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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.
| 101. |
Compressibility of water is `5xx10^(-10)m^(2)//N`. Find the decrease in volume of `100 mL` of water when subjected to a pressure of `15 MPa`.A. `7.5xx10^(-3)`B. `5xx10^(-3)`C. `2.5xx10^(-3)`D. `1.25xx10^(-3)` |
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Answer» Correct Answer - A `(-Delta V)/(V)` = compressibiltiy `xx Delta P` |
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| 102. |
Two equal and opposite forces `F` and `-F` act on a rod of unifrom cross-sectional ara `A`, as shown in the figure. Find the (i) shearing (ii) longitual stress on the section `AB`. A. `(F sin x cos x)/(A)`B. `(F sin x)/(A)`C. `(F cos x)/(A)`D. `(F sin^(2) x)/(A)` |
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Answer» Correct Answer - A stress `= (F_(tan theta))/(A) = (F^(1) cos theta)/(A^(1)) = (A sin theta cos theta)/(A)` |
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| 103. |
A cubical ball is taken to a depth in volume observed to be `0.1%` the bulk modulus of the ball is `(g = 10m//s^(2))`A. `2xx10^(7) Pa`B. `2xx10^(6) Pa`C. `2xx10^(9) Pa`D. `1.2xx10^(9) Pa` |
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Answer» Correct Answer - C `K = (p V)/(Delta V), p = hdg` |
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| 104. |
Two wires A and B of same material have radii in the ratio `2:1` and lengths in the ratio `4:1`. The ratio of the normal forces required to produce the same change in the lengths of these two wires isA. `1:1`B. `2:1`C. `1:2`D. `1:4` |
| Answer» Correct Answer - A | |
| 105. |
If the Bulk modulus of lead is `8.0xx10^(9)N//m^(2)` and the initial density of the lead is `11.4g//"cc"`, then under the pressure of `2.0xx10^(8)N//m^(2)`, the density of the lead isA. 11.3 g/ccB. 11.5 g/ccC. 11.6 g/ccD. 11.7 g/cc |
| Answer» Correct Answer - D | |
| 106. |
One end of a wire of length L and weight w is attached rigidly to a point in roof and a weight `w_(1)` is suspended from its lower end. If A 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))/(A)`B. `(w_(1)+(w)/(4))/(A)`C. `(w_(1)+(3w)/(4))/(A)`D. `(w_(1)+w_(2))/(A)` |
| Answer» Correct Answer - C | |
| 107. |
Two wires of same material and same diameter have lenghts in the ratio `2:5`. They are strechted by same force. The ratio of work done in strechting them isA. `5:2`B. `2:5`C. `1:3`D. `3:1` |
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Answer» Correct Answer - B `E alpha l, E = (1)/(2) (F^(2) l)/(AY)` |
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| 108. |
Is stress a vector quantity ? |
| Answer» Stress = `("Magnitude of internal reaction force")/("Area of cross-section")` Therefore, stress is a scalar quantity not a vector quantity. | |
| 109. |
The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, willA. be doubleB. be halfC. be four timesD. remain same. |
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Answer» Correct Answer - D Breaking force = Breaking stress `xx` Area of cross-section . When the wires is cut to half, there is no change in the area of the area of cross-section of their wire. Therefore , there is not change in the area of the maximum load (i.e., braking force ) the wire can support. |
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| 110. |
The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, willA. be doubleB. be halfC. be four timesD. remain same |
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Answer» Correct Answer - D We know that, Breaking stress = `("Breaking force")/("Area of cross-section") " "(i)` When length of the wire changes, area of cross-section remains same. Hence, breaking force will be same when length changes. |
