367619
Assertion : The number 1.203 has four significant figures and the number 0.0024 has two significant figures. Reason : All the non zero digits are significant.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The number of significant figures in 1.203 are four and the number of significant figures in 0.0024 are two. All non zero digits are significant. So, option (2) is correct.
PHXI02:UNITS AND MEASUREMENTS
367620
The number of significant figures in 0.06900 is
1 \(3\)
2 \(5\)
3 \(2\)
4 \(4\)
Explanation:
If the number is less than 1, the zero (\(s\)) on the right of decimal point and before the first non - zero digits are not significant. In 0.06900, the underlined zeros are not significant. Hence, number of significant figures are four.
PHXI02:UNITS AND MEASUREMENTS
367621
The external and internal diameters of a hollow cylinder are measured to be \(\left( {4.23 \pm 0.01} \right)cm\) and \(\left( {3.89 \pm 0.01} \right)cm\). The thickness of the wall of the cylinder is
367622
A cube has a side of length \(1.2 \times {10^{ - 2}}m.\) Its volume up to correct significant figures is
1 \(1.732 \times {10^{ - 6}}{m^3}\)
2 \(1.73 \times {10^{ - 6}}{m^3}\)
3 \(1.70 \times {10^{ - 6}}{m^3}\)
4 \(1.7 \times {10^{ - 6}}{m^3}\)
Explanation:
Here Length of the cube, \(L = 1.2 \times {10^{ - 2}}m\) Volume of the cube, \(V = {(1.2 \times {10^{ - 2}}m)^3}\) \( = 1.728 \times {10^{ - 6}}m\) as the result can have only two significant figures therefore, on rounding off, we get \(V = 1.7 \times {10^{ - 6}}{m^3}\)
PHXI02:UNITS AND MEASUREMENTS
367623
The mass of a box measured by a grocer’s balance is 2.3 \(kg\). Two gold pieces of masses 20.15 \(g\) and 20.17 \(g\) are added to the box. The total mass of the box is
1 \(2.34\,kg\)
2 \(2.30\,kg\)
3 \(2.340\,kg\)
4 \(2.3\,kg\)
Explanation:
Total mass \( = (2.3 + 0.02015 + 0.02017)\,kg\) \( = 2.34032,\) upto one decimal place \( = 2.3\,kg\).
367619
Assertion : The number 1.203 has four significant figures and the number 0.0024 has two significant figures. Reason : All the non zero digits are significant.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The number of significant figures in 1.203 are four and the number of significant figures in 0.0024 are two. All non zero digits are significant. So, option (2) is correct.
PHXI02:UNITS AND MEASUREMENTS
367620
The number of significant figures in 0.06900 is
1 \(3\)
2 \(5\)
3 \(2\)
4 \(4\)
Explanation:
If the number is less than 1, the zero (\(s\)) on the right of decimal point and before the first non - zero digits are not significant. In 0.06900, the underlined zeros are not significant. Hence, number of significant figures are four.
PHXI02:UNITS AND MEASUREMENTS
367621
The external and internal diameters of a hollow cylinder are measured to be \(\left( {4.23 \pm 0.01} \right)cm\) and \(\left( {3.89 \pm 0.01} \right)cm\). The thickness of the wall of the cylinder is
367622
A cube has a side of length \(1.2 \times {10^{ - 2}}m.\) Its volume up to correct significant figures is
1 \(1.732 \times {10^{ - 6}}{m^3}\)
2 \(1.73 \times {10^{ - 6}}{m^3}\)
3 \(1.70 \times {10^{ - 6}}{m^3}\)
4 \(1.7 \times {10^{ - 6}}{m^3}\)
Explanation:
Here Length of the cube, \(L = 1.2 \times {10^{ - 2}}m\) Volume of the cube, \(V = {(1.2 \times {10^{ - 2}}m)^3}\) \( = 1.728 \times {10^{ - 6}}m\) as the result can have only two significant figures therefore, on rounding off, we get \(V = 1.7 \times {10^{ - 6}}{m^3}\)
PHXI02:UNITS AND MEASUREMENTS
367623
The mass of a box measured by a grocer’s balance is 2.3 \(kg\). Two gold pieces of masses 20.15 \(g\) and 20.17 \(g\) are added to the box. The total mass of the box is
1 \(2.34\,kg\)
2 \(2.30\,kg\)
3 \(2.340\,kg\)
4 \(2.3\,kg\)
Explanation:
Total mass \( = (2.3 + 0.02015 + 0.02017)\,kg\) \( = 2.34032,\) upto one decimal place \( = 2.3\,kg\).
367619
Assertion : The number 1.203 has four significant figures and the number 0.0024 has two significant figures. Reason : All the non zero digits are significant.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The number of significant figures in 1.203 are four and the number of significant figures in 0.0024 are two. All non zero digits are significant. So, option (2) is correct.
PHXI02:UNITS AND MEASUREMENTS
367620
The number of significant figures in 0.06900 is
1 \(3\)
2 \(5\)
3 \(2\)
4 \(4\)
Explanation:
If the number is less than 1, the zero (\(s\)) on the right of decimal point and before the first non - zero digits are not significant. In 0.06900, the underlined zeros are not significant. Hence, number of significant figures are four.
