Answer:
The jug drowns because the density of the jug is more than that of the density of water.
0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.
<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>
Position of n the order fringe = n λ D / d
for n = 2
position = 2 λ D / d
λ = 710 nm , D = 1.5m
d = .65 x 10⁻³
position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3276.92 x 10⁻⁶ m
= 3.276x 10⁻³ m
= 3.276mm .
For λ = 660 nm
position = 2 λ D / d
λ = 660 nm , D = 1.5 m
d = .65 x 10⁻³
position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3046.15 x 10⁻⁶ m
= 3.046 x 10⁻³ m
= 3.046 mm .
Difference between their position
= 3.276mm ₋ 3.046 mm
= 0.23 mm .
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Answer:
Option (2)
Explanation:
From the figure attached,
Horizontal component, 
![A_x=12[\text{Sin}(37)]](https://tex.z-dn.net/?f=A_x%3D12%5B%5Ctext%7BSin%7D%2837%29%5D)
= 7.22 m
Vertical component, ![A_y=A[\text{Cos}(37)]](https://tex.z-dn.net/?f=A_y%3DA%5B%5Ctext%7BCos%7D%2837%29%5D)
= 9.58 m
Similarly, Horizontal component of vector C,
= C[Cos(60)]
= 6[Cos(60)]
= 
= 3 m
![C_y=6[\text{Sin}(60)]](https://tex.z-dn.net/?f=C_y%3D6%5B%5Ctext%7BSin%7D%2860%29%5D)
= 5.20 m
Resultant Horizontal component of the vectors A + C,
m
= 4.38 m
Now magnitude of the resultant will be,
From ΔOBC,
= 
= 
= 6.1 m
Direction of the resultant will be towards vector A.
tan(∠COB) = 
= 
= 
m∠COB = 
= 46°
Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.
Option (2) will be the answer.
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