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11 months ago

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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11 months ago

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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11 months ago

Class 11th Physics

Following statements are given:

(A) The average kinetic energy of a gas molecule decreases when the temperature is reduced.

(B) The average kinetic energy of a gas molecule increase with increase in pressure at constant temperature.

(C) The average kinetic energy of a gas molecule decreases with increase in volume.

(D) Pressure of a gas increase with increase in temperature at constant pressure.

(E) The volume of gas decreases with increase in temperature .

Choose the correct answer from the options given below:

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Vishal Baghel

Contributor-Level 10

$K E_{a v g} = \frac{3}{2} K T$

$P = \frac{1}{3} ρ V_{r m s}^{2}$

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11 months ago

Class 11th Thermodynamics

A certain amount of gas of volume V at $27 ° C$ temperature and pressure $2 * 1 0^{7} N m^{- 2}$ expands isothermally until its volume gets doubled. Late it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be

(Use $γ$ = 1.5):

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Vishal Baghel

Contributor-Level 10

$P_{1} = 2 * 1 0^{7} P a$

$P_{1} v_{1} = P_{2} v_{2}$

Since v₂ = 2v₁ Hence $P_{2} = \frac{P_{1}}{2} (I s o t h e r m a l E x p a n s i o n)$

$P_{2} = 1 * 1 0^{7} P a$

$P_{2} {(v_{2})}^{y} = P_{3} {(2 v_{2})}^{y}$

$P_{3} = \frac{1 * 1 0^{7}}{2^{1.5}} = 3.536 * 1 0^{6} P a$

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11 months ago

Class 11th Physics

Three identical particles A, B and C of mass 100kg each are placed in a straight line with AB = BC = 13m. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately:

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Vishal Baghel

Contributor-Level 10

F on P = $\frac{}{^{}} \frac{}{^{}}$ $\frac{G M M}{r^{2}} + \sqrt[]{2} \frac{G M M}{{(\sqrt[]{2} r)}^{2}}$

$F o n P = \frac{G * 1 0^{4}}{1 3^{2}} (1 + \frac{1}{\sqrt[]{2}}) \approx 100 G$

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11 months ago

Class 11th System of Particles and Rotational Motion

A solid cylinder and a solid sphere, having same mass M and radius R, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they the ground will be:

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Vishal Baghel

Contributor-Level 10

$v = \sqrt[]{\frac{2 g H}{1 + k^{2} / R^{2}}}$

$\frac{V_{C y l i n d e r}}{V_{S p h e r e}} = \sqrt[]{\frac{{(1 + k^{2} / R^{2})}_{S p h e r e}}{{(1 + k^{2} / R^{2})}_{C y l i n d e r}}}$

= $\sqrt[]{\frac{1 + 2 / 5}{1 + 1 / 2}} = \sqrt[]{\frac{14}{15}}$

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11 months ago

Class 11th Structure of Atom

Given below are two statements. One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A: Energy of 2s orbital of hydrogen atom is greater that that of 2s orbital of lithium.

Reason R: Energies of the orbitals in the same subshell decrease with increase in the atomic.

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

Energy of orbitals in same subshell decrease on increasing the atomic number.

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11 months ago

Class 11th Work, Energy and Power

A body of mass 0.5 kg travels on straight line path velocity u = $(3 x^{3} + 4)$ $^{}$ m/s. The net work done by the force during its displacement from x = 0 to x = 2m is:

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Vishal Baghel

Contributor-Level 10

$v_{i} = 3 * {(0)}^{2} + 4 = 4 m / s$

$v_{F} = 3 * 2^{2} + 4 = 16 m / s$

$ω = Δ k = \frac{1}{2} m [v_{F}^{2} - v_{i}^{2}]$

= $\frac{1}{2} * 0.5 [1 6^{2} - 4^{2}] = \frac{240}{4} = 60 J$

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11 months ago

Class 11th Maths

The statement $(p \Rightarrow q) \lor (p \Rightarrow r)$ is NOT equivalent to

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alok kumar singh

Contributor-Level 10

$(A) \sim (p \land (\sim r)) \lor q = (\sim p) \lor (r) \lor q$

(B) $q \lor (- r \lor p)$

(C) $(\sim p) \lor (q \lor r)$

(D) $(\sim p \lor q) \lor r$

Using Venn diagram we get B as the correct option.

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11 months ago

Class 11th Motion in a Plane

A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is 60m, then the magnitude of displacement would be:

(Given $c o s 135 °$ = -0.7)

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Vishal Baghel

Contributor-Level 10

d = R

$60 = R [\frac{3 π}{4}]$

$R = \frac{60 * 4}{3 π} = \frac{80}{π} m$

Displacement = $\sqrt[]{R^{2} + R^{2} - 2 R^{2} c o s 135}$ $^{}^{}^{}$

= $\sqrt[]{2 R^{2} - 2 R^{2} (- 0.7)}$ $^{}^{}$

= $\sqrt[]{3.4 R^{2}}$ $^{}$

= $\sqrt[]{3.4 {(\frac{80}{4})}^{2}}$ ${\frac{}{}}^{}$

47 m

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