448 + Interview Questions in ELASTICITY IN GENERAL AWARENESS Page 6 InterviewSolution

251.	A closed steel cylinder is completely filled with water at `0^(@)C`. The water is made to freeze at `0^(@)C`. Calculate the rise in pressure on the cylinder wall. It is known that density of water at `0^(@)C` is `1000 kg//m^(3)` and the density of ice at `0^(@)C` is `910 kg//m^(3)`. Bulk modulus of ice at `0^(@)C` is nearly `9 xx 10^(9) Pa`. [Compare this pressure to the atmospheric pressure. Now you can easily understand why water pipelines burst in cold regions as the winter sets in.]
Answer» Correct Answer - `8.1 xx 10^(8) Pa`

Discussion

252.	The energy stored per unit volume of a strained wire isA. `(1)/(2)xx"(load)"xx"(extension)"`B. `(1)/(2) (Y)/(("Strain")^(2))`C. `(1)/(2)y ("strain")^(2)`D. stress `xx` strain
Answer» Correct Answer - C

Discussion

253.	The potential energy of a stretched spring is proportional toA. the square of force constantB. the square of the stretching length of the wireC. the square of original length of wireD. the cube of the stretch of the wire
Answer» Correct Answer - B `u prop x^(2)`

Discussion

254.	K is the force constant of a spring. The work donein increasing its extension from `l_(1)` to `l_(2)` will beA. `(k)/(2) (l_(2)-l_(1))`B. `(k)/(2)(l_(2)^(2)-l_(1)^(2))`C. `k(l_(2)^(2)-l_(1)^(2))`D. `(k)/(2)(l_(2)^(3)-l_(1)^(3))`
Answer» Correct Answer - B `u_(1)=(1)/(2)K l_(1)^(2) and u_(2) =(1)/(2)Kl_(2)^(2)` `u=(1)/(2) K(l_(2)^(2) -l_(1)^(2))`

Discussion

255.	A very stiff bar (AB) of negligible mass is suspended horizontally by two vertical rods as shown in figure. Length of the bar is 2.5 L. The steel rod has length L and cross sectional radius of r and the brass rod has length 2L and cross sectional radius of 2r . A vertically downward force F is applied to the bar at a distance x from the steel rod and the bar remains horizontal. Find the value of x if it is given that ratio of Young’s modulus of steel and brass is `(Y_(s))/(Y_(B)) = 2`.
Answer» Correct Answer - `x = 1.25 L`

Discussion

256.	K is the force constant of a spring. The work donein increasing its extension from `l_(1)` to `l_(2)` will beA. `K(l_(2)-l_(1))`B. `(K)/(2)(l_(2)+l_(1))`C. `K(l_(2)^(2)-l_(1)^(2))`D. `(K)/(2)(l_(2)^(2)-l_(1)^(2))`
Answer» Correct Answer - D

Discussion

257.	Modulus of rigidity of ideal liquids isA. infinityB. zeroC. unityD. some finite small non-zero constant value
Answer» Correct Answer - B (b) No frictional (viscous) force exists in case of ideal fluid, hence tangential force are zero so there is no stress developed.

Discussion

258.	The force constant of a wire does not depend onA. Nature of the materialB. Radius of the wireC. Length of the wireD. None of the above
Answer» Correct Answer - D

Discussion

259.	Two wires of copper having the length in the ratio `4 : 1` and their radii ratio as `1 : 4` are stretched by the same force. The ratio of longitudinal strain in the two will beA. `1:16`B. `16:1`C. `1:64`D. `64:1`
Answer» Correct Answer - B

Discussion

260.	When a force is applied on a wire of uniform cross-sectional area `3xx10^-6m^2` and length 4 m, the increase in length is 1mm. Energy stored in it will be `(Y=2xx10^(11)(N)/(m^2)`)A. 6250 JB. 0.177 JC. 0.075 JD. 0.150 J
Answer» Correct Answer - C

Discussion

261.	A ring of lead (relative density =11.3 and breaking stress ` sigma _(b)=1.5xx10^(7) N m^(-2))` of radius r=25 cm is rotated about its axis .What is the number of rps at which the ring will rupture?
Answer» Correct Answer - 23rps

