357964
if \(10^{10}\) electrons are acquired by a body every second, then the time required for the body to get a total charge of \(C\) will be
1 \(2\;h\)
2 2 days
3 \(2yr\)
4 \(20yr\)
Explanation:
1 electron has a charge of \((1.6 \times {10^{ - 19}}C).\) \(10^{10}\) electrons would have a charge of \(q=n e\) \( = 1.6 \times {10^{ - 19}} \times {10^{10}} = 1.6 \times {10^{ - 9}}C\) Thus in \(1\,\;s,\) charge accumulated \( = 1.6 \times {10^{ - 9}}\,C\) So, time taken to accumulate \(1\,C\) \( = \frac{1}{{1.6 \times {{10}^{ - 9}}}} = 0.625 \times {10^9}\;s\) \( = 173611\;h = 7233\,{\rm{days}}\) \( \approx 20{\rm{yr}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
357965
A body has a charge of \({-3.2 \mu C}\). The number of excess electrons it has is
1 \({5.12 \times 10^{25}}\)
2 \({5 \times 10^{12}}\)
3 \({2 \times 10^{13}}\)
4 \({5.12 \times 10^{13}}\)
Explanation:
Charge on any body is quantized. \({\therefore Q=n e}\) \({n=\dfrac{Q}{e}=\dfrac{3.2 \times 10^{-6}}{1.6 \times 10^{-19}}}\) \({\Rightarrow n=2 \times 10^{13}}\)
KCET - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
357966
The number of electrons in 2 coulomb of charge is
1 \(5 \times 10^{29}\)
2 \(12.5 \times 10^{18}\)
3 \(1.6 \times 10^{19}\)
4 \(9 \times 10^{11}\)
Explanation:
Since \(q=n e\) \(\therefore \quad 2=n \times 1.6 \times 10^{-19}\) or \(n=12.5 \times 10^{18}\).
PHXII01:ELECTRIC CHARGES AND FIELDS
357967
A cup contains 250 \(g\) of water. Find the number of positive charges present in the cup of water.
1 \(1.34 \times {10^{19}}C\)
2 \(1.34 \times {10^7}C\)
3 \(2.43 \times {10^{19}}C\)
4 \(2.43 \times {10^7}C\)
Explanation:
Mass of water \( = 250\,g\) Molecular mass of water \( = 18\,g\) Number of molecules in 18 g of water (Avogadro’s Number) \( = 6.02 \times {10^{23}}\) \(\therefore \) Number of molecules in one cup of water \( = \frac{{250}}{{18}} \times 6.02 \times {10^{23}}\) Each molecule of water contains two hydrogen atoms and one oxygen atom, \(i\).\(e\), 10 electrons and 10 protons. \(\therefore \) Total positive and negative charge has the same magnitude and is \(\frac{{250}}{{18}} \times 6.02 \times {10^{23}} \times 10 \times 1.6 \times {10^{ - 19}}C\) \( = 1.34 \times {10^7}C\)
357964
if \(10^{10}\) electrons are acquired by a body every second, then the time required for the body to get a total charge of \(C\) will be
1 \(2\;h\)
2 2 days
3 \(2yr\)
4 \(20yr\)
Explanation:
1 electron has a charge of \((1.6 \times {10^{ - 19}}C).\) \(10^{10}\) electrons would have a charge of \(q=n e\) \( = 1.6 \times {10^{ - 19}} \times {10^{10}} = 1.6 \times {10^{ - 9}}C\) Thus in \(1\,\;s,\) charge accumulated \( = 1.6 \times {10^{ - 9}}\,C\) So, time taken to accumulate \(1\,C\) \( = \frac{1}{{1.6 \times {{10}^{ - 9}}}} = 0.625 \times {10^9}\;s\) \( = 173611\;h = 7233\,{\rm{days}}\) \( \approx 20{\rm{yr}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
357965
A body has a charge of \({-3.2 \mu C}\). The number of excess electrons it has is
1 \({5.12 \times 10^{25}}\)
2 \({5 \times 10^{12}}\)
3 \({2 \times 10^{13}}\)
4 \({5.12 \times 10^{13}}\)
Explanation:
Charge on any body is quantized. \({\therefore Q=n e}\) \({n=\dfrac{Q}{e}=\dfrac{3.2 \times 10^{-6}}{1.6 \times 10^{-19}}}\) \({\Rightarrow n=2 \times 10^{13}}\)
KCET - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
357966
The number of electrons in 2 coulomb of charge is
1 \(5 \times 10^{29}\)
2 \(12.5 \times 10^{18}\)
3 \(1.6 \times 10^{19}\)
4 \(9 \times 10^{11}\)
Explanation:
Since \(q=n e\) \(\therefore \quad 2=n \times 1.6 \times 10^{-19}\) or \(n=12.5 \times 10^{18}\).
PHXII01:ELECTRIC CHARGES AND FIELDS
357967
A cup contains 250 \(g\) of water. Find the number of positive charges present in the cup of water.
