Units And MeasurementClass 11 Physics NCERT Solutions

Calculation: Side, $a = 1 \text{ cm} = 10^{-2} \text{ m}$. Volume, $V = a^3 = (10^{-2} \text{ m})^3 = 10^{-6} \text{ m}^3$.

Given: Radius, $r = 2.0 \text{ cm}$ Height, $h = 10.0 \text{ cm}$

Formula: Total surface area of a cylinder, $A = 2\pi r(r+h)$.

Calculation: $A = 2 \times \pi \times 2.0 \text{ cm} \times (2.0 \text{ cm} + 10.0 \text{ cm})$ $A = 2 \times \pi \times 2.0 \times (12.0) \text{ cm}^2 = 48\pi \text{ cm}^2 \approx 150.8 \text{ cm}^2$. Since the radius (2.0 cm) has only two significant figures, the result must be rounded to two significant figures. $A = 1.5 \times 10^2 \text{ cm}^2$. Converting to $(\text{mm})^2$: $1 \text{ cm} = 10 \text{ mm}$, so $1 \text{ cm}^2 = 100 \text{ mm}^2$. $A = 1.5 \times 10^2 \times 100 \text{ mm}^2 = 1.5 \times 10^4 \text{ mm}^2$.

Calculation: Speed, $v = 18 \text{ km h}^{-1} = 18 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 5 \text{ m/s}$. Distance covered in 1 s is 5 m.

Calculation: Density of water is $1 \text{ g cm}^{-3}$ or $1000 \text{ kg m}^{-3}$. Density of lead = Relative density $\times$ Density of water. In $\text{g cm}^{-3}$: Density = $11.3 \times 1 \text{ g cm}^{-3} = 11.3 \text{ g cm}^{-3}$. In $\text{kg m}^{-3}$: Density = $11.3 \times 1000 \text{ kg m}^{-3} = 11300 \text{ kg m}^{-3} = 1.13 \times 10^4 \text{ kg m}^{-3}$.

Calculation: $1 \text{ kg} = 10^3 \text{ g}$ $1 \text{ m} = 10^2 \text{ cm} \implies 1 \text{ m}^2 = (10^2 \text{ cm})^2 = 10^4 \text{ cm}^2$ $1 \text{ kg m}^2 \text{ s}^{-2} = (10^3 \text{ g}) \times (10^4 \text{ cm}^2) \times \text{s}^{-2} = 10^7 \text{ g cm}^2 \text{ s}^{-2}$.

Calculation: 1 light year (ly) = Speed of light $\times$ 1 year $1 \text{ ly} = (3 \times 10^8 \text{ m/s}) \times (3.156 \times 10^7 \text{ s}) = 9.468 \times 10^{15} \text{ m}$. $1 \text{ m} = \frac{1}{9.468 \times 10^{15}} \text{ ly} = 1.057 \times 10^{-16} \text{ ly}$.

Calculation: $1 \text{ m} = 10^{-3} \text{ km}$ $1 \text{ s} = \frac{1}{3600} \text{ h} \implies 1 \text{ s}^{-2} = (3600)^2 \text{ h}^{-2} = 12960000 \text{ h}^{-2}$ $3.0 \text{ m s}^{-2} = 3.0 \times (10^{-3} \text{ km}) \times (12960000 \text{ h}^{-2}) = 3.0 \times 12960 \text{ km h}^{-2} = 38880 \text{ km h}^{-2}$. Rounding to two significant figures, we get $3.9 \times 10^4 \text{ km h}^{-2}$.

Calculation: First, express N in base units: $1 \text{ N} = 1 \text{ kg m s}^{-2}$. So, the unit of G is $(\text{kg m s}^{-2})(\text{m}^2)(\text{kg}^{-2}) = \text{m}^3 \text{ s}^{-2} \text{ kg}^{-1}$. Now, convert units: $1 \text{ m}^3 = (100 \text{ cm})^3 = 10^6 \text{ cm}^3$. $1 \text{ kg}^{-1} = (1000 \text{ g})^{-1} = 10^{-3} \text{ g}^{-1}$. So, $\frac{\text{m}^3}{\text{kg s}^2} = \frac{10^6 \text{ cm}^3}{10^3 \text{ g s}^2} = 10^3 \frac{\text{cm}^3}{\text{g s}^2}$. $G = 6.67 \times 10^{-11} \times 10^3 \text{ cm}^3 \text{ s}^{-2} \text{ g}^{-1} = 6.67 \times 10^{-8} \text{ cm}^3 \text{ s}^{-2} \text{ g}^{-1}$.

