Class 12th

As shown in the figure, after passing through the medium 1. The speed of light $υ_{2}$ in medium 2 will be:

(Given c = 3 * 10⁸ ms^-¹)

0 Follower 10 Views

Contributor-Level 10

In a medium, speed of light, $V = \frac{C}{\sqrt[]{μ_{r} ε_{r}}} = \frac{3 * 1 0^{8} m / s}{\sqrt[]{1 * 9}} = 1 * 1 0^{8} m / s$

0 0

New answer posted

11 months ago

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 If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

0 Follower 3 Views

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 $ω t$ &nb

...more

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 $ω t$

tanθ = 1 = q = 45°

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

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

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

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

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

less

0 0

New answer posted

11 months ago

The equation of current in a purely inductive circuit is 5sin(49pt – 30°). If the inductance is 30 mH then the equation for the voltage across the inductor, will be: ${L e t π = \frac{22}{7}}$

0 Follower 10 Views

Class 12th Physics Moving Charges and Magnetism

Contributor-Level 10

$i = 5 s i n (49 π t - 30 °)$

$L = 30 m H, π = \frac{22}{7}$

$X_{L} = ω L = 49 π * 30 * 1 0^{- 3}$

$= 49 \frac{22}{7} * 30 * 1 0^{- 3}$

$V_{L} = 23.1 s i n (49 π t + 60 °)$

0 0

New answer posted

11 months ago

As shown in the figure, a metallic rod of linear density 0.45 km^-¹ is lying horizontally on a smooth incline plane which makes an angle of 45° 5with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T magnetic field is acting on it in the vertical upward direction, will be:

{Use g = m/s²}

0 Follower 8 Views

Class 12th Physics Moving Charges and Magnetism

Contributor-Level 10

Linear density, $λ =$ 0.45 kg/m

Let length = $l$ $∴$ = m = 0.45 $l$

B = 0.15 T

For equilibrium of rod :-

mg sin 45° = FB sin 45°

$\Rightarrow (0.45 l) g = l l B$

So, $l = \frac{0.45 * 10}{0.15} = 30 A$ $\frac{}{}$

0 0

New answer posted

11 months ago

The current sensitivity of a galvanometer can be increased by:

(a) decreasing the number of turns

(b) increasing the magnetic field

(c) decreasing the area of the coil

(d) decreasing the torsional constant of the spring

Choose the most appropriate answer from the options given below:

0 Follower 2 Views

Contributor-Level 10

Current sensitivity, $\frac{}{}$ $S i = \frac{N A B}{k}$

$N i A B = k θ$

$θ = (\frac{N A B}{k}) i$

0 0

New answer posted

11 months ago

A wire of resistance R₁ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

0 Follower 10 Views

Class 12th Physics Moving Charges and Magnetism

Contributor-Level 10

$R_{2} = ρ \frac{2 l}{A / 2} = \frac{4 ρ l}{A} = 4 R_{1}$
$\frac{R_{2}}{R_{1}} = \frac{4}{1} \to 4 : 1$

0 0

New answer posted

11 months 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

0 Follower 25 Views

Class 12th Electrostatic Potential and Capacitance

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 * 10^-5N towards 'x'

0 0

New answer posted

11 months ago

Two capacitors, each having capacitance $40 μ F$ are connected in series. The space between one of the capacitors is filled with dielectric material of constant K such that the equivalence capacitance of the system became $24 μ F$ . The value of K will be:

0 Follower 3 Views

Contributor-Level 10

$C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{40 * 40 k}{40 (1 + k)} = 24$

$\begin{array}{l} \Rightarrow 40 k = 24 + 24 k \\ \Rightarrow 16 k = 24 \end{array}$

$k = \frac{24}{16} = \frac{3}{2} = 1.5$

0 0

New answer posted

11 months ago

A 1m long wire is broken into two unequal parts X and Y. The X part of the wire is stretched into another wire W. Length of W is twice the length of X and the resistance of W is twice that of Y. Find the ratio of length of X and Y.

0 Follower 8 Views

Contributor-Level 10

R_w = 2R_y

$\Rightarrow ρ (\frac{2 x}{\frac{A}{2}}) = \frac{2 ρ (1 - x)}{A}$

$\frac{4 ρ x}{A} = \frac{2 ρ (1 - x)}{A}$

->4x = 2 – 2x Þ 6x = 2

x = $\frac{2}{6} = \frac{1}{3}$ $∴ \frac{L_{x}}{L_{y}} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{3} * \frac{3}{2}$

$= \frac{1}{2}$

0 0

New answer posted

11 months ago

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

Assertion A : Alloys such as constantan and manganin are used in making standard resistance coils.

Reason R : Constantan and maganin have very small value of temperature coefficient of resistance.

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

0 Follower 1 View