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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.
401. |
Prove that if two bubbles of radii `r_(1)` and `r_(2)` coalesce isothermally in vacuum then the radius of new bubble will be `r=sqrt(r_(1)^(2)+r_(2)^(2))` |
Answer» When two bubbles coalesce then number of molecules of air will remains constant and temperature also constant so `n_(1)+n_(2)=nimpliesP_(1)V_(1)+P_(2)V_(2)=Pvimplies(4T)/(r_(1))((4)/(3)pir_(1)^(3))+(4T)/(r_(2))((4)/(3)pir_(2)^(3))=(4T)/(r)((4)/(3)pir^(3))impliesr=sqrt(r_(1)^(2)+r_(2)^(2))` |
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402. |
A space `2.5cm` wide between two large plane surfaces is filled with oil. Force required to drag a very thin plate of area `0.5m^2` just midway the surfaces at a speed of `0.5(m)/(sec)` is `1N`. The coefficient of viscosity in `kg-(s)/(m^2)` is:A. `5xx10^(-2)`B. `2.5xx10^(-2)`C. `1xx10^(-2)`D. `7.5xx10^(-2)` |
Answer» Correct Answer - B | |
403. |
Calculate the height to which water will rise in a capillary tube of diameter `1xx10^(-3)`m [given surface tension of water is `0.072Nm^(-1)` angle of contact is `0^(@),g=9.8ms^(-2)` and density of water `=1000kgm^(-3)`] |
Answer» Height of capillar rise `h=(2Tcostheta)/(r rhog)=(2xx0.072xxcos0^(0))/(5xx10^(-4)xx1000xx9.8)m=2.94xx10^(-2)m` | |
404. |
A thin horizontal disc of radius R=10cm is located with in a cylindrical cavity filled with oil whose viscosity `eta=0.08`P (figure) The distance between the disc and the horizontal planes of the cavity is equal to `h=1.0` mm find the power developed by the viscous forces acting ont he disc when it rotates with the angular velocity `omega=60rad//s`. The end effect are to be neglected. |
Answer» Correct Answer - 9 W | |
405. |
There is a 1mm thick layer of glycerine between a flat plate of area `100cm^(2)` and and a big plate. If te coefficient of viscosity of glycerine is `1.0kg//m-sec`, then how much force is required to move the plate with a velocity of 7 cm/sec. |
Answer» Correct Answer - 0.7 N | |
406. |
Statement-1: In gravity free space, the liqid in a capillary tube will rise to infinite height. Statement-2: In the absence of gravity there will be no force to prevent the rise of liquid due to surface tension.A. Statement-1, is true, Statement-2 is true, Statement-2 is correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explantion for statement-1C. statement-1 is true, statement-2 is falseD. statement-1 is false statement-2 is true |
Answer» Correct Answer - A | |
407. |
Statement-1: Machine parts are jammed in winter. Statement-2: The viscosity of lubricant used in machine part decrease at low temperature.A. Statement-1, is true, Statement-2 is true, Statement-2 is correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explantion for statement-1C. statement-1 is true, statement-2 is falseD. statement-1 is false statement-2 is true |
Answer» Correct Answer - C | |
408. |
Statement-1: The angle of contact of a liquid decrease with increase in temeperature Statement-2: With increase in temperature the surface tension of liquid increase.A. Statement-1, is true, Statement-2 is true, Statement-2 is correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explantion for statement-1C. statement-1 is true, statement-2 is falseD. statement-1 is false statement-2 is true |
