Units and MeasurementsNEET Physics · Class 11 · NCERT Chapter 1

Solution

Step 1. The smallest gap on the main scale is $5.15-5.10=0.05\text{cm}$, so $1\text{MSD}=0.05\text{cm}$.

Step 2. With 50 vernier divisions covering $2.45\text{cm}$:

Step 3. Least count $\text{LC}=1\text{MSD}-1\text{VSD}=0.05-0.049=0.001\text{cm}$.

Step 4. Reading $=\text{MSR}+(\text{VSR}\times \text{LC})=5.10+(24\times 0.001)=5.124\text{cm}$.

Solution

If $N$ VSDs $=(N-1)$ MSDs, then

So

Solution

$1\text{VSD}=\frac{19}{20}\text{mm}=0.95\text{mm}$.

$\text{LC}=1\text{MSD}-1\text{VSD}=1-0.95=0.05\text{mm}$.

Length $=35+(5\times 0.05)=35.25\text{mm}$.

Solution

$\text{LC}=\frac{\text{Pitch}}{N}=\frac{0.5}{50}=0.01\text{mm}$.

Each main scale division is the pitch ($0.5\text{mm}$) so MSR $=3\times 0.5=1.5\text{mm}$.

Diameter $=\text{MSR}+(\text{CSR}\times \text{LC})=1.5+(35\times 0.01)=1.85\text{mm}$.

(Re-reading the problem with main scale reading interpreted as $3.5\text{mm}$ on the linear scale, the answer becomes $3.5+0.35=3.85\text{mm}$. Always confirm what "main scale reads $n$ divisions" means in context.)

Solution

$\text{LC}=\frac{0.5\text{mm}}{100}=0.005\text{mm}=0.0005\text{cm}$.

Diameter $=0+(52\times 0.005)=0.26\text{mm}=0.026\text{cm}$.

Solution

Observed reading $=0.20+(25\times 0.001)=0.225\text{cm}$.

Positive zero error means the instrument reads $5\times 0.001=0.005\text{cm}$ too high, so subtract:

Corrected diameter $=0.225-0.005=0.220\text{cm}$.

Solution

$=\frac{0.01}{1.00}\times 100+2\times \frac{0.1}{2.0}\times 100=1+10=11\%$.

Solution

Apply the powers rule:

$=1(1)+2(2)+3(3)=1+4+9=14\%$.

Solution

Powers in the formula: $a^2,b^3,c^1$ in the denominator, $d^{1/2}$ under the root.

$=2(1)+3(2)+1(3)+\frac{1}{2}(4)=2+6+3+2=13\%$.

Solution

Mean $\bar{a}=\frac{5.42+5.46+5.40+5.47+5.45}{5}=5.44\text{cm}$.

Absolute deviations: $|0.02|,|0.02|,|0.04|,|0.03|,|0.01|$.

Solution

$V\propto R^3$, so

Solution

Area $A=\ell b$, so

Solution

Leading zeros do not count. The significant digits are $5$, $8$ and the trailing $0$ (after the decimal). That gives 3 significant figures.

Solution

Arithmetic sum $=9.9999$. Addition keeps the fewest decimal places of any term. $9.99$ has 2 decimal places. So we round to 2 decimal places: $10.00$. But by significant-figure rules, the number of sig figs cannot exceed the precision of the least precise input, which has 3 sig figs ($9.99$). The correctly reported value is $10.0$ (3 sig figs).

Solution

Multiplication keeps the fewest sig figs. $4.5$ has 2 sig figs, so the answer must have 2 sig figs.

$1.236\times 4.5=5.562$, rounded to $5.6$.

Solution

In scientific notation, only the digits in the mantissa count. $6.022$ has 4 significant figures.

Solution

Pressure $=\frac{\text{Force}}{\text{Area}}=\frac{[ML{T}^{-2}]}{[L^2]}=[M{L}^{-1}{T}^{-2}]$.

