A wire of length 50 cm and cross-sectional area of 1 mm 2 is extended by 1 mm. The required work (in joule) will be ( Y =2 × 1010 Nm -2)

Question

A wire of length 50 cm and cross-sectional area of 1 mm 2 is extended by 1 mm. The required work (in joule) will be ( Y =2 × 1010 Nm -2) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

If side of the cube is L then V=L3 ⇒ (d V/V)=3 (d L/L) ∴ % change in volume =3 ×( % change in length ) =3 × 1 %=3 % ∴ Bulk strain (Δ V/V)=0.03

Q. A wire of length $50 c m$ and cross-sectional area of $1 m m^{2}$ is extended by $1 mm$ . The required work (in joule) will be $(Y = 2 \times 1 0^{10} N m^{- 2})$

If side of the cube is $L$ then $V = L^{3} \Rightarrow \frac{d V}{V} = 3 \frac{d L}{L}$
$∴ %$ change in volume $= 3 \times (%$ change in length $)$
$= 3 \times 1% = 3%$
$∴$ Bulk strain $\frac{Δ V}{V} = 0.03$

Q. A wire of length 50cm and cross-sectional area of 1mm2 is extended by 1mm. The required work (in joule) will be (Y=2×1010Nm−2)

If side of the cube is L then V=L3⇒VdV​=3LdL​ ∴% change in volume =3×(% change in length ) =3×1%=3% ∴ Bulk strain VΔV​=0.03

Q. A wire of length $50 c m$ and cross-sectional area of $1 m m^{2}$ is extended by $1 mm$ . The required work (in joule) will be $(Y = 2 \times 1 0^{10} N m^{- 2})$

If side of the cube is $L$ then $V = L^{3} \Rightarrow \frac{d V}{V} = 3 \frac{d L}{L}$
$∴ %$ change in volume $= 3 \times (%$ change in length $)$
$= 3 \times 1% = 3%$
$∴$ Bulk strain $\frac{Δ V}{V} = 0.03$