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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of `30^@`. Then angle of shear isA. `12^(@)`B. `0.12^(@)`C. `1.2^(@)`D. `0.012^(@)` |
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Answer» Correct Answer - B |
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| 2. |
When a weight of `10 kg` is suspended from a copper wire of length 3 metres and diameters `0.4 mm`, its length increases by `2.4 cm`. If the diameter of the wire is doubled, then the extension in its length will beA. `9.6 cm`B. `4.8 cm`C. `1.2 cm`D. `0.6 cm` |
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Answer» Correct Answer - D |
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| 3. |
In a wire of length L, the increase in its length is `l`. If the length is reduced to half, the increase in its length will beA. Reduced by halfB. Increase by halfC. Remains sameD. None of these |
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Answer» Correct Answer - B (a)Here, L=10 cm `=10xx10^(-2) m` F=100 kgf`=100xx9.8 N` `therefore`Shearing stress `=(F)/(L^(2))=(100xx9.8)/((10xx10^(-2))^(2))=9.8xx10^(4) Nm^(-2)` |
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| 4. |
A wire of length 2.5 m has a percentage strain of 0.012% under a tensile force. The extension produce in the wire will beA. 0.03 mmB. 0.3 mmC. 0.3 mmD. 0.03 m |
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Answer» Correct Answer - B (b) Here, original length, L=2.5 m ltrbgt Strain `=(DeltaL)/(L)=0.012% =(0.012)/(100)` `therefore DeltaL`= Strain `xx` L or `DeltaL`= Extension `=(0.012)/(100)xxL` `=(0.012xx2.5)/(100)=3xx10^(-4) m =0.3 mm` |
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| 5. |
Two bars A and B are stuck using an adhesive. The contact surface of the bars make an angle q with the length. Area of cross section of each bar is `S_(0)`. It is known that the adhesive yields if normal stress at the contact surface exceeds `sigma_(0)`. Find the maximum pulling force F that can be applied without detaching the bars. |
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Answer» Correct Answer - `(sigma_(0)S_(0))/(sin^(2) theta)` |
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| 6. |
Find the greatest length of steel wire that can hang vertically without breaking.Breaking stress of steel `=8.0xx10^(8) N//m^(2)`. Density of steel `=8.0xx10^(3) kg//m^(3)`. Take `g =10 m//s^(2)`. |
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Answer» Correct Answer - `1.02xx10^(4)m` |
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| 7. |
The breaking stress of a wire is `7.2xx10^(8)N//m^(2)` and density is `78xx10^(2) kg//m^(3)`. The wire is held vertically to the rigid support. The maximum length so that the wire does not break isA. `9.94xx10^(3)m`B. `9.4xx10^(3)m`C. `94xx10^(3)`mD. `940xx10^(3)`m |
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Answer» Correct Answer - B Breaking stress `=rho L g` `therefore L=(7.2xx10^(8))/(7.8xx10^(3)xx10)=9.4xx10^(3)m.` |
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| 8. |
Elasticity is the property of material body by virtue of which a bodyA. occupies minimum surface areaB. is in equilibriumC. opposes its deformationD. attracts other bodies |
| Answer» Correct Answer - C | |
| 9. |
The property of a body by virtue of which it changes its dimensions permanently is calledA. plasticityB. elasticityC. rigidityD. None of these |
| Answer» Correct Answer - A | |
| 10. |
The nature of elastic forces by virtue of which a body possess the property of elasticity isA. electromagneticB. gravitationalC. weakD. all of the above |
| Answer» Correct Answer - A | |
| 11. |
When strain is produced in a body within elastic limit, its internal energyA. Remains constantB. DecreasesC. IncreasesD. None of the above |
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Answer» Correct Answer - C |
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| 12. |
Breaking stress does not depend uponA. area of cross-section of the wireB. the dimensions of wireC. the material of the wireD. all of these |
