NCERT Solutions: Class 11 Physics Chapter 1 - Units and Measurement

a. $1\,\text{cm}=10^{-2}\,\text{m}$, so volume $=(10^{-2})^3=10^{-6}\,\text{m}^3$. b. In mm, $r=20\,\text{mm}$ and $h=100\,\text{mm}$. Total surface area $=2\pi r(r+h)=2\pi(20)(120)=4.8\times10^3\pi\approx1.5\times10^4\,\text{mm}^2$. c. $18\,\text{km h}^{-1}=18\times\frac{1000}{3600}=5\,\text{m s}^{-1}$, so distance in $1\,\text{s}$ is $5\,\text{m}$. d. Relative density equals density in $\text{g cm}^{-3}$ relative to water, so density is $11.3\,\text{g cm}^{-3}=11.3\times10^3\,\text{kg m}^{-3}$.

a. $1\,\text{kg}=10^3\,\text{g}$ and $1\,\text{m}^2=10^4\,\text{cm}^2$, so $1\,\text{kg m}^2\text{s}^{-2}=10^7\,\text{g cm}^2\text{s}^{-2}$. b. $1\,\text{ly}\approx9.46\times10^{15}\,\text{m}$, hence $1\,\text{m}=1/(9.46\times10^{15})\approx1.06\times10^{-16}\,\text{ly}$. c. $3.0\,\text{m s}^{-2}=3.0\times10^{-3}\,\text{km s}^{-2}=3.0\times10^{-3}\times(3600)^2\,\text{km h}^{-2}=3.9\times10^4\,\text{km h}^{-2}$. d. $G=6.67\times10^{-11}\,\text{m}^3\text{kg}^{-1}\text{s}^{-2}=6.67\times10^{-11}\times10^6\times10^{-3}=6.67\times10^{-8}\,\text{cm}^3\text{g}^{-1}\text{s}^{-2}$.

The new unit of energy is $(\alpha\,\text{kg})(\beta\,\text{m})^2(\gamma\,\text{s})^{-2}=\alpha\beta^2\gamma^{-2}\,\text{J}$. Since $1\,\text{cal}=4.2\,\text{J}$, its numerical value in the new unit is $4.2/(\alpha\beta^2\gamma^{-2})=4.2\alpha^{-1}\beta^{-2}\gamma^2$.

A physical quantity with dimensions must be compared with another quantity of the same kind. Reframed examples: (a) Atoms are very small compared with ordinary objects such as a metre-scale object. (b) A jet plane moves with great speed compared with a car or train. (c) The mass of Jupiter is very large compared with the mass of the Earth. (d) This is meaningful because number of molecules is a pure count. (e) This is meaningful because it compares two masses. (f) This is meaningful because it compares two speeds.

If the speed of light is unity in the new unit system, light travels $1$ new length unit per second. The given time is $8\,\text{min}\,20\,\text{s}=8\times60+20=500\,\text{s}$. Hence the Sun-Earth distance is $500$ new length units.

The vernier callipers typically has least count about $1\,\text{mm}/20=0.05\,\text{mm}$. The screw gauge has least count $1\,\text{mm}/100=0.01\,\text{mm}$. An optical instrument measuring within a wavelength of light can measure to about $10^{-7}\,\text{m}$, which is smaller than both mechanical least counts.

Magnification $=100$, so actual thickness $=$ observed width divided by magnification $=3.5\,\text{mm}/100=0.035\,\text{mm}=3.5\times10^{-5}\,\text{m}$.

a. Make $N$ close, non-overlapping turns of the thread around a cylindrical support, measure the total width $L$ of the turns with the metre scale, and estimate diameter $d=L/N$. b. Increasing circular-scale divisions reduces least count only up to practical limits; mechanical imperfections, backlash and reading errors prevent arbitrary accuracy. c. Repeated measurements reduce random errors because positive and negative deviations tend to cancel, so the mean of $100$ readings is more reliable than the mean of $5$ readings.

Area on screen $=1.55\,\text{m}^2=1.55\times10^4\,\text{cm}^2$. Area magnification $=\frac{1.55\times10^4}{1.75}=8.857\times10^3$. Linear magnification is the square root of area magnification: $m=\sqrt{8.857\times10^3}\approx94.1$.

Leading zeros are not significant. Non-zero digits are significant. Zeros between non-zero digits are significant, and trailing zeros after a decimal point are significant.

Area of the sheet $=4.234\times1.005=4.25517\,\text{m}^2$, which to $4$ significant figures is $4.255\,\text{m}^2$. Thickness $=2.01\,\text{cm}=0.0201\,\text{m}$. Volume $=4.234\times1.005\times0.0201=0.085528917\,\text{m}^3$, which to $3$ significant figures is $8.55\times10^{-2}\,\text{m}^3$.

Gold mass total $=20.15+20.17=40.32\,\text{g}=0.04032\,\text{kg}$. Total mass $=2.30+0.04032=2.34032\,\text{kg}$, rounded to the hundredth of kg as $2.34\,\text{kg}$. Difference $=20.17-20.15=0.02\,\text{g}$, kept to the hundredth of gram.

The term subtracted from $1$ must be dimensionless. Since $v$ and $c$ both have dimensions of speed, $v^2/c^2$ is dimensionless. Therefore the correct placement is inside the square root as $1-v^2/c^2$.

Taking the atom as a sphere of radius $0.5\,\text{Å}=0.5\times10^{-10}\,\text{m}$, volume of one atom is $\frac{4}{3}\pi r^3=\frac{4}{3}\pi(0.5\times10^{-10})^3\approx5.24\times10^{-31}\,\text{m}^3$. For one mole, multiply by $N_A\approx6.02\times10^{23}$: total volume $\approx3.2\times10^{-7}\,\text{m}^3$.

Molar volume of gas $=22.4\,\text{L}=2.24\times10^{-2}\,\text{m}^3$. Taking the molecular size $1\,\text{Å}$ as diameter gives radius $0.5\times10^{-10}\,\text{m}$. Volume of one molecule $=\frac{4}{3}\pi r^3\approx5.24\times10^{-31}\,\text{m}^3$. Volume for one mole $\approx5.24\times10^{-31}\times6.02\times10^{23}=3.15\times10^{-7}\,\text{m}^3$. Ratio $=2.24\times10^{-2}/3.15\times10^{-7}\approx7.1\times10^4$. The ratio is large because gas molecules are separated by large empty spaces compared with their own sizes.

The apparent motion depends on angular displacement. For the same train displacement, a nearby tree subtends a large changing angle at the eye, so it appears to sweep backwards rapidly. A distant hill, the Moon, or stars subtend almost the same direction because their distance is extremely large compared with the train's displacement, so they appear nearly stationary or moving with the observer.

Assuming the Sun is spherical, $\rho=\frac{M}{(4/3)\pi R^3}$. With $M=2.0\times10^{30}\,\text{kg}$ and $R=7.0\times10^8\,\text{m}$, volume $=\frac{4}{3}\pi(7.0\times10^8)^3\approx1.44\times10^{27}\,\text{m}^3$. Thus $\rho\approx\frac{2.0\times10^{30}}{1.44\times10^{27}}\approx1.4\times10^3\,\text{kg m}^{-3}$. This is comparable to liquid/solid densities, not ordinary gas density.