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Multiple Choice questions
Question 1
A wire whose cross-section area is A
_{1} is stretched by L
_{1} by a certain weight. How far will a wire of same material and same length and cross-section area A
_{2} stretch if same weight is applied to it
Solution
We know that
$\Delta L= \frac {FL}{AY}$
Or
$A\Delta L = \frac {FL}{Y}$
As F,L and Y are same So ,
$A\Delta L =constant$
$A_1 L_1 = A_2 L_2$
Or
$L_2 =\frac {A_1 L_1}{A_2}$
Question 2
A wire is stretched by x mm when a load F is hanged on it.If the same wire goes over a pulley and two weight F each are hung at the two end’s the what will be the elongation in the wire
(a) x
(b) 2x
(c) x/2
(d) 0
Solution
When wire is stretched by F
$Y =\frac {F/A}{x/L}$ --- (1)
When the wire goes over the pulley, the length on each side will be L/2 .And each side will experience the same elongation .Let it be y
Then
$Y=\frac {F/A}{y/5L}$ ---(2)
From equation 1 and 2
We get
y=x/2
Total elongation=2y=x
Linked Type Comprehensions
(A) One end of a uniform wire of length L and μ mass per unit length is attached rigidly to a point in a ceiling. A Mass M is suspended from its lower end. Area of cross-section of the wire is A.
Question 3
Find the stress at a distance x from the ceiling point
Solution
Weight of wire below the point =(L-x)μg +Mg
Stress =F/A
So $S= \frac {(L-x)\mu g + Mg}{A}$
Question 4
Find the elongation of the wire. If Young Modulus of wire is Y
Solution
Let us consider a small element of length dx at distance x from ceiling point.
Let dy be the elongation of that element due dx
We know from previous question that
Stress will be
$S= \frac {(L-x)\mu g + Mg}{A}$
Strain =dy/dx
Now
$Y = \frac {Stress}{Strain}$
$Y=\frac {[(L-x) \mu g + Mg]dx}{Ady}$
Or
$dx= \frac {[(L-x) \mu g + Mg]dx}{AY}$
Total elongation of the wire can be found by integrating from 0 to L
Multiple Choice questions
Question 5
A block of gelatin is 60 mm by 60 mm by 20 mm when unstressed. A force of .245 N is applied tangentially to the upper surface causing a 5mm displacement relative to the lower surface. The block is placed such that 60X60 comes on the lower and upper surface. Find the shearing stress, shearing strain and shear modulus
(a) (68.1 N/m
^{2},.25,272.4 N/m
^{2})
(b) (68 N/m
^{2},.25,272 N/m
^{2})
(c) (67 N/m
^{2},.26,270.4 N/m
^{2})
(d) (68.5 N/m
^{2},.27,272.4 N/m
^{2})
Solution
Shear stress=F/A=.245/36*10^{-4} ==68.1 N/m^{2}
Shear strain= tanθ= d/h=5/20=.25
Shear modulus (S) =shear stress/shear strain=272.4 N/m^{2}
Paragraph Based Questions
(B) A wire of radius r is stretched without sag and tension between two point separated by distance 2L.A weight is hanged at the middle of the wire which displaced the point by a distance d. Young modulus of elasticity is Y.
Question 6
Find the tension in the wire
Solution
Length of wire after weight is hanged
$=2\sqrt {L^2 + d^2} $
Length before hanging=2L
Change in length
$=2\sqrt {L^2 + d^2} -2L $
So
Strain = $ \frac {2\sqrt {L^2 + d^2} -2L }{2L}$
Or
$strain = \sqrt {1+ (\frac {d}{L})^2} -1$
Now
$Y =\frac {stress}{strain}$
Let F be the tension then
Question 7
If in previous question d<<<<L, what will be the value of Tension
Solution
$strain = \sqrt {1+ (\frac {d}{L})^2} -1$
$=(1+ \frac {d^2}{L^2} )^ 1/2 -1 $
Expanding bionomically and as d<<<L ,neglecting higher terms
$=(1+ \frac {d^2}{2L^2} -1 $
$=\frac {d^2}{2L^2}$
Then tension
$F= Y \pi r^2 \frac {d^2}{2L^2}$
$F=\frac {Y \pi r^2 d^2}{2L^2}$
Multiple Choice Questions
Question 8
A spring is stretched by applying a load to its free end. The strain produced in the spring is
(a) volumetric.
