Verified Solution • 2025–26 Session moderate

QAIPM-2012-Ph-1767: NEET 2012 Physics - Class 11 AIPM-2012-Ph-1767 Solution: When a string is divided into three segments of length \(l_1…

When a string is divided into three segments of length l1l2l_1 l_2 and l3l_3 the fundamental frequencies of these three segments are ν1ν2\nu_1 \nu_2 and ν3\nu_3 respectively. The original fundamental frequency (ν\nu) of the string is ν=ν1+ν2+ν3\sqrt{\nu} = \sqrt{\nu_1} + \sqrt{\nu_2} + \sqrt{\nu_3} ν=ν1+ν2+ν3\nu = \nu_1 + \nu_2 + \nu_3 1ν=1ν1+1ν2+1ν3\frac{1}{\nu} = \frac{1}{\nu_1} + \frac{1}{\nu_2} + \frac{1}{\nu_3} 1ν=1ν1+1ν2+1ν3\frac{1}{\sqrt{\nu}} = \frac{1}{\sqrt{\nu_1}} + \frac{1}{\sqrt{\nu_2}} + \frac{1}{\sqrt{\nu_3}}

Multiple Choice Options

  • A 2PV
  • B 4PV
  • C 12{1 \over 2} 2 1 _ PV
  • D PV

Step-by-Step Verified Solution

In a cyclic process, _ U = 0 In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise. _ _ W = � Area of rectangle ABCD = � P(2V) = � 2PV According to first law of thermodynamics _ Q = _ U + _ W _ _ Q = _ W     (As _ U = 0 ) i.e., heat supplied to the system is equal to the work done So heat absorbed, _ Q = _ W = � 2PV _ Heat rejected by the gas = 2PV

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