| Answer» Correct Answer - B | |
| 13. |
The graph between applied force and change in the length of wire within elastic limit is aA. straight line with positive slopeB. straight line with negative slopeC. curve with positive slopeD. curve with negative slope |
| Answer» Correct Answer - A | |
| 14. |
Under elastic limit the stess isA. independent of strainB. square root of the strainC. directly proportional to strainD. indirectly proportional to strain |
| Answer» Correct Answer - C | |
| 15. |
The metal which breaks immediately after elastic limit isA. ductile metalB. reptile metalC. brittle metalD. elastic metal |
| Answer» Correct Answer - C | |
| 16. |
A body of mass m = 0 kg is attached to a wire of length 0.3 m. Calculate the maximum angular velcoity with wicch it can be rotated in a horizontal circle (Breaking stress of wire `= 4.8 x 10^(7)N//m^(2)` and area of cross-section of wire = `10^(-6)m^(2)`)A. `4 rad//s`B. `8 rad// s`C. `1 rad// s`D. `2 rad// s` |
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Answer» Correct Answer - A (a) Breaking strength=tension in the wire =`mromega^(2)` `4.8xx10^(7)xx10^(-6)=10xx0.3xxomega^(2)` `omega^(2)=(48)/(0.3 xx 10)=16 implies omega=4` rad/s |
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| 17. |
A rubber cord catapult has cross-sectional area `25 mm^(2)` and initial length of rubber cord is `10 cm`. It is stretched to `5 cm`. And then released to protect a missle of mass `5 gm` Taking `Y_("nibber")=5xx10^(7)N//m^(2)` velocity of projected missle isA. `20 ms^(-1)`B. `100 ms^(-1)`C. `250ms^(-1)`D. `200 ms^(-1)` |
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Answer» Correct Answer - C |
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| 18. |
A rubber cord catapult has cross-sectional area `25 mm^(2)` and initial length of rubber cord is `10 cm`. It is stretched to `5 cm`. And then released to protect a missle of mass `5 gm` Taking `Y_("nibber")=5xx10^(7)N//m^(2)` velocity of projected missle isA. ` 20 ms^(-1)`B. `100 ms^(-1)`C. ` 250 ms^(-1)`D. ` 200 ms^(-1)` |
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Answer» Correct Answer - C (c) Elastic potential energy stored in the catapult `(U=(1)/(2)FDeltaL=(1)/(2)(YA)/(L)DeltaL^(2))` is converted into the kinetic energy of the projectile `(K=(1)/(2)mv^(2))implies v=sqrt((YADeltaL^(2))/(Lm))` `=[((5xx10^(8))xx(25xx10^(-6))xx(5xx 10^(-2))^(2))/((10xx10^(-2))xx(5xx10^(-3)))]^(1//2)=250 ms^(-1)` |
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| 19. |
An alumminium rod and steel wire of same length and cross-section are attached end to end. Then compound wire is hung fron a rigid support and load is suspended from the free end. Y for steel is `((20)/(7))` times of aluminium. The ratio of increase in length of steel wire to the aluminium wire isA. `7:10`B. `20:7`C. `10:7`D. `7:20` |
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Answer» Correct Answer - D `(Y_(L))/(Y_(s))=(l_(s))/(l_(L))` |
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| 20. |
The longitudinal extention of any elastic material is very small.In order to have an appericiable change ,the material must be in the form ofA. short and thin wireB. thick block with any cross sectionC. a long thin wireD. breaking stress must be very small |
| Answer» Correct Answer - C | |
| 21. |
Consider a solid cube which is subjected to a pressure of `6xx10^(5) Nm^(-2)` . Due to this pressure, the each side of the cube is shortened by 2% . Find out the volumetric strain of the cube. |
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Answer» Let L be the initial length of the each side of the cube. Volumn, `V=LxxLxxL=L^(3)`=Initial volume (`V_(i)` say) If each side of the cube is shortened by 2% ,then final length of the cube `=L-2%` of L `=(L-(2L)/(100))=L(1-(2)/(100))` `therefore` Final volume, `V_(f)=L^(3)(1-(2)/(100))^(3)=V=(1-(2)/(100))^(3)` Change in volume, `DeltaV=V_(f)-V_(i)` `=V(1-(2)/(100))^(3)-V=V[(1-(2)/(100))^(3)-1]` `(DeltaV)/(V)=(1-(2)/(100))^(3)-1~= [1-(2xx3)/(100)]-1` `" "[because (1-x)^(n)~=1-nx` for `x lt lt 1`] therefore Volumetric strain, `(DeltaV)/(V)=1-0.06-1=0.06` `" "`[take positive sign] |