Answer» Correct Answer - C | |
409. |
Assertion : Bulk modulus of elasticity ( K ) represents incompressibility of the material. Reason : Bulk modulus of elasticity is proportional to change in pressure.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is fa lse.D. If the assertion and reason both are false. |
Answer» Correct Answer - A |
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410. |
Wires `A` and `B` are made from the same material. `A` has twice the diameter and three times the length of `B`. If the elastic limits are not reached, when each is stretched by the same tension, the ratio of energy stored in `A` to that in `B` isA. `2:3`B. `3:4`C. `3:2`D. `6:1` |
Answer» Correct Answer - B |
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411. |
As an air bubble rises from the bottom of a lake to the surface, its volume is doubled. Find the depth of the lake. Take atmospheric pressure = 76 cm of Hg. |
Answer» Correct Answer - 10,34m | |
412. |
A steel rod of cross-sectional area `16 cm^(2)` and two brass rods each of cross-sectional area `10 cm^(2)` together support a load of `5000 kg ` as shown in the figure. ( Given, `Y_(steel) = 2xx10^(6) kg cm^(-2) and Y_(brass) = 10 ^(6) kg cm^(-2))`. Choose the correct option(s). A. Stress in brass rod`=121 kg cm^(-2)`B. Stress in steel rod`=161 kg cm^(-2)`C. Stress in brass rod `=141 kg cm^(-2)`D. Stress in steel rod `=141 kg cm^(-2)` |
Answer» Correct Answer - A::B Area of steel rod , `A_(S) = 16 cm^(2) ` ltbr. Area of two brass rods , `A_(B) = 2xx10 = 20 cm^(2)` `F = 5000 kg` ` sigma_(S) =` Stress in steel and ` sigma_(B) =` Stress in brass Decrease in length of steel rod = Decrease in length of brass rod ` (sigma_(S))/(Y_(S)).L_(S) = (sigma_(B))/(Y_(B)).L_(B)` `rArr sigma_(S) = Y_(S)/(Y_(B)).(L_(B))/(L_(S)). sigma_(B)` or ` sigma_(S) = ((2xx10^(6))/10^(6))(20/30)sigma_(B)` `sigma_(S) = 4/3sigma_(B)` ...(i) Now, ` F = sigma_(S)A_(S) +sigma_(B)A_(B)` or `5000 = sigma_(S)xx16+sigma_(B)xx20` ...(ii) From Eqs. (i) and (ii) , we get `sigma _(B) = 120.9 kg cm^(-2)` ` sigma_(S) = 161.2 kg cm^(-2)` |
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413. |
Two wires of same diameter of the same material having the length `l` and `2l` If the force `F` is applied on each, the ratio of the work done in two wires will beA. `1 : 2`B. `1 : 4`C. `2 : 1`D. `1 : 1` |
Answer» Correct Answer - a `W = (1)/(2) F I` `:. W prop I` `:. (W_(1))/(W_(2)) = (I_(1))/(I_(2)) = (I)/(2I) = (1)/(2)` |
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414. |
A steel wire of length 20 cm and uniform cross-sectional area of `1 mm^(2)` is tied rigidly at both the ends at `45^(@)C`. If the temperature of the wire is decreased to `20^(@)C`, then the change in the tension of the wire will be [Y for steel `= 2 xx 10^(11 Nm^(-2)`, the coefficient of linear expansion for steel `= 1.1 xx 10^(-5)//.^(@)CC^(-1)`]A. 22 NB. 32 NC. 55 ND. 60 N |
Answer» Correct Answer - c Coefficient of linear expansion `alpha = (dL)/(L(Delta T))` `:. (Delta L)/(L) = alpha Delta T` and `Y = (F.L)/(A.Delta L)` From Eq. 1 `F = YA (Delta L)/(L) = YA alpha Delta T` `:. F = 2 xx 10^(11) xx 1 xx 10^(-6) xx 1.1 xx 10^(-5) xx 25` `:. F = 55 N` |
