Question:
A stationary object at $4^\circ C$ and weighing 3.5 kg falls from a height of 2000 m on snow mountain at $0^\circ C$. If the temperature of the object just before hitting the snow is $0^\circ C$ and the object comes to rest immediately then the quantity of ice that melts is (Acceleration due to gravity = $10 m/s^2$, Latent heat of ice = $3.5 \times 10^5 J/kg$)
A stationary object at $4^\circ C$ and weighing 3.5 kg falls from a height of 2000 m on snow mountain at $0^\circ C$. If the temperature of the object just before hitting the snow is $0^\circ C$ and the object comes to rest immediately then the quantity of ice that melts is (Acceleration due to gravity = $10 m/s^2$, Latent heat of ice = $3.5 \times 10^5 J/kg$)
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Ensure units are consistent ($J$, $kg$, $m$) before solving heat-work problems.
Updated On: Apr 30, 2026
- 2 gram
- 20 gram
- 200 gram
- 2 kg
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The Correct Option is C
Solution and Explanation
Step 1: Energy Conservation
Mechanical energy (Potential Energy) is converted into heat to melt ice: $mgh = m_{ice}L$.
Step 2: Potential Energy
$P.E. = 3.5 \times 10 \times 2000 = 70,000 \text{ J}$
Step 3: Melting Calculation
$70,000 = m_{ice} \times (3.5 \times 10^5)$
$m_{ice} = \frac{70,000}{350,000} = \frac{7}{35} = 0.2 \text{ kg}$
Step 4: Unit Conversion
$0.2 \text{ kg} = 200 \text{ grams}$
Final Answer:(C)
Mechanical energy (Potential Energy) is converted into heat to melt ice: $mgh = m_{ice}L$.
Step 2: Potential Energy
$P.E. = 3.5 \times 10 \times 2000 = 70,000 \text{ J}$
Step 3: Melting Calculation
$70,000 = m_{ice} \times (3.5 \times 10^5)$
$m_{ice} = \frac{70,000}{350,000} = \frac{7}{35} = 0.2 \text{ kg}$
Step 4: Unit Conversion
$0.2 \text{ kg} = 200 \text{ grams}$
Final Answer:(C)
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