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7 months ago

Class 11th Chemistry

The ionization enthalpy of Na⁺ formation from Na_(g)is 495.8 kJ mol^-1, while the electron gain enthalpy of Br is - 325.0 kJ mol^-1. Given the lattice enthalpy of NaBr is - 728.4 kJ mol^-1. The energy for the formation of NaBr ionic solid is (-) -------------- * 10^-1kJ mol^-1.

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alok kumar singh

Contributor-Level 10

Δ H_{s u b} o f N a m e t a l = 496 k J / m o l e

Bond dissociation energy of Br₂ = 228 kJ/mole

$Δ H_{f} = 495.8 + - 325 - 728.4 = - 557.6 k J / m o l e = - 5576 * 1 0^{- 1} k J / m o l e$

Note : In question is neglect $Δ_{a} H^{0} o f N a, Δ_{v a p} H^{O} o f B r_{2} (l) a n d Δ_{b o n d} H^{0} o f B r_{2} (g)$

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7 months ago

Class 11th Gravitation

A solid sphere of radius R gravitationally attracts a particle placed at 3R form its centre with a force F₁,. Now a spherical cavity of radius $(\frac{R}{2})$ is made in the sphere (as shown in figure) and the force becomes F₂. The value of F₁: F₂ is:

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Vishal Baghel

Contributor-Level 10

$F_{1} = \frac{G M m}{9 R^{2}}$

$F_{2} = \frac{G M m}{9 R^{2}} - \frac{G \frac{M}{8} m}{{(\frac{5 R}{2})}^{2}} = \frac{G M m}{R^{2}} (\frac{1}{9} - \frac{1}{50})$

$\frac{F_{1}}{F_{2}} = \frac{1}{1 - \frac{9}{50}} = \frac{50}{41}$

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7 months ago

Class 11th Chemistry

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alok kumar singh

Contributor-Level 10

Bond dissociation energy of Br₂ = 228 kJ/mole

$? H_{f} = 495.8 + ? 325 ? 728.4 = ? 557.6 k J / m o l e = ? 5576 * 1 0^{? 1} k J / m o l e$

Note : $?_{a} H^{0} ? ? o f ? ? N a, ?_{v a p} H^{O} o f ? ? B r_{2} (l) a n d ? ? ?_{b o n d} H^{0} ? ? o f ? ? B r_{2} (g)$ In question is neglected

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7 months ago

Class 11th Physics Motion in Plane

If $\vec{P} * \vec{Q} = \vec{Q} * \vec{P},$ the angle between $\vec{P} a n d \vec{Q} i s θ (0 ° < θ < 360 °) .$ The value of 'θ' will be………

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Payal Gupta

Contributor-Level 10

If angle between them is 180° then

$\vec{P} * \vec{Q} = \vec{q} * \vec{P} = 0$

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7 months ago

Class 11th Physics

The initial velocity vi required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity v_e such that $v_{i} = \sqrt[]{\frac{x}{y}} * v_{e} .$ The value of x will be………

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Payal Gupta

Contributor-Level 10

Using conservation of energy :

$\frac{1}{2} m v_{i}^{2} - \frac{G m m}{R} = 0 - \frac{G m m}{11 R}$

$v_{i}^{2} = \frac{20}{11} \frac{G m}{R}$

Escape velocity is defined as $\sqrt[]{\frac{2 G M}{R}}$

$v_{i} = \sqrt[]{\frac{10}{11} v_{e}}$

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7 months ago

Class 11th Physics Waves

The percentage increase in the speed of transverse waves produced in a stretched string if the tension is increased by 4%, will be………….%.

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Payal Gupta

Contributor-Level 10

$v = \sqrt[]{\frac{T}{μ}}$

new v' = $\sqrt[]{\frac{1.04 T}{μ}}$

% change = $(\sqrt[]{1.04} - 1) * 100$

= 2%

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7 months ago

Class 11th Physics Thermodynamics

A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52K, its efficiency is doubled. The temperature is Kelvin of the source will be…………

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Payal Gupta

Contributor-Level 10

Efficiency in the Carnot engine is defined as :

$η = \frac{w o r k}{h e a t} = \frac{1}{4}$

$η = 1 - \frac{T_{c}}{T_{h}}, 2 η = 1 - \frac{T_{c} - 52}{T_{h}} \Rightarrow T_{h} = 208 K$

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7 months ago

Class 11th Physics

Two particles having masses 4g and 16g respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is n : 2. The value of n will be……….

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Payal Gupta

Contributor-Level 10

$P = \sqrt[]{2 K m}$

$\frac{n}{2} = \sqrt[]{\frac{m_{1}}{m_{2}}} = \frac{1}{2}$

n = 1

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7 months ago

Class 11th Physics Oscillations

Two identical springs of spring constant '2k' are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is

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Payal Gupta

Contributor-Level 10

Since both spring applied force in same direction.

Equivalent spring constant = 4k

$T = 2 π \sqrt[]{\frac{m}{4 k}}$

$T = π \sqrt[]{\frac{m}{k}}$

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7 months ago

Class 11th Physics

Y = A sin $(ω t + ?_{0})$ is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is y = $\frac{A}{2}$ and it moving along negative x-direction. Then the initial phase angle $?_{0}$ will be

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Payal Gupta

Contributor-Level 10

Using general equation of SHM :

$y = A s i n (ω t + ?_{0}) = \frac{A}{2}$

$ω t + ?_{0} = \frac{π}{6}, \frac{5 π}{6}$

At t = 0

$?_{0} = \frac{π}{6}, \frac{5 π}{6}$

Since it is moving in – x direction

$? = \frac{5 π}{6}$

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