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Class 12th Physics

As shown in the figure an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is $\frac{\sqrt[]{a}}{π} A .$ The value of a is __________.

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Vishal Baghel

Contributor-Level 10

Let l = l₀ cos wt

Then v = v₀ sinwt

at t = 0, v = 0

but l = l₀

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

$V_{r m s} = l_{r m s} z$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (X_{L})$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (2 π * 50 * 200 * 1 0^{- 3})$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (20 π)$

$l_{0} = \frac{220 \sqrt[]{2}}{20 π}$

$l_{0} = \frac{11 \sqrt[]{2}}{π} = \frac{\sqrt[]{a}}{π}$

a = 121 * 2

a = 242

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Class 12th Chemistry

The major product of the following reaction contains _____________ bromine atom(s).

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alok kumar singh

Contributor-Level 10

Bromination through free radical mechanism occurs at allylic carbon.

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Class 12th Physics

A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10µA and the collector current changes by 1.5 mA. The load resistance is 5 kΩ. The voltage gain of the transistor will be ________.

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Vishal Baghel

Contributor-Level 10

I_b = 10 µA

I_C = 1.5 mA

R_L = 50 kW or (Rc)

Base – emitter voltage = 10 mv

$R_{B} = \frac{V_{B}}{I_{B}} = \frac{10 * 1 0^{- 3}}{10 * 1 0^{- 6}} = 1 0^{3} Ω$

$A_{v} = (\frac{Δ l_{C}}{Δ l_{B}}) * (\frac{R_{C}}{R_{B}})$

= $\frac{1.5 * 1 0^{- 3}}{10 * 1 0^{- 6}} * \frac{5 * 1 0^{3}}{1 0^{3}}$

A_v = 750

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11 months ago

Class 12th Physics

Two identical thin biconvex lenses of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is ________ cm.

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Vishal Baghel

Contributor-Level 10

f₁ = 15 cm

$P_{1} = \frac{1}{15} * 100 = \frac{100}{15} D$

$P_{3} = \frac{100}{15} D$

$\frac{1}{f^{2}} = (μ_{2} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$= (1.25 - 1) * \frac{- 2}{R}$

$\frac{1}{f_{2}} = \frac{0.25 * - 2}{15}$

$\frac{1}{f_{2}} = \frac{- 1}{30} c m^{- 1}$

$\begin{array}{l} p_{2} = \frac{1}{f_{2}} = - \frac{100}{30} \\ P_{R} = P_{1} + P_{2} + P_{3} \\ P_{R} = \frac{100}{15} - \frac{100}{30} + \frac{100}{15} \\ P_{R} = 10 D \end{array}$

$P_{R} = \frac{1}{f_{R}}$

$f_{R} = \frac{1}{10} * 100 c m$

$f_{R} = 10 c m$

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11 months ago

Class 12th Maths

Let f(x) and g(x) be two real polynomials of degree 2 and 1 respectively. If $f (g (x)) = 8 x^{2} - 2 x,$ and $g (f (x)) = 4 x^{2} + 6 x + 1,$ then the value of f(2) + g(2) is

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Payal Gupta

Contributor-Level 10

$f (x) = a x^{2} + b x + c$

g (x) = px + q

$f (g (x)) = a {(p x + q)}^{2} + b (p) (+ q) + c$

$\Rightarrow 8 x^{2} - 2 x = a {(p x + q)}^{2} + b (p x + q) + c$

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

$9 (f (x)) = p (a x^{2} + b x + c) + q$

? 4x² + 6x + 1 = apx² + bpx + cp + q

? Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

? b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

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Class 12th Chemistry

(a) CoCl₃.4NH₃, (b) CoCl₃.5NH₃, (c) CoCl₃. 6NH₃ and (d) CoCl(NO₃)₂. 5NH₃.

Number of complex(es) which will exist in cis-trans form is /are__________.

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alok kumar singh

Contributor-Level 10

(a) $C o C l_{3} . 4 N H_{3} i s [C o C l_{2} {(N H_{3})}_{4}] C l$ will show cis- trans isomerism as;

(b) $C o C l_{3} . 5 N H_{3} i s [C o {(N H_{3})}_{5} C l] C l_{2}$ will not show cis- trans isomerism.

(c) $C o C l_{3} . 6 N H_{3} i s [C o {(N H_{3})}_{6}] C l_{3}$ will not show cis- trans isomerism.

(d) $C o C l {(N O_{3})}_{2} . 5 N H_{3} i s [C o {(N H_{3})}_{6} C l] {(N O_{3})}_{2}$ will not show cis-trans isomerism.

