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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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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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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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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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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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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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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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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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Class 11th Physics

It momentum [P], area [A] and time [T] are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is:

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Vishal Baghel

Contributor-Level 10

Viscosity = Pascal . Second

$P^{X} A^{Y} T^{Z} = [M^{1} L^{- 1} T^{- 1}]$

= ${[M^{1} L^{+ 1} T^{- 1}]}^{x} [L^{2}] y {[T^{1}]}^{z} = M^{- 1} L^{- 1} T^{- 1}$

= x = 1, x + 2y = -1, -x + z = 1

y = -1, Z = 0

Viscosity = [P¹A^-¹T⁰]

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Class 11th Bond Parameters

Explain why the bond angle in ${N H}_{3}$ is larger than in ${P H}_{3}$ ?

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Himanshi Singh

Beginner-Level 5

The reason why bond angle is larger in ${N H}_{3}$ than ${P H}_{3}$ are given below.

In ${N H}_{3}$ , there is one lone pair on the nitrogen atom increases repulsion, while the lone pair on phosphorus is in a higher energy orbital and causes less repulsion.
Nitrogen is a small and electronegative atom whereas phosphorus is larger and less electronegative than nitrogen.
In ${P H}_{3}$ , the bonding orbitals are nearly pure p-orbitals, which are less directionals–p hybridization, on the other hand, the bond pairs remain fairly directed, leading to a larger bond angle.

Hence, the bond angle in ${N H}_{3}$ is about 107° ( less than the ideal tetrahedral 109.5°) due to lone pair pres

...more

The reason why bond angle is larger in ${N H}_{3}$ than ${P H}_{3}$ are given below.

In ${N H}_{3}$ , there is one lone pair on the nitrogen atom increases repulsion, while the lone pair on phosphorus is in a higher energy orbital and causes less repulsion.
Nitrogen is a small and electronegative atom whereas phosphorus is larger and less electronegative than nitrogen.
In ${P H}_{3}$ , the bonding orbitals are nearly pure p-orbitals, which are less directionals–p hybridization, on the other hand, the bond pairs remain fairly directed, leading to a larger bond angle.

Hence, the bond angle in ${N H}_{3}$ is about 107° ( less than the ideal tetrahedral 109.5°) due to lone pair presence and highly electronegative atom, and the bond angle in PH? ( 93.5°) is much smaller due to less repulsive lone pair, and less electronegative atom.

Hence, the bond angle in ${N H}_{3}$ (107°) is larger than in ${P H}_{3}$ (93.5°).

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