Answer:
a
The focal length of the lens in water is 
b
The focal length of the mirror in water is 
Explanation:
From the question we are told that
The index of refraction of the lens material = 
The index of refraction of the medium surrounding the lens = 
The lens maker's formula is mathematically represented as
![\frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%28n%20-1%29%20%5B%5Cfrac%7B1%7D%7BR_1%7D%20-%20%5Cfrac%7B1%7D%7BR_2%7D%20%20%5D)
Where
is the focal length
is the index of refraction
are the radius of curvature of sphere 1 and 2 of the lens
From the question When the lens in air we have
![\frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf_%7Bair%7D%7D%20%3D%20%28n-1%29%20%5B%5Cfrac%7B1%7D%7BR_1%7D%20-%20%5Cfrac%7B1%7D%7BR_2%7D%20%20%5D)
When immersed in liquid the formula becomes
![\frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf_%7Bwater%7D%7D%20%3D%20%5B%5Cfrac%7Bn_2%7D%7Bn_1%7D%20-%201%20%5D%20%5B%5Cfrac%7B1%7D%7BR_1%7D%20-%20%5Cfrac%7B1%7D%7BR_2%7D%20%20%5D)
The ratio of the focal length of the the two medium is mathematically evaluated as
![\frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }](https://tex.z-dn.net/?f=%5Cfrac%7Bf_water%7D%7Bf_%7Bair%7D%7D%20%3D%20%5Cfrac%7Bn_2%20-1%7D%7B%5B%5Cfrac%7Bn_2%7D%7Bn_1%7D%20-%201%5D%20%7D)
From the question
= 79.0 cm

and the refractive index of water(material surrounding the lens) has a constant value of 


b
The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.
Answer : The power absorbed by the bulb is, 0.600 W
Explanation :
As we know that,
Power = Voltage × Current
Given:
Voltage = 3 V
Current = 200 mA = 0.200 A
Conversion used : (1 mA = 0.001 A)
Now put all the given values in the above formula, we get:
Power = Voltage × Current
Power = 3V × 0.200 A
Power = 0.600 W
Thus, the power absorbed by the bulb is, 0.600 W
Answer:
A
Explanation:
Initial gravitational energy = final kinetic energy + heat
mgh = KE + Q
(50 kg) (9.81 m/s²) h = 78400 J + 884000 J
h = 1960 meters
Answer:-81.29 J
Explanation:
Given
mass of ball 
Length of string 
height of Room 
Gravitational Potential Energy is given by

where h=distance between datum and object
here Reference is ceiling
therefore 
Potential Energy of ball w.r.t ceiling

i.e. 81.29 J of Energy is required to lift a ball of mass 2.35 kg to the ceiling