Given: Energy, $Q = 4.2 \text{ J} = 4.2 \text{ kg m}^2 \text{ s}^{-2}$. New unit of mass, $M' = \alpha \text{ kg}$. New unit of length, $L' = \beta \text{ m}$. New unit of time, $T' = \gamma \text{ s}$.

To Show: The magnitude of a calorie in the new system is $4.2 \alpha^{-1} \beta^{-2} \gamma^{2}$.

Formula: The dimensional formula for energy is $[E] = [M^1 L^2 T^{-2}]$. For unit conversion, we use the relation $n_1 U_1 = n_2 U_2$, where $n$ is the numerical value and $U$ is the unit. $n_2 = n_1 \frac{U_1}{U_2} = n_1 \left[\frac{M_1}{M_2}\right]^a \left[\frac{L_1}{L_2}\right]^b \left[\frac{T_1}{T_2}\right]^c$.

Calculation: Here, $n_1 = 4.2$. The old units are $M_1 = 1 \text{ kg}$, $L_1 = 1 \text{ m}$, $T_1 = 1 \text{ s}$. The new units are $M_2 = \alpha \text{ kg}$, $L_2 = \beta \text{ m}$, $T_2 = \gamma \text{ s}$. The exponents from the dimensional formula are $a=1, b=2, c=-2$.

Substituting these values into the conversion formula: $n_2 = 4.2 \left[\frac{1 \text{ kg}}{\alpha \text{ kg}}\right]^1 \left[\frac{1 \text{ m}}{\beta \text{ m}}\right]^2 \left[\frac{1 \text{ s}}{\gamma \text{ s}}\right]^{-2}$ $n_2 = 4.2 \left[\frac{1}{\alpha}\right]^1 \left[\frac{1}{\beta}\right]^2 \left[\frac{1}{\gamma}\right]^{-2}$ $n_2 = 4.2 (\alpha^{-1}) (\beta^{-2}) (\gamma^{2})$ $n_2 = 4.2 \alpha^{-1} \beta^{-2} \gamma^{2}$

Conclusion: Thus, a calorie has a magnitude of $4.2 \alpha^{-1} \beta^{-2} \gamma^{2}$ in the new system of units. This is shown as required.

The statement "To call a dimensional quantity 'large' or 'small' is meaningless without specifying a standard for comparison" means that the attributes 'large' or 'small' are relative. A physical quantity can only be judged as large or small when it is compared with another physical quantity of the same kind, which acts as a reference or standard. For example, the height of a mountain is large compared to the height of a building, but it is small compared to the radius of the Earth.

Reframed statements:

(a) atoms are very small objects Reframe: The size of an atom is much smaller than the size of the tip of a pin.

(b) a jet plane moves with great speed Reframe: The speed of a jet plane is much greater than the speed of a bicycle.

(c) the mass of Jupiter is very large Reframe: The mass of Jupiter is very large compared to the mass of the Earth.

(d) the air inside this room contains a large number of molecules Reframe: The number of molecules in the air inside this room is much larger than the number of people that can fit in the room.

(e) a proton is much more massive than an electron No reframing needed. This statement is already a comparison between the mass of a proton and the mass of an electron. It is a well-defined and meaningful statement.

(f) the speed of sound is much smaller than the speed of light No reframing needed. This statement explicitly compares the speed of sound to the speed of light, which is a standard for comparison. It is a meaningful statement.

Given: Speed of light in the new system, $c = 1$ new unit of length per second. Time taken for light to travel from the Sun to the Earth, $t = 8 \text{ min} 20 \text{ s}$.

To Find: The distance between the Sun and the Earth in terms of the new unit of length.

Calculation: First, convert the time into seconds: $t = 8 \text{ min} \times 60 \text{ s/min} + 20 \text{ s} = 480 \text{ s} + 20 \text{ s} = 500 \text{ s}$.

