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1

### JEE Main 2021 (Online) 27th August Evening Shift

Curved surfaces of a plano-convex lens of refractive index $$\mu$$1 and a plano-concave lens of refractive index $$\mu$$2 have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses. A
$${1 \over {{\mu _2} - {\mu _1}}}$$
B
$${\mu _1} - {\mu _2}$$
C
$${1 \over {{\mu _1} - {\mu _2}}}$$
D
$${\mu _2} - {\mu _1}$$

## Explanation $${1 \over {{f_1}}} = ({\mu _1} - 1)\left( {{1 \over R}} \right)$$

$${1 \over {{f_2}}} = ({\mu _2} - 1)\left( {{1 \over R}} \right)$$

$${1 \over {{f_1}}} + {1 \over {{f_2}}} = {1 \over {{f_{eq}}}} = {{({\mu _1} - 1) - ({\mu _2} - 1)} \over R}$$

$${1 \over {{f_{eq}}}} = {{({\mu _1} - {\mu _2})} \over R}$$

$${R \over {{f_{eq}}}} = ({\mu _1} - {\mu _2})$$
2

### JEE Main 2021 (Online) 27th August Morning Shift

Find the distance of the image from object O, formed by the combination of lenses in the figure : A
75 cm
B
10 cm
C
20 cm
D
infinity

## Explanation

$${1 \over {{V_1}}} + {1 \over {30}} = {1 \over {10}}$$

$${1 \over {{V_1}}} = {2 \over {30}} \Rightarrow {V_1} = 15$$ cm

$${1 \over {{V_2}}} - {1 \over {10}} = - {1 \over {10}}$$

$${1 \over {{V_2}}} = 0$$

V2 = $$\infty$$

V3 = 30 cm

OV3 = 75 cm
3

### JEE Main 2021 (Online) 27th August Morning Shift

An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d1 from C and the distance of the image formed is d2 from C, the radius of curvature of this mirror is :
A
$${{2{d_1}{d_2}} \over {{d_1} - {d_2}}}$$
B
$${{2{d_1}{d_2}} \over {{d_1} + {d_2}}}$$
C
$${{{d_1}{d_2}} \over {{d_1} + {d_2}}}$$
D
$${{{d_1}{d_2}} \over {{d_1} - {d_2}}}$$

## Explanation

Using Newton's formula

$$(f + {d_1})(f - {d_2}) = {f^2}$$

$${f^2} + f{d_1} - f{d_2} - {d_1}{d_2} = {f^2}$$

$$f = {{{d_1}{d_2}} \over {{d_1} - {d_2}}}$$

$$\therefore$$ $$R = {{2{d_1}{d_2}} \over {{d_1} - {d_2}}}$$
4

### JEE Main 2021 (Online) 26th August Morning Shift

Car B overtakes another car A at a relative speed of 40 ms$$-$$1. How fast will the image of car B appear to move in the mirror of focal length 10 cm fitted in car A, when the car B is 1.9 m away from the car A?
A
4 ms$$-$$1
B
0.2 ms$$-$$1
C
40 ms$$-$$1
D
0.1 ms$$-$$1

## Explanation

Here, $${1 \over v} + {1 \over u} = {1 \over f}$$ .......(1)

$$\Rightarrow$$ $${1 \over v} + {1 \over { - 190}} = {1 \over {10}}$$

$$\Rightarrow$$ v = $${{19} \over 2}$$

Differentiating equation (1) w.r.t t we get

$$- {1 \over {{v^2}}}\left( {{{dv} \over {dt}}} \right) - {1 \over {{u^2}}}\left( {{{du} \over {dt}}} \right) = 0$$

$$\Rightarrow$$ $$\left( {{{dv} \over {dt}}} \right) = - {{{v^2}} \over {{u^2}}}\left( {{{du} \over {dt}}} \right)$$

$$\Rightarrow$$ $$\left( {{{dv} \over {dt}}} \right) = - {\left( {{{{{19} \over 2}} \over {190}}} \right)^2} \times 40$$

$$= {1 \over {400}} \times 40 = 0.1$$ m/s

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