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415. |
A steel wire of length 20 cm and uniform cross-sectional `1 mm^(2)` is tied rigidly at both the ends. The temperature of the wire is altered from `40^(@)C` to `20^(@)C`. Coefficient of linear expansion of steel is `alpha = 1.1 xx 10^(-5) .^(@)C^(-1)` and Y for steel is `2.0 xx 10^(11) Nm^(2)`, the tension in the wire isA. `2.2 xx 10^(8) N`B. 16 NC. 8 ND. 44N |
Answer» Correct Answer - d Increases in length due to rise in temperature `Delta L = a L Delta T` As,` Y = (FL)/(A Delta L)` so, `F = (YA Delta L)/(L) = (YA xx a L Delta T)/(L) = YA a Delta T` `:. F = 2 xx 10^(11) xx 10^(-6) xx 1.1 xx 10^(-5) xx 20 = 44N` |
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416. |
Two wires, one made of capper and other of steel are joined end to end. (as show in figure). The area of cross-section of copper wire is twice that of steel wire, They ae placed under compressive force of magnitude F. Find the ratio of their lenghts such that change in lenght of both wire are same `(Y_(S)= 2 xx 10 ^(11) N//m^(2) and Y_(C) = 1.1 xx 10 ^(11) N//m^(2))` A. 2.1B. 1.1C. 1.2D. 2 |
Answer» Correct Answer - B `Y_(s)= (FL_(s))/(A_(s)Deltal_(s)) and Y_(C)= (FL_(c))/(A_(c)Deltal_(c))` `therefore " "(L_(c))/(L_(s))= (((Y_(c)A_(c)Deltal_(c)))/(F))/(((Y_(s)A_(s)Deltal_(s))/(F)))=((Y_(c))/(Y_(s)))(A_(C)/(A_(s)))((Deltal_(c))/(Deltal_(s)))` ` Here (A_(C))/(A_(S))= 2, (Deltal_(c))/(Deltal_(s))=1, (Y_(C))/(Y_(S))= (1.1)/(2)` `therefore" " (L_(C))/(L_(s))=((1.1)/(2))(2)(1)=1.1` |
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417. |
If the volume of the given mass of a gas is increased four times, the temperature is raised from `27^(@)C` to `127^(@)C`. The elasticity will becomeA. 4 timesB. `1//4` timesC. 3 timesD. `1//3` times |
Answer» Correct Answer - D |
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418. |
A spherical ball contracts in volume by `0.02%` when subjected to a normal uniform pressure of 200 atmospheres. Then Bulk modulus (in `N//m^(2)`) of the material of the ball is (Atomospheric pressure `=10^(5)N//m^(2)`)A. `10^(9)`B. `10^(10)`C. `10^(11)`D. `10^(12)` |
Answer» Correct Answer - B `K=(dp)/((dv)/(v))=(200xx10^(5))/(2xx10^(-3))=100xx10^(8) =10^(10)`. |
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419. |
A spherical ball contracts in volume by 0.01% when subjected to a normal uniform pressure of 100 atmospheres. Calculate the bulk modulus of the meterial.A. `10xx10^(12)"dyne"//cm^(2)`B. `1.096 xx 10^(11) N//m^(2)`C. `0.196xx10^(11)N//m^(2)`D. `10.96 xx 10^(11) N//m^(2)` |
Answer» Correct Answer - C `K=(P)/(Delta v//v)=(100xx1.01xx10^(6))/(0.01//100)` `=10^(12)"dyne"//cm^(2)` |
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420. |
When a sphere is taken to bottom of sea 1 km deep, it contracts by 0.01%. The bulk modulus of elasticity of the material of sphere is (Given density of water = 1 g `cm^(-3)`)A. `9.8 xx 10^(10) Nm^(-2)`B. `10.2 xx 10^(10) Nm^(-2)`C. `0.98 xx 10^(10) Nm^(-2)`D. `8.4 xx 10^(10) Nm^(-2)` |
Answer» Correct Answer - a Bulk modulus , B `= (Delta p)/((-Delta V)/(V)) (-Delta V)/(V) = (0.01)/(100) = 10^(-4)` `Delta p` = pressure of water `= h rho g = 10^(3) xx 1 xx 10^(3) xx 9.8` `Delta p = 9.8 xx 10^(6)` `B = (9.8 xx 10^(6))/(10^(-4)) = 9.8 xx 10^(10) Nm^(-2)` |
