

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.
301. |
One end of uniform wire of length `L` and of weight `W` is attached rigidly to a point in the roof and a weight `W_(1)` is suspended from its lower end. If `s` is the area of cross section of the wire, the stress in the wire at a height (`3L//4`) from its lower end isA. `(w_(1))/(S)`B. `(w_(1)+(w)/(4))/(S)`C. `((w_(1)+(3w)/(4))/(S))`D. `(w_(1)+w)/(S)` |
Answer» Correct Answer - C (c) At length `(3L)/(4)` from lower end, tension in the wire, T=suspended load `+(3)/(4)xx` weight of wire `=w_(1)+(3w)/(4)` `therefore` Stress `=(T)/(S)=(w_(1)+(3w)/(4))/(S)` |
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302. |
To what depth below the surface of sea should a rubber ball be taken as to decreases its volume by 0.1% (Given denisty of sea water = 1000 kg `m^(-3)`, Bulk modulus of rubber `= 9 xx 10^(8) Nm^(-2)`, acceleration due to gravity `= 10 ms^(-2)`)A. 9mB. 18 mC. 180 mD. 90 m |
Answer» Correct Answer - d Bulk modulus, `B = ("Normal stress")/("Volume strain")` `:. B = (p)/(Delta V//V)` Given, `(Delta V)/(V) = (0.1)/(100), B = 9 xx 10^(8) Nm^(-2)` `p = h d g = h xx 1000 xx 10` `:. (B Delta V)/(V) = p = hdg` `implies h = (B Delta V)/(Vdg) = (9 xx 10^(8) xx 0.1)/(1000 xx 100 xx 10) = 90 m` |
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303. |
The elastic limit and ultimate strength for steel is `2.48 xx 10^(8) Pa` and `4.89 xx 10^(8) Pa` respectively. A steel wire of 10 m length and 2 mm cross sectional diameter is subjected to longitudinal tensile stress. Young’s modulus of steel is `Y = 2 xx 10^(11) Pa` (a) Calculate the maximum elongation that can be produced in the wire without permanently deforming it. How much force is needed to produce this extension? (b) Calculate the maximum stretching force that can be applied without breaking the wire. |
Answer» Correct Answer - (a) `1.24 cm, 779N` (b) `1535 N` |
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304. |
One end of a unifrom wire of length L and weigth W is attached rigidly to a point in the roof and a weight `W_(1)` is suspended from its lower end. If S is the area of cross-section of the wire then the stress in the wire at a height `(3L)/(4)` from its lower end isA. `((W_(1) + W))/(S)`B. `(W_(1))/(S)`C. `((W_(1) + (3W)/(4)))/(S)`D. `((W_(1) + (W)/(4)))/(S)` |
Answer» Correct Answer - c Downward force acting at A = weight of `(3)/(4)` length of the wire + weight attached `W_(1)` `= (3)/(4) W + W_(1)` `:.` Stress `= (F)/(A) = (W_(1) + (3)/(4) W)/(S)` |
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305. |
What will be the bulk modulus, if the compressibility of water is `4 xx 10^(-5)` per unit atmospheric pressure?A. `2.533 xx 10^(9) Nm^(-2)`B. `3.354 xx 10^(9) Nm^(-2)`C. `2.233 xx 10^(9) Nm^(-2)`D. `4.562 xx 10^(9) Nm^(-2)` |
Answer» Bulk modulus (B) `= (1)/("Compressibility") = (1)/(K)` `= (1)/(4 xx 10^(-5)) = 0.25 xx 10^(5) atm` `= 0.25 xx 10^(5) xx 1.013 xx 10^(5) Nm^(-2)` `= 2.533 xx 10^(9) Nm^(-2)` |
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306. |
A and B are two wire. The radius of A is twice that of B. If they are stretched by the same load, then the stress on B isA. equal to that of AB. two times that of AC. four times that of AD. half that of A |
Answer» Correct Answer - c stress `= (F)/(S)` Let r be the radius of B For `A, (F)/(pi (2 r)^(2)) = (F)/(pi (4 r^(2)))` For B, stress `= (F)/(pi r^(2))` `:. (F)/(pi r^(2)) = 4 [(F)/(pi r^(2))]` stress on B = 4 (stress on A) |
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307. |
Compressibility of water is `5 xx 10^(-10) m^(2) N^(-1)`. The change in volume of 100 mL water subjected to `15 xx 10^(6)` Pa pressure will beA. increases by 0.75 mLB. decreases by 1.50 mLC. increases by 1.50 mLD. decreases by 0.74 mL |
