A uniform wire of length l of weight w is suspended from the roof with a weight of W at the other end. The stress in the wire at 3l distance from the top is (AW+γ2Aw), where A is the cross sectional area of the wire. The value of γ is ________.
Numerical answer type. Work it out, then open the answer.
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Answer: 3
Tension at a point = load W + weight of the wire below that point.
At 3l from the top, the wire below is 32l long, weighing 32w.
One end of a metal wire is fixed to a ceiling and a load of 2 kg hangs from the other end. A similar wire is attached to the bottom of the load and another load of 1 kg hangs from this lower wire. Then the ratio of longitudinal strain of upper wire to that of the lower wire will be ________. [Area of cross section of wire =0.005 cm2,Y=2×1011 N m−2 and g=10 m s−2 ]
Numerical answer type. Work it out, then open the answer.
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Answer: 3
Upper wire holds both loads: T1=(2+1)×10=30N.
Lower wire holds only the 1 kg load: T2=1×10=10N.
Two blocks of mass 2 kg and 4 kg are connected by a metal wire going over a smooth pulley as shown in figure. The radius of wire is 4.0×10−5 m and Young's modulus of the metal is 2.0×1011 N/m2. The longitudinal strain developed in the wire is απ1. The value of α is _________. [Use g=10 m/s2]
Numerical answer type. Work it out, then open the answer.
Each of three blocks P,Q andR shown in figure has a mass of 3 kg. Each of the wires A and B has cross-sectional area 0.005 cm2 and Young's modulus 2×1011 N m−2. Neglecting friction, the longitudinal strain on wire B is ________ ×10−4. (Take g=10 m/s2)
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 2
Whole system: a=mP+mQ+mRmRg=930=310m/s2
Wire B pulls both P and Q: TB=(mP+mQ)a=6×310=20N
Let a wire be suspended from the ceiling (rigid support) and stretched by a weight W attached at its free end. The longitudinal stress at any point of cross-sectional area A of the wire is:
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Answer: (a)
Wire mass is neglected, so the tension is W at every point of the wire.
A metal block of mass m is suspended from a rigid support through a metal wire of diameter 14 mm. The tensile stress developed in the wire under equilibrium state is 7×105 N m−2. The value of mass m is _________ kg. (Take, g=9.8 m s−2 and π=722)
Numerical answer type. Work it out, then open the answer.
A steel rod has a radius of 20 mm and a length of 2.0 m. A force of 62.8 kN stretches it along its length. Young's modulus of steel is 2.0×1011 N/m2. The longitudinal strain produced in the wire is _____________ ×10−5
Numerical answer type. Work it out, then open the answer.
A block of mass M is attached to a wire. The upper part of wire is made of steel and lower part of wire of copper ( as shown in figure ). Both parts have the same area of cross section. Neglecting the mass of wires, the ratio of longitudinal strain developed in steel and copper wires is: ( Given, Young's modulus of steel =2×1011 N/m2 and Young's modulus of copper =1.5×1011 N/m2)
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Answer: (d)
Massless wire: both parts carry the same tension Mg and have the same area, so the same stress.
A string of area of cross-section 4 mm2 and length 0.5 m is connected with a rigid body of mass 2 kg. The body is rotated in a vertical circular path of radius 0.5 m. The body acquires a speed of 5 m/s at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is _________ ×10−5.
(use Young's modulus 1011 N/m2 and g=10 m/s2)
Numerical answer type. Work it out, then open the answer.
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 × 103 kg. The inner and outer radii of each column are 50 cm and 100 cm respectively. Assuming uniform local distribution, calculate the compression strain of each column. [Use Y = 2.0 × 1011 Pa, g = 9.8 m/s2]
Two wires A and B are stretched by the same load. The radius of wire A is double the radius of wire B. The stress on the wire B as compared to the stress on the wire A is
Two wires of different materials have same length L and same diameter d. The second wire is connected at the end of the first wire and forms one single wire of double the length. This wire is subjected to stretching force F to produce the elongation I. The two wires have
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Answer: (c)
In series the same force F acts through both wires, and both have the same area 4πd2.
Stress =AF is therefore the same in both.
Strain =Ystress differs because the materials have different Y.
So the answer is (c) same stress but different strain.
Two identical wires of substances ' P ' and ' Q ' are subjected to equal stretching force along the length. If the elongation of ' Q ' is more than that of ' P ', then
Show answer and solution
Answer: (b)
Same force on identical wires: ΔL=AYFL∝Y1.
ΔLQ>ΔLP⇒YP>YQ.
A larger Young's modulus means the material resists deformation more, i.e. it is more elastic.
A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g = 3.1π ms–2, what will be the tensile stress that would be developed in the wire?
The stress-strain curves are drawn for two different materials X and Y. It is observed that the ultimate strength point and the fracture point are close to each other for material X but are far apart for material Y.
We can say that materials X and Y are likely to be (respectively),
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Answer: (c)
Brittle material: fracture occurs soon after the ultimate strength point, so the two points are close (X).
Ductile material: large plastic deformation between ultimate strength and fracture, so the points are far apart (Y).
