367645
The radius of a sphere is 1.41 \(cm\). Its volume to an appropriate number of significant figures is
1 \(11.73\,c{m^3}\)
2 \(117\;c{m^3}\)
3 \(11.736\;c{m^3}\)
4 \(11.7\;c{m^3}\)
Explanation:
Radius of the sphere, \(r = 1.41\,cm\) (3 significant figures) Volume of the sphere, \(V = \frac{4}{3}\pi {r^3} = \frac{4}{3} \times 3.14 \times {\left( {1.41} \right)^3}\) \( = 11.736\,c{m^3}\) Round off up to 3 significant figure \( = 11.7\,c{m^3}\)
PHXI02:UNITS AND MEASUREMENTS
367646
The result of multiplication of 107.88 and 0.610 is:
1 65.8068
2 65.807
3 65.81
4 65.8
Explanation:
\({107.88 \times 0.610=65.8068}\) Considering significant units, the minimum number of significant units in the given data is 3 . \({\Rightarrow}\) Product is 65.8 So correct option is(4)
PHXI02:UNITS AND MEASUREMENTS
367647
The length and breath of a metal sheet are 3.124 \(m\) and 3.002 \(m\) respectively. What is the area of this sheet up to four correct significant figures?
1 \(9.3{m^2}\)
2 \(9.378{m^2}\)
3 \(9.43{m^2}\)
4 \(9.4{m^2}\)
Explanation:
Given length \((l) = 3.124\,m\) breath \((b) = 3.002 = 9.378248\,{m^2}\) We know that area of the sheet \((A) = l \times b\) \( = 3.124 \times 3.002 = 9.378248\,{m^2}\) Since both length and breadth have four significant figures, therefore area of the sheet after rounding off to four significant figures is \(9.378{m^2}\)
PHXI02:UNITS AND MEASUREMENTS
367648
A jeweller inserts a gem weighing 3.48 \(g\) into a box weighing 1.8 \(kg\). Find the total weight of the box and the gem to correct number of significant figures.
1 \(1.80\,kg\)
2 \(1.8\,kg\)
3 \(1.85\,kg\)
4 \(1.9\,kg\)
Explanation:
Total weight \( = 1.8\,kg + 0.00348\,kg = 1.80348\,kg\) In finding the sum or difference, the result should have the number of decimal places equal to the smallest number of decimal places in individual quantities. Therefore, total weight \( = 1.8\,kg\) since only one decimal places should be used.
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PHXI02:UNITS AND MEASUREMENTS
367645
The radius of a sphere is 1.41 \(cm\). Its volume to an appropriate number of significant figures is
1 \(11.73\,c{m^3}\)
2 \(117\;c{m^3}\)
3 \(11.736\;c{m^3}\)
4 \(11.7\;c{m^3}\)
Explanation:
Radius of the sphere, \(r = 1.41\,cm\) (3 significant figures) Volume of the sphere, \(V = \frac{4}{3}\pi {r^3} = \frac{4}{3} \times 3.14 \times {\left( {1.41} \right)^3}\) \( = 11.736\,c{m^3}\) Round off up to 3 significant figure \( = 11.7\,c{m^3}\)
PHXI02:UNITS AND MEASUREMENTS
367646
The result of multiplication of 107.88 and 0.610 is:
1 65.8068
2 65.807
3 65.81
4 65.8
Explanation:
\({107.88 \times 0.610=65.8068}\) Considering significant units, the minimum number of significant units in the given data is 3 . \({\Rightarrow}\) Product is 65.8 So correct option is(4)
PHXI02:UNITS AND MEASUREMENTS
367647
The length and breath of a metal sheet are 3.124 \(m\) and 3.002 \(m\) respectively. What is the area of this sheet up to four correct significant figures?
1 \(9.3{m^2}\)
2 \(9.378{m^2}\)
3 \(9.43{m^2}\)
4 \(9.4{m^2}\)
Explanation:
Given length \((l) = 3.124\,m\) breath \((b) = 3.002 = 9.378248\,{m^2}\) We know that area of the sheet \((A) = l \times b\) \( = 3.124 \times 3.002 = 9.378248\,{m^2}\) Since both length and breadth have four significant figures, therefore area of the sheet after rounding off to four significant figures is \(9.378{m^2}\)
PHXI02:UNITS AND MEASUREMENTS
367648
A jeweller inserts a gem weighing 3.48 \(g\) into a box weighing 1.8 \(kg\). Find the total weight of the box and the gem to correct number of significant figures.
1 \(1.80\,kg\)
2 \(1.8\,kg\)
3 \(1.85\,kg\)
4 \(1.9\,kg\)
Explanation:
Total weight \( = 1.8\,kg + 0.00348\,kg = 1.80348\,kg\) In finding the sum or difference, the result should have the number of decimal places equal to the smallest number of decimal places in individual quantities. Therefore, total weight \( = 1.8\,kg\) since only one decimal places should be used.