Discussion

262.	What work has to be done to make a hoop out of a copper band of length `l=3m,`width h=6 cm and thickness `delta =1mm?` (Y of copper `=1.3xx10^(11) ""Nm^(-2)` ) [Hint: Consider strain energy in a a thin layer and intergraate from `r-delta//2 "to" r+delta//2]`
Answer» Correct Answer - `(1)/(6)pi^(2)hdelta^(3)Y//=80J`

Discussion

263.	A wire of length 1 m and radius 1 mm is welded to another wire of length 2 m and radius 2 mm the free end of the first is clamped and aload of 5kg ispplied at the free end of the second wire . What is the total increase of the compound wire ? (Y of both wires `=2xx10^(11) ""Nm^(-2)`)
Answer» Correct Answer - `11.7xx10^(-5)m`

Discussion

264.	A wire of length 2 ,m and radius 2 mm is stretched by 2.5 mm when a load of 5 kg is suspended from it Another wire of the same material put of length 1 m and radius 1 mm is attached below the load and a load of 3kg is suspended from this wire what is the total increase in length and what are the individual exensions of the wirs ?
Answer» Correct Answer - `7mm ,4mm,3mm`

Discussion

265.	If a load of `9 kg` is suspended on a wire, the increase in length is `4.5 mm`. The force constant of the wire isA. `0.49 xx 10^(4) N//m`B. `1.96 xx 10^(4) N//m`C. `4.9 xx 10^(4) N//m`D. `0.196 xx 10^(4) N//m`
Answer» Correct Answer - B

Discussion

266.	When wire is stretched, a work is performed on the wire, This work done on wire isA. simply wastedB. lost in the form of heatC. stored in the form of elastic potential energyD. used up to overcome the fall in the gravitational potential energy
Answer» Correct Answer - C

Discussion

267.	In the above question, the ratio of the increase in energy of the wire to the decrease in gravitational potential energy when load moves downwards by `1mm`, will beA. 1B. `(1)/(4)`C. `(1)/(3)`D. `(1)/(2)`
Answer» Correct Answer - D

Discussion

268.	In above question, the work done in the two wire isA. 0.5 J, 0.03 JB. 0.25 J, 0 JC. 0.03 J, 0.25 JD. 0J, 0j
Answer» Correct Answer - a As work done, `W = (F^(2)I)/(2((pi D^(2))/(4))y)` Where, Y,I and F are constants. `implies W prop (1)/(D^(2))` or `(W_(1))/(W_(2)) = (D_(2)^(2))/(D_(1)^(2)) = 16` Now, `W_(1) = (1)/(2) xx 10^(3 xx 1 xx 10^(-3) = 0.5 J` `:. W_(2) = (1)/(2) xx 10^(3) xx (10^(-3))/(16) = (1)/(32) = 0.03125` `(w_(1))/(W_(2)) = (0.5)/(0.3125) = 16`

Discussion

269.	Increase in length of a wire is 1 mm when suspended by a weight. If the same weight is suspended on a wire of double its length and double its radius, the increase in length will beA. `2 mm`B. `0.5 mm`C. `4 mm`D. `0.25 mm`
Answer» Correct Answer - B

Discussion

270.	A wire is suspended by one end. At the other end a weight equivalent to 20 N force is applied. If the increase in length is 1.0 mm, the increase in energy of the wire will beA. 0.01 JB. 0.02 JC. 0.04 JD. 1.00 J
Answer» Correct Answer - A

Discussion

271.	The wires `A` and `B` shown in Fig. are made of the same material and have radii `r_(A)` and `r_(B)`, respectively. The block between them has a mass `m`. When the force `F` is `mg//3`, one of the wires breaks. Then A. A will break before B is `r_(A)=r_(B)`B. A will break before B is `r_(A)lt2r_(B)`C. Either A or B may break if `r_(A)=2r_(B)`D. The length of A and B must be known to predict which wire will break.
Answer» Correct Answer - A::B::C

Discussion

272.	Two wires A and B of same length and of the same material have the respective radii `r_(1)` and `r_(2)`. Their one end is fixed with a rigid support, and at other end equal twisting couple is applied. Then the ratio of the angle of twist at the end of A and the angle of twist at the end of B will beA. `(r_(2)^(4))/(r_(1)^(4))`B. `(r_(1)^(4))/(r_(2)^(4))`C. `(r_(2)^(2))/(r_(1)^(2))`D. `(R_(1)^(2))/(r_(2)^(2))`
Answer» Correct Answer - a As torque, `tau = (pi eta r^(4))/(2l) theta` In the given problem, `r^(4) theta` = constant `implies theta prop (1)/(r^(4)) implies (theta_(A))/(theta_(B)) = (r_(2)^(4))/(r_(1)^(4))`