A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of:
Young's Modulus & Elongation · Comparing different materials
The two wires A and B of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire A and wire B is 20/11. When the joined wire is kept under certain tension the elongations in the wires A and B are equal. If the length of wire A is 2.2 m, then the length of wire B is
______ m.
Show answer and solution
Answer: (c)
Series wires carry the same tension T and have equal area: Δl=AYTL
Equal elongations give YALA=YBLB, so LB=LAYAYB
Young's Modulus & Elongation · Comparing different materials
Two wires A and B made of different materials of lengths 6.0 cm and 5.4 cm, respectively and area of cross sections 3.0×10−5 m2 and 4.5×10−5 m2, respectively are stretched by the same magnitude under a given load. The ratio of the Young's modulus of A to that of B is x:3. The value of x is ______
Young's Modulus & Elongation · Young's modulus & its nature
A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is P×1011 Nm−2, where the value of P is: (Take g=3π m/s2)
Young's Modulus & Elongation · Elongation: wires of the same material
Two wires A and B are made of same material having ratio of lengths LBLA=31 and their diameters ratio dBdA=2. If both the wires are stretched using same force, what would be the ratio of their respective elongations?
Young's Modulus & Elongation · Young's modulus & its nature
Young's modulus of material of a wire of length 'L' and cross-sectional area A is Y. If the length of the wire is doubled and cross-sectional area is halved then Young's modulus will be:
Show answer and solution
Answer: (d)
Young's modulus depends only on the material, not on length or area.
Young's Modulus & Elongation · Elongation: wires of the same material
A wire of length L and radius r is clamped at one end. If its other end is pulled by a force F, its length increases by l. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become:
Young's Modulus & Elongation · Young's modulus & its nature
The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's modulus, respectively, are 8×108 N m−2 and 2×1011 N m−2, is:
Show answer and solution
Answer: (a)
Maximum stress is the elastic limit, so maximum strain =2×10118×108=4×10−3
Young's Modulus & Elongation · Elongation: wires of the same material
Two metallic wires P and Q have same volume and are made up of same material. If their area of cross sections are in the ratio 4:1 and force F1 is applied to P, an extension of Δl is produced. The force which is required to produce same extension in Q is F2.
The value of F2F1 is _________.
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 16
Same volume and material: F=VYA2Δl∝A2 for the same Δl.
Young's Modulus & Elongation · Young's modulus & its nature
For a solid rod, the Young's modulus of elasticity is 3.2×1011 N m−2 and density is 8×103 kg m−3. The velocity of longitudinal wave in the rod will be.
Young's Modulus & Elongation · Young's modulus & its nature
A 100 m long wire having cross-sectional area 6.25×10−4 m2 and Young's modulus is 1010 N m−2 is subjected to a load of 250 N, then the elongation in the wire will be:
Young's Modulus & Elongation · Elongation: wires of the same material
A wire of length ' L ' and radius ' r ' is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by ' l '. Another wire of same material of length ' 2L ' and radius ' 2r ' is pulled by a force ' 2f '. Then the increase in its length will be:
Young's Modulus & Elongation · Elongation: wires of the same material
A force is applied to a steel wire 'A', rigidly clamped at one end. As a result elongation in the wire is 0.2 mm. If same force is applied to another steel wire ' B ' of double the length and a diameter 2.4 times that of the wire ' A ', the elongation in the wire ' B ' will be (wires having uniform circular cross sections)
Young's Modulus & Elongation · Comparing different materials
Young's moduli of the material of wires A and B are in the ratio of 1:4, while its area of cross sections are in the ratio of 1:3. If the same amount of load is applied to both the wires, the amount of elongation produced in the wires A and B will be in the ratio of
Young's Modulus & Elongation · Comparing different materials
The Young's modulus of a steel wire of length 6 m and cross-sectional area 3 mm2, is 2×1011 N/m2. The wire is suspended from its support on a given planet. A block of mass 4 kg is attached to the free end of the wire. The acceleration due to gravity on the planet is 41 of its value on the earth. The elongation of wire is (Take g on the earth =10 m/s2):
Young's Modulus & Elongation · Comparing different materials
Under the same load, wire A having length 5.0 m and cross section 2.5×10−5 m2 stretches uniformly by the same amount as another wire B of length 6.0 m and a cross section of 3.0×10−5 m2 stretches. The ratio of the Young's modulus of wire A to that of wire B will be:
Young's Modulus & Elongation · Elongation: wires of the same material
A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force F, its length increases by 5 cm. Another wire of the same material of length 4L and radius 4r is pulled by a force 4F under same conditions. The increase in length of this wire is __________________ cm.
Numerical answer type. Work it out, then open the answer.
Young's Modulus & Elongation · Comparing different materials
The elongation of a wire on the surface of the earth is 10−4 m. The same wire of same dimensions is elongated by 6 × 10−5 m on another planet. The acceleration due to gravity on the planet will be ____________ ms−2. (Take acceleration due to gravity on the surface of earth = 10 ms−2)
Numerical answer type. Work it out, then open the answer.
Young's Modulus & Elongation · Elongation: wires of the same material
A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be ________ cm.
Numerical answer type. Work it out, then open the answer.
Young's Modulus & Elongation · Elongation: wires of the same material
Two separate wires A and B are stretched by 2 mm and 4 mm respectively, when they are subjected to a force of 2 N. Assume that both the wires are made up of same material and the radius of wire B is 4 times that of the radius of wire A. The length of the wires A and B are in the ratio of a: b. Then a/b can be expressed as 1/x where x is _________.
Numerical answer type. Work it out, then open the answer.
Young's Modulus & Elongation · Young's modulus & its nature
A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changed to L1 when mass M is suspended from its free end. The expression for Young's modulus is:
Young's Modulus & Elongation · Elongation: wires of the same material
Two rods of same material and volume having circular cross-section are subjected to tension T. Within the elastic limit, same force is applied to both the rods. Diameter of the first rod is half of the second rod, then the extensions of first rod to second rod will be in the ratio
Young's Modulus & Elongation · Elongation: wires of the same material
A wire of length L and radius r is rigidly fixed at one end. On stretching the other end of the wire with a force F, the increase in length is l. If another wire of the same material but double the length and radius is stretched with a force 2F, then increase in length is
Young's Modulus & Elongation · Comparing different materials
Young's moduli of two wires A and B are in the ratio 7: 4. Wire A is 2 m long and has radius R. Wire B is 1.5 m long and has radius 2 mm. If the two wires stretch by the same length for a given load, then the value of R is close to :-
Young's Modulus & Elongation · Elongation: wires of the same material
Two wires are made of the same material and have the same volume. The first wire has cross- sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Δl on applying a force F, how much force is needed to stretch the second wire by the same amount?
Show answer and solution
Answer: (c)
Same volume: L=AV, so F=LYAΔl=VYA2Δl∝A2 for the same Δl.
Young's Modulus & Elongation · Comparing different materials
The Young's modulus of steel is twice that of brass. Two wires of same length and of 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
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Answer: (d)
Lower ends at the same level means equal extensions: AYsWsL=AYbWbL.
Young's Modulus & Elongation · Elongation: wires of the same material
Copper of fixed volume V is drawn into wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is Δl. Which of the following graphs is a straight line?
Show answer and solution
Answer: (b)
Fixed volume: A=lV
Δl=AYFl=VYFl2, so Δl∝l2
So the graph of Δl versus l2 is a straight line: answer (b).
Young's Modulus & Elongation · Elongation: wires of the same material
Two wires are made of the same material and have the same volume. However wire 1 has cross- sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Δx on applying force F, how much force is needed to stretch wire 2 by the same amount?
Show answer and solution
Answer: (c)
Same volume: L=AV, so F=LYAΔx=VYA2Δx∝A2 for the same Δx.
Wire Combinations & Special Cases · Wires joined end to end & loaded
A metal string A is suspended from a rigid support and its free end is attached to a block of mass M. Second block having mass 2 M is suspended at the bottom of the first block using a string B. The area of cross sections of strings A and B are same. The ratio of lengths of strings of A to B is 2 and the ratio of their Young's moduli (YA/YB) is 0.5. The ratio of elongations in A to B is ______.
Show answer and solution
Answer: (d)
Tension in A (holds both blocks) =3Mg; tension in B =2Mg.
Wire Combinations & Special Cases · Wires joined end to end & loaded
A string A of length 0.314 m and Young's modulus 2×1010 N/m2 is connected to another string B of length and Young's modulus both twice of those of A. This series combination of strings is then suspended from a rigid support and its free end is fixed to a load of mass 0.8 kg. The net change in length of the combination is ______ mm.
(radius of both the strings is 0.2 mm and acceleration due to gravity =10 m/s2) (Mass of both strings is to be neglected as compared to the mass of load)
Show answer and solution
Answer: (b)
Both strings carry F=0.8×10=8 N; A=π(0.2×10−3)2=1.256×10−7 m2
ΔlA=AYFL=1.256×10−7×2×10108×0.314=10−3 m=1 mm
For B, L and Y are both doubled, so YL is unchanged and ΔlB=1 mm.
Wire Combinations & Special Cases · Searle's experiment & errors
Figure represents the extension (Δl) of a wire of length 1 meter, suspended from the ceiling of the room at one end with a load W connected to the other end. If the cross-sectional area of the wire is 10−5 m2 then the Young's modulus of the wire is N/m2.
Wire Combinations & Special Cases · Natural length from two tensions
The length of a light string is 1.4 m when the tension on it is 5 N. If the tension increases to 7 N, the length of the string is 1.56 m. The original length of the string is__________m.
Numerical answer type. Work it out, then open the answer.
Wire Combinations & Special Cases · Own weight, pulleys & sag
Two persons pull a wire towards themselves. Each person exerts a force of 200 N on the wire. Young's modulus of the material of wire is 1×1011 N m−2. Original length of the wire is 2 m and the area of cross section is 2 cm2. The wire will extend in length by _________ μm.