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| 111. |
A steel rod 2.0 m long has a cross-sectional area of `0.30cm^(2)`. The rod is now hung by one end from a support structure, and a 550 kg milling machine is hung from the strain, and the elongation of the rod. |
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Answer» `"Stress"=(F_(_|_))/(A)=((500kg)(9.8m//s^(2)))/(3.0xx10^(-5)m^(2))` `=1.8xx10^(8)Pa` `"Strain "=(Deltal)/(l_(0))=("Stress")/(Y)=(1.8xx10^(8)Pa)/(20xx10^(10)Pa)` `=9.0xx10^(-4)` Elongation = `Deltal` `="(strain)"xxI_(0)` `=(9.0xx10^(-4))(2.0m)` `=1.8 mm` |
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| 112. |
A 4 cm cube has its upper face displaced by 0.1 mm by a tangential force of 8 kN. Calculate the shear modulus of the cube. |
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Answer» Here, each side of the cube L = 4 cm Area of the face over which the force is applied, `a=L^(2)="16 sp. Cm"` Displacement, `DeltaL=0.1 mm = 0.01 cm` Force applied, `F=8kN=8000xx10^(5)=8xx10^(8)` dyne As, `G=(FL)/(aDeltaL)` `G=(8xx10^(8)xx4)/(16xx0.01)=2xx10^(10)"dyne/sq. cm"` |
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| 113. |
Assuming that shear stress at the base of a mountain is equal to the force per unit area due to its weight. Calculate the maximum possible height of a mountain on the earth if breaking stress of a typical rock is `3xx10^8Nm^-3` and its density `3xx10^-3kgm^-3`. (Take `g=10ms^-2`)A. 4 KmB. 8 KmC. 10 KmD. 16 Km |
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Answer» Correct Answer - C For a mountain of height h and base area A, weight `W = Ahrhog` So, pressure at the base sue to its own weight is `P = (W)/(A) = (Ahrhog)/(A) = hrhog` The mountain will exist, if `hrhog le` Breaking stress `therefore h le ("Breaking stress")/(rhog)` `h le (30xx 10^(7))/(3xx 10^(3) xx 10) = 10^(4) m = 10 Km` `therefore h_("max") = 10 km` |
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| 114. |
One end of a nylon rope of length 4.5 m and diameter 6 mmis fixed to a tree limb. A monkey weighing 100 N jumps to catch the free end and stays there. Find the elongation of the rope and the corresponding change in the diameter. Young modulus of nylon =0.2.A. `1.327 mu m`B. `0.151 um m`C. `0.625 mu m`D. ` 0.425 mu m` |
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Answer» Correct Answer - A Here l = 9 m, D= 6 mm `= 6xx10^(-3) m`, F = 200 N, `Y= 4.8xx 10^(11) N m^(-2), sigma = 0.2` As ` Y = (F)/(L) .(l)/(Deltal)` `therefore Deltal = (F)/(A) .(l)/(Y) = (200 xx 9)/(3.14 xx (6xx 10^(-3))^(2) xx 4.8 xx 10^(11))` `= 13.269 xx 10^(-5)m` |
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| 115. |
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6mm. What is the maximum tension that can be exterted by the riveted strip if the shearing stress on the rivet is not to exceed `6.9 xx 10^(7) Pa`? Assume that each rivet is to carry one quarter of the load .A. `2xx10^(3)N`B. `3.9xx10^(3)N`C. `7.8xx10^(3)N`D. `15.6xx10^(3)N` |
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Answer» Correct Answer - C Radius of a rivet `r= (6)/(2) mm = 3 mm = 3 xx 10^(-3) m` Maximum stress `= 6.9 xx 10^(7) Pa` Maximum load on a river = Maximum stress `xx` Area of cross-section `= 6.9 xx 10^(7) xx pi xx (3 xx 10^(-3))^(2) = 1950 N` `therefore` Maximum tension that can be exerted by rivet strip `= 4 xx 1950 N = 7800 N = 7.8 xx 10^(3)N` |
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| 116. |