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| 22. |
The material which practically does not exhibit elastic after effect isA. CopperB. RubberC. SteelD. Quartz |
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Answer» Correct Answer - D |
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| 23. |
Modulus of rigidity of a liquidA. Non-zero constantB. InfiniteC. ZeroD. Cannot be predicted |
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Answer» Correct Answer - B (b) Elongation `Deltal=(FL)/(AY)therefore Deltal prop (1)/(A)` [F,L and Y are constant] or `(A_(2))/(A_(1))=(Deltal_(1))/(Deltal_(2))` (Area of cross-section of second wire) `implies A_(2)=A_(1)((0.1)/(0.05))=2A_(1)=2xx4=8 mm^(2)` |
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| 24. |
Modulus of rigidity of a liquidA. zeroB. oneC. infiniteD. non zero values |
| Answer» Correct Answer - A | |
| 25. |
The shearing strain produced in a block of metal subjected to a shearing stress of `10^(8)N//m^(2)` is ( Modulus of rigidity `eta=8 xx 10^(10)N//m^(2)`)A. `1.1xx10^(-3)`B. `1.5xx10^(-3)`C. `1.25 xx 10^(-3)`D. `1.6xxO^(-3)` |
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Answer» Correct Answer - C `eta=("shearing stress")/("shearing strain")` `"shearing strain"= ("shearing stress")/(eta)` `=(10^(8))/(8xx10^(10))=1.25xx10^(-3)`. |
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| 26. |
The dimensional formula for the modulus of rigidity isA. `[L^(-1)M^(1)T^(2)]`B. `[L^(1)M^(1)T^(-2)]`C. `[L^(-1)M^(1)T^(-2)]`D. `[L^(-2)M^(-1)T^(2)]` |
| Answer» Correct Answer - C | |
| 27. |
Modulus of rigidity of diamond isA. Too lessB. Greater than all mattersC. Less than all mattersD. Zero |
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Answer» Correct Answer - B |
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| 28. |
An elastic spring of unstretched length L and spring constant K is stretched by a small length x. It is further stretched by another small length y. the work done in second stretcing isA. `(1)/(2) ky^(2)`B. `(1)/(2) K (x^(2) + y^(2))`C. `(1)/(2 Ky (2x + y))`D. `(1)/(2) Ky (2x + y)` |
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Answer» Correct Answer - d P.E stored in the spring when it is stretched by a length x is `U_(1) = (1)/(2) Kx^(2)` when it is stretched by a further distance y, the P.E `(1)/(2) K (x + y)^(2)` `:.` work done `= U_(2) - U_(1) = (1)/(2) K [(x + y)^(2) - x^(2)]` `= (1)/(2) k [x^(2) + y^(2) + 2xy - x^(2)]` `W = (1)/(2) Ky (2x + y)` |
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| 29. |
There is a circular tube in a vertical plane. Two liquids which do not mix and of densities `d_1 and d_2` are filled in the tube. Each liquid subtends `90^@` angle at centre. Radius joining their interface make an angle `alpha` with vertical. Ratio `(d_1)/(d_2)` is : A. `(1+tanalpha)/(1-tanlpha)`B. `(1+sinalpha)/(1-cosalpha)`C. `(1+sinalpha)/(1-sinalpha)`D. `(1+cosalpha)/(1-cosalpha)` |
| Answer» Correct Answer - A | |
| 30. |
A spherical solild of volume V is made of a material of density `rho_(1)`. It is falling through a liquid of density `rho_(2)(rho_(2) lt rho_(1))`. Assume that the liquid applies a viscous froce on the ball that is proportional ti the its speed v, i.e., `F_(viscous)=-kv^(2)(kgt0)`. The terminal speed of the ball isA. `sqrt((Vg(rho_(1)-rho_(2))/(k))`B. `(Vgrho_(1))/(k)`C. `sqrt((Vgrho_(1))/(k))`D. `(V(rho_(1)-rho_(2)))/(k)` |
| Answer» Correct Answer - A | |
| 31. |
A uniform wire having mass per unit length `lambda` is placed over a liquid surface. The wire causes the liquid to depress by `y(y lt lt a)` as shown in figure. Find surface tension of liquid. Neglect end effect. |
| Answer» Correct Answer - `(lamdaa)/(2y)` | |
| 32. |
A simple pendulum made of bob of mass `m` and a metallic wire of negligible mass has time period `2 s` at `T = 0^(@)C`. If the temperature of the wire is increased and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope `S`. If the coefficient of linear expansion of metal is `alpha` then value of `S` is :A. `(1)/(alpha)`B. `alpha`C. `(alpha)/(2)`D. `2alpha` |