Discussion

273.	If the force constant of a wire is K, the work done in increasing the length of the wire by `l` isA. `K//2`B. `Kl`C. `Kl^(2)//2`D. `Kl^(2)`
Answer» Correct Answer - C

Discussion

274.	When shearing force is applied on a body, then the elastic potential energy is stored in it. On removing the force, this energyA. Converts into kinetic energyB. Coverts into heat energyC. Remains as potential energyD. None of the above
Answer» Correct Answer - B

Discussion

275.	If one end of a wire is fixed with a rigid support and the other end is stretched by a force of 10 N, the increase in length is 0.5 mm. The ratio of enegry of the wire and the work done in displacing it through 1.5 mm by the weight isA. `(1)/(3)`B. `(1)/(4)`C. `(1)/(2)`D. `1`
Answer» Correct Answer - C

Discussion

276.	A cup filled with water has a hole in the side, through which the liquid is flowing out. If the cup is dropped from a height which of the following is/are correct?A. Water keeps coming out, flowing at the same rate as beforeB. Water keeps coming out, but is will flow a bit slower than before.C. Water keeps coming out, but start to flow upward relative to the cupD. water stops flowing
Answer» Correct Answer - A::B::C

Discussion

277.	A vertical `U-tube` contains a liquid. The total length of the liquid column inside the tube is `1`. When the liquid is in equilibrium, the liquid surface in one of the arms of the `U-tube` is pushed down slightly and released. The entire liquid column will undergo a periodic motion.A. The motion is not simple harmonic motion.B. the motion is simple harmonic motionC. if it undergoes simple harmonic motion the time period will be `2pisqrt((l)/(g))`D. It is undergoes simple harmonic motion the time period will be `2pisqrt((l)/(2g))`
Answer» Correct Answer - B::D

Discussion

278.	An L shaped glass tube is kept inside a bus that is moving with constant acceleration during the motion the level of the liquid in the left arm is at 12 cm whereas in the right arm it is at 8 cm when the orientation of the tube is as shown assuming that the diameter of the tube is much smaller than levels of the liquid and neglecting effect of surface tension, acceleration of the bus find the `(g=10m//s^(2))`.
Answer» `tantheta=(a)/(g)=(h_(2)-h_(1))/(h_(2)tan45^(@)+h_(1)tan45^(@))=(4cm)/(20cm)impliesa=2m//s^(2)`

Discussion

279.	The greatest length of a steel wire that can hang without breaking is 10 km. I the breaking stress for steel is `8.25xx10^(8)N//m^(2)`. Then the density of the steel is `(g=10m//s^(2))`A. `7200kg//m^(3)`B. `8250kg//m^(3)`C. `6200kg//m^(3)`D. `5200kg//m^(3)`
Answer» Correct Answer - B Breaking stress `=Lrhog`

Discussion

280.	An open pan P filled with water of density `(rho_w)` is placed on a vertical rod, maintaining equilibrium. A block of density `rho` is placed on one side of the pan as shown. Water depth is more then height of the block. A. Equilibrium will be maintained only if `rholtrho_(W)`B. equilibrium will be maintained only `rholerho_(W)`C. Equilibrium will be maintained for all relations between `rho` and `rho_(W)`D. Equilibrium will not be maintained in all cases.
Answer» Correct Answer - B

Discussion

281.	A mercury manometer is connected as shown in the figure. The difference in height `Deltah`: (symbols have usual meaning)A. `(rhodcottheta)/(rho_(Hg))`B. `(rhodtantheta)/(rho_(Hg))`C. `(rhodsintheta)/(rho_(Hg))`D. none of these
Answer» Correct Answer - B

Discussion

282.	Assertion Modulus of elasticity does not depend upon the dimensions of material Reason Modulus of elasticity is a material property.A. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If both Assertion and Reason are false.
Answer» Correct Answer - A

