Ncert Exercises & Solutions

9.7 Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

We know:

Young’s modulus of steel, Y = 2 × 10 $^{11}$ Pa

Total force exerted, F = Mg = 50000 × 9.8 N

Force exerted on each column = $\frac{50000 \times 9.8}{4}$ = 122500 N

Young’s modulus, Y = $\frac{Stress}{Strain}$ $\Rightarrow Strain = \frac{Stress}{Y}$

Area of each column, A = π (R $^{2}$ – r $^{2}$ ) = π ((0.6) $^{2}$ – (0.3) $^{2}$ )

$Stain in each column = \frac{122500}{π [{(0.6)}^{2} - {(0.3)}^{2} \times 2 \times 10^{11}]} =$ 7.22 × 10 $^{- 7}$

Hence, the compressional strain of each column is 7.22 × 10 $^{- 7}$ .

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