165653
$2 \mu \mathrm{F}$ capacitor is connected with $50 \mathrm{~V}$ supply \& $3 \mu \mathrm{F}$ capacitor is connected with $100 \mathrm{~V}$ supply. Now after removing battery if two plates of same type of charges are placed to form new capacitor then potential difference is
165654
Assertion : Two concentric charged shells are given. The potential difference between the shells depends on charge of inner shell. Reason : Potential due to charge of outer shell remains same at every point inside the sphere.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion
2 If both Assertion and Reason are correct, but Reason is not a correct explanation of the Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect.
Explanation:
: \(\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{Q}_1+\mathrm{Q}_2}{\mathrm{R}_2}\) $\mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{Q}_{1}}{\mathrm{R}_{1}}+\frac{\mathrm{Q}_{2}}{\mathrm{R}_{2}}\right)$ Potential difference, $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}$ $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{Q}_{1}}{\mathrm{R}_{1}}+\frac{\mathrm{Q}_{2}}{\mathrm{R}_{2}}\right)-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{Q}_{1}+\mathrm{Q}_{2}}{\mathrm{R}_{2}}\right)$ $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \mathrm{Q}_{1}\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$ Therefore Assertion and reason are correct and the reason is a correct explanation of the assertion.
[AIIMS-2010]
Capacitance
165656
If the separation between the plates of a capacitor is $5 \mathrm{~mm}$. then the area of the plate of a $3 F$ parallel plate capacitor is :
1 $4.259 \times 10^{9} \mathrm{~m}^{2}$
2 $1.964 \times 10^{9} \mathrm{~m}^{2}$
3 $12.81 \times 10^{9} \mathrm{~m}^{2}$
4 $1.694 \times 10^{9} \mathrm{~m}^{2}$
Explanation:
: Given, The separation between the plates, $\mathrm{d}=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m}, \mathrm{C}=3 \mathrm{~F}$ The Expression of parallel plate capacitor is given by , $\mathrm{C}=\frac{\mathrm{A} \varepsilon_{0}}{\mathrm{~d}}$ So, area of plate, $\mathrm{A}=\frac{\mathrm{Cd}}{\varepsilon_{0}}=\frac{3 \times 5 \times 10^{-3}}{8.85 \times 10^{-12}} \mathrm{~m}^{2}$ $\mathrm{~A}=1.694 \times 10^{-3} \times 10^{12}$ $\mathrm{~A}=1.694 \times 10^{9} \mathrm{~m}^{2}$
[AIIMS-1998]
Capacitance
165657
An air parallel plate capacitor has capacity $C$. The capacity and distance between plates are doubled when immersed in a liquid then dielectric constant of the liquid is :
1 1
2 2
3 3
4 4
Explanation:
: Given, $\mathrm{C}^{\prime}=2 \mathrm{C}$ Capacity of air parallel plate capacitor, $\mathrm{C}=\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}$ When immersed in liquid, then distance is doubled $\mathrm{C}^{\prime}=\frac{\mathrm{KA} \varepsilon_{\mathrm{o}}}{2 \mathrm{~d}}$ or, $2 \mathrm{C} =\frac{\mathrm{K}}{2} \mathrm{C}$ $\mathrm{K} =4$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Capacitance
165653
$2 \mu \mathrm{F}$ capacitor is connected with $50 \mathrm{~V}$ supply \& $3 \mu \mathrm{F}$ capacitor is connected with $100 \mathrm{~V}$ supply. Now after removing battery if two plates of same type of charges are placed to form new capacitor then potential difference is
165654
Assertion : Two concentric charged shells are given. The potential difference between the shells depends on charge of inner shell. Reason : Potential due to charge of outer shell remains same at every point inside the sphere.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion
2 If both Assertion and Reason are correct, but Reason is not a correct explanation of the Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect.
Explanation:
: \(\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{Q}_1+\mathrm{Q}_2}{\mathrm{R}_2}\) $\mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{Q}_{1}}{\mathrm{R}_{1}}+\frac{\mathrm{Q}_{2}}{\mathrm{R}_{2}}\right)$ Potential difference, $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}$ $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{Q}_{1}}{\mathrm{R}_{1}}+\frac{\mathrm{Q}_{2}}{\mathrm{R}_{2}}\right)-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{Q}_{1}+\mathrm{Q}_{2}}{\mathrm{R}_{2}}\right)$ $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \mathrm{Q}_{1}\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$ Therefore Assertion and reason are correct and the reason is a correct explanation of the assertion.