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421. |
When a pressure of 100 atmosphere is applied on a spherical ball, then its volume reduces to `0.01%`. The bulk modulus of the material of the rubber in dyne/`cm^2` isA. `10xx10^(12)`B. `100xx10^(12)`C. `1xx10^(12)`D. `20xx10^(12)` |
Answer» Correct Answer - C (c) `because` Bulk modulus of the material of the rubber `B=(Deltap)/(DeltaV//V)=(100)/(0.01//100)=10^(6) "atm" =10^(11) Nm^(-2)` `=10^(12)` dyne `cm^(-2)` |
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422. |
Coefficient of isothermal elasticity `E_(theta)` and coefficient of adiabatic elasticity `E_(phi)` are related by `(gamma = C_(p) //C_(v))`A. `E_(theta) = gamma E_(phi)`B. `E_(phi) = gamma E_(theta)`C. `E_(theta) = gamma//E_(phi)`D. `E_(theta) = gamma^(2)E_(phi)` |
Answer» Correct Answer - B |
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423. |
A ball falling in a lake of depth 200 m shows 0.1% decrease in its volume at the bottom. What is the bulk moduls |
Answer» Correct Answer - A | |
424. |
A uniform cube is subjected to volume compression. If each side is decreased by 2% , then bulk strain isA. 0.02B. 0.03C. 0.04D. 0.06 |
Answer» Correct Answer - D (d) We have `V=l^(3)` `(DeltaV)/(V)=(3Deltal)/(l)=3((2)/(100))=0.06` |
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425. |
The bulk modulus of water is `2.1xx10^(9) Nm^(-2)`. The pressure required ot increase the density of water by 0.1% isA. `2.1 xx 10^(3) Nm^(-2)`B. `2.1 xx 10^(6) Nm^(-2)`C. `2.1 xx 10^(5) Nm^(-2)`D. `2.1 xx 10^(7) Nm^(-2)` |
Answer» Correct Answer - B (b) We have`drho=(rho)/(B).dp` or `(drho)/(rho)=(dp)/(B)implies(0.1)/(100)=(dp)/(2.1xx10^(9))` `therefore` The pressure required to increase the density `dp=2.1xx10^(6) Nm^(-2)` |
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426. |
If the volume of a block os aluminium is decreased be `1%` the pressure (stress) on is surface is increased by (Bulk moduals) of `Al=7.5xx10^(10)Nm^(-2))`A. `7.5 xx 10^(10) Nm^(-2)`B. `7.5 xx 10^(8) Nm^(-2)`C. `7.5 xx 10^(6) Nm^(-2)`D. `7.5 xx 10^(4) Nm^(-2)` |
Answer» Correct Answer - b Given, `(Delta V)/(V) xx 100 = 1% = (1)/(100)` Bulk modulus, `B = (P)/(Delta V //V) = (pV)/(Delta V)` or `P = (B Delta V)/(V) = 7.5 xx 10^(10) xx (1)/(100)` `= 7.5 xx 10^(8) Nm^(-2)` |
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427. |
When the load on a wire is increased slowly from 1 kg wt to 2 kg wt the elongation increases from 0.2 mm to 0.3 mm. How much work is done during the extension ? `(g=9.8m//s^(2))`A. `1.96xx10^(-3)J`B. `19.6xx10^(-3)J`C. `0.196xx10^(-3)J`D. `16xx10^(-3)J` |
Answer» Correct Answer - A `W_(2)-W_(1)=(1)/(2)(F_(2)l_(2)-F_(1)l_(1))` `=(1)/(2) g[l_(2)m_(2)-m_(1)l_(1)]` `=(1)/(2) g[2xx0.3 xx 10^(-3)-1xx0.2 xx 10^(-3)]` `=(1)/(2) xx 9.8 xx 10^(-3) [ 0.6 -0.2]` `=4.9xx10^(-3) [ 0.4]` `=19.6 xx10^(-3)J`. |
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428. |
A 5m aluminium wire `(Y = 7 xx 10^(10) Nm^(-2)`) of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire `(Y = 12 xx 10^(10) Nm^(-2)`) of the same length under the same weight, the diameter (in mm) should beA. 1.75B. 1.5C. 2.3D. 5 |