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 20
Both ends pulled with 200 N, so the tension in the wire is T=200 N (not 400 N).
Wire Combinations & Special Cases · Natural length from two tensions
An elastic spring under tension of 3 N has a length a. Its length is b under tension 2 N. For its length (3a−2b), the value of tension will be _______ N.
Numerical answer type. Work it out, then open the answer.
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Answer: 5
a=L+3c and b=L+2c, so c=a−b and L=3b−2a
For length 3a−2b: Tc=(3a−2b)−L=(3a−2b)−(3b−2a)=5(a−b)
Wire Combinations & Special Cases · Force constant of a wire
A spring has length L and force constant K. It is cut into two springs of length L1 and L2 such that L1=NL2 ( N is an integer). The force constant of spring of length L1 is
Wire Combinations & Special Cases · Own weight, pulleys & sag
A wire of cross sectional area A, modulus of elasticity 2×1011 Nm−2 and length 2 m is stretched between two vertical rigid supports. When a mass of 2 kg is suspended at the middle it sags lower from its original position making angle θ=1001 radian on the points of support. The value of A is ________ ×10−4 m2 (consider x<<L ).
(given: g=10 m/s2)
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 1
At the middle point, the vertical parts of the two tensions hold the weight: 2Tsinθ=mg.
T=2θmg=2×0.0120=1000 N
Each half of length L becomes Lsecθ, so strain =secθ−1≈2θ2=5×10−5.
Wire Combinations & Special Cases · Searle's experiment & errors
Young's modulus is determined by the equation given by Y=49000lmcm2dyne where M is the mass and l is the extension of wire used in the experiment. Now error in Young modules (Y) is estimated by taking data from M−l plot in graph paper. The smallest scale divisions are 5 g and 0.02 cm along load axis and extension axis respectively. If the value of M and l are 500 g and 2 cm respectively then percentage error of Y is:
Wire Combinations & Special Cases · Wires joined end to end & loaded
Two wires each of radius 0.2 cm and negligible mass, one made of steel and the other made of brass are loaded as shown in the figure. The elongation of the steel wire is __________ × 10 −6 m. [Young's modulus for steel = 2 × 1011 Nm−2 and g = 10 ms−2 ]
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 20
Steel wire holds both blocks: T=(2+1.14)×10=31.4 N
Wire Combinations & Special Cases · Natural length from two tensions
The length of a wire becomes l1 and l2 when 100 N and 120 N tensions are applied respectively. If 10l2=11l1, the natural length of wire will be x1l1. Here the value of x is _____________.
Numerical answer type. Work it out, then open the answer.
Wire Combinations & Special Cases · Searle's experiment & errors
As shown in the figure, in an experiment to determine Young's modulus of a wire, the extension-load curve is plotted. The curve is a straight line passing through the origin and makes an angle of 45 ∘ with the load axis. The length of wire is 62.8 cm and its diameter is 4 mm. The Young's modulus is found to be x×104 Nm−2. The value of x is ___________.
Numerical answer type. Work it out, then open the answer.
Wire Combinations & Special Cases · Wires joined end to end & loaded
A steel wire of length 3.2 m(Ys=2.0×1011 N m−2) and a copper wire of length 4.4 m(Yc=1.1×1011 N m−2), both of radius 1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm. The load applied, in Newton, will be: ( Given π=722)
Show answer and solution
Answer: (d)
Same load F and area A=722×(1.4×10−3)2=6.16×10−6 m2
Wire Combinations & Special Cases · Own weight, pulleys & sag
A uniform heavy rod of mass 20 kg, cross sectional area 0.4 m2 and length 20 m is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is x×10−9 m. The value of x is _______________.
(Given, young modulus Y = 2 × 1011 Nm−2 and g = 10 ms−2)
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 25
Own-weight elongation: Δl=2AYWL with W=20×10=200 N
Wire Combinations & Special Cases · Natural length from two tensions
A wire of length L is hanging from a fixed support. The length changes to L1 and L2 when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to:
Wire Combinations & Special Cases · Searle's experiment & errors
In an experiment to determine the Young's modulus, steel wires of five different lengths (1,2,3,4, and 5 m) but of same cross section (2 mm2) were taken and curves between extension and load were obtained. The slope (extension/load) of the curves were plotted with the wire length and the following graph is obtained. If the Young's modulus of given steel wires is x×1011 Nm−2, then the value of x is __________.
Numerical answer type. Work it out, then open the answer.
