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11 months ago

Class 12th English

Two traders, T1 and T2, were involved in the buying and selling of MCS shares over five trading days. At the beginning of the first day, the MCS share was priced at Rs. 100, while at the end of the fifth day it was priced at Rs. 110. At the end of each day, the MCS share price either went up by Rs. 10, or else, it came down by Rs. 10. Both T1 and T2 took buying and selling decisions at the end of each trading day.

The beginning price of MCS share on a given day was the same as the ending price of the previous day.

T1 and T2 started with the same number of shares and amount of cash, and had enough of both. Below are some addition

...more

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Payal Gupta

Contributor-Level 10

	Day 1	Day 2	Day 3	Day 4	Day 5
Start Price	100	90	80	90	100
End Price	90	80	90	100	110

Assume initial number of share with T1 and T2 is X. In the above table T1 by 10 share each on day 1, day 2 and sold 10 shares and day 3, day 4 and day 5.

Total share with T1 is X – 10.

In the above T2 buy share only on day 2.

Total shares with T2 is X + 10.

T2 had 20 shares more than T1.

Therefore at the end of day 3 the price of share is Rs. 90

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11 months ago

Class 12th Atoms

An particular hydrogen like ion emits radiation of frequency 2.92 * 10¹⁵ Hz when it makes transition from n = 3 to n = 1. The frequency in Hz of radiation emitted in transition from n = 2 to n = 1 will be:

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Vishal Baghel

Contributor-Level 10

$h f_{1} = R (\frac{1}{1^{2}} - \frac{1}{3^{2}}) = R * \frac{8}{9}$

$h f_{2} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = R * \frac{3}{4}$

$\frac{f_{1}}{f_{2}} = \frac{8 / 9}{3 / 4} = \frac{32}{27}$

f₂ = f₁x $(\frac{27}{32})$

$= (2.92 * 1 0^{15}) * (\frac{27}{32})$

$= 2.46 * 1 0^{15} H z$

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11 months ago

Class 12th English

The beginning price of MCS share on a given day was the same as the ending price of the previous day.

T1 and T2 started with the same number of shares and amount of cash, and had enough of both. Below are some addition

...more

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Payal Gupta

Contributor-Level 10

	Day 1	Day 2	Day 3	Day 4	Day 5
Start Price	100	110	120	130	120
End Price	110	120	130	120	110

Let initial amount with T1 and Michal is .

T1 sold shares on Day 1, Day 2, Day 3, whereas buys shares on Day 4 and Day 5.

Total money with T1 is = Y + 110 * 10 + 120 * 10 + 130 * 10 – 120 * 10 – 110 * 10 = Y + 1300

Total money with T2 = Y + 1200

Total money with T2 = Y + 120 * 10 + 130 * 10 + 120 * 10 = Y + 3700

Total money with T2 & T1 = 2Y + 5000.

Therefore maximum possible increase is 5000.

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11 months ago

Class 12th Dual Nature of Radiation and Matter

The de-Broglie wavelength of a particle having kinetic energy E is $λ .$ How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?

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alok kumar singh

Contributor-Level 10

$λ' = λ * \frac{75}{100} = \frac{3}{4} λ$

$λ = \frac{h}{p} = \frac{h}{\sqrt[]{2 m E}}$

$\Rightarrow \frac{E'}{E} = {(\frac{λ}{λ'})}^{2} = {(\frac{4}{3})}^{2}$

$\Rightarrow E' = \frac{16}{9} E$

Extra energy required,

$Δ E = E' - E = \frac{7}{9} E$

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11 months ago

Class 12th English

Directions questions: Answer the following questions based on the information given below.

The beginning price of MCS share on a given day was the same as the ending price of the previous day.

T1 and T2 started with th

...more

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Payal Gupta

Contributor-Level 10

	Day 1	Day 2	Day 3	Day 4	Day 5
Start Price	100	90	100	110	120
End Price	90	80	110	120	110

Let initial amount with T1 and T2 is Y.

Total money with T1 = Y – 900 + 1000 + 1100 + 1200 – 1100 = Y + 1300

Total money with T2 = Y + 1200

Therefore difference between T1 and T2 is Rs. 100 and Number of shares with T2 and T1 is same.

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11 months ago

Class 12th Physics

If you are provided a set of resistance 2 $Ω$ , 4 $Ω$ , 6 $Ω$ and 8 $Ω$ . Connected these resistances so as to obtain an equivalent resistance of $\frac{46}{3} Ω .$

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alok kumar singh

Contributor-Level 10

$R_{e q} = \frac{2 * 4}{2 + 4} + 6 + 8 = \frac{46}{3} Ω$

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11 months ago

Class 12th Physics Ray Optics and Optical Instruments

Car B overtakes another car A at a relative speed of 40ms^-¹. How fast will the image of car B appear to move in the mirror of focal length 10cm fitted in car A, when the car B is 1.9m away from the car A?

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Vishal Baghel

Contributor-Level 10

For mirror

${\vec{V}}_{l_{1} M} = - m^{2} {\vec{V}}_{O_{1} M}$

${\vec{V}}_{l_{1} M}$ Velocity of image w.r.t. mirror

${\vec{V}}_{0 M}$ Velocity of object w.r.t. mirror

m magnification

$\frac{1}{v} + \frac{1}{- 1.9} = \frac{1}{- 0.1}$

$\frac{1}{v} = \frac{1}{1.9} - \frac{1}{0.1}$

$= \frac{0.1 - 1.9}{1.9 * 0.1}$

$v = \frac{1.9 * 0.1}{- 1.8} = \frac{0.19}{- 1.8} = - 0.105$

$m = - \frac{v}{u} = - (\frac{- 0.1}{- 1.9})$

$| {\vec{V}}_{l_{1} M} | = {(\frac{1}{19})}^{2} | {\vec{V}}_{O_{1} M} |$

$= \frac{1}{{(19)}^{2}} * 40 = 0.1 m / s$

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11 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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11 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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11 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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