Q: Write the dimensions of \$a/b\$ in the relation \$P = \dfrac{a - t^2}{bx}\$; where \$P\$ is the pressure, \$x\$ is the distance, and \$t\$ is the time.

\$P=\dfrac { a-{ t }^{ 2 } }{ bx } \$\$P=\left[ { ML }^{ -1 }{ T }^{ -2 } \right] \$\$a\$ will have same unit as \${ t }^{ 2 }\$. So, \$a=\left[ { T }^{ 2 } \right] \$\$b=\left[ \dfrac { a-{ t }^{ 2 } }{ px } \right] =\dfrac { \left[ { T }^{ 2 } \right] }{ \left[ { ML }^{ -1 }{ T }^{ -2 }L \right] } =\left[ { M }^{ -1 }{ L }^{ 0 }{ T }^{ 4 } \right] \$\$\dfrac { a }{ b } =\dfrac { \left[ { T }^{ 2 } \right] }{ \left[ { M }^{ -1 }{ L }^{ 0 }{ T }^{ 4 } \right] } =\left[ { ML }^{ 0 }{ T }^{ -2 } \right] \$

Q: The dimensional formula for permittivity of free space \$(\epsilon_{0})\$ in the equation \$F = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {q_{1}q_{2}}{r^{2}}\$ where, symbols have their usual meaning is

Permittivity of free space \$\epsilon_o = \dfrac{q_1q_2}{4\pi Fr^2}\$Dimensions of \$[F] = [MLT^{-2}]\$Dimensions of \$[q] = [AT]\$Dimensions of \$[r] = [L]\$Thus dimensional formula of \$[\epsilon_o] = \dfrac{[AT][AT]}{[MLT^{-2}][L^2]}\$\$\implies\$ \$[\epsilon_o] = [M^{-1}L^{-3} T^4A^2]\$

Q: Find dimensions of constants \$a\$ and \$b\$ in equation.\$\left (P + \dfrac {a}{V^{2}}\right ) (V - b) = RT\$, where \$P\$ is pressure and \$V\$ is volume, \$R\$ is universal gas constant, \$T\$ is temperature.

We can add and structure only like quantities.\$\Rightarrow\$ Dimensions of \$P\$ = Dimensions of \$\dfrac{a}{V^2}\$ ...... (i) and dimension of \$V\$ = Dimensions of b ............ (ii)From (i), Dimensions of a = Dimensions of \$P \times\$ Dimensions of \$V^2 [a]\$ \$ = [M^1L^{-1}T^{-2}] \times [L^3]^2\$ \$ = [M^1 L^5 T^{-2}]\$Unit of \$a =\$ Unit of \$p \times Unit \,\,of \,\, v^2 \$ \$= \dfrac{N}{m^2} \times m^6 = N m^4\$From (ii), [b] = \$[V] = [M^0L^3T^0]\$So unit of b = Unit of V = \$m^3\$

Q: A wire has a mass (0.3 \$\pm\$ 0.003)g, radius (0.5 \$\pm\$ 0.005)mm and length (6 \$\pm\$ 0.06)cm. The maximum percentage error in the measurement of its density is -

The volume of the wire is given by, \$V = \pi r^2 L\$Thus density of the wire, \$\rho = \dfrac{m}{\pi r^2 L}\$Taking log and then differentiate: \$\implies \dfrac{\Delta \rho}{\rho} \times 100 = \dfrac{\Delta m}{m} \times 100 + 2 \dfrac{\Delta r}{r} \times 100 + \dfrac{\Delta L}{L} \times 100\$\$\therefore \$ \$ \dfrac{\Delta \rho}{\rho} \times 100 = \dfrac{0.003}{0.3} \times 100 + 2 \dfrac{0.005}{0.5} \times 100 + \dfrac{0.06}{6} \times 100 = 4\$%.\$\implies \$ Percentage error in density is \$4\$%.

Q: A physical quantity P is related to four observables a, b, c and d as follows :P \$a^3\, b^2\$ / \$\sqrt{c}\$ d The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?

\$ P = \dfrac{a^3 b^2}{\sqrt{c} d} \$Maximum fractional error in P is given by\$\dfrac{\Delta P}{P} = \pm \left[3\dfrac{\Delta a}{a} + 2\dfrac{\Delta b}{b}+ \dfrac{1}{2}\dfrac{\Delta c}{c}+ \dfrac{\Delta d}{d}\right]\$On putting above value: \$= \pm \dfrac{13}{100} = 0.13 \$Percentage error in \$P = 13\$ %Value of P is given as 3.763.By rounding off the given value to the first decimal place, we get \$P = 3.8\$.

Q: The radius of a sphere is measured to be \$5.3 \pm 0.1 cm\$. Calculate the percentage error in the measurement of its volume.