[AIIMS-2010]
Capacitance
165656
If the separation between the plates of a capacitor is $5 \mathrm{~mm}$. then the area of the plate of a $3 F$ parallel plate capacitor is :
1 $4.259 \times 10^{9} \mathrm{~m}^{2}$
2 $1.964 \times 10^{9} \mathrm{~m}^{2}$
3 $12.81 \times 10^{9} \mathrm{~m}^{2}$
4 $1.694 \times 10^{9} \mathrm{~m}^{2}$
Explanation:
: Given, The separation between the plates, $\mathrm{d}=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m}, \mathrm{C}=3 \mathrm{~F}$ The Expression of parallel plate capacitor is given by , $\mathrm{C}=\frac{\mathrm{A} \varepsilon_{0}}{\mathrm{~d}}$ So, area of plate, $\mathrm{A}=\frac{\mathrm{Cd}}{\varepsilon_{0}}=\frac{3 \times 5 \times 10^{-3}}{8.85 \times 10^{-12}} \mathrm{~m}^{2}$ $\mathrm{~A}=1.694 \times 10^{-3} \times 10^{12}$ $\mathrm{~A}=1.694 \times 10^{9} \mathrm{~m}^{2}$
[AIIMS-1998]
Capacitance
165657
An air parallel plate capacitor has capacity $C$. The capacity and distance between plates are doubled when immersed in a liquid then dielectric constant of the liquid is :
1 1
2 2
3 3
4 4
Explanation:
: Given, $\mathrm{C}^{\prime}=2 \mathrm{C}$ Capacity of air parallel plate capacitor, $\mathrm{C}=\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}$ When immersed in liquid, then distance is doubled $\mathrm{C}^{\prime}=\frac{\mathrm{KA} \varepsilon_{\mathrm{o}}}{2 \mathrm{~d}}$ or, $2 \mathrm{C} =\frac{\mathrm{K}}{2} \mathrm{C}$ $\mathrm{K} =4$
165653
$2 \mu \mathrm{F}$ capacitor is connected with $50 \mathrm{~V}$ supply \& $3 \mu \mathrm{F}$ capacitor is connected with $100 \mathrm{~V}$ supply. Now after removing battery if two plates of same type of charges are placed to form new capacitor then potential difference is
165654
Assertion : Two concentric charged shells are given. The potential difference between the shells depends on charge of inner shell. Reason : Potential due to charge of outer shell remains same at every point inside the sphere.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion
2 If both Assertion and Reason are correct, but Reason is not a correct explanation of the Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect.
Explanation:
: \(\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{Q}_1+\mathrm{Q}_2}{\mathrm{R}_2}\) $\mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{Q}_{1}}{\mathrm{R}_{1}}+\frac{\mathrm{Q}_{2}}{\mathrm{R}_{2}}\right)$ Potential difference, $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}$ $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{Q}_{1}}{\mathrm{R}_{1}}+\frac{\mathrm{Q}_{2}}{\mathrm{R}_{2}}\right)-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{Q}_{1}+\mathrm{Q}_{2}}{\mathrm{R}_{2}}\right)$ $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \mathrm{Q}_{1}\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$ Therefore Assertion and reason are correct and the reason is a correct explanation of the assertion.