1 \(1.34 \times {10^{19}}C\)
2 \(1.34 \times {10^7}C\)
3 \(2.43 \times {10^{19}}C\)
4 \(2.43 \times {10^7}C\)
Explanation:
Mass of water \( = 250\,g\) Molecular mass of water \( = 18\,g\) Number of molecules in 18 g of water (Avogadro’s Number) \( = 6.02 \times {10^{23}}\) \(\therefore \) Number of molecules in one cup of water \( = \frac{{250}}{{18}} \times 6.02 \times {10^{23}}\) Each molecule of water contains two hydrogen atoms and one oxygen atom, \(i\).\(e\), 10 electrons and 10 protons. \(\therefore \) Total positive and negative charge has the same magnitude and is \(\frac{{250}}{{18}} \times 6.02 \times {10^{23}} \times 10 \times 1.6 \times {10^{ - 19}}C\) \( = 1.34 \times {10^7}C\)
357964
if \(10^{10}\) electrons are acquired by a body every second, then the time required for the body to get a total charge of \(C\) will be
1 \(2\;h\)
2 2 days
3 \(2yr\)
4 \(20yr\)
Explanation:
1 electron has a charge of \((1.6 \times {10^{ - 19}}C).\) \(10^{10}\) electrons would have a charge of \(q=n e\) \( = 1.6 \times {10^{ - 19}} \times {10^{10}} = 1.6 \times {10^{ - 9}}C\) Thus in \(1\,\;s,\) charge accumulated \( = 1.6 \times {10^{ - 9}}\,C\) So, time taken to accumulate \(1\,C\) \( = \frac{1}{{1.6 \times {{10}^{ - 9}}}} = 0.625 \times {10^9}\;s\) \( = 173611\;h = 7233\,{\rm{days}}\) \( \approx 20{\rm{yr}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
357965
A body has a charge of \({-3.2 \mu C}\). The number of excess electrons it has is
1 \({5.12 \times 10^{25}}\)
2 \({5 \times 10^{12}}\)
3 \({2 \times 10^{13}}\)
4 \({5.12 \times 10^{13}}\)
Explanation:
Charge on any body is quantized. \({\therefore Q=n e}\) \({n=\dfrac{Q}{e}=\dfrac{3.2 \times 10^{-6}}{1.6 \times 10^{-19}}}\) \({\Rightarrow n=2 \times 10^{13}}\)
KCET - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
357966
The number of electrons in 2 coulomb of charge is
1 \(5 \times 10^{29}\)
2 \(12.5 \times 10^{18}\)
3 \(1.6 \times 10^{19}\)
4 \(9 \times 10^{11}\)
Explanation:
Since \(q=n e\) \(\therefore \quad 2=n \times 1.6 \times 10^{-19}\) or \(n=12.5 \times 10^{18}\).
PHXII01:ELECTRIC CHARGES AND FIELDS
357967
A cup contains 250 \(g\) of water. Find the number of positive charges present in the cup of water.
1 \(1.34 \times {10^{19}}C\)
2 \(1.34 \times {10^7}C\)
3 \(2.43 \times {10^{19}}C\)
4 \(2.43 \times {10^7}C\)
Explanation:
Mass of water \( = 250\,g\) Molecular mass of water \( = 18\,g\) Number of molecules in 18 g of water (Avogadro’s Number) \( = 6.02 \times {10^{23}}\) \(\therefore \) Number of molecules in one cup of water \( = \frac{{250}}{{18}} \times 6.02 \times {10^{23}}\) Each molecule of water contains two hydrogen atoms and one oxygen atom, \(i\).\(e\), 10 electrons and 10 protons. \(\therefore \) Total positive and negative charge has the same magnitude and is \(\frac{{250}}{{18}} \times 6.02 \times {10^{23}} \times 10 \times 1.6 \times {10^{ - 19}}C\) \( = 1.34 \times {10^7}C\)
357964
if \(10^{10}\) electrons are acquired by a body every second, then the time required for the body to get a total charge of \(C\) will be
1 \(2\;h\)
2 2 days
3 \(2yr\)
4 \(20yr\)
Explanation:
1 electron has a charge of \((1.6 \times {10^{ - 19}}C).\) \(10^{10}\) electrons would have a charge of \(q=n e\) \( = 1.6 \times {10^{ - 19}} \times {10^{10}} = 1.6 \times {10^{ - 9}}C\) Thus in \(1\,\;s,\) charge accumulated \( = 1.6 \times {10^{ - 9}}\,C\) So, time taken to accumulate \(1\,C\) \( = \frac{1}{{1.6 \times {{10}^{ - 9}}}} = 0.625 \times {10^9}\;s\) \( = 173611\;h = 7233\,{\rm{days}}\) \( \approx 20{\rm{yr}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
357965
A body has a charge of \({-3.2 \mu C}\). The number of excess electrons it has is
1 \({5.12 \times 10^{25}}\)
2 \({5 \times 10^{12}}\)
3 \({2 \times 10^{13}}\)
4 \({5.12 \times 10^{13}}\)
Explanation:
Charge on any body is quantized. \({\therefore Q=n e}\) \({n=\dfrac{Q}{e}=\dfrac{3.2 \times 10^{-6}}{1.6 \times 10^{-19}}}\) \({\Rightarrow n=2 \times 10^{13}}\)
KCET - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
357966
The number of electrons in 2 coulomb of charge is
1 \(5 \times 10^{29}\)
2 \(12.5 \times 10^{18}\)
3 \(1.6 \times 10^{19}\)
4 \(9 \times 10^{11}\)
Explanation:
Since \(q=n e\) \(\therefore \quad 2=n \times 1.6 \times 10^{-19}\) or \(n=12.5 \times 10^{18}\).
