139941
Find the value of \(\frac{1.53 \times 0.9995}{1.592}\) with due regard for significant figures
1 0.961
2 0.123
3 0.921
4 0.913
Explanation:
A Significant figure rule \(\frac{1.53 \times 0.9995}{1.592}=\frac{1.529}{1.592}=\frac{1.53}{1.592}=0.961\) \((9,6,1)\) There are three significant figures.
BITSAT-2007
Units and Measurements
139942
How many significant figures are there in 0.30100?
1 1
2 3
3 5
4 None of these
Explanation:
C No. of significant figure in 0.30100 \(=(3,0,1,0,0)\) \(=(5 \text { significant figure })\)
BITSAT-2010
Units and Measurements
139945
The number of significant figures in 0.002305 is
1 6
2 4
3 7
4 2
5 3
Explanation:
B The number of significant figures in 0.002305 Digits \(=(2,3,0,5)=4\) significant figures \(0 . \underline{002305}\) Non-significant figure (Left zeros)
Kerala CEE - 2010
Units and Measurements
139947
The mass of a metal cube is \(5.74 \mathrm{~g}\) and its volume is \(1.2 \mathrm{~cm}^{3}\). Then its density expressed up to appropriate significant figures will be
1 \(4.8 \mathrm{~g} \mathrm{~cm}^{-3}\)
2 \(4.78 \mathrm{~g} \mathrm{~cm}^{-3}\)
3 \(4.783 \mathrm{~g} \mathrm{~cm}^{-3}\)
4 \(5.0 \mathrm{~g} \mathrm{~cm}^{-3}\)
Explanation:
A As we know that, density \(\rho=\frac{\mathrm{m}}{\mathrm{V}}\) Given, \(\mathrm{m}=5.74 \mathrm{~g}, \mathrm{~V}=1.2 \mathrm{~cm}^{3}\) \(\begin{aligned} \therefore \quad \rho =\frac{5.74}{1.2}=4.78333 \mathrm{~g} / \mathrm{cm}^{3} \\ \approx 4.8 \mathrm{~g} / \mathrm{cm}^{3}\end{aligned}\)
SCRA-2013
Units and Measurements
139948
The length and breadth of a metal sheet are \(3.124 \mathrm{~m}\) and \(3.002 \mathrm{~m}\) respectively. The area of this sheet upto four correct significant figure is:
1 \(9.378 \mathrm{~m}^{2}\)
2 \(9.37 \mathrm{~m}^{2}\)
3 \(9.378248 \mathrm{~m}^{2}\)
4 \(9.3782 \mathrm{~m}^{2}\)
Explanation:
A Given, Length \(=3.124 \mathrm{~m}\) \(\text { Breadth }=3.002 \mathrm{~m} \) \(=\mathrm{L} \times \mathrm{B} \\ \text { Area } \) \(=3.124 \times 3.002=9.378 \mathrm{~m}^2\) F. measuring Instrument
139941
Find the value of \(\frac{1.53 \times 0.9995}{1.592}\) with due regard for significant figures
1 0.961
2 0.123
3 0.921
4 0.913
Explanation:
A Significant figure rule \(\frac{1.53 \times 0.9995}{1.592}=\frac{1.529}{1.592}=\frac{1.53}{1.592}=0.961\) \((9,6,1)\) There are three significant figures.
BITSAT-2007
Units and Measurements
139942
How many significant figures are there in 0.30100?
1 1
2 3
3 5
4 None of these
Explanation:
C No. of significant figure in 0.30100 \(=(3,0,1,0,0)\) \(=(5 \text { significant figure })\)
BITSAT-2010
Units and Measurements
139945
The number of significant figures in 0.002305 is
1 6
2 4
3 7
4 2
5 3
Explanation:
B The number of significant figures in 0.002305 Digits \(=(2,3,0,5)=4\) significant figures \(0 . \underline{002305}\) Non-significant figure (Left zeros)
Kerala CEE - 2010
Units and Measurements
139947
The mass of a metal cube is \(5.74 \mathrm{~g}\) and its volume is \(1.2 \mathrm{~cm}^{3}\). Then its density expressed up to appropriate significant figures will be
1 \(4.8 \mathrm{~g} \mathrm{~cm}^{-3}\)
2 \(4.78 \mathrm{~g} \mathrm{~cm}^{-3}\)
3 \(4.783 \mathrm{~g} \mathrm{~cm}^{-3}\)
4 \(5.0 \mathrm{~g} \mathrm{~cm}^{-3}\)
Explanation:
A As we know that, density \(\rho=\frac{\mathrm{m}}{\mathrm{V}}\) Given, \(\mathrm{m}=5.74 \mathrm{~g}, \mathrm{~V}=1.2 \mathrm{~cm}^{3}\) \(\begin{aligned} \therefore \quad \rho =\frac{5.74}{1.2}=4.78333 \mathrm{~g} / \mathrm{cm}^{3} \\ \approx 4.8 \mathrm{~g} / \mathrm{cm}^{3}\end{aligned}\)
SCRA-2013
Units and Measurements
139948
The length and breadth of a metal sheet are \(3.124 \mathrm{~m}\) and \(3.002 \mathrm{~m}\) respectively. The area of this sheet upto four correct significant figure is:
1 \(9.378 \mathrm{~m}^{2}\)
2 \(9.37 \mathrm{~m}^{2}\)
3 \(9.378248 \mathrm{~m}^{2}\)
4 \(9.3782 \mathrm{~m}^{2}\)
Explanation:
A Given, Length \(=3.124 \mathrm{~m}\) \(\text { Breadth }=3.002 \mathrm{~m} \) \(=\mathrm{L} \times \mathrm{B} \\ \text { Area } \) \(=3.124 \times 3.002=9.378 \mathrm{~m}^2\) F. measuring Instrument
139941
Find the value of \(\frac{1.53 \times 0.9995}{1.592}\) with due regard for significant figures
1 0.961
2 0.123
3 0.921
4 0.913
Explanation:
A Significant figure rule \(\frac{1.53 \times 0.9995}{1.592}=\frac{1.529}{1.592}=\frac{1.53}{1.592}=0.961\) \((9,6,1)\) There are three significant figures.
