00 Elasticity Mechanical Properties of olids tress and train. When a weight of 0kg is suspended fro a copper wire of length 3 and diaeter 0.4. Its length increases by.4c. If the diaeter of the wire is doubled, then the extension in its length will be a) 7.6 c b) 4.8 c c). c d) 0.6 c. A force of 008 6 0 N 6 required for breaking a aterial. The density ρ of the 3 3 aterial is3 0 kg. If the wire is to beak under its own weight, the length of the wire ade of that aterial should be (taking g = 0s ) a) 0 b) 00 c) 00 d) 000 3. The breaking force for a wire of diaeter D of a aterial is F. The breaking force 00 for a wire of the sae aterial of radius D is a) F b) F c) 4 F Young s, Bulk and Rigidity Modulus 4. The Young s odulus of brass and steel are respectively d) 4F.0 0 N and.0 0 N. A brass wire and a steel wire of the sae length are extended by each under the sae force. If radii of brass and steel wires are RB and R respectively, then

a) R = R b) B RB R = c) R = 4RB d) R = R B 5. Figure shows a unifor rod of length 30cn having a ass of 3.0kg. The strings shown in the figure are pulled by constant forces of 0N and 3N. Find the force exerted by 0c part of the rod on the 0c part. All the surfaces are sooth and the strings and pulleys are light a) 36N b) N c) 64N d) 4N 6. A particle of ass is attached to three identical assless springs of spring constant k as shown in the figure. The tie period of vertical oscillation of the particle is a) π b) π c) π d) π k k 3k k 7. For a given aterial, the young s odulus is.4 ties that of rigidity odulus, then Poisson s ratio is a) 0. b) 0.4 c). d).4 8. One-fourth length of a spring of force constant k is cut away. The force constant of the reaining spring will be a) 3 4 k b) 4 k c) k d) 4k 3 008 9. The adjacent graph shows the extension ( Δl) of a wire of length suspended fro the top of a roof at one end with a load W connected to the other end. If the cross-

6 sectional area of the wire is 0, calculate the Young s odulus of the aterial of the wire a) 0 N 007 b) 0 N c) 3 0 N d) 0 N 3 0. The Young s odulus of the aterial of the wire of length L and radius r is Y N. If the length is reduced to L/ and radius r/, the Young s odulus will be a) Y/ b) Y c) Y d) 4Y. An iron rod of length and cross-sectional area of 50 is stretched by 0.5, when a ass of 50kg is hung fro its lower end. Young s odulus of iron rod is 0 a) 9.6 0 N 8 b) 9.6 0 N 0 c) 9.6 0 N d) 9.6 0 N 5 006. When a sphere is taken to botto of sea k deep, it contracts by 0.0%. The bulk odulus of elasticity of the aterial of sphere is (given, density of water = gc 3 ) a) 9.8 0 N 0 b) 0. 0 N 0 c) 0.98 0 N 0 d) 8.4 0 N 0 003 3. The Young s odulus of a wire of length (L) and radius (r) is Y. If the length is L r reduced to and radius to, then its Young s odulus will be a) Y b) Y c) Y d) 4Y

Work Done in tretching Wire 008 4. When a etal wire elongates by hanging a load Mg on it, the gravitational potential energy of ass M decreases by Mgl. This energy appears 007 a) As elastic potential energy copletely b) As theral energy copletely c) Half as elastic potential energy and half as theral energy d) As kinetic energy of the load copletely 5. If the tension on a wire is reoved at once, then a) It will break b) Its teperature will reduce c) There will be no change in its teperature d) Its teperature increases 6. A wire is suspended by one end. At the other end a weight equivalent to 0N force is applied. If the increase in length is, the increase in the energy of the wire will be 006 a) 0.0J b) 0.0J c) 0.04J d).00j 7. Young s odulus of the aterial of a wire is Y. On a pulling the wire by a force F, the increase in its length is x. The potential energy of the stretched wire is a) Fx b) Yx c) Fx d) None of these 8. What will be energy stored in a strained wire a) load x extension b) stress x strain c) load x strain d) load x stress

Poisson s Ratio and Theral tress 008 9. When a rod is heated but prevented fro expanding, the stress developed is independent of a) Material of the rod b) Rise in teperature c) Length of rod d) None of these 004 0. Poisson s ratio cannot have the value a) 0. b) 0.7 c) 0. d) 0.5 Key ) d ) b 3) d 4) b 5) d 6) b 7) a 8) b 9) a 0) b ) c ) a 3) b 4) c 5) d 6) a 7) a 8) a 9) c 0) b Hints tress and train. l r l r = = l r l.4 4 4 l = = l = 0.6c. tress Force g V ρ g LAρ g = = = = Area A A A tress = Lρ g

tress = 6 6 0 N 3 3 ; ρ = 3 0 kg ; g = 0s stress L = ρ g 6 6 0 = = 0 = 00 3 3 0 0 Force Force 3. tress = = Area π r Young s Bulk and Rigidity Modulus FL FL 4. Δ L = = YA Yπ R YR = Constant R = R B 5. Net force on the rod, f = 3 0 = N 6. Acceleration f = = = 4s 3 F 0 = x a = x 4 F = 4 + 0 = 4N iilarly, 3 F = a = 4 F = 3 8 = 4N F = F + F cos 45 + F cos 45 0 0 net A B C 0 0 = ky + k( y cos45 )cos45 = ky = ky + ky cos 45 0 Also, F = k y k y = ky net k = k

T = π = π k k 7. Δd / d Y σ = = Δ L / L η 8. 9. But,, Y =.4η.4η σ = = 0. η k l 4 k = k. 3 l 4 = 0, F = 0N ; FL 0 Y = = 6 4 Al 0 0 0. Concept. longitudinal stress Mgl Y = = longitudinal strain Al L =, L = 0.5 A 6 = 0, L = = 0 N A = 50 = 50 0 50 9.8 = 50 0 0.5 0 6 3 = 0.5 0, M = 50kg Y 6 3. Bulk odulus Δ v 0.0 = = 0 V 00 ΔpV. B = Δ v 4 Δ p = hρ g = Δ p = 9.8 0 9.8 0 B = 4 0 3. Concept 6 3 3 0 0 9.8 6 0 = 9.8 0 N = 9.6 0 N 0

Work Done in tretching Wire 4. Decrease in potential energy = Mgl Elastic potential energy stored in stretched wire = Mgl Energy appearing as heat 5. Concept 6. Increase in energy 7. U = stress strain U = F x U = Fx 8. Concept 0 0 = Mgl Mgl = Mgl 3 = = 0.0J