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Class 12th Maths

Let f(x) be a quadratic polynomial with leading coefficient 1 such that f(0) = p, p $\neq 0$ , and f(1) = $\frac{1}{3} .$ If the equations f(x) = 0 and fo fo fo f(x) = 0 have a common real root , then f(3) is equal to

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Payal Gupta

Contributor-Level 10

Let f (x) = (x -α) (x - β)

It is given that f (0) = p = p

and

$f (1) = \frac{1}{3} \Rightarrow (1 - α) (1 - β) = \frac{1}{3}$

Now, let us assume that is the common root of f (x) = 0 and fofofof (x) = 0

fofofof (x) = 0

$f (- 3) = (- 3 - \frac{7}{6}) (3 - 3) = 25$

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Class 12th Chemistry

The total number of monobromo derivatives formed by the alkanes with molecular formula C₅H₁₂ is (excluding stereo isomers)___________.

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Payal Gupta

Contributor-Level 10

Consider the image below

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Class 12th Chemistry

The separation of two coloured substances was done by paper chromatography. The distance travelled by solvent front, substance A and substance B from the base line are 3.25cm, 2.08cmand 1.05cm, respectively. The ratio of R_f values of A to B is ________.

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Payal Gupta

Contributor-Level 10

$R_{f}^{0} = distance travelled by the solute$ $/$ $/ distance travelled by the solvent$

${(R_{f})}_{A} = \frac{2.08}{3.25} {(R_{f})}_{B} = \frac{1.05}{3.25}$

$\frac{{(R_{f})}_{A}}{{(R_{f})}_{B}} = \frac{2}{1}$

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Class 12th Chemistry

A sample of 4.5mg of an unknown monohydric alcohol, R-OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be 3.1 mL. The molecular weight of the unknown alcohol is ____________ g/mol.

[Nearest integer]

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Payal Gupta

Contributor-Level 10

(Calculation is done considering STP condition)

$R O H + C H_{3} M g l \to C H_{4} + R O M g l$

No. of moles ROH = no. of moles of CH4

$\frac{4.5 * 1 0^{- 3}}{M} = \frac{31}{22400} \Rightarrow 32.52 \approx 33 g m / m o l e$

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Class 12th Chemistry

The spin-only magnetic moment value of M³+ ion (in gaseous state) from the pairs Cr³⁺/Cr²⁺, Mn³⁺/Mn²⁺, Fe³⁺/Fe²⁺ and Co³⁺/Co²⁺ that has negative standard electrode potential, is________ B.M [Nearest integer]

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Payal Gupta

Contributor-Level 10

In given electrodes, only $E_{C r^{3 +} / C r^{2 +}}^{0}$ is negative

Cr (24) – [Ar] 4s¹3d⁵4p⁰

Number of unpaired electrons = 3

$μ = \sqrt[]{n (n + 2)} B M = \sqrt[]{3 (3 + 2)} B M = 3.97 B M \approx θ 4 B M$

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Class 12th Chemistry

The sum of number of lone pairs of electrons present on the central atoms of XeO₃, XeOF₄ and XeF₆, is___________.

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Payal Gupta

Contributor-Level 10

XeO₃ ⇒ Number of lone pair = 1

XeOF₄ ⇒ Number of lone pair = 1

XeF₆ ⇒ Number of lone pair = 1

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Class 12th Chemistry Chemical Kinetics

For the decomposition of azomethane,

CH₃N₂CH₃(g) → CH₃CH₃(g) + N₂(g), a first order reaction, the variation in partial pressure with time at 600 K is given as.

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Payal Gupta

Contributor-Level 10

For first order reaction,

$l n (\frac{P}{P_{0}}) = - k . t$

$K = 3.465 * 1 0^{4}$

$t_{1 / 2} = \frac{0.693}{3.465 * 1 0^{4}} = 2 * 1 0^{- 5}$ sec

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Class 12th Chemistry

When the excited electron of a H atom from n = 5 drops to the ground state, the maximum number of emission lines observed are____________.

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Payal Gupta

Contributor-Level 10

For single H- atom, maximum number of spectral lines = (n – 1), n = orbit number of excited electron.

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Class 12th Physics

Magnetic flux (in weber) in a closed circuit of resistance $20 Ω$ varies with time t(s) as $?$ = 8t² – 9t + 5. The magnitude of the induced current at t = 0.25s will be _______mA.

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Payal Gupta

Contributor-Level 10

$e_{i n} = | \frac{d ?}{d t} | = | 16 t - 9 | e_{i n} = | 16 * \frac{1}{4} - 9 | = 5 v o l t$

$l = \frac{e_{i n}}{R} = \frac{5}{20} A = \frac{5}{20} * 1000 m A = 250 m A$

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Class 12th Physics

A convex lens of focal length 20cm is placed in front of a convex mirror with principal axis coinciding each other. The distance between the lens and mirror is 10 cm. A point object is placed on principal axis at a distance of 60cm from the convex lens. The image formed by combination coincides the object itself. The focal length of the convex mirror is ______cm.

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Payal Gupta

Contributor-Level 10

For refraction of light at lens

$\frac{1}{v} - \frac{1}{- 60} = \frac{1}{20}$

$\Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{60} \Rightarrow \frac{1}{v} = \frac{3 - 1}{60} = \frac{1}{30}$

v = 30 cm

If image has to be formed at object itself then light ray should retrace its path. Hence after refraction at lens, it must strikes normally to the mirror

RM = 20 cm Radius of curvature of mirror

Fm = 10 cm focal length of mirror

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