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6 months ago

Class 12th Chemistry

Assuming 1 $μ g$ of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is __________* 10^-1 $μ g$

[Given: In 10 = 2.303; log 2 = 0.30]

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Payal Gupta

Contributor-Level 10

$λ t = l n \frac{N_{o}}{N_{t}}$

$\frac{0.693}{t_{1 / 2}} * t = l n \frac{1}{N_{t}}$

Or $\frac{0.693 * 100}{30} = l n \frac{1}{N_{t}}$

$\frac{1}{N_{t}} = 10$

$∴ N_{t} = \frac{1}{10} = 0.1$

Or N_t = 1 * 101 $μ g$

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6 months ago

Class 12th Chemistry

1.80g of solute A was dissolved in 62.5cm³ of ethanol and freezing point of the solution was found to be 155.1 K . The molar mass of solute A is __________g mol^-1.

[Given: Freezing point of ethanol is 156.0 K. Density of ethanol is 0.80 g cm^-3. Freezing point depression constant of ethanol is 2.00 K kg mol^-1]

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Payal Gupta

Contributor-Level 10

$Δ T_{f} = K_{f} * m$

$(156 - 155.1) = 2 * \frac{1.8}{M W} * \frac{1000}{(62.5 * 0.8)}$

0.9 = $\frac{2 * 1.8 * 1000}{M W * 50}$

$∴ M W = 80 g m / m o l$

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6 months ago

Class 12th Physics

A capacitor of capacitance 500 $μ F$ is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. Then maximum current in LC circuit will be_______ A.

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Payal Gupta

Contributor-Level 10

C = 500 $μ F,$ V = 100 v, L = 50 mH

In this LC – oscillation

q = q₀ cos $ω t$

$i = \frac{- d q}{d t} = q_{0} ω s i n ω t ω = \frac{1}{\sqrt[]{2 c}} = \frac{1}{\sqrt[]{50 * 1 0^{- 3} * 5 * 1 0^{- 4}}}$

= $\frac{1000}{5} = 200$

So, i_max = $q_{0} ω = 500 * 1 0^{6} * 100 * 200$

= 10A

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6 months ago

Class 12th Physics Nuclei

Two radioactive materials A and B have decay constants 25 $λ$ and 16 $λ$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of B to that of A will be “e” after a time $\frac{1}{a λ} .$ The value of a is__________.

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Payal Gupta

Contributor-Level 10

$λ_{A} = 25 λ,$ $λ_{B} = 16 λ$

At t = 0 N_A = N_B = N₀

after t = $\frac{1}{a λ} : - \frac{N_{B}}{N_{A}} = \frac{N_{0} e^{- 16 λ t}}{N_{0} e^{- 25 λ t}} = e^{(25 λ - 16 λ) t}$

e = $e^{(9 λ \frac{1}{a λ})}$

$\Rightarrow \frac{9}{a} = 1 \Rightarrow a = 9$

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6 months ago

Class 12th Physics

A 8 V Zener diode along with a series resistance R is connected across a 20V supply (as shown in the figure). If the maximum Zener current is 25 mA, then the minimum value of R will be_______ $Ω$ .

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Payal Gupta

Contributor-Level 10

$i_{(z e n e r - m a x)} = 25 m A$

20 – i_max R – 8 = 0

i_max R = 12

At minimum zener current $(μ A) : -$

$20 - i_{m i n} R - i_{m i n} R_{L} = 0$

$\frac{R}{R_{L}} = \frac{12}{8} = \frac{3}{2}$

$l_{m i n} R = 12$

$i_{m i n} R_{L} = 8$

At maxm zener current –

$20 - i_{m a x} R - 8 = 0$

$i_{L} = O {a s i_{z} m a x^{m} = 25 m A}$

i_maxR = 12v

25 * 103 R = 12

$R = \frac{12 * 1 0^{3}}{25} = 12 * 40 = 480 Ω$

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6 months ago

Class 12th Physics

The metallic bob of simple pendulum has the relative density. 5 The time period of this pendulum is 10 s. If the metallic bob is immersed in water, then the new time period becomes $5 \sqrt[]{x} s .$ The value of x will be_______.

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Payal Gupta

Contributor-Level 10

g_eff = g – $(\frac{ρ_{w}}{ρ_{b}}) g$

$T = 2 π \sqrt[]{\frac{l}{g}}$

$T' = 2 π \sqrt[]{\frac{l}{g_{e f f}}}$

$\Rightarrow T' = \sqrt[]{\frac{5}{4}} T$

= $\sqrt[]{\frac{5}{4}} * 10$

= $5 \sqrt[]{5} s e c$

So, x = 5

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6 months ago

Class 12th Guru Gobind Singh Indraprastha University

Can I do BCA at IPU University if I don't have computer science or math in class 12th?

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Khushi Jaiswal

Contributor-Level 10

Yes you can pursue a BCA at IPU University even if you did not have computer Science or math in class 12 provided you have studied one of the specified subject related to computer Science or alternatively have completed a 3 year Diploma in Engineering with a minimum of 50% aggregate

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6 months ago

Class 12th Physics Wave Optics

Nearly 10% of the power of a 110 W light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of 1 m from the bulb to a distance of 5 m is a * 10^?2 W/m². The value of 'a' will be________.

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Payal Gupta

Contributor-Level 10

P = 10% of 110W = 11W

(Visible)

$I_{1} - I_{2} = \frac{ρ}{4 π} (\frac{1}{r_{1}^{2}} - \frac{1}{r_{2}^{2}})$

$= \frac{11}{4 * \frac{22}{7}} (\frac{1}{1} - \frac{1}{25})$

$\frac{7}{8} * (\frac{24}{25}) = a * 1 0^{- 2} W / m^{2}$

$\frac{28 * 24}{8} * 1 0^{- 2} = a * 1 0^{- 2}$

⇒ 84 * 10^-2 = a * 10^-2 ⇒ a = 84

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6 months ago

Class 12th Mahatma Jyoti Rao Phoole University

Can I get admission in Mahatma Jyoti Rao Phoole University for BPharma courses with 60% in Class 12th?

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Saurabh Khanduri

Contributor-Level 10

Yes, candidateswho have secured 60% or higher aggregate in Class 12 board exams are eligible for admissions into the BPharma courses of Mahatma Jyoti Rao Phoole University. The admission is done on merit basis, taking into account the total aggregate of scored by a candidate in their Class 12 board exams. The Mahatma Jyoti Rao Phoole University accepts candidates who have scored a minimum of 50% aggregate in their Class 12 board exams.

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