Units And Measurements Practice Questions

The pitch of the screw is the distance moved per rotation: $P = \frac{0.25 \text{ cm}}{5} = 0.05 \text{ cm}$.
The least count (LC) is calculated as $\text{LC} = \frac{P}{\text{Total divisions}} = \frac{0.05 \text{ cm}}{100} = 0.0005 \text{ cm}$.
Since $1 \text{ Main Scale Division (MSD)} = \text{Pitch}$, the main scale reading is $\text{MSR} = 4 \times 0.05 \text{ cm} = 0.20 \text{ cm}$.
The circular scale reading is $\text{CSR} = 30 \times \text{LC} = 30 \times 0.0005 \text{ cm} = 0.0150 \text{ cm}$.
The total thickness of the wire is $\text{MSR} + \text{CSR} = 0.20 \text{ cm} + 0.0150 \text{ cm} = 0.2150 \text{ cm}$.

For a sphere of radius $r$, the surface area is $S = 4\pi r^2$ and the volume is $V = \frac{4}{3}\pi r^3$.
Taking the relative error for surface area: $\frac{\Delta S}{S} = 2\frac{\Delta r}{r}$. Given $\frac{\Delta S}{S} = \alpha$, we have $2\frac{\Delta r}{r} = \alpha$, which implies $\frac{\Delta r}{r} = \frac{\alpha}{2}$.
Taking the relative error for volume: $\frac{\Delta V}{V} = 3\frac{\Delta r}{r}$.
Substituting the value of $\frac{\Delta r}{r}$ into the volume expression:
$\frac{\Delta V}{V} = 3 \left( \frac{\alpha}{2} \right) = \frac{3}{2}\alpha$

To find the Planck length $l_p$ (dimension $L$), we assume it is proportional to the product of the constants: $l_p \propto G^x \hbar^y c^z$.
The dimensional formulas are $[G] = M^{-1} L^3 T^{-2}$, $[\hbar] = M L^2 T^{-1}$, and $[c] = L T^{-1}$.
Substituting these into the expression for $l_p$ gives $L^1 = (M^{-1} L^3 T^{-2})^x (M L^2 T^{-1})^y (L T^{-1})^z$.
Equating the exponents: for $M: -x + y = 0 \implies x = y$; for $T: -2x - y - z = 0 \implies z = -3x$; for $L: 3x + 2y + z = 1$.
Substituting $y = x$ and $z = -3x$ into the equation for $L$ yields $3x + 2x - 3x = 1$, which simplifies to $2x = 1$, or $x = 1/2$.
Thus, $x = 1/2$, $y = 1/2$, and $z = -3/2$, resulting in the expression $l_p = \left( \frac{G \hbar}{c^3} \right)^{1/2}$.

Percentage error $\;\frac{\Delta A}{A} \times 1 0 0=3 \;\frac{\Delta P}{P} \times 1 0 0+2 \;\frac{\Delta Q}{Q}+\;\frac{1}{2} \;\frac{\Delta R}{R} \times 1 0 0+\;\frac{\Delta S}{S} \times 1 0 0$ $=3 \times 0 . 5+2 \times 1+0 . 5 \times 3+1 . 5=6 . 5 \%$

density $=M/V$ ${\Delta p}/p={\Delta M}/V+3{\Delta r}/r$ $=1.5$ $=4.5$

The orbital period $T$ of a satellite is given by Kepler's third law: $T^2 = \frac{4\pi^2 r^3}{GM}$.
Rearranging this to solve for the mass of the Earth $M$, we get $M = \frac{4\pi^2 r^3}{G T^2}$.
Using the principle of relative uncertainty, we differentiate the expression: $\frac{\Delta M}{M} = 3\frac{\Delta r}{r} + 2\frac{\Delta T}{T}$.
Given that $\frac{\Delta r}{r} \approx 0$ and $\frac{\Delta T}{T} = 10^{-2}$, we substitute these values into the equation:
$\frac{\Delta M}{M} = 3(0) + 2(10^{-2}) = 2 \times 10^{-2}$.

