Class 12th

Contributor-Level 10

Power is maximum at resonance

So, $X_{L} = X_{C} = \frac{1}{ω C}$ $_{}_{} \frac{}{}$

$\Rightarrow C = \frac{1}{ω * X_{L}}$

$= \frac{1}{250 * 100 * π^{2}}$

$= 4 μ F [π^{2} = 10]$

0 0

New answer posted

11 months ago

Directions for question : Answer the questions on the basis of the information given below.

Honda Motors manufactures cars in their Chennai plant. The performance of the plant for 10 consecutive years is given below. The plant when started the operations in year 2001 has a cash surplus of 10 million rupees and no stock was there in the plant. It is also known that for every unit of car manufacture 2 units of body and 3 units of engine are used.

P Production (units)

S Sales (units)

SP Selling Price (Rs./unit)

FC Fixed Cost (Million Rs.)

CB Cost of Body (per unit of body)

CE Cost of Engine (per unit of engine)

0 Follower 1 View

Contributor-Level 10

Please find the answer below

0 0

New answer posted

11 months ago

Class 12th Physics

An inductor coil stores 64J of magnetic field energy nad dissipates energy at the rate of 640W when a current of 8A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds:

0 Follower 2 Views

Contributor-Level 10

$U_{B} = \frac{1}{2} L l^{2} . . . . . . . . . . . . (i)$

P_R = l²R

=> 640 = (8)² R

R = $10 Ω$

from (i) 64 = $\frac{1}{2} * L * {(8)}^{2}$

L = 2H

$l = \frac{L}{R} = \frac{2}{10} = 0.2 s e c o n d$

0 0

New answer posted

11 months ago

Class 12th Physics

In the given figure, the emf of the cell is 2.2V and if internal resistance is 0.6 $Ω$ . Calculate the power dissipated in the whole circuit:

0 Follower 33 Views

Class 12th Physics Moving Charges and Magnetism

Contributor-Level 10

$R_{e q} = 0.6 Ω + \frac{\frac{4 * 12}{4 + 12} * \frac{6 * 8}{6 + 8}}{\frac{4 * 12}{4 + 12} + \frac{6 * 8}{6 + 8}} Ω$

$= 0.6 Ω + \frac{3 * \frac{48}{14}}{3 + \frac{48}{14}} Ω$

$= 0.6 Ω + \frac{(144 / 14) Ω}{(\frac{90}{14})}$

$\begin{array}{l} = 0.6 Ω + 1.6 Ω \\ = 2.2 Ω \end{array}$

p = $\frac{v^{2}}{R_{e q}} = \frac{{(2.2 V)}^{2}}{2.2 Ω} = 2.2 W$

0 0

New answer posted

11 months ago

The fractional change in the magnetic field intensity at a distance 'r' from centre on the axis of current carrying coil of radius 'a' to the magnetic field intensity at the centre of the same coil is: (Take r < a)

0 Follower 3 Views

Contributor-Level 10

$B_{a x i s} = \frac{μ_{0} N l a^{2}}{2 {(a^{2} + r^{2})}^{3 / 2}}$

$B_{c e n t} = \frac{μ_{0} N I}{2 a}$

Fractional change = $\frac{B_{c e n t r e} - B_{a x i s}}{B_{c e n t r e}}$

$= 1 - \frac{B_{a x i s}}{B_{c e n t r e}}$

$= 1 - (1 - \frac{3 r^{2}}{2 a^{2}}) [r < < a]$

$= \frac{3 r^{2}}{2 a^{2}}$

0 0

New answer posted

11 months ago

In an ancient game of trading “Pirate Bay”, 6 players took part. The game is played in a manner that the players select a port each and then they trade with each other. Everything that they sent out to other player is known as exports and anything that they receive from other players is known as imports. The table below shows details of imports and exports of between those six players. The entry corresponding to a particular row i and column j gives the total amount of exports by a player i to a player j in all the previous trade rounds. For example the entry in row 2 column 4 means P2 has exported to P4 a total of 58 times.

0 Follower 4 Views

Contributor-Level 10

From the table above.

0 0

New answer posted

11 months ago

0 Follower 1 View

Contributor-Level 10

From the table above.

0 0

New answer posted

11 months ago

0 Follower 1 View

Contributor-Level 10

Player 1	Player 2	Export of Player 1 to 2	Export of Player 2 to 1	Winner
P1	P2	50	35	P1
P1	P4	40	17	P1
P1	P4	44	70	P4
P1	P5	57	98	P5
P1	P6	68	35	P1
P2	P3	35	9	P2
P2	P4	58	32	P2
P2	P5	43	64	P5
P2	P6	59	25	P2
P3	P4	15	28	P4
P3	P5	7	53	P5
P3	P6	24	57	P6
P3	P5	68	45	P4
P3	P6	35	62	P6
P5	P6	52	35	P5

Rank	Player	Number of times Winner	Points
1	P5	4	8
2	P1	3	6
2	P2	3	6
2	P4	3	6
4	P6	2	4
5	P3	0	0

We cannot determine the answer as there is a tie between 3 players and we do not know how to break that tie.

0 0

New answer posted

11 months ago

Directions for question: Answer the questions on the basis of the information give below.

0 Follower 1 View