Wire Combinations & Special Cases · Searle's experiment & errors
In an experiment to determine the Young's modulus of wire of a length exactly 1 m, the extension in the length of the wire is measured as 0.4 mm with an uncertainty of ±0.02 mm when a load of 1 kg is applied. The diameter of the wire is measured as 0.4 mm with an uncertainty of ±0.01 mm. The error in the measurement of Young's modulus (ΔY) is found to be x×1010 Nm−2. The value of x is _________________. (take g=10 ms−2)
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 2
Y=πd2l4MgL=π(0.4×10−3)2(0.4×10−3)4×10×1≈2×1011 N m−2
Wire Combinations & Special Cases · Wires joined end to end & loaded
Two wires of same length and radius are joined end to end and loaded. The Young's moduli of the materials of the two wires are Y1 and Y2. The combination behaves as a single wire then its Young's modulus is:
Show answer and solution
Answer: (b)
Same load F, area A, each of length L: Δl=AY1FL+AY2FL
Wire Combinations & Special Cases · Own weight, pulleys & sag
A uniform heavy rod of weight 10 kg ms−2, cross-sectional area 100 cm2 and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 × 1011 Nm−2. Neglecting the lateral contraction, find the elongation of rod due to its own weight.
Show answer and solution
Answer: (d)
A rod hanging under its own weight stretches by Δl=2AYWL (average tension is W/2).
Wire Combinations & Special Cases · Natural length from two tensions
The value of tension in a long thin metal wire has been changed from T1 to T2. The lengths of the metal wire at two different values of tension T1 and T2 are l1 and l2 respectively. The actual length of the metal wire is:
Wire Combinations & Special Cases · Force constant of a wire
An object of mass m is suspended at the end of a massless wire of length L and area of crosssection A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is:
Wire Combinations & Special Cases · Wires joined end to end & loaded
In an experiment, brass and steel wires of length 1 m each with areas of cross section 1mm2 are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is, [Given, the Young's Modulus for steel and brass are, respectively, 120 × 109 N/m2 and 60 × 109 N/m2]
Show answer and solution
Answer: (a)
Series wires with equal area have the same stress σ; Δl=σ(YsLs+YbLb)
Wire Combinations & Special Cases · Searle's experiment & errors
A load of mass M kg is suspended from a steel wire of length 2m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is:
Show answer and solution
Answer: (c)
In liquid the tension becomes Mg− buoyancy =Mg(1−ρρl)=Mg(1−82)=43Mg
Wire Combinations & Special Cases · Own weight, pulleys & sag
A uniformly tapering conical wire is made from a material of Young’s modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values R and 3 R, respectively. The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower end. The equilibrium extended length, of this wire, would equal:
Show answer and solution
Answer: (b)
Radius at distance x from the top: r=R(1+L2x); tension is Mg throughout.
ΔL=∫0Lπr2YMgdx=πYR2Mg∫0L(1+2x/L)2dx
∫0L(1+2x/L)2dx=2L(1−31)=3L (same as ΔL=πYR1R2MgL with R1R2=3R2)
Extended length =L(1+31πYR2Mg), so the answer is (b).
Wire Combinations & Special Cases · Searle's experiment & errors
A thin 1 m long rod has a radius of 5 mm. A force of 50 πkN is applied at one end to determine its Young’s modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01 mm, which of the following statements is false?
Show answer and solution
Answer: (c)
Stress =πr2F=π(5×10−3)250π×103=2×109 N/m2.
Smallest measurable extension is the least count 0.01 mm=10−5 m, so minimum strain =10−5/1=10−5.
Maximum Y=10−52×109=2×1014 N/m2, not 1014, so (c) is false.
Fractional errors: length 10−5 (smallest, so largest figure of merit), radius 2×50.01=4×10−3, extension largest. So (a), (b), (d) are true.
Wire Combinations & Special Cases · Wires joined end to end & loaded
If the ratio of diameters, lengths and Young's modulus of steel and copper wires shown in the figure are p, q and s respectively, then the corresponding ratio of increase in their lengths would be
Show answer and solution
Answer: (b)
Tension in steel (holds both blocks) =7mg; tension in copper =5mg.
Δl=(πd2/4)YTL, so ΔlcΔls=TcTs⋅LcLs⋅ds2dc2⋅YsYc
Wire Combinations & Special Cases · Wires joined end to end & loaded
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 moduli of copper and steel are respectively 1.0×1011 Nm −2 and 2.0×1011 Nm−2, the total extension of the composite wire is:
Show answer and solution
Answer: (d)
Same force and area: Δl∝YL
ΔlcΔls=1.00.5×2.0×10111.0×1011=41⇒Δls=0.25 mm
Wire Combinations & Special Cases · Wires joined end to end & loaded
If the ratio of lengths, radii and Young's moduli of steel and brass wires in the figure are a, b and c respectively, then the corresponding ratio of increase in their lengths is:
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Answer: (c)
Tension in steel (holds both masses) =3Mg; tension in brass =2Mg.
Wire Combinations & Special Cases · Own weight, pulleys & sag
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 are hung at the two ends, the elongation of the wire will be (in mm)
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Answer: (a)
With W hanging from a fixed support, the tension in the wire is W.
Over a pulley with W at each end, each side balances the other, so the tension is still W.
Same tension gives the same elongation: answer (a) l.
A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is 4×108 N/m2 and its radius is 4 mm, then maximum acceleration the lift can take is m/s2.
(take g=10 m/s2 and π=3.14)
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Answer: (a)
Maximum tension T=Sπr2=4×108×3.14×(4×10−3)2=20096 N.
