Two opposite forces $F_{1} = 120 N a n d F_{2} = 80 N$ act on an heavy elastic plank of modulus of elasticity $y = 2 \times 10^{11} N / m^{2}$ and length $L = 1 m$ placed over a smooth horizontal surface. The cross-sectional area of plank is $A = 0.5 m^{2}$ . If the change in the length of plank is (in nm )

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Solution

The correct option is A 1
We use the formula
$Δ L = \frac{F L}{A y}$
The average force acting on the plank is $\frac{120 + 80}{2} = 100 N$ , Thus
$Δ L = \frac{100 \times 1}{0.5 \times 2 \times 10^{1} 1}$
or
$Δ L = 1 \times 10^{- 9} = 1 n m$

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Q. Two opposite forces

F_{1} = 120 N

and

F_{2} = 80 N

act on an elastic plank of modulus of elasticity

Y = 2 \times 10^{11} N m^{2}

and length

l = 1 m

placed over a smooth horizontal surface. The cross-sectional area of the planck is

S = 0.5 m^{2}

, the change in length of the plank is

x \times 10^{- 9} m

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Two opposite forces F1=120NandF2=80N act on an heavy elastic plank of modulus of elasticity y=2×1011N/m2 and length L=1m placed over a smooth horizontal surface. The cross-sectional area of plank is A=0.5m2. If the change in the length of plank is (in nm )

The correct option is A 1We use the formulaΔL=FLAyThe average force acting on the plank is 120+802=100N, ThusΔL=100×10.5×2×1011orΔL=1×10−9=1nm

The correct option is A 1
We use the formula
$Δ L = \frac{F L}{A y}$
The average force acting on the plank is $\frac{120 + 80}{2} = 100 N$ , Thus
$Δ L = \frac{100 \times 1}{0.5 \times 2 \times 10^{1} 1}$
or
$Δ L = 1 \times 10^{- 9} = 1 n m$