367654
The length, breadth and thickness of a block are given by \(l = 12\,cm\), \(b = 6\,cm\) and \(t = 2.45\,cm\). The volume of the block according to the idea of significant figures should be
Volume \(V = l \times b \times t\) \( = 12 \times 6 \times 2.45 = 176.4c{m^3}\) \(V = 1.764 \times {10^2}c{m^3}\) Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure.
PHXI02:UNITS AND MEASUREMENTS
367655
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:
\(\frac{{1.53 \times 0.9995}}{{1.592}} = \frac{{1.529235}}{{1.592}}\) \( = 0.9605747 = 0.961\) (Rounding off to three significant digits)
PHXI02:UNITS AND MEASUREMENTS
367656
More the number of significant figures shows more the
1 Accuracy
2 Error
3 Number of figures
4 Value
Explanation:
Conceptual Question
PHXI02:UNITS AND MEASUREMENTS
367657
If \({x=0.72+0.8+3.87-1.089}\), then number of significant digits in the value of \({x}\) is
1 1
2 4
3 2
4 6
Explanation:
\(x = 0.72 + 0.8 + 3.87 - 1.089\) \( = 0.7 + 0.8 + 3.9 - 1.1 = 4.3\) Thus, the number of significant digit is 2.
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PHXI02:UNITS AND MEASUREMENTS
367654
The length, breadth and thickness of a block are given by \(l = 12\,cm\), \(b = 6\,cm\) and \(t = 2.45\,cm\). The volume of the block according to the idea of significant figures should be
Volume \(V = l \times b \times t\) \( = 12 \times 6 \times 2.45 = 176.4c{m^3}\) \(V = 1.764 \times {10^2}c{m^3}\) Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure.
PHXI02:UNITS AND MEASUREMENTS
367655
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:
\(\frac{{1.53 \times 0.9995}}{{1.592}} = \frac{{1.529235}}{{1.592}}\) \( = 0.9605747 = 0.961\) (Rounding off to three significant digits)
PHXI02:UNITS AND MEASUREMENTS
367656
More the number of significant figures shows more the
1 Accuracy
2 Error
3 Number of figures
4 Value
Explanation:
Conceptual Question
PHXI02:UNITS AND MEASUREMENTS
367657
If \({x=0.72+0.8+3.87-1.089}\), then number of significant digits in the value of \({x}\) is
1 1
2 4
3 2
4 6
Explanation:
\(x = 0.72 + 0.8 + 3.87 - 1.089\) \( = 0.7 + 0.8 + 3.9 - 1.1 = 4.3\) Thus, the number of significant digit is 2.
367654
The length, breadth and thickness of a block are given by \(l = 12\,cm\), \(b = 6\,cm\) and \(t = 2.45\,cm\). The volume of the block according to the idea of significant figures should be
Volume \(V = l \times b \times t\) \( = 12 \times 6 \times 2.45 = 176.4c{m^3}\) \(V = 1.764 \times {10^2}c{m^3}\) Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure.
PHXI02:UNITS AND MEASUREMENTS
367655
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:
\(\frac{{1.53 \times 0.9995}}{{1.592}} = \frac{{1.529235}}{{1.592}}\) \( = 0.9605747 = 0.961\) (Rounding off to three significant digits)
PHXI02:UNITS AND MEASUREMENTS
367656
More the number of significant figures shows more the
1 Accuracy
2 Error
3 Number of figures
4 Value
Explanation:
Conceptual Question
PHXI02:UNITS AND MEASUREMENTS
367657
If \({x=0.72+0.8+3.87-1.089}\), then number of significant digits in the value of \({x}\) is
1 1
2 4
3 2
4 6
Explanation:
\(x = 0.72 + 0.8 + 3.87 - 1.089\) \( = 0.7 + 0.8 + 3.9 - 1.1 = 4.3\) Thus, the number of significant digit is 2.
367654
The length, breadth and thickness of a block are given by \(l = 12\,cm\), \(b = 6\,cm\) and \(t = 2.45\,cm\). The volume of the block according to the idea of significant figures should be
Volume \(V = l \times b \times t\) \( = 12 \times 6 \times 2.45 = 176.4c{m^3}\) \(V = 1.764 \times {10^2}c{m^3}\) Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure.
PHXI02:UNITS AND MEASUREMENTS
367655
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:
\(\frac{{1.53 \times 0.9995}}{{1.592}} = \frac{{1.529235}}{{1.592}}\) \( = 0.9605747 = 0.961\) (Rounding off to three significant digits)
PHXI02:UNITS AND MEASUREMENTS
367656
More the number of significant figures shows more the
1 Accuracy
2 Error
3 Number of figures
4 Value
Explanation:
Conceptual Question
PHXI02:UNITS AND MEASUREMENTS
367657
If \({x=0.72+0.8+3.87-1.089}\), then number of significant digits in the value of \({x}\) is
1 1
2 4
3 2
4 6
Explanation:
\(x = 0.72 + 0.8 + 3.87 - 1.089\) \( = 0.7 + 0.8 + 3.9 - 1.1 = 4.3\) Thus, the number of significant digit is 2.