By | September 23, 2020

CBSE Test Paper 01

Chapter 9 Mechanical Properties of Solids

1. A solid brass sphere is initially surrounded by air, and the air pressure exerted on it is 1.0 $×$ $1{0}^{5}$ N/ ${\mathrm{m}}^{2}$ (normal atmospheric pressure). The sphere is lowered into the ocean to a depth at which the pressure is 2.0 $×$ $1{0}^{7}$ N/ ${\mathrm{m}}^{2}$. The volume of the sphere in air is 0.50 m3. By how much does this volume change once the sphere is submerged? modulus of brass as 61 GPa 1
1. -1.7 $×$ $1{0}^{-4}$ ${\mathrm{m}}^{3}$
2. -1.4 $×$ $1{0}^{-4}$ ${\mathrm{m}}^{3}$
3. -1.5 $×$ $1{0}^{-4}$ ${\mathrm{m}}^{3}$
4. -1.6 $×$ $1{0}^{-4}$ ${\mathrm{m}}^{3}$
2. For a rope of yield strength ${\mathrm{S}}_{\mathrm{y}}$ loaded in tension with weight Mg the minimum area A of the rope should be 1
1. $\ge$ Mg ${\mathrm{S}}_{\mathrm{y}}$
2. $\ge$ Mg/ ${\mathrm{S}}_{\mathrm{y}}$
3. $\ge$ Mg/3 ${\mathrm{S}}_{\mathrm{y}}$
4. $\ge$ Mg/2 ${\mathrm{S}}_{\mathrm{y}}$
3. A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminum (wire B) of equal lengths as shown in Figure. The cross-sectional areas of wires A and B are 1.0 $\mathrm{m}{\mathrm{m}}^{2}$ and 2.0 $\mathrm{m}{\mathrm{m}}^{2}$, respectively. At what point along the rod should a mass m be suspended in order to produce equal strains in both steel and aluminum wires. Take Young’s modulus of steel as 200 GPa, for aluminum 70 GPa 1
1. 0.42 m from steel wire
2. 0.40 m from steel wire
3. 0.43 m from steel wire
4. 0.44 m from steel wire
4. Material is said to be brittle if 1
1. material cross section is significantly reduced at failure
2. fracture occurs soon after the elastic limit is passed
3. a large amount of plastic deformation takes place between the elastic limit and the fracture point
4. material elongates a lot before finally breaking
5. A piece of copper having a rectangular cross-section of 15.2 mm $×$ 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? Take Young’s modulus of copper as 11 $×$ $1{0}^{10}$Pa 1
1. $0.06×{10}^{-2}$
2. $0.11×{10}^{-2}$
3. $0.04×{10}^{-2}$
4. $0.14×{10}^{-2}$
6. Define Poisson’s ratio? What is its unit? 1
7. What are ductile and brittle materials? 1
8. What is the Young’s modulus for a perfect rigid body? 1
9. Why do we prefer a spring made of steel and not of copper? 2
10. When a load of a wire is increased from 3 kg wt to 5 kg wt, the length of that wire changes from 0.61 mm to 1.02 mm. calculate the change in the elastic potential energy of the wire. 2
11. A wire elongates by l mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each is hung at the two ends, then what will be the elongation of the wire in mm? 2
12. The stress-strain graph for a metal wire is given in the figure. Up to the point B, the wire returns to its original state O along the curve BAO, when it is gradually unloaded. Point E corresponds to the fracture point of the wire. 3
1. Up to which point of curve, is Hooke’s law obeyed? This point is also called ‘Proportionality limit’.
2. Which point on the curve corresponds to elastic limit and yield point of the wire?
3. Indicate the elastic and plastic regions of the stress-strain curve.
4. What change happens when the wire is loaded up to a stress corresponding to point C on curve, and then unloaded gradually?
13. Establish $s=ut+\frac{1}{2}a{t}^{2}$ from velocity-time graph for a uniform accelerated motion? 3
14. A man carrying mass M = 125 kg makes a flying tackle at Vj = 4 m/s on a stationary quarterback of mass m = 85 kg and his helmet makes solid contact with quarterback’s femur. 3
1. What is the final speed of two athletes immediately after contact and also determine the average force exerted on the quarterback’s femur, when the last collision occur at 0.100 s?
2. If the area of cross-section of quarterback’s femur is 5 $×$ 104 m2, then estimate the shear stress exerted on the femur in the collision.
15. Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires. 5

CBSE Test Paper 01
Chapter 9
Mechanical Properties of Solids

1. (d) -1.6 $×$ $1{0}^{-4}$ ${\mathrm{m}}^{3}$
Explanation: given initial pressure p1 = $1.0×{10}^{5}\phantom{\rule{thickmathspace}{0ex}}N/{m}^{2}$
final pressure p$2.0×{10}^{7}\phantom{\rule{thickmathspace}{0ex}}N/{m}^{2}$
volume of sphere V = 0.50 m3
bulk modulus B of brass = 61 Gpa = 61 $×$ 109pa
bulk modulus is given by B = $-\frac{\mathrm{\Delta }p}{\frac{\mathrm{\Delta }V}{V}}$$DeltaV=-\frac{\mathrm{\Delta }p×V}{B}=-\frac{\left({p}_{2}-{p}_{1}\right)×V}{B}$$=-\frac{\left(2.0×{10}^{7}-1.0×{10}^{5}\right)×0.50}{61×{10}^{9}}$
$DeltaV=-1.6×{10}^{-4}\phantom{\rule{thickmathspace}{0ex}}{m}^{3}$
1. $\ge$ Mg/ ${\mathrm{S}}_{\mathrm{y}}$
Explanation: We know that Yield strength times the Area give the weight of the body.
As Yield strength = weight of body/area
For safety purposes
Yield strength $\ge$ weight of body/area
${\mathrm{S}}_{\mathrm{y}}$ $\ge$ Mg/A
$\ge$ Mg/ ${\mathrm{S}}_{\mathrm{y}}$