With what minimum acceleration can a fireman slide down a rope whose breaking strength is `(2//3)` of his weight?A. `(g)/(2)`B. `(2)/(3)g`C. `(3)/(2)g`D. `(g)/(3)` |
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Answer» Correct Answer - D Net downward force in the string, ma = mg - T But `T_("min") = (2)/(3) mg " " therefore " "ma_("min") = mg - (2)/(3) mg a_("min") = g - (2)/(3) g = (g)/(3)` |
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| 117. |
Find out longitudinal stress and tangential stress on a fixed block. |
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Answer» Logitudinal or normal stress `rArr sigma_(1) = (100sin 30^(@))/(5xx2) = 5N//m^(2)` Tangential stress `rArr_(1) = (100cos 30^(@))/(5xx2) = 5 sqrt(3)N//m^(2)` |
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| 118. |
The marina Trench is located in the pacific ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the Trench is about `1.1 xx 10^(8) Pa`. A steel ball of initial volume `0.32 m^(3)` is dropped into the ocean and falls to the bottom of the Trench. what is the change in the volume of the ball when it reaches to the bottom? Bulk modulus for steel =`1.6 xx 10^(11) Nm^(-2)`. |
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Answer» Water pressure at the bottom, p `=1.1xx10^(8)` Pa Initial volume of the steel ball, `V=0.32"m"^(3)` Bulk modulus of steel, B = `1.6xx10^(11)"Nm"^(-2)` The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface. Let the change in the volume of the ball on reaching the bottom of the trench be `DeltaV`. Bulk modulus, `B=((p)/(DeltaV))/(V)` `DeltaV=(B)/(pV)` `=(1.1xx10^(8)xx0.32)/(1.6xx10^(11))` `=2.2xx10^(-4)"m"^(3)` Therefore, the change in volume of the ball on reaching the bottom of the trench is `2.2xx10^(-4)"m"^(3)`. |
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| 119. |
A wire can sustain the weight of `40kg` before breaking. If the wire is cut into `4-`n equal parts, each part can sustain a weight of `...kfg`A. `40`B. `160`C. `10`D. `20` |
| Answer» Correct Answer - A | |
| 120. |
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. be doubleB. be halfC. be four timesD. remain same |
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Answer» Correct Answer - D Breaking force `prop A` Braking stress is a constant for a given material and it does not depedning upont length or thickness of wire. |
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| 121. |
Assertion: The stress-strain behaviour varies from material to material. Reason: A rubber can be pulled to several time its original length and still returns to its originshape. |
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Answer» Correct Answer - B The stress-strain curves vary from material to material. These curves deforms with increasing loads. Rubber is an elastomer. |
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| 122. |
The value of force constant between the applied elastic force `F` and displacement will be A. `sqrt(3)`B. `(1)/(sqrt(3))`C. `(1)/(2)`D. `(sqrt(3))/(2)` |
| Answer» Correct Answer - B | |
| 123. |
What happens to the elastic property of a substance after annealing (cooling slowly after heating)A. increasesB. decreasesC. remains constantD. become zero |
| Answer» Correct Answer - B | |
| 124. |
The stress strain curves for brass, steel and rubber are shown in the figure. The lines A, B and C are forA. rubber, brass and steel respectivelyB. brass, Steel and rubber respectivelyC. steel, brass and rubber respectivelyD. steel, rubber and brass respectively. |
| Answer» Correct Answer - C | |
| 125. |
The diagram shoes stress v/s strain curve for the materials A and B. From the curves we infer thatA. `A` is brittle but `B` is ductieB. `A` is ductile and `B` is britteC. both `A` and `B` are ductileD. both `A` and `B` are brittle |
