367624
The radius of the earth is \(6.37 \times {10^6}m\) and its mass is \(5.975 \times {10^{24}}kg\). Find the earth’s average density to appropriate significant figures.
1 \(5.54 \times {10^3}kg{m^{ - 3}}\)
2 \(5.52 \times {10^3}kg{m^{ - 3}}\)
3 \(5.53 \times {10^3}kg{m^{ - 3}}\)
4 \(5.517 \times {10^3}kg{m^{ - 3}}\)
Explanation:
From the given informations \(M = 5.975 \times {10^{24}}kg\) \(R = 6.37 \times {10^6}m\) Average density \((D) = \frac{M}{V}\) \( = \frac{{5.975 \times {{10}^{24}}}}{{\frac{4}{3} \times (3.142) \times {{(6.37 \times {{10}^6})}^3}}}\) \( = 0.005517 \times {10^6}kg{m^{ - 3}}\) \( = 5.52 \times {10^3}kg{m^{ - 3}}\) (to three significant figures)
PHXI02:UNITS AND MEASUREMENTS
367625
The sum of numbers \(436.32,227.2\) and \(0.301\) in appropriate significant figures is
1 \(664\)
2 \(663.8\)
3 \(663.821\)
4 \(663.82\)
Explanation:
Sum of 436.32, 227.2 and 0.301 is 663.821. Rounding to one decimal place leads to 663.8
NCERT Exemplar
PHXI02:UNITS AND MEASUREMENTS
367626
The result after adding \(3.8 \times {10^{ - 6}}\) to \(4.2 \times {10^{ - 5}}\) with due regard to significant figure is
1 \(45.8 \times {10^{ - 6}}\)
2 \(4.58 \times {10^{ - 5}}\)
3 \(0.458 \times {10^{ - 4}}\)
4 \(4.6 \times {10^{ - 5}}\)
Explanation:
\(3.8 \times {10^{ - 6}} + 4.2 \times {10^{ - 5}}\) \( = \left( {3.8 \times {{10}^{ - 1}} + 4.2} \right) \times {10^{ - 5}}\) \( = \left( {0.38 + 4.2} \right) \times {10^{ - 5}}\) \( = \left( {4.58} \right){10^{ - 5}}\) Rounding off to one place of decimal. The sum \( = 4.6 \times {10^{ - 5}}\)
PHXI02:UNITS AND MEASUREMENTS
367627
What is the number of significant figure in \(\left( {3.20 + 4.80} \right) \times {10^5}?\)
1 \(4\)
2 \(5\)
3 \(2\)
4 \(3\)
Explanation:
\(\left( {3.20 + 4.80} \right) \times {10^5} = 8.00 \times {10^5}\) Number of significant figures are 3.
367624
The radius of the earth is \(6.37 \times {10^6}m\) and its mass is \(5.975 \times {10^{24}}kg\). Find the earth’s average density to appropriate significant figures.
1 \(5.54 \times {10^3}kg{m^{ - 3}}\)
2 \(5.52 \times {10^3}kg{m^{ - 3}}\)
3 \(5.53 \times {10^3}kg{m^{ - 3}}\)
4 \(5.517 \times {10^3}kg{m^{ - 3}}\)
Explanation:
From the given informations \(M = 5.975 \times {10^{24}}kg\) \(R = 6.37 \times {10^6}m\) Average density \((D) = \frac{M}{V}\) \( = \frac{{5.975 \times {{10}^{24}}}}{{\frac{4}{3} \times (3.142) \times {{(6.37 \times {{10}^6})}^3}}}\) \( = 0.005517 \times {10^6}kg{m^{ - 3}}\) \( = 5.52 \times {10^3}kg{m^{ - 3}}\) (to three significant figures)
PHXI02:UNITS AND MEASUREMENTS
367625
The sum of numbers \(436.32,227.2\) and \(0.301\) in appropriate significant figures is
1 \(664\)
2 \(663.8\)
3 \(663.821\)
4 \(663.82\)
Explanation:
Sum of 436.32, 227.2 and 0.301 is 663.821. Rounding to one decimal place leads to 663.8
NCERT Exemplar
PHXI02:UNITS AND MEASUREMENTS
367626
The result after adding \(3.8 \times {10^{ - 6}}\) to \(4.2 \times {10^{ - 5}}\) with due regard to significant figure is
1 \(45.8 \times {10^{ - 6}}\)
2 \(4.58 \times {10^{ - 5}}\)
3 \(0.458 \times {10^{ - 4}}\)
4 \(4.6 \times {10^{ - 5}}\)
Explanation:
\(3.8 \times {10^{ - 6}} + 4.2 \times {10^{ - 5}}\) \( = \left( {3.8 \times {{10}^{ - 1}} + 4.2} \right) \times {10^{ - 5}}\) \( = \left( {0.38 + 4.2} \right) \times {10^{ - 5}}\) \( = \left( {4.58} \right){10^{ - 5}}\) Rounding off to one place of decimal. The sum \( = 4.6 \times {10^{ - 5}}\)
PHXI02:UNITS AND MEASUREMENTS
367627
What is the number of significant figure in \(\left( {3.20 + 4.80} \right) \times {10^5}?\)
1 \(4\)
2 \(5\)
3 \(2\)
4 \(3\)
Explanation:
\(\left( {3.20 + 4.80} \right) \times {10^5} = 8.00 \times {10^5}\) Number of significant figures are 3.