Two wires as shown in the figure below, made of steel and have breaking stress of 12×108 N/m2. Area of cross-section of upper wire is 0.008 cm2 and of lower wire is 0.004 cm2. The maximum mass that can be added to pan without breaking any wire is ______ kg.
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Answer: (b)
Upper wire holds (30+10+m)g: (40+m)×10≤12×108×8×10−7=960 N⇒m≤56 kg.
Lower wire holds (10+m)g: (10+m)×10≤12×108×4×10−7=480 N⇒m≤38 kg.
Both wires must survive, so take the smaller value.
A brass wire of length 2 m and radius 1 mm at 27∘C is held taut between two rigid supports. Initially it was cooled to a temperature of −43∘C creating a tension T in the wire. The temperature to which the wire has to be cooled in order to increase the tension in it to 1.4T, is ∘C.
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Answer: (a)
Tension in a wire held between rigid supports: T=YAαΔθ, so T∝Δθ.
First cooling: Δθ=27−(−43)=70∘C gives T.
For 1.4T: Δθ=1.4×70=98∘C, so final temperature =27−98=−71∘C.
A metallic bar of Young's modulus, 0.5×1011 N m−2 and coefficient of linear thermal expansion 10−5∘C−1, length 1 m and area of cross-section 10−3 m2 is heated from 0∘C to 100∘C without expansion or bending. The compressive force developed in it is:
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Answer: (b)
Thermal force when expansion is prevented: F=YAαΔT.
The density and breaking stress of a wire are 6×104 kg/m3 and 1.2×108 N/m2 respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity is 31rd of the value on the surface of earth. The maximum length of the wire without breaking is _______ m (take, g=10 m/s2 ).
Numerical answer type. Work it out, then open the answer.
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Answer: 600
Wire breaks at the top under its own weight: ρLg′=S⇒L=ρg′S.
A thin rod having a length of 1 m and area of cross-section 3×10−6 m2 is suspended vertically from one end. The rod is cooled from 210∘C to 160∘C. After cooling, a mass M is attached at the lower end of the rod such that the length of rod again becomes 1 m. Young's modulus and coefficient of linear expansion of the rod are 2×1011 N m−2 and 2×10−5 K−1, respectively. The value of M is __________ kg.
(Take g=10 m s−2)
Numerical answer type. Work it out, then open the answer.
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Answer: 60
Contraction on cooling: ΔL=LαΔT=1×2×10−5×50=10−3 m.
The load must stretch it back by the same amount: Mg=LYAΔL
The area of cross section of the rope used to lift a load by a crane is 2.5×10−4 m2. The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be:
(take g=10 ms−2)
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Answer: (a)
Breaking stress is fixed, so the required area is proportional to the load: A∝W.
A metal wire of length 0.5 m and cross-sectional area 10−4 m2 has breaking stress 5×108 Nm−2. A block of 10 kg is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be _________ ms−1.
Numerical answer type. Work it out, then open the answer.
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Answer: 50
Maximum tension T=SA=5×108×10−4=5×104 N.
Tension gives the centripetal force: rmv2=T⇒v2=mTr=105×104×0.5=2500
A steel rod with y = 2.0 × 1011 Nm−2 and α = 10−5 ∘C−1 of length 4 m and area of cross-section 10 cm2 is heated from 0∘ C to 400∘C without being allowed to extend. The tension produced in the rod is x × 105 N where the value of x is ____________.
Numerical answer type. Work it out, then open the answer.
Two blocks of masses 3 kg and 5 kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is π24×102 Nm-2. What is the minimum radius of the wire? (Take g = 10 ms-2)
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Answer: (c)
Tension in an Atwood machine: T=m1+m22m1m2g=82×3×5×10=37.5 N.
Wires W1 and W2 are made of same material having the breaking stress of 1.25 × 109 N/m2. W1 and W2 have cross-sectional area of 8 × 10−7 m2 and 4 × 10−7 m2, respectively. Masses of 20 kg and 10 kg hang from them as shown in the figure. The maximum mass that can be placed in the pan without breaking the wires is ____________ kg. (Use g = 10 m/s2)
Numerical answer type. Work it out, then open the answer.
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Answer: 40
W1 holds (20+10+m)g: (30+m)×10≤1.25×109×8×10−7=1000 N⇒m≤70 kg.
W2 holds (10+m)g: (10+m)×10≤1.25×109×4×10−7=500 N⇒m≤40 kg.
Both wires must survive, so take the smaller value.
A body of mass m = 10 kg is attached to one end of a wire of length 0.3 m. The maximum angular speed (in rad s–1) with which it can be rotated about its other end in space station is: (Breaking stress of wire = 4.8 × 107 Nm–2 and
area of cross-section of the wire = 10–2 cm2) is:
Numerical answer type. Work it out, then open the answer.
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Answer: 4
In a space station there is no weight; tension only provides the centripetal force: T=mω2L.
Maximum tension T=4.8×107×10−6=48 N (as 10−2 cm2=10−6 m2).
The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit?
A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to:
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Answer: (a)
Length unchanged, so the compressive stress equals the thermal stress: πr2F=YαT.