Solution

Planck's constant: $E=h\nu$, so $[h]=[E][T]=[M{L}^2{T}^{-2}][T]=[M{L}^2{T}^{-1}]$.

Angular momentum $L=mvr$ has $[L]=[M][L{T}^{-1}][L]=[M{L}^2{T}^{-1}]$. Same dimensions.

Solution

From electromagnetism, $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$, where $c$ is the speed of light.

Therefore the expression has the dimensions of speed: $[L{T}^{-1}]$.

Solution

Work $W=F\cdot d$ has $[M{L}^2{T}^{-2}]$.

Torque $\tau =r\times F$ also has $[M{L}^2{T}^{-2}]$.

The two are dimensionally identical though physically different (work is a scalar, torque is a vector).

Solution

$\sigma =\frac{P/A}{T^4}$.

$P/A$ has dimensions $\frac{[M{L}^2{T}^{-3}]}{[L^2]}=[M{T}^{-3}]$.

Dividing by $[T^4]$ (here $T$ is temperature, denoted $\Theta$): $[\sigma ]=[M{T}^{-3}{\Theta }^{-4}]$.

Solution

From Newton's law of gravitation $F=\frac{G{m}_1{m}_2}{r^2}$, so $G=\frac{F{r}^2}{m_1{m}_2}$.

$[G]=\frac{[ML{T}^{-2}][L^2]}{[M][M]}=[M^{-1}{L}^3{T}^{-2}]$.

Solution

The expression $\frac{1}{2}C{V}^2$ is the energy stored in a capacitor, so its dimensions are those of energy: $[M{L}^2{T}^{-2}]$.

Solution

$\sin \theta$ is dimensionless, so $\frac{a{t}^2}{x}$ must be dimensionless.

$[a]=\frac{[x]}{[t^2]}=\frac{[L]}{[T^2]}=[L{T}^{-2}]$.

Solution

Assume $\nu =k{F}^a{\ell }^b{\mu }^c$. Comparing dimensions:

$[T^{-1}]=[ML{T}^{-2}{]}^a[L{]}^b[M{L}^{-1}{]}^c$.

Mass: $a+c=0$. Length: $a+b-c=0$. Time: $-2a=-1\Rightarrow a=1/2$.

Solving: $c=-1/2$, $b=-1$. So $\nu \propto \frac{1}{\ell }\sqrt{\frac{F}{\mu }}$.

Solution

Femto = $10^{-15}$. So $1\text{fs}=10^{-15}\text{s}$.

Solution

Luminous intensity is one of the seven SI base quantities. Its unit is the candela (cd).

Solution

Distance $=c\times t=(3\times 10^8\text{m/s})\times (3.156\times 10^7\text{s/year})\approx 9.46\times 10^{15}\text{m}$.

Solution

Velocity numerical value is inversely proportional to the unit of velocity. New unit of velocity $=\frac{2L}{T/2}=\frac{4L}{T}$, four times the old unit.

If the unit grows 4×, the numerical value shrinks to 1/4.

Solution

$C=2\pi R$, so $\frac{\Delta C}{C}=\frac{\Delta R}{R}$.

$\frac{0.02}{3.05}\times 100\approx 0.656\%\approx 0.66\%$.

Solution

Strain $=\frac{\text{change in length}}{\text{original length}}$ — same dimensions in numerator and denominator, so dimensionless. Stress, modulus of elasticity and force per unit area all have dimensions $[M{L}^{-1}{T}^{-2}]$.

Solution

Solving for ${\epsilon }_0$: ${\epsilon }_0=\frac{Q^2}{4\pi ER}$. Replacing dimensions:

$[{\epsilon }_0]=\frac{[AT{]}^2}{[M{L}^2{T}^{-2}][L]}=\frac{[A^2{T}^2]}{[M{L}^3{T}^{-2}]}=[M^{-1}{L}^{-3}{T}^4{A}^2]$.