| Answer» Correct Answer - C | |
| 33. |
Two immiscible of densities `2.5g//cm^(3)` and `0.8g//cm^(3)` are taken in the ratio of their masses as `2:3` respectively find the average density of the liquid conbination. |
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Answer» Let masses be 2m g & 3m, g, then `V=V_(1)+V_(2)=((2m)/(2.5)+(3m)/(0.8))cm^(3)` total mass `=2m+3m+5mg` therefore the average desity `rho=(5m)/(V)=(5m)/((2m)/(2.5)+(3m)/(0.8))=(5)/((2)/(2.5)+(3)/(0.8))=(10)/(9.1)gm//cm^(3)=1.09gm//cm^(3)` |
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| 34. |
A hollow metallic sphere has inner and outer radii, respectively as 5 cm and 10 cm if the mass of the sphere is 2.5 kg, find (a) density of the material, (b) relative density of the material of the sphere. |
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Answer» The volume of the material of the sphere is `V=((4)/(3))pi(r_(2)^(3)-r_(1)^(3))=(4)/(3)xx3.14xx[((10)/(100))^(3)-((5)/(100))^(3)]=(4)/(3)xx3.14xx[0.001-0.000125]` `=(4)/(3)xx3.14xx0.000875m^(3)=0.00367m^(3)` (a) therefore density of material of the sphere is `rho=(M)/(V)=(2.5)/(0.00367)kg//m^(3)=682kg//m^(3)` (b). Relative density of the material of the sphere `rho=(681.2)/(1000)=0.6812` |
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| 35. |
During blood transfusion the needle is inserted in a veinn where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein? [density of hole blood `=1.06xx10^(3)kgm^(-3)`] (a). 0.192 (b). 0.182 (c). 0.172 (d). 0.162 |
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Answer» Correct Answer - A Pressure `P=hrhogimpliesh=(2000)/(1.06xx10^(3)xx9.8)=0.192m` |
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| 36. |
when a metal wire elongates by hanging a load Mg on it , the gravitational poyential energy of mass M decreases by Mgl. This energy appearsA. as elastic potential energy completelyB. as thermal energy completelyC. half as potential energy and half as thermal energyD. as kinetic energy of the load completely |
| Answer» Correct Answer - C | |
| 37. |
A stress of `10^(6) N//m^(2)` is required for breaking a material. If the density of the material is `3 xx 10^(3) Kg//m^(3)`, then what should be the minimum length of the wire made of the same material so that it breaks by its own weight `(g = 10m//s^(2))`A. 66.6 mB. 60.0 mC. 33.3 mD. 30.0 m |
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Answer» Correct Answer - C Breakign stress `=(mg)/("Area") =(Al rho g)/(A)=rho l g` `therefore l=("Breaking stress")/(rho g)` `=(10^(6))/(3xx10^(3)xx10)=33.33m` |
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| 38. |
A stress of `10^(6) N//m^(2)` is required for breaking a material. If the density of the material is `3 xx 10^(3) Kg//m^(3)`, then what should be the minimum length of the wire made of the same material so that it breaks by its own weight `(g = 10m//s^(2))`A. 10 mB. 33.3 mC. 5 mD. 66.6 m |
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Answer» Correct Answer - B (b) Maximum breaking stress `=(F)/(A)=(mg)/(A)` or `((Alrho)g)/(A)=sigma_("max")` `therefore l_("max")=(sigma_("max"))/(rhog)=(10^(6))/((3xx10^(3))(10))=33.33 m` |
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| 39. |
A stress of `10^(6) Nm^(-2)` is required for breaking a material. If the density of the material is `3 xx 10^(3) kg m^(-3)`., then what should be the length of the wire made of this material, so that it breakes under its own weight?A. 17 mB. 34 mC. 22 mD. 43 m |
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Answer» Correct Answer - B Breading stress `=rho l g` `therefore l=("Breading stress")/(rho g)` `=(10^(6))/(3xx110^(3)xx10)=(100)/(3)=33.3 m`. |
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| 40. |
The breaking stress of a wire depends onA. material of wireB. length of wireC. radius of wireD. shape of cross-section |
