NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
151716
In an atom electron revolve around the nucleus along a path of radius $0.72 \AA$ making $9.4 \times 10^{18}$ revolutions per second. The equivalent current is [Given $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ ]
1 $1.4 \mathrm{~A}$
2 $1.8 \mathrm{~A}$
3 $1.2 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
D Given, Radius of electron orbit $(\mathrm{r})=0.72 \AA=0.72 \times 10^{-10} \mathrm{~m}$ Frequency $(\mathrm{f})=9.4 \times 10^{18} \mathrm{rev} / \mathrm{s}$. $\because$ Equivalent current, $\mathrm{I}=\frac{\mathrm{e}}{\mathrm{t}}=$ e.f $\quad\left(\therefore \mathrm{f}=\frac{1}{\mathrm{t}}\right)$ $\mathrm{I}=1.6 \times 10^{-19} \times 9.4 \times 10^{18} .$ $\mathrm{I}=1.504 \square 1.5 \mathrm{~A}$
Karnataka CET-2022
Current Electricity
151732
Identify the correct variation of drift velocity $\left(v_{d}\right)$ with electric field strength $(E)$ :
A We know that, Current density, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ $\because$ Current, $\quad \mathrm{I}=\mathrm{neAv} v_{\mathrm{d}}$ So, $\quad J=\frac{n e A v_{d}}{A}$ $\mathrm{J}=$ nev $_{\mathrm{d}}$ Also, current density $\mathrm{J}$ is related to electric field $\mathrm{E}$ $\mathrm{J}=\sigma \mathrm{E}$ $n v_{\mathrm{d}}=\sigma \mathrm{E} \quad\left[\because \mathrm{J}=\mathrm{nev}_{\mathrm{d}}\right]$ Hence, $v_{d} \propto E$
AP EAMCET-06.09.2021
Current Electricity
151747
When a potential difference $V$ is applied across a conductor at a temperature $T$, the drift velocity of electrons is proportional to
1 $\sqrt{\mathrm{V}}$
2 $\mathrm{V}$
3 $\sqrt{\mathrm{T}}$
4 $\mathrm{T}$
Explanation:
B When a potential difference $\mathrm{V}$ is applied across a conductor at a temperature $\mathrm{T}$, the drift velocity of electrons is proportional to $\mathrm{V}$. According to ohm's law- $\therefore \quad \text { current, } \mathrm{I}=\text { neA } \mathrm{v}_{\mathrm{d}}$ Where, $\mathrm{v}_{\mathrm{d}}=$ drift velocity From equation (i), we get- $\mathrm{V}=\mathrm{neAv}_{\mathrm{d}} \times \mathrm{R}$ $\mathrm{V}_{\mathrm{d}} \propto \mathrm{V}$
BITSAT-2019
Current Electricity
151750
A cylindrical conductor of diameter $0.1 \mathrm{~mm}$ carries a current of $90 \mathrm{~mA}$. The current density (in $\mathrm{Am}^{-2}$ ) is $(\pi \square 3)$ :
1 $1.2 \times 10^{7}$
2 $3 \times 10^{6}$
3 $6 \times 10^{6}$
4 $2.4 \times 10^{7}$
Explanation:
A Given, current (I) $=90 \mathrm{~mA}=90 \times 10^{-3} \mathrm{~A}$ Diameter of cylindrical conductor $(\mathrm{d})=0.1 \mathrm{~mm}$ Current density, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ $=\frac{\mathrm{I}}{\left(\frac{\pi \mathrm{d}^{2}}{4}\right)}=\frac{90 \times 10^{-3}}{\frac{\pi \times\left(0.1 \times 10^{-3}\right)^{2}}{4}}$ $=\frac{120 \times 10^{-3}}{10^{-8}}$ $\mathrm{~J}=120 \times 10^{5}=1.2 \times 10^{7} \mathrm{Am}^{-2}$
151716
In an atom electron revolve around the nucleus along a path of radius $0.72 \AA$ making $9.4 \times 10^{18}$ revolutions per second. The equivalent current is [Given $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ ]
1 $1.4 \mathrm{~A}$
2 $1.8 \mathrm{~A}$
3 $1.2 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
D Given, Radius of electron orbit $(\mathrm{r})=0.72 \AA=0.72 \times 10^{-10} \mathrm{~m}$ Frequency $(\mathrm{f})=9.4 \times 10^{18} \mathrm{rev} / \mathrm{s}$. $\because$ Equivalent current, $\mathrm{I}=\frac{\mathrm{e}}{\mathrm{t}}=$ e.f $\quad\left(\therefore \mathrm{f}=\frac{1}{\mathrm{t}}\right)$ $\mathrm{I}=1.6 \times 10^{-19} \times 9.4 \times 10^{18} .$ $\mathrm{I}=1.504 \square 1.5 \mathrm{~A}$
Karnataka CET-2022
Current Electricity
151732
Identify the correct variation of drift velocity $\left(v_{d}\right)$ with electric field strength $(E)$ :
