367490
The side of a cubical block when measured with a vernier calipers is 2.50 \(cm\) . The vernier constant is 0.01 \(cm\) . The maximum possible error in the area of the side of the block is
1 \({\pm 0.01 {~cm}^{2}}\)
2 \({\pm 0.02 {~cm}^{2}}\)
3 \({\pm 0.05 {~cm}^{2}}\)
4 \({\pm 0.10 {~cm}^{2}}\)
Explanation:
\({: a=2.50 {~cm}}\) Vernier constant \({=0.01 {~cm}}\) \({\therefore}\) Maximum possible error in measurement of \({a=0.01 {~cm}}\) \( \Rightarrow Area = {a^2}\) \(\,\,\,\,\,\,\,\frac{{\Delta A}}{A} = \frac{{2\Delta a}}{a}\) \( \Rightarrow \Delta A = \frac{{2(0.01)}}{{(2.50)}} \times {\left( {2.50} \right)^2} = 0.5{\rm{ c}}{{\rm{m}}^2}\) Maximum possible error in measurement of area \({=0.05 {~cm}^{2}}\). So correct option is (3)
PHXI02:UNITS AND MEASUREMENTS
367491
The errors that always occur in the measurement with screw gauge is
1 Random errors
2 Systematic errors
3 Gross errors
4 Negligible errors
Explanation:
Conceptual Question
PHXI02:UNITS AND MEASUREMENTS
367492
A cylindrical wire of mass \((0.4 \pm 0.01) g\) has length \((8 \pm 0.04)cm\) and radius \((6 \pm 0.03)mm\). The maximum error in its density will be
1 \(3.5 \%\)
2 \(5 \%\)
3 \(4 \%\)
4 \(1 \%\)
Explanation:
Given, mass of cylinder, \(m = (0.4 \pm 0.01)g\) Length, \(l = (8 \pm 0.04)cm\) Radius, \(r = (6 \pm 0.03)mm\) Density, \(d=\dfrac{\text { mass }}{\text { volume }}=\dfrac{\text { mass }}{\pi r^{2} l}\) \(\therefore\) Maximum error in density, using error formula is \(\dfrac{\Delta d}{d}=\left(\dfrac{\Delta m}{m}+2 \dfrac{\Delta r}{r}+\dfrac{\Delta l}{l}\right)\) \(=\left(\dfrac{0.01}{0.4}+2 \times \dfrac{0.03}{6}+\dfrac{0.04}{8}\right)\) \(=\left(\dfrac{1}{40}+\dfrac{1}{100}+\dfrac{1}{200}\right)=0.04\) Percentage error in density \(=4 \%\) So, correct option is (3).
JEE - 2023
PHXI02:UNITS AND MEASUREMENTS
367493
The radius of a sphere is \(\left( {5.3 \pm 0.1} \right)cm\). The percentage error in its volume is
367490
The side of a cubical block when measured with a vernier calipers is 2.50 \(cm\) . The vernier constant is 0.01 \(cm\) . The maximum possible error in the area of the side of the block is
1 \({\pm 0.01 {~cm}^{2}}\)
2 \({\pm 0.02 {~cm}^{2}}\)
3 \({\pm 0.05 {~cm}^{2}}\)
4 \({\pm 0.10 {~cm}^{2}}\)
Explanation:
\({: a=2.50 {~cm}}\) Vernier constant \({=0.01 {~cm}}\) \({\therefore}\) Maximum possible error in measurement of \({a=0.01 {~cm}}\) \( \Rightarrow Area = {a^2}\) \(\,\,\,\,\,\,\,\frac{{\Delta A}}{A} = \frac{{2\Delta a}}{a}\) \( \Rightarrow \Delta A = \frac{{2(0.01)}}{{(2.50)}} \times {\left( {2.50} \right)^2} = 0.5{\rm{ c}}{{\rm{m}}^2}\) Maximum possible error in measurement of area \({=0.05 {~cm}^{2}}\). So correct option is (3)
PHXI02:UNITS AND MEASUREMENTS
367491
The errors that always occur in the measurement with screw gauge is
1 Random errors
2 Systematic errors
3 Gross errors
4 Negligible errors
Explanation:
Conceptual Question
PHXI02:UNITS AND MEASUREMENTS
367492
A cylindrical wire of mass \((0.4 \pm 0.01) g\) has length \((8 \pm 0.04)cm\) and radius \((6 \pm 0.03)mm\). The maximum error in its density will be
1 \(3.5 \%\)
2 \(5 \%\)
3 \(4 \%\)
4 \(1 \%\)
Explanation:
Given, mass of cylinder, \(m = (0.4 \pm 0.01)g\) Length, \(l = (8 \pm 0.04)cm\) Radius, \(r = (6 \pm 0.03)mm\) Density, \(d=\dfrac{\text { mass }}{\text { volume }}=\dfrac{\text { mass }}{\pi r^{2} l}\) \(\therefore\) Maximum error in density, using error formula is \(\dfrac{\Delta d}{d}=\left(\dfrac{\Delta m}{m}+2 \dfrac{\Delta r}{r}+\dfrac{\Delta l}{l}\right)\) \(=\left(\dfrac{0.01}{0.4}+2 \times \dfrac{0.03}{6}+\dfrac{0.04}{8}\right)\) \(=\left(\dfrac{1}{40}+\dfrac{1}{100}+\dfrac{1}{200}\right)=0.04\) Percentage error in density \(=4 \%\) So, correct option is (3).
