The following questions which appeared in All India Institute of Medical Sciences (AIIMS) 2005 entrance question paper for admitting students to the MBBS Degree course are simple as usual. They are meant for checking your knowledge and understanding of fundamentals.

**(1) The apparent depth of water in a cylindrical water tank of diameter 2R cm is reducing at the rate of x cm/minute when water is being drained out at a constant rate. The amount of water drained in c.c. per minute is (n _{1}= refractive index of air, n_{2 }= refractive index of water)**

**(a) x π R ^{2 }n_{1}/n_{2} (b) x π R^{2 }n_{2}/n_{1} (c) 2 π R^{ }n_{1}/n_{2} (d) π R^{2}x **

Since the refractive index is the ratio of real depth to the apparent depth, we have

Real depth = Apparent depth × refractive index.

Therefore, the rate at which the real depth is decreasing = xn_{2}/n_{1} cm per minute.

The amount of water drained in c.c. per minute is therefore equal to **x π R ^{2 }n_{2}/n_{1}**,

_{ }given in option (b).

**(2) A telescope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm. If the telescope is used to see a 50 m tall building at a distance of 2 km, what is the length of the image of the building formed by the objective lens?**

**(a) 5 cm (b) 10 cm (c) 1 cm (d) 2cm**

At the first glance this question may seem to be one involving the magnification produced by a telescope; but, this is quite simple since you are asked to consider the objective only.

The objective will produce the image of the building at the focus (which is at 2 m from the lens) and hence from the expression for magnification (M) we have

M = Distance of image/ Distance of object = Height of image/ Height of object

so that ** **2/ 2000 = x/50 where ‘x’ is the height of image in metre.

Therefore, x = 2×50/2000 = 0.05 m = 5 cm.

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