JEE Main 2014 · 12 Apr · Shift 1 · Physics Q8

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PhysicsSingle correct+4 / -1Question 8 of 90

Steel ruptures when a shear of $3.5\times10^8$ N m $^{-2}$ is applied. The force needed to punch a 1 cm diameter hole in a steel sheet 0.3 cm thick is nearly :

1.4\times10^4

2.7\times10^4

3.3\times10^4

1.1\times10^4

Answer and worked solution

Official answer: C

The sheared area is the side of the punched cylinder: $\pi dt=\pi(0.01)(0.003)=9.42\times10^{-5}$ m².
$F=3.5\times10^8\times9.42\times10^{-5}\approx3.3\times10^4\text{ N}.$

Topics:Stress and strain

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JEE Main 2014 · 12 Apr · Shift 1, Physics question 8