A lift is tied with thick iron ropes having mass ' M '. The maximum acceleration of the lift is ' a ' m/s2 and maximum safe stress is ' S ' N/m2. The minimum diameter of the rope is
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Answer: (b)
Maximum tension in the rope (lift accelerating up): T=M(g+a).
The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100∘C is: (For steel Young's modulus is 2×1011 Nm−2 and coefficient of thermal expansion is 1.1×10−5 K−1)
Elastic Potential Energy · Work done & energy stored
A copper wire of length 3 m is stretched by 3 mm by applying an external force. The volume of the wire is 600×10−6 m3. The elastic potential energy stored in the wire in stretched condition would be ______ J.
(Given Young modulus of copper =1.1×1011 N/m2)
Numerical answer type. Work it out, then open the answer.
A steel wire of length 2 m and Young's modulus 2.0×1011 N m−2 is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10−3 respectively, then the elastic potential energy density of the wire is __________ ×105 (in SI units).
Numerical answer type. Work it out, then open the answer.
The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length 100 cm to stretch it by 1 mm is (if Young's modulus of the wire =2.0×1011 Nm−2):
Elastic Potential Energy · Work done & energy stored
The elastic potential energy stored in a steel wire of length 20 m stretched through 2 cm is 80 J. The cross sectional area of the wire is __________ mm2.
( Given, Y=2.0×1011 Nm−2)
Numerical answer type. Work it out, then open the answer.
The elastic behaviour of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of 5 × 10−4 is __________ kJ/m3. Assume that material is elastic upto the linear strain of 5 × 10−4.
Numerical answer type. Work it out, then open the answer.
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Answer: 25
From the graph, strain 10−10 at stress 20 N/m2, so Y=10−1020=2×1011 N/m2.
Elastic Potential Energy · Catapult: energy to speed
A stone of mass 20 g is projected from a rubber catapult of length 0.1 m and area of cross section 10−6 m2 stretched by an amount 0.04 m. The velocity of the projected stone is ______________ m/s. (Young's modulus of rubber = 0.5 × 109 N/m2)
Numerical answer type. Work it out, then open the answer.
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Answer: 20
Elastic energy becomes kinetic energy: 2LYAx2=21mv2.
Elastic Potential Energy · Work done & energy stored
A metal rod has length, cross-sectional area and Young's modulus as L,A and Y, respectively. If the elongation in the rod produced is ℓ, then work done is proportional to
Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is 1: 4, the ratio of their diameters is:
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Answer: (d)
u=2YS2 with S=πd2/4F, so for the same load and material u∝d41.
Elastic Potential Energy · Work done & energy stored
A wire of length ' L ' and area of cross section ' A ' is made of material of Young's modulus ' Y. It is stretched by an amount ' x '. The work done in stretching the wire is
Elastic Potential Energy · Work done & energy stored
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:
Elastic Potential Energy · Work done & energy stored
Work done in stretching a wire through 1 mm is 2 J. What amount of work will be done for elongating another wire of same material, with half the length and double the radius of cross section, by 1 mm?
Elastic Potential Energy · Catapult: energy to speed
A boy's catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross-section and of negligible mass. The boy keeps a stone weighing 0.02kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms–1. Neglect the change in the area of cross-section of the cord while stretched. The Young's modulus of rubber is closest to:
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Answer: (b)
Energy stored in the cord becomes kinetic energy: 21mv2=21×0.02×400=4 J.
Stored energy =2LYAx2, with A=π(3×10−3)2=2.83×10−5 m2.
Y=2.83×10−5×(0.2)22×4×0.42≈3×106 N m−2
This is of order 106. So the answer is (b) 106 N m−2.
Elastic Potential Energy · Work done & energy stored
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 certain gas is isothermally compressed to (31)rd of its initial volume (Vo=3 litre) by applying required pressure. If the bulk modulus of the gas is 3×105 N/m2, the magnitude of work done on the gas is ______ J.
Numerical answer type. Work it out, then open the answer.
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Answer: 989
For an isothermal process the bulk modulus equals the pressure: B=P, so initial P0=3×105Pa.
The increase in pressure required to decrease the volume of a water sample by 0.2% is P×105 Nm−2. Bulk modulus of water is 2.15×109 Nm−2. The value of P is _________.
Numerical answer type. Work it out, then open the answer.
A sample of a liquid is kept at 1 atm. It is compressed to 5 atm which leads to a change of volume of 0.8 cm3. If the bulk modulus of the liquid is 2 GPa, the initial volume of the liquid was _______ litre. (Take 1 atm = 105 Pa)
Numerical answer type. Work it out, then open the answer.
Bulk Modulus & Compressibility · Fractional compression at depth
The fractional compression (VΔV) of water at the depth of 2.5 km below the sea level is __________ %. Given, the Bulk modulus of water = 2×109 N m−2, density of water = 103 kg m−3, acceleration due to gravity g=10 m s−2.