A We know that, Current density, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ $\because$ Current, $\quad \mathrm{I}=\mathrm{neAv} v_{\mathrm{d}}$ So, $\quad J=\frac{n e A v_{d}}{A}$ $\mathrm{J}=$ nev $_{\mathrm{d}}$ Also, current density $\mathrm{J}$ is related to electric field $\mathrm{E}$ $\mathrm{J}=\sigma \mathrm{E}$ $n v_{\mathrm{d}}=\sigma \mathrm{E} \quad\left[\because \mathrm{J}=\mathrm{nev}_{\mathrm{d}}\right]$ Hence, $v_{d} \propto E$
AP EAMCET-06.09.2021
Current Electricity
151747
When a potential difference $V$ is applied across a conductor at a temperature $T$, the drift velocity of electrons is proportional to
1 $\sqrt{\mathrm{V}}$
2 $\mathrm{V}$
3 $\sqrt{\mathrm{T}}$
4 $\mathrm{T}$
Explanation:
B When a potential difference $\mathrm{V}$ is applied across a conductor at a temperature $\mathrm{T}$, the drift velocity of electrons is proportional to $\mathrm{V}$. According to ohm's law- $\therefore \quad \text { current, } \mathrm{I}=\text { neA } \mathrm{v}_{\mathrm{d}}$ Where, $\mathrm{v}_{\mathrm{d}}=$ drift velocity From equation (i), we get- $\mathrm{V}=\mathrm{neAv}_{\mathrm{d}} \times \mathrm{R}$ $\mathrm{V}_{\mathrm{d}} \propto \mathrm{V}$
BITSAT-2019
Current Electricity
151750
A cylindrical conductor of diameter $0.1 \mathrm{~mm}$ carries a current of $90 \mathrm{~mA}$. The current density (in $\mathrm{Am}^{-2}$ ) is $(\pi \square 3)$ :
1 $1.2 \times 10^{7}$
2 $3 \times 10^{6}$
3 $6 \times 10^{6}$
4 $2.4 \times 10^{7}$
Explanation:
A Given, current (I) $=90 \mathrm{~mA}=90 \times 10^{-3} \mathrm{~A}$ Diameter of cylindrical conductor $(\mathrm{d})=0.1 \mathrm{~mm}$ Current density, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ $=\frac{\mathrm{I}}{\left(\frac{\pi \mathrm{d}^{2}}{4}\right)}=\frac{90 \times 10^{-3}}{\frac{\pi \times\left(0.1 \times 10^{-3}\right)^{2}}{4}}$ $=\frac{120 \times 10^{-3}}{10^{-8}}$ $\mathrm{~J}=120 \times 10^{5}=1.2 \times 10^{7} \mathrm{Am}^{-2}$
151716
In an atom electron revolve around the nucleus along a path of radius $0.72 \AA$ making $9.4 \times 10^{18}$ revolutions per second. The equivalent current is [Given $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ ]
1 $1.4 \mathrm{~A}$
2 $1.8 \mathrm{~A}$
3 $1.2 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
D Given, Radius of electron orbit $(\mathrm{r})=0.72 \AA=0.72 \times 10^{-10} \mathrm{~m}$ Frequency $(\mathrm{f})=9.4 \times 10^{18} \mathrm{rev} / \mathrm{s}$. $\because$ Equivalent current, $\mathrm{I}=\frac{\mathrm{e}}{\mathrm{t}}=$ e.f $\quad\left(\therefore \mathrm{f}=\frac{1}{\mathrm{t}}\right)$ $\mathrm{I}=1.6 \times 10^{-19} \times 9.4 \times 10^{18} .$ $\mathrm{I}=1.504 \square 1.5 \mathrm{~A}$
Karnataka CET-2022
Current Electricity
151732
Identify the correct variation of drift velocity $\left(v_{d}\right)$ with electric field strength $(E)$ :
A We know that, Current density, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ $\because$ Current, $\quad \mathrm{I}=\mathrm{neAv} v_{\mathrm{d}}$ So, $\quad J=\frac{n e A v_{d}}{A}$ $\mathrm{J}=$ nev $_{\mathrm{d}}$ Also, current density $\mathrm{J}$ is related to electric field $\mathrm{E}$ $\mathrm{J}=\sigma \mathrm{E}$ $n v_{\mathrm{d}}=\sigma \mathrm{E} \quad\left[\because \mathrm{J}=\mathrm{nev}_{\mathrm{d}}\right]$ Hence, $v_{d} \propto E$
AP EAMCET-06.09.2021
Current Electricity
151747
When a potential difference $V$ is applied across a conductor at a temperature $T$, the drift velocity of electrons is proportional to
1 $\sqrt{\mathrm{V}}$
2 $\mathrm{V}$
3 $\sqrt{\mathrm{T}}$
4 $\mathrm{T}$
Explanation:
B When a potential difference $\mathrm{V}$ is applied across a conductor at a temperature $\mathrm{T}$, the drift velocity of electrons is proportional to $\mathrm{V}$. According to ohm's law- $\therefore \quad \text { current, } \mathrm{I}=\text { neA } \mathrm{v}_{\mathrm{d}}$ Where, $\mathrm{v}_{\mathrm{d}}=$ drift velocity From equation (i), we get- $\mathrm{V}=\mathrm{neAv}_{\mathrm{d}} \times \mathrm{R}$ $\mathrm{V}_{\mathrm{d}} \propto \mathrm{V}$
