B When mass number increases, surface area also increases. We know that, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ Where, $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ $\mathrm{A}=$ mass number $\mathrm{R} \propto \mathrm{A}^{1 / 3}$
J and K CET- 2007
ATOMS
145293
In a Thomson set up, $\mathrm{E}=30 \mathrm{~V} \mathrm{~cm}^{-1}$ and $\mathrm{B}=6$ G. Then speed of the electron that goes un deflected in the common region of the two fields will be
1 $15 \times 10^{6} \mathrm{~m} / \mathrm{s}$
2 $25 \times 10^{6} \mathrm{~m} / \mathrm{s}$
3 $5 \times 10^{6} \mathrm{~m} / \mathrm{s}$
4 $1 \times 10^{6} \mathrm{~m} / \mathrm{s}$
Explanation:
C For no deflection of electron, force due to magnetic field $=$ force due to electric field Or $\quad \mathrm{Bev}=\mathrm{Ee}$ $\mathrm{v} =\frac{\mathrm{E}}{\mathrm{B}}$ $\mathrm{v} =\frac{30 \times 100}{6 \times 10^{-4}}$ $=5 \times 10^{6} \mathrm{~m} / \mathrm{sec} \quad\left(\because 1 \text { tesla }=10^{4} \text { gauss }\right)$
J and K CET- 2002
ATOMS
145296
Match the following |(A) Michael Faraday|(e) Quantum model of Hydrogen atom| | |(B) Niels Bohr|(f) Laws of electromagnetic induction| |(C) J.J Thomson|(g) Discovery of Neutron| |(D) Chadwick|(h) Discovery of Electron|
1 **A** -(h),**B** -(g),**C** -(e),**D** -(f)
2 **A** -(c),**B** -(f),**C** -(h),**D** -(g)
3 **A** -(f),**B** -(e),**C** -(h),**D** -(g)
4 **A** -(g),**B** -(e),**C** -(h),**D** -(f)
Explanation:
C | | Name of Scientist | | Discoveries | | :--- | :--- | :---: | :--- | | A. | Michael faraday | f. | Laws of electro- \lt br> magnetic induction | | B. | Niels Bohr | e. | Quantum model of \lt br> Hydrogen atom | | C. | J.J. Thomson | h. | Discovery of electron | | D. | Chadwick | g. | Discovery of neutron |
TSEAMCET(Medical)-2015
ATOMS
145297
In Millikan's oil drop experiment, the charge on oil drop is calculated to $6.35 \times 10^{-19} \mathrm{C}$. The number of excess electrons on the drop is
1 3.9
2 4
3 4.2
4 6
Explanation:
B We know that, $\text { Total charge }(\mathrm{Q})=6.35 \times 10^{-19} \mathrm{C}$ $\because \quad \mathrm{Q}=\mathrm{ne} \quad(\mathrm{n}=\text { number of electrons })$ $\therefore \quad \mathrm{n}=\frac{\mathrm{Q}}{\mathrm{e}}=\frac{6.35 \times 10^{-19}}{1.6 \times 10^{-19}}=4$
B When mass number increases, surface area also increases. We know that, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ Where, $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ $\mathrm{A}=$ mass number $\mathrm{R} \propto \mathrm{A}^{1 / 3}$
J and K CET- 2007
ATOMS
145293
In a Thomson set up, $\mathrm{E}=30 \mathrm{~V} \mathrm{~cm}^{-1}$ and $\mathrm{B}=6$ G. Then speed of the electron that goes un deflected in the common region of the two fields will be
1 $15 \times 10^{6} \mathrm{~m} / \mathrm{s}$
2 $25 \times 10^{6} \mathrm{~m} / \mathrm{s}$
3 $5 \times 10^{6} \mathrm{~m} / \mathrm{s}$
4 $1 \times 10^{6} \mathrm{~m} / \mathrm{s}$
Explanation:
C For no deflection of electron, force due to magnetic field $=$ force due to electric field Or $\quad \mathrm{Bev}=\mathrm{Ee}$ $\mathrm{v} =\frac{\mathrm{E}}{\mathrm{B}}$ $\mathrm{v} =\frac{30 \times 100}{6 \times 10^{-4}}$ $=5 \times 10^{6} \mathrm{~m} / \mathrm{sec} \quad\left(\because 1 \text { tesla }=10^{4} \text { gauss }\right)$
J and K CET- 2002
ATOMS
145296
Match the following |(A) Michael Faraday|(e) Quantum model of Hydrogen atom| | |(B) Niels Bohr|(f) Laws of electromagnetic induction| |(C) J.J Thomson|(g) Discovery of Neutron| |(D) Chadwick|(h) Discovery of Electron|
1 **A** -(h),**B** -(g),**C** -(e),**D** -(f)
2 **A** -(c),**B** -(f),**C** -(h),**D** -(g)
3 **A** -(f),**B** -(e),**C** -(h),**D** -(g)
4 **A** -(g),**B** -(e),**C** -(h),**D** -(f)
Explanation:
C | | Name of Scientist | | Discoveries | | :--- | :--- | :---: | :--- | | A. | Michael faraday | f. | Laws of electro- \lt br> magnetic induction | | B. | Niels Bohr | e. | Quantum model of \lt br> Hydrogen atom | | C. | J.J. Thomson | h. | Discovery of electron | | D. | Chadwick | g. | Discovery of neutron |
TSEAMCET(Medical)-2015
ATOMS
145297
In Millikan's oil drop experiment, the charge on oil drop is calculated to $6.35 \times 10^{-19} \mathrm{C}$. The number of excess electrons on the drop is
1 3.9
2 4
3 4.2
4 6
Explanation:
B We know that, $\text { Total charge }(\mathrm{Q})=6.35 \times 10^{-19} \mathrm{C}$ $\because \quad \mathrm{Q}=\mathrm{ne} \quad(\mathrm{n}=\text { number of electrons })$ $\therefore \quad \mathrm{n}=\frac{\mathrm{Q}}{\mathrm{e}}=\frac{6.35 \times 10^{-19}}{1.6 \times 10^{-19}}=4$
B When mass number increases, surface area also increases. We know that, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ Where, $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ $\mathrm{A}=$ mass number $\mathrm{R} \propto \mathrm{A}^{1 / 3}$
J and K CET- 2007
ATOMS
145293
In a Thomson set up, $\mathrm{E}=30 \mathrm{~V} \mathrm{~cm}^{-1}$ and $\mathrm{B}=6$ G. Then speed of the electron that goes un deflected in the common region of the two fields will be
1 $15 \times 10^{6} \mathrm{~m} / \mathrm{s}$
2 $25 \times 10^{6} \mathrm{~m} / \mathrm{s}$
3 $5 \times 10^{6} \mathrm{~m} / \mathrm{s}$
4 $1 \times 10^{6} \mathrm{~m} / \mathrm{s}$
Explanation:
C For no deflection of electron, force due to magnetic field $=$ force due to electric field Or $\quad \mathrm{Bev}=\mathrm{Ee}$ $\mathrm{v} =\frac{\mathrm{E}}{\mathrm{B}}$ $\mathrm{v} =\frac{30 \times 100}{6 \times 10^{-4}}$ $=5 \times 10^{6} \mathrm{~m} / \mathrm{sec} \quad\left(\because 1 \text { tesla }=10^{4} \text { gauss }\right)$
J and K CET- 2002
ATOMS
145296
Match the following |(A) Michael Faraday|(e) Quantum model of Hydrogen atom| | |(B) Niels Bohr|(f) Laws of electromagnetic induction| |(C) J.J Thomson|(g) Discovery of Neutron| |(D) Chadwick|(h) Discovery of Electron|
1 **A** -(h),**B** -(g),**C** -(e),**D** -(f)
2 **A** -(c),**B** -(f),**C** -(h),**D** -(g)
3 **A** -(f),**B** -(e),**C** -(h),**D** -(g)
4 **A** -(g),**B** -(e),**C** -(h),**D** -(f)
Explanation:
