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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main · Physics · Mechanical Properties of Solids

Two metal strips are riveted together at their ends by four rivets,...

Two metal strips are riveted together at their ends by four rivets, each of diameter 6mm6\text{mm}. Assuming that each of the rivets carries one-quarter of the load, the maximum tension that can be produced in the riveted strip is [Breaking shear stress of the rivet = 6.9×107Pa6 .9 \times {10}^{7}Pa]

Options

  1. A7.8×103N7.8 \times {10}^{3} \text{N}
  2. B6.9×103N6.9 \times {10}^{3} \text{N}
  3. C3.14×103N3.14 \times {10}^{3} \text{N}
  4. Dnone of these

Correct answer

A. 7.8×103N7.8 \times {10}^{3} \text{N}

Step-by-step solution

Diameter of each rivet (D)=6mm( \text{D} ) = 6\text{mm}

∴Radius (r)=D2=3 mm=3×10−3m∴ \text{Radius (r)} = \dfrac{\text{D}}{2} = \text{3 mm} = 3 \times {10}^{- 3} \text{m}

Maximum shearing stress on each rivet = 6.9 × 107 Pa

Let w be the maximum load that can be subjected to the riveted strip.
As each rivet carry one-quarter of the load, therefore

Load on each rivet=w4\text{Load on each rivet} = \dfrac{\text{w}}{4}

Maximum shearing stress=Maximum shearing forceArea\text{Maximum shearing stress} = \dfrac{\text{Maximum shearing force}}{\text{Area}}

∴6.9×107=w / 4\pir2∴ \text{6.9} \times {10}^{7} = \dfrac{\text{w / 4}}{\text{\pi} r^{2}}

or w=6.9×1074\pir2\text{w} = 6.9 \times {10}^{7} 4{\text{\pi r}}^{2}

or w=6.9×107×4×3.14×(3×10−3)2\text{w} = 6.9 \times {10}^{7} \times 4 \times 3.14 \times {( 3 \times {10}^{- 3} )}^{2}

=6.9×4×3.14×9×10= 6.9 \times 4 \times 3.14 \times 9 \times 10

=7.8×103N= 7.8 \times {10}^{3} \text{N}
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