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IIT-JEE 2011 Paper 2 Offline

Numerical

Water (with refractive index = 4/3) in a tank is 18 cm deep. Oil of refractive index 7/4 lies on water making a convex surface of radius of curvature R = 6 cm as shown. Consider oil to act a thin lens. An object S is placed 24 cm above water surface. The location of its image is at x cm above the bottom of the tank. Then x is __________.

Your Input ________

Answer

Correct Answer is 2

Explanation

We have

$${{{n_2}} \over v} - {{{n_1}} \over u} = {{{n_1} - {n_2}} \over R}$$

For the first refracting surface (air-oil), we have n2 = 7/4; n1 = 1; R = 6 cm. Therefore,

$${7 \over {4{v_1}}} - {1 \over {24}} = {{ - (7/4)} \over 6}$$

or v1 = 21 cm and for the second interface (water-oil), we have

$${n_1} = {7 \over 4};{n_2} = {4 \over 3}u = {v_1};R = \infty $$

Therefore,

$${4 \over {3{v_2}}} - {7 \over {4 \times 21}} = 0$$

v2 = 16 cm and v2 + x = height of water.

Therefore,

$$x = 18 - 16 = 2$$

2

IIT-JEE 2010 Paper 2 Offline

Numerical

Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from $${{25} \over 3}$$ m to $${{50} \over 7}$$ m in 30 s. What is the speed of the object in km per hour?

Your Input ________

Answer

Correct Answer is 3

Explanation

Focal length of a convex mirror, $$f = {R \over 2} = {{20} \over 2}$$ m = 10 m

For first object, $${v_1} = + {{25} \over 3}$$ m, $$f = + 10$$ m

Using mirror formula $${1 \over v} + {1 \over u} = {1 \over f}$$

$$\therefore$$ $${1 \over {(25/3)}} + {1 \over {{u_1}}} = {1 \over {10}}$$ or $${1 \over {{u_1}}} = {1 \over {10}} - {3 \over {25}}$$

or $${u_1} = - 50$$ m

For second object,

$${v_2} = + {{50} \over 7}$$ m, $$f = + 10$$ m

$$\therefore$$ $${1 \over {{v_2}}} + {1 \over {{u_2}}} = {1 \over f}$$

$${1 \over {(50/7)}} + {1 \over {{u_2}}} = {1 \over {10}}$$ or $${1 \over {{u_2}}} = {1 \over {10}} - {7 \over {50}}$$ or $${u_2} = - 25$$ m

Speed of the object $$ = {{25} \over {30}}$$ m s$$-$$1

$$ = {{25} \over {30}} \times {{18} \over 5}$$ km h$$-$$1 = 3 km h$$-$$1

3

IIT-JEE 2010 Paper 2 Offline

Numerical

A large glass slab ($$\mu$$ = 5/3) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R?

Your Input ________

Answer

Correct Answer is 6

Explanation

From the figure shown here, we have

$$\tan {i_c} = {R \over t}$$

$$\sin {i_c} = {R \over {\sqrt {{R^2} + {t^2}} }} = {1 \over \mu } = {3 \over 5}$$

$$25{R^2} = 9{R^2} + 9{t^2}$$

$$16{R^2} = 9{t^2} \Rightarrow R = {{3t} \over 4} = {{3 \times 8} \over 4} = 6$$ cm

4

IIT-JEE 2010 Paper 1 Offline

Numerical

The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from m25 to m50. The ratio $${{{m_{25}}} \over {{m_{50}}}}$$ is __________.

Your Input ________

Answer

Correct Answer is 6

Explanation

$${1 \over v} - {1 \over u} = {1 \over f}$$

or, $${u \over v} - 1 = {u \over f}$$

or, $${u \over v} = \left( {{{u + f} \over f}} \right)$$

$$\therefore$$ $$m = {v \over u} = \left( {{f \over {u + f}}} \right)$$

$${{{m_{25}}} \over {{m_{50}}}} = {{\left( {{{20} \over { - 25 + 20}}} \right)} \over {\left( {{{20} \over { - 50 + 20}}} \right)}} = 6$$

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