Answer
(a) $1.027N$
(b) same
(c) $3.96m/s^2$
Work Step by Step
(a) We can find the required tension as follows:
$F_b=V\rho_wg$
$\implies F_b=(0.82\times 10^{-5}m^3)(1000Kkg/m^3)(9.8m/s^2)$
$F_b=0.08N$
The mass of lead is given as
$m=V\rho_{total}$
$m=(0.82\times 10^{-5}m^3)(11340Kg/m^3)$
$m=0.093Kg$
Now $T=m(g+a)-F_b$
We plug in the known values to obtain:
$T=(0.093Kg)(9.8m/s^2+2.1m/s^2)-0.08N$
$T=1.027N$
(b) We know that the tension will stay the same because the buoyant force does not change as long as the mas is completely inside the fluid.
(c) We know that
$a=\frac{T+F_b-w}{m}$
We plug in the known values to obtain:
$a=\frac{1.2N+0.08N-0.912N}{0.093Kg}$
$\implies a=3.96m/s^2$
