State the number of significant figures in the following: $2.64 \times 10^{24} \mathrm{~kg}$
Answer
3 Explanation: The given quantity is $2.64 \times 10^{24} \mathrm{~kg}$. 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
Assuming that the mass (m) of the largest stone that can be moved by a flowing river depends only upon the velocity v, the density $\rho$ of water and the acceleration due to gravity g. Show that m varies, with the sixth power of the velocity of the flow.
Answer
$\text{Let}\text{ m}\propto\text{v}^{\text{a}}\rho^{\text{b}}\text{g}^\text{c}$ $\therefore\text{m}=\text{k}\text{v}^{\text{a}}\rho^{\text{b}}\text{g}^{\text{c}},$ where k is constant. where k is constant Taking the dimensions of various physical quantities on both the sides, We have, $[\text{M}]=-[\text{LT}^{-1}]^{\text{a}}[\text{ML}^{-3}]^{\text{b}}[\text{LT}^{-2}]^{\text{c}}$ $=[\text{M}^{\text{b}}\text{L}^{\text{a}-3\text{b}+\text{c}}\text{T}^{-\text{a}-2\text{c}}]$ Comparing the powers of M, L and T on both the sides, We have: $\text{b} = 1 \dots(\text{i})$ $\text{a} - 3\text{b} + \text{c} = 0 ....(\text{ii})$ $-\text{a} - 2\text{c} = 0 \dots(\text{iii})$ Solving these equation We get: $\text{b} = 1, \text{a} = 6 \text{ and c} = -3$ $\therefore\text{m}=\text{k}\text{v}^{6}\rho^{1}\text{g}^{-3}$ $\text{m}\propto\text{v}^6$
nm stands for nanometre, $1 \mathrm{~nm}=10^{-9} \mathrm{~m}, \mathrm{mN}$ stands for milli-newton, $1 \mathrm{mN}=10^{-3} \mathrm{~N}, \mathrm{Nm}$ stands for newton metre.
Obtain the dimensional formula for coefficient of viscosity.
Answer
Coefficient of viscosity $(\eta)=\frac{\text{fdx}}{\text{A.dv}}=\frac{[\text{MLT}]^{-2}[\text{L}]}{[\text{L}^2][\text{LT}^{-1}]}$ $=[\text{M}^{-1}\text{L}^{-1}\text{T}^{-1}]$
The rotational kinetic energy of a body is given by $\text{F}=\frac{1}{2}\text{l}\omega^2$ where w is the angular velocity of the body. Use the equation to obtain dimensional formula for moment of inertia I. Also write its SI unit
Answer
The given relation is $\text{F}=\frac{1}{2}\text{l}\omega^2$ $\text{I}=\frac{[\text{E}]}{[\omega]^2}=\frac{[\text{ML}^2\text{T}^{-2}]}{[\text{T}^{-1}]^2}\Big[\frac{\text{ML}^2\text{T}^{-2}}{\text{T}^{-2}}\Big]=[\text{ML}^2]$ Its si unit is joule.
If velocity of light $c$, Planck’s constant $h$ and gravitational contant $G$ are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.
Answer
We have to apply principle of homogeneity to solve this problem. Principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be same, i.e., dimensions of $\text{LHS}$ and $\text{RHS}$ should be equal, We know that, dimensions of, $[\text{h}]=[\text{ML}^2\text{T}^{-1}],[\text{c}]=[\text{LT}^{-1}],\text[{G}]=[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$
Let $\text{m}\propto\text{c}^\text{x}\text{h}^\text{v}\text{G}^\text{z}$
$\Rightarrow\text{m}=\text{kc}^\text{a}\text{h}^\text{b}\text{G}^\text{c}\ \ \ \ ...(\text{i})$
Where, $k$ is a dimensionless constant of proportionality.
Substituting dimensions of each term in Eq. $(i)$, we get,
$[\text{ML}^0\text{T}^0]=[\text{LT}^{-1}]^\text{a}\times[\text{ML}^2\text{T}^{-1}]^\text{b}[\text{M}^{-1}\text{L}^3\text{T}^{-2}]^\text{c}$
Comparing powers of same terms on both sides, we get,
$\text{b}-\text{c}=1\ \ \ ...(\text{ii})$
$\text{a}+2\text{b}+3\text{c}=0\ \ \ \ ...(\text{iii})$
$-\text{a}-\text{b}-2\text{c}=0\ \ \ \ ...(\text{iv})$
Adding Eqs. $(ii), (iii)$ and $(iv)$, we get,
$2\text{b}=1\Rightarrow\text{b}=\frac{1}{2}$
Substituting value of $b$ in Eq. $(ii),$ we get,
$\text{c}=-\frac{1}{2}$
From Eq. $(iv)$
$\text{a}=-\text{b}-2\text{c}$
Substituting values of $b$ and $c,$ we get,
$\text{a}=-\frac{1}{2}-2\Big(-\frac{1}{2}\Big)=\frac{1}{2}$
Putting values of $a, b$ and $c$ in Eq. $(i)$, we get,
$\text{m}=\text{kc}^\frac{1}{2}\text{h}^\frac{1}{2}\text{G}^{-\frac{1}{2}}=\text{k}\sqrt{\frac{\text{ch}}{\text{G}}}$
Let $\text{L}\propto\text{c}^\text{a}\text{h}^\text{b}\text{G}^\text{c}$
$\Rightarrow\text{L}=\text{kc}^\text{a}\text{h}^\text{b}\text{G}^\text{c}\ \ \ \ ...(\text{v})$
Where $k$ is a dimensionless constant.
