Class 12th

A capacitor is discharging through a resistor R. Consider in time t₁ the energy stored in the capacitor reduces to half of its initial value and in time t₂, the charge stored in the capacitor reduces to half of its initial value and in time t₂ the charge stored reduces to one eight of its initial value. The ratio t₁/t₂ will be

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Contributor-Level 10

According to Law of discharging of capacitor, we can write

$Q = Q_{0} e^{- \frac{t}{τ}} \Rightarrow \frac{Q_{2}}{8} = Q_{0} e^{- \frac{t_{2}}{τ}} \Rightarrow t_{2} 3 τ l n (2), a n d$

$U = \frac{Q^{2}}{2 C} = \frac{Q_{0}^{2}}{2 C} e^{- \frac{2 t}{τ}} \Rightarrow \frac{1}{2} (\frac{Q_{0}^{2}}{2 C}) = \frac{Q_{0}^{2}}{2 C} e^{- \frac{2 t}{τ}} \Rightarrow t_{1} = \frac{τ}{2} l n (2)$

$\Rightarrow \frac{t_{1}}{t_{2}} = \frac{1}{6}$

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3 months ago

The electric field at a point associated with a light wave is given by

E = 200 $[s i n (6 * 1 0^{15}) t + s i n (9 * 1 0^{15}) t] V m^{- 1}$

Given : h = 4.14 * 10^-¹⁵ eVs

If the light fall on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be

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Contributor-Level 10

The light wave contains two lights of different frequencies, so

$E_{1} = h υ_{1} = \frac{4.14 * 1 0^{- 15} * 6 * 1 0^{15}}{2 π} = 3.96 e V, a n d$

$E_{2} = h υ_{2} = \frac{4.14 * 1 0^{- 15} * 9 * 1 0^{15}}{2 π} = 5.92 e V$

Maximum kinetic energy of the photoelectron = 5.9-2.5 = 3.42 eV

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New answer posted

3 months ago

$t a n (2 t a n^{- 1} \frac{1}{5} + s e c^{- 1} \frac{\sqrt[]{5}}{2} + 2 t a n^{- 1} \frac{1}{8})$ is equal to:

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Contributor-Level 9

$t a n (2 t a n^{- 1} \frac{1}{5} + s e c^{- 1} \frac{\sqrt[]{5}}{2} + 2 t a n^{- 1} \frac{1}{8}) t a n (2 t a n^{- 1} \frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{5} * \frac{1}{8}} + s e c^{- 1} \frac{\sqrt[]{5}}{2})$

$= t a n (t a n^{- 1} \frac{3}{4} + t a n^{- 1} \frac{1}{2}) = t a n (t a n^{- 1} \frac{\frac{3}{4} + \frac{1}{2}}{1 - \frac{3}{8}})$

$= t a n (t a n^{- 1} \frac{\frac{5}{4}}{\frac{5}{8}}) = 2$

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New answer posted

3 months ago

In the following unclear reaction,

D $\overset{α}{\to} D_{1} \overset{β^{-}}{\to} D_{2} \overset{α}{\to} D_{3} \overset{γ}{\to} D_{4}$

Mass number of D is 182 and atomic number is 74. Mass number and atomic number of D₄ respectively will be________

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Contributor-Level 10

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3 months ago

Let S = ${θ \in [0, 2 π] : 8^{2 s i n^{2} θ} + 8^{2 c o s^{2} θ} = 16} .$ Then $n (S) + \sum_{θ \in S}^{} (s e c (\frac{π}{4} + 2 θ) c o s e c (\frac{π}{4} + 2 θ))$ is equal to:

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Class 12th Physics Ray Optics and Optical Instruments

Contributor-Level 9

$S = {θ \in [0, 2 π] : 8^{2 s i n^{2} x} + 8^{2 c o s^{2} x} = 16}$

Now apply AM $\geq G M$ for $\frac{8^{2 s i n^{2} x} + 8^{2 c o s^{2} x}}{2} \leq {(8^{2 s i n^{2} x + 2 c o s^{2} x})}^{\frac{1}{2}} \Rightarrow 8^{2 s i n^{2} x} = 8^{2 c o s^{2} x}$

Þ $s i n^{2} θ = c o s^{2} θ$ $∴ θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$

$= 4 + [c o s e c (\frac{π}{2} + π) + c o s e c (\frac{π}{2} + 3 π) + c o s e c (\frac{π}{2} + 5 π) + c o s e c (\frac{π}{2} + 7 π)]$

$= 4 - 2 (4) = - 4$

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3 months ago

The speed of light in media 'A' and 'B' are 2.0 * 10¹⁰ cm/s and 1.5 * 10¹⁰ cm/s respectively. A ray of light enters from the medium B to an incident angle 'q'. If the ray suffers total internal reflection, then

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Contributor-Level 10

$\frac{s i n θ_{C}}{V_{B}} = \frac{s i n 90 °}{V_{A}} \Rightarrow s i n θ_{C} = \frac{V_{B}}{V_{A}} = \frac{1.5 * 1 0^{10}}{2.0 * 1 0^{10}} = \frac{3}{4}$

According to question, we can write

$θ > θ = s i n^{- 1} (\frac{3}{4})$

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3 months ago

Let E₁, E₂, E₃ be three mutually exclusive events such that P(E₁) = $\frac{2 + 3 p}{6}$ , P(E₂) = $\frac{2 - p}{8}$ and P(E₃) = $\frac{1 - p}{2} .$ If the maximum and minimum values of p are p₁ and p₂, then (p₁ + p₂) is equal to:

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Contributor-Level 9

$0 \leq \frac{2 + 3 p}{6} \leq 1 \Rightarrow p \in [- \frac{2}{3}, \frac{4}{3}], 0 \leq \frac{2 - p}{8} \leq 1 \Rightarrow p \in [- 6, 2] a n d 0 \leq \frac{1 - p}{2} \leq 1 \Rightarrow p \in [- 1, 1]$

$0 < P (E_{1}) + P (E_{2}) + P (E_{3}) \leq 10 \leq \frac{13}{12} - \frac{p}{8} \leq 1 \Rightarrow p \in [\frac{2}{3}, \frac{26}{3}]$ Taking intersection to all $p \in [\frac{2}{3}, 1]$

$∴ p_{1} + p_{2} = \frac{5}{3}$

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3 months ago

The mean and variance of a binomial distribution are a and $\frac{α}{3}$ respectively. If P(X = 1) = $\frac{4}{243}$ then P(X = 4 or 5) is equal to:

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Contributor-Level 9

Given, mean = np = a. and variance = npq = $\frac{α}{3} \Rightarrow q = \frac{1}{3} a n d p = \frac{2}{3}$

$P (X = 1) = n p^{1} q^{n - 1} = \frac{4}{243} \Rightarrow n {(\frac{2}{3})}^{1} {(\frac{1}{3})}^{n - 1} = \frac{4}{243} \Rightarrow n = 6$

$P (X = 4 o r 5) =^{6} C_{4} {(\frac{2}{3})}^{4} {(\frac{1}{3})}^{2} +^{6} C_{5} {(\frac{2}{3})}^{5} {(\frac{1}{3})}^{1} = \frac{16}{27}$

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3 months ago

Two point charge Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is:

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Contributor-Level 10

According to question, we can write

^{$F_{C A} = F_{C B} = \frac{K q Q}{x^{2} + {(\frac{d}{2})}^{2}} \Rightarrow F = 2 F_{C A} c o s θ = \frac{2 K q Q x}{{[x^{2} + {(\frac{d}{2})}^{2}]}^{\frac{3}{2}}}$}

For maxima of force $\frac{d F}{d x} = 0, s o$

$x = \frac{d}{2 \sqrt[]{2}}$

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New answer posted

3 months ago

Class 12th Vector Algebra

Let $\vec{a} = α \hat{i} + \hat{j} - \hat{k} a n d \vec{b} = 2 \hat{i} + \hat{j} - α \hat{k}, α > 0 .$ If the projection of $\vec{a} * \vec{b}$ on the vector $- \hat{i} + 2 \hat{j} - 2 \hat{k}$ is 30, then a equal to:

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