Answer» Correct Answer - d The reciprocal of bulk modulus B is called compressibility Since, `B = ("Change in pressure")/("volume strain")` `implies K = ("Volume strain")/("Change in pressure")` Given, `K = 5 xx 10^(-10) m^(2) N^(-1), Delta P = 15 xx 10^(6) Pa` volume strain `= K xx` change in pressure `:.` volume strain `= 5 xx 10^(-10) xx 15 xx 10^(8)` `implies` volume strain `= 75 xx 10^(-4) mL` `(Delta V)/(V) = 75 xx 10^(-4)` `implies (Delta V)/(100) = 75 xx 10^(-4)` `implies Delta V = 0.7 mL` so. decreases in volume by 0.75 mL |
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308. |
Human bones remain elastic if strain is less than 0.5%. However, the young’s modulus for compression `(Y_(c))` and stretch `(Y_(s))` are different. The typical values are `Y_(c) = 9.4 xx 10^(9) Pa` and `Y_(s) = 16 xx 10^(9) Pa`. The shear modulus of elasticity for the bone is `h = 10^(10) Pa` Answer following questions with regard to a leg bone of length `20 cm` and cross sectional area `3 cm^(2)` (a) Calculate the maximum stretching force that the bone can sustain and still remain elastic. (b) A man of mass 60 kg jumps from a height of 10 m on a concrete floor. Half his momentum is absorbed by the impact of the floor on the particular bone we are talking about. The impact lasts for 0.02 s. Will the compressive stress exceed the elastic limit? (c) How much shearing force will be needed to break the bone if breaking strain is `5^(@)C`. |
Answer» Correct Answer - (a) `2.4 xx 10^(4)N` , (b) Yes , (c) `2.6 xx 10^(5) N` |
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309. |
The presssure of a medium is changed from `1.01xx10^(5) Pa` to ` 1.165xx10^(5) Pa` and change in volume is `10 % ` keeping temperature constant . The bulk modulus of the medium is (a) `204.8 xx 10^(5) Pa` (b) `102.4xx10^(5) Pa` (c ) `5.12xx10^(5) Pa` (d) `1.55xx10^(5) Pa` |
Answer» Correct Answer - A From the definition of bulk modulus , ` B = - (Deltap)/((Delta V//V))` Substituting the values , we have ` B = ((1.165 - 1.01)xx10^(5)) /((10//100)) = 1.55xx10^(5) Pa` Therfore , the correct option is (d). |
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310. |
A wire can support a load Mg without breaking. It is cut into two equal parts. The maximum load that each part an support isA. Mg/4B. Mg/2C. MgD. 2 Mg |
Answer» Correct Answer - C Breaking load depends upon radius of the wire but it is independent of length. There force load = mg. |
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311. |
If the thickness of the wire is doubled, then the breaking force in the above question will beA. 6 FB. 4 FC. 8 FD. 2 F |
Answer» Correct Answer - B |
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312. |
Assertion : Steel is more elastic than rubber. Reason : Under given deforming force, steel is deformed less than rubber.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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313. |
The deforming force produces a change in the shape of the body without changing its volume, strain produced is calledA. longitudinal strainB. shearing strainC. volumetic strainD. normal stress |
Answer» Correct Answer - B (b) The deforming force produces a change in the shape of the body without changing its volume , strain produced is called shearing strain. |
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314. |
The strain stress curves of three wires of different materials are shown in the figure. P, Q and R are the elastic limits of the wires. The figure shown thatA. elasticity of wire P is maximumB. elasticity of wire Q is maximumC. tensile strength of wire R is maximumD. None of the above |
Answer» Correct Answer - c As stress is shown `Y = cot theta = (1)/(tan theta) = (1)/("slope")` So, elasticity of wire P is minimum and or wire R is maximum. |
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315. |
The strain stress curves of three wires of different materials are shown in the figure. P, Q and R are the elastic limits of the wires. The figure shown that |
