CBSE Class 11 Physics Exemplar Chapter 1: Units and Measurements NCERT Solutions

Solution: To determine the number of significant figures in 0.06900, we apply the rules:

1. Non-zero digits are always significant.
2. Zeros between non-zero digits are significant.
3. Leading zeros (zeros to the left of the first non-zero digit) are not significant.
4. Trailing zeros (zeros to the right of the last non-zero digit) are significant if the number contains a decimal point.

\nIn the number 0.06900:

• The zeros before '6' (0.0) are leading zeros and are not significant.
• The digits '6' and '9' are non-zero and are significant.
• The two zeros after '9' are trailing zeros and appear after the decimal point, so they are significant.

\nTherefore, the significant figures are 6, 9, 0, 0. The total number of significant figures is 4.

Answer: (b) 4

Solution: When adding or subtracting numbers, the result should be rounded off to the same number of decimal places as the number with the fewest decimal places.

The given numbers are:

• 436.32 (2 decimal places)
• 227.2 (1 decimal place)
• 0.301 (3 decimal places)

\nFirst, we find the arithmetic sum:

$436.32\; +\; 227.2\; +\; 0.301\; =\; 663.821$

\nThe number with the least number of decimal places is 227.2, which has only one decimal place. Therefore, the sum should be rounded to one decimal place.

Rounding 663.821 to one decimal place gives 663.8.

Answer: (c) 663.8

Note: In addition and subtraction, the precision is determined by the number of decimal places, not the total number of significant figures.

Solution: Density is calculated using the formula: Density = Mass / Volume.

Given:

• Mass = 4.237 g (This number has 4 significant figures)
• Volume = 2.5 cm³ (This number has 2 significant figures)

\nWhen multiplying or dividing numbers, the result should have the same number of significant figures as the number with the fewest significant figures in the original data.

Calculate the density:

Density = $\backslash frac\{4.237\; \backslash text\{\; g\}\}\{2.5\; \backslash text\{\; cm\}^3\}$

Density = $1.6948\; \backslash text\{\; g\; cm\}^\{-3\}$

\nSince the volume (2.5 cm³) has the fewest significant figures (two), the density should be rounded to two significant figures.

Rounding 1.6948 to two significant figures gives 1.7.

Answer: (c) $1.7\; \backslash text\{\; g\; cm\}^\{-3\}$

Solution: We need to round both numbers to three significant figures.

For 2.745:

• The first three significant figures are 2, 7, and 4.
• The digit following the third significant figure (4) is 5.
• According to the rounding rule (round to the nearest even digit when the digit to be rounded is followed by 5), since the preceding digit '4' is even, it remains unchanged.

\nSo, 2.745 rounded to 3 significant figures is 2.74.

For 2.735:

• The first three significant figures are 2, 7, and 3.
• The digit following the third significant figure (3) is 5.
• According to the rounding rule, since the preceding digit '3' is odd, it is increased by one.

\nSo, 2.735 rounded to 3 significant figures is 2.74.

Answer: (d) 2.74 and 2.74

Solution: We are given the length and breadth of a rectangular sheet with their uncertainties.

Length, $l\; =\; (16.2\; \backslash pm\; 0.1)\; \backslash text\{\; cm\}$

Breadth, $b\; =\; (10.1\; \backslash pm\; 0.1)\; \backslash text\{\; cm\}$

1. Calculate the Area:

The area of a rectangle is given by $A\; =\; l\; \backslash times\; b$.

Nominal Area = $16.2\; \backslash text\{\; cm\}\; \backslash times\; 10.1\; \backslash text\{\; cm\}\; =\; 163.62\; \backslash text\{\; cm\}^2$

2. Determine Significant Figures for Area:

The length (16.2) has 3 significant figures, and the breadth (10.1) has 3 significant figures. Therefore, the area should be reported to 3 significant figures.

Rounding 163.62 to 3 significant figures gives $A\; =\; 164\; \backslash text\{\; cm\}^2$.