| Answer» Correct Answer - B | |
| 126. |
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. `0.25`B. `0.50`C. `2.00`D. `4.00` |
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Answer» Correct Answer - C `k_(1) = (pi4R^(2)x)/(2L), k_(2) = (piR^(2)y)/(L)` `F = k_(1) x = k_(2) y` `rArr (y)/(x) = (k_(1))/(k_(2)) = 2` |
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| 127. |
A wire can be broken by `400kg.wt`. The load required to break the wire of double the thickness of the same material will be (in `kgwt`.)A. `800`B. `1600`C. `3200`D. `6400` |
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Answer» Correct Answer - B `B.S = (F)/(A), F alpha r^(2)` |
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| 128. |
A spring is stretched by applying a load to its free end. The strain produced in the spring isA. volume strainB. longitudinal strianC. shearing strainD. all the above |
| Answer» Correct Answer - C | |
| 129. |
A massless rod of length `l` is hinged at one end and is held horizontal by two identical vertical wires, which are tied at distances `a` and `b` form the hinged end. `A` load `P` si applied at the free end of the rod. The tension in the secound wire isA. `(Pl)/(a^(2) + b^(2))a`B. `(Pl)/(a^(2) + b^(2))b`C. `(Pl)/(a + b)a`D. `(Pl)/(a + b)b` |
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Answer» Correct Answer - B The rod is in rotational equilibrium taking torque about the hinge. sum tau `= T_(1) a + T_(2)b - Pl = 0......(1)` As the wires are elastic, `T_(1) = k( Delta l_(1))` and `T_(2) - k(Delta l_(2))` Hence, `(T_(1))/(T_(2)) = (Delta l_(1))/(Delta l_(2)) = (a)/(b)` On solving equs (1) and (2) for `T_(1)` and `T_(2)`, we get `T_(1) = (Pl)/(a^(2) + b^(2))a, T_(2) = (pl)/(a^(2) + b^(2))b` |
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| 130. |
The bulk modulus for an incompresssible liquid isA. infinityB. unityC. zeroD. between `0` and `1` |
| Answer» Correct Answer - A | |
| 131. |
A rubber ball of bulk modulus `B` is taken to a depth h of a liquid of density `rho`. Find the fractional change in the radius of the ball.A. `(delta r)/(r) = (rho gh)/(3B)`B. `(delta r)/(r) = (rho gh)/(2B)`C. `(delta r)/(r) = (3rho gh)/(B)`D. `(delta r)/(r) = (2rho gh)/(B)` |
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Answer» Correct Answer - A The volumetric strain `(delta v) = - (rho)/(B)`, where `P = rho gh` Then, `- (delta v)/(v) = (pgh)/(B)` Since the volume of the sphere is `v = (4)/(3) pi r^(3)`, We have `(delta v)/(v) = (3delta r)/(r)` Using eqs(i) and (ii) we have `(delta r)/(r) = (rho gh)/(3B)` |
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| 132. |
Bulk modulus for rubber is `9.8xx10^(8)Nm^(-2)`. To what depth should a rubber ball be taken in a lake so that its volume is decreased by `0.1%`A. `0.1m`B. `1m`C. `10m`D. `100m` |
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Answer» Correct Answer - D `B = (Delta P)/((Delta V//V)) rArr Delta P = B xx (Delta V)/(V)` `rArr Delta P = 9.8xx10^(8) xx10^(-3) = 9.8xx10^(5) Nm^(-2)` `Delta P = rho gh` `h = (Delta P)/(rho g) = (9.8xx10^(5))/(10^(3)xx9.8) rArr h = 10^(2) m = 100m` |
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| 133. |
When a rubber ball of volume `v`, bulk modulus `K` is at a depth `h` in water then decreases in its volume isA. `(h rho gv)/(K)`B. `(h rho gv)/(2K)`C. `(h rho gv)/(3K)`D. `(h rho gv)/(4K)` |
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Answer» Correct Answer - A `K = - (Delta pV)/(Delta V)` |
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| 134. |
