The ultimate tensile strength of structural mild steel is about (N/sqmm):
a) 160
b) 260
c) 520
d) 420.
Answers
Answer:
ok
Explanation:
Let F be the foot and S be the summit of the
mountain FOS.
Then, ∠OFS = 45º and therefore ∠OSF = 45º.
Consequently,
OF = OS = h km(say).
Let FP = 1000 m = 1 km be the slope so that
∠OFP = 30º. Draw PM ⊥OS and PL ⊥OF.
Join PS. It is given that ∠MPS = 60º.
In ∆FPL, We have
`\text{sin }30^\text{o}=\frac{PL}{PF}`
`\Rightarrow PL=PF\text{ sin }30^\text{o}=( 1\times
\frac{1}{2})=\frac{1}{2}km.`
`\therefore OM=PL=\frac{1}{2}km`
`\Rightarrow MS=OS-OM=( h-\frac{1}{2})km`
`\text{Also, cos }30^\text{o}=\frac{FL}{PF}`
`\Rightarrow FL=PF\text{ cos }30^{o}=( 1\times\frac
{\sqrt{3}}{2})=\frac{\sqrt{3}}{2}km`
Now, h = OS = OF = OL + LF
`\Rightarrow h=OL+\frac{\sqrt{3}}{2}`
`\Rightarrow OL=( h-\frac{\sqrt{3}}{2})km`
`\Rightarrow PM=( h-\frac{\sqrt{3}}{2})km`
In ∆PSM, we have
`\text{tan }60^\text{o}=\frac{SM}{PM}`
⇒ SM = PM. tan 60º …..(ii)
`\Rightarrow ( h-\frac{1}{2})=( h-\frac{\sqrt{3}}
{2})\sqrt{3}`
[Using equations (i) and (ii)]
`\Rightarrow h-\frac{1}{2}=h\sqrt{3}-\frac{3}{2}`
⇒ h(√3 – 1) = 1
`\Rightarrow h=\frac{1}{\sqrt{3}-1}`
`\Rightarrowh=\frac{\sqrt{3}+1}{(\sqrt{3}-1)(\sqrt
{3}+1)}=\frac{\sqrt{3}+1}{2}`
`=\frac{2.732}{2}=1.336km`
Hence, the height of the mountain is 1.366 km.