You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Longitudinal Stress

If Young's mo...

Question

If Young's modulus for steel is $2.0 \times 10^{11} N m^{- 2}$ , then how much weight be suspended from a steel wire of length $2.0 m$ and diameter $1.0 m m$ so that the length of the wire be increased by $1.0 m m$ ? ( $g = 9.8 m s^{- 2}$ )

Open in App

Solution

The change in length $l = \frac{F L}{π r^{2} Y}$
where F is weight to be suspended , L=2.0 m is initial length of wire, and r is radius of wire= .5 mm with Y as Young's modulus of material
putting these values we get $F = 78.5 N e w t o n$ or mass as $\frac{78.5}{9.8} = 8.01 k i l o g r a m$

Suggest Corrections

Similar questions

Q. A composite wire consits of a steel wire of length

1.5 m

and a copper wire of length

2.0 m

, with a uniform cross-sectional area of

2.5 \times 10^{- 5} m^{2}

. It is loaded with a mass of

200 k g

. Find the extension produced. Young's modulus of copper is

1.0 \times 10^{11} N m^{- 2}

and that of steel is

2.0 \times 10^{11} N m^{- 2}

.
Take

g = 9.8 m s^{- 2}

A load of 4 kg is suspended from a ceiling through a steel wire of length 20 m and radius 2 mm. It is found that the length of the wire increases by 0.031 mm, as equilibrium is achieved. If g = 3.1 $\times π m s^{- 2}$ the value of

Young’s modulus is

Q. A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the students measures the extension in the length of the wire to be 0.8 mm with an uncertainty of +/- 0.05 mm at a load of exactly 1kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of +/-0.01 mm.The Young’s modulus obtained from the reading is

(g = 9.8 m s^{- 2})

Q. A composite wire of a uniform cross-section

5.5 \times 10^{- 5} m^{2}

consists of a steel wire of length

1.5 m

and a copper wire of length with a mass of

200 k g

is [Young's modulus of steel is

2 \times 10^{11} N m^{- 2}

and that of copper is

1 \times 10^{11} N m^{- 2}

. Take

g = 10 m s^{- 2}]

Q. A load of

4.0 k g

is suspended from a ceiling through a steel wire of length

20 m

and radius

2.0 m m

. It is found that the length of the wire increases by

0.031 m m

as equilibrium is achieved. If

g = 3.1 x π m s^{- 2}

, the value of young's modulus in

N m^{- 2}

Join BYJU'S Learning Program

If Young's modulus for steel is 2.0×1011Nm−2, then how much weight be suspended from a steel wire of length 2.0m and diameter 1.0mm so that the length of the wire be increased by 1.0mm? (g=9.8ms−2)

The change in lengthl=FLπr2Y where F is weight to be suspended , L=2.0 m is initial length of wire, and r is radius of wire= .5 mm with Y as Young's modulus of materialputting these values we get F=78.5Newton or mass as 78.59.8=8.01kilogram

If Young's modulus for steel is $2.0 \times 10^{11} N m^{- 2}$ , then how much weight be suspended from a steel wire of length $2.0 m$ and diameter $1.0 m m$ so that the length of the wire be increased by $1.0 m m$ ? ( $g = 9.8 m s^{- 2}$ )

The change in length $l = \frac{F L}{π r^{2} Y}$
where F is weight to be suspended , L=2.0 m is initial length of wire, and r is radius of wire= .5 mm with Y as Young's modulus of material
putting these values we get $F = 78.5 N e w t o n$ or mass as $\frac{78.5}{9.8} = 8.01 k i l o g r a m$