Discussion

283.	Calculate the excess pressure withing a bubble of air of radius 0.1 mm in water. If te bubble had been formed 10 cm below the water surface on a day when the atmospheric pressure was `1.013xx10^(3)Pa` then what would have been the total pressure inside the bubble? surface tension of water `=73xx10^(-3)N//m`
Answer» Excess pressure `P_("excess")=(2T)/(r)=(2xx73xx10^(-3))/(0.1xx10^(-3))=1460Pa` The pressure at a depth d, in liquid is `P=hdg`. Therefore, the total pressure inside the air bubble is `P_("in")=P_(atm)+hdg+(2T)/(r)=(1.013xx10^(5))+(10xx10^(-2)xx10^(3)xx9.8)+1460` `=101300+980+1460=103740=1.037xx10^(5)Pa`

Discussion

284.	The limbs of a manometer consist of uniform capillary tubes of radii are `10^(-3)&7.2xx10^(-4)`m find out the correct pressure difference if the level of the liquid in narrower tube stands 0.2 m above that in the broader tube. (density of liquid `=10^(3)kg//m^(3)`, surface tension `=72xx10^(-3)N//m`)
Answer» If `P_(1)` and `P_(2)` are the pressures in the broader and narrower tubes of radii `r_(1)` and `r_(2)` respectively, the pressure just below the meniscus in the respective tubes will be `P_(1)-(2T)/(r_(1)) and P_(2)-(2T)/(r_(2))` So that `[P_(1)-(2T)/(r_(1))]-[P_(2)-(2T)/(r_(2))]=hrhog or P_(1)-P_(2)=hrhog-2T[(1)/(r_(2))-(1)/(r_(1))]` `P_(1)-P_(2)=0.2xx10^(3)xx9.8-2xx72xx10^(-3)[(1)/(7.2xx10^(-4))-(1)/(14xx10^(-4))]=1960-97=1863Pa`

Discussion

285.	Find out pressure at points A and B also find angle `theta`
Answer» Pressure at A-`P_(A)=P_(atm)-rhoglsintheta` Pressure at B `P_(B)=P_(atm)+rho_(2)ghtheta` But `P_(B)` is also equal to `P_(B)=P_(A)+rho_(3)glsintheta` Hence `P_(atm)+rho_(2)gh=P_(A)+rho_(3)glsintheta` `P_(atm)+rho_(2)gh=P_(atm)-rho_(1)glsintheta+rho_(3)glsintheta` `sintheta=(rho_(2)h)/((rho_(3)-rho_(1))l)`

Discussion

286.	A solid right cylinder of length l stands upright at rest on the bottom of a large tub filled with water up to height h as shown in the figure-I density of material of the cylinder equals to that of water. Now the cylinder is pulled slowly out of water with the help of a thin light inextensible thread as shown in figure-II. Find the work done by the tension force develop in the thread. (a). mgh (b). mgl (c). 0.5 mgl (d). mg(0.5l+h)
Answer» Correct Answer - C Work done by tension force=work done by gravity =0.5mgl

Discussion

287.	A spherical ball is moving with terminal velocity inside a liquid determine the relationship of rate of heat loss with the radius of ball.
Answer» Rate of heat loss =power `=Fxxv=6pietarvxxv=6pietarv^(2)=6rhoetar[(2)/(9)(gr^(2)(rho_(0)-rho_(l)))/(eta)]^(2)` Therefore rate of heat loss `propr^(5)`

Discussion

288.	A metal cylinder of length L and radius R is fixed rigidly to ground with its axis vertical. A twisting torque `tau_(0)` is applied along the circumference at the top of the cylinder. This causes an angular twist of `theta_(0)` (rad) in the top surface. Calculate the shear modulus of elasticity `(eta)` of the material of the cylinder.
Answer» Correct Answer - `(2l tau_(0))/(pi R^(4) theta_(0))`

Discussion

289.	Bulk modulus of elasticity of rubber is `10^(9)N//m^(2)`. If it is taken down to a 100 m deep lake, then decrease in its volume will be ` " "(g=10m//s^(2))`A. `0.1%`B. `0.2%`C. `1%`D. `2%`
Answer» Correct Answer - A `K=v (dp)/(dv)` `therefore (dv)/(v)=(dp)/(k)=(h rho g)/(k)` `=(100xx10^(3)xx10)/(10^(9)) =10^(-3)=0.1%`.

