Blocks of masses  $m$, $2m$ , $4m$ and $8m$ are arranged in a line on a frictionless floor. Another block of mass $m$, moving with speed $v$  along the same line (see figure) collides with mass m in perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass  $8m$ starts moving the total energy loss is  $p\%$ of the original energy. Value of '$p$'  is close to:

According to the question, all collisions are perfectly inelastic, so after the final collision, all blocks are moving together.
 

Let the final velocity be  $v^{\text{'}}$, using momentum conservation  $mv=16m{v}^{\text{'}}\Rightarrow v^{\text{'}}=\frac{v}{16}$
Now initial energy  $E_f=\frac{1}{2}m{v}^2$
Final energy :  $E_f=\frac{1}{2}\times 16m\times {\left(\frac{v}{16}\right)}^2=\frac{1}{2}\frac{m{v}^2}{16}$
 $\Rightarrow \frac{1}{2}m{v}^2[1-\frac{1}{16}]\Rightarrow \frac{1}{2}m{v}^2\left[\frac{15}{16}\right]$
The total energy loss is $P\%$  of the original energy.
 $\therefore \%P=\frac{Energyloss}{Originalenergy}\times 100$
 $=\frac{\frac{1}{2}m{v}^2\left[\frac{15}{16}\right]}{\frac{1}{2}m{v}^2}\times 100=93.75\%$
Hence, value of P is close to  $94$.