A lead bullet just melts when stopped by an obstacle. Assuming that 25 cal

of heat is absorbed by the obstacle, the velocity of the bullet if its initial

temperature is $27°\text{C}$ is. Given melting point of lead$=327°\text{C}$  specific heat

of lead$=0.03\text{cal/g/°C}$ . Latent heat of fusion of lead $=6\text{cal/g}$

given that

bullet penetrated in to an obstacle with speed = v

K.E of bullet $=\frac{1}{2}{\text{mv}}^{\text{2}}$

initial temperature of bullet$=27°\text{C}$

if $\text{3/4}$ of K.E is converted into heat

so heat$\text{Q}=\frac{3}{4}\left(\frac{1}{2}{\text{mv}}^{\text{2}}\right)\text{J}$

Q is utilized to melt the bullet and 25 lost for obstacle material

${\text{t}}_{\text{1}}=27°\text{C}$

melting point to lead$=327°\text{C}$

${\text{L}}_{\text{lead}}=6\text{cal}$

let mass of bullet = m

specific heat of lead $=0.03\text{cal/g°c}$

heat utilized $\text{Q}=\left(\text{m}\times 0.03\left(327-27\right)+\text{m}\times 6\right)4200\text{J}$

$\frac{3}{4}\times \frac{1}{2}{\text{mv}}^{\text{2}}=\text{m}\left(0.03\times 300+6\right)4200\text{J}$

$\frac{{\text{3v}}^{\text{2}}}{\text{8}}=15\times 4.2\times {10}^3$

${\text{v}}^{\text{2}}=\frac{15\times 8\times 4.2\times {10}^3}{3}\Rightarrow v^2=\sqrt{168000}$

$\therefore \text{v}=409.87=410\text{m/s}$