Mechanical Properties of Solids and Fluids MCQ Questions & Answers in Basic Physics | Physics
Learn Mechanical Properties of Solids and Fluids MCQ questions & answers in Basic Physics are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
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
A
$$0.2\,J$$
B
$$10\,J$$
C
$$20\,J$$
D
$$0.1\,J$$
Answer :
$$0.1\,J$$
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
A
$$\frac{1}{{2\sqrt 2 }}$$
B
$$\frac{1}{2}$$
C
$$\frac{1}{4}$$
D
$$\frac{1}{{\sqrt 2 }}$$
Answer :
$$\frac{1}{{\sqrt 2 }}$$
$${v_1} = \sqrt {2g\left( {\frac{h}{2}} \right)} .$$
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
$$\left( {g = 9.8\,m/{{\sec }^2}} \right)$$
15.
A uniform cube is subjected to volume compression. If each side is decreased by $$1\% ,$$ then bulk strain is
A
0.01
B
0.06
C
0.02
D
0.03
Answer :
0.03
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
A
$$v$$
B
$$\frac{v}{4}$$
C
$$\frac{v}{2}$$
D
$$2v$$
Answer :
$$\frac{v}{2}$$
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 $$g = 10\,m{s^{ - 2}}$$ )
A
$$20\,m$$
B
$$200\,m$$
C
$$100\,m$$
D
$$2000\,m$$
Answer :
$$200\,m$$
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
A
$$6\pi \eta r{v_0}$$
B
$$12\pi \eta r{v_0}$$
C
$$18\pi \eta r{v_0}$$
D
$$24\pi \eta r{v_0}$$
Answer :
$$12\pi \eta r{v_0}$$
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 $$ = 1.01 \times {10^5}\,N/{m^2}.$$
A
$$3.81 \times {10^5}\,N/{m^2}$$
B
$$6 \times {10^6}\,N/{m^2}$$
C
$$5 \times {10^7}\,N/{m^2}$$
D
$$4.6 \times {10^2}\,N/{m^2}$$
Answer :
$$3.81 \times {10^5}\,N/{m^2}$$
Suppose $${P_{{\text{gas}}}}$$ is the pressure of the gas on the oil. As the points $$A$$ and $$B$$ are at the same level in the mercury columns, so
$$\eqalign{
& {P_A} = {P_B} \cr
& {\text{or}}\,\,{P_{{\text{gas}}}} + {P_{{\text{oil}}}}g{h_{{\text{oil}}}} = {P_a} + {\rho _{Hg}}g{h_{Hg}} \cr
& {\text{or}}\,\,{P_{{\text{gas}}}} + 820 \times 9.8 \times \left( {1 + 1.50} \right) = {P_a} + 13.6 \times {10^3} \times 9.8 \times \left( {1.5 + 0.75} \right) \cr
& {\text{or}}\,\,{P_{{\text{gas}}}} + 20.09 \times {10^3} = {P_a} + 299.88 \times {10^3} \cr
& \therefore {P_{{\text{gas}}}} - Pa = 299.88 \times {10^3} - 20.09 \times {10^3} \cr
& {\text{or}}\,\,{\left[ {{P_{{\text{gas}}}}} \right]_{{\text{gauge}}}} = 279.8 \times {10^3}\,N/{m^2} = 2.8 \times {10^5}\,N/{m^2} \cr
& {\text{Absolute pressure of gas}} \cr
& {\left[ {{P_{{\text{gas}}}}} \right]_{{\text{absolute}}}} = {\left[ {{P_{{\text{gas}}}}} \right]_{{\text{gauge}}}} + {P_a} = 2.8 \times {10^5} + 1.01 \times {10^5} = 3.81 \times {10^5}N/{m^2} \cr} $$