Out of solids , liquids and gases, which one has all the three types of modulus ofelasticity and why gases have only bulk modulus of elasticity.A. SolidsB. LiquidsC. GasesD. Both solids and liquids |
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Answer» Correct Answer - A Solids |
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| 135. |
A glass slab is subjected to a pressure of 10 atm. The fractional change in its volume is (Bulk modalus of glass `=37 xx 10^(9) Nm^(-2), 1 atm= 1 xx 10^(5) N m^(-2)`)A. `2.7 xx10^(-2)`B. `2.7 xx 10^(-3)`C. `2.7 xx 10^(-4)`D. `2.7 xx 10^(-5)` |
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Answer» Correct Answer - D Bulk modulus, `B = (P)/(DeltaV//V)` `therefore` Fractional change in volume, `(DeltaV)/(V)= (P)/(B)` Here, `P = 10 "atm" = 10 xx 1 xx 10^(5) Nm^(2) , B = 37 xx 10^(9) N m^(-2)` `therefore (DeltaV)/(V) = (1xx10^(6) N m^(-2))/( 37 xx 10^(9)N m^(-2))= 0.027 xx 10^(-3) = 2.7 xx 10^(-5)` |
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| 136. |
what is the Bulk modulus for a perfect rigin body ? |
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Answer» Bulk modulus (K) `=(p)/(DeltaV//V)=(pV)/(DeltaV)` For perfectly rigid body, change in volume `DeltaV = 0` `therefore" "K = (pV)/(0) = oo` Therefore, bulk modulus for a perfectly rigid body is infinity `(oo)`. |
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| 137. |
To What depth must a rubber ball be taken in deep sea so that its volume is decreased y 0.1 %. (The bulk modulus of rubber is `9.8 xx 10^8 Nm^(-2)` , and the density of seac water is `10^3 kg m^(-3).)` |
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Answer» Given, Bulk modulus of rubber `(K) = 9.8 xx 10^(8) N//m^(2)` Density of sea water `(rho) = 10^(3) kg//m^(3)` Percentage decrease in volume, `((DeltaV)/(V) xx 100) = 0.1 implies (DeltaV)/(V) = (0.1)/(100)` `implies" "(DeltaV)/(V) = (1)/(1000)` Let the rubber ball be taken upto depth h. `therefore` Change in pressure (p) = `hrhog` `therefore` Bulk modulus `(K) = |(p)/(DeltaV//V)|=(hrhog)/((DeltaV//V))` `implies " "h=(Kxx(DeltaV//V))/(rhog)=(9.8xx10^(8)xx(1)/(1000))/(10^(3)xx9.8)=100m` |
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| 138. |
A spring of force constant `800N//m` has an extension of 5cm. The work done in extending it from 5cm to 15cm is |
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Answer» `W = (1)/(2) K (x_(2)^(2) - x_(1)^(2)) = (1)/(2) xx 800 [15^(2) - 5^(2)] xx 10^(-4)` `= 400xx [225-25] xx 10^(-4) = 4xx10^(-2) xx 200 = 8J` |
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| 139. |
A sphere of mass `m` attached with the free end of a steel wire of length `l` swings in the veritical plane form the horizontal positon. Elongation of the wire in the vericle positon isA. `(mgl)/(Y(pi r^(2)))`B. `(2mgl)/(Y(pi r^(2)))`C. `(mgl)/(3Y(pi r^(2)))`D. `(3mgl)/(Y(pi r^(2)))` |
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Answer» Correct Answer - D `Delta l = ((F_("net")l))/(YA) , V = sqrt(2gl)` `= ((mg + (mv^(2))/(l))l)/(YA) = ((mg + 2mg)l)/(YA)` `Deltas L = (3mgl)/(Y(pi r^(2))), v = (1)/(2)Fe, U = (YA)/(2l) xx (Delta l)^(2) = (9m^(2) g^(2) l)/(2Y pi r^(2))` |
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| 140. |
A thin cylindrical rod of length `2.5m` and radius `5mm` is firmly fixed at upper end when lower end is twiced, the shear angle is found to be `0.06^(@)` then angle of twising isA. `10^(@)`B. `20^(@)`C. `30^(@)`D. `40^(@)` |
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Answer» Correct Answer - C `phi = (r theta)/(l)` |
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| 141. |
