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3 months ago

Class 11th Hydrogen Bonding

Why HF has a higher boiling point than HCl?

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Jaya Sinha

Beginner-Level 5

The difference in boiling or melting point, even after nearly the same molecular geometry, is because opf the presence of hydrogen bond. HF forms intermolecular hydrogen bonds due to fluorine's high electronegativity. This intermolecular hydrogen bonding requires more energy due to break the hydrogen bonds and melt or boil. So, stronger intermolecular forces of HF result in a higher boiling point.

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Class 11th Physics

If K₁ and K₂ are the thermal conductivities, L₁ and L₂ are the lengths and A₁ and A₂ are the cross sectional areas of steel and copper rods respectively such that $\frac{K_{2}}{K_{1}} = 9, \frac{A_{1}}{A_{2}} = 2, \frac{L_{1}}{L_{2}} = 2 .$ Then, for the arrangement as shown in the figure, the value of temperature T of the steel – copper junction in the steady state will be:

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alok kumar singh

Contributor-Level 10

$\frac{K_{2}}{k_{1}} = 9 \frac{A_{1}}{A_{2}} = 2, \frac{L_{1}}{L_{2}} = 2$

$\frac{450 - T}{\frac{L_{1}}{A_{1} k_{1}}} = \frac{T}{\frac{L_{2}}{k_{2} A_{2}}} \Rightarrow (450 - T) \frac{L_{2}}{K_{2} A_{2}} = (T) \frac{L_{1}}{A_{1} k_{1}}$

$(450 - T) = T (\frac{L_{1}}{L_{2}} \frac{k_{2} A_{2}}{k_{1} A_{1}})$ $\Rightarrow 450 - T = T (2 * 9 * \frac{1}{2}) = 9 T$

10 T = 450

T = 45° C

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3 months ago

Class 11th Physics

Two satellites A and B, having masses in the ration 4 : 3, are revolving in circular orbits of radii 3r and 4r respectively around the earth. The ratio of total mechanical energy of A to B is:

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alok kumar singh

Contributor-Level 10

$\frac{E_{A}}{E_{B}} = \frac{- \frac{G M m_{A}}{2 * 3 r}}{\frac{- G M m_{B}}{2 * 4 r}} = \frac{m_{A}}{m_{B}} * \frac{4}{3} = \frac{4}{3} * \frac{4}{3} = \frac{16}{9}$

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3 months ago

Class 11th Physics

A ball is projected from the ground with a speed 15 ms^-1 at an angle θ with horizontal so that its range and maximum height are equal. Then 'tanθ' will be equal to

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Payal Gupta

Contributor-Level 10

$H = \frac{u^{2} s i n^{2} θ}{2 g} a n d R = \frac{u^{2} s i n 2 θ}{g}$

According to question

$R = H \Rightarrow \frac{u^{2} s i n^{2} θ}{2 g} = \frac{2 u^{2} s n θ c o s θ}{g} \Rightarrow t a n θ = 4$

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3 months ago

Class 11th Physics Mechanical Properties of Fluids

Two cylindrical of equal cross-sectional are 16 cm² contain water upto height 100cm and 150 cm respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, it [Take, density of water = 10³ kg/m³ and g = 10 ms^-2]:

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alok kumar singh

Contributor-Level 10

$ρ = 1 0^{3} k g / m^{3}$

g = 10 m/s²

Final height in both vessels

$= \frac{100 + 150}{2} = 125 c m$

So, less in U = $\int A * 0.25 * g * 0.25$

$1 0^{3} * 625 * 1 0^{- 4} * 10 * 16 * 1 0^{- 4}$

= 625 * 16 * 104

= 1J.

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3 months ago

Class 11th Work, Energy and Power

A bag of sand of mass 9.8kg is suspended by a rope. A bullet of 200g travelling with speed 10 ms^-1 gets embedded in it, then loss of kinetic energy will be

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Payal Gupta

Contributor-Level 10

$v = \frac{m v_{0}}{M + m} \Rightarrow K_{f} = \frac{1}{2} (M + m) v^{2} = \frac{m v_{0}^{2}}{2 (M + m)}$

$K_{i} = \frac{m v_{0}^{2}}{2}$

Loss in Kinetic energy = $Δ K = K_{i} - K_{f}$

$Δ K = \frac{1}{2} m v_{0}^{2} (\frac{M}{M + m}) = \frac{1}{2} * 0.2 * 100 = [\frac{9.8}{10}] = 9.8 J$

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3 months ago

Class 11th Physics Gravitation

An object is taken to a height above the surface of earth at a distance $\frac{5}{4} R$ from the centre of the earth. Where radius of earth, R = 6400km. The percentage decrease in the weight of the object will be

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Payal Gupta

Contributor-Level 10

As we know that $g_{P} = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{\frac{5}{4} R})}^{2} = \frac{16 g}{25}$

$\Rightarrow Δ g = g - g_{P} = \frac{9 g}{25}$

The percentage decrease in the weight of the object = $\frac{Δ g}{g} * 100 = 36 %$

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3 months ago

Class 11th Physics

Let $η_{1}$ is the efficiency of an engine at T₁ = $447 ° C$ and T₂ = $147 ° C$ while $η_{2}$ is the efficiency at T₁ = $947 ° C$ and T₂ $47 ° C$ . The ratio $\frac{η_{1}}{η_{2}}$ will be

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Payal Gupta

Contributor-Level 10

As we know that

$η = 1 - \frac{T_{2}}{T_{1}}, s o$

$\frac{n_{1}}{n_{2}} = \frac{0.42}{0.74} \approx 0.56$

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3 months ago

Class 11th Physics

Sound travels in a mixture of two moles of helium and n moles of hydrogen. If rms speed of gas molecules in the mixture is $\sqrt[]{2}$ time the speed of sound, then the value of n will be

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Payal Gupta

Contributor-Level 10

As we know that

$v = \sqrt[]{\frac{γ R T}{M}} \Rightarrow$ Speed of sound in a gas, and

$v_{r m s} = \sqrt[]{\frac{3 R T}{M}} \Rightarrow$ Rms speed of gas molecules, so

$v_{r m s} = \sqrt[]{2} v \Rightarrow \sqrt[]{\frac{3 R T}{M}} = \sqrt[]{2} \sqrt[]{\frac{γ R T}{M}} \Rightarrow γ = \frac{3}{2}$

According formula of specific heat ratio of mixture, we can write

$\Rightarrow$ 5n = 14 n = 2.8

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New answer posted

3 months ago

Class 11th Physics Oscillations

The length of a seconds pendulum at a height h = 2R from earth surface will be:

(Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = $π^{2} m s^{- 2}$ )

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Payal Gupta

Contributor-Level 10

As we know that

$g_{P} = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{R + 2 R})}^{2} = \frac{g}{9}, s o$

$T = 2 π \sqrt[]{\frac{l}{g_{P}}} \Rightarrow l = \frac{g_{p} T^{2}}{4 π^{2}} = \frac{1}{9} m$

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