139877
An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increases in the apparent frequency?
1 Zero
2 \(0.5 \%\)
3 \(5 \%\)
4 \(20 \%\)
Explanation:
D When the observer is moving towards a stationary source. Apparent frequency, \(\mathrm{V}^{\prime}=\mathrm{V}\left[\frac{\mathrm{V}+\mathrm{V}_{0}}{\mathrm{~V}}\right]\) \(\mathrm{V}^{\prime}=\mathrm{V}\left[\frac{\mathrm{V}+\frac{\mathrm{V}}{5}}{\mathrm{~V}}\right]\) \(\text { Or, } \quad \mathrm{V}^{\prime}=\frac{6}{5} \mathrm{~V}\) \(\therefore\) Percentage increase in frequency is \(=\frac{\mathrm{V}^{\prime}-\mathrm{V}}{\mathrm{V}} \times 100\) \(=\frac{\frac{6}{5} \mathrm{~V}-\mathrm{V}}{\mathrm{V}} \times 100\) \(=\frac{\frac{1}{5} \mathrm{~V}}{\mathrm{~V}} \times 100\) \(\therefore \quad \%\) change \(=20 \%\)
UP CPMT-2008
Units and Measurements
139878
The side of a cubical block when measured with a vernier calipers is \(2.50 \mathrm{~cm}\). The vernier constant is \(0.01 \mathrm{~cm}\). The maximum possible error in the area of the side of the block is
1 \(\pm 0.01 \mathrm{~cm}^{2}\)
2 \(\pm 0.02 \mathrm{~cm}^{2}\)
3 \(\pm 0.05 \mathrm{~cm}^{2}\)
4 \(\pm 0.10 \mathrm{~cm}^{2}\)
Explanation:
C Given, \(l=2.50 \mathrm{~cm}\) Vernier constant \(\Delta l=0.01 \mathrm{~cm}\) \(\therefore\) Maximum possible error in measurement of \(\Delta l=0.01\) cm \(\Rightarrow\) Area \(=l^{2}\) \(\frac{\Delta \mathrm{A}}{\mathrm{A}}=\frac{2 \Delta l}{l}\) \(\Rightarrow \Delta \mathrm{A}=\frac{2(0.01)}{(2.50)} \times(2.50)^{2}=0.05 \mathrm{~cm}^{2}\) Maximum possible error in measurement of area \(=\) \(\pm 0.05 \mathrm{~cm}^{2}\)
SRMJEE - 2009
Units and Measurements
139879
The percentage errors in the measurement of mass and speed are \(1 \%\) and \(2 \%\) respectively. What is the percentage error in the kinetic energy
1 \(5 \%\)
2 \(2.5 \%\)
3 \(3 \%\)
4 \(1.5 \%\)
Explanation:
A Given that, \(\frac{\Delta \mathrm{m}}{\mathrm{m}} \times 100=1 \%, \frac{\Delta \mathrm{v}}{\mathrm{v}} \times 100=2 \%\) \(\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2}\) taking \(\log\) of both side \(\log \mathrm{k}=\log \mathrm{m}+2 \log \mathrm{v}\) Differentiating both side \(\therefore \frac{\Delta \mathrm{K}}{\mathrm{K}} \times 100=\left(\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{2 \Delta \mathrm{v}}{\mathrm{v}}\right) \times 100\) \(\%\) error in \(\mathrm{K}=(\%\) error in \(\mathrm{m})+2 \times(\%\) error in \(\mathrm{v})\) \(=1 \%+2 \times 2 \%=5 \%\)
SRMJEE - 2010
Units and Measurements
139880
A Force \(F\) is applied on a square plate side \(L\). If the percentage error in the determination of \(L\) is \(2 \%\) and that in \(\mathrm{F}\) is \(4 \%\), what is the permissible error in pressure
139877
An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increases in the apparent frequency?
