367658
Subtract 0.2 \(J\) from 7.26 \(J\) and express the result with correct number of significant figure.
1 \(7\,J\)
2 \(7.1\,J\)
3 \(7.06\,J\)
4 \(7.0\,J\)
Explanation:
Subtraction is correct upto one place of decimal, corresponding to the least number of decimal places.\(7.26 - 0.2 = 7.06 = 7.1\,J\)
PHXI02:UNITS AND MEASUREMENTS
367659
The length, breadth and thickness of a rectangular sheet of metal are 4.234 \(m\), 1.005 \(m\) and 2.01 \(cm\) respectively. The area and volume of the sheet to correct significant figures are
Given, length \((l) = 4.234\,m\) Breadth \((2) = 1.005\,m\) Thickness \((t) = 2.01\,cm = 0.0201\,m\) Area of sheet (1) \( = 2\left( {l \times b + b \times t + t \times l} \right)\) \( = \left[ {\left( {4.234 \times 1.005} \right) \times \left( {1.005 \times 0.0201} \right)} \right.\left. { + \left( {0.0201 \times 4.234} \right)} \right]\) \( = 2 \times 4.3604739\) \( = 8.7209478\,{m^2}\) As thickness has least number of significant figure 3, therefore rounding -off area up to three significant figure, we get Area of sheet \((1) = 8.72\,{m^2}\) Volume of sheet \((V) = l \times b \times t\) \( = 4.234 \times 1.005 \times 0.0201\) \( = 0.0855289\) Rounding-off up to three significant figures, we get Volume of the sheet \( = 0.0855{m^3}\)
NCERT Exemplar
PHXI02:UNITS AND MEASUREMENTS
367660
If \(L = 2.331\,cm\), \(B = 2.1\,cm\), then \(L + B = \)
1 \(4.4\,\,cm\)
2 \(4.43\,\,cm\)
3 \(4\,\,cm\)
4 \(4.431\,\,cm\)
Explanation:
Given, \(L = 2.331\,cm\) \( = 2.33\) (correct up to two decimal places) and \(B = 2.1\,cm = 2.10\,cm\) \(\therefore \,L + B = 2.33 + 2.10 = 4.43\,cm = 4.4\,cm\) Since minimum significant figure is 2.
367658
Subtract 0.2 \(J\) from 7.26 \(J\) and express the result with correct number of significant figure.
1 \(7\,J\)
2 \(7.1\,J\)
3 \(7.06\,J\)
4 \(7.0\,J\)
Explanation:
Subtraction is correct upto one place of decimal, corresponding to the least number of decimal places.\(7.26 - 0.2 = 7.06 = 7.1\,J\)
PHXI02:UNITS AND MEASUREMENTS
367659
The length, breadth and thickness of a rectangular sheet of metal are 4.234 \(m\), 1.005 \(m\) and 2.01 \(cm\) respectively. The area and volume of the sheet to correct significant figures are
Given, length \((l) = 4.234\,m\) Breadth \((2) = 1.005\,m\) Thickness \((t) = 2.01\,cm = 0.0201\,m\) Area of sheet (1) \( = 2\left( {l \times b + b \times t + t \times l} \right)\) \( = \left[ {\left( {4.234 \times 1.005} \right) \times \left( {1.005 \times 0.0201} \right)} \right.\left. { + \left( {0.0201 \times 4.234} \right)} \right]\) \( = 2 \times 4.3604739\) \( = 8.7209478\,{m^2}\) As thickness has least number of significant figure 3, therefore rounding -off area up to three significant figure, we get Area of sheet \((1) = 8.72\,{m^2}\) Volume of sheet \((V) = l \times b \times t\) \( = 4.234 \times 1.005 \times 0.0201\) \( = 0.0855289\) Rounding-off up to three significant figures, we get Volume of the sheet \( = 0.0855{m^3}\)
NCERT Exemplar
PHXI02:UNITS AND MEASUREMENTS
367660
If \(L = 2.331\,cm\), \(B = 2.1\,cm\), then \(L + B = \)
1 \(4.4\,\,cm\)
2 \(4.43\,\,cm\)
3 \(4\,\,cm\)
4 \(4.431\,\,cm\)
Explanation:
Given, \(L = 2.331\,cm\) \( = 2.33\) (correct up to two decimal places) and \(B = 2.1\,cm = 2.10\,cm\) \(\therefore \,L + B = 2.33 + 2.10 = 4.43\,cm = 4.4\,cm\) Since minimum significant figure is 2.
367658
Subtract 0.2 \(J\) from 7.26 \(J\) and express the result with correct number of significant figure.
1 \(7\,J\)
2 \(7.1\,J\)
3 \(7.06\,J\)
4 \(7.0\,J\)
Explanation:
Subtraction is correct upto one place of decimal, corresponding to the least number of decimal places.\(7.26 - 0.2 = 7.06 = 7.1\,J\)
PHXI02:UNITS AND MEASUREMENTS
367659
The length, breadth and thickness of a rectangular sheet of metal are 4.234 \(m\), 1.005 \(m\) and 2.01 \(cm\) respectively. The area and volume of the sheet to correct significant figures are
Given, length \((l) = 4.234\,m\) Breadth \((2) = 1.005\,m\) Thickness \((t) = 2.01\,cm = 0.0201\,m\) Area of sheet (1) \( = 2\left( {l \times b + b \times t + t \times l} \right)\) \( = \left[ {\left( {4.234 \times 1.005} \right) \times \left( {1.005 \times 0.0201} \right)} \right.\left. { + \left( {0.0201 \times 4.234} \right)} \right]\) \( = 2 \times 4.3604739\) \( = 8.7209478\,{m^2}\) As thickness has least number of significant figure 3, therefore rounding -off area up to three significant figure, we get Area of sheet \((1) = 8.72\,{m^2}\) Volume of sheet \((V) = l \times b \times t\) \( = 4.234 \times 1.005 \times 0.0201\) \( = 0.0855289\) Rounding-off up to three significant figures, we get Volume of the sheet \( = 0.0855{m^3}\)
NCERT Exemplar
PHXI02:UNITS AND MEASUREMENTS
367660
If \(L = 2.331\,cm\), \(B = 2.1\,cm\), then \(L + B = \)
1 \(4.4\,\,cm\)
2 \(4.43\,\,cm\)
3 \(4\,\,cm\)
4 \(4.431\,\,cm\)
Explanation:
Given, \(L = 2.331\,cm\) \( = 2.33\) (correct up to two decimal places) and \(B = 2.1\,cm = 2.10\,cm\) \(\therefore \,L + B = 2.33 + 2.10 = 4.43\,cm = 4.4\,cm\) Since minimum significant figure is 2.
