The bulk modulus for an incompressible liquid is:
1. zero
2. unity
3. infinity
4. between \(0\) and \(1\)

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:
1. \(\frac{1}{2}MgL\)
2. \(Mgl\)
3. \(MgL\)
4. \(\frac{1}{2}Mgl\)

A wire can sustain a weight of 10 kg before breaking. If the wire is cut into two equal parts, then each part can sustain a weight of:

Young’s modulus of steel is twice that of brass. Two wires of the same length and of the 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:
1. \(1:2\)
2. \(2:1\)
3. \(4:1\)
4. \(1:1\)

The bulk modulus of a spherical object is \(B.\) If it is subjected to uniform pressure \(P,\) the fractional decrease in radius is:

Two wires are made of the same material and have the same volume. The first wire has a cross-sectional area \(A\) and the second wire has a 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?

A light rod of length \(2~\text{m}\) is suspended from the ceiling horizontally by means of two vertical wires of equal length. A weight \(W\) is hung from the light rod as shown in the figure. The rod is hung by means of a steel wire of cross-sectional area \(A_1 = 0.1~\text{cm}^2\) and brass wire of cross-sectional area \(A_2 = 0.2~\text{cm}^2.\) To have equal stress in both wires, \(\frac{T_1}{T_2}?\)${\mathrm{}}_{}$

One end of a uniform wire of length \(L\) and of weight \(W\) is attached rigidly to a point in the roof and a weight \(W_1\) is suspended from its lower end. If \(A\) is the area of the cross-section of the wire, the stress in the wire at a height \(\frac{3L}{4}\) from its lower end is:
1. \(\frac{W+W_1}{A}\)
2. \(\frac{4W+W_1}{3A}\)
3. \(\frac{3W+W_1}{4A}\)
4. \(\frac{\frac{3}{4}W+W_1}{A}\)

To break a wire, a force of \(10^6~\text{N/m}^{2}\) is required. If the density of the material is \(3\times 10^{3}~\text{kg/m}^3,\) then the length of the wire which will break by its own weight will be:
1. \(34~\text m\) 
2. \(30~\text m\) 
3. \(300~\text m\) 
4. \(3~\text m\)