A lead bullet at 27,^{circ}C just melts when | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the mass of the bullet be $m\,g$.The heat required for the bullet to reach its melting point and melt is given by $Q_1 = mc\Delta	heta + mL$.

Vedclass · Accepted Answer

$410$

Vedclass · Answer

(A) Let the mass of the bullet be $m\,g$.The heat required for the bullet to reach its melting point and melt is given by $Q_1 = mc\Delta	heta + mL$.Substituting the values: $Q_1 = m 	imes 0.03 	imes (327 - 27) + m 	imes 6 = m 	imes 0.03 	imes 300 + 6m = 9m + 6m = 15m\,cal$.Converting this to Joules: $Q_1 = 15m 	imes 4.2 = 63m\,J$.The kinetic energy of the bullet is $K.E. = \frac{1}{2}mv^2$,where $m$ is in $kg$. Since $m$ is in $g$,$m_{kg} = m 	imes 10^{-3}$.So,$K.E. = \frac{1}{2} 	imes (m 	imes 10^{-3}) 	imes v^2 = 0.5 	imes 10^{-3} mv^2\,J$.Since $25\%$ of the heat is absorbed by the obstacle,$75\%$ of the kinetic energy is absorbed by the bullet to melt it.Therefore,$0.75 	imes (0.5 	imes 10^{-3} mv^2) = 63m$.$0.375 	imes 10^{-3} v^2 = 63$.$v^2 = \frac{63}{0.375 	imes 10^{-3}} = \frac{63}{0.000375} = 168000$.$v = \sqrt{168000} \approx 409.88\,m/s \approx 410\,m/s$.

Similar Questions

The abscissae of $A$ and $B$ are the roots of the equation $x^2+2ax-b^2=0$ and their ordinates are the roots of the equation $y^2+2py-q^2=0$. Find the equation of the circle with $AB$ as diameter.

If $\vec F$ is the force acting on a particle having position vector $\vec r$ and $\vec \tau$ is the torque of this force about the origin,then:

Which one of the following equations represents the Freundlich adsorption isotherm?

If the tangent to the curve $y = \frac{x}{x^2 - 3}$,$x \in R$,$(x \neq \pm \sqrt{3})$ at a point $(\alpha, \beta) \neq (0, 0)$ on it is parallel to the line $2x + 6y - 11 = 0$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module