Class 12th

Class 12th Physics Electrostatic Potential and Capacitance

Contributor-Level 10

Efficiency $η = 50 %$

$\Rightarrow η = (1 - \frac{T_{2}}{T_{1}}) * 100 = 50$

$\Rightarrow 1 - \frac{T_{2}}{T_{1}} = \frac{50}{100} \Rightarrow 1 - \frac{T_{2}}{T_{1}} = \frac{1}{2}$ -(i)

After the change :- $1 - \frac{T_{2}^{1}}{T_{1}} = \frac{50 * (1.3)}{100} = \frac{65}{100} = . 65$ $\frac{_{}^{}}{_{}} \frac{}{} \frac{}{}$ -(ii)

$\Rightarrow 1 - (\frac{T_{2} - 40}{T_{1}}) = . 65$ --(ii)

$\Rightarrow \frac{40}{T_{1}} = . 65 - . 50 = . 15 \Rightarrow T_{1} = \frac{4000}{15} = 266.7 k$

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Two identical thin metal plates has charge q₁ and q₂ respectively such that q₁ > q₂. The plates were brought close to each other to from a parallel plate capacitor of capacitance C. The potential difference between them is:

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Class 12th Physics Electrostatic Potential and Capacitance

Contributor-Level 10

V = $\frac{q_{(c a p a c i t a n c e)}}{C} = \frac{q_{1} - q_{2}}{2 c}$

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Two identical metallic spheres A and B when placed at certain distance in air repel each other with force of F. Another identical uncharged sphere C is first place in contact with A and then in contact with B and finally placed at midpoint between sphere A and B. The force experienced by sphere C will be:

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Contributor-Level 10

For sphere 'C' -> after contacting with 'A'. q_A = q_C = $\frac{q}{2} .$

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For sphere 'C' -> after contacting with 'A'. q_A = q_C = $\frac{q}{2} .$

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contacting with 'B'. q_B = q_C = $\frac{3 q}{4}$

$F_{N e t} = | F_{1} - F_{2} |$

F₂ = $\frac{K \cdot 3 q^{2} * 4}{r^{2} 8} = \frac{3}{2} \frac{k q^{2}}{r^{2}}$

$F_{1} = \frac{9 k q^{2} * 4}{16 r^{2}} = \frac{9 k q^{2}}{4 r^{2}} = \frac{9}{4} F$

$∴ F_{N e t} = | \frac{9}{4} F - \frac{3}{2} F |$

= $\frac{9 - 6}{4} F = \frac{3}{4} F$

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11 months ago

The dimensions of $(\frac{B^{2}}{μ_{0}}) w i l l b e$ $\frac{^{}}{_{}}$

(if $_{}$ $μ_{0}$ : permeability of free space and B : magnetic field)

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Where can we get set 2 question paper of class 12th Physics?

Contributor-Level 10

$D i m e n s i o n o f \frac{B^{2}}{μ_{0}} : -$ Magnetic energy density

$u_{B} = \frac{B^{2}}{2 μ_{0}}$

Dimension $\frac{J}{m^{3}} \to [M L^{2} T^{- 2} L^{- 3}]$ $\frac{}{^{}}^{}^{}^{}$

$= [M L^{- 1} T^{- 2}]$

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Class 12th Science

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Varshitha M

Contributor-Level 7

You can obtain Class 12 Physics Set 2 question paper from the school, teachers, or by checking your previous year's board exam paper booklet. Coaching institutes may have printed versions available, or Libraries might hold copies. There are several educational apps and booklet such as CBSE sample paper which contains every set for practice, there is also generally included Set 2 with marking schemes.

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A beam of electromagnetic radiation of intensity $6.4 * 10^{- 5} W / {c m}^{2}$ is comprised of wavelength, $λ = 310 n m$ . It falls normally on a metal (work function $φ = 2 e V$ ) of surface area $1 {c m}^{2}$ . If one in $10^{3}$ photons ejects an electron, total number of electrons ejected in is $10^{x} . (h_{c} = 1240 e V n m, 1 e V = 1.6 * 10^{- 19} J)$ , then $x$ is

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Contributor-Level 10

Energy of photon, $E = \frac{1240}{310} = 4 e V > 2 e V$ $_{}_{}_{}_{}$ (so photoelectric effect will take place)

$= 4 * 1.6 * 10^{- 19} = 6.4 * 10^{- 19} Joule$

Number of photons falling per second

$= \frac{6.4 * 10^{- 5} * 1}{6.4 * 10^{- 19}} = 10^{14}$

Number of photoelectron emitted per second

$= \frac{10^{14}}{10^{3}} = 10^{11}$

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11 months ago

The balancing length for a cell is $560 c m$ in a potentiometer experiment. When an external resistance of $10 Ω$ is connected in parallel to the cell, the balancing length changes by $60 c m$ . If the internal resistance of the cell is $\frac{N}{10} Ω$ , where $N$ is an integer then value of $N$ is

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Class 12th Physics Electrostatic Potential and Capacitance

Contributor-Level 10

Loss of energy $= \frac{1}{2} {C V}^{2} - \frac{1}{2} (2 C) * V_{Common}^{2}$ $\frac{}{}^{} \frac{}{}_{}^{}$

$\begin{array}{r} = \frac{1}{2} * 60 * 10^{- 12} * (20)^{2} - 60 * 10^{- 12} * (10)^{2} \\ = 60 * 10^{- 12} (200 - 100) \\ = 6000 * 10^{- 12} \\ = 6 n J \end{array}$

n=12

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11 months ago

A $60 p F$ capacitor is fully charged by a $20 V$ supply. It is then disconnected from the supply and is connected to another uncharged $60 p F$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $n J$ )

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Contributor-Level 10

Common potential after connection.

$V_{common} = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} = \frac{60 * 20 + 0}{120} = 10 V o l t$

Loss of energy $= \frac{1}{2} {C V}^{2} - \frac{1}{2} (2 C) * V_{Common}^{2}$

\begin{array}{r} = \frac{1}{2} * 60 * 10^{- 12} * (20)^{2} - 60 * 10^{- 12} * (10)^{2} \\ = 60 * 10^{- 12} (200 - 100) \\ = 6000 * 10^{- 12} \\ = 6 n J \end{array}

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11 months ago

A loop ABCDEFA of straight edges has six corner points $A (0,0, 0), B (5,0, 0), C (5,5$ , $0), D (0,5, 0), E (0,5, 5)$ and $F (0,0, 5)$ . The magnetic field in this region in $\overset{?}{B} = (3 \hat{i} + 4 \hat{k}) T$ . The quantity of flux through the loop ABCDEFA (in Wb) is

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Contributor-Level 10

$? = \overset{?}{B} \cdot \overset{?}{A} = (3 \hat{i} + 4 \hat{k}) \cdot (25 \hat{i} + 25 \hat{k})$

$? = (3 * 25) + (4 * 25) = 175 weber$

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11 months ago

Class 12th Maths

Let $S = {(x, y) \in N * N : 9 {(x - 3)}^{2} + 16 {(y - 4)}^{2} \leq 144}$ and $T = {(x, y) \in R * R : {(x - 7)}^{2} + {(y - 4)}^{2} \leq 36} .$

Then $n (S \cap T)$ is equal to………….

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