question_answer 19)

A mild steel wire of length 1.0 m and cross-sectional area \[{{h}_{1}},\] is stretched, well within its elastic 'limit, horizontally between two pillars. A mass of 100 g is suspended from' the mid point of the wire. Calculate the depression at the mid point. \[B,{{P}_{B}}=58+{{h}_{1}}\] \[{{P}_{A}}={{P}_{B'}}\]

Answer:

Refer Fig. 6, let x be the depression at the mid point i.e. \[CD=x\] In Fig.\[\therefore \] \[77=58+{{h}_{1}}\] \[{{h}_{1}}=77-58=19\] Increase in length,\[1\cdot 06\times {{10}^{3}}kg/{{m}^{3}}.\] \[h=\frac{p}{\rho g}=\frac{2000}{1\cdot 06\times {{10}^{3}}\times 9\cdot 8}=0\cdot 1925m\] \[0\cdot 1925mi.e.0\cdot 2m.\] \[2\times {{10}^{-3}}\] Strain \[r=2\times {{10}^{-3}}m;\] If r is the tension in the wire, then \[D=2r=2\times 2\times {{10}^{-3}}\]or \[=4\times {{10}^{-3}}m;\] Here, \[\eta =2\cdot 084\times {{10}^{-3}}pa-s;\] \[1\cdot 06\times {{10}^{-3}}kg{{m}^{-3}}\] and \[{{N}_{R}}=2000\]\[{{\text{ }\!\!\upsilon\!\!\text{ }}_{\text{c}}}\text{=}\frac{{{\text{N}}_{\text{R}}}\text{ }\!\!\eta\!\!\text{ }}{\text{ }\!\!\rho\!\!\text{ D}}=\frac{2000\times \left( 2\cdot 084\times {{10}^{-3}} \right)}{\left( 1\cdot 06\times {{10}^{3}} \right)\times \left( 4\times {{10}^{-3}} \right)}=0\cdot 98m/s.\] Hence, \[\pi {{r}^{2}}{{\upsilon }_{c}}=\frac{22}{7}\times {{\left( 2\times {{10}^{-3}} \right)}^{2}}\times 0\cdot 98=1\cdot 23\times \text{1}{{\text{0}}^{\text{-5}}}{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{-1}}}\] Stress \[25{{m}^{2}}.\] \[{{m}^{3}}\] \[9\cdot 8m/{{s}^{2}}.\] \[{{\text{ }\!\!\upsilon\!\!\text{ }}_{\text{1}}}\text{=180 km/h = 50 m/s; }\]