The relationship between distance, speed, and time is: Distance = Speed $\times$ Time

Using the values in the new system of units: Distance = $c \times t$ Distance = $(1 \text{ new unit of length/s}) \times (500 \text{ s})$ Distance = $500$ new units of length.

Final Answer: The distance between the Sun and the Earth is 500 new units of length.

The precision of a measuring device is determined by its least count. A smaller least count implies a more precise measurement.

Let's calculate the least count for each device:

(a) A vernier callipers with 20 divisions on the sliding scale: The least count (LC) of a vernier callipers is given by the value of one main scale division (MSD) divided by the number of vernier scale divisions (VSD). Assuming 1 MSD = 1 mm, and 20 VSD coincide with 19 MSD: LC = $\frac{1 \text{ MSD}}{20} = \frac{1 \text{ mm}}{20} = 0.05 \text{ mm}$.

(b) A screw gauge of pitch 1 mm and 100 divisions on the circular scale: The least count (LC) of a screw gauge is given by the pitch divided by the number of divisions on the circular scale. LC = $\frac{\text{Pitch}}{\text{No. of circular scale divisions}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}$.

(c) An optical instrument that can measure length to within a wavelength of light: The wavelength of visible light is in the range of 400 nm to 700 nm. Let's take an average wavelength, $\lambda \approx 500 \text{ nm}$. $1 \text{ nm} = 10^{-6} \text{ mm}$. So, the least count is approximately $500 \times 10^{-6} \text{ mm} = 0.0005 \text{ mm}$.

Conclusion: The optical instrument has the smallest least count. Therefore, it is the most precise device for measuring length among the three options.

Given: Magnification of the microscope, $M = 100$. Observed average width of the hair (image size), $w_{observed} = 3.5 \text{ mm}$.

To Find: The estimate on the thickness of the hair (actual size).

Formula: Magnification is the ratio of the size of the image to the size of the object. $M = \frac{\text{Observed width}}{\text{Actual thickness}}$

Calculation: Rearranging the formula to find the actual thickness: Actual thickness = $\frac{\text{Observed width}}{M}$ Actual thickness = $\frac{3.5 \text{ mm}}{100} = 0.035 \text{ mm}$.

Final Answer: The estimate on the thickness of the hair is $0.035 \text{ mm}$.

(a) Estimating the diameter of a thread: To estimate the diameter of the thread using a metre scale, one can wind the thread closely on a uniform cylindrical object, like a pencil or a rod, ensuring that the turns touch each other without overlapping. Let's say we make 'n' such turns (e.g., n=50). Then, measure the total length 'L' of the winding along the axis of the cylinder using the metre scale. The diameter of the thread 'd' can then be estimated as: $d = \frac{L}{n}$ By making 'n' large, the measurement error in 'L' is distributed, leading to a more accurate estimate of 'd'.

(b) Increasing the accuracy of a screw gauge: No, it is not possible to increase the accuracy of a screw gauge arbitrarily by increasing the number of divisions on the circular scale. While theoretically, increasing the number of divisions decreases the least count (LC = Pitch / No. of divisions) and thus increases precision, there are practical limitations:

(c) Reliability of multiple measurements: A set of 100 measurements is expected to yield a more reliable estimate than a set of 5 measurements because it significantly reduces the effect of random errors. Random errors are unpredictable variations in measurements that can be positive or negative. When a large number of readings are taken, these random errors tend to cancel each other out upon averaging. The mean of a larger set of data is statistically more likely to be closer to the true value of the quantity being measured. This principle, based on the central limit theorem, makes the estimate more reliable and reduces the uncertainty in the final result.

Given: Area of the object (house on the slide), $A_o = 1.75 \text{ cm}^2$. Area of the image (house on the screen), $A_i = 1.55 \text{ m}^2$.

To Find: The linear magnification, $M_L$.

Formula: Areal magnification, $M_A = \frac{\text{Area of image}}{\text{Area of object}} = \frac{A_i}{A_o}$. Linear magnification, $M_L$, is related to areal magnification by $M_A = (M_L)^2$. Therefore, $M_L = \sqrt{M_A}$.

Calculation: First, we need to express both areas in the same unit. Let's convert $A_i$ to $\text{cm}^2$. $1 \text{ m} = 100 \text{ cm}$, so $1 \text{ m}^2 = (100 \text{ cm})^2 = 10000 \text{ cm}^2 = 10^4 \text{ cm}^2$. $A_i = 1.55 \text{ m}^2 = 1.55 \times 10^4 \text{ cm}^2$.

