A uniform rope of density \(rho\) and length \(L\) is hanging from roof. If young’s modulus of material of rope is \(Y\), then elongation produced in rope due to its own weight is:
The elongation of a uniform rope under its own weight is given by \(\Delta L = \frac{MgL}{2AY}\). Substituting mass \(M = \rho A L\), we obtain \(\Delta L = \frac{\rho g L^2}{2Y}\).
A rubber sphere is taken in a lake to a depth \(1800\text{ m}\). If bulk modulus of rubber is \(6 \times 10^8\text{ N/m}^2\), then radius of this rubber sphere will decrease by:
The pressure change is \(dP = \rho g h = 10^3 \times 10 \times 1800 = 1.8 \times 10^7\text{ N/m}^2\). The fractional volume change is \(\frac{dV}{V} = \frac{dP}{B} = \frac{1.8 \times 10^7}{6 \times 10^8} = 3\%\). Since \(\frac{dV}{V} = 3\frac{dr}{r}\), the radius decreases by \(\frac{3\%}{3} = 1\%\).
The approximate depth of an ocean is $2700 \text{ m}$. The compressibility of water is $45.4 \times 10^{-11} \text{ Pa}^{-1}$ and density of water is $10^3 \text{ kg/m}^3$. What fractional compression of water will be obtained at the bottom of the ocean?
(2015)
Pressure at depth $h$ is $$P = \rho gh = 10^3 \times 9.8 \times 2700 \approx 26.4 \times 10^6 \text{ Pa}$$. Fractional compression is $$\frac{\Delta V}{V} = P \times K = (26.4 \times 10^6) \times (45.4 \times 10^{-11}) \approx 1.2 \times 10^{-2}$$.
When a block of mass $M$ is suspended by a long wire of length $L$, the length of the wire becomes $(L + l)$. The elastic potential energy stored in the extended wire is :
(2019)
The elastic potential energy stored in a stretched wire is given by $U = \frac{1}{2} \times \text{Force} \times \text{Extension}$. Here, the applied force is the weight of the block $Mg$ and the extension is $l$. Therefore, $U = \frac{1}{2} Mgl$.
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area $A$ and the second wire has cross-sectional area $3A$. If the length of the first wire is increased by $\Delta l$ on applying a force $F$, how much force is needed to stretch the second wire by the same amount?
(2018)
Volume $V = A_1 L_1 = A_2 L_2 \Rightarrow A L_1 = 3A L_2 \Rightarrow L_2 = L_1/3$. Force $F = \frac{Y A \Delta l}{L_1}$. For the second wire, $F' = \frac{Y (3A) \Delta l}{L_1/3} = 9 \left( \frac{Y A \Delta l}{L_1} \right) = 9F$.
The Young’s modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of:
(2015 Re)
We know $\Delta L = \frac{FL}{AY}$. Since $L$, $A$, and $\Delta L$ are the same for both wires, $F \propto Y$. Therefore, $$\frac{F_s}{F_b} = \frac{Y_s}{Y_b} = \frac{2}{1} = 2:1$$.
The bulk modulus of a spherical objects is ‘$B$’. If it is subjected to uniform pressure ‘$P$’, the fractional decrease in radius is:
(2017-Delhi)
Bulk modulus $B = \frac{P}{\Delta V/V} \Rightarrow \frac{\Delta V}{V} = \frac{P}{B}$. For a sphere, $V = \frac{4}{3}\pi r^3$, so the fractional change in volume is $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}$. Therefore, $$\frac{\Delta r}{r} = \frac{1}{3} \frac{\Delta V}{V} = \frac{P}{3B}$$.