[AIIMS-2010]
Capacitance
165656
If the separation between the plates of a capacitor is $5 \mathrm{~mm}$. then the area of the plate of a $3 F$ parallel plate capacitor is :
1 $4.259 \times 10^{9} \mathrm{~m}^{2}$
2 $1.964 \times 10^{9} \mathrm{~m}^{2}$
3 $12.81 \times 10^{9} \mathrm{~m}^{2}$
4 $1.694 \times 10^{9} \mathrm{~m}^{2}$
Explanation:
: Given, The separation between the plates, $\mathrm{d}=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m}, \mathrm{C}=3 \mathrm{~F}$ The Expression of parallel plate capacitor is given by , $\mathrm{C}=\frac{\mathrm{A} \varepsilon_{0}}{\mathrm{~d}}$ So, area of plate, $\mathrm{A}=\frac{\mathrm{Cd}}{\varepsilon_{0}}=\frac{3 \times 5 \times 10^{-3}}{8.85 \times 10^{-12}} \mathrm{~m}^{2}$ $\mathrm{~A}=1.694 \times 10^{-3} \times 10^{12}$ $\mathrm{~A}=1.694 \times 10^{9} \mathrm{~m}^{2}$
[AIIMS-1998]
Capacitance
165657
An air parallel plate capacitor has capacity $C$. The capacity and distance between plates are doubled when immersed in a liquid then dielectric constant of the liquid is :
1 1
2 2
3 3
4 4
Explanation:
: Given, $\mathrm{C}^{\prime}=2 \mathrm{C}$ Capacity of air parallel plate capacitor, $\mathrm{C}=\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}$ When immersed in liquid, then distance is doubled $\mathrm{C}^{\prime}=\frac{\mathrm{KA} \varepsilon_{\mathrm{o}}}{2 \mathrm{~d}}$ or, $2 \mathrm{C} =\frac{\mathrm{K}}{2} \mathrm{C}$ $\mathrm{K} =4$
165653
$2 \mu \mathrm{F}$ capacitor is connected with $50 \mathrm{~V}$ supply \& $3 \mu \mathrm{F}$ capacitor is connected with $100 \mathrm{~V}$ supply. Now after removing battery if two plates of same type of charges are placed to form new capacitor then potential difference is
165654
Assertion : Two concentric charged shells are given. The potential difference between the shells depends on charge of inner shell. Reason : Potential due to charge of outer shell remains same at every point inside the sphere.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion
2 If both Assertion and Reason are correct, but Reason is not a correct explanation of the Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect.
Explanation:
: \(\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{Q}_1+\mathrm{Q}_2}{\mathrm{R}_2}\) $\mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{Q}_{1}}{\mathrm{R}_{1}}+\frac{\mathrm{Q}_{2}}{\mathrm{R}_{2}}\right)$ Potential difference, $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}$ $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{Q}_{1}}{\mathrm{R}_{1}}+\frac{\mathrm{Q}_{2}}{\mathrm{R}_{2}}\right)-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{Q}_{1}+\mathrm{Q}_{2}}{\mathrm{R}_{2}}\right)$ $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \mathrm{Q}_{1}\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$ Therefore Assertion and reason are correct and the reason is a correct explanation of the assertion.
[AIIMS-2010]
Capacitance
165656
If the separation between the plates of a capacitor is $5 \mathrm{~mm}$. then the area of the plate of a $3 F$ parallel plate capacitor is :
1 $4.259 \times 10^{9} \mathrm{~m}^{2}$
2 $1.964 \times 10^{9} \mathrm{~m}^{2}$
3 $12.81 \times 10^{9} \mathrm{~m}^{2}$
4 $1.694 \times 10^{9} \mathrm{~m}^{2}$
Explanation:
: Given, The separation between the plates, $\mathrm{d}=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m}, \mathrm{C}=3 \mathrm{~F}$ The Expression of parallel plate capacitor is given by , $\mathrm{C}=\frac{\mathrm{A} \varepsilon_{0}}{\mathrm{~d}}$ So, area of plate, $\mathrm{A}=\frac{\mathrm{Cd}}{\varepsilon_{0}}=\frac{3 \times 5 \times 10^{-3}}{8.85 \times 10^{-12}} \mathrm{~m}^{2}$ $\mathrm{~A}=1.694 \times 10^{-3} \times 10^{12}$ $\mathrm{~A}=1.694 \times 10^{9} \mathrm{~m}^{2}$
[AIIMS-1998]
Capacitance
165657
An air parallel plate capacitor has capacity $C$. The capacity and distance between plates are doubled when immersed in a liquid then dielectric constant of the liquid is :
1 1
2 2
3 3
4 4
Explanation:
: Given, $\mathrm{C}^{\prime}=2 \mathrm{C}$ Capacity of air parallel plate capacitor, $\mathrm{C}=\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}$ When immersed in liquid, then distance is doubled $\mathrm{C}^{\prime}=\frac{\mathrm{KA} \varepsilon_{\mathrm{o}}}{2 \mathrm{~d}}$ or, $2 \mathrm{C} =\frac{\mathrm{K}}{2} \mathrm{C}$ $\mathrm{K} =4$
