Class 12th

A small square loop of wire of side $l$ is placed inside a large square loop of wire. L(L >> $l$ ). Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is:

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Contributor-Level 10

Assuming current 'i' In outer loop magnetic field at centre

$= 4 * \frac{μ_{0}}{4 π} \frac{i}{L / 2} * [2 s i n 45 °)$

$= \frac{2 \sqrt[]{2} μ_{0} i}{π L}$

M = $\frac{F m α t h r o u g h i n n e r l o o p}{i}$ $\frac{}{}$

= $\frac{2 \sqrt[]{2} μ_{0} i l^{2}}{π L} = \frac{2 \sqrt[]{2} μ_{0} l^{2}}{π L}$ $\frac{_{}^{}}{} \frac{_{}^{}}{}$

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In a building there are 15 bulbs of $45 W, 15$ bulbs of $100 W, 15$ small fans of $10 W$ and 2 heaters of $1 k W$ . The voltage of electric main is $220 V$ . The minimum fuse capacity (rated value) of the building will be

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Contributor-Level 10

Total power is $(15 * 45) + (15 * 100) + (15 * 10) + (2 * 1000) = 4325 W$

So, current is $\frac{4325}{220} = 19.66 A$

Answer is $20 A m p$ .

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To increase the resonant frequency in series LCR circuit,

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Class 12th Physics Moving Charges and Magnetism

Contributor-Level 10

$f = \frac{1}{2 π \sqrt[]{L C}}$

$f α \frac{1}{\sqrt[]{C}}$

Another capacitor should be added in series with the first capacitor.

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A particle of mass $m$ and charge $q$ has an initial velocity $\vec{v} = v_{0} \hat{j}$ . If an electric field $\vec{E} = E_{0} \hat{i}$ and magnetic field $\vec{B} = B_{0} \hat{i}$ act on the particle, its speed will double after a time:

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Class 12th Physics Moving Charges and Magnetism

Contributor-Level 10

Since magnetic force cannot change the speed. So only electric field which is along -direction will change the speed along -direction only.

$v_{x} = \frac{E_{0} q}{m} t$ but $v_{y} = v_{0}$

$2 v_{0} = \sqrt[]{v_{x}^{2} + v_{y}^{2}}$

4 v_{0}^{2} = \frac{E_{o}^{2} q^{2} t^{2}}{m^{2}} + v_{0}^{2}

t = \frac{\sqrt[]{3} m v_{o}}{q E_{o}}

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Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. It the ratio of radii of their circular paths is 6 : 5 and their respective masses ratio is 9 : 4. Then, the ratio of their charges will be:

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Contributor-Level 10

Radius of circular path R = $\sqrt[]{\frac{2 m k}{q B}}$ $\frac{}{}$

$q = \frac{\sqrt[]{2 m k}}{R B}$

$\frac{q_{1}}{q_{2}} = \sqrt[]{\frac{m_{1}}{m_{2}}} * \frac{R_{2}}{R_{1}} = \sqrt[]{\frac{9}{4}} * \frac{5}{6} = \frac{5}{4}$

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A condenser of $2 μ F$ capacitance is charged steadily from 0 to 5 C. Which of the following graph represents correctly the variation of potential difference (V) across it's plates with respect to the charge (Q) on the condenser?

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Contributor-Level 10

Q = CV

$V = \frac{Q}{C}$

Straight line with slope = $\frac{1}{C ?}$ $\frac{}{}$

$S l o p e = \frac{1}{C} = \frac{1}{2 * 1 0^{- 6}} = 5 * 1 0^{5}$

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An emf of $20 V$ is applied at time $t = 0$ to a circuit containing in series $10 m H$ inductor and $5 Ω$ resistor. The ratio of the currents at time $t = \infty$ and at $t = 40 s$ is close to: (take $e^{2} = 7.389)$

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Contributor-Level 10

$\begin{array}{r} i & = \frac{V}{r} (1 - e^{- R t / L}) \\ = 4 (1 - e^{- 500 t}) \end{array}$

At $t = \infty$

$i_{1} = 4 A$

at $t = 40 s$

$i_{2} = 4 (1 - e^{- 20000})$

= 4 [1 - \frac{1}{{(e^{2})}^{10000}}]

= 4 [1 - \frac{1}{(7.389)^{10000}}]

\frac{i_{1}}{i_{2}}

is slightly greater than 1 .

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Class 12th NCERT Solutions

Boiling point of a 2% aqueous solution of a non-volatile solute A is equal to the boiling point of 8% aqueous of non-volatile solute B. The relation between molecular weights of A and B is

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Contributor-Level 10

For A : 100 gm solution 2 gm solute A

$∴ M o l a l i t y = \frac{2 / M_{A}}{0.098}$

For B : 100 gm solution 8 gm solute B

$∴ M o l a l i t y = \frac{8 / M_{B}}{0.092}$

$\frac{1}{4.261} = \frac{M_{A}}{M_{B}}$

$∴ M_{B} = 4.261 * M_{A}$

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Class 12th Chemistry

Given below are two statements. One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A: Activated charcoal adsorbs SO₂ more efficiency than CH₄.

Reason R: Gases with lower critical temperature are readily adsorbed by activated charcoal.

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

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Contributor-Level 10

SO₂ is absorbed to a greater extent then CH₄ on activated charcoal under same Gases with higher critical temperature are readily absorbed by activated charcoal.

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Class 12th Chemistry

Given below are two statements.

Statement I: O₂, Cu²⁺, and Fe³⁺ are weakly attracted by magnetic field and are magnetized in the same direction as magnetic field.

Statement II: NaCl and H₂O are weakly magnetized in opposite direction to magnetic field.

In the light of the above statements, choose the most appropriate answer from the options given below.

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