Answer:
346.66 Hz
Explanation:
= Length of string which is unfingered = l
= Length of string which is vibrate when fingered = 
= Unfingered frequency = 260 Hz
= Fingered frequency
Frequency is inversely proportional to length

So,

The frequency of the fingered string is 346.66 Hz
Kinetic energy is the energy possessed by a body while in motion. It is calculated by 1/2mv², where m is the mass of the body and v is the velocity.
Therefore, kinetic energy is dependent on both mass of the body and the velocity. An increase in mass increases the kinetic energy, an increase in velocity also increases kinetic energy of the body. Thus, doubling the mass and doubling the velocity will both increase the kinetic energy of the body.
Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
That type of bending is called "diffraction" of waves.