If the volume of a block of aluminium is decreased by 1%, the pressure (stress) on its surface is increased by (Bulk modulus of Al=7.5× 1010Nm-2)

Question

If the volume of a block of aluminium is decreased by 1%, the pressure (stress) on its surface is increased by (Bulk modulus of Al=7.5× 1010Nm-2)  (A)  7.5× 1010Nm-2  (B)  7.5× 108 text Nm-2  (C)  7.5× 106Nm-2  (D)  7.5× 104Nm-2

Tardigrade Admin · Accepted Answer

Given, (Δ V/V)× 100=1%=(1/100) Bulk modulus, B=(p/Δ V/V)=(pV/Δ V) Or p=(BΔ V/V)=7.5× 1010× (1/100) =7.5× 108Nm-2

Q. If the volume of a block of aluminium is decreased by 1%, the pressure (stress) on its surface is increased by (Bulk modulus of $A l = 7.5 \times 10^{10} N m^{- 2})$

$7.5 \times 10^{10} N m^{- 2}$

$7.5 \times 10^{8} N m^{- 2}$

$7.5 \times 10^{6} N m^{- 2}$

$7.5 \times 10^{4} N m^{- 2}$

$7.5 \times 10^{2} N m^{- 2}$

Given, $\frac{Δ V}{V} \times 100 = 1$
Bulk modulus, $B = \frac{p}{Δ V / V} = \frac{p V}{Δ V}$
Or $p = \frac{B Δ V}{V} = 7.5 \times 10^{10} \times \frac{1}{100}$
$= 7.5 \times 10^{8} N m^{- 2}$

Q. If the volume of a block of aluminium is decreased by 1%, the pressure (stress) on its surface is increased by (Bulk modulus of Al=7.5×1010Nm−2)

7.5×1010Nm−2

7.5×108Nm−2

7.5×106Nm−2

7.5×104Nm−2

7.5×102Nm−2

Given, VΔV​×100=1 Bulk modulus, B=ΔV/Vp​=ΔVpV​ Or p=VBΔV​=7.5×1010×1001​ =7.5×108Nm−2