Answer» Correct Answer - D (d) As stress is shown on X-axis and strain on Y-axis. So, we can say that `Y="cot"theta=(1)/("tan"theta)=(1)/("slope")` So, elasticity of wire P is minimum and of wire R is maximum. |
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316. |
The strain-stress curves of three wire of different materials are shown in the figure. P,Q and R are the elastic limits of the wires. The figures shows that A. Elasticity of wire P is maximumB. Elastiity of wire Q is maximumC. Tensile strength of R is maximumD. None of the above is true |
Answer» Correct Answer - D |
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317. |
A cube of aluminium of sides 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would beA. 0.02B. 0.1C. 0.005D. 0.002 |
Answer» Correct Answer - D |
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318. |
One end of a wire 2 m long and dameter 2 mm is fixed in a celling . A naughty boy of mass 10 kg jumps to catch the free end and stays, there. The chamge is length of wire is `(Take r = 10 m//s^(2), Y= 2 xx 10^(11) N//m^(2)`)A. `3.1 85 xx 10^(-5)`B. 2 mmC. 3 mmD. 4mm |
Answer» Correct Answer - A (a) For equilibrium of the boy, Tension in wire, T=weight of boy= mg = 100N `=(100xx10^(6))/(3.14) N//m^(2)implies "strain"=(DeltaL)/(L)=(DeltaL)/(2)` `therefore Y=("stress")/("strain")=(100xx10^(6)xx2)/(3.14 xx Delta2` `therefore DeltaL=(100xx10^(6)xx2)/(3.14xx2xx10^(11))=31.85xx10^(-5) m` |
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319. |
Statement I: Smaller drops of liquid resist deforming forces better than the larger drops. Statement II: Excess pressure inside a drop is directly proportional to its surface area.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 | |
320. |
A cable that can support a load W is cut into two equal parts .T he maximum load that can be supported by either part isA. `(W)/(4)`B. `(W)/(2)`C. WD. 2W |
Answer» Correct Answer - C (c) `("Stress")_(1)=("Stress")_(2)` Maximum load does not depend on the length. The maximum load than can be supported by either part is W. |
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321. |
A steel wire of length 1m and diameter 4mm is stretched horizontally between two rigit supports attached to its end. What load would be required to be hung from the mid-point of the wire to produce a depression of 1cm? `(Y= 2 xx 10^(11)Nm^(-2))` |
Answer» Correct Answer - `1.02 xx 10^(4)` |
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322. |
If a wax coated capillary tube is dipped in water, then water in it will-A. rise upB. depressC. sometimes rise and sometimes failD. rise up and come out as a fountain |
Answer» Correct Answer - B | |
323. |
An aeroplane of mass `3xx10^(4)kg` and total wing area of 120 `m^(2)` is in a level flight at some height the difference in pressure between the upper and lower surfaces of its wings in kilo pascal is `(g=10m//s^(2))`A. 2.5B. 5C. 10D. 12.5 |
Answer» Correct Answer - A | |
324. |
The gauge pressure of `3xx10^(5)N//m^(2)` must be maintained in the main water pipes of a city how much work must be done to pump 50,000 `m^(3)` of water at a pressure of `1.0xx10^(5)N//m^(2)`A. `10^(11)J`B. `10^(10)J`C. `10^(9)J`D. `10^(8)J` |
Answer» Correct Answer - B | |
325. |
Water rises to a height `h` in a capillary at the surface of earth. On the surface of the moon the height of water column in the same capillary will be-A. `6h`B. `(1)/(6)h`C. `h`D. zero |
Answer» Correct Answer - A | |
326. |
The baromatric pressure and height on the earth are `10^(5)` Pa and 760 mm respectively if it is taken to moon then barometric height will be-A. 76 mmB. 126.6mmC. zeroD. 760 mm |
Answer» Correct Answer - C | |
327. |
Assertion : A hollow shaft is found to be stronger than a solid shaft made of same material. Reason : The torque required to produce a given twist in hollow cylinder is greater than that required to twist a solid cylinder of same size and material.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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328. |