Discussion

290.	Atmospheric pressure is `P_(0)` and density of water at the sea level is `rho_(0)`. If the bulk modulus of water is B, calculate the pressure deep inside the sea at a depth h below the surface.
Answer» Correct Answer - `P = P_(0)-B ln (1-(rho_(0) gh)/(B))`

Discussion

291.	A glass full of water is placed on a weighing scale, which reds 10 N, A coin with weight 1 N is gently released into the water. At finrst the coin accelerates as it falls and about halfway down the glass, the coin reaches terminal velocity eventually the coin rests on te bottom of the glass, accelerationdue to gravity is `10m//s^(2)`. The scale roads (a). 10.5N when the coin accelerates at `5m//s^(2))` (b). 11.5 N when the coin decelerates at `5m//s^(2)` (c). 11N, when the coin moves with terminal velocity (d). 11N when the coin rests on the bottom.
Answer» Correct Answer - A::C::D When coin moves with terminal velocity or rests on the bottom Reading `=10+1=11N` When coin moves with acceleration of `5m//s^(2)` Reading `=10+1(1//2)=10.5N`

Discussion

292.	The isothermal Bulk modulus of an ideal gas at pressure P isA. PB. `P//gamma`C. `P gamma`D. `P^(gamma)`
Answer» Correct Answer - A The perfect gas equation for isothermal gas is, `PV=RT` `Pdv+Vdp=0` `Pdv= -Vdp` `(-(dv)/(v))P=dP` `P=-(dPv)/(dv)=K.`

Discussion

293.	To compress a liquid by `10%` of its original volume, the pressure required is `2xx10^(5) N//m^(2)`. The bulk modulus of the liquid isA. `2xx10^(5)N//m^(2)`B. `2xx10^(7)N//m^(2)`C. `2xx10^(4)N//m^(2)`D. `2xx10^(6)N//m^(2)`
Answer» Correct Answer - D

Discussion

294.	The ratio of adiabatic and isothermal bulk modulus of elasticity for a perfect gas isA. `gamma`B. `gamma^(2)`C. `1//gamma`D. `1//gamma^(2)`
Answer» Correct Answer - A

Discussion

295.	The bulk modulus of a perfectly rigid body isA. zeroB. infiniteC. finiteD. unity
Answer» Correct Answer - B

Discussion

296.	For the system shown in the figure the cylinder on the left at L has a mass of 600 kg and a cross sectional area of `800cm^(2)` the piston on the right at S has cross sectional area `25cm^(2)` and negligible weight if the apparatus is filled with oil `(rho=0.75gm//cm^(3))` find thhe force F required to hold the system in equilibrium.
Answer» Correct Answer - 37.5 N

Discussion

297.	A crown made of gold and copper weights 210g in air and 198 g in water the weight of gold in crown is [Given, Density of gold `=19.3g//cm^(2)` and density of copper `=8.5g//cm^(3)`]A. 93 gB. 100gC. 150 gD. 193g
Answer» Correct Answer - D

Discussion

298.	If a rubber ball is taken at the depth of 200 m in a pool, its volume decreases by`0.1%` If the density of the water is `1xx10^(3)Kg//m^(3)"and"g=10ms^(2), then the volume elasticity in N//m2 will beA. `10^(8)`B. `2xx10^(8)`C. `10^(9)`D. `2xx10^(9)`
Answer» Correct Answer - D

Discussion

299.	Figure shows the graph of elastic potential energy `(U)` stored versus extension, for a steel wire `(Y = 2xx10^(11) Pa)` of volume `200 c c`. If area of cross- section `A` and original length L, then A. `A = 10^(-4) m^(2)`B. `A= 10 ^(-3) m^(2)`C. `L=1.5 m`D. `L = 2 m`
Answer» Correct Answer - A::D `U =1/2 kx^(2)` . It is a parabola symmetric about U- axis . At `x = 0.2 mm`, `U =0.2 J` ( from the figure ) `:. 0.2 = 1/2 k(2xx10^(-4))^(2)` `rArr k = (YA)/(L) ` `(A)/L = (K)/(Y) = (10^(7)) Nm^(-1)` `k = (YA)/(L)` `rArr (A)/(L) = (k)/(Y) = (10^(7))/(2xx10^(11)) = 5xx10^(-5)` ...(i) `AL = "Volume" = 200xx10^(-6) m^(3)` ...(ii) On solving Eqs. (i) and (ii) , We get `A= 10^(-4) m^(2)` and `L = 2 m`

