You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Young's Modulus of Elasticity

The Y of a ma...

Question

The Y of a material having a cross sectional area of $1 c m$ $^{2}$ is 2 x 10 $^{12}$ dynes/cm $^{2}$ . The force required to double the length of the wire is:

1 x 10

^{12}

dynes

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 x 10

^{12}

dynes

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.5 x 10

^{12}

dynes

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 x 10

^{12}

dynes

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D 2 x 10 $^{12}$ dynes
$Y = \frac{F l}{A Δ l}$
$F = \frac{Y A Δ l}{l}$
$= 2 \times 10^{12} \times 1$ $(∴ \frac{Δ l}{l} = 1)$
$= 2 \times 10^{12}$ dynes.

Suggest Corrections

Similar questions

Q. The Yogng's modulus of a material having area of

1 c m

is a cross-sectional

2 \times 10^{12} {dyne/cm}^{2}

. If the length is to be doubled the force required is

Q. What force is required to stretch a copper wire

1 c m^{2}

in cross-section to double its length?

Y

for copper is

1.26 \times 10^{12} d y n e c m^{- 2}

Q. Two exactly similar wires of steel (y

=

20 x 10

^{11}

dyne/cm

^{2}

) and copper (y

=

12 x 10

^{11}

dyne/cm

^{2}

)are stretched by equal forces. If the total elongation is 1cm, elongation of copper wire is

Q. An iron bar whose cross sectional area is

4 c m^{2}

is heated from

0^{\circ} C

and

100^{\circ} C

. The force required to prevent the expansion of the rod is :
(Y of Iron =

2 \times 10^{12} d y n e / c m^{2}

α

of Iron =

12 \times 10^{6} /^{\circ} C

)

Q. A tensile force of 2 x 10

^{5}

dynes doubles the length of an elastic cord whose area of cross-section is 2 sq. cm. Young's modulus of the material of the cord is

Join BYJU'S Learning Program

The Y of a material having a cross sectional area of 1cm 2 is 2 x 1012 dynes/cm2. The force required to double the length of the wire is:

The correct option is D 2 x 1012dynesY=FlAΔlF=YAΔll=2×1012×1 (∴Δll=1)=2×1012 dynes.

The Y of a material having a cross sectional area of $1 c m$ $^{2}$ is 2 x 10 $^{12}$ dynes/cm $^{2}$ . The force required to double the length of the wire is:

The correct option is D 2 x 10 $^{12}$ dynes
$Y = \frac{F l}{A Δ l}$
$F = \frac{Y A Δ l}{l}$
$= 2 \times 10^{12} \times 1$ $(∴ \frac{Δ l}{l} = 1)$
$= 2 \times 10^{12}$ dynes.