367645
The radius of a sphere is 1.41 \(cm\). Its volume to an appropriate number of significant figures is
1 \(11.73\,c{m^3}\)
2 \(117\;c{m^3}\)
3 \(11.736\;c{m^3}\)
4 \(11.7\;c{m^3}\)
Explanation:
Radius of the sphere, \(r = 1.41\,cm\) (3 significant figures) Volume of the sphere, \(V = \frac{4}{3}\pi {r^3} = \frac{4}{3} \times 3.14 \times {\left( {1.41} \right)^3}\) \( = 11.736\,c{m^3}\) Round off up to 3 significant figure \( = 11.7\,c{m^3}\)
PHXI02:UNITS AND MEASUREMENTS
367646
The result of multiplication of 107.88 and 0.610 is:
1 65.8068
2 65.807
3 65.81
4 65.8
Explanation:
\({107.88 \times 0.610=65.8068}\) Considering significant units, the minimum number of significant units in the given data is 3 . \({\Rightarrow}\) Product is 65.8 So correct option is(4)
PHXI02:UNITS AND MEASUREMENTS
367647
The length and breath of a metal sheet are 3.124 \(m\) and 3.002 \(m\) respectively. What is the area of this sheet up to four correct significant figures?
1 \(9.3{m^2}\)
2 \(9.378{m^2}\)
3 \(9.43{m^2}\)
4 \(9.4{m^2}\)
Explanation:
Given length \((l) = 3.124\,m\) breath \((b) = 3.002 = 9.378248\,{m^2}\) We know that area of the sheet \((A) = l \times b\) \( = 3.124 \times 3.002 = 9.378248\,{m^2}\) Since both length and breadth have four significant figures, therefore area of the sheet after rounding off to four significant figures is \(9.378{m^2}\)
PHXI02:UNITS AND MEASUREMENTS
367648
A jeweller inserts a gem weighing 3.48 \(g\) into a box weighing 1.8 \(kg\). Find the total weight of the box and the gem to correct number of significant figures.
1 \(1.80\,kg\)
2 \(1.8\,kg\)
3 \(1.85\,kg\)
4 \(1.9\,kg\)
Explanation:
Total weight \( = 1.8\,kg + 0.00348\,kg = 1.80348\,kg\) In finding the sum or difference, the result should have the number of decimal places equal to the smallest number of decimal places in individual quantities. Therefore, total weight \( = 1.8\,kg\) since only one decimal places should be used.
367645
The radius of a sphere is 1.41 \(cm\). Its volume to an appropriate number of significant figures is
1 \(11.73\,c{m^3}\)
2 \(117\;c{m^3}\)
3 \(11.736\;c{m^3}\)
4 \(11.7\;c{m^3}\)
Explanation:
Radius of the sphere, \(r = 1.41\,cm\) (3 significant figures) Volume of the sphere, \(V = \frac{4}{3}\pi {r^3} = \frac{4}{3} \times 3.14 \times {\left( {1.41} \right)^3}\) \( = 11.736\,c{m^3}\) Round off up to 3 significant figure \( = 11.7\,c{m^3}\)
PHXI02:UNITS AND MEASUREMENTS
367646
The result of multiplication of 107.88 and 0.610 is:
1 65.8068
2 65.807
3 65.81
4 65.8
Explanation:
\({107.88 \times 0.610=65.8068}\) Considering significant units, the minimum number of significant units in the given data is 3 . \({\Rightarrow}\) Product is 65.8 So correct option is(4)
PHXI02:UNITS AND MEASUREMENTS
367647
The length and breath of a metal sheet are 3.124 \(m\) and 3.002 \(m\) respectively. What is the area of this sheet up to four correct significant figures?
1 \(9.3{m^2}\)
2 \(9.378{m^2}\)
3 \(9.43{m^2}\)
4 \(9.4{m^2}\)
Explanation:
Given length \((l) = 3.124\,m\) breath \((b) = 3.002 = 9.378248\,{m^2}\) We know that area of the sheet \((A) = l \times b\) \( = 3.124 \times 3.002 = 9.378248\,{m^2}\) Since both length and breadth have four significant figures, therefore area of the sheet after rounding off to four significant figures is \(9.378{m^2}\)
PHXI02:UNITS AND MEASUREMENTS
367648
A jeweller inserts a gem weighing 3.48 \(g\) into a box weighing 1.8 \(kg\). Find the total weight of the box and the gem to correct number of significant figures.
1 \(1.80\,kg\)
2 \(1.8\,kg\)
3 \(1.85\,kg\)
4 \(1.9\,kg\)
Explanation:
Total weight \( = 1.8\,kg + 0.00348\,kg = 1.80348\,kg\) In finding the sum or difference, the result should have the number of decimal places equal to the smallest number of decimal places in individual quantities. Therefore, total weight \( = 1.8\,kg\) since only one decimal places should be used.