PHXII01:ELECTRIC CHARGES AND FIELDS
357967
A cup contains 250 \(g\) of water. Find the number of positive charges present in the cup of water.
1 \(1.34 \times {10^{19}}C\)
2 \(1.34 \times {10^7}C\)
3 \(2.43 \times {10^{19}}C\)
4 \(2.43 \times {10^7}C\)
Explanation:
Mass of water \( = 250\,g\) Molecular mass of water \( = 18\,g\) Number of molecules in 18 g of water (Avogadro’s Number) \( = 6.02 \times {10^{23}}\) \(\therefore \) Number of molecules in one cup of water \( = \frac{{250}}{{18}} \times 6.02 \times {10^{23}}\) Each molecule of water contains two hydrogen atoms and one oxygen atom, \(i\).\(e\), 10 electrons and 10 protons. \(\therefore \) Total positive and negative charge has the same magnitude and is \(\frac{{250}}{{18}} \times 6.02 \times {10^{23}} \times 10 \times 1.6 \times {10^{ - 19}}C\) \( = 1.34 \times {10^7}C\)
357964
if \(10^{10}\) electrons are acquired by a body every second, then the time required for the body to get a total charge of \(C\) will be
1 \(2\;h\)
2 2 days
3 \(2yr\)
4 \(20yr\)
Explanation:
1 electron has a charge of \((1.6 \times {10^{ - 19}}C).\) \(10^{10}\) electrons would have a charge of \(q=n e\) \( = 1.6 \times {10^{ - 19}} \times {10^{10}} = 1.6 \times {10^{ - 9}}C\) Thus in \(1\,\;s,\) charge accumulated \( = 1.6 \times {10^{ - 9}}\,C\) So, time taken to accumulate \(1\,C\) \( = \frac{1}{{1.6 \times {{10}^{ - 9}}}} = 0.625 \times {10^9}\;s\) \( = 173611\;h = 7233\,{\rm{days}}\) \( \approx 20{\rm{yr}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
357965
A body has a charge of \({-3.2 \mu C}\). The number of excess electrons it has is
1 \({5.12 \times 10^{25}}\)
2 \({5 \times 10^{12}}\)
3 \({2 \times 10^{13}}\)
4 \({5.12 \times 10^{13}}\)
Explanation:
Charge on any body is quantized. \({\therefore Q=n e}\) \({n=\dfrac{Q}{e}=\dfrac{3.2 \times 10^{-6}}{1.6 \times 10^{-19}}}\) \({\Rightarrow n=2 \times 10^{13}}\)
KCET - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
357966
The number of electrons in 2 coulomb of charge is
1 \(5 \times 10^{29}\)
2 \(12.5 \times 10^{18}\)
3 \(1.6 \times 10^{19}\)
4 \(9 \times 10^{11}\)
Explanation:
Since \(q=n e\) \(\therefore \quad 2=n \times 1.6 \times 10^{-19}\) or \(n=12.5 \times 10^{18}\).
PHXII01:ELECTRIC CHARGES AND FIELDS
357967
A cup contains 250 \(g\) of water. Find the number of positive charges present in the cup of water.
1 \(1.34 \times {10^{19}}C\)
2 \(1.34 \times {10^7}C\)
3 \(2.43 \times {10^{19}}C\)
4 \(2.43 \times {10^7}C\)
Explanation:
Mass of water \( = 250\,g\) Molecular mass of water \( = 18\,g\) Number of molecules in 18 g of water (Avogadro’s Number) \( = 6.02 \times {10^{23}}\) \(\therefore \) Number of molecules in one cup of water \( = \frac{{250}}{{18}} \times 6.02 \times {10^{23}}\) Each molecule of water contains two hydrogen atoms and one oxygen atom, \(i\).\(e\), 10 electrons and 10 protons. \(\therefore \) Total positive and negative charge has the same magnitude and is \(\frac{{250}}{{18}} \times 6.02 \times {10^{23}} \times 10 \times 1.6 \times {10^{ - 19}}C\) \( = 1.34 \times {10^7}C\)