(b) shear.
(c) longitudinal and shear.
(d) longitudinal.
Solution
(c)
Question 9
A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillar. A mass m is suspended from the mid point of the wire.Strain in the wire is
(a) $\frac {x}{L}$
(b) $\frac {x^2}{2L^2}$
(c) $\frac {x^2}{2L}$
(d) $\frac {x^2}{L}$
Solution
Total Change in length
$\Delta L = 2\sqrt {L^2 + x^2} - 2L$
$= 2L (1 + \frac {x^2}{L^2})^{1/2} - 2L $
Expanding by Binomial theorem and ignoring high order term as x <<
$=2L ( 1 + \frac {x^2}{2L^2}) - 2L$
$= \frac {x^2}{L}$
Now Strain = $\frac {\Delta L}{2L} = \frac {x^2}{2L^2}$
Hence (b) is the correct option
Question 10
The stress versus strain graphs for wires of two materials A and B are as shown in the figure. If $Y_A$ and $Y_B$ are the Young 's modulii of the materials, then
(a) $Y_A = 2 Y_B$
(b) $Y_B= 3 Y_A$
(c) $Y_A= 3 Y_B$
(c) $Y_A= 2 Y_B$
Solution
$\frac {Y_A}{Y_B} = \frac {tan 60}{tan 30} = 3$
or $Y_A= 3 Y_B$
Question 11
The adjacent graph shows the extension ($\Delta L$) of a wire of length 1m suspended from the top of a roof at one end with a load W connected to the other end. If the cross sectional area of the wire is $10^{-6} \ m^2$, calculate the young?s modulus of the material of the wire
(a) $2 \times 10^{11} \ (N / m^2)$
(b) $2 \times 10^{-11} \ (N / m^2)$
(c) $2 \times 10^{-12} \ (N / m^2) $
(d) $2 \times 10^{-12} \ (N / m^2)$
Solution
(A) is the correct option
$l = 1 \times 10^{-4} \ m$
F = 20 N
$A = 10^{-6} \ m^2$
L = 1 m
$Y = \frac {FL }{ Al} = \frac {20 \times 1}{10^{-6} \times 10^{-4}} = 2 \times 10^{11} \ N/m^2$
Question 12
A wire is suspended from the ceiling and stretched under the action of a weight F suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight.
(a) Tensile stress at any cross section A of the wire is F/A.
(b) Tensile stress at any cross section is zero.
(c) Tensile stress at any cross section A of the wire is 2F/A.
(d) Tension at any cross section A of the wire is F
Solution
Answer is (a) and (d)
Question 13
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 W1 is suspended from its lower end. If S is the area of cross-section of the wire, the stress in the wire at a height 3L/4 from its lower end is
(a) $\frac {W_1}{S}$
(b) $\frac {W_1 + W/4}{S}$
(c) $\frac {W_1 + 3W/4}{S}$
(d) $\frac {W_1 + W}{S}$
Solution
Total force at height 3L/4 from its lower end = Weight suspended + Weight of 3/4 of the chain = $W_1+(3W/4)$
Hence stress is given by
$=\frac {W_1 + 3W/4}{S}$
Question 14
For an ideal liquid
(a) the bulk modulus is infinite.
(b) the bulk modulus is zero.
(c) the shear modulus is infinite.
(d) the shear modulus is zero.
Solution
(a) and (d)
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