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Class 12th Chemistry

Given below are two statements :

Statement I : Emulsions of oil in water are unstable and sometimes they separate into two layers on standing.

Statement II : For stabilization of an emulsion, excess of electrolyte is added. In the light of the above statements, choose the most appropriate answer from the options given below :

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Vishal Baghel

Contributor-Level 10

Emulsions gets separated into two layers on standing.

For stabilization of emulsion. Emulsifying agents added into it but not electrolyte.

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Class 12th Physics

Statement – I : The reactance of an ac circuit is zero. It is possible that the circuit contains a capacitor and an inductor.

Statement – II : In ac circuit, the average power delivered by the source never becomes zero.

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

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Payal Gupta

Contributor-Level 10

Z = $\sqrt[]{R^{2} + {(x_{C} - x_{L})}^{2}},$ if only L & C are present then R = 0 then p = 0

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Class 12th Physics Moving Charges and Magnetism

A proton, a deuteron and an α - particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is

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Payal Gupta

Contributor-Level 10

$r = \frac{m v}{q B} = \frac{\sqrt[]{2 m k . E}}{q B}$

$r \propto \frac{\sqrt[]{m}}{q}$

m_p = m

q_p = q

m_d = 2m

q_d = q

m_a = 4m

q_a = 2q

$\frac{r_{p}}{r_{d}} = \frac{\sqrt[]{m_{p}} * q_{d}}{q_{P} * \sqrt[]{m_{d}}}$

$\frac{r_{d}}{r_{α}} = \sqrt[]{2}$

$r_{p} : r_{d} : r_{α} = 1 : \sqrt[]{2} : 1$

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Class 12th Physics

The soft-iron is a suitable material for making an electromagnet. This is because soft iron has

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Payal Gupta

Contributor-Level 10

High permeability and low retentivity

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Class 12th Physics

What will be the most suitable combination of three resistors A = $2 Ω$ , B = $4 Ω$ , C = $6 Ω$ so that $(\frac{22}{3}) Ω$ is equivalent resistance of combination

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Payal Gupta

Contributor-Level 10

Given

$\begin{array}{l} A = 2 Ω \\ B = 4 Ω \\ C = 6 Ω \\ R_{e q} = \frac{2 * 4}{2 + 4} + 6 \end{array}$

$R_{e q} = \frac{22}{3} Ω$

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Class 12th Maths

let f(x) be a quadratic polynomial such that f(-2) + f(3) = 0. If one of the roots of f(x) = 0 is -1, then the sum of the roots of f(x) = 0 is equal to:

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alok kumar singh

Contributor-Level 10

$f (- 2) + f (3) = 0 .$ One root of f (x) = 0 is (-1)

Let's assume other root to be a

$∴ f (x) = a (x + 1) (x - α)$

Given that f (-2) + f (3) = 0

a (-2 + 1) (-2 -a) + a (3 + 1) (3 - a) = 0

Þ 14 -3a = 0 Þ a = $\frac{14}{3}$

$∴$ sum of roots = $\frac{14}{3} - 1 = \frac{11}{3}$

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Class 12th Physics

In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emef's of two cells respectively is 3:2, the difference in the balancing length of the potentiometer wire in above two cases will be __________ cm.

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Vishal Baghel

Contributor-Level 10

$E α l$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$

$\Rightarrow \frac{3}{2} = \frac{75}{l_{2}}$

$l_{2} = \frac{75 * 2}{3}$

$l_{2} = 50 c m$

$& l_{1} = 75 c m$

$Δ l = l_{1} - l_{2} = 25 c m$

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Class 12th Maths

Let R₁ = ${(a, b) \in N * N : | a - b | \leq 13} a n d$

R₂ = ${(a, b) \in N * N : | a - b | \neq 13} .$ Then on N :

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alok kumar singh

Contributor-Level 10

R₁ = ${(a, 1) \in N * N : | a - b | \leq 13}$

$(a, a) \in R_{1} a s | a - a | \leq 13 (R e f l e x i v e)$

$(a, b) & (b, a) \in R_{1} a s | a - b | = | b - a | \to (s y m m e t r i c) .$

But it is not necessary that if (a,b) & (b, c) $\in R, t h e n (a, c) \in R$

Eg $- (21, 10) \in R & (10, 1) \in R b u t (21, 1) \notin R_{1}$

R₂ = ${(a, b) \in N * N : | a - b | \neq 13 ∴ R_{2} \to N o t e q u i v a l e n c e s o l u t o i n .}$

$(a, b) & (b, a) \in R_{2} a s | a - b | = | b - a |$

But it is not necessary that if (a, b) & (b, c) $\in R_{2}$ then (a, c) also $\in R_{2}$ .

Eg – (21, 1) $\in R_{2} & (1, 8) \in R_{2} b u t (21, 8) \notin R_{2}$

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Class 12th Physics

Two identical charged particles each having a mass 10g and charge 2.0 * 10^-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10ms^-2]

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Payal Gupta

Contributor-Level 10

$μ m g = \frac{k q_{1} q_{2}}{L^{2}}$

$\Rightarrow 0.25 * 0.01 * 10 = \frac{9 * 1 0^{9} * 2 * 1 0^{- 7} * 2 * 1 0^{- 7}}{L^{2}}$

$= \frac{9 * 2^{2} * 1 0^{- 5 + 4}}{25 * 10}$

L = 12 cm

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Class 12th Physics

When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is $υ_{1} .$ When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes $υ_{2} .$ If $υ_{2} = x υ_{1},$ the value of x will be __________.

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Vishal Baghel

Contributor-Level 10

hu = hu₀ + K.E

Cases u = 2u₀

h2u₀ = hu₀ + K.E₁

K.E₁ = hu₀

$\frac{1}{2} m v_{1}^{2} = h υ_{0}$

$v_{1}^{2} = \frac{2 h υ_{0}}{m}$

$v_{1} = \sqrt[]{\frac{2 h υ_{0}}{m}}$ - (1)

Now, cases 2

h 5u₀ = hu₀ + k.E₂

k.E₂ = 4hu₀

$\frac{1}{2} m v_{2}^{2} = 4 h υ_{0}$

v₂ = $\sqrt[]{\frac{8 h υ_{0}}{m}}$ - (2)

$\frac{v_{2}}{v_{1}} = \sqrt[]{\frac{8}{2} = 2}$

v₂ = 2v₁

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