NEET-XII-Physics
14: Some Mechanical Properties of Matter
- #1A load of 10 kg is suspended by a metal wire 3 m long and having a cross-sectional area 4 mm2. Find (a) the stress (b) the strain and (c) the elongation. Young modulus of the metal is 2.0 × 1011 N m-2.Ans : Given:
Mass of the load (m) = 10 kg
Length of wire (L) = 3 m
Area of cross-section of the wire (A) = 4 mm2 = 4.0 × 10-6 m2
Young's modulus of the metal Y = 2.0 × 1011 N m-2 (a) Stress = F/A
F = mg
= `` 10\times 10`` = 100 N (g = 10 m/s2)
`` \therefore \frac{F}{A}=\frac{100}{4\times {10}^{-6}}``
`` =2.5\times {10}^{7}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}`` (b) Strain = `` \frac{\Delta L}{L}``
​
Or, `` \,\mathrm{\,Strain\,}=\frac{\,\mathrm{\,Stress\,}}{Y}``
`` \,\mathrm{\,Strain\,}=\frac{2.5\times {10}^{7}}{2\times {10}^{11}}``
`` =1.25\times {10}^{-4}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}`` (c) Let the elongation in the wire be `` ∆L``.
`` \,\mathrm{\,Strain\,}=\frac{\Delta L}{L}``
`` \Rightarrow \,\mathrm{\,\Delta \,}L=\left(\,\mathrm{\,Strain\,}\right)\times L``
`` =1.25\times {10}^{-4}\times 3``
`` =3.75\times {10}^{-4}\,\mathrm{\,m\,}``
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- #1-athe stressAns : Stress = F/A
F = mg
= `` 10\times 10`` = 100 N (g = 10 m/s2)
`` \therefore \frac{F}{A}=\frac{100}{4\times {10}^{-6}}``
`` =2.5\times {10}^{7}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
- #1-bthe strain andAns : Strain = `` \frac{\Delta L}{L}``
​
Or, `` \,\mathrm{\,Strain\,}=\frac{\,\mathrm{\,Stress\,}}{Y}``
`` \,\mathrm{\,Strain\,}=\frac{2.5\times {10}^{7}}{2\times {10}^{11}}``
`` =1.25\times {10}^{-4}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
- #1-cthe elongation. Young modulus of the metal is 2.0 × 1011 N m-2.Ans : Let the elongation in the wire be `` ∆L``.
`` \,\mathrm{\,Strain\,}=\frac{\Delta L}{L}``
`` \Rightarrow \,\mathrm{\,\Delta \,}L=\left(\,\mathrm{\,Strain\,}\right)\times L``
`` =1.25\times {10}^{-4}\times 3``
`` =3.75\times {10}^{-4}\,\mathrm{\,m\,}``
Page No 300: