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Class 11th Chemistry

The molar heat capacity for an ideal gas at constant pressure is 20.785JK^-1 mol^-1. The change in internal energy is 5000 J upon heating it from 300 K to 500 K. The number of moles of the gas at constant volume is __________. (Nearest Integer)

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Payal Gupta

Contributor-Level 10

$C_{P, m} = C_{v, m} + R$

$C_{v, m} = 20.785 - 8.314 = 12.471 J K^{- 1} m o l^{- 1}$

$n = \frac{5000}{12.471 * 200} = \frac{25}{12.471} \approx 2$

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Class 11th Chemistry

Amongst the following, the number of oxide(s) which are paramagnetic in nature is Na₂O, KO₂, NO₂, N₂O, ClO₂, NO, SO₂, Cl₂O (Nearest Integer)

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Payal Gupta

Contributor-Level 10

$K O_{2}, N O_{2}, C l O_{2}, N O$ are paramagnetic.

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Class 11th Physics

A ball of mass m is thrown vertically upwards. Another ball of mass 2 is thrown at an angle q with vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is $\frac{1}{x} .$ The value of x is _______.

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Payal Gupta

Contributor-Level 10

$T_{1} = \frac{2 u}{g} T_{2} = \frac{2 V s i n θ}{g}$

$A s, T_{1} = T_{2} \Rightarrow \frac{2 u}{g} = \frac{2 V s i n θ}{g} \Rightarrow u = v s i n θ$ - (i)

$\frac{H_{1}}{H_{2}} = \frac{u^{2}}{2 g} * \frac{2 g}{v^{2} s i n^{2} θ} = {(\frac{u}{v s i n θ})}^{2} = 1$

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Class 11th Physics

A pulley of radius 1.5m is rotated about its axis by a force F = (12t – 3t²)N applied tangentially (while t is measured in seconds). If moment of inertia of the pulley about its axis of rotation is 4.5 kg m², the number of rotation made by the pulley before its direction of motion is reversed, will be $\frac{K}{π} .$ The value of K is ________.

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Payal Gupta

Contributor-Level 10

r = 1.5 m

F = 12t – 3t2 N

l = 4.5 kgm²

$α = \frac{T}{l} = \frac{\frac{2}{3} (12 t - 3 t^{2})}{4.5} = \frac{1}{3} (12 t - 3 t^{2})$

$\Rightarrow α = 4 t - t^{2}$

$\Rightarrow \frac{d w}{d t} = u t - t^{2} = \int_{0}^{0} d w = \int {(4 t - t^{2})}_{0}^{t} d t$

No. of rev = $\frac{36}{2 π} = \frac{18}{π} = \frac{k}{π} \Rightarrow k = 18$

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8 months ago

Class 11th Physics Mechanical Properties of Solids

A square aluminium (shear modulus is 25 * 10⁹ Nm^-2) slab of side 60 cm and thickness 15 cm is subjected to a shearing force (on its narrow face) of 18.0 * 10⁴ N. The lower edge is riveted to the floor. The displacement of the upper edge is ________ $μ m .$

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Payal Gupta

Contributor-Level 10

Shear modulus = 25 * 109 N/m2

$\frac{F}{A} = τ \frac{Δ x}{h} \Rightarrow Δ x = \frac{F h}{A τ} = \frac{18 * 1 0^{4} * 0.15}{3600 * 1 0^{- 4} * 25 * 1 0^{9}}$

$= \frac{15 * 1 0^{2}}{200 * 25 * 1 0^{5}}$

$= \frac{3 * 1 0^{2}}{5 * 200} * 1 0^{- 5}$

$= \frac{60}{200} * 1 0^{- 5} = 3 μ m$

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8 months ago

Class 11th Physics Oscillations

A mass 0.9kg, attached to a horizontal spring, executes SMH with an amplitude A₁. When this mass passes through its mean position, then a smaller mass of 124 g is placed over it both masses move together with amplitude A₂. If the ratio $\frac{A_{1}}{A_{2}} i s \frac{α}{α - 1},$ then the value of a will be______.

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Payal Gupta

Contributor-Level 10

$m_{1} = 900 g m, w_{1} = \sqrt[]{\frac{k}{. 9}}$

$m_{2} = 1024 g m, ω_{2} = \sqrt[]{\frac{k}{1.024}}$

$m_{1} v_{1} = m_{2} V_{2}$

$\Rightarrow m_{1} ω_{1} A_{1} = m_{2} ω_{2} A_{2}$

$\Rightarrow \frac{A_{1}}{A_{2}} = \sqrt[]{\frac{1024}{900}} = \frac{32}{30} = \frac{16}{15} = \frac{16}{16 - 1}$

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8 months ago

Class 11th Physics

The one division of main scale of Vernier calipers reads 1mm and 10 divisions of Vernier scale is equal to the 9 divisions on main scale, When the two jaws of the instrument touch each other, the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies in between 4.1 cm and 4.2cm and 6^th Vernier division coincides with a main scale division. The diameter of the bob will be ________* 10^-2 cm.

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Payal Gupta

Contributor-Level 10

1 msD = 1mm

10 vsD = 9msD

1vsD = 0.9 MsD

L.C. = 1MSD – 1VSD = 1 – 0.9 = 0.1 mm

Zero error = 4LC = 0.4 mm

Reading = MSR + VSR + correction

= 4.1 cm + 6 * .01 cm + (-0.04 cm) = (4.1 + 0.06 – 0.04) cm

= 4.12 cm = 412 * 10^-² cm

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8 months ago

Class 11th Maths

If the circles x² + y² + $2 (3 - \sqrt[]{3})$ x + $2 (4 - \sqrt[]{6}) y = k + 6 \sqrt[]{3} + 8 \sqrt[]{6}, k > 0,$ touch internally at the point P(, ), then ${(α + \sqrt[]{3})}^{2} + {(β + \sqrt[]{6})}^{2}$ is equal to

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Payal Gupta

Contributor-Level 10

The circle $x^{2} + y^{2} + 6 x + 8 y + 16 = 0$ has centre (3, 4) and radius 3 units

The circle

$x^{2} + y^{2} + 2 (3 - \sqrt[]{3}) x + 2 (4 - \sqrt[]{6}) y = k + 6 \sqrt[]{3} + 8 \sqrt[]{6}, k > 0$ has centre $(\sqrt[]{3} - 3, \sqrt[]{6} - 4)$ and radius $\sqrt[]{k + 34}$

$?$ These two circles touch internally hence

$\sqrt[]{3 + 6} = | \sqrt[]{k + 34} - 3 |$

here, k = 2 is only possible $(? k > 0)$

Equation of common tangent to two circle is

$2 \sqrt[]{3} x + 2 \sqrt[]{6} y + 16 + 6 \sqrt[]{3} + 8 \sqrt[]{6} + k = 0$

$? k = 2$ then equation is

$x + \sqrt[]{2} y + 3 + 4 \sqrt[]{2} + 3 \sqrt[]{3} = 0$ ….(i)

$? (α, β)$ are foot of perpendicular from (3, 4) to line (i) then

$\frac{α + 3}{1} = \frac{β + 4}{\sqrt[]{2}} = \frac{- 3 - 4 \sqrt[]{2} + 3 + 4 \sqrt[]{2} + 3 + \sqrt[]{3}}{1 + 2}$

$∴ {(α + \sqrt[]{3})}^{2} + {(β + \sqrt[]{6})}^{2} = 25$

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8 months ago

Class 11th Maths

Let an = $\int_{- 1}^{n} (1 + \frac{x}{2} + \frac{x^{2}}{3} + . . . . + \frac{x^{n - 1}}{n}) d x$ for every n $\in$ N. Then the sum of all the elements of the set ${n \in N : a_{n} \in (2, 30)} i s . . . . . . . .$

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Payal Gupta

Contributor-Level 10

$? a_{n} = \int_{- 1}^{0} (1 + \frac{x}{2} + \frac{x^{2}}{2} + . . . . + \frac{x^{n - 1}}{n}) d x$

$= {[x + \frac{x^{2}}{2^{2}} + \frac{x^{3}}{3^{2}} + . . . . . + \frac{x^{n}}{n^{2}}]}_{- 1}^{n}$

$a_{n} = \frac{n + 1}{1^{2}} + \frac{n^{2} - 1}{2^{2}} = \frac{n^{3} + 1}{3^{2}} + \frac{n^{4} - 1}{4^{2}} + . . . . . + \frac{n^{n} + {(- 1)}^{n + 1}}{n^{2}}$

Here a₁ = 2, $a_{2} = \frac{2 + 1}{1} + \frac{2^{2} - 1}{2} = 3 + \frac{3}{2} = \frac{9}{2}$

$a_{4} = 5 + \frac{15}{4} + \frac{65}{9} + \frac{255}{16} > 31$

$∴$ The required set is {2, 3}

$? a_{n} \in (2, 30)$

$∴$ Sum of elements = 5.

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8 months ago

Class 11th Sets

Let A = {1, 2, 3, 4, 5, 6, 7}. Define B = $B = {T \subseteq A : e i t h e r 1 \notin T o r 2 \in T}$ and $C = {T \subseteq A : T t h e s u m o f a l l t h e e l e m e n t s o f T i s a p r i m e n u m b e r} .$ Then the number of elements in the set $B \cup C$ is

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Payal Gupta

Contributor-Level 10

$? (B \cup C)' = B' \cup C'$

B' is a set containing sub sets of A containing element 1 and not containing 2.

And C is a set containing subsets of A whose sum of elements is not prime.

so, we need to calculate number of subsets of

{3, 4, 5, 6, 7} whose sum of elements plus 1 is composite.

Number of such 5 elements subset = 1

Number of such 4 elements subset = 3 (except selecting 3 or 7)

Number of such 3 elements subset = 6 (except selecting {3, 4, 5}, {3, 6, 7}, {4, 5, 7} or {5, 6, 7})

Number of such 2 elements subset = 7 (except selecting {3, 7}, {4, 6}, {5, 7})

Number of such 1 elements subset = 3 (except selecting {3, 7}, (4, 6}, {5, 7})

Number

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$? (B \cup C)' = B' \cup C'$

B' is a set containing sub sets of A containing element 1 and not containing 2.

And C is a set containing subsets of A whose sum of elements is not prime.

so, we need to calculate number of subsets of

{3, 4, 5, 6, 7} whose sum of elements plus 1 is composite.

Number of such 5 elements subset = 1

Number of such 4 elements subset = 3 (except selecting 3 or 7)

Number of such 3 elements subset = 6 (except selecting {3, 4, 5}, {3, 6, 7}, {4, 5, 7} or {5, 6, 7})

Number of such 2 elements subset = 7 (except selecting {3, 7}, {4, 6}, {5, 7})

Number of such 1 elements subset = 3 (except selecting {3, 7}, (4, 6}, {5, 7})

Number of such 0 elements subset = 1

$n (B \cap C) = 21 \Rightarrow n (B \cup C) = 2^{7} - 21 = 107$

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