Using the concept of significant figures, (1) 1001 has 4 significant figure. (2) 010.1 has 3 significant figure. (3) 100.100 has 6 significant figure. (4) 0.0010010 has 5 significant figure. Correct answer is (1).
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367650
Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as \(4.62\,\,s,\) \(4.632\,\,s,\) \(4.6\,\,s,\) and \(4.64 s\). The arithmetic mean of these readings in correct significant figure is
1 \(4.6 s\)
2 \(4.62 s\)
3 \(5 s\)
4 \(4.623\,\,s\)
Explanation:
Arithmetic mean of time periods of oscillation of the pendulum, \(T=\dfrac{(4.62+4.632+4.6+4.64) s}{4}=4.623 s\) So, final answer with correct significant figure, \(T=4.6 s\). So, option (1) is correct.
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367651
The decimal equivalent of 1/20 up to three significant figures is
1 \(0.0050\)
2 \(0.0500\)
3 \(0.05000\)
4 \(5.0 \times {10^{ - 2}}\)
Explanation:
\(\frac{1}{{20}} = 0.05 = 0.0500\)
PHXI02:UNITS AND MEASUREMENTS
367652
The mass and volume of a body are 4.237 \(g\) and \(2.5\,c{m^3},\) respectively. The density of the material of the body in correct significant figures is
1 \(1.7g\,c{m^{ - 3}}\)
2 \(1.695\,g\,c{m^{ - 3}}\)
3 \(1.6048\,g\,c{m^{ - 3}}\)
4 \(1.69\,g\,c{m^{ - 3}}\)
Explanation:
In multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures. In this question, density should be reported to two significant figures. \({\mathop{\rm Density}\nolimits} = \frac{{4.237g}}{{2.5\,c{m^3}}} = 1.6498\) As rounding off the number, we get density \( = 1.7\,g\,c{m^{ - 3}}\)
PHXI02:UNITS AND MEASUREMENTS
367653
A long thread of length 1.23 \(m\) is cut to obtain a small thread of length 12.3 \(mm\). What is the new length of the long thread ?
1 \(1.23\,m\)
2 \(1.20\,m\)
3 \(1.30\,m\)
4 \(1.22\,m\)
Explanation:
New length \( = 1.23 - 12.3 \times {10^{ - 3}} = 1.1277\,m\) Least number of decimal places in any of the individual quantities is 2. Hence, new length \( = 1.22\,m\).
Using the concept of significant figures, (1) 1001 has 4 significant figure. (2) 010.1 has 3 significant figure. (3) 100.100 has 6 significant figure. (4) 0.0010010 has 5 significant figure. Correct answer is (1).
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367650
Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as \(4.62\,\,s,\) \(4.632\,\,s,\) \(4.6\,\,s,\) and \(4.64 s\). The arithmetic mean of these readings in correct significant figure is
1 \(4.6 s\)
2 \(4.62 s\)
3 \(5 s\)
4 \(4.623\,\,s\)
Explanation:
Arithmetic mean of time periods of oscillation of the pendulum, \(T=\dfrac{(4.62+4.632+4.6+4.64) s}{4}=4.623 s\) So, final answer with correct significant figure, \(T=4.6 s\). So, option (1) is correct.
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367651
The decimal equivalent of 1/20 up to three significant figures is
1 \(0.0050\)
2 \(0.0500\)
3 \(0.05000\)
4 \(5.0 \times {10^{ - 2}}\)
Explanation:
\(\frac{1}{{20}} = 0.05 = 0.0500\)
PHXI02:UNITS AND MEASUREMENTS
367652
The mass and volume of a body are 4.237 \(g\) and \(2.5\,c{m^3},\) respectively. The density of the material of the body in correct significant figures is
1 \(1.7g\,c{m^{ - 3}}\)
2 \(1.695\,g\,c{m^{ - 3}}\)
3 \(1.6048\,g\,c{m^{ - 3}}\)
4 \(1.69\,g\,c{m^{ - 3}}\)
Explanation:
In multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures. In this question, density should be reported to two significant figures. \({\mathop{\rm Density}\nolimits} = \frac{{4.237g}}{{2.5\,c{m^3}}} = 1.6498\) As rounding off the number, we get density \( = 1.7\,g\,c{m^{ - 3}}\)
PHXI02:UNITS AND MEASUREMENTS
367653
A long thread of length 1.23 \(m\) is cut to obtain a small thread of length 12.3 \(mm\). What is the new length of the long thread ?
