Class 12th

A planar loop of wire rotates in a uniform magnetic field. Initially, at $t = 0$ , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10 s$ about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at:

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alok kumar singh

Contributor-Level 10

At any time $t$

$? = B A c o s ? \frac{π t}{5} (? ω = \frac{2 π}{T} = \frac{2 π}{10} = \frac{π}{5})$

$\frac{d ?}{d t} = - \frac{π}{5} B A s i n ? (\frac{π t}{5})$

$e = \frac{π}{5} B A s i n ? (\frac{π t}{5})$

e will be maximum when $\frac{π t}{5}$ is $\frac{π}{2}$ .

$t = 2.5 s e c$
e will be minimum when $\frac{π t}{5}$ is $π, 0$

$t = 0,5 s e c$

So, answer will be (1).

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a year ago

Class 12th Maths Matrices

$2 s i n (\frac{π}{22}) s i n (\frac{3 π}{22}) s i n (\frac{5 π}{22}) s i n (\frac{7 π}{22}) s i n (\frac{9 π}{22}) i s e q u a l t o :$

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$2 s i n (\frac{π}{22}) s i n (\frac{3 π}{22}) s i n (\frac{5 π}{22}) s i n (\frac{7 π}{22}) s i n (\frac{9 π}{22})$

$= \frac{2 s i n \frac{32 π}{11}}{2^{5} s i n \frac{π}{11}} = \frac{1}{16}$

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a year ago

Class 12th Maths

Let $\vec{a} = \hat{i} - \hat{j} + 2 \hat{k} a n d l e t \hat{b}$ be a vector such that $\vec{a} * \vec{b} = 2 \hat{i} - \hat{k} a n d \vec{a}, \vec{b} = 3 .$ Then the projection of $\vec{b}$ on the vector $\vec{a} - \vec{b} i s :$

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$\vec{a} = \hat{i} - \hat{j} + 2 \hat{k}$

$\vec{a} * \vec{b} = 2 \hat{i} - \hat{k}$

$\vec{a} . \vec{b} = 3$

${| \vec{a} * \vec{b} |}^{2} + {| \vec{a} . \vec{b} |}^{2} = {| \vec{a} |}^{2} . {| \vec{b} |}^{2}$

$= \frac{\vec{b} . \vec{a} - {| \vec{b} |}^{2}}{| \vec{a} - \vec{b} |} = \frac{3 - \frac{7}{3}}{\sqrt[]{\frac{7}{3}}} = \frac{2}{\sqrt[]{21}}$

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a year ago

In the circuit, the logical value of A = 1 or B = 1 when potential at A or B is 5 V and the logical value of A = 0 or B = 0 when potential at A or B is 0 V. The truth table of the given circuit will be:

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When both A and B have logical value 'l' both diode are reverse bias and current will flow in resistor hence output will be 5 volt i.e. logical value '1'.

In all other case conduction will take place hence output will be zero value i.e. logical value 'o'.

So truth table is

A B Y

0 0

0 1 0 (AND gate)

...more

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a year ago

Class 12th Maths

The shortest distance between the lines $\frac{x + 7}{- 6} = \frac{y - 6}{7} = Z a n d \frac{7 - x}{2} = y - 2 = z - 6$ is

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$A (- 7 \hat{i} + 6 \hat{j}) C (7 \hat{i} + 2 \hat{j} + 6 \hat{k})$

$\vec{b} = - 6 \hat{i} + 7 \hat{j} + \hat{k}$ $\vec{d} = - 2 \hat{i} + \hat{j} + \hat{k}$

$\vec{b} * \vec{d} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 6 & 7 & 1 \\ - 2 & 1 & 1 \end{matrix} | = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$

the shortest distance between the lines

$=$ $| \frac{42 - 8 + 24}{29} |$ $=$ $\frac{58}{29}$ $=$ $2$

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a year ago

The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by:

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Contributor-Level 10

Kindly consider the following answer

$\vec{μ} = - \frac{e}{2 m} \vec{L}$

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a year ago

Class 12th Atoms

The momentum of an electron revolving in nth orbit is given by:

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Contributor-Level 10

mvr = $\frac{n h}{2 π}$ $\frac{}{}$

So momentum (mv) = $\frac{n h}{2 π r}$

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a year ago

Class 12th Maths

A plane E is perpendicular to the two planes 2x – 2y + z = 0 and x – y + 2z = 4, and passes through the point P(1, -1, 1). If the distance of the plane E from the point Q(a, a, 2) is $3 \sqrt[]{2},$ then (PQ)² is equal to

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First plane, P₁ = 2x – 2y + z = 0,

normal vector $\equiv n_{1} = (2, - 2, 1)$

Second plane, P₂ = x – y + 2z = 4,

normal vector $\equiv n_{2} = (1, - 1, 2)$

Plane perpendicular to P₁ and P₂ will have normal vector n₃ where n₃ = (n₁* n₂)

Hence,

n₃ = (3, 0)

Distance PQ

$= \sqrt[]{21} \Rightarrow {(P Q)}^{2} = 21$

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a year ago

A metal exposed to light of wavelength 800nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500nm is used. The work function of the metal is: (Take hc = 1230 eV –nm).

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$_{} \frac{}{}$ $K . E_{1} = \frac{1230}{800} - ?$ - (i)

$_{}_{} \frac{}{}$ $K . E_{2} = 2 K . E_{1} = \frac{1230}{500} - ?$ - (ii)

from (i) & (ii)

$? = 0.615$ ev

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a year ago