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11 months ago

Class 12th physics ncert exemplar solutions class 12th chapter two

If p is the density and $η$ is coefficient of viscosity of fluid which flows with a speed v in the pipe of diameter d, the correct formula for Reynolds number R_e is:

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11 months ago

Class 12th Mechanical Properties of Fluids

Two capacitors having C₁ and C₂ respectively are connected as shown in figure. Initially, capacitor C₁ is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C₁ is not connected to uncharged capacitor C₂ by closing the switch S. The amount of charge on the capacitor C₂, after equilibrium, is

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Payal Gupta

Contributor-Level 10

After switch 'S' is closed

$Q_{1} + Q_{2} = C_{1} V$ - (1)

Using KVL

$\frac{Q_{1}}{C_{1}} - \frac{Q_{2}}{C_{2}} = 0$

$Q_{1} = Q_{2} \frac{C_{1}}{C_{2}}$ - (2)

from (1) & (2)

$\Rightarrow Q_{2} [\frac{C_{1} + C_{2}}{C_{2}}] = C_{1} V \Rightarrow Q_{2} = (\frac{C_{1} C_{2}}{C_{1} + C_{2}}) V$

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11 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Four

Arrange the following in the decreasing order of their covalent character:

(A) LiCl
(B) NaCl
(C) KCl
(D) CsCl

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alok kumar singh

Contributor-Level 10

Covalent character is $L i C l > N a C l > K C l > C s C l .$ As the cationic size increases polarization decreases.

$∴$ Covalent character decreases.

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11 months ago

Class 12th physics ncert exemplar solutions class 12th chapter two

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

Assertion A : If we move form poles to equator, the direction of acceleration due to gravity of earth always points towards the centre of earth without any variation in its magnitude.

Reason R : At equator, the direction of acceleration due to the gravity is towards the centre of earth.

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

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Vishal Baghel

Contributor-Level 10

$g' = g - ω^{2} R s i n θ$

$g' (a t θ = 0 p o l e) = g$

$g' (a t θ = 90 °, e q u a t o r) = g - ω^{2} R$

g decreases as we move bole to equator

So, A is false statement

But 'R' statement is true

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11 months ago

Class 12th Mechanical Properties of Solids

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats 1.4. Vessel is moving with speed $υ$ and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by:

(R = universal gas constant)

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Payal Gupta

Contributor-Level 10

$K E = Δ U$

$Δ U = \frac{1}{2} m v^{2}$

$n C_{v} Δ T = \frac{m v^{2}}{2}$

$\Rightarrow \frac{m}{M} \frac{R}{y - 1} Δ T = \frac{m v^{2}}{2}$

$Δ T = \frac{0.4}{2} \frac{M v^{2}}{R}$

$Δ T = \frac{M v^{2}}{5 R}$

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11 months ago

Class 12th physics ncert exemplar solutions class 12th chapter two

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is-

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Vishal Baghel

Contributor-Level 10

K.ER (Rotational) = $\frac{1}{2} I_{C M} ω^{2} = \frac{1}{2} * \frac{2}{5} M R^{2} ω^{2} = \frac{M R^{2} ω^{2}}{5}$ - (1)

$K . E_{T o t a l} = K . E_{R} + K . E_{T}$

$= \frac{M R^{2} ω^{2}}{5} + \frac{1}{2} M V_{c m^{2}}$

$= \frac{7 M R^{2} ω^{2}}{10}$ - (2)

$\frac{k . E_{R}}{K . E_{T o t a l}} = \frac{1}{5} * \frac{10}{7} = \frac{2}{7}$

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11 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Which of the following statements are correct?

(A) The electronic configuration of Cr is [Ar] 3d⁵4s¹.

(B) The magnetic quantum number may a negative value.

(C) In the ground state of an atom, the orbitals are filled in order of their increasing energies.

(D) The total number of nodes are given by (n-2)

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alok kumar singh

Contributor-Level 10

(A) Cr = [Ar]3d⁵4s¹

(B) m = $- l t o + l$

(C) According to Aufbau principle, orbital are filled in order of their increasing energies.

(D) Total nodes = n – 1

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11 months ago

Class 12th physics ncert exemplar solutions class 12th chapter two

Arrange the four graphs in descending order of total work done; where W₁, W₂, W₃ and W₄ are the work done corresponding to figure a, b, c and d respectively.

(a) W₃ > W₂ > W₁ >W₄ (b) W₃ > W₂ > W₄ > W₁ (c) (d) W₂> W₃ > W₁ > W₄

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Vishal Baghel

Contributor-Level 10

Work done represents are under

F vs x diagram

w₃ > w₂ > w₁ > w₄

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11 months ago

Class 12th physics ncert exemplar solutions class 12th chapter two

In the arrangement shown in figure a₁, a₂, a₃ and a₄ are the accelerations of masses m₁, m₂, m₃ and m₄ respectively. Which of the following relation is true for this arrangement?

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Vishal Baghel

Contributor-Level 10

$\sum_{}^{} \vec{T} . \vec{a} = 0$

$\frac{T}{2} a_{1} + \frac{T}{4} a_{2} + \frac{T}{8} a_{3} + \frac{T}{8} a_{4} = 0$

$\Rightarrow 4 a_{1} + 2 a_{2} + a_{3} + a_{4} = 0$

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11 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Six

Production of iron in blast furnace follows the following equation

$F e_{3} O_{4} (s) + 4 C O (g) \to 3 F e (I) + 4 C O_{2} (g)$

When 4.640 kg of Fe₃O₄ and 2.520 kg of CO are allowed to react then the amount of iron (in g) produced is:

[Given : Molar Atomic mass (g mo^-1); Fe = 56

Molar Atomic mass (g mol^-1); O = 16

Molar Atomic mass (g mol^-1); C = 12]

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alok kumar singh

Contributor-Level 10

Moles of Fe₃O₄ = $\frac{4.64 * 1 0^{3}}{232} = 20$

Moles of CO = $\frac{2.52 * 1 0^{3}}{28} = 90$

So limiting reagent = Fe₃O₄

So moles of Fe formed = 60

Weight of Fe = 60 * 56 = 3360

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