139891
A metallic bar of coefficient of linear expansion \(10^{-5} \mathrm{~K}^{-1}\) is heated from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). The percentage increase in its length is
139892
In a simple pendulum experiment, the maximum percentage error in the measurement of length is \(2 \%\) and that in the observation of the time-period is \(3 \%\) then, the maximum percentage error in determination of the acceleration due to gravity \(g\) is
1 \(5 \%\)
2 \(6 \%\)
3 \(7 \%\)
4 \(8 \%\)
5 \(10 \%\)
Explanation:
D Given, \(\frac{\Delta l}{l} \times 100=2 \%, \frac{\Delta \mathrm{T}}{\mathrm{T}} \times 100=3 \%, \frac{\triangle \mathrm{g}}{\mathrm{g}} \times 100=?\) We know that, Time period \((\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\) \(\mathrm{g}=\frac{4 \pi^{2} l}{\mathrm{~T}^{2}}\) Taking \(\log\) and differentiating of both side \(\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta l}{l}-2 \frac{\Delta \mathrm{T}}{\mathrm{T}}\) For maximum percentage error is +ve sign \(\frac{\Delta \mathrm{g}}{\mathrm{g}} =\frac{\Delta l}{l}+2 \frac{\Delta \mathrm{T}}{\mathrm{T}}\) \(\frac{\Delta \mathrm{g}}{\mathrm{g}} =2 \%+2 \times 3 \%\) \(=2 \%+6 \%\) \(=8 \%\)
Kerala CEE- 2014
Units and Measurements
139893
The percentage error in measuring \(M, L\) and \(T\) are \(1 \%, 1.5 \%\) and \(3 \%\) respectively. Then the percentage error in measuring the physical quantity with dimensions \(\left[\mathrm{ML}^{-1} \mathbf{T}^{-1}\right]\) is
139895
The mass and volume of a body are found to be \(5.00 \pm 0.05 \mathrm{~kg}\) and \(1.00 \pm 0.05 \mathrm{~m}^{3}\) respectively. Then the maximum possible percentage error in its density is
1 \(6 \%\)
2 \(3 \%\)
3 \(10 \%\)
4 \(5 \%\)
5 \(7 \%\)
Explanation:
A Given, \((\mathrm{m} \pm \Delta \mathrm{m})=(5 \pm 0.05)\) \((v \pm \Delta v)=(1.00 \pm 0.05)\) We know, \(\mathrm{P}=\frac{\mathrm{m}}{\mathrm{V}}\) taking log and differenting of both side \(\frac{\Delta \mathrm{P}}{\mathrm{P}}=\frac{\Delta \mathrm{M}}{\mathrm{M}}-\frac{\Delta \mathrm{V}}{\mathrm{V}}\) for maximum percentage error is \(+\mathrm{ve}\) sign \(\frac{\Delta \mathrm{P}}{\mathrm{P}}=\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{\Delta \mathrm{v}}{\mathrm{V}}\) \(\frac{\Delta \mathrm{P}}{\mathrm{P}} \quad =\frac{0.05}{5} \times 100+\frac{0.05}{1} \times 100\) \(=6 \%\) \(\%\) error in density \(=6 \%\)
139891
A metallic bar of coefficient of linear expansion \(10^{-5} \mathrm{~K}^{-1}\) is heated from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). The percentage increase in its length is
139892
In a simple pendulum experiment, the maximum percentage error in the measurement of length is \(2 \%\) and that in the observation of the time-period is \(3 \%\) then, the maximum percentage error in determination of the acceleration due to gravity \(g\) is
1 \(5 \%\)
2 \(6 \%\)
3 \(7 \%\)
4 \(8 \%\)
5 \(10 \%\)
Explanation:
D Given, \(\frac{\Delta l}{l} \times 100=2 \%, \frac{\Delta \mathrm{T}}{\mathrm{T}} \times 100=3 \%, \frac{\triangle \mathrm{g}}{\mathrm{g}} \times 100=?\) We know that, Time period \((\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\) \(\mathrm{g}=\frac{4 \pi^{2} l}{\mathrm{~T}^{2}}\) Taking \(\log\) and differentiating of both side \(\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta l}{l}-2 \frac{\Delta \mathrm{T}}{\mathrm{T}}\) For maximum percentage error is +ve sign \(\frac{\Delta \mathrm{g}}{\mathrm{g}} =\frac{\Delta l}{l}+2 \frac{\Delta \mathrm{T}}{\mathrm{T}}\) \(\frac{\Delta \mathrm{g}}{\mathrm{g}} =2 \%+2 \times 3 \%\) \(=2 \%+6 \%\) \(=8 \%\)
Kerala CEE- 2014
Units and Measurements
139893
The percentage error in measuring \(M, L\) and \(T\) are \(1 \%, 1.5 \%\) and \(3 \%\) respectively. Then the percentage error in measuring the physical quantity with dimensions \(\left[\mathrm{ML}^{-1} \mathbf{T}^{-1}\right]\) is
139895
The mass and volume of a body are found to be \(5.00 \pm 0.05 \mathrm{~kg}\) and \(1.00 \pm 0.05 \mathrm{~m}^{3}\) respectively. Then the maximum possible percentage error in its density is
1 \(6 \%\)
2 \(3 \%\)
3 \(10 \%\)
4 \(5 \%\)
5 \(7 \%\)
Explanation:
A Given, \((\mathrm{m} \pm \Delta \mathrm{m})=(5 \pm 0.05)\) \((v \pm \Delta v)=(1.00 \pm 0.05)\) We know, \(\mathrm{P}=\frac{\mathrm{m}}{\mathrm{V}}\) taking log and differenting of both side \(\frac{\Delta \mathrm{P}}{\mathrm{P}}=\frac{\Delta \mathrm{M}}{\mathrm{M}}-\frac{\Delta \mathrm{V}}{\mathrm{V}}\) for maximum percentage error is \(+\mathrm{ve}\) sign \(\frac{\Delta \mathrm{P}}{\mathrm{P}}=\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{\Delta \mathrm{v}}{\mathrm{V}}\) \(\frac{\Delta \mathrm{P}}{\mathrm{P}} \quad =\frac{0.05}{5} \times 100+\frac{0.05}{1} \times 100\) \(=6 \%\) \(\%\) error in density \(=6 \%\)
139891
A metallic bar of coefficient of linear expansion \(10^{-5} \mathrm{~K}^{-1}\) is heated from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). The percentage increase in its length is
139892
In a simple pendulum experiment, the maximum percentage error in the measurement of length is \(2 \%\) and that in the observation of the time-period is \(3 \%\) then, the maximum percentage error in determination of the acceleration due to gravity \(g\) is
1 \(5 \%\)
2 \(6 \%\)
3 \(7 \%\)
4 \(8 \%\)
5 \(10 \%\)
Explanation:
D Given, \(\frac{\Delta l}{l} \times 100=2 \%, \frac{\Delta \mathrm{T}}{\mathrm{T}} \times 100=3 \%, \frac{\triangle \mathrm{g}}{\mathrm{g}} \times 100=?\) We know that, Time period \((\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\) \(\mathrm{g}=\frac{4 \pi^{2} l}{\mathrm{~T}^{2}}\) Taking \(\log\) and differentiating of both side \(\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta l}{l}-2 \frac{\Delta \mathrm{T}}{\mathrm{T}}\) For maximum percentage error is +ve sign \(\frac{\Delta \mathrm{g}}{\mathrm{g}} =\frac{\Delta l}{l}+2 \frac{\Delta \mathrm{T}}{\mathrm{T}}\) \(\frac{\Delta \mathrm{g}}{\mathrm{g}} =2 \%+2 \times 3 \%\) \(=2 \%+6 \%\) \(=8 \%\)
Kerala CEE- 2014
Units and Measurements
139893
The percentage error in measuring \(M, L\) and \(T\) are \(1 \%, 1.5 \%\) and \(3 \%\) respectively. Then the percentage error in measuring the physical quantity with dimensions \(\left[\mathrm{ML}^{-1} \mathbf{T}^{-1}\right]\) is
139895
The mass and volume of a body are found to be \(5.00 \pm 0.05 \mathrm{~kg}\) and \(1.00 \pm 0.05 \mathrm{~m}^{3}\) respectively. Then the maximum possible percentage error in its density is