1. 0.43 m from steel wire
Explanation: $\frac{{F}_{1}}{{A}_{steel}×{y}_{steel}}=\frac{{F}_{2}}{{A}_{al}×{y}_{aluminium}}$ let L be the length of each wire.
cross sectional area of steel Asteel = 2mm2
young modulus of steel ysteel = 200 Gpa= 2 $×$ 1011 pa
cross sectional area of aluminium Aal = 1mm2
young modulus of aluminium yal = 70 Gpa =7 $×$ 1010 pa
Let after placing the mass m wight on lower ends of wire be F1 and F2 then stress on wires A and B will be $\frac{{f}_{1}}{{A}_{steel}}$ and $\frac{{f}_{2}}{{A}_{al}}$
Now given condition is strain should be equal thus
Young modulus y = $\frac{stress}{strain}$ $⇒$ strain = $\frac{stress}{y}$
Strainst = $\frac{{\mathrm{stress}}_{st}}{{y}_{steel}}=\frac{{F}_{1}}{{A}_{steel}×{y}_{steel}}$ and strainal = $\frac{stres{s}_{al}}{{y}_{al}}$
for equal strains $⇒$ strainst = strainal
$\frac{{F}_{1}}{{A}_{steel}×{y}_{steel}}=\frac{{F}_{2}}{{A}_{al}×{y}_{aluminium}}$ $⇒$
if mass m is placed at a distance x and y from two wires then
F1x = F2y
$\frac{{F}_{1}}{{F}_{2}}$ = $\frac{y}{x}$ $\to$(2)
From equation 1 and 2
$\frac{y}{x}$ = $\frac{{A}_{\text{steel}}×{y}_{\text{steel}}}{{A}_{\text{al}}×{y}_{al}}$ $⇒$ x = $\frac{{A}_{al}×{y}_{al}}{{A}_{steel}×{y}_{steel}}y$ $\to$ (3)
also given x + y = 1.05 (total length of rod)
y = 1.05 – x $⇒$ (4)
thus from 3 and 4
x = $\frac{{A}_{al}×{y}_{al}}{{A}_{steel}×{y}_{steel}}\left(1.05-x\right)$ $⇒$ $x{A}_{steel}×{y}_{steel}={A}_{al}{y}_{al}1.05-{A}_{al}{y}_{al}x$
$x\left({A}_{steel}{y}_{steel}+{A}_{al}{y}_{al}\right)={A}_{al}×{y}_{al}×1.05$$⇒x=\frac{2×{10}^{-6}×7×{10}^{10}×7×{10}^{10}×1.05}{\left(2×2×{10}^{11}+1×7×{10}^{10}\right)×{10}^{-6}}$
x = 0.43 m
Thus mass should be placed 0.43 m from steel wire.
1. fracture occurs soon after the elastic limit is passed
Explanation: If the ultimate strength and fracture points are close, it means very small plastic range beyond elastic limit, the material is said to be brittle.
1. $0.14×{10}^{-2}$
Explanation: given for copper
cross section = 15.2mm $×$ 19.1mm
thus cross sectional area (A) = 15.2 $×$ 19.1 mm= 2.9 $×$ 10-4m2
restoring force tension (T) = 44500 N
young modulus of copper (y) = 11 $×$ 1010 pa
also y = $\frac{stress}{strain}⇒strain=\frac{stress}{y}$
$strain=\frac{44500}{2.9×{10}^{-4}×11×{10}^{10}}$$strain=0.1394×{10}^{-2}\approx 0.14×{10}^{-2}$

1. Poisson’s ratio is the ratio of transverse contraction strain to longitudinal extension strain in the direction of stretching force . It has no units.

2. Ductile materials: It is the ability of a material to withstand tensile force when it is applied upon it as it undergoes plastic deformation .For eg:- copper, Iron
Brittle materials: A material is brittle if, when subjected to stress, it breaks without significant plastic deformation. For eg:- Cast Iron, Glass.

3. $Y=\frac{FL}{A\mathrm{\Delta }L}$ a rigid body cannot be re-shaped by applying any deforming force.

$\therefore \mathrm{\Delta }L=0$

$Y=\frac{Fl}{A\left(0\right)}$= infinity i.e. for a perfect rigid body. Hence, Young’s modulus is infinity.

4. We prefer to have a spring made of steel because Young’s modulus of copper is less than that of the steel. As a result of the same shearing strain the stress i.e., the restoring force developed in the spring will be more and the spring will have more strength. In other words, we can say that in copper it does not rebound back to its original shape whereas steel comes back to its original shape.

5. Here, ${F}_{1}=3kgf=3×9.8N=29.4N$
$\mathrm{\Delta }$${l}_{1}=0.61mm=6.1×{10}^{-4}m$
${F}_{2}=5kgf=5×9.8=49N$