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A steel rod is projecting out of a rigid wall. The shearing strength of steel is $345 {MNm}^{- 2}$ . The dimensions AB =5 cm, BC = BE=2 cm.The maximum load that can be put on the face ABCD is: (neglect bending of the rod) ( $g = 10 {ms}^{- 2}$ )

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3450 kg

1380 kg

13800 kg

345 kg.

answer is C.

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Detailed Solution

F=Shear strength area on which shear stress acts
$= 345 \times 10^{6} \times 4 \times 10^{- 4} = 138000 N$
So the mass that can be placed = 13800 Kg.

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