If the potential energy is minimum at r = ro = 0.74Ao, is the force attractive or repulsive at r = 0.5Ao; 1.9Ao and α? |
Answer» Since, potential energy is minimum at ro = 0.74Ao. therefore interatomic force between two atoms is zero for ro = 0.74Ao . 1) At r = 0.5 Ao (Which is less than ro), the force is repulsive. 2) At r = 1.9Ao (Which is greater than ro), the force is attractive. 3) At r = α, the force is zero. |
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329. |
A hollow shaft is found to be stronger than a solid shaft made of same equal material? Why? |
Answer» A hollow shaft is found to be stronger than a solid shaft made of equal material because the torque required to produce a given twist in hollow cylinder is greater than that required to produce in solid cylinder of same length and material through same angle. |
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330. |
Water is more elastic than air. Why? |
Answer» Since volume elasticity is the reciprocal of compressibility and since air is more compressible than water hence water in more elastic than air. |
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331. |
A water tank is supported by four pillars. The pillars are strong enough to sustain ten times the stress developed in them when the tank is completely full. An engineer decides to increase the every dimension of the tank and the pillars by hundred times so as to store more water. Do you think he has taken a right decision? Assume that material used in construction of the tank and pillars remain same. |
Answer» Correct Answer - No |
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332. |
The maximum load that a wire can sustain is W. If the wire is cut to half its value, the maximum load it can sustain isA. `W `B. `W/2`C. `W/4`D. `2W` |
Answer» Correct Answer - A `sigma_(max)= F_(max)/A` `:. F_(max) = F_(max)/A` , which is independent of length of wire. |
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333. |
The potential energy U of diatomic molecules as a function of separation `r` is shown in figure. Identify the correct stateement. A. The atoms are in equilibrium if `r = OA`B. The force is repulsive only if `r` lies between A and BC. The force is attractive if `r` lies between A and BD. The atoms are in equilibrium if `r =OB` |
Answer» Correct Answer - D `F = - (dU)/(dr)` = - slope of U - r graph At B, `"slope" = 0 ` `:. F = 0` Hence , atoms are in equilibrium at B. From A to B or even before A slope is negative . Hence , force is positive force mean repulsion. |
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334. |
The potential energy U between two molecules as a function of the distance X between them has been shown in the figure. The two molecules are A. Attracted when x lies between A and B and are repelled when X lies between B and CB. Attracted when x lies between B and C and are repelled when X lies between A and BC. Attracted when they reach BD. Repelled when they each B |
Answer» Correct Answer - B |
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335. |
The diagram shows stress v/s strain curve for the materials A and B. From the curves we infer that A. A is brittle but B is ductileB. A is ductile and B is brittleC. Both A and B are ductileD. Both A and B are brittle |
Answer» Correct Answer - B |
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336. |
The figure shows a soap film in which a closed elastic thread is lying. The film inside the thread is pricked. Now the sliding wire is moved out so that the surface area increases. The radius circle of the circle formed by elastic thread willA. increaseB. decreasesC. remains sameD. data insufficient |
Answer» Correct Answer - C | |
337. |
The following four wires are made of the same material which of these will have the largest extension when the same tension is appliedA. length 50 cm and diameter 0.5 mmB. length 100 cm and diameter 1 mmC. length 200 cm and diameter 2 mmD. length 300 cm and diameter 3mm |
