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Class 11th Maths

Consider the following statements:

P : Ramu is intelligent.

Q : Ramu is rich.

R : Ramu is not honest.

The negation of the statement “Ramu is intelligent and honest if and only if Ramu is not rich” can be expressed as

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Payal Gupta

Contributor-Level 10

P : Ramu is intelligent

Q : Ramu is rich

R : Ramu is not honest

Give statement, “Ramu is intelligent and honest if any only if Ramu is not rich”

$= (P \land \sim R) \Leftrightarrow \sim Q$

So, negation of the statement is

$\sim [(P \land \sim R) \Rightarrow \sim Q]$

$= ((P \land \sim R) \land Q) \lor (\sim Q \land (\sim P \lor R))$

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Class 11th NCERT Class 12 Chemistry

Given below are two statements.

Statement I: The chlorides of Be and Al have Cl-bridged structure. Both are soluble in organic solvents and act as Lewis bases.

Statement II: Hydroxides of Be and Al dissolve in excess alkali to give beryllate and aluminate ions.

In the light of the above statements, choose the correct answer form the options given below.

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alok kumar singh

Contributor-Level 10

Be₂Cl₄ is Lewis acid and Al₂Cl₆ has complete octet. Be and Al are amphoteric metal therefore dissolve in acid as well as alkaline solution and form beryllate and aluminate ions in excess alkali.

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Class 11th Maths Statistics

If the mean deviation about median for the numbers 3, 5, 7, 2k, 12, 16, 21, 24, arranged in the ascending order, is 6 then the median is

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Payal Gupta

Contributor-Level 10

Median = $\frac{2 k + 12}{2} = k + 6$

Mean deviation = $\sum \frac{| x_{i} - M |}{n} = 6$

$\Rightarrow \frac{(k + 3) + (k + 1) + (k - 1) + (6 - k) + (6 - k) + (10 - k) + (15 - k) + (18 - k)}{8}$

$∴ \frac{58 - 2 k}{8} = 6$

k = 5

$M e d i a n = \frac{2 * 5 + 12}{2} = 11$

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Class 11th Physics System of Particles and Rotational Motion

Mass per unit are of a circular disc of radius a depends on the distance $r$ from its centre as $σ (r) = A + B r$ . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is:

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alok kumar singh

Contributor-Level 10

Moment of inertia of ring

$d l = {d m r}^{2}$

I = \int_{0}^{a} ? (A + B r) 2 π r d r π r^{2}

= 2 π A \int_{0}^{a} ? r^{3} d r + 2 π B \int_{0}^{a} ? r^{4} d r

= 2 π  [A \frac{a^{4}}{4} + B \frac{a^{5}}{5}]

= 2 π a^{4}  [\frac{A}{4} + \frac{B a}{5}]

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Class 11th Waves

A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much les than the speed of sound). The observers hears 2 beats/sec. The oscillation frequency of each tuning fork is $v_{0} = 1400 H z$ and the velocity of sound in air is $350 m / s$ . The speed of each tuning fork is close to:

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alok kumar singh

Contributor-Level 10

$f_{1} = f_{o} (\frac{v}{v - v_{s}}) f_{1} - f_{2} = 2$

$f_{2} = f_{o} (\frac{v}{v + v_{s}}) f_{0} v [\frac{v + v_{s} - (v - v_{s})}{v^{2} - v_{s}^{2}}] = 2; \frac{2 f_{0} v v_{s}}{v^{2} - v_{s}^{2}} = 2$

$? v_{s} ? v ∴ v_{s} = \frac{1}{4} m / s$

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Class 11th Chemistry

Given below are two statements.

Statement I: Hydrogen peroxide can act as an oxidizing agent in both acidic and basic conditions.

Statement II: Density of hydrogen peroxide at 298 K is lower than that of D₂O.

In the light of the above statements, choose the correct answer from the options given below.

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alok kumar singh

Contributor-Level 10

Depending on the nature of reducing agent H₂O₂ can act as oxidizing agent in both acidic as well as basic medium.

Density of D₂O = 1.1 gm/cc

Density of D₂O₂ = 1.45 gm/cc

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Class 11th Maths

Let the foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$ and the hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is

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Payal Gupta

Contributor-Level 10

Ellipse : $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$

Eccentricity = $\sqrt[]{1 - \frac{7}{10}} = \frac{3}{4}$

Foci $\equiv (\pm a e, 0)$ $\equiv (\pm 3, 0)$

Hyperbola : $\frac{x^{2}}{(\frac{144}{25})} - \frac{y^{2}}{(\frac{α}{25})} = 1$

Eccentricity = $\sqrt[]{1 + \frac{α}{144}} = \frac{1}{12} \sqrt[]{144 + α}$

$= 2 . \frac{\frac{81}{25}}{\frac{12}{5}} = \frac{27}{10}$

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Class 11th Maths

The tangents at the points A(1, 3) and B(1, -1) on the parabola y² – 2x – 2y = 1 meet at the point P. Then the area (in unit²) of the triangle PAB is

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Payal Gupta

Contributor-Level 10

$y^{2} - 2 x - 2 y = 1$

$\Rightarrow {(y - 1)}^{2} = 2 (x + 1)$ …. (i)

Equation of tangent at A is 2x – y – 5 = 0 ………. (ii)

D is mid point of AB solving (ii) with y = 1 P (3, 1)

$∴ P D = 4, A D = 2$

$A r e a o f Δ A P D = \frac{1}{2} (P D) (A D) = 4$

$∴ A r e a o f Δ A P B = 8 s q . u n i t s$

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Class 11th Chemistry

The incorrect statement is

(a) The first ionization enthalpy of K is less than that of Na and Li

(b) Xe does not have the lowest first ionization enthalpy in its group

(c) The first ionization enthalpy of element with atomic number 37 is lower than that of the element with atomic number 38.

(d) The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30.

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alok kumar singh

Contributor-Level 10

Ionisation enthalpy order :

Li > Na > K

He > Ne > Ar > Kr > Xe > Rn

Sr > Rb

Zn > Ga

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Class 11th Oscillations

In figure (A), mass '2 m' is fixed on mass 'm' which is attached to two spring of spring constant k. In figure (B), mass 'm' is attached to two springs of spring constant 'k' and '2k'. If mass 'm' in (A) and in (B) are displaced by distance 'x' horizontally and then released, then time period T₁ and T₂ corresponding to (A) and (B) respectively follow the relation.

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Vishal Baghel

Contributor-Level 10

$T_{1} = 2 π \sqrt[]{\frac{3 m}{2 k}}$

$T_{2} = 2 π \sqrt[]{\frac{m}{3 k}}$

$\frac{T_{1}}{T_{2}} = \frac{2 π \sqrt[]{\frac{3 m}{2 k}}}{2 π \sqrt[]{\frac{m}{3 k}}} = \frac{3}{\sqrt[]{2}}$

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