Discussion

300.	The elastic limit of a steel cable is `3.0xx10^(8) N//m^(2)` and the cross-section area is `4 cm^(2)`. Find the maximum upward acceleration that can be given to a `900 kg` elevator supported by the cable if the stress is not to exceed one - third of the elastic limit.
Answer» Correct Answer - B::C::D `(m(g+a))/A = 1/3sigma_(max)` `:. a = (sigma_(max)A)/(3 m) -g` `= ((3xx10^(8))(4xx10^(-4)))/(3xx900)-9.8` ` = 34.64 m//s^(2)`

Discussion

Explore topic-wise InterviewSolutions in .

The energy stored per unit volume of a strained wire isA. `(1)/(2)xx"(load)"xx"(extension)"`B. `(1)/(2) (Y)/(("Strain")^(2))`C. `(1)/(2)y ("strain")^(2)`D. stress `xx` strain

The potential energy of a stretched spring is proportional toA. the square of force constantB. the square of the stretching length of the wireC. the square of original length of wireD. the cube of the stretch of the wire

K is the force constant of a spring. The work donein increasing its extension from `l_(1)` to `l_(2)` will beA. `(k)/(2) (l_(2)-l_(1))`B. `(k)/(2)(l_(2)^(2)-l_(1)^(2))`C. `k(l_(2)^(2)-l_(1)^(2))`D. `(k)/(2)(l_(2)^(3)-l_(1)^(3))`

K is the force constant of a spring. The work donein increasing its extension from `l_(1)` to `l_(2)` will beA. `K(l_(2)-l_(1))`B. `(K)/(2)(l_(2)+l_(1))`C. `K(l_(2)^(2)-l_(1)^(2))`D. `(K)/(2)(l_(2)^(2)-l_(1)^(2))`

Modulus of rigidity of ideal liquids isA. infinityB. zeroC. unityD. some finite small non-zero constant value

The force constant of a wire does not depend onA. Nature of the materialB. Radius of the wireC. Length of the wireD. None of the above

Two wires of copper having the length in the ratio `4 : 1` and their radii ratio as `1 : 4` are stretched by the same force. The ratio of longitudinal strain in the two will beA. `1:16`B. `16:1`C. `1:64`D. `64:1`

When a force is applied on a wire of uniform cross-sectional area `3xx10^-6m^2` and length 4 m, the increase in length is 1mm. Energy stored in it will be `(Y=2xx10^(11)(N)/(m^2)`)A. 6250 JB. 0.177 JC. 0.075 JD. 0.150 J

A ring of lead (relative density =11.3 and breaking stress ` sigma _(b)=1.5xx10^(7) N m^(-2))` of radius r=25 cm is rotated about its axis .What is the number of rps at which the ring will rupture?

What work has to be done to make a hoop out of a copper band of length `l=3m,`width h=6 cm and thickness `delta =1mm?` (Y of copper `=1.3xx10^(11) ""Nm^(-2)` ) [Hint: Consider strain energy in a a thin layer and intergraate from `r-delta//2 "to" r+delta//2]`

A wire of length 1 m and radius 1 mm is welded to another wire of length 2 m and radius 2 mm the free end of the first is clamped and aload of 5kg ispplied at the free end of the second wire . What is the total increase of the compound wire ? (Y of both wires `=2xx10^(11) ""Nm^(-2)`)

If a load of `9 kg` is suspended on a wire, the increase in length is `4.5 mm`. The force constant of the wire isA. `0.49 xx 10^(4) N//m`B. `1.96 xx 10^(4) N//m`C. `4.9 xx 10^(4) N//m`D. `0.196 xx 10^(4) N//m`

When wire is stretched, a work is performed on the wire, This work done on wire isA. simply wastedB. lost in the form of heatC. stored in the form of elastic potential energyD. used up to overcome the fall in the gravitational potential energy

In the above question, the ratio of the increase in energy of the wire to the decrease in gravitational potential energy when load moves downwards by `1mm`, will beA. 1B. `(1)/(4)`C. `(1)/(3)`D. `(1)/(2)`

In above question, the work done in the two wire isA. 0.5 J, 0.03 JB. 0.25 J, 0 JC. 0.03 J, 0.25 JD. 0J, 0j