BITSAT-2019
Current Electricity
151750
A cylindrical conductor of diameter $0.1 \mathrm{~mm}$ carries a current of $90 \mathrm{~mA}$. The current density (in $\mathrm{Am}^{-2}$ ) is $(\pi \square 3)$ :
1 $1.2 \times 10^{7}$
2 $3 \times 10^{6}$
3 $6 \times 10^{6}$
4 $2.4 \times 10^{7}$
Explanation:
A Given, current (I) $=90 \mathrm{~mA}=90 \times 10^{-3} \mathrm{~A}$ Diameter of cylindrical conductor $(\mathrm{d})=0.1 \mathrm{~mm}$ Current density, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ $=\frac{\mathrm{I}}{\left(\frac{\pi \mathrm{d}^{2}}{4}\right)}=\frac{90 \times 10^{-3}}{\frac{\pi \times\left(0.1 \times 10^{-3}\right)^{2}}{4}}$ $=\frac{120 \times 10^{-3}}{10^{-8}}$ $\mathrm{~J}=120 \times 10^{5}=1.2 \times 10^{7} \mathrm{Am}^{-2}$
151716
In an atom electron revolve around the nucleus along a path of radius $0.72 \AA$ making $9.4 \times 10^{18}$ revolutions per second. The equivalent current is [Given $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ ]
1 $1.4 \mathrm{~A}$
2 $1.8 \mathrm{~A}$
3 $1.2 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
D Given, Radius of electron orbit $(\mathrm{r})=0.72 \AA=0.72 \times 10^{-10} \mathrm{~m}$ Frequency $(\mathrm{f})=9.4 \times 10^{18} \mathrm{rev} / \mathrm{s}$. $\because$ Equivalent current, $\mathrm{I}=\frac{\mathrm{e}}{\mathrm{t}}=$ e.f $\quad\left(\therefore \mathrm{f}=\frac{1}{\mathrm{t}}\right)$ $\mathrm{I}=1.6 \times 10^{-19} \times 9.4 \times 10^{18} .$ $\mathrm{I}=1.504 \square 1.5 \mathrm{~A}$
Karnataka CET-2022
Current Electricity
151732
Identify the correct variation of drift velocity $\left(v_{d}\right)$ with electric field strength $(E)$ :
A We know that, Current density, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ $\because$ Current, $\quad \mathrm{I}=\mathrm{neAv} v_{\mathrm{d}}$ So, $\quad J=\frac{n e A v_{d}}{A}$ $\mathrm{J}=$ nev $_{\mathrm{d}}$ Also, current density $\mathrm{J}$ is related to electric field $\mathrm{E}$ $\mathrm{J}=\sigma \mathrm{E}$ $n v_{\mathrm{d}}=\sigma \mathrm{E} \quad\left[\because \mathrm{J}=\mathrm{nev}_{\mathrm{d}}\right]$ Hence, $v_{d} \propto E$
AP EAMCET-06.09.2021
Current Electricity
151747
When a potential difference $V$ is applied across a conductor at a temperature $T$, the drift velocity of electrons is proportional to
1 $\sqrt{\mathrm{V}}$
2 $\mathrm{V}$
3 $\sqrt{\mathrm{T}}$
4 $\mathrm{T}$
Explanation:
B When a potential difference $\mathrm{V}$ is applied across a conductor at a temperature $\mathrm{T}$, the drift velocity of electrons is proportional to $\mathrm{V}$. According to ohm's law- $\therefore \quad \text { current, } \mathrm{I}=\text { neA } \mathrm{v}_{\mathrm{d}}$ Where, $\mathrm{v}_{\mathrm{d}}=$ drift velocity From equation (i), we get- $\mathrm{V}=\mathrm{neAv}_{\mathrm{d}} \times \mathrm{R}$ $\mathrm{V}_{\mathrm{d}} \propto \mathrm{V}$
BITSAT-2019
Current Electricity
151750
A cylindrical conductor of diameter $0.1 \mathrm{~mm}$ carries a current of $90 \mathrm{~mA}$. The current density (in $\mathrm{Am}^{-2}$ ) is $(\pi \square 3)$ :
1 $1.2 \times 10^{7}$
2 $3 \times 10^{6}$
3 $6 \times 10^{6}$
4 $2.4 \times 10^{7}$
Explanation:
A Given, current (I) $=90 \mathrm{~mA}=90 \times 10^{-3} \mathrm{~A}$ Diameter of cylindrical conductor $(\mathrm{d})=0.1 \mathrm{~mm}$ Current density, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ $=\frac{\mathrm{I}}{\left(\frac{\pi \mathrm{d}^{2}}{4}\right)}=\frac{90 \times 10^{-3}}{\frac{\pi \times\left(0.1 \times 10^{-3}\right)^{2}}{4}}$ $=\frac{120 \times 10^{-3}}{10^{-8}}$ $\mathrm{~J}=120 \times 10^{5}=1.2 \times 10^{7} \mathrm{Am}^{-2}$
