JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Heat & Thermodynamics

A bullet of mass 5 g , traveling with a speed of

210 m / s

, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is

0.030 cal / (g -^{0} C) (1 cal = 4.2 \times 1 0^{7} ergs)

close to:

(A) $83. 3^{\circ} C$
(B) $38. 4^{\circ} C$
(C) $87. 5^{\circ} C$ ✓ Correct
(D) $119. 2^{\circ} C$

Answer: (C)

Solution

The bullet carries kinetic energy

\frac{1}{2} m v^{2}

, of which half stays in the bullet as heat, so

Q = \frac{1}{4} m v^{2}

. Setting this equal to

m s Δ T

, the mass cancels and

Δ T = \frac{v ^{2}}{4 s}

. With

v = 210

m/s

= 2.1 \times 1 0^{4}

cm/s and

s = 0.030 \times 4.2 \times 1 0^{7}

erg/(g

\cdot^{\circ}

C),

Δ T = \frac{( 2.1 \times 1 0 ^{4} ) ^{2}}{4 \times 0.030 \times 4.2 \times 1 0 ^{7}} \approx 87.5^{\circ} C

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Heat & Thermodynamics

Solution

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