Steel ruptures when a shear of 3.5 × 108 Nm-2 is applied. The force needed to punch a 1 cm diameter hole in a steel sheet 0.3 cm thick is nearly :

Question

Steel ruptures when a shear of 3.5 × 108 Nm-2 is applied. The force needed to punch a 1 cm diameter hole in a steel sheet 0.3 cm thick is nearly : (A) 1.4 × 104 N (B) 2.7 × 104 N (C) 3.3 × 104 N (D) 1.1 × 104 N

Tardigrade Admin · Accepted Answer

( F / A ) = stress =3.5 × 108 N / m 2 A =2 π rt =2 π × (1/2) × 0.3 =0.3 π × 10-4 m 2 F = A × stress =0.3 π × 10-4 × 3.5 × 108 =1.05 π × 104 =3.3 × 104 N

Q. Steel ruptures when a shear of $3.5 \times 1 0^{8} N m^{- 2}$ is applied. The force needed to punch a $1 c m$ diameter hole in a steel sheet $0.3 c m$ thick is nearly :

$1.4 \times 1 0^{4} N$

$2.7 \times 1 0^{4} N$

$3.3 \times 1 0^{4} N$

$1.1 \times 1 0^{4} N$

$\frac{F}{A} =$ stress $= 3.5 \times 1 0^{8} N / m^{2}$
$A = 2 π r t$
$= 2 π \times \frac{1}{2} \times 0.3$
$= 0.3 π \times 1 0^{- 4} m^{2}$
$F = A \times$ stress
$= 0.3 π \times 1 0^{- 4} \times 3.5 \times 1 0^{8}$
$= 1.05 π \times 1 0^{4}$
$= 3.3 \times 1 0^{4} N$

Q. Steel ruptures when a shear of 3.5×108Nm−2 is applied. The force needed to punch a 1cm diameter hole in a steel sheet 0.3cm thick is nearly :

1.4×104N

2.7×104N

3.3×104N

1.1×104N