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10-1.Thermometry, Thermal Expansion and Calorimetry
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A lead bullet of $10\, g$ travelling at $300\, m/s$ strikes against a block of wood and comes to rest. Assuming $50\%$ of heat is absorbed by the bullet, the increase in its temperature is ........ $^oC$
$($Specific heat of lead $= 150\,J/kg, K)$
A$100$
B$125$
C$150$
D$200$
Solution
(c) Since specific heat of lead is given in Joules, hence use $W = Q$ instead of $W = JQ$.
$\Rightarrow$ $\frac{1}{2} \times \left( {\frac{1}{2}m{v^2}} \right) = m.c.\Delta \theta $==>$\Delta \theta = \frac{{{v^2}}}{{4c}} = \frac{{{{(300)}^2}}}{{4 \times 150}} = 150^\circ C$.
$\Rightarrow$ $\frac{1}{2} \times \left( {\frac{1}{2}m{v^2}} \right) = m.c.\Delta \theta $==>$\Delta \theta = \frac{{{v^2}}}{{4c}} = \frac{{{{(300)}^2}}}{{4 \times 150}} = 150^\circ C$.
Standard 11
Physics
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