139936
The result of multiplication of \(\mathbf{1 0 7 . 8 8}\) and \(\mathbf{0 . 6 1 0}\) is :
1 65.8068
2 65.807
3 65.81
4 65.8
Explanation:
D Given that, \(107.88 \times 0.610=65.8068\) Considering significant units, the minimum number of significant units in the given data is 3 . \(\Rightarrow\) Product is 65.8 .
SRMJEE - 2008
Units and Measurements
139916
The final result of the sum of the numbers 523.32, 1.21525 and 107.3 rounded to correct significant figures is :
1 631.8
2 631.835
3 631.83
4 631.8352
5 631.83524
Explanation:
A from the question - \(523.32+1.21524+107.3\) \(=523.32000+1.21524+107.30000\) \(=631.83524=631.8\) The final result of the sum of the numbers rounded to correct significant figures is 631.8
Kerala CEE 04.07.2022
Units and Measurements
139917
The number of significant figures in quantity \(0.00005041 \mathrm{~J}\) is
1 9
2 4
3 3
4 10
Explanation:
B \(0.00005041=5041 \times 10^{-8}\) Significant digit \(=5041\) Only 4 significant figures
AP EAMCET-07.07.2022
Units and Measurements
139918
If \(N_{A}, N_{B}\) and \(N_{C}\) are the number of significant figures in \(A=0.001204 \mathrm{~m}, B=43120000 \mathrm{~m}\) and \(\mathrm{C}=1.200 \mathrm{~m}\) respectively then
139936
The result of multiplication of \(\mathbf{1 0 7 . 8 8}\) and \(\mathbf{0 . 6 1 0}\) is :
1 65.8068
2 65.807
3 65.81
4 65.8
Explanation:
D Given that, \(107.88 \times 0.610=65.8068\) Considering significant units, the minimum number of significant units in the given data is 3 . \(\Rightarrow\) Product is 65.8 .
SRMJEE - 2008
Units and Measurements
139916
The final result of the sum of the numbers 523.32, 1.21525 and 107.3 rounded to correct significant figures is :
1 631.8
2 631.835
3 631.83
4 631.8352
5 631.83524
Explanation:
A from the question - \(523.32+1.21524+107.3\) \(=523.32000+1.21524+107.30000\) \(=631.83524=631.8\) The final result of the sum of the numbers rounded to correct significant figures is 631.8
Kerala CEE 04.07.2022
Units and Measurements
139917
The number of significant figures in quantity \(0.00005041 \mathrm{~J}\) is
1 9
2 4
3 3
4 10
Explanation:
B \(0.00005041=5041 \times 10^{-8}\) Significant digit \(=5041\) Only 4 significant figures
AP EAMCET-07.07.2022
Units and Measurements
139918
If \(N_{A}, N_{B}\) and \(N_{C}\) are the number of significant figures in \(A=0.001204 \mathrm{~m}, B=43120000 \mathrm{~m}\) and \(\mathrm{C}=1.200 \mathrm{~m}\) respectively then
139936
The result of multiplication of \(\mathbf{1 0 7 . 8 8}\) and \(\mathbf{0 . 6 1 0}\) is :
1 65.8068
2 65.807
3 65.81
4 65.8
Explanation:
D Given that, \(107.88 \times 0.610=65.8068\) Considering significant units, the minimum number of significant units in the given data is 3 . \(\Rightarrow\) Product is 65.8 .
SRMJEE - 2008
Units and Measurements
139916
The final result of the sum of the numbers 523.32, 1.21525 and 107.3 rounded to correct significant figures is :
1 631.8
2 631.835
3 631.83
4 631.8352
5 631.83524
Explanation:
A from the question - \(523.32+1.21524+107.3\) \(=523.32000+1.21524+107.30000\) \(=631.83524=631.8\) The final result of the sum of the numbers rounded to correct significant figures is 631.8
Kerala CEE 04.07.2022
Units and Measurements
139917
The number of significant figures in quantity \(0.00005041 \mathrm{~J}\) is
1 9
2 4
3 3
4 10
Explanation:
B \(0.00005041=5041 \times 10^{-8}\) Significant digit \(=5041\) Only 4 significant figures
AP EAMCET-07.07.2022
Units and Measurements
139918
If \(N_{A}, N_{B}\) and \(N_{C}\) are the number of significant figures in \(A=0.001204 \mathrm{~m}, B=43120000 \mathrm{~m}\) and \(\mathrm{C}=1.200 \mathrm{~m}\) respectively then
139936
The result of multiplication of \(\mathbf{1 0 7 . 8 8}\) and \(\mathbf{0 . 6 1 0}\) is :
1 65.8068
2 65.807
3 65.81
4 65.8
Explanation:
D Given that, \(107.88 \times 0.610=65.8068\) Considering significant units, the minimum number of significant units in the given data is 3 . \(\Rightarrow\) Product is 65.8 .
SRMJEE - 2008
Units and Measurements
139916
The final result of the sum of the numbers 523.32, 1.21525 and 107.3 rounded to correct significant figures is :
1 631.8
2 631.835
3 631.83
4 631.8352
5 631.83524
Explanation:
A from the question - \(523.32+1.21524+107.3\) \(=523.32000+1.21524+107.30000\) \(=631.83524=631.8\) The final result of the sum of the numbers rounded to correct significant figures is 631.8
Kerala CEE 04.07.2022
Units and Measurements
139917
The number of significant figures in quantity \(0.00005041 \mathrm{~J}\) is
1 9
2 4
3 3
4 10
Explanation:
B \(0.00005041=5041 \times 10^{-8}\) Significant digit \(=5041\) Only 4 significant figures
AP EAMCET-07.07.2022
Units and Measurements
139918
If \(N_{A}, N_{B}\) and \(N_{C}\) are the number of significant figures in \(A=0.001204 \mathrm{~m}, B=43120000 \mathrm{~m}\) and \(\mathrm{C}=1.200 \mathrm{~m}\) respectively then