1 \(1.23\,m\)
2 \(1.20\,m\)
3 \(1.30\,m\)
4 \(1.22\,m\)
Explanation:
New length \( = 1.23 - 12.3 \times {10^{ - 3}} = 1.1277\,m\) Least number of decimal places in any of the individual quantities is 2. Hence, new length \( = 1.22\,m\).
Using the concept of significant figures, (1) 1001 has 4 significant figure. (2) 010.1 has 3 significant figure. (3) 100.100 has 6 significant figure. (4) 0.0010010 has 5 significant figure. Correct answer is (1).
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367650
Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as \(4.62\,\,s,\) \(4.632\,\,s,\) \(4.6\,\,s,\) and \(4.64 s\). The arithmetic mean of these readings in correct significant figure is
1 \(4.6 s\)
2 \(4.62 s\)
3 \(5 s\)
4 \(4.623\,\,s\)
Explanation:
Arithmetic mean of time periods of oscillation of the pendulum, \(T=\dfrac{(4.62+4.632+4.6+4.64) s}{4}=4.623 s\) So, final answer with correct significant figure, \(T=4.6 s\). So, option (1) is correct.
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367651
The decimal equivalent of 1/20 up to three significant figures is
1 \(0.0050\)
2 \(0.0500\)
3 \(0.05000\)
4 \(5.0 \times {10^{ - 2}}\)
Explanation:
\(\frac{1}{{20}} = 0.05 = 0.0500\)
PHXI02:UNITS AND MEASUREMENTS
367652
The mass and volume of a body are 4.237 \(g\) and \(2.5\,c{m^3},\) respectively. The density of the material of the body in correct significant figures is
1 \(1.7g\,c{m^{ - 3}}\)
2 \(1.695\,g\,c{m^{ - 3}}\)
3 \(1.6048\,g\,c{m^{ - 3}}\)
4 \(1.69\,g\,c{m^{ - 3}}\)
Explanation:
In multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures. In this question, density should be reported to two significant figures. \({\mathop{\rm Density}\nolimits} = \frac{{4.237g}}{{2.5\,c{m^3}}} = 1.6498\) As rounding off the number, we get density \( = 1.7\,g\,c{m^{ - 3}}\)
PHXI02:UNITS AND MEASUREMENTS
367653
A long thread of length 1.23 \(m\) is cut to obtain a small thread of length 12.3 \(mm\). What is the new length of the long thread ?
1 \(1.23\,m\)
2 \(1.20\,m\)
3 \(1.30\,m\)
4 \(1.22\,m\)
Explanation:
New length \( = 1.23 - 12.3 \times {10^{ - 3}} = 1.1277\,m\) Least number of decimal places in any of the individual quantities is 2. Hence, new length \( = 1.22\,m\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI02:UNITS AND MEASUREMENTS
367649
Match the Column I and Column II
1 A – Q, B – P, C – S, D – R
2 A – S, B – R, C – P, D – Q
3 A – P, B – Q, C – R, D – S
4 A – R, B – S, C – Q, D – P
Explanation:
Using the concept of significant figures, (1) 1001 has 4 significant figure. (2) 010.1 has 3 significant figure. (3) 100.100 has 6 significant figure. (4) 0.0010010 has 5 significant figure. Correct answer is (1).
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367650
Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as \(4.62\,\,s,\) \(4.632\,\,s,\) \(4.6\,\,s,\) and \(4.64 s\). The arithmetic mean of these readings in correct significant figure is
1 \(4.6 s\)
2 \(4.62 s\)
3 \(5 s\)
4 \(4.623\,\,s\)
Explanation:
Arithmetic mean of time periods of oscillation of the pendulum, \(T=\dfrac{(4.62+4.632+4.6+4.64) s}{4}=4.623 s\) So, final answer with correct significant figure, \(T=4.6 s\). So, option (1) is correct.
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367651
The decimal equivalent of 1/20 up to three significant figures is
1 \(0.0050\)
2 \(0.0500\)
3 \(0.05000\)
4 \(5.0 \times {10^{ - 2}}\)
Explanation:
\(\frac{1}{{20}} = 0.05 = 0.0500\)
PHXI02:UNITS AND MEASUREMENTS
367652
The mass and volume of a body are 4.237 \(g\) and \(2.5\,c{m^3},\) respectively. The density of the material of the body in correct significant figures is
1 \(1.7g\,c{m^{ - 3}}\)
2 \(1.695\,g\,c{m^{ - 3}}\)
3 \(1.6048\,g\,c{m^{ - 3}}\)
4 \(1.69\,g\,c{m^{ - 3}}\)
Explanation:
In multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures. In this question, density should be reported to two significant figures. \({\mathop{\rm Density}\nolimits} = \frac{{4.237g}}{{2.5\,c{m^3}}} = 1.6498\) As rounding off the number, we get density \( = 1.7\,g\,c{m^{ - 3}}\)
PHXI02:UNITS AND MEASUREMENTS
367653
A long thread of length 1.23 \(m\) is cut to obtain a small thread of length 12.3 \(mm\). What is the new length of the long thread ?
1 \(1.23\,m\)
2 \(1.20\,m\)
3 \(1.30\,m\)
4 \(1.22\,m\)
Explanation:
New length \( = 1.23 - 12.3 \times {10^{ - 3}} = 1.1277\,m\) Least number of decimal places in any of the individual quantities is 2. Hence, new length \( = 1.22\,m\).
Using the concept of significant figures, (1) 1001 has 4 significant figure. (2) 010.1 has 3 significant figure. (3) 100.100 has 6 significant figure. (4) 0.0010010 has 5 significant figure. Correct answer is (1).
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367650
Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as \(4.62\,\,s,\) \(4.632\,\,s,\) \(4.6\,\,s,\) and \(4.64 s\). The arithmetic mean of these readings in correct significant figure is
1 \(4.6 s\)
2 \(4.62 s\)
3 \(5 s\)
4 \(4.623\,\,s\)
Explanation:
Arithmetic mean of time periods of oscillation of the pendulum, \(T=\dfrac{(4.62+4.632+4.6+4.64) s}{4}=4.623 s\) So, final answer with correct significant figure, \(T=4.6 s\). So, option (1) is correct.
JEE - 2024
PHXI02:UNITS AND MEASUREMENTS
367651
The decimal equivalent of 1/20 up to three significant figures is
1 \(0.0050\)
2 \(0.0500\)
3 \(0.05000\)
4 \(5.0 \times {10^{ - 2}}\)
Explanation:
\(\frac{1}{{20}} = 0.05 = 0.0500\)
PHXI02:UNITS AND MEASUREMENTS
367652
The mass and volume of a body are 4.237 \(g\) and \(2.5\,c{m^3},\) respectively. The density of the material of the body in correct significant figures is
1 \(1.7g\,c{m^{ - 3}}\)
2 \(1.695\,g\,c{m^{ - 3}}\)
3 \(1.6048\,g\,c{m^{ - 3}}\)
4 \(1.69\,g\,c{m^{ - 3}}\)
Explanation:
In multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures. In this question, density should be reported to two significant figures. \({\mathop{\rm Density}\nolimits} = \frac{{4.237g}}{{2.5\,c{m^3}}} = 1.6498\) As rounding off the number, we get density \( = 1.7\,g\,c{m^{ - 3}}\)
PHXI02:UNITS AND MEASUREMENTS
367653
A long thread of length 1.23 \(m\) is cut to obtain a small thread of length 12.3 \(mm\). What is the new length of the long thread ?
1 \(1.23\,m\)
2 \(1.20\,m\)
3 \(1.30\,m\)
4 \(1.22\,m\)
Explanation:
New length \( = 1.23 - 12.3 \times {10^{ - 3}} = 1.1277\,m\) Least number of decimal places in any of the individual quantities is 2. Hence, new length \( = 1.22\,m\).
