A block of ice falls from a height and completely melts. If only$\frac{3}{4}$ of the energy is retained by the block, the height of the fall in km should be$\left(g=10{\text{ms}}^{-2}\right)$

Given that A block of Ice having mass = m

Falls from height = h

Initial P.E of Ice block = mgh

During the fall total ice melted so heat utilized Q = mL

This heat energy is availed by conversion of P.E into heat energy while falling against air friction.

It is said to be retained ${\text{3/4}}^{\text{th}}$ of P.E, the amount of P.E converted into heat is

$\frac{3}{4}\text{of}\text{P}\text{.E}=\frac{\text{3mgh}}{\text{4}}$

We get $\frac{\text{3mgh}}{\text{4}}=\text{mL}$

We know $\text{g}=10{\text{m/s}}^{\text{2}}\text{,}\text{h}=?,\text{L}=\text{80}\text{cal/g}$

$\text{L}=80000\times 4.2\text{J/kg}$

$\frac{3\times 10\times \text{h}}{4}=80,000\times 4.2$

$\text{h}=\frac{80,000\times 4.2\times 4}{10\times 3}$

$\text{h}=44,800\text{m}$

$\therefore \text{h}=\text{44}\text{.8}\text{km}$