Answer:
<h3>
Young modulus of elasticity for a gas is</h3><h2>
<em>Zero</em></h2>
Explanation:
<em>As</em><em> </em><em>the</em><em> </em><em>gas</em><em> </em><em>doesn't</em><em> </em><em>undergo</em><em> </em><em>any</em><em> </em><em>chan</em><em>g</em><em>es</em><em> </em>
<em>so</em><em> </em><em>the</em><em> </em><em>young</em><em> </em><em>modules</em><em> </em><em>of</em><em> </em><em>gas</em><em> </em><em>is</em><em> </em><em>not</em><em> </em><em>defined</em><em>.</em><em>.</em><em>.</em>
The diagram is missing; however, we know that the intensity of a sound wave is inversely proportional to the square of the distance from the source:

where I is the intensity and r is the distance from the source.
We can assume for instance that the initial distance from the source is r=1 m, so that we put

The intensity at r=3 m will be

Therefore, the sound intensity has decreased by a factor

.
Answer:
Minimum thickness will be 100 nm
Explanation:
We have given refractive index is n = 1.5
Wavelength of the light incidence
= 600 nm
We have to find the smallest thickness of the film so that there will be minimum light reflect
For minimum thickness of non reflecting film
, here t is thickness,
is wavelength and n is refractive index
Putting all values 
So minimum thickness will be 100 nm
Explanation:
(i)
O is the object and I is the image.
The image formed is enlarged and it is erect. So the magnification will be positive (+) and greater than 1.
Refer above image. 1
(ii)
O is the object and I is the image.
The image formed is diminished and erect. So the magnification will be positive (+) and less than1.
Refer above image. 2
(iii)
The image will be formed as the 2F on the other side of the lens and it will be of same of the object.