When a force is applied on a wire of uniform cross-sectional area 3- 10-6m2 and length 4 m- the increase in length is 1 mm- Energy stored in it will be Y-2- 1011N-m2-

The correct option is C 0.075 #160;J U = 12 F Δ l = 12×Y Δ lAL×Δ l = 12×2 × 10^11× 3× 10^ 6× (10^ 3)^24 = 0.075 #160;J ...

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Standard VIII

Chemistry

Surface Tension

When a force ...

Question

When a force is applied on a wire of uniform cross-sectional area $3 \times 10^{- 6} m^{2}$ and length $4 m$ , the increase in length is $1 m m$ . Energy stored in it will be ( $Y = 2 \times 10^{11} N / m^{2}$ ):

6250 J

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0.177 J

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0.075 J

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0.150 J

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Solution

The correct option is C $0.075 J$
$U = \frac{1}{2} F Δ l = \frac{1}{2} \times \frac{Y Δ l A}{L} \times Δ l = \frac{1}{2} \times \frac{2 \times 10^{11} \times 3 \times 10^{- 6} \times (10^{- 3})^{2}}{4} = 0.075 J$

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Similar questions

When a force is applied on a wire of uniform cross-sectional area $3 \times 10^{- 6} m^{2}$ and length $4 m$ , the increase in length is $1 m m$ . Energy stored in it will be

$(Y = 2 \times 10^{11} N m^{- 2})$

When a force is applied on a wire of uniform cross-sectional area $3 \times 10^{- 6} m^{2}$ and length $4 m$ , the increase in length is $1 m m$ . Energy stored in it will be

$(Y = 2 \times 10^{11} N m^{- 2})$

Q. When a force is applied on a wire of uniform cross-sectional area

3 \times 10^{- 6} m^{2}

and length 4 m, the increase in length is 1 mm. Energy stored in it will be

(Y = 2 \times 10^{11} N / m^{2})

When a force is applied on a wire of uniform cross-sectional area $3 \times 10^{- 6} m^{2}$ and length $4 m$ , the increase in length is $1 m m$ . Energy stored in it will be

$(Y = 2 \times 10^{11} N m^{- 2})$

Q. A steel wire of length

2

m is hanging from a rigid horizontal support. How much energy is stored in it when a load of

5

kg is suspended on it? If the load is increased to

10

kg then by how much the energy stored will increase? (

Y = 2 \times 10^{11} N / m^{2}

, area of cross-section

= 10^{- 6} m^{2}

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When a force is applied on a wire of uniform cross-sectional area 3×10−6m2 and length 4m, the increase in length is 1 mm. Energy stored in it will be (Y=2×1011N/m2):

The correct option is C 0.075 JU=12FΔl=12×YΔlAL×Δl=12×2×1011×3×10−6×(10−3)24=0.075J

When a force is applied on a wire of uniform cross-sectional area $3 \times 10^{- 6} m^{2}$ and length $4 m$ , the increase in length is $1 m m$ . Energy stored in it will be ( $Y = 2 \times 10^{11} N / m^{2}$ ):

The correct option is C $0.075 J$
$U = \frac{1}{2} F Δ l = \frac{1}{2} \times \frac{Y Δ l A}{L} \times Δ l = \frac{1}{2} \times \frac{2 \times 10^{11} \times 3 \times 10^{- 6} \times (10^{- 3})^{2}}{4} = 0.075 J$