Answer» Correct Answer - A | |
338. |
A 2m long rod of radius 1 cm which is fixed from one end is given a twist of 0.8 radians the shear strain developed will beA. 0.002B. 0.004C. 0.008D. 0.016 |
Answer» Correct Answer - B | |
339. |
One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end fo another horizontal thin copper wire of lenth L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire isA. 0.25B. 0.5C. 2D. 4 |
Answer» Correct Answer - C | |
340. |
PQRS is a rectangular frame of copper wire shown in figure the side RS of the frame is movable. If a soap film is formed on it then what is the diameter of the wire to maintain equilibirum (given surface tension of soap solution `=0.045N//m` and density of copper `=8.96xx10^(3)kg//m^(3))` |
Answer» Correct Answer - 1.14 mm | |
341. |
A force F is needed to break a copper wire having radius R. The force needed to break a copper wire of same length and radius 2R will beA. `F//2`B. 2 FC. 4 FD. `F//4` |
Answer» Correct Answer - C |
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342. |
A copper ball of radius `r` travels with a uniform speed v in a viscous fluid if the ball is changed withh another ball of radius 2r then new uniform speed will beA. vB. 2vC. 4vD. 8v |
Answer» Correct Answer - C | |
343. |
A force F is needed to break a copper wire having radius R. The force needed to break a copper wire of same length and radius 2R will beA. `(F)/(2)`B. `2F`C. `4F`D. `(F)/(4)` |
Answer» Correct Answer - C | |
344. |
The breaking stress of a wire of length L and radius r is `5 kg-wt//m^(2)`. The wire of length `2l` and radius `2r` of the same material will have breaking stress in `kg-wt//m^(2)`A. 5B. 10C. 20D. 80 |
Answer» Correct Answer - A |
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345. |
If a wire having initial diameter of 2 mm produced the longitudinal strain of 0.1%, then the final diameter of wire is `(sigma = 0.5)`A. 2.002 mmB. 1.998 mmC. 1.999 mmD. 2.001 mm |
Answer» Correct Answer - C `sigma=(Delta d)/(D)xx(L)/(l)` `(Delta d)/(D)=-D xx 5 xx 10^(-4) =-2 xx 10^(-3) xx 5 xx10^(-4)` `=-1xx10^(-6)` `d_(2)-d_(1)=-0.001 mm` `d_(2)=d_(1)-0.001 =2-0.001=2.0000-0.001` `d_(2)=1.999 mm`. |
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346. |
To break a wire of one meter length, minimum 40 kg wt. is required. Then the wire of the same material of double radius and 6 m length will require breaking weightA. 40 kg-wtB. 80 kg -wtC. 160 kg-wtD. 320 kg-wt |
Answer» Correct Answer - C (c) Breaking force=Breaking stress `xx` area of cross-section of wire `therefore` Breaking force `prop r^(2)` If radius becomes doubled, then breaking force becomes 4 times i.e. `40xx4=160 kg-wt` |
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347. |
A steel wire of diameter 1 mm, and length 2 m is stretched by applying a force of 2.2 kg wt. the strain isA. `0.7xx10^(-4)`B. `1.4xx10^(-4)`C. `1.9xx10^(-5)`D. `2.8xx10^(-4)` |
Answer» Correct Answer - B `(l)/(L)=(F)/(Ay)=(2.2xx10)/(3.14xx0.25xx10^(-6)xx2xx10^(11))` `=(4.4)/(3.14)xx10^(-4)=1.4xx10^(-4)` |
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348. |
The length of the wire is increased by 2% by applying a load of 2.5 kg-wt. what is the linear strain produced in the wire?A. 0.1B. 0.01C. 0.2D. 0.02 |
Answer» Correct Answer - d Percentage change in length of wire `= (Delta I)/(I) xx 100%` `(Delta I)/(I) xx 100 = 2% implies (Delta I)/(I) = (2)/(100) = 0.02` |
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349. |
The change in dimensions per unit original dimensions isA. stressB. deformationC. strainD. formation |
Answer» Correct Answer - C | |
350. |
Within elastic limit, which of the following graphs correctly represents the variation of extension in the length of wire with the external load?A. B. C. D. |
Answer» Correct Answer - D | |