

Area of the house on the slide $=1.75 cm^{2}$

Area of the image of the house formed on the screen $=1.55 m^{2}$

$=1.55 \times 10^{4} cm^{2}$

Arial magnification, $m_a=\frac{\text{ Area of image }}{\text{ Area of object }}=\frac{1.55}{1.75} \times 10^{4}$

$\therefore$ Linear magnifications, $m_l=\sqrt{m_a}$

$=\sqrt{\frac{1.55}{1.75} \times 10^{4}}=94.11$

Answer

(a)

The given quantity is $0.007 ~m^{2}$.

If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant. This means that here, two zeros after the decimal are not significant. Hence, only 7 is a significant figure in this quantity.

(b)

The given quantity is $2.64 \times 10^{24} ~kg$.

Here, the power of 10 is irrelevant for the determination of significant figures. Hence, all digits i.e., 2, 6 and 4 are significant figures.

(c)

The given quantity is $0.2370 ~g cm^{-3}$.

For a number with decimals, the trailing zeroes are significant. Hence, besides digits 2, 3 and 7, 0 that appears after the decimal point is also a significant figure.

(d)

The given quantity is $6.320 ~J$.

For a number with decimals, the trailing zeroes are significant. Hence, all four digits appearing in the given quantity are significant figures.

(e)

The given quantity is $6.032 ~Nm^{-2}$.

All zeroes between two non-zero digits are always significant.

(f)

The given quantity is $0.0006032 ~m^{2}$.

If the number is less than one, then the zeroes on the right of the decimal point (but left to the first non-zero) are insignificant. Hence, all three zeroes appearing before 6 are not significant figures. All zeros between two non-zero digits are always significant. Hence, the remaining four digits are significant figures.

Answer

Length of sheet, $l=4.234 ~m$

Breadth of sheet, $b=1.005 ~m$

Thickness of sheet, $h=2.01 cm=0.0201 ~m$

The given table lists the respective significant figures:

Hence, area and volume both must have least significant figures i.e., 3 .

Surface area of the sheet $=2(l \times b+b \times h+h \times l)$

$=2(4.234 \times 1.005+1.005 \times 0.0201+0.0201 \times 4.234)$

$=2(4.25517+0.02620+0.08510)$

$=2 \times 4.360$

$=8.72 ~m^{2}$

Volume of the sheet $=l \times b \times h$

$=4.234 \times 1.005 \times 0.0201$

$=0.0855 ~m^{3}$

This number has only 3 significant figures i.e., 8 , 5, and 5.

Answer

Mass of grocer’s box $=2.300 ~kg$

Mass of gold piece $\mathbf{I}=20.15 ~g=0.02015 ~kg$

Mass of gold piece $\mathbf{I I}=20.17 ~g=0.02017 ~kg$

Total mass of the box $=2.3+0.02015+0.02017=2.34032 ~kg$

In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is $2.3 ~kg$.

Difference in masses $=20.17-20.15=0.02 ~g$

In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

Answer

Given the relation,

$ m=\frac{m_0}{(1-v^{2})^{\frac{1}{2}}} $

Dimension of $m=M^{1} L^{0} T^{0}$

Dimension of $m_0=M^{1} L^{0} T^{0}$

Dimension of $v=M^{0} L^{1} T^{-1}$

Dimension of $v^{2}=M^{0} L^{2} T^{-2}$

Dimension of $c=M^{0} L^{1} T^{-1}$

The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S. This is only possible when the factor, $(1-v^{2})^{\frac{1}{2}}$ is dimensionless i.e., $(1-v^{2})$ is dimensionless. This is only possible if $v^{2}$ is divided by $c^{2}$. Hence, the correct relation is

$ m=\frac{m_0}{(1-\frac{v^{2}}{c^{2}})^{\frac{1}{2}}} $

Answer

Radius of hydrogen atom, $r=0.5 \quad \dot{A}=0.5 \times 10^{-10} m$

Volume of hydrogen atom $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times(0.5 \times 10^{-10})^{3}$

$=0.524 \times 10^{-30} m^{3}$

1 mole of hydrogen contains $6.023 \times 10^{23}$ hydrogen atoms.

$\therefore$ Volume of 1 mole of hydrogen atoms $=6.023 \times 10^{23} \times 0.524 \times 10^{-30}$

$=3.16 \times 10^{-7} m^{3}$

Answer

Radius of hydrogen atom, $r=0.5 \quad \dot{A}=0.5 \times 10^{-10} m$

Volume of hydrogen atom $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times(0.5 \times 10^{-10})^{3}$

$=0.524 \times 10^{-30} m^{3}$

Now, 1 mole of hydrogen contains $6.023 \times 10^{23}$ hydrogen atoms.

$\therefore$ Volume of 1 mole of hydrogen atoms, $V_a=6.023 \times 10^{23} \times 0.524 \times 10^{-30}$

$=3.16 \times 10^{-7} m^{3}$

Molar volume of 1 mole of hydrogen atoms at STP,

$V_m=22.4 L=22.4 \times 10^{-3} m^{3}$

$\therefore \frac{V_m}{V_a}=\frac{22.4 \times 10^{-3}}{3.16 \times 10^{-7}}=7.08 \times 10^{4}$

Hence, the molar volume is $7.08 \times 10^{4}$ times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.

Answer

Line of sight is defined as an imaginary line joining an object and an observer’s eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly.

On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.

Answer

Mass of the Sun, $M=2.0 \times 10^{30} kg$

Radius of the Sun, $R=7.0 \times 10^{8} m$

Volume of the Sun, $V=\frac{4}{3} \pi R^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times(7.0 \times 10^{8})^{3}$

$=\frac{88}{21} \times 343 \times 10^{24}=1437.3 \times 10^{24} m^{3}$

Density of the Sun $=\frac{\text{ Mass }}{\text{ Volume }}=\frac{2.0 \times 10^{30}}{1437.3 \times 10^{24}} \sim 1.4 \times 10^{3} kg / m^{5}$

The density of the Sun is in the density range of solids and liquids. This high density is attributed to the intense gravitational attraction of the inner layers on the outer layer of the Sun.