3. Calculate the Error in Area:

For multiplication, the relative error in the area is the sum of the relative errors in length and breadth.

$$\backslash frac\{\backslash Delta\; A\}\{A\}\; =\; \backslash frac\{\backslash Delta\; l\}\{l\}\; +\; \backslash frac\{\backslash Delta\; b\}\{b\}$$

Substitute the values:

$$\backslash frac\{\backslash Delta\; A\}\{A\}\; =\; \backslash frac\{0.1\}\{16.2\}\; +\; \backslash frac\{0.1\}\{10.1\}$$

To add these fractions, find a common denominator:

$$\backslash frac\{\backslash Delta\; A\}\{A\}\; =\; \backslash frac\{0.1\; \backslash times\; 10.1\; +\; 0.1\; \backslash times\; 16.2\}\{16.2\; \backslash times\; 10.1\}$$

$$\backslash frac\{\backslash Delta\; A\}\{A\}\; =\; \backslash frac\{1.01\; +\; 1.62\}\{163.62\}$$

$$\backslash frac\{\backslash Delta\; A\}\{A\}\; =\; \backslash frac\{2.63\}\{163.62\}$$

Now, calculate the absolute error $\backslash Delta\; A$:

$\backslash Delta\; A\; =\; A\; \backslash times\; \backslash frac\{2.63\}\{163.62\}$

$\backslash Delta\; A\; =\; 163.62\; \backslash times\; \backslash frac\{2.63\}\{163.62\}$

$\backslash Delta\; A\; =\; 2.63\; \backslash text\{\; cm\}^2$

4. Round the Error:

The absolute error should be rounded to one significant figure. Rounding 2.63 to one significant figure gives $\backslash Delta\; A\; =\; 3\; \backslash text\{\; cm\}^2$.

5. Express the Area with Error:

The area with error is expressed as $A\; \backslash pm\; \backslash Delta\; A$.

Area = $(164\; \backslash pm\; 3)\; \backslash text\{\; cm\}^2$.

Answer: (a) $164\; \backslash pm\; 3\; \backslash text\{\; cm\}^2$

Solution: We need to find the dimensional formula for each quantity in the given pairs.

Pair (a): Work and Torque

• Work = Force × Distance. Dimensions: $[MLT^\{-2\}]\; \backslash times\; [L]\; =\; [ML^2T^\{-2\}]$
• Torque = Force × Perpendicular distance. Dimensions: $[MLT^\{-2\}]\; \backslash times\; [L]\; =\; [ML^2T^\{-2\}]$

\nBoth have the same dimensions: $[ML^2T^\{-2\}]$.

Pair (b): Angular momentum and Planck's constant

• Angular momentum = $mvr$. Dimensions: $[M]\; \backslash times\; [LT^\{-1\}]\; \backslash times\; [L]\; =\; [ML^2T^\{-1\}]$
• Planck's constant ($h$) is related to energy ($E$) and frequency ($v$) by $E\; =\; hv$, so $h\; =\; E/v$. Dimensions: $\backslash frac\{[ML^2T^\{-2\}]\}\{[T^\{-1\}]\}\; =\; [ML^2T^\{-1\}]$

\nBoth have the same dimensions: $[ML^2T^\{-1\}]$.

Pair (c): Tension and Surface tension

• Tension is a force. Dimensions: $[MLT^\{-2\}]$
• Surface tension = Force / Length. Dimensions: $\backslash frac\{[MLT^\{-2\}]\}\{[L]\}\; =\; [MT^\{-2\}]$

\nTheir dimensions are different: $[MLT^\{-2\}]$ for tension and $[MT^\{-2\}]$ for surface tension.

Pair (d): Impulse and Linear momentum

• Impulse = Force × Time. Dimensions: $[MLT^\{-2\}]\; \backslash times\; [T]\; =\; [MLT^\{-1\}]$
• Linear momentum = Mass × Velocity. Dimensions: $[M]\; \backslash times\; [LT^\{-1\}]\; =\; [MLT^\{-1\}]$

\nBoth have the same dimensions: $[MLT^\{-1\}]$. \nTherefore, the pair that does not have the same dimensional formula is Tension and Surface tension.

Answer: (c) Tension and surface tension