Question

A piece of metal of 2kg is suspended from one end of a vertical wire whose other end is fixed and the piece if fully emersedin oil of density 0.7x103 kg/m3.the length of wire increases by 1mm.if diameter of wire is 0.6mm,young's module is 2.0x1011 N/m2 and volume of metal piece is 800cm3,thn calculate d initial length of wire

Answer 1

Answer: Initial length of wire is 4 m

Explanation:

Given that,

Mass of piece $m = 2\ kg$

Oil density $\rho = 0.7\times 10^{3}\ kg/m^{3}$

Increases length $\Delta d = 1 mm$

Diameter of wire $D = 0.6\ mm$

Young's modulus $Y = 2.0\times 10^{11}\ N/m^{2}$

Volume of metal piece $V = 800\ cm^{3}$

According to buoyancy force

If the metal piece fully emerged then the force,

$F = mg - \rho gV$

Now, the formula of Young's modulus

$Y = \dfrac{F\times d}{A\times \Delta d}$

Now, put the value of F in Young's modulus

$d = \dfrac{Y\times A\times \Delta d}{m\times g - \rho \times g \times V}$

$d = \dfrac{2.0\times 10^{11}\ N/m^{2}\times3.14\times 0.3\times 0.3\times10^{-6}\ m^{2}\times1\times10^{-3}\ m}{2\ kg\times 9.8\ m/s^{2} - 0.7\times10^{3}\ kg/m^{3}\times9.8\ m/s^{2}\times 800\times10^{-6}\ m}$

$d = \dfrac{0.562\times10^{2}}{14.112}\ m$

$d = 0.04005\times10^{2}\ m$

$d = 4\ m$

Hence, this is the required solution.

Answer 2

Explanation:

,

Mass of piece m = 2\ kgm=2 kg

Oil density \rho = 0.7\times 10^{3}\ kg/m^{3}ρ=0.7×10

3

kg/m

3

Increases length \Delta d = 1 mmΔd=1mm

Diameter of wire D = 0.6\ mmD=0.6 mm

Young's modulus Y = 2.0\times 10^{11}\ N/m^{2}Y=2.0×10

11

N/m

2

Volume of metal piece V = 800\ cm^{3}V=800 cm

3

According to buoyancy force

If the metal piece fully emerged then the force,

F = mg - \rho gVF=mg−ρgV

Now, the formula of Young's modulus

Y = \dfrac{F\times d}{A\times \Delta d}Y=

A×Δd

F×d

Now, put the value of F in Young's modulus

d = \dfrac{Y\times A\times \Delta d}{m\times g - \rho \times g \times V}d=

m×g−ρ×g×V

Y×A×Δd

d = \dfrac{2.0\times 10^{11}\ N/m^{2}\times3.14\times 0.3\times 0.3\times10^{-6}\ m^{2}\times1\times10^{-3}\ m}{2\ kg\times 9.8\ m/s^{2} - 0.7\times10^{3}\ kg/m^{3}\times9.8\ m/s^{2}\times 800\times10^{-6}\ m}d=

2 kg×9.8 m/s

2

−0.7×10

3

kg/m

3

×9.8 m/s

2

×800×10

−6

m

2.0×10

11

N/m

2

×3.14×0.3×0.3×10

−6

m

2

×1×10

−3

m

d = \dfrac{0.562\times10^{2}}{14.112}\ md=

14.112

0.562×10

2

m

d = 0.04005\times10^{2}\ md=0.04005×10

2

m

d = 4\ md=4 m