A lead bullet flying with a velocity of 100 ms- 1 pierces a fixed board after which its velocity becomes 60 ms- 1. If 40 % of energy lost to increases the temperature of bullet, specific heat capacity of lead =125 J kg- 1K- 1. Find change in its temperature (in ° C ). (Approximate the answer to nearest integer)

Question

A lead bullet flying with a velocity of 100 ms- 1 pierces a fixed board after which its velocity becomes 60 ms- 1. If 40 % of energy lost to increases the temperature of bullet, specific heat capacity of lead =125 J kg- 1K- 1. Find change in its temperature (in ° C ). (Approximate the answer to nearest integer) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

vb=200 ms- 1 Δ K text loss =-(1/2) m (vf2-vi2)=-(1/2) m (602-1002) msΔ T=0.4× (1/2)m(.160× 40.) Δ T=(0 . 2/125)× 160× 40 Δ T=10.2° C

vb​=200ms−1 ΔKloss ​=−21​m(vf2​−vi2​)=−21​m(602−1002) msΔT=0.4×21​m(160×40) ΔT=1250.2​×160×40 ΔT=10.2∘C

$v_{b} = 200 m s^{- 1}$
$Δ K_{loss} = - \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) = - \frac{1}{2} m (6 0^{2} - 10 0^{2})$
$m s Δ T = 0.4 \times \frac{1}{2} m (160 \times 40)$
$Δ T = \frac{0.2}{125} \times 160 \times 40$
$Δ T = 10. 2^{\circ} C$