PHXI02:UNITS AND MEASUREMENTS
367621
The external and internal diameters of a hollow cylinder are measured to be \(\left( {4.23 \pm 0.01} \right)cm\) and \(\left( {3.89 \pm 0.01} \right)cm\). The thickness of the wall of the cylinder is
367622
A cube has a side of length \(1.2 \times {10^{ - 2}}m.\) Its volume up to correct significant figures is
1 \(1.732 \times {10^{ - 6}}{m^3}\)
2 \(1.73 \times {10^{ - 6}}{m^3}\)
3 \(1.70 \times {10^{ - 6}}{m^3}\)
4 \(1.7 \times {10^{ - 6}}{m^3}\)
Explanation:
Here Length of the cube, \(L = 1.2 \times {10^{ - 2}}m\) Volume of the cube, \(V = {(1.2 \times {10^{ - 2}}m)^3}\) \( = 1.728 \times {10^{ - 6}}m\) as the result can have only two significant figures therefore, on rounding off, we get \(V = 1.7 \times {10^{ - 6}}{m^3}\)
PHXI02:UNITS AND MEASUREMENTS
367623
The mass of a box measured by a grocer’s balance is 2.3 \(kg\). Two gold pieces of masses 20.15 \(g\) and 20.17 \(g\) are added to the box. The total mass of the box is
1 \(2.34\,kg\)
2 \(2.30\,kg\)
3 \(2.340\,kg\)
4 \(2.3\,kg\)
Explanation:
Total mass \( = (2.3 + 0.02015 + 0.02017)\,kg\) \( = 2.34032,\) upto one decimal place \( = 2.3\,kg\).
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PHXI02:UNITS AND MEASUREMENTS
367619
Assertion : The number 1.203 has four significant figures and the number 0.0024 has two significant figures. Reason : All the non zero digits are significant.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The number of significant figures in 1.203 are four and the number of significant figures in 0.0024 are two. All non zero digits are significant. So, option (2) is correct.
PHXI02:UNITS AND MEASUREMENTS
367620
The number of significant figures in 0.06900 is
1 \(3\)
2 \(5\)
3 \(2\)
4 \(4\)
Explanation:
If the number is less than 1, the zero (\(s\)) on the right of decimal point and before the first non - zero digits are not significant. In 0.06900, the underlined zeros are not significant. Hence, number of significant figures are four.
PHXI02:UNITS AND MEASUREMENTS
367621
The external and internal diameters of a hollow cylinder are measured to be \(\left( {4.23 \pm 0.01} \right)cm\) and \(\left( {3.89 \pm 0.01} \right)cm\). The thickness of the wall of the cylinder is
367622
A cube has a side of length \(1.2 \times {10^{ - 2}}m.\) Its volume up to correct significant figures is
1 \(1.732 \times {10^{ - 6}}{m^3}\)
2 \(1.73 \times {10^{ - 6}}{m^3}\)
3 \(1.70 \times {10^{ - 6}}{m^3}\)
4 \(1.7 \times {10^{ - 6}}{m^3}\)
Explanation:
Here Length of the cube, \(L = 1.2 \times {10^{ - 2}}m\) Volume of the cube, \(V = {(1.2 \times {10^{ - 2}}m)^3}\) \( = 1.728 \times {10^{ - 6}}m\) as the result can have only two significant figures therefore, on rounding off, we get \(V = 1.7 \times {10^{ - 6}}{m^3}\)
PHXI02:UNITS AND MEASUREMENTS
367623
The mass of a box measured by a grocer’s balance is 2.3 \(kg\). Two gold pieces of masses 20.15 \(g\) and 20.17 \(g\) are added to the box. The total mass of the box is
1 \(2.34\,kg\)
2 \(2.30\,kg\)
3 \(2.340\,kg\)
4 \(2.3\,kg\)
Explanation:
Total mass \( = (2.3 + 0.02015 + 0.02017)\,kg\) \( = 2.34032,\) upto one decimal place \( = 2.3\,kg\).
367619
Assertion : The number 1.203 has four significant figures and the number 0.0024 has two significant figures. Reason : All the non zero digits are significant.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The number of significant figures in 1.203 are four and the number of significant figures in 0.0024 are two. All non zero digits are significant. So, option (2) is correct.
PHXI02:UNITS AND MEASUREMENTS
367620
The number of significant figures in 0.06900 is
1 \(3\)
2 \(5\)
3 \(2\)
4 \(4\)
Explanation:
If the number is less than 1, the zero (\(s\)) on the right of decimal point and before the first non - zero digits are not significant. In 0.06900, the underlined zeros are not significant. Hence, number of significant figures are four.
PHXI02:UNITS AND MEASUREMENTS
367621
The external and internal diameters of a hollow cylinder are measured to be \(\left( {4.23 \pm 0.01} \right)cm\) and \(\left( {3.89 \pm 0.01} \right)cm\). The thickness of the wall of the cylinder is
367622
A cube has a side of length \(1.2 \times {10^{ - 2}}m.\) Its volume up to correct significant figures is
1 \(1.732 \times {10^{ - 6}}{m^3}\)
2 \(1.73 \times {10^{ - 6}}{m^3}\)
3 \(1.70 \times {10^{ - 6}}{m^3}\)
4 \(1.7 \times {10^{ - 6}}{m^3}\)
Explanation:
Here Length of the cube, \(L = 1.2 \times {10^{ - 2}}m\) Volume of the cube, \(V = {(1.2 \times {10^{ - 2}}m)^3}\) \( = 1.728 \times {10^{ - 6}}m\) as the result can have only two significant figures therefore, on rounding off, we get \(V = 1.7 \times {10^{ - 6}}{m^3}\)
PHXI02:UNITS AND MEASUREMENTS
367623
The mass of a box measured by a grocer’s balance is 2.3 \(kg\). Two gold pieces of masses 20.15 \(g\) and 20.17 \(g\) are added to the box. The total mass of the box is
1 \(2.34\,kg\)
2 \(2.30\,kg\)
3 \(2.340\,kg\)
4 \(2.3\,kg\)
Explanation:
Total mass \( = (2.3 + 0.02015 + 0.02017)\,kg\) \( = 2.34032,\) upto one decimal place \( = 2.3\,kg\).