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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry

જ્યારે \(-10\,^oC\) એ રહેલ \(M_1\) ગ્રામ બરફને (વિશિષ્ટ ઉષ્મા \(= 0.5\, cal\, g^{-1}\,^oC^{-1}\)) \(50\,^oC\) એ રહેલ \(M_2\) ગ્રામ પાણીમાં મિશ્ર કરતાં અંતે બરફ રહેતો નથી અને પાણીનું તાપમાન \(0\,^oC\) થાય છે. બરફની ગલનગુપ્ત ઉષ્મા \(cal\, g^{-1}\) માં કેટલી થાય?

A \(\frac{{50{M_2}}}{{{M_1}}} - 5\)
B \(\frac{{5{M_2}}}{{{M_1}}} - 5\)
C \(\frac{{50{M_2}}}{{{M_1}}}\)
D \(\frac{{5{M_1}}}{{{M_2}}} - 50\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{{50{M_2}}}{{{M_1}}} - 5\)

Step-by-step Solution

Detailed explanation

\(Heat\,lost=Heat\,gain\) \( \Rightarrow \,{M_2} \times 1 \times 50 = {M_1} \times 0.5 \times 10 + {M_1}.{L_f}\) \( \Rightarrow \,\,{L_f} = \frac{{500{M_2} - 5{M_1}}}{{{M_1}}}\) \( = \frac{{500{M_2}}}{{{M_1}}} - 5\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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