Increase in length of a wire is 1 mm when suspended by a weight. If the same weight is suspended on a wire of double its length and double its radius, the increase in length will beA. `2 mm`B. `0.5 mm`C. `4 mm`D. `0.25 mm`

A wire is suspended by one end. At the other end a weight equivalent to 20 N force is applied. If the increase in length is 1.0 mm, the increase in energy of the wire will beA. 0.01 JB. 0.02 JC. 0.04 JD. 1.00 J

If the force constant of a wire is K, the work done in increasing the length of the wire by `l` isA. `K//2`B. `Kl`C. `Kl^(2)//2`D. `Kl^(2)`

When shearing force is applied on a body, then the elastic potential energy is stored in it. On removing the force, this energyA. Converts into kinetic energyB. Coverts into heat energyC. Remains as potential energyD. None of the above

If one end of a wire is fixed with a rigid support and the other end is stretched by a force of 10 N, the increase in length is 0.5 mm. The ratio of enegry of the wire and the work done in displacing it through 1.5 mm by the weight isA. `(1)/(3)`B. `(1)/(4)`C. `(1)/(2)`D. `1`

The greatest length of a steel wire that can hang without breaking is 10 km. I the breaking stress for steel is `8.25xx10^(8)N//m^(2)`. Then the density of the steel is `(g=10m//s^(2))`A. `7200kg//m^(3)`B. `8250kg//m^(3)`C. `6200kg//m^(3)`D. `5200kg//m^(3)`

A mercury manometer is connected as shown in the figure. The difference in height `Deltah`: (symbols have usual meaning)A. `(rhodcottheta)/(rho_(Hg))`B. `(rhodtantheta)/(rho_(Hg))`C. `(rhodsintheta)/(rho_(Hg))`D. none of these

Find out pressure at points A and B also find angle `theta`

A spherical ball is moving with terminal velocity inside a liquid determine the relationship of rate of heat loss with the radius of ball.

Bulk modulus of elasticity of rubber is `10^(9)N//m^(2)`. If it is taken down to a 100 m deep lake, then decrease in its volume will be ` " "(g=10m//s^(2))`A. `0.1%`B. `0.2%`C. `1%`D. `2%`

Atmospheric pressure is `P_(0)` and density of water at the sea level is `rho_(0)`. If the bulk modulus of water is B, calculate the pressure deep inside the sea at a depth h below the surface.

The isothermal Bulk modulus of an ideal gas at pressure P isA. PB. `P//gamma`C. `P gamma`D. `P^(gamma)`

To compress a liquid by `10%` of its original volume, the pressure required is `2xx10^(5) N//m^(2)`. The bulk modulus of the liquid isA. `2xx10^(5)N//m^(2)`B. `2xx10^(7)N//m^(2)`C. `2xx10^(4)N//m^(2)`D. `2xx10^(6)N//m^(2)`

The ratio of adiabatic and isothermal bulk modulus of elasticity for a perfect gas isA. `gamma`B. `gamma^(2)`C. `1//gamma`D. `1//gamma^(2)`

The bulk modulus of a perfectly rigid body isA. zeroB. infiniteC. finiteD. unity

A crown made of gold and copper weights 210g in air and 198 g in water the weight of gold in crown is [Given, Density of gold `=19.3g//cm^(2)` and density of copper `=8.5g//cm^(3)`]A. 93 gB. 100gC. 150 gD. 193g

If a rubber ball is taken at the depth of 200 m in a pool, its volume decreases by`0.1%` If the density of the water is `1xx10^(3)Kg//m^(3)"and"g=10ms^(2), then the volume elasticity in N//m2 will beA. `10^(8)`B. `2xx10^(8)`C. `10^(9)`D. `2xx10^(9)`

Figure shows the graph of elastic potential energy `(U)` stored versus extension, for a steel wire `(Y = 2xx10^(11) Pa)` of volume `200 c c`. If area of cross- section `A` and original length L, then A. `A = 10^(-4) m^(2)`B. `A= 10 ^(-3) m^(2)`C. `L=1.5 m`D. `L = 2 m`

The elastic limit of a steel cable is `3.0xx10^(8) N//m^(2)` and the cross-section area is `4 cm^(2)`. Find the maximum upward acceleration that can be given to a `900 kg` elevator supported by the cable if the stress is not to exceed one - third of the elastic limit.