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NEET Physics Solved Question Paper 2009

Short Answer Type

41.

The displacement x of a particle at time t moving under a constant force is t = $\sqrt{x}$ + 3, x in metre, t in second. Find the work done by the interval from t = 0 to t= 6 s.

42.
Calculate the distance below and above the surface of the earth, at which the value of acceleration due to gravity becomes 1/4th that at earth's surface ?

The acceleration due to gravity at depth x below the surface of earth is

$g_{x} = g (1 - \frac{x}{R_{e}}) or \frac{g_{x}}{g} = (1 - \frac{x}{R_{e}}) Given \frac{g_{x}}{g} = \frac{1}{4} ∴ \frac{1}{4} = 1 - \frac{x}{R_{e}} or \frac{x}{R_{e}} = 1 - \frac{1}{4} = \frac{3}{4} or x = \frac{3}{4} R_{e} The acceleration due to gravity at height h is g_{h} = \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}} or \frac{g_{h}}{g} = \frac{1}{{(1 + \frac{h}{R_{e}})}^{2}} Given, \frac{g_{h}}{g} = \frac{1}{4} ∴ \frac{1}{4} = \frac{1}{{(1 + \frac{h}{R_{e}})}^{2}} \Rightarrow 1 + \frac{h}{R_{e}} = 2 or \frac{h}{R_{e}} = 1 or h = R_{e}$

43.

Two cells each of same emf e but of internal resistances r₁ and r₂ are connected in series through an external resistance R. If the potential difference between the ends of the first cell is zero, what will be the value of R in terms of r₁ and r₂ ?

44.

A ray of light travelling inside a rectangular glass block of refractive index $\sqrt{2}$ is incident on the glass-air surface at an angle of incidence of 45°. Show that the ray will emerge into the air at an angle of refraction equal to 90°.