BITSAT-2007
Units and Measurements
139942
How many significant figures are there in 0.30100?
1 1
2 3
3 5
4 None of these
Explanation:
C No. of significant figure in 0.30100 \(=(3,0,1,0,0)\) \(=(5 \text { significant figure })\)
BITSAT-2010
Units and Measurements
139945
The number of significant figures in 0.002305 is
1 6
2 4
3 7
4 2
5 3
Explanation:
B The number of significant figures in 0.002305 Digits \(=(2,3,0,5)=4\) significant figures \(0 . \underline{002305}\) Non-significant figure (Left zeros)
Kerala CEE - 2010
Units and Measurements
139947
The mass of a metal cube is \(5.74 \mathrm{~g}\) and its volume is \(1.2 \mathrm{~cm}^{3}\). Then its density expressed up to appropriate significant figures will be
1 \(4.8 \mathrm{~g} \mathrm{~cm}^{-3}\)
2 \(4.78 \mathrm{~g} \mathrm{~cm}^{-3}\)
3 \(4.783 \mathrm{~g} \mathrm{~cm}^{-3}\)
4 \(5.0 \mathrm{~g} \mathrm{~cm}^{-3}\)
Explanation:
A As we know that, density \(\rho=\frac{\mathrm{m}}{\mathrm{V}}\) Given, \(\mathrm{m}=5.74 \mathrm{~g}, \mathrm{~V}=1.2 \mathrm{~cm}^{3}\) \(\begin{aligned} \therefore \quad \rho =\frac{5.74}{1.2}=4.78333 \mathrm{~g} / \mathrm{cm}^{3} \\ \approx 4.8 \mathrm{~g} / \mathrm{cm}^{3}\end{aligned}\)
SCRA-2013
Units and Measurements
139948
The length and breadth of a metal sheet are \(3.124 \mathrm{~m}\) and \(3.002 \mathrm{~m}\) respectively. The area of this sheet upto four correct significant figure is:
1 \(9.378 \mathrm{~m}^{2}\)
2 \(9.37 \mathrm{~m}^{2}\)
3 \(9.378248 \mathrm{~m}^{2}\)
4 \(9.3782 \mathrm{~m}^{2}\)
Explanation:
A Given, Length \(=3.124 \mathrm{~m}\) \(\text { Breadth }=3.002 \mathrm{~m} \) \(=\mathrm{L} \times \mathrm{B} \\ \text { Area } \) \(=3.124 \times 3.002=9.378 \mathrm{~m}^2\) F. measuring Instrument
139941
Find the value of \(\frac{1.53 \times 0.9995}{1.592}\) with due regard for significant figures
1 0.961
2 0.123
3 0.921
4 0.913
Explanation:
A Significant figure rule \(\frac{1.53 \times 0.9995}{1.592}=\frac{1.529}{1.592}=\frac{1.53}{1.592}=0.961\) \((9,6,1)\) There are three significant figures.
BITSAT-2007
Units and Measurements
139942
How many significant figures are there in 0.30100?
