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8 months ago

Class 12th Dual Nature of Radiation and Matter

In a photoelectric experiment ultraviolet light o wavelength 280nm is used with lithium cathode having work function $?$ = 2.5eV. If the wavelength of incident light is switched to 400nm, find out the change in the stopping potential. (h = 6.63 * 10^-³⁴ Js, c = 3 * 10⁸ ms^-¹)

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Vishal Baghel

Contributor-Level 10

$\frac{h c}{λ_{1}} = W + e V_{1}$

$\frac{h c}{λ_{2}} = W + e V_{2}$

$= 12400 (\frac{1}{2800} - \frac{1}{4000}) [λ_{1} & λ_{2} a r e t a k e n i n A °]$

$\begin{array}{l} = 4.4 - 3.1 \\ = 1.3 V \end{array}$

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8 months ago

Class 12th Physics

An electric bulb of 500 watt at 100 volt is used in a circuit having a 200V supply. Calculate resistance R to be connected in series with the bulb so that the power delivered by the bulb is 500W.

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alok kumar singh

Contributor-Level 10

As power delivered by the bulb is 500 W, potential difference across it should be 00 V. Thus,

$R = R_{b u l b} = \frac{10 0^{2}}{500} = 20 Ω$

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8 months ago

Class 12th English

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)

Years	P	S	SP	FC	CB	CE
2001	15000	12000	20000	60	1500	2000
2002	13000	10000	20000	65	2000	3000
2003	20000	22000	22000	65	3000	4000
2004	18000	15000	23000	67	3750	4100
2005	25000	24000	24000	70	3800	4150
2006	12000	19000	30000	70	6000	7000
2007	17000	15000	28000	73	4000	5000
2008	18000	15000	28000	75	4200	4700
2009	16000	15000	28000	75	4100	4500
2010	15000	19000	28000	80	3500	3750

At end of which year did Honda motors earned the highest profit?

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Payal Gupta

Contributor-Level 10

Please find the answers below

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8 months ago

Class 12th Physics

Four NOR gates are connected as shown in figure. The truth table for the given figure is:

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alok kumar singh

Contributor-Level 10

$Y = \bar{(\bar{\bar{A + B}} + A) + (\bar{\bar{A + B}} + B)} + (\bar{A + B} + A) . (\bar{A + B} + B)$

A B Y

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0 1 0&nbs

...more

$Y = \bar{(\bar{\bar{A + B}} + A) + (\bar{\bar{A + B}} + B)} + (\bar{A + B} + A) . (\bar{A + B} + B)$

A B Y

0 1

0 1 0

1 0 0

1 1

less

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New answer posted

8 months ago

Class 12th Alternating Current

A series LCR circuit driven by 300V at a frequency of 50 Hz contains a resistance R = $3 k Ω,$ an inductor of inductive reactances X_L = 250π $Ω$ and an unknown capacitor. The value of capacitance to maximize the average power should be:

(Take π² = 10)

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Vishal Baghel

Contributor-Level 10

Power is maximum at resonance

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

$\Rightarrow C = \frac{1}{ω * X_{L}}$

$= \frac{1}{250 * 100 * π^{2}}$

$= 4 μ F [π^{2} = 10]$

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8 months ago

Class 12th English

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

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)

...more

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Payal Gupta

Contributor-Level 10

Please find the answer below

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New answer posted

8 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:

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Vishal Baghel

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

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

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8 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:

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Vishal Baghel

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

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8 months ago

Class 12th Physics Moving Charges and Magnetism

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)

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Vishal Baghel

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

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

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

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

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8 months ago

Class 12th English

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.

...more

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Payal Gupta

Contributor-Level 10

From the table above.

0 0

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	P1	P2	P3	P4	P5	P6
P1	*	50	40	44	57	68
P2	35	*	35	58	43	59
P3	17	9	*	15	7	24
P4	70	32	28	*	68	35
P5	98	64	53	45	*	52
P6	35	25	57	62	35	*

	P1	P2	P3	P4	P5	P6
P1	*	50	40	44	57	68
P2	35	*	35	58	43	59
P3	17	9	*	15	7	24
P4	70	32	28	*	68	35
P5	98	64	53	45	*	52
P6	35	25	57	62	35	*

	P1	P2	P3	P4	P5	P6
P1	*	50	40	44	57	68
P2	35	*	35	58	43	59
P3	17	9	*	15	7	24
P4	70	32	28	*	68	35
P5	98	64	53	45	*	52
P6	35	25	57	62	35	*