Given, $P=a^{\;\frac{1}{2}} b^{2} c^{3} d^{-4}$ Relative error = $\;\frac{\Delta P}{P} \times 100=(\;\frac{1}{2} \times \;\frac{\Delta a}{a}+2 \;\frac{\Delta b}{b}+3 \;\frac{\Delta c}{c}+4 \;\frac{\Delta d}{d}) \times 100$ $\;\; =\;\frac{1}{2} \times 2+2 \times 1+3 \times 3+4 \times 5$ $\;\; =32 \%$

$T={rhg}/2\times 10^3$ ${\Delta T}/T={\Delta r}/r+{\Delta h}/h+0$ $100\times {\Delta T}/T$ $=(\frac{10^{-2}\times .01}{1.25\times 10^ {-2}}+\frac{10^{-2}\times .01}{1.45\times 10^{-2}})100$ $=(0.8+0.689)$ $=(1.489)$ $100\times {\Delta T}/T= 1.489$ $≃1.5$

$M\propto T^xv^yh^z$ $M^1L^0T^0=(T^1)^x(L^1T^{-1})^y(M^1L^2T^{-1})^z$ $M^1L^0T^0=M^zL^{y+2z}+T^{x-y-z}$ $z=1$ $y+2z=0\;\;\;\;\;\;x-y-z=0$ $y=-2\;\;\;\;\;\;\;\;\;\;\;x+2-1=0$ $x=-1$ $M\Rightarrow T^{-1}C^{-2}h^1$

AD = Cln(BD)(B)(D)→dimensionless$[AD]=[C]$Checking options one by one${A-C}/D$ dimensions of A are not same as dimensions of C. meaningless

Readinng$=0.5+25(0.5/50)+5(0.5/50)=80$mm

$t={t_1+t-2+t_3+t_4}/4={90+91+95+92}/4=92$Now mean deviation is equal to $({2+1+3+0}/4)=1.5$Since least count of clock is one second, so $\Delta t=2$sec

$J=\sigma E$ $\sigma =J/E=\frac{I^{1}L^{-2}}{M^1L^1T^{-2}}/{I^1T^1}$ $\sigma =M^{-1}L^{-3}T^3I^2$

Least count of vernier calliper is $1/10$mm = 0.1mm = 0.01cm

$1/{4\pi \epsilon _0}q^2/r^2=F$ \epsilon _0=\[A^2T^2]$/$[MLT^{-2}L^2]$=[M^{-1}L^{-3}A^2T^4]$

We know ${\text{R}}=\;\frac{V}{I}$ $\therefore \;\frac{\Delta R}{R}=\;\frac{\Delta V}{V}+\;\frac{\Delta I}{I}$Percentage error in ${\text{R}}=$ $\;\;\frac{\Delta R}{R} \times 100=\;\frac{\Delta V}{V} \times 100+\;\frac{\Delta I}{I} \times 100$ $\; \therefore \;\frac{\Delta R}{R} \times 100=3 \%+3 \%=6 \%$

30 vernier scale divisions coincide with 29 main scale divisions.Therefore 1 V.S.D $=\;\frac{29}{30}$ M.S.DLeast count $=1 {\text{M}} . S . D-1$ V.S.D$=1 {\text{M}} \cdot {\text{S}} \cdot {\text{D}}-\;\frac{29}{30} {\text{M}} \cdot {\text{S}} \cdot {\text{D}}$ $=\;\frac{1}{30} M \cdot S \cdot D$ $=\;\frac{1}{30} \times 0.5^0$ Reading of Vernier $=$ Main Scale Reading $+$ Vernier scale reading $\times$ Least countGiven that,Main Scale Reading $=58.5$Vernier scale reading $=09$ division$\therefore$ Reading of Vernier $=58.5^{\circ}+9 \times \;\frac{0.5^{\circ}}{30}$ $=58.65^{\circ}$

Least count of screw gauge$\;=\;\frac{Pitch}{\;\text{ Number of division on circular scale }\;}$ $\;=\;\frac{1}{100} {\text{mm}}$ $\;=0.01 {\text{mm}}$Diameter of the wire $=$ M.S.R $+$ C.S.R $\times$ L.C$\;=0+52 \times 0.01$ $\;=0.52 {\text{mm}}$ $\;=0.052 {\text{cm}}$