11.
A wire suspended vertically from one of its ends is stretched by attaching a weight of $$200\,N$$ to the lower end. The weight stretches the wire by $$1\,mm.$$ Then the elastic energy stored in the wire is
Elastic energy $$ = \frac{1}{2} \times F \times x$$
$$\eqalign{
& F = 200\,N,x = 1\,mm = {10^{ - 3}}\,m \cr
& \therefore E = \frac{1}{2} \times 200 \times 1 \times {10^{ - 3}} = 0.1\,J \cr} $$
12.
A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of the stresses developed in the two wires and $$Y$$ of steel $$ = 2 \times {10^{11}}\,N/{m^2}.$$ $$Y$$ of copper $$ = 1.3 \times {10^{11}}\,N/{m^2}.$$
13.
Equal volumes of two immiscible liquids of densities $$\rho $$ and $$2\rho $$ are filled in a vessel as shown in figure. Two small holes are punched at depth $$\frac{h}{2}$$ and $$\frac{3h}{2}$$ from the surface of lighter liquid. If $${v_1}$$ and $${v_2}$$ are the velocities of a flux at these two holes, then $$\frac{{{v_1}}}{{{v_2}}}$$ is
$${v_1} = \sqrt {2g\left( {\frac{h}{2}} \right)} .$$
For $${v_2}$$ we can replace the liquid of density $$\rho $$ and height $$h$$ to a liquid of density $$2\rho $$ and height $$\frac{h}{2}.$$
Thus $${v_2} = \sqrt {2gh} = \sqrt 2 v.$$
14.
When a $$4\,kg$$ mass is hung vertically on a light spring that obeys Hooke’s law, the spring stretches by $$2\,cms.$$ The work required to be done by an external agent in stretching this spring by $$5\,cms$$ will be
$$\left( {g = 9.8\,m/{{\sec }^2}} \right)$$
If side of the cube is $$L.$$ then $$V = {L^3}$$
$$\eqalign{
& \Rightarrow \frac{{dV}}{V} = 3\frac{{dL}}{L} \cr
& \therefore \% \,{\text{change in volume}} = 3 \times \left( {\% \,{\text{change in length}}} \right) \cr
& = 3 \times 1\% = 3\% \cr
& \therefore {\text{Bulk}}\,{\text{strain}}\frac{{\Delta V}}{V} = 0.03 \cr} $$
16.
A small spherical ball falling through a viscous medium of negligible density has terminal velocity $$v.$$ Another ball of the same mass but of radius twice that of the earlier falling through the same viscous medium will have terminal velocity
If $$\rho $$ is the density of the ball and $${\rho '}$$ that of the another ball, $$m$$ for the balls are the same, but $$r' = 2r$$
$$\eqalign{
& \therefore mg = 6\pi r\eta v\left( {{\text{by}}\,{\text{Stoke's}}\,{\text{law}}} \right) \cr
& {\text{or,}}\,\,6\pi r\eta v = 6\pi 2r\eta v' \cr
& {\text{So,}}\,\,v' = \frac{v}{2} \cr} $$
17.
A force of $$6 \times {10^6}\,N{m^{ - 2}}$$ is required for breaking a material. Then density $$\rho $$ of the material is $$3 \times {10^3}\,kg\,{m^{ - 3}}.$$ If the wire is to break under its own weight, the length of the wire made of that material should be (take $$g = 10\,m{s^{ - 2}}$$ )
Breaking stress $$ = \frac{{{\text{Force}}}}{{{\text{area}}}}$$
The breaking force will be its own weight.
$$F = mg = V\rho g = {\text{area}} \times \ell \rho g$$
Breaking stress $$ = 6 \times {10^6} = \frac{{{\text{area}} \times \ell \times \rho g}}{{{\text{area}}}}$$
$${\text{or}}\,\,\ell = \frac{{6 \times {{10}^6}}}{{3 \times {{10}^3} \times 10}} = 200\,m.$$
18.
A copper wire of length $$1.0\,m$$ and a steel wire of length $$0.5\,m$$ having equal cross-sectional areas are joined end to end. The composite wire is stretched by a certain load which stretches the copper wire by $$1\,mm.$$ If the Young’s modulii of copper and steel are respectively $$1.0 \times {10^{11}}N{m^{ - 2}}$$ and $$2.0 \times {10^{11}}N{m^{ - 2}},$$ the total extension of the composite wire is :
19.
A container filled with viscous liquid is moving vertically downwards with constant speed $$3{v_0}.$$ At the instant shown, a sphere of radius $$r$$ is moving vertically downwards (in liquid) has speed $${v_0}.$$ The coefficient of viscosity is $$\eta .$$ There is no relative motion between the liquid and the container. Then at the shown instant, the magnitude of viscous force acting on sphere is
Relative to liquid, the velocity of sphere is $$2{v_0}$$ upwards.
$$\therefore $$ Viscous force on sphere $$ = 6\pi \eta r2{v_0}\,{\text{downwards}}$$
$$ = 12\pi \eta r{v_0}\,{\text{downwards}}$$
20.
What is the absolute pressure of the gas above the liquid surface in the tank shown in fig. Density of oil $$ = 820\,kg/{m^3},$$ density of mercury $$ = 13.6 \times {10^3}\,kg/{m^3}.$$ Given 1 atmospheric pressure $$ = 1.01 \times {10^5}\,N/{m^2}.$$