JEE - 2023
PHXI02:UNITS AND MEASUREMENTS
367493
The radius of a sphere is \(\left( {5.3 \pm 0.1} \right)cm\). The percentage error in its volume is
367490
The side of a cubical block when measured with a vernier calipers is 2.50 \(cm\) . The vernier constant is 0.01 \(cm\) . The maximum possible error in the area of the side of the block is
1 \({\pm 0.01 {~cm}^{2}}\)
2 \({\pm 0.02 {~cm}^{2}}\)
3 \({\pm 0.05 {~cm}^{2}}\)
4 \({\pm 0.10 {~cm}^{2}}\)
Explanation:
\({: a=2.50 {~cm}}\) Vernier constant \({=0.01 {~cm}}\) \({\therefore}\) Maximum possible error in measurement of \({a=0.01 {~cm}}\) \( \Rightarrow Area = {a^2}\) \(\,\,\,\,\,\,\,\frac{{\Delta A}}{A} = \frac{{2\Delta a}}{a}\) \( \Rightarrow \Delta A = \frac{{2(0.01)}}{{(2.50)}} \times {\left( {2.50} \right)^2} = 0.5{\rm{ c}}{{\rm{m}}^2}\) Maximum possible error in measurement of area \({=0.05 {~cm}^{2}}\). So correct option is (3)
PHXI02:UNITS AND MEASUREMENTS
367491
The errors that always occur in the measurement with screw gauge is
1 Random errors
2 Systematic errors
3 Gross errors
4 Negligible errors
Explanation:
Conceptual Question
PHXI02:UNITS AND MEASUREMENTS
367492
A cylindrical wire of mass \((0.4 \pm 0.01) g\) has length \((8 \pm 0.04)cm\) and radius \((6 \pm 0.03)mm\). The maximum error in its density will be
1 \(3.5 \%\)
2 \(5 \%\)
3 \(4 \%\)
4 \(1 \%\)
Explanation:
Given, mass of cylinder, \(m = (0.4 \pm 0.01)g\) Length, \(l = (8 \pm 0.04)cm\) Radius, \(r = (6 \pm 0.03)mm\) Density, \(d=\dfrac{\text { mass }}{\text { volume }}=\dfrac{\text { mass }}{\pi r^{2} l}\) \(\therefore\) Maximum error in density, using error formula is \(\dfrac{\Delta d}{d}=\left(\dfrac{\Delta m}{m}+2 \dfrac{\Delta r}{r}+\dfrac{\Delta l}{l}\right)\) \(=\left(\dfrac{0.01}{0.4}+2 \times \dfrac{0.03}{6}+\dfrac{0.04}{8}\right)\) \(=\left(\dfrac{1}{40}+\dfrac{1}{100}+\dfrac{1}{200}\right)=0.04\) Percentage error in density \(=4 \%\) So, correct option is (3).
JEE - 2023
PHXI02:UNITS AND MEASUREMENTS
367493
The radius of a sphere is \(\left( {5.3 \pm 0.1} \right)cm\). The percentage error in its volume is
367490
The side of a cubical block when measured with a vernier calipers is 2.50 \(cm\) . The vernier constant is 0.01 \(cm\) . The maximum possible error in the area of the side of the block is
1 \({\pm 0.01 {~cm}^{2}}\)
2 \({\pm 0.02 {~cm}^{2}}\)
3 \({\pm 0.05 {~cm}^{2}}\)
4 \({\pm 0.10 {~cm}^{2}}\)
Explanation:
\({: a=2.50 {~cm}}\) Vernier constant \({=0.01 {~cm}}\) \({\therefore}\) Maximum possible error in measurement of \({a=0.01 {~cm}}\) \( \Rightarrow Area = {a^2}\) \(\,\,\,\,\,\,\,\frac{{\Delta A}}{A} = \frac{{2\Delta a}}{a}\) \( \Rightarrow \Delta A = \frac{{2(0.01)}}{{(2.50)}} \times {\left( {2.50} \right)^2} = 0.5{\rm{ c}}{{\rm{m}}^2}\) Maximum possible error in measurement of area \({=0.05 {~cm}^{2}}\). So correct option is (3)
PHXI02:UNITS AND MEASUREMENTS
367491
The errors that always occur in the measurement with screw gauge is
1 Random errors
2 Systematic errors
3 Gross errors
4 Negligible errors
Explanation:
Conceptual Question
PHXI02:UNITS AND MEASUREMENTS
367492
A cylindrical wire of mass \((0.4 \pm 0.01) g\) has length \((8 \pm 0.04)cm\) and radius \((6 \pm 0.03)mm\). The maximum error in its density will be
1 \(3.5 \%\)
2 \(5 \%\)
3 \(4 \%\)
4 \(1 \%\)
Explanation:
Given, mass of cylinder, \(m = (0.4 \pm 0.01)g\) Length, \(l = (8 \pm 0.04)cm\) Radius, \(r = (6 \pm 0.03)mm\) Density, \(d=\dfrac{\text { mass }}{\text { volume }}=\dfrac{\text { mass }}{\pi r^{2} l}\) \(\therefore\) Maximum error in density, using error formula is \(\dfrac{\Delta d}{d}=\left(\dfrac{\Delta m}{m}+2 \dfrac{\Delta r}{r}+\dfrac{\Delta l}{l}\right)\) \(=\left(\dfrac{0.01}{0.4}+2 \times \dfrac{0.03}{6}+\dfrac{0.04}{8}\right)\) \(=\left(\dfrac{1}{40}+\dfrac{1}{100}+\dfrac{1}{200}\right)=0.04\) Percentage error in density \(=4 \%\) So, correct option is (3).
JEE - 2023
PHXI02:UNITS AND MEASUREMENTS
367493
The radius of a sphere is \(\left( {5.3 \pm 0.1} \right)cm\). The percentage error in its volume is