A body of mass M is attached to lower end of a metal wire whose upper and is fixed. The elongation is l. The ratio of loss of gravitational potential energy to the energy stored in the wire isA. 1B. 2C. `(1)/(2)`D. `(4)/(3)` |
| Answer» Correct Answer - B | |
| 142. |
Assertion : Strain is a unitless quantity. Reason : Strain is equivalent to force |
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Answer» Correct Answer - C Strain is the ratio of change in dimensions of the body to the original dimension . Because it is a ratio, it is a dimensionless quantity. |
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| 143. |
The stress-strain graphs for two materials are shown in Fig. 7(EP).3 (assume scale). A. Material (ii) is more elastic than material (i) and hence material (ii) is more brittleB. Material (i) and (ii) have the same elasticity and the same brittlenessC. Material (ii) is elastic over a larger region of strain as compared to (i)D. Material (ii) is more brittle than material (i) |
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Answer» Correct Answer - C::D It is clear from the two graphs, the ultimate tensile strength for material (ii) is greater, hence material (ii) is elastic over larger region as compared to material (i). For material (ii) fracture point is nearer, hence it is more brittle. |
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| 144. |
The length of a rubber cord is `l_(1)` m when the tension is 4 N and `l_(2)` m when the tension is 6 N.The length when the tension is 9 N, isA. `(2.5l_(2) - 1.5l_(1))m`B. `(6l_(2)- 1.5l_(1))m`C. `(3l_(1) - l_(2))m`D. `(3.5l_(2) =- 2.5l_(1))m` |
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Answer» Correct Answer - A Let the original unstrectched length be l. `therefore Y=(T)/(A)(l)/(Deltal)` Now `Y =(4)/(A)(l)/((l_(1)-l))=(6)/(A)(l)/((l_(2) -l)) =(9)/(A)(l)/((l_(3)-l))` `therefore 4(l_(3)-l) = 9(l_(1)-l) or 4l_(3)+ 5l = 9l_(1)" "…..(i)` Again, `6(l_(3)-l) =9(l_(2) -l) or 2l_(3) + l = 3l_(2)" "...(ii)` To obtain `l_(3)`,solve (i) and (ii) simultaneously. `therefore l_(3) = (2.5l_(2) - 1.2l_(1))m` |
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| 145. |
When the tension on a wire is `4N` its length is `l_(1)`. When the tension on the wire is `4N` length is `l_(1)`. When the tension on the wire is `5N` its length is `l_(2)`. Find its natural length.A. `5l_(1) - 4l_(2)`B. `4l_(1) - 5l_(2)`C. `10l_(1) - 8l_(2)`D. `8l_(1) - 10l_(2)` |
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Answer» Correct Answer - A `(l_(1) - l)/(l_(2 - l) = (T_(1))/(T_(2))` |
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| 146. |
A block of weight `15N` slides on a horizontal table the co-efficient of siliding fricition is `0.4`. The area of the block in contact with the table is `0.5m^(2)`. The shearing stress will beA. `120 Nm^(-2)`B. `140 Nm^(-2)`C. `160 Nm^(-2)`D. `180 Nm^(-2)` |
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Answer» Correct Answer - A Stress `= (mu mg)/(A) = F = f = mu mg` |
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| 147. |
A steel cable with a radius of 1.5 cm support a chairlift at a ski area.if the maximum stress is not to exceed `10^(8) Nm^(-2)`, what is the maximum load the cable can support? |
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Answer» Radius of the steel cable, r = 1.5 cm = 0.015 m Maximum allowable stress = `10^(8)"N m"^(-2)` Maximun stress= `("Maximum force")/("Area of cross-section")` `therefore` Maximum force = Maximum stress `xx` Area of cross-section `=10^(8)xxpi(0.015)^(2)` `7.065xx10^(4)` N Hence, the cable can support the maximum load of `7.065xx10^(4)` N. |
