A lead bullet strikes at target with a velocity of 480 m $s^{- 1}$ . If the bullet falls dead, the rise in temperature of bullet (c = 0.03cal/gram), assuming that heat produced is equally shared between the bullet and target is

557°C

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457°C

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857°C

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754°C

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Solution

The correct option is B 457°C
mc $△ T$ = $\frac{\frac{1}{2} (\frac{1}{2} m v^{2})}{J}$
$△ T$ = $\frac{v^{2}}{4 J c}$ = $\frac{480 \times 480}{4 \times 4.2 \times (0.003 \times 10^{8})}$ = $457^{\circ}$ C

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A lead bullet strikes at target with a velocity of 480 ms−1. If the bullet falls dead, the rise in temperature of bullet (c = 0.03cal/gram), assuming that heat produced is equally shared between the bullet and target is

The correct option is B 457°C mc△T = 12(12mv2)J △T = v24Jc = 480×4804×4.2×(0.003×108) = 457∘C

A lead bullet strikes at target with a velocity of 480 m $s^{- 1}$ . If the bullet falls dead, the rise in temperature of bullet (c = 0.03cal/gram), assuming that heat produced is equally shared between the bullet and target is

The correct option is B 457°C
mc $△ T$ = $\frac{\frac{1}{2} (\frac{1}{2} m v^{2})}{J}$
$△ T$ = $\frac{v^{2}}{4 J c}$ = $\frac{480 \times 480}{4 \times 4.2 \times (0.003 \times 10^{8})}$ = $457^{\circ}$ C