Class 12th

Contributor-Level 10

$C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{40 * 40 k}{40 (1 + k)} = 24$

$\begin{array}{l} \Rightarrow 40 k = 24 + 24 k \\ \Rightarrow 16 k = 24 \end{array}$

$k = \frac{24}{16} = \frac{3}{2} = 1.5$

0 0

New answer posted

8 months ago

A 1m long wire is broken into two unequal parts X and Y. The X part of the wire is stretched into another wire W. Length of W is twice the length of X and the resistance of W is twice that of Y. Find the ratio of length of X and Y.

0 Follower 4 Views

Contributor-Level 10

R_w = 2R_y

$\Rightarrow ρ (\frac{2 x}{\frac{A}{2}}) = \frac{2 ρ (1 - x)}{A}$

$\frac{4 ρ x}{A} = \frac{2 ρ (1 - x)}{A}$

->4x = 2 – 2x Þ 6x = 2

x = $\frac{2}{6} = \frac{1}{3}$ $∴ \frac{L_{x}}{L_{y}} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{3} * \frac{3}{2}$

$= \frac{1}{2}$

0 0

New answer posted

8 months ago

Given below are two statements: one is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : Alloys such as constantan and manganin are used in making standard resistance coils.

Reason R : Constantan and maganin have very small value of temperature coefficient of resistance.

In the light of the above statements, choose the correct answer from the options given below:

0 Follower 1 View

Contributor-Level 10

Based on theoretical data.

0 0

New answer posted

8 months ago

Class 12th Thermodynamics

A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by 40°C, its efficiency increases by 30%. The temperature of the source will be:

0 Follower 6 Views

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)

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

0 0

New answer posted

8 months ago

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:

0 Follower 3 Views

Class 12th Physics Electrostatic Potential and Capacitance

Contributor-Level 10

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

0 0

New answer posted

8 months ago

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:

0 Follower 4 Views

Contributor-Level 10

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

...more

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

offer

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

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

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

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

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

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

less

0 0

New answer posted

8 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)

0 Follower 2 Views

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

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

0 0

New answer posted

8 months ago

Class 12th Science

0 Follower 12 Views

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.

0 0

New answer posted

8 months ago

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

0 Follower 1 View

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

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

0 0

New answer posted

8 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

0 Follower 2 Views