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| 148. |
A steel cable with a radius 2 cm supports a chairlift at a ski area. If the maximum stress is not to exceed `108 N m^(-2)`, the maximam load the cable can supportA. `4pi xx 10^(5)N`B. `4pi xx 10^(4) N`C. `2pi xx 10^(5) N`D. `2pi xx 10^(4)` |
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Answer» Correct Answer - B Here, `r = 2 cm = 2xx 10^(-2) m` Maximum load = Maximum stress`xx`Area of the cross-section `= 10^(8) N m^(-2) xx pixx(2xx 10^(-2)m)^(2)` `= 4 pi xx 10^(4)N` |
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| 149. |
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1m, is whirled in a vertical circle with an angular velocity of `2 rev.//s` at the bottom of the circle. The cross-sectional area of the wire is `0.065 cm^(2)` . Calculate the elongaton of the wire when the mass is at the lowest point of its path `Y_(steel) = 2 xx 10^(11) Nm^(-2)`.A. 0.52 mmB. 1.52 mmC. 2.52 mmD. 3.52 mm |
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Answer» Correct Answer - C Here, m = 15 kg , r = L = 1 m `v = 2 "rev s"^(-1), A = 0.05 cm^(2) = 0.05 xx 10^(-4) m^(2)` When the mass is the lowest point of the vertical circle, the stretching force is `F = mg + mromega^(2) = mg + mr (2piv)^(2)` `= 15 xx 10 + 15 xx (1)xx (2pixx2)^(2) = 2516 N` As ` Y= (F//A)/(DeltaL//L) therefore DeltaL = (FL)/(AY)` `DeltaL = (2526N xx 1 m)/(0.05 xx 10^(-4)m^(2) xx 2 xx 10^(11))= 2.52 xx 10^(-3)m = 2.52 mm` |
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| 150. |
A rod of length 1.05 m having negliaible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in fig. The cross-sectional area of wire A and B are `1 mm^(2)` and 2` mm^(2)`, respectively . At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires. Given, `Y_(steel) = 2 xx 10^(11) Nm^(-2) and Y-(alumi n i um) = 7.0 xx 10^(10)N^(-2)` A. Mass `m` should be suspended close to wire `A` to have equal stresses in both the wires.B. Mass `m` should be suspended close to `B` to have equal stresses in both the wires.C. Mass `m` should be suspended at the middle of the wires to have equal stresses in both the wires.D. Mass `m` should be suspended close to wire `A` to have equal strain in both wires. |
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Answer» Correct Answer - B::D Let the mass is suspended at `x` from the end `B`, which develop equal stress in wires. Let `T_(A)` and `T_(B)` be the tension in wire `A` and wire `B` respectively. Stess in steel wire `A, S_(A) = (T_(A))/(A_(B)) = (T_(A))/(10^(-6))` Stress in `Al` wire , `S_(B) = (T_(B))/(A_(B)) = (T_(B))/(2xx10^(-6))` where `A_(A)` and `A_(B)` are cross-sectional areas of wire `A` and `B` respectively. Also from rotational equilibrium, net torque is zero, i.r, `T_(B)x - T_(A) (l-x) = 0` `rArr (T_(B))/(T_(A)) = (l-x)/(x)` For equal stress, `S_(A) = S_(B)` `rArr S_(A) = S_(B) rArr (T_(A))/(10^(-6)) = (T_(B))/(2xx10^(-6))` `rArr (l-x)/(x) = 2 rArr (l)/(x) - 1 = 2 rArr x = (l)/(3)` `:. l - x = l - (l)/(3) = (2l)/(3)` Hence, mass `m` should be suspended close to wrie `B`(`Al` wire) We know, Strian `= ("stress")/(Y)` So, for equal strain in the wires, `rArr (S_(A))/(Y_("steel")) = (S_(B))/(Y_(Al)` `rArr (Y_("steel"))/(T_(A)//a_(A)) = (Y_(Al))/(T_(B)//a_(B))` `rArr (Y_("steel"))/(Y_(Al)) = (T_(A))/(T_(B)) xx (a_(B))/(a_(A)) = ((x)/(l-x)) ((2a_(A))/(a_(A)))` `rArr (200xx10^(9))/(70xx10^(9)) = (2x)/(l-x) rArr (20)/(7) = (2x)/(l-x)` `rArr 17x = 10l rArr x = (10l)/(17)` `rArr l-x = l - (10l)/(17) = (7l)/(17)` Hence, mass `m` should be suspended close to wire `A`(steel wire). |
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