Answer» Correct Answer - c `I = (FL)/(pi r^(2) Y) implies r^(2) prop (1)/(Y)` (F,L and I are constants) `(r_(2))/(r_(1)) = [(Y_(1))/(Y_(2))]^(1//2) = [(7 xx 10^(10))/(12 xx 10^(10))]^(1//2)` `implies r_(2) = 1.5 xx ((7)/(12))^(1//2) = 1.145 mm` `:.` Diameter = 2.29 mm. |
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429. |
A 5 m long aluminium wire `(Y=7xx10^10(N)/(m^2))` of diameter 3 mm supprts a 40 kg mass. In order to have the same elongation in a copper wire `(Y=12xx10^10(N)/(m^2))` of the same length under the same weight, the diameter should now, in mmA. 1.75B. 1.5C. 2.5D. `5.0` |
Answer» Correct Answer - C |
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430. |
Find out longitudinal stress and tangential stress on a fixed block. |
Answer» Longitudinal or normal stress `sigma_(1)=(100sin30^(@))/(5xx2)=5N//m^(2)` Tangential stress `sigma_(2)=(100cos30^(0))/(5xx2)=5sqrt(3)m//s^(2)` |
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431. |
A rod has a radius of 100 mm and a length of 10 cm. A 100 N force compresses along its length . Calculate the longitudinal stress developed in the rod. |
Answer» Given, radius of the rod, r 100 mm `=100xx10^(-3)m,` Length of the rod, `l=10 cm =10xx10^(-2)m` and force, `F_(n)=100 N` Longitude stress (developed in the rod) `(F_(n))/(A)=(100)/(pi(100xx10^(-3))^(2))` `=3183 Nm^(-2)` |
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432. |
When composite rod is free, then composite length increases to 2.002 m for temperature `20^(@)C` to `120^(@)C` when composite rod is fixed between the support there is no change in component length find y and `alpha` of steel, if `y_(c)=1.5xx10^(13)N//m^(2)alpha_(c)=1.6xx10^(-5)//.^(@)C`. |
Answer» `Deltal=l_(S)alpha_(S)DeltaT+l_(C)alpha_(C)DeltaTimplies0.002=[1.5alpha_(S)+0.5xx1.6xx10^(-5)]xx100` `impliesalpha_(S)=(1.2xx10^(-5))/(1.5)=8xx10^(-6)// .^(@)C` there is no change in compoenent length |
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433. |
`[L^(-1)M^(1)T^(-2)]` are the dimensions ofA. modulus of elasticityB. modulus of rigidityC. constant of elasticityD. all of the above |
Answer» Correct Answer - D | |
434. |
The quality of the material which opposes the changes in shape, volumer or length is calledA. Intermolecular repulsionB. Intermolecular behaviourC. ViscosityD. Elasticity |
Answer» Correct Answer - D |
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435. |
A rod of length `l` and radius `r` is joined to a rod of length `l//2` and radius `r//2` of same material. The free end of small rod is fixed to a rigid base and the free end of larger rod is given a twist of `theta^(@)`, the twist angle at the joint will beA. `theta//4`B. `theta//2`C. `5 theta//6`D. `8 theta//9` |
Answer» Correct Answer - D |
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436. |
The point on the stress-strain curve at which the strain begins to increase even without any increase in the stress isA. elastic limit pointB. breaking pointC. plasticity limit pointD. yield point |
Answer» Correct Answer - D | |
437. |
Square meter per newton is the SI unit ofA. Bulk modulusB. compressibilityC. stressD. strain |
Answer» Correct Answer - B | |
438. |