C | | Name of Scientist | | Discoveries | | :--- | :--- | :---: | :--- | | A. | Michael faraday | f. | Laws of electro- \lt br> magnetic induction | | B. | Niels Bohr | e. | Quantum model of \lt br> Hydrogen atom | | C. | J.J. Thomson | h. | Discovery of electron | | D. | Chadwick | g. | Discovery of neutron |
TSEAMCET(Medical)-2015
ATOMS
145297
In Millikan's oil drop experiment, the charge on oil drop is calculated to $6.35 \times 10^{-19} \mathrm{C}$. The number of excess electrons on the drop is
1 3.9
2 4
3 4.2
4 6
Explanation:
B We know that, $\text { Total charge }(\mathrm{Q})=6.35 \times 10^{-19} \mathrm{C}$ $\because \quad \mathrm{Q}=\mathrm{ne} \quad(\mathrm{n}=\text { number of electrons })$ $\therefore \quad \mathrm{n}=\frac{\mathrm{Q}}{\mathrm{e}}=\frac{6.35 \times 10^{-19}}{1.6 \times 10^{-19}}=4$
B When mass number increases, surface area also increases. We know that, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ Where, $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ $\mathrm{A}=$ mass number $\mathrm{R} \propto \mathrm{A}^{1 / 3}$
J and K CET- 2007
ATOMS
145293
In a Thomson set up, $\mathrm{E}=30 \mathrm{~V} \mathrm{~cm}^{-1}$ and $\mathrm{B}=6$ G. Then speed of the electron that goes un deflected in the common region of the two fields will be
1 $15 \times 10^{6} \mathrm{~m} / \mathrm{s}$
2 $25 \times 10^{6} \mathrm{~m} / \mathrm{s}$
3 $5 \times 10^{6} \mathrm{~m} / \mathrm{s}$
4 $1 \times 10^{6} \mathrm{~m} / \mathrm{s}$
Explanation:
C For no deflection of electron, force due to magnetic field $=$ force due to electric field Or $\quad \mathrm{Bev}=\mathrm{Ee}$ $\mathrm{v} =\frac{\mathrm{E}}{\mathrm{B}}$ $\mathrm{v} =\frac{30 \times 100}{6 \times 10^{-4}}$ $=5 \times 10^{6} \mathrm{~m} / \mathrm{sec} \quad\left(\because 1 \text { tesla }=10^{4} \text { gauss }\right)$
J and K CET- 2002
ATOMS
145296
Match the following |(A) Michael Faraday|(e) Quantum model of Hydrogen atom| | |(B) Niels Bohr|(f) Laws of electromagnetic induction| |(C) J.J Thomson|(g) Discovery of Neutron| |(D) Chadwick|(h) Discovery of Electron|
1 **A** -(h),**B** -(g),**C** -(e),**D** -(f)
2 **A** -(c),**B** -(f),**C** -(h),**D** -(g)
3 **A** -(f),**B** -(e),**C** -(h),**D** -(g)
4 **A** -(g),**B** -(e),**C** -(h),**D** -(f)
Explanation:
C | | Name of Scientist | | Discoveries | | :--- | :--- | :---: | :--- | | A. | Michael faraday | f. | Laws of electro- \lt br> magnetic induction | | B. | Niels Bohr | e. | Quantum model of \lt br> Hydrogen atom | | C. | J.J. Thomson | h. | Discovery of electron | | D. | Chadwick | g. | Discovery of neutron |
TSEAMCET(Medical)-2015
ATOMS
145297
In Millikan's oil drop experiment, the charge on oil drop is calculated to $6.35 \times 10^{-19} \mathrm{C}$. The number of excess electrons on the drop is
1 3.9
2 4
3 4.2
4 6
Explanation:
B We know that, $\text { Total charge }(\mathrm{Q})=6.35 \times 10^{-19} \mathrm{C}$ $\because \quad \mathrm{Q}=\mathrm{ne} \quad(\mathrm{n}=\text { number of electrons })$ $\therefore \quad \mathrm{n}=\frac{\mathrm{Q}}{\mathrm{e}}=\frac{6.35 \times 10^{-19}}{1.6 \times 10^{-19}}=4$