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Class 12th Moving Charges and Magnetism

The magnitude field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point along the axis at a distance $\frac{r}{2}$ from the centre is :

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Vishal Baghel

Contributor-Level 10

$B_{C} = \frac{μ_{0} l}{2 r} = B$ $_{} \frac{_{}}{}$ -(1)

$\begin{array}{l} B_{P} = \frac{μ_{0} l r^{2}}{2 {(r^{2} + \frac{r^{2}}{4})}^{3 / 2}} \\ B_{P} = \frac{μ_{0} l r^{2}}{2 * r^{3} {(\frac{5}{4})}^{3 / 2}} \end{array}$

$B_{P} = \frac{μ_{0} l}{2 r {(\frac{5}{4})}^{3 / 2}}$

$\Rightarrow B_{P} = \frac{B 4^{3 / 2}}{5^{3 / 2}}$

$B_{P} = B \frac{2^{3}}{{(\sqrt[]{5})}^{3}}$

$B_{P} = {(\frac{2}{\sqrt[]{5}})}^{3} B$

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11 months ago

Class 12th Electrostatics

A parallel plate capacitor is formed by two plates each of area 30πcm² separated by 1 mm. A material of dielectric strength 3.6 * 10⁷ Vm^-¹ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7 * 10^-⁶ C, the value of dielectric constant of the material is [Use ] $\frac{1}{4 π ε_{0}} = 9 * 1 0^{9} N m^{2} C^{- 2}$

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Vishal Baghel

Contributor-Level 10

A = 30p cm² = 30p * 10^-⁴ m²

d = 1mm = 10^-³ m

E = (dielectric strength for breakdown)

= 3.6 * 10⁷ v/m

Q = 7 * 10^-⁶C

$\frac{σ}{\in_{0}} = E$

$\Rightarrow \frac{Q}{k A \in_{0}} = E$

$k = \frac{Q}{A \in_{0} E}$

$= \frac{7 * 1 0^{- 6} * 4 π * 9 * 1 0^{9}}{30 π * 1 0^{- 4} * 1 * 3.6 * 1 0^{7}}$

$= \frac{7 * 4 π * 9}{30 π * 3.6}$

$= \frac{7 * 4 * 9 * 10}{30 * 36}$

$= \frac{70}{30}$

$k = \frac{7}{3} = 2.33$

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Class 12th Chemical Kinetics

${[F e {(C N)}_{6}]}^{4 -}$

${[F e {(C N)}_{6}]}^{3 -}$

${[T i {(C N)}_{6}]}^{3 -}$

${[N i {(C N)}_{4}]}^{2 -}$

${[C o {(C N)}_{6}]}^{3 -}$

Among the given complexes, number of paramagnetic complexes in__________.

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alok kumar singh

Contributor-Level 10

CN is a strong field ligand, so pairing occurs.

1. ${[F e {(C N)}_{6}]}^{4 -}$

So; it is diamagnetic

2. ${[F e {(C N)}_{6}]}^{3 -}$

So; it is paramagnetic.

3. ${[T i {(C N)}_{6}]}^{3 -}$

Ti³⁺ = 4s⁰3d¹

It is paramagnetic

4. ${[N i {(C N)}_{4}]}^{2 -}$

It is diamagnetic

5. ${[C o {(C n)}_{6}]}^{3 -}$

Hence, ${[F e {(C N)}_{6}]}^{3 -} a n d {[T i {(C N)}_{6}]}^{3 -}$ ${_{}}^{} {_{}}^{}$ are paramagnetic.

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11 months ago

Class 12th Physics

A baseband signal of 3.5 MHz frequency is modulated with a carrier signal of 3.5 GHz frequency using amplitude modulation method. What should be the minimum size of antenna required to transmit the modulated signal?

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Vishal Baghel

Contributor-Level 10

Baseband Signal frequency 3.5 MHz

& Carrier signal frequency = 3.5 GHz

$υ_{C} = 3.5 G H z = 3.5 * 1 0^{9}$

$λ = \frac{C}{υ_{c}} = \frac{3 * 1 0^{8}}{3.5 * 1 0^{9}} = \frac{3}{3.5} * 10$

$= \frac{30}{35} * 10$

= $\frac{60}{7}$ $\frac{}{}$

Size of antenna = $\frac{λ}{4} = \frac{60}{7 * 4}$ $\frac{}{} \frac{}{}$

= 21.4 mm

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