1 1
2 3
3 5
4 None of these
Explanation:
C No. of significant figure in 0.30100 \(=(3,0,1,0,0)\) \(=(5 \text { significant figure })\)
BITSAT-2010
Units and Measurements
139945
The number of significant figures in 0.002305 is
1 6
2 4
3 7
4 2
5 3
Explanation:
B The number of significant figures in 0.002305 Digits \(=(2,3,0,5)=4\) significant figures \(0 . \underline{002305}\) Non-significant figure (Left zeros)
Kerala CEE - 2010
Units and Measurements
139947
The mass of a metal cube is \(5.74 \mathrm{~g}\) and its volume is \(1.2 \mathrm{~cm}^{3}\). Then its density expressed up to appropriate significant figures will be
1 \(4.8 \mathrm{~g} \mathrm{~cm}^{-3}\)
2 \(4.78 \mathrm{~g} \mathrm{~cm}^{-3}\)
3 \(4.783 \mathrm{~g} \mathrm{~cm}^{-3}\)
4 \(5.0 \mathrm{~g} \mathrm{~cm}^{-3}\)
Explanation:
A As we know that, density \(\rho=\frac{\mathrm{m}}{\mathrm{V}}\) Given, \(\mathrm{m}=5.74 \mathrm{~g}, \mathrm{~V}=1.2 \mathrm{~cm}^{3}\) \(\begin{aligned} \therefore \quad \rho =\frac{5.74}{1.2}=4.78333 \mathrm{~g} / \mathrm{cm}^{3} \\ \approx 4.8 \mathrm{~g} / \mathrm{cm}^{3}\end{aligned}\)
SCRA-2013
Units and Measurements
139948
The length and breadth of a metal sheet are \(3.124 \mathrm{~m}\) and \(3.002 \mathrm{~m}\) respectively. The area of this sheet upto four correct significant figure is:
1 \(9.378 \mathrm{~m}^{2}\)
2 \(9.37 \mathrm{~m}^{2}\)
3 \(9.378248 \mathrm{~m}^{2}\)
4 \(9.3782 \mathrm{~m}^{2}\)
Explanation:
A Given, Length \(=3.124 \mathrm{~m}\) \(\text { Breadth }=3.002 \mathrm{~m} \) \(=\mathrm{L} \times \mathrm{B} \\ \text { Area } \) \(=3.124 \times 3.002=9.378 \mathrm{~m}^2\) F. measuring Instrument
139941
Find the value of \(\frac{1.53 \times 0.9995}{1.592}\) with due regard for significant figures
1 0.961
2 0.123
3 0.921
4 0.913
Explanation:
A Significant figure rule \(\frac{1.53 \times 0.9995}{1.592}=\frac{1.529}{1.592}=\frac{1.53}{1.592}=0.961\) \((9,6,1)\) There are three significant figures.
BITSAT-2007
Units and Measurements
139942
How many significant figures are there in 0.30100?
1 1
2 3
3 5
4 None of these
Explanation:
C No. of significant figure in 0.30100 \(=(3,0,1,0,0)\) \(=(5 \text { significant figure })\)
BITSAT-2010
Units and Measurements
139945
The number of significant figures in 0.002305 is
1 6
2 4
3 7
4 2
5 3
Explanation:
B The number of significant figures in 0.002305 Digits \(=(2,3,0,5)=4\) significant figures \(0 . \underline{002305}\) Non-significant figure (Left zeros)
Kerala CEE - 2010
Units and Measurements
139947
The mass of a metal cube is \(5.74 \mathrm{~g}\) and its volume is \(1.2 \mathrm{~cm}^{3}\). Then its density expressed up to appropriate significant figures will be
1 \(4.8 \mathrm{~g} \mathrm{~cm}^{-3}\)
2 \(4.78 \mathrm{~g} \mathrm{~cm}^{-3}\)
3 \(4.783 \mathrm{~g} \mathrm{~cm}^{-3}\)
4 \(5.0 \mathrm{~g} \mathrm{~cm}^{-3}\)
Explanation:
A As we know that, density \(\rho=\frac{\mathrm{m}}{\mathrm{V}}\) Given, \(\mathrm{m}=5.74 \mathrm{~g}, \mathrm{~V}=1.2 \mathrm{~cm}^{3}\) \(\begin{aligned} \therefore \quad \rho =\frac{5.74}{1.2}=4.78333 \mathrm{~g} / \mathrm{cm}^{3} \\ \approx 4.8 \mathrm{~g} / \mathrm{cm}^{3}\end{aligned}\)
SCRA-2013
Units and Measurements
139948
The length and breadth of a metal sheet are \(3.124 \mathrm{~m}\) and \(3.002 \mathrm{~m}\) respectively. The area of this sheet upto four correct significant figure is:
1 \(9.378 \mathrm{~m}^{2}\)
2 \(9.37 \mathrm{~m}^{2}\)
3 \(9.378248 \mathrm{~m}^{2}\)
4 \(9.3782 \mathrm{~m}^{2}\)
Explanation:
A Given, Length \(=3.124 \mathrm{~m}\) \(\text { Breadth }=3.002 \mathrm{~m} \) \(=\mathrm{L} \times \mathrm{B} \\ \text { Area } \) \(=3.124 \times 3.002=9.378 \mathrm{~m}^2\) F. measuring Instrument
