JEE Main 2026 — Jan 23, Shift 1 — Physics solution

JEE Main2026Jan 23, Shift 1Physics

In a perfectly inelastic collision, two spheres made of the same material with masses $15 kg$ and $25 kg$ , moving in opposite directions with speeds of $10 m/s$ and $30 m/s$ , respectively, strike each other and stick together. The rise in temperature (in $^{\circ} C$ ), if all the heat produced during the collision is retained by these spheres, is : (specific heat of sphere material $31 cal/kg \cdot^{\circ} C$ and $1 cal = 4.2 J$ )

A
$1.75$
B
$1.44$
C
$1.15$
D
$1.95$

Solution

Step 1 — KE lost in a perfectly inelastic collision with relative speed $v_{rel} = 10 + 30 = 40 m/s$ (opposite directions):

$Δ K E = \frac{1}{2} μ v_{rel}^{2}, μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} = \frac{15 \times 25}{40} = \frac{375}{40} = 9.375 kg$

$Δ K E = \frac{1}{2} (9.375) (40)^{2} = 7500 J$

Step 2 — Equate to heat absorbed. Specific heat $S = 31 cal/kg \cdot^{\circ} C = 31 \times 4.2 J/kg \cdot^{\circ} C$ , total mass $40 kg$ :

$Δ T = \frac{7500}{40 \times 31 \times 4.2} = \frac{7500}{5208} \approx 1.44^{\circ} C$

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