## A cubical solid aluminium (bulk modulus = -VdP/dV = 70 GPa) block has an edge length 1 m on the surface of earth .

Q: A cubical solid aluminium (bulk modulus = -VdP/dV = 70 GPa) block has an edge length 1 m on the surface of earth . It is kept on the floor of a 5 km deep ocean . Taking the average density of water and the acceleration due to gravity to be 103 kg/m^3 and 10 m/s^2 respectively , the change in edge length of the block in mm is …..

Click to See Solution :
Ans: (0.24)
Sol: $\displaystyle \frac{\Delta V}{V} = -\frac{\Delta p}{B}$ ; Where B = Bulk modulus

V = l3

$\frac{\Delta V}{V} = 3 \frac{\Delta l}{l}$

$\displaystyle 3\frac{\Delta l}{l} = |-\frac{\Delta p}{B}|=\frac{\rho g h}{B}$

$\displaystyle \Delta l = \frac{\rho g h l}{3 B}$

## A body of mass m = 10 kg is attached to one end of a wire of length 0.3 m . The maximum angular speed …..

Q: A body of mass m = 10 kg is attached to one end of a wire of length 0.3 m . The maximum angular speed (in rad/s) with which it can be rotated about its other end in space station is (Breaking stress of wire = 4.8 × 107 N/m^2 and area of cross section of the wire = 10-2 cm2) is …..

Click to See Solution :
Ans: (4)

Sol: Breaking stress of wire = 4.8 × 107 N/m^2

A = 10-2 cm2 = 10-6 m2

l = 0.3 m

T = m ω2 l

$\displaystyle \frac{T}{A} = \frac{m \omega^2 l}{A}$

$\displaystyle Breaking \; Stress = \frac{10 \times 0.3 \omega^2}{10^{-6}}$

$\displaystyle 4.8 \times 10^7 = \frac{10 \times 0.3 \omega^2}{10^{-6}}$

## A cube of metal is subjected to a hydrostatic pressure of 4 GPa . The percentage change in the length …

Q: A cube of metal is subjected to a hydrostatic pressure of 4 GPa . The percentage change in the length of the side of the cube is close to (Given bulk modulus of metal , B = 8 × 1010)

(a) 1.67

(b) 5

(c) 20

(d) 0.6

Click to See Solution :
Ans: (a)
Sol: P = 4 GPa = 4 × 109 Pa , B = 8 × 109 Pa

As Bulk Modulus $\displaystyle B = \frac{P V}{\Delta V}$

$\displaystyle \frac{\Delta V}{V} = \frac{P}{B} = \frac{4 \times 10^9}{8 \times 10^{10}}$

$\displaystyle \frac{\Delta V}{V} = \frac{1}{20}$

$\displaystyle V = a^3$

$\displaystyle \frac{\Delta V}{V} = \frac{3a^2}{a^3} \Delta a$

$\displaystyle \frac{\Delta a}{a} \times 100 = \frac{1}{3} \times \frac{\Delta V}{V} \times 100$

$\displaystyle \frac{\Delta a}{a} \times 100 = \frac{1}{3} \times \frac{1}{20} \times 100$

= 1.67 %

## Two steel wires having same length are suspended from a ceiling under the same load . If the ratio of their energy stored ….

Q: Two steel wires having same length are suspended from a ceiling under the same load . If the ratio of their energy stored per unit volume is 1:4 , the ratio of their diameters is

(a) √2:1

(b) 1:√2

(c) 2:1

(d) 1:2

Click to See Solution :
Ans: (a)
Sol: Energy stored per unit volume, $\displaystyle u = \frac{U}{V} = \frac{1}{2}Stress \times Strain$

$\displaystyle u = \frac{1}{2} Stress \times \frac{Stress}{Y}$

$\displaystyle u = \frac{1}{2} \frac{F}{A} \times \frac{F}{A Y}$

$\displaystyle u = \frac{1}{2} \frac{F^2}{A^2 Y}$

$\displaystyle u \propto \frac{1}{D^4 }$ (Since A ∝ D2)

$\displaystyle \frac{u_1}{u_2} = \frac{1}{4}$

$\displaystyle \frac{D_2^4}{D_1^4} = \frac{1}{4}$

$\displaystyle \frac{D_2}{D_1} = \frac{1}{\sqrt{2}}$

$\displaystyle \frac{D_1}{D_2} = \frac{\sqrt{2}}{1}$