\$\textbf{Step 1: Differentiating the volume} \$Volume of the sphere: \$V = \dfrac{4}{3} \pi r^3 \$ Differentiating with respect to \$r\$\$\dfrac{dV}{dr} = \dfrac{4}{3} \pi . 3r^2 \$ \$\Rightarrow \dfrac{dV}{dr} = \dfrac{4}{3} \pi r^3 . \dfrac{3}{r} \$ \$\Rightarrow \dfrac{dV}{V} = \dfrac{3 dr}{r} \$ Or\$\Rightarrow \dfrac{\Delta V}{V} = \dfrac{3 \Delta r}{r} \$ \$\textbf{Step 2: Percentage error in the volume} \$Given: \$r = 5.3 \pm 0.1\ cm \$So, \$\Delta r=0.1 cm\$ and \$r=5.3 cm\$Percentage error in volume= \$\dfrac{\Delta V}{V} \times 100\% = \dfrac{3 \times 0.1\ cm }{5.3\ cm}\times 100\% = 5.66\%\$Hence, Option B is correct.

Q: Check the correctness of the equation\$ F=mv^2/r\$ where F is a force, m is mass, v is velocity and r is a radius.

We know Force\$F= mass\quad \times \quad acceleration\$----------(1)And given Force\$F=\dfrac{m{v}^{2}}{r}\$----------(2)Equating 1 and 2 dimensionaly we get ,\$ma=\dfrac{m{v}^{2}}{r}\$\$ML{T}^{-2}=M{L}^{2}{T}^{-2}{L}^{-1}\$\$ML{T}^{-2}=ML{T}^{-2}\$

Question 1

Write the dimensions of $a/b$ in the relation $P = \dfrac{a - t^2}{bx}$; where $P$ is the pressure, $x$ is the distance, and $t$ is the time.

Accepted Answer

$P=\dfrac { a-{ t }^{ 2 } }{ bx } $$P=\left[ { ML }^{ -1 }{ T }^{ -2 } \right] $$a$ will have same unit as ${ t }^{ 2 }$. So, $a=\left[ { T }^{ 2 } \right] $$b=\left[ \dfrac { a-{ t }^{ 2 } }{ px }  \right] =\dfrac { \left[ { T }^{ 2 } \right]  }{ \left[ { ML }^{ -1 }{ T }^{ -2 }L \right]  } =\left[ { M }^{ -1 }{ L }^{ 0 }{ T }^{ 4 } \right] $$\dfrac { a }{ b } =\dfrac { \left[ { T }^{ 2 } \right]  }{ \left[ { M }^{ -1 }{ L }^{ 0 }{ T }^{ 4 } \right]  } =\left[ { ML }^{ 0 }{ T }^{ -2 } \right] $

Question 2

The period of a simple pendulum is given by $T = 2\pi \sqrt {\frac{l}{g}} $, where $l$ is length of the pendulum and g is acceleration due to gravity. Show that this equation is dimensionally correct.

Accepted Answer

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Question 3

The dimensional formula for permittivity of free space $(\epsilon_{0})$ in the equation $F = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {q_{1}q_{2}}{r^{2}}$ where, symbols have their usual meaning is

Accepted Answer

Permittivity of free space  $\epsilon_o = \dfrac{q_1q_2}{4\pi  Fr^2}$Dimensions of  $[F] = [MLT^{-2}]$Dimensions of  $[q] = [AT]$Dimensions of  $[r] = [L]$Thus dimensional formula of  $[\epsilon_o] = \dfrac{[AT][AT]}{[MLT^{-2}][L^2]}$$\implies$  $[\epsilon_o] = [M^{-1}L^{-3} T^4A^2]$

Question 4

Find dimensions of constants $a$ and $b$ in equation.$\left (P + \dfrac {a}{V^{2}}\right ) (V - b) = RT$, where $P$ is pressure and $V$ is volume, $R$ is universal gas constant, $T$ is temperature.

Accepted Answer

We can add and structure only like quantities.$\Rightarrow$ Dimensions of $P$ = Dimensions of $\dfrac{a}{V^2}$                  ......  (i) and dimension of $V$ = Dimensions of b                     ............ (ii)From (i), Dimensions of a = Dimensions of $P \times$ Dimensions of $V^2 [a]$                        $ = [M^1L^{-1}T^{-2}] \times [L^3]^2$                        $ = [M^1 L^5 T^{-2}]$Unit of $a =$ Unit of $p \times Unit \,\,of \,\, v^2 $                 $= \dfrac{N}{m^2} \times m^6 = N m^4$From (ii), [b] = $[V] = [M^0L^3T^0]$So unit of b = Unit of V = $m^3$

Question 5

A wire has a mass (0.3 $\pm$ 0.003)g, radius (0.5 $\pm$ 0.005)mm and length (6 $\pm$ 0.06)cm. The maximum percentage error in the measurement of its density is -

