3 Marks Question

Question 13 Marks

The kinetic energy K of a rotating body depends on its moment of inertia I and its angular speed $\omega.$ Assuming the relation to be $\text{K}=\text{kI}^{\text{a}}\omega^{\text{b}}$ where k is a dimensionless constant, find a and b. Moment of inertia of a sphere about its diameter is $\frac{2}{5}\text{Mr}^2.$

Answer

$\text{K = kI}^{\text{a}}\omega^{\text{b}}$ where k = Kinetic energy of rotating body and k = dimensionless constantDimensions of left side are,
$\text{K}=[\text{ML}^2\text{T}^{-2}]$
Dimensions of right side are,
$\text{I}^{\text{a}}=[\text{ML}^2]^{\text{a}},\omega^{\text{b}}=[\text{T}^{-1}]^{\text{b}}$
According to principle of homogeneity of dimension,
$[\text{ML}^2\text{T}^{-2}]=[\text{ML}^2\text{T}^{-2}][\text{T}^{-2}]^{\text{b}}$
Equating the dimension of both sides,
2 = 2a and -2 = -b ⇒ a = 1 and b = 2

View full question & answer→

Question 23 Marks

What are the dimensions of$:$

Volume of $A$ cube of edge $A.$
Volume of $A$ sphere of radius $A.$
The ratio of the volume of $A$ cube of edge $A$ to the volume of $A$ sphere of radius $A$?

Answer

$L^3$
$L^3$
$M^0 L^0 T^0.$

View full question & answer→

Question 33 Marks

Let x and a stand for distance. Is $\int\frac{\text{dx}}{\sqrt{\text{a}^2-\text{x}^2}}=\frac{1}{\text{a}}\sin^{-1}\frac{\text{a}}{\text{x}}$ dimensionally correct?

Answer

Dimension of the left side $=\int\frac{\text{dx}}{\sqrt{(\text{a}^2-\text{x}^2})}=\int\frac{\text{L}}{\sqrt{(\text{L}^2-\text{L}^2)}}=[\text{L}^0]$
Dimension of the right side $=\frac{1}{\text{a}}\sin^{-1}\Big(\frac{\text{a}}{\text{x}}\Big)=[\text{L}^{-1}]$
So, the dimension of $\int\frac{\text{dx}}{\sqrt{(\text{a}^2-\text{x}^2)}}\neq\frac{1}{\text{a}}\sin^{-1}\Big(\frac{\text{a}}{\text{x}}\Big)$
So, the equation is dimensionally incorrect.

View full question & answer→

Question 43 Marks

Theory of relativity reveals that mass can be converted into energy. The energy E so obtained is proportional to certain powers of mass m and the speed c of light. Guess a relation among the quantities using the method of dimensions.

Answer

Let energy $\text{E}\propto\text{M}^{\text{a}}\text{C}^{\text{b}}$ where M = Mass, C = speed of light$\Rightarrow\text{E}= \text{KM}^{\text{a}}\text{C}^{\text{b}}$ (K = proportionality constant)
Dimension of left side$\text{E}=[\text{ML}^2\text{T}^{-2}]$
Dimension of right side$\text{M}^{\text{a}}=[\text{M}]^{\text{a}},[\text{C}]^{\text{b}}=[\text{LT}^{-1}]^{\text{b}}$
$\therefore[\text{ML}^2\text{T}^{-2}]=[\text{M}]^{\text{a}}[\text{LT}^{-1}]^{\text{b}}$
$\Rightarrow\text{a}=1;\text{b}=2$
So, the relation is $\text{E = KMC}^2$

View full question & answer→

Physics 3 Marks Question

Take a timed test