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### JEE Main 2021 (Online) 1st September Evening Shift

MCQ (Single Correct Answer)
A glass tumbler having inner depth of 17.5 cm is kept on a table. A student starts pouring water ($$\mu$$ = 4/3) into it while looking at the surface of water from the above. When he feels that the tumbler is half filled, he stops pouring water. Up to what height, the tumbler is actually filled?
A
11.7 cm
B
10 cm
C
7.5 cm
D
8.75 cm

## Explanation

Consider the actual height of the tumbler be H.

The refractive index of the water, $$\mu$$ = 4/3

The refractive index of the air, $$\mu$$ = 1

As we know that,

$${\mu _{water}} = {{{H_{real}}} \over {{H_{apparent}}}} \Rightarrow {4 \over 3} = {H \over {{H_{apparent}}}}$$

$$\Rightarrow {H_{apparent}} = {{3H} \over 4}$$

Height of air observed by observer = 17.5 $$-$$ H

According to question, both height observed by observer is same.

$${{3H} \over 4}$$ = 17.5 $$-$$ H

$$\Rightarrow$$ H = 10 cm

Option (b)
2

### JEE Main 2021 (Online) 31st August Morning Shift

MCQ (Single Correct Answer)
Two plane mirrors M1 and M2 are at right angle to each other shown. A point source 'P' is placed at 'a' and '2a' meter away from M1 and M2 respectively. The shortest distance between the images thus formed is : (Take $$\sqrt 5$$ = 2.3)

A
3a
B
4.6a
C
2.3a
D
2$$\sqrt {10}$$a

## Explanation

Shortest distance is 2a between I1 & I3

But answer given is for I1 & I2

= $$\sqrt {{{(4a)}^2} + {{(2a)}^2}}$$

= $$a\sqrt {20}$$

= 4.47 a

Option (b)
3

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

MCQ (Single Correct Answer)
The light waves from two coherent sources have same intensity I1 = I2 = I0. In interference pattern the intensity of light at minima is zero. What will be the intensity of light at maxima?
A
I0
B
2 I0
C
5 I0
D
4 I0

## Explanation

$${I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}$$

= 4 I0
4

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

MCQ (Single Correct Answer)
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

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

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

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

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

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

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