Accepted Answer

The volume of the wire is given by, $V = \pi  r^2  L$Thus density of the wire, $\rho  =  \dfrac{m}{\pi  r^2  L}$Taking log and then differentiate:              $\implies  \dfrac{\Delta \rho}{\rho}  \times 100  =  \dfrac{\Delta m}{m} \times 100  +  2  \dfrac{\Delta r}{r}  \times 100  +  \dfrac{\Delta L}{L}  \times 100$$\therefore $  $  \dfrac{\Delta \rho}{\rho}  \times 100  =  \dfrac{0.003}{0.3} \times 100  +  2  \dfrac{0.005}{0.5}  \times 100  +  \dfrac{0.06}{6}  \times 100  =  4$%.$\implies $ Percentage error in density is $4$%.

Question 6

State the number of significant figures in the following:$(a)  0.007  m^2$$(b)  2.64 \times 10^{24}  kg$$(c)  0.2370  g cm^{-3}$$(d)  6.320  J $$(e)  6.032  Nm^{-2} $$(f)  0.0006032  m^2 $

Accepted Answer

The significant figures of a number are those digits that carry meaning contributing to its measurement resolution. This includes all digits except:1. All leading zeros;2. Trailing zeros when they are merely placeholders to indicate the scale of the number (exact rules are explained at identifying significant figures); 3. Spurious digits introduced, for example, by calculations carried out to greater precision than that of the original data, or measurements reported to a greater precision than the equipment supports.(a)$0.007 = $ $ 7 \times 10^{-3}$Significant digit $=7$Only one significant digit. (b) Number of significant digit $=3$(c)All non-zero numbers are always significant.All zeroes before a non zero number are insignificant. All zeroes which are simultaneously to the right of the decimal point and at the end of the number are significant. The only non significant digit here is the zero before the decimal point.Hence, number of significant digits is $4$.(d)All non-zero numbers are always significant. All zeroes which are simultaneously to the right of the decimal point and at the end of the number are significant.Hence, number of significant digits is $4$.(e)All the digits are significant. (no leading and trailing zero)No. of significant digits $= 4$(f)Significant digits $= 6032$No. of sig. digits $= 4$

Question 7

A physical quantity P is related to four observables a, b, c and d as follows :P $a^3\, b^2$ / $\sqrt{c}$ d The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?

Accepted Answer

$  P = \dfrac{a^3 b^2}{\sqrt{c}  d} $Maximum fractional error in P is given by$\dfrac{\Delta P}{P} = \pm \left[3\dfrac{\Delta a}{a} + 2\dfrac{\Delta b}{b}+ \dfrac{1}{2}\dfrac{\Delta c}{c}+ \dfrac{\Delta d}{d}\right]$On putting above value:        $= \pm \dfrac{13}{100} = 0.13 $Percentage error in $P = 13$ %Value of P is given as 3.763.By rounding off the given value to the first decimal place, we get $P = 3.8$.

Question 8

The radius of a sphere is measured to be $5.3 \pm 0.1 cm$. Calculate the percentage error in the measurement of its volume.

Accepted Answer

$\textbf{Step 1: Differentiating the volume} $Volume of the sphere: $V = \dfrac{4}{3} \pi r^3 $ Differentiating with respect to $r$$\dfrac{dV}{dr} = \dfrac{4}{3} \pi . 3r^2 $ $\Rightarrow  \dfrac{dV}{dr} = \dfrac{4}{3} \pi r^3 . \dfrac{3}{r} $ $\Rightarrow \dfrac{dV}{V} = \dfrac{3 dr}{r} $  Or$\Rightarrow \dfrac{\Delta V}{V} = \dfrac{3 \Delta r}{r} $ $\textbf{Step 2: Percentage error in the volume} $Given: $r = 5.3 \pm 0.1\ cm $So, $\Delta r=0.1 cm$ and $r=5.3 cm$Percentage error in volume= $\dfrac{\Delta V}{V} \times 100\% = \dfrac{3 \times 0.1\ cm }{5.3\ cm}\times 100\% = 5.66\%$Hence, Option B is correct.

Question 9

Check the correctness of the equation$ F=mv^2/r$ where F is a force, m is mass, v is velocity and r is a radius.

Accepted Answer

We know Force$F= mass\quad \times \quad acceleration$----------(1)And given Force$F=\dfrac{m{v}^{2}}{r}$----------(2)Equating 1 and 2 dimensionaly we get ,$ma=\dfrac{m{v}^{2}}{r}$$ML{T}^{-2}=M{L}^{2}{T}^{-2}{L}^{-1}$$ML{T}^{-2}=ML{T}^{-2}$

Question 10

The difference between the measured value and true value is :

Accepted Answer

Error is  a measure of the estimated difference between the observed or calculated value of a quantity and its true value.Error $=|$true value $-$ measured value$|$