A steel ring of radius 10 cm and cross-section area
1 cm2 is fitted to a wooden disc of radius 10.5 cm
If Young's modulus of steel is 2 x 10^11 Nm-2, then
the force with which the steel ring is expanded is
(in N)
(2) 10^5
(1) 10^4
(3) 10^6
(4) 10^7
Answers
Answered by
10
Answer:
The answer will be 10 ^8 N
Explanation:
According to the problem the radius of the ring , r = 10 cm
The cross sectional area , A = 1 cm^2
The radius of the disc ,R = 10.5 cm
As mentioned the Young's modulus of steel is 2 x 10^11 Nm^(-2)
Let the force be F
As we know,
F = AE [(R – r) / r] [ where E is the Young's modulus]
= 0.01 x 2 x 10^11 x [ (0.105 -0.1)/0.1]
= 10 ^8
Answered by
5
Answer:
See the attachment
Good luck
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