Now, calculate the areal magnification: $M_A = \frac{1.55 \times 10^4 \text{ cm}^2}{1.75 \text{ cm}^2} \approx 8857.14$.

Finally, calculate the linear magnification: $M_L = \sqrt{M_A} = \sqrt{8857.14} \approx 94.11$.

Rounding to three significant figures (as in the given data): $M_L = 94.1$.

Final Answer: The linear magnification of the projector-screen arrangement is 94.1.

Given: Length, $l = 4.234 \text{ m}$ (4 significant figures) Breadth, $b = 1.005 \text{ m}$ (4 significant figures) Thickness, $t = 2.01 \text{ cm} = 0.0201 \text{ m}$ (3 significant figures)

The least number of significant figures in the given measurements is 3. Therefore, the final results for area and volume must be rounded to 3 significant figures.

Area of the sheet: Formula: Total surface area $A = 2(lb + bt + tl)$.

Calculation: $lb = 4.234 \text{ m} \times 1.005 \text{ m} = 4.25517 \text{ m}^2$ $bt = 1.005 \text{ m} \times 0.0201 \text{ m} = 0.0202005 \text{ m}^2$ $tl = 0.0201 \text{ m} \times 4.234 \text{ m} = 0.0851034 \text{ m}^2$ $A = 2(4.25517 + 0.0202005 + 0.0851034) = 2(4.3604739) = 8.7209478 \text{ m}^2$. Rounding the final answer to 3 significant figures: $A = 8.72 \text{ m}^2$.

Volume of the sheet: Formula: Volume $V = l \times b \times t$.

Calculation: $V = 4.234 \text{ m} \times 1.005 \text{ m} \times 0.0201 \text{ m} = 0.0855289 \text{ m}^3$. Rounding the final answer to 3 significant figures: $V = 0.0855 \text{ m}^3$.

Final Answer: The area of the sheet is $8.72 \text{ m}^2$ and the volume is $0.0855 \text{ m}^3$.

Given: Mass of the box, $m_{box} = 2.30 \text{ kg}$. Mass of the first gold piece, $m_1 = 20.15 \text{ g} = 0.02015 \text{ kg}$. Mass of the second gold piece, $m_2 = 20.17 \text{ g} = 0.02017 \text{ kg}$.

(a) The total mass of the box: Calculation: Total mass, $M_{total} = m_{box} + m_1 + m_2$. $M_{total} = 2.30 \text{ kg} + 0.02015 \text{ kg} + 0.02017 \text{ kg} = 2.34032 \text{ kg}$.

According to the rule for addition, the final result should be rounded to the same number of decimal places as the number with the least number of decimal places. Here, $m_{box} = 2.30 \text{ kg}$ has two decimal places. So, we round the total mass to two decimal places. $M_{total} = 2.34 \text{ kg}$.

Final Answer (a): The total mass of the box is $2.34 \text{ kg}$.

(b) The difference in the masses of the pieces: Calculation: Difference in masses, $\Delta m = m_2 - m_1$. $\Delta m = 20.17 \text{ g} - 20.15 \text{ g} = 0.02 \text{ g}$.

According to the rule for subtraction, the final result should be rounded to the same number of decimal places as the numbers involved. Both masses have two decimal places, so the result should also have two decimal places. The result $0.02 \text{ g}$ is already correct.

Final Answer (b): The difference in the masses of the pieces is $0.02 \text{ g}$.

The given relation is $m = \frac{m_{o}}{(1-v^{2})^{1/2}}$. We can use the principle of dimensional homogeneity to correct the formula. According to this principle, quantities can be added or subtracted only if they have the same dimensions. Also, the arguments of functions like square root must be dimensionless.

In the denominator, the term $(1 - v^2)$ involves subtracting $v^2$ from the number 1.

Let's check the dimensions of $v^2$:

To make $v^2$ dimensionless, we must divide it by another quantity that has the same dimensions, i.e., $[L^2 T^{-2}]$. The only other relevant physical constant with the dimensions of speed is the speed of light, $c$, which also has dimensions $[L T^{-1}]$. Therefore, $c^2$ has dimensions $[L^2 T^{-2}]$.

