NEET-XII-Physics
14: Some Mechanical Properties of Matter
- #7Two persons pull a rope towards themselves. Each person exerts a force of 100 N on the rope. Find the Young modulus of the material of the rope if it extends in length by 1 cm. Original length of the rope = 2 m and the area of cross section = 2 cm2.Ans : Given:
Force (F) applied by two persons on the rope = 100 N
`` \,\mathrm{\,Original\,}\,\mathrm{\,length\,}\,\mathrm{\,of\,}\,\mathrm{\,rope\,}L=2\,\mathrm{\,m\,}``
`` \,\mathrm{\,Extension\,}\,\mathrm{\,in\,}\,\mathrm{\,the\,}\,\mathrm{\,rope\,}∆L=0.01\,\mathrm{\,m\,}``
`` \,\mathrm{\,Area\,}\,\mathrm{\,of\,}\,\mathrm{\,cross\,}-section\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,rope\,}A=2\times {10}^{-4}``
`` \,\mathrm{\,We\,}\,\mathrm{\,know\,}\,\mathrm{\,that\,}:``
`` \,\mathrm{\,Young\,}\text{'}\,\mathrm{\,s\,}\,\mathrm{\,modulus\,}Y=\frac{F}{A}\times \frac{L}{∆L}``
`` =\frac{100}{2\times {10}^{-4}}\times \frac{2}{0.01}``
`` \Rightarrow Y=1\times {10}^{8}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
Hence, the required Young's modulus for the rope is `` 1\times {10}^{8}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``.
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