JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
At \(40\,^oC\), a brass wire of \(1\, mm\) is hung from the ceiling. A small mass, \(M\) is hung from the free end of the wire. When the wire is cooled down from \(40\,^oC\) to \(20\,^oC\) it regains its original length of \(0.2\, m\). The value of \(M\) is close to ........\(kg\) (Coefficient of linear expansion and Young's modulus of brass are \(10^{-5}/^oC\) and \(10^{11}\, N/m^2\), respectively; \(g = 10\, ms^{-2}\))
- A \(0.5\)
- B \(9\)
- C \(0.9\)
- D \(1.5\)
Answer & Solution
Correct Answer
(B) \(9\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l} Mg = \left( {\frac{{Ay}}{l}} \right)\,\Delta l\\ Mg = \left( {Ay} \right)\,\alpha \Delta T = 2\pi \end{array}\) It is closest to \(9\)
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