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10-1.Thermometry, Thermal Expansion and Calorimetry
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A lead ball moving with a velocity $V$ strikes a wall and stops. If $50\%$ of its energy is converted into heat, then what will be the increase in temperature (Specific heat of lead is $S$)
A
$\frac{{2{V^2}}}{{JS}}$
B
$\frac{{{V^2}}}{{4JS}}$
C
$\frac{{{V^2}}}{J}$
D
$\frac{{{V^2}S}}{{2J}}$
Solution
(b) $W = JQ$ $\Rightarrow$ $\frac{1}{2}\left( {\frac{1}{2}m{V^2}} \right) = J \times mS\Delta \theta $ $\Rightarrow$ $\Delta \theta = \frac{{{V^2}}}{{4\,JS}}$
Standard 11
Physics
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