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Electric Current Physics Current Electricity Physics Class 12th

A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in V) shown in the figure is:

(Take $\frac{1}{4 π ε_{0}} = 9 \times 1 0^{9}$ SI units)

Option 1 -

$3 * 1 0^{5}$

Option 2 -

$1 * 1 0^{5}$

Option 3 -

$0.5 * 1 0^{5}$

Option 4 -

Zero

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

For uniformly charged spherical shell,

$V = \frac{k q}{R} (For r \leq R)$

$∴ V_{C} = V_{P}$

$V_{C} - V_{P} = Zero$

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A wire of length 'l' and resistance $100 Ω$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

alok kumar singh

Divided into 10 parts

$R = \frac{ρ l}{A}$

$R^{'} = \frac{ρ l}{10 A} = \frac{R}{10}$

$R_{S} = 5 \times \frac{R}{10} [series]$

$R_{S} = 50$

$R_{P} = \frac{R}{50} [parallel]$

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$= 52 Ω$

A uniform heating wire of resistance 36 $Ω$ is connected across a potential difference of 240V. The wire is then cut into half and a potential difference of 240 V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1 : x, where x is________.

alok kumar singh

In first condition R₁ = 36 $Ω$

In second condition R₂ = 18 $Ω$

$P_{1} = \frac{V^{2}}{R_{1}} = \frac{{(240)}^{2}}{36}$

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$P_{2} = \frac{{(240)}^{2}}{9}$

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$Current in circuit i = \frac{10}{4 + 1} = 2 A$

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$l = J A c o s θ \Rightarrow 5 = J \times 0.04 \times c o s 60 ° \Rightarrow J = \frac{5}{0.02} = 250 A / m^{2}$

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$l_{1} = \frac{50}{2000} = 25 m A, a n d l = \frac{V_{i} - V_{Z}}{1000} = \frac{100 - 50}{1000} = 50 m A$

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A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in V) shown in the figure is:

(Take $\frac{1}{4 π ε_{0}} = 9 \times 1 0^{9}$ SI units)

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A thin spherical shell is charged by some source. The potential difference between the two points C and P (in V) shown in the figure is: (Take 1 4 π ε 0 = 9 × 1 0 9 SI units)

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A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in V) shown in the figure is:

(Take $\frac{1}{4 π ε_{0}} = 9 \times 1 0^{9}$ SI units)