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Answer» Correct Answer - a Breaking stress for a wire depends only on material. |
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| 41. |
Estimate the change in the density of water in ocean at a depth of 400 m below the surface. The density of water at the surface `=1030kgm^-3` and the bulk modulus of water `=2x10^9Nm^-2`.A. `0.5 kg //m^(3)`B. `0.5 kg//m^(3)`C. `1.5 kg//m^(3)`D. `2.0kg//m^(3)` |
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Answer» Correct Answer - D `k=(rho dp)/(d rho)` `therefore d rho=(1000 xx h rho g)/(2xx10^(9)) = (10^(3) xx 4 xx 10^(2) xx 10^(3) xx 10)/(2xx10^(9)) = 2` |
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| 42. |
Two parallel forces of 4000 N are applied tangentially in opposite direction to the opposite faces of a metallic cube of side length 0.25 m. The shear modulus of the material is `80xx10^(9)` Pa. Then the displacement of the upper surface relative to the lower surface isA. `2xx10^(-5)m `B. `2xx10^(-6)m `C. `2xx10^(-7)m`D. `2xx10^(-8)m ` |
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Answer» Correct Answer - C `x=(Fh)/(A eta) = (4000 xx 0.25)/(625xx10^(-4) xx 80xx10^(9)) = 2xx10^(-7)m.` |
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| 43. |
There are two wires of same material. Their radii and lengths are both in the ratio `1:2` If the extensions produced are equal, the ratio of the loads isA. `1:2`B. `4:1`C. `2:1`D. `1:4` |
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Answer» Correct Answer - A `(F_(1))/(F_(2))=(y r_(1)^(2)l)/(L_(1))xx(L_(2))/(y r_(2)^(2)l)=(1)/(4) xx (2)/(1) = 1:2`. |
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| 44. |
3 m long copper wire is stretched to increase its length by 0.3 cm. The lateral strain produced in the wire is,A. `0.26xx10^(-2)`B. `0.26xx10^(-3)`C. `0.26 xx10^(-4)`D. `0.26xx10^(-5)` |
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Answer» Correct Answer - B `sigma=("Lateral strain")/("Longitudinal strain")` `therefore " Lateral strain"=sigma xx "Longitudinal strain" ` `=0.26 xx (0.3)/(300) = 0.26 xx 10^(-3)` |
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| 45. |
Steel is preferred for making springs over copper becauseA. steel is cheaperB. steel does not react with atmosphereC. elasticity of steel is moreD. steel also has magnetic property |
| Answer» Correct Answer - C | |
| 46. |
A steel plate of face area `4 cm^2` and thickness 0.5 cm is fixed rigidly at the lower surface. A tangential force of 10 N is applied on the upper surface. Find the lateral displacement of the upper surface with respect to the lower surface. Rigidity modulus of steel `=8.4xx10^10Nm^-2`.A. `1.5xx10^(-6)m`B. `1.5xx10^(-7)m `C. `1.5xx10^(-8)m`D. `1.5xx10^(-9)m` |
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Answer» Correct Answer - D `eta =(Fh)/(Ax)` `x=(Fh)/(Aeta)=(10xx 0.5)/(4xx10^(-4)xx8.4xx10^(10)) =(5xx10^(-6))/(4xx8.4)` `=1.5xx10^(-7) m`. |
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| 47. |
Steel is preferred for making springs over copper becauseA. it is coventionalB. steel is easy available than copperC. steel is less elastic than copperD. steel is more elastic than copper |
| Answer» Correct Answer - D | |
| 48. |
A tangential force 2100 N is applied on a surface area `3xx10^(-6)m^(2)` which is 0.1 m from a fixed face. If the force produces a shift of `7xx10^(-3)` m of the upper surface with respect to the bottom. Then the modulus of rigidity of the material will be,A. `10^(8)N//m^(2)`B. `10^(9)N//m^(2)`C. `10^(10)N//m^(2)`D. `10^(11)N//m^(2)` |
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Answer» Correct Answer - C `h=(F)/(A theta)=(Fh)/(Ax) = (2100 xx 0.1)/(3xx10^(-6)xx7xx10^(-3))` `=10^(10)N//m^(2)` |
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| 49. |
The applied force produces changes in dimensions is aA. deformationB. contractionC. formationD. alteration |
| Answer» Correct Answer - A | |
| 50. |
The force which responds to produced deformation is a body isA. deforming forceB. forceC. restoring forceD. intermolecular force |
| Answer» Correct Answer - A | |