Using lens equation;
1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)
Substituting;
1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm
Therefore, the object should be place 99.23 cm from the lens.
Answer:
11300 kgm3
Hope this helps
Object true weight is given as

now we know that g = 9.8 m/s^2


now when it is complete submerged in water its apparent weight is given as 82 N
apparent weight = weight - buoyancy force
apparent weight = 82 N
weight = 123 N
now we have
82 = 123 - buoyancy force
buoyancy force = 123 - 82 = 41 N
now we also know that buoyancy force is given as




now as we know that mass of the object is 12.55 kg
its volume is 4.18 * 10^-3 m^3
now we know that density will be given as mass per unit volume



so here density of object is 3002.4 kg/m^3
Q = m.Lf.
Here, m = 18.6 g
Lf = 333 J/g
Substitute their values,
Q = 18.6 × 333 = 6193.8 J
We know, 1 KJ = 1000 J
So, 6193.8 J = 6.193 KJ
In short, Your Answer would be Option D
Hope this helps!