367624
The radius of the earth is \(6.37 \times {10^6}m\) and its mass is \(5.975 \times {10^{24}}kg\). Find the earth’s average density to appropriate significant figures.
1 \(5.54 \times {10^3}kg{m^{ - 3}}\)
2 \(5.52 \times {10^3}kg{m^{ - 3}}\)
3 \(5.53 \times {10^3}kg{m^{ - 3}}\)
4 \(5.517 \times {10^3}kg{m^{ - 3}}\)
Explanation:
From the given informations \(M = 5.975 \times {10^{24}}kg\) \(R = 6.37 \times {10^6}m\) Average density \((D) = \frac{M}{V}\) \( = \frac{{5.975 \times {{10}^{24}}}}{{\frac{4}{3} \times (3.142) \times {{(6.37 \times {{10}^6})}^3}}}\) \( = 0.005517 \times {10^6}kg{m^{ - 3}}\) \( = 5.52 \times {10^3}kg{m^{ - 3}}\) (to three significant figures)
PHXI02:UNITS AND MEASUREMENTS
367625
The sum of numbers \(436.32,227.2\) and \(0.301\) in appropriate significant figures is
1 \(664\)
2 \(663.8\)
3 \(663.821\)
4 \(663.82\)
Explanation:
Sum of 436.32, 227.2 and 0.301 is 663.821. Rounding to one decimal place leads to 663.8
NCERT Exemplar
PHXI02:UNITS AND MEASUREMENTS
367626
The result after adding \(3.8 \times {10^{ - 6}}\) to \(4.2 \times {10^{ - 5}}\) with due regard to significant figure is
1 \(45.8 \times {10^{ - 6}}\)
2 \(4.58 \times {10^{ - 5}}\)
3 \(0.458 \times {10^{ - 4}}\)
4 \(4.6 \times {10^{ - 5}}\)
Explanation:
\(3.8 \times {10^{ - 6}} + 4.2 \times {10^{ - 5}}\) \( = \left( {3.8 \times {{10}^{ - 1}} + 4.2} \right) \times {10^{ - 5}}\) \( = \left( {0.38 + 4.2} \right) \times {10^{ - 5}}\) \( = \left( {4.58} \right){10^{ - 5}}\) Rounding off to one place of decimal. The sum \( = 4.6 \times {10^{ - 5}}\)
PHXI02:UNITS AND MEASUREMENTS
367627
What is the number of significant figure in \(\left( {3.20 + 4.80} \right) \times {10^5}?\)
1 \(4\)
2 \(5\)
3 \(2\)
4 \(3\)
Explanation:
\(\left( {3.20 + 4.80} \right) \times {10^5} = 8.00 \times {10^5}\) Number of significant figures are 3.
367624
The radius of the earth is \(6.37 \times {10^6}m\) and its mass is \(5.975 \times {10^{24}}kg\). Find the earth’s average density to appropriate significant figures.
1 \(5.54 \times {10^3}kg{m^{ - 3}}\)
2 \(5.52 \times {10^3}kg{m^{ - 3}}\)
3 \(5.53 \times {10^3}kg{m^{ - 3}}\)
4 \(5.517 \times {10^3}kg{m^{ - 3}}\)
Explanation:
From the given informations \(M = 5.975 \times {10^{24}}kg\) \(R = 6.37 \times {10^6}m\) Average density \((D) = \frac{M}{V}\) \( = \frac{{5.975 \times {{10}^{24}}}}{{\frac{4}{3} \times (3.142) \times {{(6.37 \times {{10}^6})}^3}}}\) \( = 0.005517 \times {10^6}kg{m^{ - 3}}\) \( = 5.52 \times {10^3}kg{m^{ - 3}}\) (to three significant figures)
PHXI02:UNITS AND MEASUREMENTS
367625
The sum of numbers \(436.32,227.2\) and \(0.301\) in appropriate significant figures is
1 \(664\)
2 \(663.8\)
3 \(663.821\)
4 \(663.82\)
Explanation:
Sum of 436.32, 227.2 and 0.301 is 663.821. Rounding to one decimal place leads to 663.8
NCERT Exemplar
PHXI02:UNITS AND MEASUREMENTS
367626
The result after adding \(3.8 \times {10^{ - 6}}\) to \(4.2 \times {10^{ - 5}}\) with due regard to significant figure is
1 \(45.8 \times {10^{ - 6}}\)
2 \(4.58 \times {10^{ - 5}}\)
3 \(0.458 \times {10^{ - 4}}\)
4 \(4.6 \times {10^{ - 5}}\)
Explanation:
\(3.8 \times {10^{ - 6}} + 4.2 \times {10^{ - 5}}\) \( = \left( {3.8 \times {{10}^{ - 1}} + 4.2} \right) \times {10^{ - 5}}\) \( = \left( {0.38 + 4.2} \right) \times {10^{ - 5}}\) \( = \left( {4.58} \right){10^{ - 5}}\) Rounding off to one place of decimal. The sum \( = 4.6 \times {10^{ - 5}}\)
PHXI02:UNITS AND MEASUREMENTS
367627
What is the number of significant figure in \(\left( {3.20 + 4.80} \right) \times {10^5}?\)
1 \(4\)
2 \(5\)
3 \(2\)
4 \(3\)
Explanation:
\(\left( {3.20 + 4.80} \right) \times {10^5} = 8.00 \times {10^5}\) Number of significant figures are 3.
