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Questions

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Practice

Questions

Pick a topic and practice with clear explanations.

Engineering

Let $\alpha > 0$ . If $\int_{0}^{\alpha} \frac{x}{\sqrt{x + \alpha} - \sqrt{x}} dx = \frac{16+20\sqrt{2}}{15}$ , then $\alpha$ is equal to :

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A metal wire of uniform mass density having length L and mass M is bent to form a semicircular arc and a particle of mass m is placed at the centre of the arc. The gravitational force on the particle by the wire is: [Diagram: Semicircular arc with mass m at center]

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Questions

Questions

Let α>0\alpha > 0α>0. If ∫0αxx+α−xdx=16+20215\int_{0}^{\alpha} \frac{x}{\sqrt{x + \alpha} - \sqrt{x}} dx = \frac{16+20\sqrt{2}}{15}∫0α​x+α​−x​x​dx=1516+202​​, then α\alphaα is equal to :

Related questions

Let $\alpha > 0$ . If $\int_{0}^{\alpha} \frac{x}{\sqrt{x + \alpha} - \sqrt{x}} dx = \frac{16+20\sqrt{2}}{15}$ , then $\alpha$ is equal to :