A) \[\sqrt{3}:1\]
B) 7 m/s
C) \[1:\sqrt{3}\]
D) \[7.9\times {{10}^{4}}K\]
Correct Answer: A
Solution :
Minimum tension \[\frac{3}{2}KT=10.2\times 1.6\times {{10}^{-19}}J\] Maximum tension \[T=\frac{2}{3}\times \frac{10.2\times 1.6\times {{10}^{-19}}}{K}\] Let \[=\frac{2}{3}\times \frac{10.2\times 1.6\times {{10}^{-19}}}{1.38\times {{10}^{-23}}}\] So, \[=7.9\times {{10}^{4}}K\] ...(1) \[\upsilon =400m/s,r=160m,a=?\] ...(2) Dividing equation (1) by (2) \[F=\frac{m{{\upsilon }^{2}}}{r}\] \[ma=\frac{m{{\upsilon }^{2}}}{r}\] \[a=\frac{{{\upsilon }^{2}}}{r}\] Or \[a=\frac{{{(400)}^{2}}}{160}=\frac{16\times {{10}^{4}}}{160}\] i.e., \[={{10}^{3}}m/{{s}^{2}}=1km/{{s}^{2}}\] \[{{T}_{1}}=\frac{m{{\upsilon }^{2}}}{r}-mg\] \[{{T}_{2}}=\frac{m{{\upsilon }^{2}}}{r}+mg\] Or \[\frac{m{{\upsilon }^{2}}}{r}=x\] Or \[{{T}_{1}}=x-mg\] \[{{T}_{2}}=x+mg\]
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