Class 12th

If the plane P passes through the intersection of two mutually perpendicular planes 2x + ky – 5z = 1 and 3kx – ky + 5, k < 3 and intercepts a unit length on positive x-axis, then the intercept made by the plane P on the y-axis is

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Contributor-Level 10

$π_{1} \equiv 2 x + k y - 5 z = 1$

$π_{2} \equiv 3 k x - k y + z = 5$

Given $π_{1} ⊥ π_{2} \Rightarrow 6 k - k^{2} - 5 = 0$

k = 1, 5

Now required plane is p₁ + $λ π_{2} = 0$

y intercept = $\frac{1 + 5 λ}{1 - λ}$

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New answer posted

a year ago

$(p \land r) \Leftrightarrow (p \land (\sim q))$ is equivalent to $(\sim p)$ when r is

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Class 12th Dual Nature of Radiation and Matter

Contributor-Level 10

Draw the truthtable

p q $p \land \sim q$ $p \land q$ $(p \land q) \Leftrightarrow (p \land \sim q)$

T F T F

...more

Draw the truthtable

p q $p \land \sim q$ $p \land q$ $(p \land q) \Leftrightarrow (p \land \sim q)$

T F T F

T F T F F

F T F T

F F T

So, r = q

less

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New answer posted

a year ago

An α particle and a proton are accelerated form rest through the same potential difference. The ratio of linear momenta acquired by above two particals will be

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Contributor-Level 10

$k_{α} = 2 e V \Rightarrow P_{α} = \sqrt[]{2 K_{α} m_{α}}$

k_P = eV P_P = $\sqrt[]{2 K_{P} m_{P}}$

$∴ \frac{P_{α}}{P_{P}} = \sqrt[]{\frac{K_{α} m_{α}}{K_{P} m_{P}}} = \sqrt[]{\frac{2 * (4 m_{P})}{m_{P}}} \begin{matrix} \sqrt[]{8} : 1 & 2 \sqrt[]{2} : 1 \end{matrix}$

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New answer posted

a year ago

Class 12th Physics

An unpolarised light beam of intensity 2l₀ is passed through a polaroid P and then through another polaroid Q which is oriented is such a way that its passing axis makes an angle of 30° relative to that of P. The intensity of the emergent light is

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Contributor-Level 10

$I = I_{0} c o s^{2} 30 °$

$= I_{0} {(\frac{\sqrt[]{3}}{2})}^{2} = \frac{3}{4} I_{0}$

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New answer posted

a year ago

Let S be the sample space of all five digit numbers. It p is the probability that a randomly selected number form S, is a multiple of 7 but not divisible by 5, then 9p is equal to

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Contributor-Level 10

Find total (5 digit) number of divisors of 7. Find total (5 digit) number of divisors of 35

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New answer posted

a year ago

Class 12th Physics

A circuit element X when connected to an a.c. supply of peak voltage 100V given a peak current of 5A which is in phase with the voltage. A second element Y when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by $\frac{π}{2} .$ If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

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Contributor-Level 10

v = 100 sin $ω t$

$i_{0} = \frac{100}{R_{x}} = 5 \Rightarrow R_{x} = 20 Ω$

i = $5 s i n (ω t - \frac{π}{2})$

i = 100 5sin $ω t$

$5 = \frac{100}{R_{y}} \Rightarrow R_{y} = \frac{100}{5} = 20 Ω$

When x and y both are connected in series :-

v = 100sin $ω t$

tan = 1 = 45°

R₀ = $\sqrt[]{R_{x}^{2} + R_{y}^{2}} = 20 \sqrt[]{2} Ω$

$∴ \overset{o}{l_{0}} = \frac{v_{0}}{R_{0}} = \frac{100}{20 \sqrt[]{2}} = \frac{5}{\sqrt[]{2}} A .$

$∴ l_{m s} = \frac{l_{0}}{\sqrt[]{2}}$

$= \frac{5}{\sqrt[]{2} \cdot \sqrt[]{2}}$

$= \frac{5}{2} A = \frac{5}{2} A$

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New answer posted

a year ago

Class 12th Determinants

Let $\vec{a} = 2 \hat{i} - \hat{j} + 5 \hat{k} a n d \vec{b} = α \hat{i} + β \hat{j} + 2 \hat{k} .$ If $((\vec{a} * \vec{b}) * \hat{i}) . \hat{k} = \frac{23}{2},$ then $| \vec{b} * 2 \hat{j} |$ is equal to

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Class 12th Physics Moving Charges and Magnetism

Contributor-Level 10

$(\vec{a} * \vec{b}) * \hat{i} = (\vec{a} . \hat{i}) \vec{b} - (\vec{b} . \hat{i}) \vec{a}$

$((\vec{a} * \vec{b}) * \hat{i}) . \hat{k} = (\vec{a} . \hat{i}) (\vec{b} . \hat{k}) - (\vec{b} . \hat{i}) (\vec{a} . \hat{k})$

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New answer posted

a year ago

A wire X of length 50cm carrying a current of 2A is placed parallel to a long wire Y of length 5m. The wire Y current of 3A. The distance between two wires is 5cm and currents flow in the same direction. The force acting on the wire Y is

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Contributor-Level 10

F_x-y = $\frac{μ_{0} i_{1} i_{2}}{2 π r} * (. 5)$

$= \frac{2 * 1 0^{- 7} * 6 * 0.5}{0.05}$

$= \frac{6}{5} * 1 0^{- 5} N = 1.2 * 1 0^{- 5}$

Force on Y = 1.2 * 105N towards 'x'

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New answer posted

a year ago

Let $\vec{a} = α \hat{i} + \hat{j} + β \hat{k} a n d \vec{b} = 3 \hat{i} - 5 \hat{j} + 4 \hat{k}$ be two vectors, such that $\vec{a} * \vec{b} = - \hat{i} + 9 \hat{i} + 12 \hat{k} .$ Then the projection of $\vec{b} - 2 \vec{a} o n \vec{b} + \vec{a}$ is equal to

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Contributor-Level 10

$\vec{a} * \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ α & 1 & β \\ 3 & - 5 & 4 \end{matrix} | = (4 + 5 β) \hat{i} + (3 β - 4 α) \hat{j} + (- 5 α - 3) \hat{k}$

Compare with $\vec{a} * \vec{b} = - \hat{i} + 9 \hat{j} + 12 \hat{k}$

b = -1, a = -3

Projection of $\vec{b} - 2 \vec{a} o n \vec{b} + \vec{a}$

$= \frac{(\vec{b} - 2 \vec{a}) . (\vec{b} + \vec{a})}{| \vec{b} + \vec{a} |}$

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New answer posted

a year ago

Let y = y₁ (x) and y = y₂ (x) be two distinct solutions of the differential equation = x + y, with y₁ (0) = 0 and y₂(0) = 1 respectively. Then, the number of points of intersection of y = y₁ (x) and y = y₂ (x) is

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