Check the accuracy of following formulas:
(i) $F =\frac{m v^2}{r}$
(ii) $T =\frac{\text { hrgd }}{2}$
(iii) $T =2 \pi \sqrt{\frac{l}{g}}$
(iv) $Y =\frac{ MgL }{\pi r^2 l}$ Here, Y is Young's modulus.
(v) $S =u t+\frac{1}{2} a t^2$
(vi) $\frac{1}{2} m v^2=m g h$
(vii) $v^2=u^2+2 a s$ where $v$ and $u$ are final and initial velociteis respectively. a is acceleration and $s$ is the distance covered.
(viii) $n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$, Here $n$ is frequency, T is tension force, $l$ is length and $m$ is mass of unit length.
✓
Answer
(i) $F =\frac{m v^2}{r}$, Here F is the centripetal force, $m$ is mass of particle, $r$ is radius of circular path and $v$ is velocity of particle. $
\begin{array}{l}
\text { Dimension of left side of equation }=\left[M^1 L^1 T^{-2}\right] \\
\text { Dimension of right side of equation }=\frac{M^1 L^2 T^{-2}}{L^1} \\
\qquad=M^1 L^1 T^{-2} \\
\text { Dimension of right side }=\left[M^1 L^1 T^{-2}\right]
\end{array}
$
The given formula is dimensionally true.
(ii) $T =\frac{h r g d}{2}$
Here, $T =$ Surface tension
$r=$ Radius of capillary
$h=$ Height of liquid in capillary
$d=$ Density of the liquid
If the units on the right and left sides of the formula have the same dimensions then the equation is considered accurate.
In the formula,
Quantity on the left side $= T =$ Surface tension $=\left[ MT ^{-2}\right]$
Quantity on right side
$
\begin{array}{l}
=\frac{h r g d}{2}=\frac{[L][L]\left[LT^{-2}\right]\left[ML^{-3}\right]}{2} \\
=\frac{1}{2}\left[MT^{-2}\right]=\left[MT^{-2}\right]
\end{array}
$
$
\left(\frac{1}{2} \text { is dimensionless quantity. }\right)
$
$\therefore$ The dimensions of M and T are the same on both sides hence the equation is true from this point of view.
(iii) $T =2 \pi \sqrt{\frac{l}{g}}$
In the above equation,
Quantity on the left side $= T =$ Time interval $=\left[ M ^0 L^0 T^1\right]$
Hence on left side of quantity, dimension of mass is zero, dimension of length is zero and dimension of unit of time is one.
Dimension of right side of equation $=2 \pi \sqrt{\frac{l}{g}}$
$
=2 \pi \sqrt{\frac{L}{L^1 T^{-2}}}=\sqrt{T^2}=T
$
We can write this as $\left[ M ^0 L^0 T^1\right]$.
Dimensions of both sides are equal so above equation is correct.
(iv) $Y =\frac{ MgL }{\pi r^2 l}$
The dimension of quantity $( Y )$ (elasticity coefficient) on the left side $=\left[ M ^1 L^{-1} T^{-2}\right]$
The dimensions on the right side of the formula are as follows:
M , mass $=[ M ]$ Acceleration due to gravity, $g=\left[ LT ^{-2}\right]$
Length ( L ) $=[ L ]$
Increase in length $(I)=[ L ]$
Here, $\pi$ is dimensionless.
Hence, dimension of term $\frac{ MgL }{\pi r^2 l}=\frac{\left[ M ^1 L^1 T^{-2} L\right]}{\left[ L ^2\right][ L ]}$
$
=\left[M^1 L^{-1} T^{-2}\right]
$
The dimensions on both sides of formula are equal,
ence this formula is true from the dimension point of view.
(v) In the formula $S =u t+\frac{1}{2} a t^2$.
The quantity $(s)$ on the left side is the distance whose dimension is L . The dimension of the quantity on the left side is $=\left[ M ^0 L^1 T^0\right]$
$u=$ velocity, $t=$ time, $a=$ acceleration
Dimension of the first quantity on the right side,
$
\begin{aligned}
u & =M^0 L^1 T^{-1} \\
t & =T \\
u t & =\left[M^0 L^1 T^0\right]
\end{aligned}
$
Dimension of second quantity on the right side,
$
\therefore \quad\left[LT^{-2}\right]\left[T^2\right]=L=\left[M^0 L^1 T^0\right]
$
In equation $S =u t+\frac{1}{2} a t^2$, the dimensions of $S , u t$ and $\frac{1}{2} a t^2$ are equal so this equation is dimensionally correct.
(vi) $\frac{1}{2} m v^2=m g h$
Dimensions of left side of equation $=\frac{1}{2} m v^2$
$
\begin{array}{l}
m=\text { mass, } v=\text { velocity } \\
\qquad m v^2=[M]\left[LT^{-1}\right]^2=\left[M^1 L^2 T^{-2}\right]
\end{array}
$
Dimensions of right side $m g h=[ M ]\left[ LT ^{-2}\right][ L ]$
$
=M^1 L^2 T^{-2}
$
Hence, the dimensions of $\frac{1}{2} m v^2$ and $m g h$ are equal so the formula is dimensionally correct.
(vii) In the formula $v^2=u^2+2 a s$
Dimension of $v^2$ present on left side
$
\begin{aligned}
& =(\text { Dimension of velocity })^2 \\
v^2 & =\left[M^0 L^1 T^{-1}\right]^2 \\
& =\left[M^0 L^2 T^{-2}\right]
\end{aligned}
$
Dimension of right side (first quantity)
$
\begin{aligned}
u^2 & =\left[M^0 L^1 T^{-1}\right]^2 \\
& =\left[M^0 L^2 T^{-2}\right]
\end{aligned}
$
Dimension of second quantity of right side $=2 a s$
$
\begin{array}{l}
=\left[LT^{-2}\right][L] \quad(2 \text { is dimensionless) } \\
=\left[L^2 T^{-2}\right]=\left[M^0 L^2 T^{-2}\right]
\end{array}
$
In equation, $v ^2=u^2+2 a s$, dimensions of $v^2, u^2$ and $2 a s$ are equal. Hence, the equation is completely true.
(viii) In the formula $n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
In equation, dimension of quantity present on left side, i.e. frequency $(n)=\left[ M ^0 L^0 T^{-1}\right]$
Dimension of quantities present on right side :
Dimension of $l=\left[ L ^1\right]$
Dimension of mass, $M =\left[ M ^1\right]$
Dimension of $g=\left[ LT ^{-2}\right]$
Dimension of $m=\frac{\text { Mass }}{\text { Length }}=\left[ ML ^{-1}\right]$
Dimension of right side $=\frac{1}{2 L^1} \sqrt{\left(\frac{ M \left[ LT ^{-2}\right]}{ M ^1 L^{-1}}\right)}$
$
=\frac{1}{2 L^1} \sqrt{L^2 T^{-2}}=\frac{1}{2} T^{-1}=\left[M^0 L^0 T^{-1}\right]
$
Here, $\frac{1}{2}$ is pure number, it has no dimensions.
Dimension of right side $= T ^{-1}=$ dimension of left side.
Hence, the formula is completely correct.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