1 Zero
2 \(0.5 \%\)
3 \(5 \%\)
4 \(20 \%\)
Explanation:
D When the observer is moving towards a stationary source. Apparent frequency, \(\mathrm{V}^{\prime}=\mathrm{V}\left[\frac{\mathrm{V}+\mathrm{V}_{0}}{\mathrm{~V}}\right]\) \(\mathrm{V}^{\prime}=\mathrm{V}\left[\frac{\mathrm{V}+\frac{\mathrm{V}}{5}}{\mathrm{~V}}\right]\) \(\text { Or, } \quad \mathrm{V}^{\prime}=\frac{6}{5} \mathrm{~V}\) \(\therefore\) Percentage increase in frequency is \(=\frac{\mathrm{V}^{\prime}-\mathrm{V}}{\mathrm{V}} \times 100\) \(=\frac{\frac{6}{5} \mathrm{~V}-\mathrm{V}}{\mathrm{V}} \times 100\) \(=\frac{\frac{1}{5} \mathrm{~V}}{\mathrm{~V}} \times 100\) \(\therefore \quad \%\) change \(=20 \%\)
UP CPMT-2008
Units and Measurements
139878
The side of a cubical block when measured with a vernier calipers is \(2.50 \mathrm{~cm}\). The vernier constant is \(0.01 \mathrm{~cm}\). The maximum possible error in the area of the side of the block is
1 \(\pm 0.01 \mathrm{~cm}^{2}\)
2 \(\pm 0.02 \mathrm{~cm}^{2}\)
3 \(\pm 0.05 \mathrm{~cm}^{2}\)
4 \(\pm 0.10 \mathrm{~cm}^{2}\)
Explanation:
C Given, \(l=2.50 \mathrm{~cm}\) Vernier constant \(\Delta l=0.01 \mathrm{~cm}\) \(\therefore\) Maximum possible error in measurement of \(\Delta l=0.01\) cm \(\Rightarrow\) Area \(=l^{2}\) \(\frac{\Delta \mathrm{A}}{\mathrm{A}}=\frac{2 \Delta l}{l}\) \(\Rightarrow \Delta \mathrm{A}=\frac{2(0.01)}{(2.50)} \times(2.50)^{2}=0.05 \mathrm{~cm}^{2}\) Maximum possible error in measurement of area \(=\) \(\pm 0.05 \mathrm{~cm}^{2}\)
SRMJEE - 2009
Units and Measurements
139879
The percentage errors in the measurement of mass and speed are \(1 \%\) and \(2 \%\) respectively. What is the percentage error in the kinetic energy
1 \(5 \%\)
2 \(2.5 \%\)
3 \(3 \%\)
4 \(1.5 \%\)
Explanation:
A Given that, \(\frac{\Delta \mathrm{m}}{\mathrm{m}} \times 100=1 \%, \frac{\Delta \mathrm{v}}{\mathrm{v}} \times 100=2 \%\) \(\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2}\) taking \(\log\) of both side \(\log \mathrm{k}=\log \mathrm{m}+2 \log \mathrm{v}\) Differentiating both side \(\therefore \frac{\Delta \mathrm{K}}{\mathrm{K}} \times 100=\left(\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{2 \Delta \mathrm{v}}{\mathrm{v}}\right) \times 100\) \(\%\) error in \(\mathrm{K}=(\%\) error in \(\mathrm{m})+2 \times(\%\) error in \(\mathrm{v})\) \(=1 \%+2 \times 2 \%=5 \%\)
SRMJEE - 2010
Units and Measurements
139880
A Force \(F\) is applied on a square plate side \(L\). If the percentage error in the determination of \(L\) is \(2 \%\) and that in \(\mathrm{F}\) is \(4 \%\), what is the permissible error in pressure
139877
An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increases in the apparent frequency?
1 Zero
2 \(0.5 \%\)
3 \(5 \%\)
4 \(20 \%\)
Explanation:
D When the observer is moving towards a stationary source. Apparent frequency, \(\mathrm{V}^{\prime}=\mathrm{V}\left[\frac{\mathrm{V}+\mathrm{V}_{0}}{\mathrm{~V}}\right]\) \(\mathrm{V}^{\prime}=\mathrm{V}\left[\frac{\mathrm{V}+\frac{\mathrm{V}}{5}}{\mathrm{~V}}\right]\) \(\text { Or, } \quad \mathrm{V}^{\prime}=\frac{6}{5} \mathrm{~V}\) \(\therefore\) Percentage increase in frequency is \(=\frac{\mathrm{V}^{\prime}-\mathrm{V}}{\mathrm{V}} \times 100\) \(=\frac{\frac{6}{5} \mathrm{~V}-\mathrm{V}}{\mathrm{V}} \times 100\) \(=\frac{\frac{1}{5} \mathrm{~V}}{\mathrm{~V}} \times 100\) \(\therefore \quad \%\) change \(=20 \%\)
UP CPMT-2008
Units and Measurements
139878
The side of a cubical block when measured with a vernier calipers is \(2.50 \mathrm{~cm}\). The vernier constant is \(0.01 \mathrm{~cm}\). The maximum possible error in the area of the side of the block is
1 \(\pm 0.01 \mathrm{~cm}^{2}\)
2 \(\pm 0.02 \mathrm{~cm}^{2}\)
3 \(\pm 0.05 \mathrm{~cm}^{2}\)
4 \(\pm 0.10 \mathrm{~cm}^{2}\)
Explanation:
C Given, \(l=2.50 \mathrm{~cm}\) Vernier constant \(\Delta l=0.01 \mathrm{~cm}\) \(\therefore\) Maximum possible error in measurement of \(\Delta l=0.01\) cm \(\Rightarrow\) Area \(=l^{2}\) \(\frac{\Delta \mathrm{A}}{\mathrm{A}}=\frac{2 \Delta l}{l}\) \(\Rightarrow \Delta \mathrm{A}=\frac{2(0.01)}{(2.50)} \times(2.50)^{2}=0.05 \mathrm{~cm}^{2}\) Maximum possible error in measurement of area \(=\) \(\pm 0.05 \mathrm{~cm}^{2}\)
SRMJEE - 2009
Units and Measurements
139879
The percentage errors in the measurement of mass and speed are \(1 \%\) and \(2 \%\) respectively. What is the percentage error in the kinetic energy
1 \(5 \%\)
2 \(2.5 \%\)
3 \(3 \%\)
4 \(1.5 \%\)
Explanation:
A Given that, \(\frac{\Delta \mathrm{m}}{\mathrm{m}} \times 100=1 \%, \frac{\Delta \mathrm{v}}{\mathrm{v}} \times 100=2 \%\) \(\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2}\) taking \(\log\) of both side \(\log \mathrm{k}=\log \mathrm{m}+2 \log \mathrm{v}\) Differentiating both side \(\therefore \frac{\Delta \mathrm{K}}{\mathrm{K}} \times 100=\left(\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{2 \Delta \mathrm{v}}{\mathrm{v}}\right) \times 100\) \(\%\) error in \(\mathrm{K}=(\%\) error in \(\mathrm{m})+2 \times(\%\) error in \(\mathrm{v})\) \(=1 \%+2 \times 2 \%=5 \%\)
SRMJEE - 2010
Units and Measurements
139880
A Force \(F\) is applied on a square plate side \(L\). If the percentage error in the determination of \(L\) is \(2 \%\) and that in \(\mathrm{F}\) is \(4 \%\), what is the permissible error in pressure
139877
An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increases in the apparent frequency?
