A metallic rod breaks when strain produced is 0.2 %. The Young's modulus of the material of the rod is 7 × 109 N / m 2. The area of cross section of support a load of 104 N is

Question

A metallic rod breaks when strain produced is 0.2 %. The Young's modulus of the material of the rod is 7 × 109 N / m 2. The area of cross section of support a load of 104 N is (A) 7.1 × 10-4 m 2 (B) 7.1 × 10-2 m 2 (C) 7.1 × 10-8 m 2 (D) 7.1 × 10-6 m 2

Tardigrade Admin · Accepted Answer

Stress = Strain × Y Thus, maximum stress =(0.2/100) × 7 × 109=1.4 × 107 Now, F orce = Stress × Area Thus, 104=1.4 × 107 × A, or, A =7.14 × 10-4 m 2

Q. A metallic rod breaks when strain produced is $0.2%$ . The Young's modulus of the material of the rod is $7 \times 1 0^{9} N / m^{2}$ . The area of cross section of support a load of $1 0^{4} N$ is

$7.1 \times 1 0^{- 4} m^{2}$

$7.1 \times 1 0^{- 2} m^{2}$

$7.1 \times 1 0^{- 8} m^{2}$

$7.1 \times 1 0^{- 6} m^{2}$

Stress $=$ Strain $\times Y$
Thus, maximum stress $= \frac{0.2}{100} \times 7 \times 1 0^{9} = 1.4 \times 1 0^{7}$
Now, $F$ orce $=$ Stress $\times$ Area
Thus, $1 0^{4} = 1.4 \times 1 0^{7} \times A$ , or, $A = 7.14 \times 1 0^{- 4} m^{2}$

Q. A metallic rod breaks when strain produced is 0.2%. The Young's modulus of the material of the rod is 7×109N/m2. The area of cross section of support a load of 104N is

7.1×10−4m2

7.1×10−2m2

7.1×10−8m2

7.1×10−6m2