Substituting dimensions of each term in Eq. $(v)$, we get
$[\text{M}^0\text{LT}^0]=[\text{LT}^{-1}]^\text{a}\times[\text{ML}^2\text{T}^{-1}]^\text{b}\times[\text{M}^{-1}\text{L}^3\text{T}^{-2}]^\text{c}$
$=[\text{M}^{\text{b}-\text{c}\ }\text{L}^{\text{a}+2\text{b}+3\text{c}}\ \text{T}^{-\text{a}-\text{b}-2\text{c}}]$
On comparing powers of same terms, we get,
$\text{b}-\text{c}=0\ \ \ ...(\text{vi})$
$\text{a}+2\text{b}+3\text{c}=1\ \ \ ...(\text{vii})$
$-\text{a}-\text{b}-2\text{c}=0\ \ \ \ ...(\text{viii})$
Adding Eqs. $(vi), (vii)$ and $(viii)$, we get,
$2\text{b}=1\Rightarrow\text{b}=\frac{1}{2}$
Substituting value of $b$ in Eq. $(vi)$, we get,
$\text{c}=\frac{1}{2}$
From Eq. $(viii), \text{a}=-\text{b}-2\text{c}$
Substituting values of $b$ and $c$, we get,
$\text{a}=-\frac{1}{2}-2\Big(\frac{1}{2}\Big)=-\frac{3}{2}$
Putting values of $a, b$ and $c$ in Eq. $(v)$, we get,
$\text{L}=\text{kc}^{-\frac{3}{2}}\text{h}^\frac{1}{2}\text{G}^\frac{1}{2}=\text{k}\sqrt{\frac{\text{hG}}{\text{c}^3}}$
Let $\text{T}\propto\text{c}^\text{a}\text{h}^\text{b}\text{G}^\text{c}$
$\Rightarrow\text{T}=\text{c}^\text{a}\text{h}^\text{b}\text{G}^\text{c}\ \ \ \ ...(\text{ix})$
Where, $k$ is a dimensionless constant.
Substituting dimensions of each term in Eq. $(ix)$, we get
$[\text{M}^0\text{L}^0\text{T}^1]=[\text{LT}^{-1}]^\text{a}\times[\text{ML}^2\text{T}^{-1}]^\text{b}\times[\text{M}^{-1}\text{L}^3\text{T}^{-2}]^\text{c}$
$=[\text{M}^{\text{b}-\text{c}}\text{ L}^{\text{a}+2\text{b}+3\text{c}}\ \text{T}^{-\text{a}-\text{b}-2\text{c}}]$
On comparing powers of same terms, we get,
$\text{b}-\text{c}=0 \ \ \ ...(\text{x})$
$\text{a}+2\text{b}+3\text{c}=1\ \ \ ...(\text{xi})$
$-\text{a}-\text{b}-2\text{c}=1\ \ \ \ ...(\text{xii})$
Adding Eqs. $(x), (xi)$ and $(xii)$, we get,
$2\text{b}=1\Rightarrow\text{b}=\frac{1}{2}$
Substituting the value of $b$ in Eq. $(x)$, we get,
$\text{c}=\text{b}=\frac{1}{2}$
From Eq. $(xii),$
$\text{a}=-\text{b}-2\text{c}-1$
Substituting values of $b$ and $c$, we get,
$\text{a}=-\frac{1}{2}-2\Big(\frac{1}{2}\Big)-1=-\frac{5}{2}$
Putting values of $a, b$ and $c$ in Eq. $(ix)$, we get,
$\text{T}=\text{kc}^{\frac{-5}{2}}\text{h}^\frac{1}{2}\text{G}^\frac{1}{2}=\text{k}\sqrt{\frac{\text{hG}}{\text{c}^5}}$
Using the relation E = hv, obtain the dimensions of Planck's constant.
Answer
We know that dimensional formula of energy E of photon is [$M^1L^2T^{-2}$) and dimensional formula of frequency v is [$T^{-1}$] $[\text{h}]=\frac{[\text{E}]}{[\text{V}]}=\frac{\text{M}^{1}\text{L}^2\text{T}^{-2}}{[\text{T}^{-2}]}=[\text{M}^1\text{L}^2\text{T}^{-1}]$
As relative density is defined as the ratio of the density of given substance and the density of standard distance (water), it is a dimensionless quantity.
Explain this statement clearly: “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:
Atoms are very small objects
A jet plane moves with great speed
The mass of Jupiter is very large
The air inside this room contains a large number of molecules
A proton is much more massive than an electron
The speed of sound is much smaller than the speed of light.
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.
An atom is a very small object in comparison to a soccer ball.
A jet plane moves with a speed greater than that of a bicycle.
Mass of Jupiter is very large as compared to the mass of a cricket ball.
The air inside this room contains a large number of molecules as compared to that present in a geometry box.
Angular diameter of moon is the angle subtended at a point on the earth, by two diameterically opposite ends of the moon. Its value is about $0.5^\circ$.
Magnitude of force F experienced by a certain object moving with speed v is given by $\text{F}=\text{k}\text{v}^2$ where k is constant. Find the dimensions of K.