1 Zero
2 \(0.5 \%\)
3 \(5 \%\)
4 \(20 \%\)
Explanation:
D When the observer is moving towards a stationary source. Apparent frequency, \(\mathrm{V}^{\prime}=\mathrm{V}\left[\frac{\mathrm{V}+\mathrm{V}_{0}}{\mathrm{~V}}\right]\) \(\mathrm{V}^{\prime}=\mathrm{V}\left[\frac{\mathrm{V}+\frac{\mathrm{V}}{5}}{\mathrm{~V}}\right]\) \(\text { Or, } \quad \mathrm{V}^{\prime}=\frac{6}{5} \mathrm{~V}\) \(\therefore\) Percentage increase in frequency is \(=\frac{\mathrm{V}^{\prime}-\mathrm{V}}{\mathrm{V}} \times 100\) \(=\frac{\frac{6}{5} \mathrm{~V}-\mathrm{V}}{\mathrm{V}} \times 100\) \(=\frac{\frac{1}{5} \mathrm{~V}}{\mathrm{~V}} \times 100\) \(\therefore \quad \%\) change \(=20 \%\)
UP CPMT-2008
Units and Measurements
139878
The side of a cubical block when measured with a vernier calipers is \(2.50 \mathrm{~cm}\). The vernier constant is \(0.01 \mathrm{~cm}\). The maximum possible error in the area of the side of the block is
1 \(\pm 0.01 \mathrm{~cm}^{2}\)
2 \(\pm 0.02 \mathrm{~cm}^{2}\)
3 \(\pm 0.05 \mathrm{~cm}^{2}\)
4 \(\pm 0.10 \mathrm{~cm}^{2}\)
Explanation:
C Given, \(l=2.50 \mathrm{~cm}\) Vernier constant \(\Delta l=0.01 \mathrm{~cm}\) \(\therefore\) Maximum possible error in measurement of \(\Delta l=0.01\) cm \(\Rightarrow\) Area \(=l^{2}\) \(\frac{\Delta \mathrm{A}}{\mathrm{A}}=\frac{2 \Delta l}{l}\) \(\Rightarrow \Delta \mathrm{A}=\frac{2(0.01)}{(2.50)} \times(2.50)^{2}=0.05 \mathrm{~cm}^{2}\) Maximum possible error in measurement of area \(=\) \(\pm 0.05 \mathrm{~cm}^{2}\)
SRMJEE - 2009
Units and Measurements
139879
The percentage errors in the measurement of mass and speed are \(1 \%\) and \(2 \%\) respectively. What is the percentage error in the kinetic energy
1 \(5 \%\)
2 \(2.5 \%\)
3 \(3 \%\)
4 \(1.5 \%\)
Explanation:
A Given that, \(\frac{\Delta \mathrm{m}}{\mathrm{m}} \times 100=1 \%, \frac{\Delta \mathrm{v}}{\mathrm{v}} \times 100=2 \%\) \(\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2}\) taking \(\log\) of both side \(\log \mathrm{k}=\log \mathrm{m}+2 \log \mathrm{v}\) Differentiating both side \(\therefore \frac{\Delta \mathrm{K}}{\mathrm{K}} \times 100=\left(\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{2 \Delta \mathrm{v}}{\mathrm{v}}\right) \times 100\) \(\%\) error in \(\mathrm{K}=(\%\) error in \(\mathrm{m})+2 \times(\%\) error in \(\mathrm{v})\) \(=1 \%+2 \times 2 \%=5 \%\)
SRMJEE - 2010
Units and Measurements
139880
A Force \(F\) is applied on a square plate side \(L\). If the percentage error in the determination of \(L\) is \(2 \%\) and that in \(\mathrm{F}\) is \(4 \%\), what is the permissible error in pressure