1 \(6 \%\)
2 \(3 \%\)
3 \(10 \%\)
4 \(5 \%\)
5 \(7 \%\)
Explanation:
A Given, \((\mathrm{m} \pm \Delta \mathrm{m})=(5 \pm 0.05)\) \((v \pm \Delta v)=(1.00 \pm 0.05)\) We know, \(\mathrm{P}=\frac{\mathrm{m}}{\mathrm{V}}\) taking log and differenting of both side \(\frac{\Delta \mathrm{P}}{\mathrm{P}}=\frac{\Delta \mathrm{M}}{\mathrm{M}}-\frac{\Delta \mathrm{V}}{\mathrm{V}}\) for maximum percentage error is \(+\mathrm{ve}\) sign \(\frac{\Delta \mathrm{P}}{\mathrm{P}}=\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{\Delta \mathrm{v}}{\mathrm{V}}\) \(\frac{\Delta \mathrm{P}}{\mathrm{P}} \quad =\frac{0.05}{5} \times 100+\frac{0.05}{1} \times 100\) \(=6 \%\) \(\%\) error in density \(=6 \%\)
139891
A metallic bar of coefficient of linear expansion \(10^{-5} \mathrm{~K}^{-1}\) is heated from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). The percentage increase in its length is
139892
In a simple pendulum experiment, the maximum percentage error in the measurement of length is \(2 \%\) and that in the observation of the time-period is \(3 \%\) then, the maximum percentage error in determination of the acceleration due to gravity \(g\) is
1 \(5 \%\)
2 \(6 \%\)
3 \(7 \%\)
4 \(8 \%\)
5 \(10 \%\)
Explanation:
D Given, \(\frac{\Delta l}{l} \times 100=2 \%, \frac{\Delta \mathrm{T}}{\mathrm{T}} \times 100=3 \%, \frac{\triangle \mathrm{g}}{\mathrm{g}} \times 100=?\) We know that, Time period \((\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\) \(\mathrm{g}=\frac{4 \pi^{2} l}{\mathrm{~T}^{2}}\) Taking \(\log\) and differentiating of both side \(\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta l}{l}-2 \frac{\Delta \mathrm{T}}{\mathrm{T}}\) For maximum percentage error is +ve sign \(\frac{\Delta \mathrm{g}}{\mathrm{g}} =\frac{\Delta l}{l}+2 \frac{\Delta \mathrm{T}}{\mathrm{T}}\) \(\frac{\Delta \mathrm{g}}{\mathrm{g}} =2 \%+2 \times 3 \%\) \(=2 \%+6 \%\) \(=8 \%\)
Kerala CEE- 2014
Units and Measurements
139893
The percentage error in measuring \(M, L\) and \(T\) are \(1 \%, 1.5 \%\) and \(3 \%\) respectively. Then the percentage error in measuring the physical quantity with dimensions \(\left[\mathrm{ML}^{-1} \mathbf{T}^{-1}\right]\) is
139895
The mass and volume of a body are found to be \(5.00 \pm 0.05 \mathrm{~kg}\) and \(1.00 \pm 0.05 \mathrm{~m}^{3}\) respectively. Then the maximum possible percentage error in its density is
1 \(6 \%\)
2 \(3 \%\)
3 \(10 \%\)
4 \(5 \%\)
5 \(7 \%\)
Explanation:
A Given, \((\mathrm{m} \pm \Delta \mathrm{m})=(5 \pm 0.05)\) \((v \pm \Delta v)=(1.00 \pm 0.05)\) We know, \(\mathrm{P}=\frac{\mathrm{m}}{\mathrm{V}}\) taking log and differenting of both side \(\frac{\Delta \mathrm{P}}{\mathrm{P}}=\frac{\Delta \mathrm{M}}{\mathrm{M}}-\frac{\Delta \mathrm{V}}{\mathrm{V}}\) for maximum percentage error is \(+\mathrm{ve}\) sign \(\frac{\Delta \mathrm{P}}{\mathrm{P}}=\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{\Delta \mathrm{v}}{\mathrm{V}}\) \(\frac{\Delta \mathrm{P}}{\mathrm{P}} \quad =\frac{0.05}{5} \times 100+\frac{0.05}{1} \times 100\) \(=6 \%\) \(\%\) error in density \(=6 \%\)