By dividing $v^2$ by $c^2$, we get a dimensionless quantity: Dimensions of $\frac{v^2}{c^2} = \frac{[L^2 T^{-2}]}{[L^2 T^{-2}]} = [M^0 L^0 T^0]$, which is dimensionless.

So, the term inside the square root should be $(1 - \frac{v^2}{c^2})$.

Corrected Relation: The correct relation is: $m = \frac{m_{o}}{\sqrt{1 - \frac{v^2}{c^2}}}$

Given: Radius of a hydrogen atom, $r \approx 0.5 \AA = 0.5 \times 10^{-10} \text{ m}$. Avogadro's number, $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$.

To Find: The total atomic volume of one mole of hydrogen atoms.

Formula: Volume of a single spherical atom, $V_{atom} = \frac{4}{3}\pi r^3$. Total volume of one mole of atoms, $V_{total} = N_A \times V_{atom}$.

Calculation: First, calculate the volume of a single hydrogen atom: $V_{atom} = \frac{4}{3} \pi (0.5 \times 10^{-10} \text{ m})^3$ $V_{atom} = \frac{4}{3} \pi (0.125 \times 10^{-30} \text{ m}^3) \approx 0.5236 \times 10^{-30} \text{ m}^3$.

Now, calculate the total volume for one mole: $V_{total} = (6.022 \times 10^{23}) \times (0.5236 \times 10^{-30} \text{ m}^3)$ $V_{total} \approx 3.153 \times 10^{-7} \text{ m}^3$.

Since the given radius has one significant figure (0.5), the final answer should be rounded to one significant figure. $V_{total} \approx 3 \times 10^{-7} \text{ m}^3$.

Final Answer: The total atomic volume of a mole of hydrogen atoms is approximately $3 \times 10^{-7} \text{ m}^3$.

Given: Molar volume of an ideal gas at STP, $V_{molar} = 22.4 \text{ L}$. Size (radius) of a hydrogen molecule, $r \approx 1 \AA = 1 \times 10^{-10} \text{ m}$.

To Find: The ratio of molar volume to the atomic volume of a mole of hydrogen, and explain why it is large.

Reason for the large ratio: The ratio is very large because the molar volume refers to the volume occupied by the gas as a whole, which is mostly empty space. The atomic volume, on the other hand, is the sum of the actual volumes of the individual molecules. In a gas at STP, the molecules are very far apart from each other compared to their own size. The large ratio signifies that the average intermolecular distance in a gas is much greater than the molecular diameter.

Final Answer: The ratio of the molar volume to the atomic volume is approximately $8.9 \times 10^3$. This ratio is so large because in a gas, the volume is primarily empty space between the widely separated molecules.

This phenomenon is explained by the concept of parallax. Parallax is the apparent shift in the position of an object with respect to a background when viewed from two different lines of sight.

Expectation/Guess: Since the Sun is composed of hot plasma, which is a gaseous state of matter, one might initially expect its density to be in the range of gases. However, the Sun's immense mass creates an extremely strong gravitational force that compresses the matter towards its core. This compression would likely increase its average density significantly, possibly into the range of liquids and solids.

Given: Mass of the Sun, $M_{Sun} = 2.0 \times 10^{30} \text{ kg}$. Radius of the Sun, $R_{Sun} = 7.0 \times 10^{8} \text{ m}$.

To Find: The average mass density of the Sun, $\rho_{Sun}$.

Formula: Density, $\rho = \frac{\text{Mass}}{\text{Volume}}$. Volume of a sphere (the Sun), $V = \frac{4}{3}\pi R^3$.

Rounding to two significant figures (as in the given data): $\rho_{Sun} \approx 1.4 \times 10^3 \text{ kg/m}^3$.

Conclusion: The average density of the Sun is approximately $1400 \text{ kg/m}^3$. The density of water is $1000 \text{ kg/m}^3$, and typical densities of solids and liquids are in the range of $10^3$ to $10^4 \text{ kg/m}^3$. The calculated density of the Sun falls squarely within this range. Therefore, the guess that the Sun's density is in the range of solids and liquids is correct. This high density is due to the extreme gravitational compression, despite its high temperature.