Edge of the aluminium cube, L= 10 cm = 0.1 m
The mass attached to the cube, m = 100 kg
Shear modulus (η) of aluminium = 25 GPa = 25 × 10 9 Pa
Shear modulus, η =\( \frac{\text{Shear stress}}{\text{ Shear strain }}= \frac{\frac{F}{A}}{\frac{L}{ ΔL}}\)
Where, F = Applied force = mg = 100 × 9.8 = 980 N
A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m 2
ΔL = Vertical deflection of the cube
\(∴ ΔL = \frac{FL }{ Aη} \)
=\( \frac{980 × 0.1 }{ 10 ^{- 2} × (25 × 10 ^9) }\)
= 3.92 × 10 - 7 m
The vertical deflection of this face of the cube is 3.92 ×10 - 7 m.
The elastic behavior of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of 5×10–4 is ____ kJ/m3. Assume that material is elastic up to the linear strain of 5×10–4
Give reasons for the following.
(i) King Tut’s body has been subjected to repeated scrutiny.
(ii) Howard Carter’s investigation was resented.
(iii) Carter had to chisel away the solidified resins to raise the king’s remains.
(iv) Tut’s body was buried along with gilded treasures.
(v) The boy king changed his name from Tutankhaten to Tutankhamun.
Find the mean deviation about the median for the data
xi | 15 | 21 | 27 | 30 | 35 |
fi | 3 | 5 | 6 | 7 | 8 |