269164
If the length and breadth of a plate are\((5.0 \pm 0.2) \mathrm{cm}\) and \((4.0 \pm 0.1) \mathrm{cm}\) then the absolute error in measurement of area is
269165
If the length of a cylinder is measured to be\(8.28 \mathrm{~cm}\) with an error of \(0.01 \mathrm{~cm}\) then the percentageerror in measured length isnearly
1 \(0.4 \%\)
2 \(0.2 \%\)
3 \(0.1 \%\)
4 \(0.5 \%\)
Explanation:
\(\frac{\Delta l}{l} \times 100 \)
Units and Measurements
269166
A student performs experiment with simple pendulum and measures time for 10 vibrations. If he measures the time for 100 vibrations, the error in measurement oftime period will be reduced by a factor of
1 10
2 90
3 100
4 1000
Explanation:
\( \frac{X_{1}}{X_{2}}=\frac{N_{2}}{N_{1}}\)
Units and Measurements
269167
If \(L_{1}=(3.03 \pm 0.02) m\) and \(L_{2}=(2.01 \pm 0.02) \mathrm{m}\) then \(L_{1}+2 L_{2}\) is (in m)
269164
If the length and breadth of a plate are\((5.0 \pm 0.2) \mathrm{cm}\) and \((4.0 \pm 0.1) \mathrm{cm}\) then the absolute error in measurement of area is
269165
If the length of a cylinder is measured to be\(8.28 \mathrm{~cm}\) with an error of \(0.01 \mathrm{~cm}\) then the percentageerror in measured length isnearly
1 \(0.4 \%\)
2 \(0.2 \%\)
3 \(0.1 \%\)
4 \(0.5 \%\)
Explanation:
\(\frac{\Delta l}{l} \times 100 \)
Units and Measurements
269166
A student performs experiment with simple pendulum and measures time for 10 vibrations. If he measures the time for 100 vibrations, the error in measurement oftime period will be reduced by a factor of
1 10
2 90
3 100
4 1000
Explanation:
\( \frac{X_{1}}{X_{2}}=\frac{N_{2}}{N_{1}}\)
Units and Measurements
269167
If \(L_{1}=(3.03 \pm 0.02) m\) and \(L_{2}=(2.01 \pm 0.02) \mathrm{m}\) then \(L_{1}+2 L_{2}\) is (in m)
269164
If the length and breadth of a plate are\((5.0 \pm 0.2) \mathrm{cm}\) and \((4.0 \pm 0.1) \mathrm{cm}\) then the absolute error in measurement of area is
269165
If the length of a cylinder is measured to be\(8.28 \mathrm{~cm}\) with an error of \(0.01 \mathrm{~cm}\) then the percentageerror in measured length isnearly
1 \(0.4 \%\)
2 \(0.2 \%\)
3 \(0.1 \%\)
4 \(0.5 \%\)
Explanation:
\(\frac{\Delta l}{l} \times 100 \)
Units and Measurements
269166
A student performs experiment with simple pendulum and measures time for 10 vibrations. If he measures the time for 100 vibrations, the error in measurement oftime period will be reduced by a factor of
1 10
2 90
3 100
4 1000
Explanation:
\( \frac{X_{1}}{X_{2}}=\frac{N_{2}}{N_{1}}\)
Units and Measurements
269167
If \(L_{1}=(3.03 \pm 0.02) m\) and \(L_{2}=(2.01 \pm 0.02) \mathrm{m}\) then \(L_{1}+2 L_{2}\) is (in m)
269164
If the length and breadth of a plate are\((5.0 \pm 0.2) \mathrm{cm}\) and \((4.0 \pm 0.1) \mathrm{cm}\) then the absolute error in measurement of area is
269165
If the length of a cylinder is measured to be\(8.28 \mathrm{~cm}\) with an error of \(0.01 \mathrm{~cm}\) then the percentageerror in measured length isnearly
1 \(0.4 \%\)
2 \(0.2 \%\)
3 \(0.1 \%\)
4 \(0.5 \%\)
Explanation:
\(\frac{\Delta l}{l} \times 100 \)
Units and Measurements
269166
A student performs experiment with simple pendulum and measures time for 10 vibrations. If he measures the time for 100 vibrations, the error in measurement oftime period will be reduced by a factor of
1 10
2 90
3 100
4 1000
Explanation:
\( \frac{X_{1}}{X_{2}}=\frac{N_{2}}{N_{1}}\)
Units and Measurements
269167
If \(L_{1}=(3.03 \pm 0.02) m\) and \(L_{2}=(2.01 \pm 0.02) \mathrm{m}\) then \(L_{1}+2 L_{2}\) is (in m)
269164
If the length and breadth of a plate are\((5.0 \pm 0.2) \mathrm{cm}\) and \((4.0 \pm 0.1) \mathrm{cm}\) then the absolute error in measurement of area is
269165
If the length of a cylinder is measured to be\(8.28 \mathrm{~cm}\) with an error of \(0.01 \mathrm{~cm}\) then the percentageerror in measured length isnearly
1 \(0.4 \%\)
2 \(0.2 \%\)
3 \(0.1 \%\)
4 \(0.5 \%\)
Explanation:
\(\frac{\Delta l}{l} \times 100 \)
Units and Measurements
269166
A student performs experiment with simple pendulum and measures time for 10 vibrations. If he measures the time for 100 vibrations, the error in measurement oftime period will be reduced by a factor of
1 10
2 90
3 100
4 1000
Explanation:
\( \frac{X_{1}}{X_{2}}=\frac{N_{2}}{N_{1}}\)
Units and Measurements
269167
If \(L_{1}=(3.03 \pm 0.02) m\) and \(L_{2}=(2.01 \pm 0.02) \mathrm{m}\) then \(L_{1}+2 L_{2}\) is (in m)