The volume stress in a body is equal toA. change in areaB. change in pressureC. change in volumeD. change is length |
Answer» Correct Answer - B | |
439. |
The force required to punch a hole of diameter 2 mm, will be (If a shearing stress `4xx10^(8)N//m^(2)`.)A. `400pi N`B. `1600 pi N`C. `1800pi N `D. `1200pi N` |
Answer» Correct Answer - A Stress `=(F)/(A)` `therefore F= "Stress" xx pi r^(2)` |
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440. |
Shearing stress causes change inA. LengthB. BreadthC. ShapeD. Volume |
Answer» Correct Answer - C |
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441. |
SI units of stress isA. `N//m^(2)`B. `Nm^(2)`C. `m^(2)//N`D. `N^(2)m` |
Answer» Correct Answer - A | |
442. |
SHEARING STRESS AND TANGENTIAL STRESSA. shape and size of a bodyB. length of a bodyC. area of a bodyD. volume of a body |
Answer» Correct Answer - A | |
443. |
The tensile stress is related to the term change inA. shape of a bodyB. length of a bodyC. size of a bodyD. volume of a body |
Answer» Correct Answer - B | |
444. |
What is the dimensional formula of tensile stress?A. `[L^(-1)M^(1)T^(-2)]`B. `[L^(1)M^(-2)T^(-1)]`C. `[L^(-2)M^(1)T^(-1)]`D. `[L^(-1)M^(-2)T^(1)]` |
Answer» Correct Answer - A | |
445. |
A cylindrical steel wire of `3 m` length is to stretch no more than `0.2 cm` When a tensile force of `400 N` is applied to each end of the wire ? What minimum diameter is required for the wire ?? `Y_(steel) = 2.1xx10^(11) N//m^2` |
Answer» Correct Answer - A::C `Deltal = (Fl)/(AY) = (Fl)/((pid^(2)//4)Y)` `:. d = sqrt((4 Fl)/(pi(Deltal)Y))` ` = sqrt((4xx400xx3)/(3.14xx0.2xx10^(-2)xx2.1xx10^(11)))` `= 1.91 xx10^(-3) m = 1.91 mm` |
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446. |
The length of a wire is increased by `1 mm` on the application of a given load. If a wire of the same material, but of length and radius twice that of the first, on application of the same load, extension isA. 2 mmB. 0.5 mmC. 4 mmD. 0.25 mm |
Answer» Correct Answer - B `Y=(F)/(pi r^(2)) xx (L)/(l_(1))` `therefore l_(1)=(FL_(1))/(pi r_(1)^(2)y) and l_(2)=(FL_(2))/(pi r_(2)^(2)y)` `(l_(2))/(l_(1))=(L_(2))/(L_(1)) xx (r_(1)^(2))/(r_(2)^(2))` |
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447. |
Equal stretching force is applied along the length of two identical wires made of different substances A and B. It is observed that the elongation of B is less than A. Then,A. P is more elastic than QB. Q is more elastic than PC. P and Q are equally elasticD. P is elastic and Q is plastic |
Answer» Correct Answer - B | |
448. |
A wire of length `L` and radius `r` is fixed at one end. When a stretching force `F` is applied at free end, the elongation in the wire is `l`. When another wire of same material but of length `2L` and radius `2r`, also fixed at one end is stretched by a force `2F` applied at free end, then elongation in the second wire will beA. `l`B. `2l`C. `l//2`D. `4l` |
Answer» Correct Answer - A `y=(F_(1)L_(1))/(pi r_(1)^(2)l_(1)) " " therefore l_(1)=(ypi r_(1)^(2))/(F_(1)L_(1))` `y=(F_(2)L_(2))/(pi r_(2)^(2)l_(2))" " therefore l_(2)=(y pi r_(2)^(2))/(F_(2)L_(2))` `therefore (l_(2))/(l_(1))=(r_(2)^(2))/(r_(1)^(2)) xx (L_(1))/(L_(2)) xx (F_(1))/(F_(2))` |
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