Question

A rubber ball of mass 10 gm and volume $15c{m}^3$ is dipped in water to a depth of 10 m. Assuming that the density of water is uniform throughout the depth, the acceleration of the ball and time taken by it to reach to the surface, if it is released from rest are

• A. $200cm/s^2$ and $10\sqrt{2}/7s$
• B. $400cm/s^2$ and $\left(8\sqrt{2}/7\right)s$
• C. $490cm/s^2$and $\left(10\sqrt{2}/7\right)s$
• D. $550cm/s^2$ and $\left(8\sqrt{2}/7\right)s$

Answer 1

The correct option is C $490cm/s^2$and $\left(10\sqrt{2}/7\right)s$


$F_{net}=F_{up}–mg$
$=Vdg–mg$
$=15\times 1\times g–10g=5g$
$ma=5g$, $\because g=980cm/s^2$
$10a=5\times 980$
$a=490cm/s^2$
depth of water, s = 10m = 1000cm
s = ut + $\frac{1}{2}a{t}^2$
as ball is realeased from rest u=0
s = $\frac{1}{2}a{t}^2$
1000 = $\frac{1}{2}\times 490\times t^2$
$t^2=\frac{200}{49}$
t = $\sqrt{\frac{200}{49}}$
t = $\left(10\sqrt{2}/7\right)s$