Bulk Modulus & Compressibility · Fractional compression at depth
If average depth of an ocean is 4000 m and the bulk modulus of water is 2×109 Nm−2, then fractional compression VΔV of water at the bottom of ocean is α×10−2. The value of α is _______ (Given, g=10 ms−2,ρ=1000 kg m−3)
Numerical answer type. Work it out, then open the answer.
Under isothermal condition, the pressure of a gas is given by P=aV−3, where a is a constant and V is the volume of the gas. The bulk modulus at constant temperature is equal to
A certain pressure 'P' is applied to 1 litre of water and 2 litre of a liquid separately. Water gets compressed to 0.01% whereas the liquid gets compressed to 0.03%. The ratio of Bulk modulus of water to that of the liquid is x3. The value of x is ____________.
Numerical answer type. Work it out, then open the answer.
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Answer: 1
B=ΔV/VP; same P, so B∝ΔV/V1 (the volumes themselves do not matter).
An object is located at 2 km beneath the surface of the water. If the fractional compression VΔV is 1.36%, the ratio of hydraulic stress to the corresponding hydraulic strain will be ____________. [Given: density of water is 1000 kgm−3 and g = 9.8 ms−2]
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Answer: (b)
Hydraulic stress at depth: P=ρgh=1000×9.8×2000=1.96×107Pa.
Bulk Modulus & Compressibility · Fractional compression at depth
When a rubber ball is taken to a depth of __________m in deep sea, its volume decreases by 0.5%. (The bulk modulus of rubber = 9.8 × 108 Nm−2, Density of sea water = 103 kgm−3, g = 9.8 m/s2)
Numerical answer type. Work it out, then open the answer.
Bulk Modulus & Compressibility · Density of compressed matter
The normal density of a material is ρ and its bulk modulus of elasticity is K. The magnitude of increase in density of material, when a pressure P is applied uniformly on all sides, will be:
Bulk Modulus & Compressibility · Density of compressed matter
The density of a metal at normal pressure p is ρ. When it is subjected to an excess pressure, the density becomes ρ′. If K is the bulk modulus of the metal, then the ratio ρρ′ is
A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, (rdr) is:
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Answer: (d)
Extra pressure in the liquid (Pascal's law): P=amg, acting on the sphere from all sides.
A bottle has an opening of radius a and length b. A cork of length b and radius (a + Δa) where (Δa < < a) is compressed to fit into the opening completely (See figure). If the bulk modulus of cork is B and frictional coefficient between the bottle and cork is μ then the force needed to push the cork into the bottle is:
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Answer: (d)
Only the radius changes (length b fixed): V=πr2b⇒VΔV=a2Δa.
Pressure on the cork: P=BVΔV=a2BΔa
Normal force on the curved surface: N=P×2πab=4πBbΔa
Bulk Modulus & Compressibility · Fractional compression at depth
The approximate depth of an ocean is 2700 m. The compressiblity of water is 45.4 × 10−11 Pa−1 and density of water is 103 kg/m3. What fractional compression of water will be obtained at the bottom of the ocean?
The Bulk moduli of Ethanol, Mercury and water are given as 0.9, 25 and 2.2 respectively in units of 109 Nm−2. For a given value of pressure, the fractional compression in volume is VΔV. Which of the following statements about VΔV for these three liquids is correct?
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Answer: (a)
Fractional compression VΔV=BP, so for the same P, VΔV∝B1.
B: ethanol 0.9 < water 2.2 < mercury 25 (in 109N m−2).
Smallest B compresses most: ethanol > water > mercury.
A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of 105 N at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement θ of the rod axis from its original position would be: (shear moduli, G=1010 N/m2)
Two slabs with square cross section of different materials (1,2) with equal sides (l) and thickness d1 and d2 such that d2=2d1 and l>d2. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is θ2=2θ1. If the shear moduli of material 1 is 4×109 N/m2, then shear moduli of material 2 is x×109 N/m2, where value of x is ________.
Numerical answer type. Work it out, then open the answer.
A square aluminum (shear modulus is 25×109 Nm−2) slab of side 60 cm and thickness 15 cm is subjected to a shearing force (on its narrow face) of 18.0×104 N. The lower edge is riveted to the floor. The displacement of the upper edge is ____________ μm.
Numerical answer type. Work it out, then open the answer.
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Answer: 48
Force acts on the narrow face: A=0.60×0.15=0.09m2.
Shear Modulus & Poisson's Ratio · Relations between elastic constants
If Y, K and η are the values of Young's modulus, bulk modulus and modulus of rigidity of any material respectively. Choose the correct relation for these parameters.
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Answer: (d)
Standard relation: Y9=K1+η3 (equivalently Y=3K+η9Kη).
As shown in the figure, forces of 105 N each are applied in opposite directions, on the upper and lower faces of a cube of side 10 cm, shifting the upper face parallel to itself by 0.5 cm. If the side of another cube of the same material is 20 cm, then under similar conditions as above, the displacement will be:
A uniform wire (Young's modulus 2×1011 Nm −2) is subjected to longitudinal tensile stress of 5×107 Nm−2. If the overall volume change in the wire is 0.02%, the fractional decrease in the radius of the wire is close to: