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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of `30^@`. Then angle of shear isA. `12^(@)`B. `1.2^(@)`C. `0.12^(@)`D. `0.012^(@)` |
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Answer» Correct Answer - C Angle of twist at free end `=30^(@)=(30)/(180)xxpi` rad `=(pip)/(6)` rad Displacement of the free surface, `Delta L=(2pi r)/(2 pi)xx(pi)/(6)=(pi r)/(6)=(pixx0.4)/(6)cm` Angle of shear or shearing strain `=(Delta L)/(L)` `=(pixx0.4)/(6xx100)` rad. `=(pixx0.4)/(6xx100)xx(180)/(pi)` degree `=0.12^(@)` |
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| 2. |
Two rod A and B of the same material and dame length have radii `r_(1) and r_(2)` respectively. When they are rigidly fixed at one end and twisted by the same couple applied at the other end, find the ratio of the angles of twist at the ends of A and B. |
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Answer» If a wire of length l, and radius r is fixed at one end and a torque `tau` is applied at the other end produces angle of twist `theta` in the wire, then `tau = (pi G r^(4)theta)/(2l) or theta = (2 tau l)/(pi G r^(4))` i.e., `theta prop 1/(r^4) so (theta_(1))/(theta_2) = (r_(2)^(4))/(r_(1)^(4))`. |
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| 3. |
Explain why it is difficult to make mercury enter a fine therometer tube? |
| Answer» Since the cohesive force between the molecules of mercury are very large as compared to adhesive forces between molecules of mercury and glass containers. Due to it, the mercury does not wet the surface of a glass container. That is why, it is difficult to make mercury enter a fine therometer tube. | |
| 4. |
A balloon filled with helium does not rise in air indefinitely but halts after a certain height. (Neglect winds). Explain |
| Answer» in the beginning the balloon filled with helium rieses in air because weight of the air displaced by balloon is more than the weight of the balloon and helium gas filled inside balloon. We know that the density of air and the value of ac celeration due to gravity decreases with height. due to it, the weight of air decrease at greater height. The balloon halts at such a height where the weight of the air displaced just equal to the weight of helium gas and the balloon. | |
| 5. |
A stone of density `2.5 xx10^3 kg m^(-3)` completely immersed in sea water is allowed to sink from rest in 2 s. Neglect the effect of vicosity. Relative density of sea water is 1.025. |
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Answer» Correct Answer - 11.6 m Let m be the mass of stone. Volume of stone. `V = (m)/(2.5xx10^3)m^3` upward thrust on stone. ` = (m)/(2.5)xx10^3 xx(1.025 xx 10^3)xxg` Weight of stonw = mg , acting vertically downwards. The resultant downward force on stone is `F = mg - (1.025)/(2.5)mg = mg[1 - (1.025)/(2.5)]` The ac celeration of stone is `a = (F)/(m) = g[1 - (1.025)/(2.5)] = 9.8 xx (1.475)/(2.5)` ` =5.782 ms^(-2)` Starting from rest the distance travelled by stone is `S = (1)/(2)a t^2 = (1)/(2)xx(5.782)xx2^2 ~~11.6 m` |
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| 6. |
A 1-L flask contains some mercury. It is found that at different temperature, the volume of air inside the flask remains the same. What is the volume of mercury in the flask, given that the coefficient of linear expansion of glass`=9xx10^(-6)//^(@)C` and the coefficient of volume expansion of `Hg=1.8xx10^(-4)//^(@)C` ? |
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Answer» For glass `gamma_(g) = 3 alpha_(g) = 3 xx 9 xx 10^(-6)` ` = 27 xx 10^(-6) .^(@)C^(-1)` Volume of air in the flask, ` V =1 litre = 1000 cm^(3)` Let `V_(m)` be the volume of mercury in the flask and `V_(a)` be the volume of air in the flask. Volume of gas flask = volume of mercury + volume of air `V =V_(m) + V+(a)` As `V_(a) ` remains constant, when temperature is raised by `Delta T` , so ` Delta V = Delta V_(m)` or, `V gamma_(g) Delta T = V_(m) gamma_(m) DeltaT` ` or V_(m) = (gamma_(g)V)/(gamma_m) = ((27 xx 10^(-6)) xx 1000)/(1.8 xx 10^(-4)) = 150 cm^(3)`. |
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| 7. |
Can you measure temperature upto `500^(@)C` with a mercury thermometer, knowing that the mercury boils at `357^(@)C`? |
| Answer» Yes, we can measure temperature upto `500^(@)C` with a mercury thermometer provided the space above mercury in thermomter is filled with nitrogen. This increases the boiling point of mercury. | |
| 8. |
A boat having a length 4m and breadth 2.5 m is floating on lake. The boat sinks by 2 cm when a load is loaded on it. What is the weight of the load ? Use `g =10 ms^(-2),` density of water ` = 10^3 kg m^(-3)` |
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Answer» Correct Answer - 200 kg f Area of the boat `A = 4xx 2.5 = 10m^2.` when load is placed on the boat. Depression of the boat = 2 cm = 0.02m. Volume of the water displaced by the load in boat ` = 10xx 0.02 m^3 = 0.20 m^3` Ac cording to principle of flotation weight of load = weight of water displaced ` = 0.20 xx 10^3 xx 10 N` ` =(0.02xx10^3xx10)/(10)kg" " f = 200 kg f` |
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| 9. |
Specific heat of argon at constant pressure is `0.125 cal.g^(-1) K^(-1)`, and at constant volume `0.075 cal.g^(-1)K^(-1)`. Calculate the density of argon at N.T.P. Given `J = 4.18 xx 10^(7)" erg "cal^(-1)` and normal pressure = `1.01 xx 10^(6)" dyne "cm^(-2)`. |
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Answer» Here, `C_(p) = 0.125 cal. G^(-1) K^(-1)`, `C_(upsilon) = 0.075 cal. G^(-1) K^(-1)` , `J = 4.18 xx 10^(7)erg cal.^(-1)` Normal pressure , `P = 1.01 xx 10^(6)dyn e//cm^(2)` Normal temperature , `T = 273^(@)K` we have to calculate density , `rho =?` As, gas constant for 1 g of gas `r = (P upsilon)/(T) = (P)/(rho T)` Where, `upsilon` = volume oc cupied by 1 gram of gas Now, `C_(p) - C_(upsilon) - r/J` or, `r = (C_(p)-C_(upsilon))J` `(P)/(rho T) = (0.125 - 0.075) 4.18 xx 10^(7)` or `(1.01 xx 10^(6))/(rho xx 273) = 0.05 xx 4.18 xx 10^(7)` `:. rho = (1.01 xx 10^(6))/(273 xx 0.05 xx 4.18 xx 10^(7))` `=1.77 xx 10^(-3) g cm^(-3)`. |
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| 10. |
Calculate the specific heat capacity at constant volume for a gas. Given specific heat capacity at constant pressure is `6.85 cal mol^(-1) K^(-1), R = 8.31 J mol^(-1)K^(-1)`. `J - 4.18 J cal^(-1)`. |
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Answer» Here, `C_(p) = 6.85 cal mol^(-1)K^(-1)`, `R=8.31 mol^(-1) K^(-1)` `:. R = 8.31/4.18 cal mol^(-1)K^(-1)` `C_(V)=C_(p) - R = 6.85 - 1.988` `=4.862 cal mol^(-1)K^(-1)`. |
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| 11. |
A light cylindrical vessel is kept on a horizontal surface. Its base area is A. a hole of cross sectional area a is made just at its bottom sdie (where a ltlt A). Find minimum coefficeint of friction necessary for sliding of the vesssel due to the impact force of the emerging liquid. |
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Answer» Let h be the height of liquid in vessel above the hole. Velocity of efflux, `upsilon = sqrt(2 gh)` Impact force of the emerging liquid on the vessel and liquid contents is `F = upsilon (dm)/(dt) = upsilon(a rho upsilon) = a rho upsilon^(2) = a rho 2 gh = 2 arho gh` The force of friction, `f = mu R = muA h rho g` For just sliding the vessel, f =f So, `mu A h rhog = 2 a rho gh or mu = (2a)/(A)` |
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| 12. |
A light clindrical vessel is kept on a horizontal surface. The base are is A. A hole of cross-sectional area a is made just at its bottom side (where alt ltA) fig. Find minimum coeffiecent of friction necessary for sliding of the vessel due to the impact of the emerging liquid. |
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Answer» Let h be the height of liquid in vessel above the hole. Velocity of efflux, `upsilon = sqrt(2 gh)` Impact force of the imerging liquid on the vessel and liquid contents is `F = upsilon (dm)/(dt) = upsilon(a rho upsilon) = a rho upsilon^(2) = a rho 2 gh` `= 2 a rho gh ` The force of friction, `f = mu R =mu A h rho g` For just sliding the vessel , `f =F` so, `mu A h rho g = 2 a rho g h or mu = (2a)/(A)`. |
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| 13. |
A cylinder containing water up to height 25 cm has a hole of cross-section `0.25 cm^(2)` at its bottom. It is counter possed in a balance. What is the initial change in the balancing weight when water begins to flow out? |
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Answer» Here, `a = 0.25 cm^(2) = 0.25 xx 10^(-4)m^(-2)`, . `h = 25 cm = 0.25 m` Initial velocity of water flowing out of hole `upsilon = sqrt(2gh)` When wate emerges out from the hole. The weight of water decrease. The decrease in weight is equal to the rate of change of linear momentum. Initially velocity of liquid flowing out `upsilon` be taken constant for a small interval of time `Delta t`. then `F = (Delta p)/(Delta t) = (Delta (m upsilon))/(Delta t) = upsilon (Delta m)/(Delta t)` `=(upsilonDelta(V rho))/(Delta t) = upsilon rho (Delta V)/(Delta t)` Where, `(Delta V)/(Delta t)` = volume flowing out per second `=aupsilon` `F = upsilon rho.(a upsilon) = a upsilon^(2) rho` `=a(2gh) rho = 2 a gh rho` `= 2 xx (0.25 xx 10^(-4)) xx 9.8 xx 0.25 xx 10^(3)` ltrgt `=12.5 xx 9.8 xx 10^(-3) N = 12.5 xx 10^(-3)kg f`. |
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| 14. |
The water level on a tank is 5m high. There is a hole of `1 cm^(2)` cross-section at the bottom of the tank. Find the initial rate with which water will leak through the hole. (`g= 10ms^(-2)`) |
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Answer» Rate of leakage of water, `V = aupsilon = asqrt(2gh) = (1xx10^(-4)) xx sqrt(2 xx 10 xx 5)` `=(10^(-3))m^(3) s^(-1)`. |
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| 15. |
What is a fluid? Show that fluid exerts a thrust. |
| Answer» Fluid is the mane given to a substance which begins to flow when external force is applied on it . When the fluid is kept in a container, the molicules of a fluid in random motion due to their thermal velocities are constantly collinding with the walls of the container and rebounding from them. They suffer a change in momentum. Due to it, they transfer some momentum to the wall. This mumentum transfered to the wall perunit time by the molecules of flluid ac counts for the thrust of liquid on the wall of the container. | |
| 16. |
A capillary tube of 1mm diameter and 20 cm long is fitted horizontally to a vessel kept full of alcohol of density 0.8 gm//c.c. The depth of centre of capillary tube below the surface of alcohol is 20 cm. If the visosity of oil is `0.12 dyn e cm^(-2)` s, find the amount of liquid that will flow in 5 minuts. |
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Answer» Correct Answer - 3.85 gms Amount of liquid flowing in time t= volume of liquid flowing per sec xx density of liquid xx time `= (pi p r^4)/(8 eta l)xxrhoxxt = (pi (h rho g)r^4 rho t)/(8 eta l)` |
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| 17. |
in a plate, a sucrose solution of coefficient of visocity `1.5 xx10^(-3) Nsm^(-2)` is driven at a velocity of `10^(-3) ms ^(-1)` throught x ylem vessel of radius `2mu` m and length`5mu`m. Find the hydrostatic pressure difference across the length of xylem vessel. |
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Answer» Correct Answer - `15 Nm^(-2)` Here, `eta = 1.5 xx10^(-3) N.s m^(-2) , upsilon = 10^(-3) ms^(-1),` `r = 2xx 10^(-6)m, l = 5xx 10^(-6)m` `V = pi r^2 upsilon = (pi p r^4)/(8 eta l)` `or p = (8 eta l upsilon)/(r^2) = 8xx1.5 xx10^(-3)xx(5xx10^(-6)xx10^(-3))/((2xx10^(-6))2)` `= 15 N m^(-2)` |
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| 18. |
A tank of square cross-section of each side is filled with a liquid of height h . Find the thrust experienced by the vertical surfaces and bottom surface of the tank. |
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Answer» Over the verticla surface, the pressure due to liquid is not same at all point . The points at Greater depth experience large pressure and hence greater thrust. Average pressure on the vertical surface of tank in contact with liquid. `P_(av) = ((0+h rhog))/(2) = 1/2 h rho g` `:.` Thrust on the vertical surfaces of tank `P_(av) xx area = ((h rho g)/(2)) xx 4 hl = 2 glh^(2)` Over the bottom surface, the pressure due to liquid is uniform at all points which is given by. `P=h rho g` `:.` Thurst on the bottom surface = `P xx l^(2)` `= h rho g l^(2)`. |
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| 19. |
A tank 5 m high is half filled with water and then is filled to top with oil of density `0.85 g//cm^3` The pressure at the bottom of the tank, due to these liquids is |
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Answer» Correct Answer - `462.5 g wt. cm^(-2)` `Pressure = ("total weight of water and liquid")/("base area of tank")` `A xx (500//2)xx1xx980` ` = (+Axx500//2)xx0.85xx(980)/(A) dyn e //cm^2` ` =(250 xx 980 + 250 xx0.85 xx 980)/(980)g" wt "cm^(-2)` ` = 462.5 g wt. cm^(-2)` Alternative method, Average density =` (0.85 xx 250 + 1 xx 250)/(500)` =` 0.925 g//c.c`. `P = h rho g = 500xx0.925 xx 980` `= 462.5 xx 980 dyn e//cm^2` ` =462.5 g wt. cm^(-2)` |
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| 20. |
What will be the length of mercury column in a barometer tube when the atmospheric pressure is `75 cm` of mercury and the tube is inclined at an angle of `60^@` with the horizontal direction? |
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Answer» Correct Answer - 86.6 cm `l =h//sin60^@ = 75//(sqrt3//2) = 75xx2//sqrt3` = 86.6 cm |
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| 21. |
A barometer kept in an alevator ac celerating upward reads 76 cm of mrecury. If the elevator is ac celerating upwards at `4.5ms^(-2)` find the air pressure in the elevator in cm of mercury. Density of mercuty ` =13.6 xx 10^3 kg m^(-3) G = 9.8 m//s^2` |
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Answer» Correct Answer - 111 cm of Hg When elevator moves upwards with ac celeration a then effection ac celeration = g+a Pressure ` = h rho (g+a)` ` =0.76 xx(13.6 xx 10^3) xx(9.8 + 4.5)` ` = (0.76xx13.6xx10^3 xx14.3)/(9.8xx(13.6xx10^3))m" of "Hg` ` = 1.11 m" of "Hg = 111 cm" of "Hg` |
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| 22. |
Reading of a barometer at the top and ground floors of a buliding are 75.000 cm. and 75.125 cm. respectively. The density of mercry is ` 13600 kg//m^2` and that of air is `1.36 kg//m^2` What is the height of the buliding ? |
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Answer» Correct Answer - 12.5 m Pressure difference ` = 75.125 - 75.000 xx 10^(-2) xx 13600 xx 9.8` ` = 0.125 xx 10^(-2) xx 13600 xx 9.8 Pa` if h is the height of the building, this pressure differecen must be equal to the pressure due to a column of air of height h, i.e., ` h rho f = 0.125 xx 10^(-2) xx 13600 xx 9.8` or `h = (0.125 xx 10^(-2) x 13600 xx9.8)/(1.36xx9.8) = 12.5m` |
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| 23. |
A brass rod length 50 cm and diamteer 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at `250^(@)C` if the original length are at `40^(@)C` ? Coefficient of linear expansion of brass and steel are `2.10xx10^(-5) .^@C^(-1)` and `1.2 xx 10^(-5) ^(@)C^(-1)` respectively. |
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Answer» `Delta L_1 = L_1 alpha_1 Delta T =50 xx(2.10xx10^(-5)) (250 - 40) = 0.2205 cm` `Delta L_2 = L_2 alpha_2 Delta T =50 (1.2xx10^(-5)) (250 - 40) =0.126 cm` :. Change in length of combined rod ` =Delta L_1 + Delta L_2 = 0.220 + 0.126 = 0.346 cm` |
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| 24. |
Copper of fixed volume `V` is drawn into wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is `trianglel`. Which of the following graphs is a straight line?A. `Delta l` versus `1//l`B. `Delta l` versus `l^(2)`C. `Delta l` versus `1//l^(2)`D. `Delta l` versus `l` |
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Answer» Correct Answer - B Volume `V=Al` or `A=V//l` `Y=(Fl)/(A Delta l)=(F l)/((V/l)Delta l)=(Fl^(2))/(V Delta l)` `Delta l=(F l^(2))/(Vy)` , So `Delta l prop l^(2)` |
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| 25. |
Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury `T = 435. 5xx10^(-3) Nm^(-1)` |
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Answer» Let R be the radius of big single drop formed when two small drops of radii `r_1 and r_2` collapse together. Then volume of bid drop = volume of two small drops `:. (4)/(3) pi R^1 = (4)/(3) pi r_1^3 + (4)/(3) pi r_2^3 or R^3 = r_1^3 + r_2^3 = (0.1)^3 + (0.2)^3 = 0.001 + 0.008 = 0.009 cm^3` `:. R = 0.21 cm` Change in surface area ` = 4pi R^2 - (4pi r_1^2 + 4pi_2^2)` Energy relased ` =ST xx` change in surface area `= T xx [ 4pi R^2 - (4pi r_1^2 + 4pi r_2^2)] = 4pi T[R^2 - (r_1^2 + r_2^2)]` `4xx3.142 xx 435.5 xx10^(-3) [(0.21)^2 - {(0.1)^2 + (0.2)^2}] = - 32.33xx 10^(-7)J` |
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| 26. |
A person weighing 60 kg takes in 2000 kcal diet in a day. If this energy were to be used in heating the person without any losses, his rise in temperature would be nearly (Given sp. Heat of human body is `0.83 cal g^(-1), .^@C^(-1)`A. `30^@C`B. `40^@C`C. `35^@C`D. `45^@C` |
| Answer» Correct Answer - B | |
| 27. |
The graph between two temperature scales `A` and `B` is shown in Fig. Between upper fixed point and lower fixed point there are `150` equal divisions on scales `A` and `100` on scale `B`. The relation between the temperature in two scales is given by_ A. `(t_(A)-180)/(100)=(t_(B))/(150)`B. `(t_(A)-30)/(150)=(t_(B))/(100)`C. `(t_(B)-180)/(150)=(t_(A))/(100)`D. `(t_(B)-40)/(100)=(t_A)/(1080)` |
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Answer» Correct Answer - B From graph, we note that for scale A, the lowest fixed point than `0^(@)A` and the highest point is `180^(@)A`. For scale B, the lowest point is `0^(@)B` and the highest point is `100^(@)B`. Therefore, the relation `(t_(A)-30)/(150)= (t_(B)-0)/(100)= (t_(B))/(100)` is correct. |
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| 28. |
An alumminium rod and steel wire of same length and cross-section are attached end to end. Then compound wire is hung fron a rigid support and load is suspended from the free end. Y for steel is `((20)/(7))` times of aluminium. The ratio of increase in length of steel wire to the aluminium wire isA. `20 : 3`B. `10 : 7`C. `7 : 20`D. `1 : 7` |
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Answer» Correct Answer - C Here, `Y_(s)=(20)/(7)Y_(a)` or `(Y_(s))/(Y_(a))=(20)/(7)` As `Y=(FL)/(A Delta L)` or `Delta L=(FL)/(AY)` `Delta L prop(1)/(Y)` as both the wires are of same length. `(Delta L_(s))/(Delta L_(a))=(Y_(a))/(Y_(s))=(1)/((Y_(s)//Y_(a)))=(1)/(20//7) =(7)/(20)` |
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| 29. |
A block of wood, specific gravity 0.6 and mass 90.0 g is floting in water. A hole is drilled in it removing 8.0 g of wood. The hole is filled with lead of density `11.43 g//cm^(3)`. What will be the effect on the block? |
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Answer» The block will sink. The same is explained below. Volume of the block, `V =(mass)/(density) = (90.0)/(0.6) = 150 cm^(3)` Mass of the wood removed` = 8.0 g`, volume of the hole from which the wood is removed = ` (8.0)/(0.6) = 40/3 cm^(3)` Volume of lead inserted in block `= 40/3 cm^(3)` Mass of lead in block = `40/3 xx 11.34 = 151.2 g` Total mass of block with lead `=(90-8)+151.2 = 233.2g` when the whole block is in water, volume of the water displaced = `150 cm^(3)` Weight of water displaced =` 150 xx 1 = 150 gf`, Which is less than the weight of the block. so the block will sink. |
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| 30. |
The electrical resistance in ohms of a certain thermometer varies with temperature ac cording to the approximate law: `R =R_(0)[1+alpha(T-T_(0))]` The resistances is `101.6 Omega` at the triple-point of water `273.16K`, and `165.5 Omega` at the normal melting point of lead `(600.5K)`. What is the temperature when the resistance is `123.4 Omega`? |
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Answer» Here, `R_(0)=101.6 Omega , T_(0)=273.16 K` Case (i) `R_(1)=165.5 Omega , T_(1)=600.5K`, Case(ii) `R_(2)=123.4 Omega , T_(2)=?` Using the relation `R =R_(0)[1+alpha(T-T_(0))]` case (i) `165.5 = 101.6[1+alpha(600.5-273.16)]` `alpha = (165.5-101.6)/(101.6xx(600.5-273.16)) = (63.9)/(101.6xx327.34)` case(ii) `123.4 = 101.6[1+(63.9)/(101.6 xx 327.34)(T_(2)-273.16)] = 101.6+(63.9)/(327.34) (T_(2)-273.16)` or, `T_(2)=((123.4-101.6)xx327.34)/(63.9)+273.16 = 111.67+273.16 = 384.83 K`. |
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| 31. |
A lift is tied with a thick iton wires and its mass is 800 kg. if the maximum ac celeration of lift is `2.2 ms^(-2)` and the maximum safe stress is `1.4 xx 10^8 Nm^(-2)` find the minimum diameter of the wire take `g =9.8 ms^(-2)` |
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Answer» Correct Answer - 9.34 mm Tension in the wire, F = m(g+a) = 800 (9.8 + 2.2) = 9600 N Stress, `S = (F)/(pi(D//2))^2 = (4F)/(pi D^2)` or `D = sqrt((4F)/(pi S)) = sqrt((4xx9600xx7)/(22xx1.4 xx 10^8)` ` =9.34 xx 10^(-3) m = 9.34 mm` |
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| 32. |
A U-shaped wire is dipped in a soap solution, and removed. A thin soap film formed between the wire and a light slider supports a weight of `1.5xx10^(-2)N` (which includes the small weigh of the slider). The length of the slider is 30cm. What is the surface tension of the film? |
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Answer» We know that soap film has two free surfaces , so total length of the film to be supported, `2l = 2xx30=60cm =0.6m` Total force on the slider due to surface tension will be , `F = S xx 2l = S xx 0.6N` In equilibrium position, the force F on slider due to surface tension must balance the weight mg `(=1.5xx10^(2)N)` , i.e, `F=1.5xx10^(-2) = Sxx0.6 or S=(1.5xx10^(-2))/(0.5) = 2.5 xx 10^(-2)Nm^(-1)`. |
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| 33. |
A cube of wood floating in water supports a 200g mass resting at the centre of its top face. When the mass is removed, the cube rises 2 cm. find the volume of the cube. |
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Answer» Let l be the length of each side of wooden cube in cm. Since the cube rises 2cm out of water when a 200 g mass placed at the centre of its top face is removed, hence. upward thrust of water on cube due to displaced volume of water , `V(=l xx l xx 2)` ust be equal to weight of mass removed. so, `l xx l xx 2 xx 1 xx g = 200 g or l^(2) =(10)^(3)` `=1000 cm^(3)`. |
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| 34. |
A balloon filled with air is just floating in water in a beaker, as shown in fig. If the balloon is submerged more by a small distance and released, then the balloon will move up or move down to bottom or remains at the same location where released. Explain. |
| Answer» When the balloon is submerged more by a small distance, the pressure on the air in the balloon will increase, consequently the volume of the balloon will decrease, which inturn will decrease the buoyant force on balloon. Now the downward force (= weigth of air in balloon + downward thurst dueto liquid column) becomes more than the buoyant force on balloon. As a result of it, the balloon sink to the balloon sinks to the botton. | |
| 35. |
A steel wire 2mm in diameter is stretched between two clamps, when its temperature is `40^@C` Calculate the tension in the wire, when its temperature falls to `30^@C` Given, coefficient Y for steel ` = 21 xx 10^(11) dyn e//cm^2` |
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Answer» Correct Answer - `7.26xx 10^6dyn e` `Delta l = l alpha(theta_2- theta_1)` ` = lxx(11 xx 10^(-6))xx (40 - 30)cm.` `Now, F = (Y A Delta l)/(l) = (Ypi r^2 Delta l)/(l)` |
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| 36. |
Two exactly similar wires of steel and copper are stretched by equal forces. If the total elogation is 1cm. Find by how much is each wire elongated ? Given Y for steel ` = 20 xx 10^(11) dyn e//cm^2 and Y for copper = 12xx10^(11) dyn e//cm^2` |
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Answer» Correct Answer - 0.375 cm and 0.625 cm `Delta l_s + Delta l_c = 1 cm………. (i)` `Y_s = (F)/(A)xx(l)/(Delta l_s) and Y_c = (F)/(A)xx(l)/(Delta l_c)` `:. (Y_s)/(Y_c) = (Delta l_c)/(Delta l_s)` `or Delta l_c = Delta l_s xx(Y_s)/(Y_c) = Delta l_s xx (20 xx 10^(11))/(12 xx 10(11)) = (5)/(3) Delta l_s` From (i), `Delta l_s = (3)/(8) cm = 0.375 cm` `Delta l_c = (5)/(3) xx (3)/(8) = (5)/(8) cm = 0.625 cm` |
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| 37. |
A cubical block of density `rho` is floating on the surface of water. Out of its height L, fraction x is submerged in water. The vessel is in an elevator ac celerating upward with ac celeration a. What is the fraction immersed ? |
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Answer» Let `rho_w` be the density of water , volume of block, `V = L^3.` Mass of block, `m =V rho = L^3 rho` Let l be the height of iceberg submerged in water. Volume of iceberg in water `= l xx L^2 :.` Weight of water displaced by iceberg `= lL^2 xx rho_wg` Weight of block of ice berg ` = L^3 rho g` As iceberg is floating, so weight of iceberg = weight of water displaced by iceberg `L^3 rho g = IL^2 rho_w g or (I)/(L) = (rho)/(rho_w) =x (given)` When vessel is in an elevator which is ac celerating upwards with ac celeration a, then effective weight of block = m(g + a). Let fraction of block `x_1` be submerged into water when elevator is ac celerating upwards. Since block is floating in the water, so `m (g + a ) = (x_1 L^3) rho_w (g+a) or x_1 = (m)/(L^3 rho_w) = (L^3 rho)/(L^3 rho_w) = (rho)/(rho_w) = x` It meas, the fraction of the block submerged is independent of any ac celeration of elevator. |
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| 38. |
A copper wire of length 2.2m and a steel wire of length 1.6m, both of diameter 3.0mm are connected end to end. When stretched by a force, the elongation in length 0.50 mm is produced in the copper wire. The streching force is `(Y_(copper)=1.1xx10^(11)Nm^(-2)`, `Y_(steel)=2.0xx10^(11)Nm^(-2))`A. `1.8xx10^(2)N`B. `2.4xx10^(2)N`C. `3.6xx10^(2)N`D. `5.4xx10^(2)N` |
| Answer» Correct Answer - A | |
| 39. |
A solid body floating in water has `(1)/(4)th` of the volume above surface of water. What fraction of its volume will project upward if it floats in a liquid of specific gravity 1.1 ? |
| Answer» Correct Answer - `7//22` | |
| 40. |
Why the walls and roof of the green house are made of glass ?A. The glass equally transmits the radiations from the sun as well as those from outsideB. The glass transmits the radiations coming from sun but not those given out by the bodies insideC. The glass absorbs most of the radiations coming from sumD. The glass neither transmits nor absorbs any radiation coming from sun |
| Answer» Correct Answer - B | |
| 41. |
Two spherre of the same material and radii 4 m and 1m respectively are at temperature 1000 K and 2000 K respectively. The ratio of energies radiated by them per second isA. `1 : 2`B. `2 : 1`C. `1 : 1`D. `1 : 4` |
| Answer» Correct Answer - C | |
| 42. |
Spherical particles of pollen are shaken up in wate are allowed to settle. The depth of water is ` 2 xx 10^(-2)m`. What is the diameter of the largest paricle remaining in suspension for one hour later? Density of pollen = `1.8 xx 10^(3) kg m^(-3)`, vescosity of water = `1 xx 10^(-2)` deca poise and density of water = `1xx 10^(3) kgm^(-3)`. |
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Answer» Terminal velocity of the spherical of pollen, falling through water is `upsilon=(2 (rho -rho_(0))gr^(2))/(9 eta)` ..(i) The paricle will not be reaching the bottom in a time t if `upsilon le d/t = (2 xx 10^(-2)m)/(1 hour) = (2 xx 10^(-2)M)/(60 xx 60s) = (10^(-4))/(18) ms^(-1)` from (i), ` r = [(9 eta upsilon)/(2(rho - rho_(0))g)]^(1//2)` `=[(9 xx 10^(-2) xx (10^(-4)//18))/(2[(1.8 xx 10^(3)-1 xx 10^(3))xx9.8])]^(1//2)` `=5.6 xx 10^(-6)m`. |
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| 43. |
What should be the average velocity of a water in a tube of radius 0.005m, so that the flow is just turbulent? The viscosity of water is 0.001 Pa-s. |
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Answer» Here `D= 2 r = (2xx0.005)m,` `rho_(water) = 10^(3)kg//m^(3), eta = 0.001 Pa-s`. The flow will be just turblent if `N_(R) = 3000` `upsilon_(c) = (N_(R)eta)/(rho D) = (3000 xx 0.001)/(10^(3) xx (2 xx 0.005)) = 0.30 m//s`. |
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| 44. |
The flow rate of water from a tap of diameter 1.25 cm is `0.48 L//mi n`. The coefficient of viscosity of water is `10^(-3) Pa-s`. After sometime, the flow rate is increased to `3L//mi n`. Characterise the flow for both the flow rates. |
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Answer» `N_(R)=(rho upsilon D)/(eta) = (rho D)/(eta) ((V)/(pi D^(2)//4))` `=(4 rho V)/(pi eta D)`…(i) In first case, `rho = 10^(3) kg//m^(3) , D = 1.25 xx 10^(-2) m` `V= (0.48 xx 10^(-3))/(60)m^(3) s^(-1) , eta = 10^(-3)Pa-s` Putting the value in teh relation (i) , we shall get, `N_(R) = 5095`, which is more than 3000. Thus, the flow of liquid is turbulent. |
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| 45. |
Water flow at a speed of `7cm s^(-1)` through a pipe of tube of radius `1.5 cm`. What is the nature of the flow ? Cofficient of viscosity of water is `10^(-3) kg m^(-1) s^(-1)` and its density is `10^(3) kg m^(-3)` |
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Answer» Here, `v_(c)=7 cm s^(-1)=0.07 ms^(-1)`, `D=2 r=2xx0.05=0.03 m` `eta=10^(-3) kg m^(-1) s^(-1), rho=10^(3) kg m^(-3)` `N_(R)=(rho v_(c)D)/(eta)=(10^(3)xx0.07xx0.02)/(10^(-3))=2100` `A_(s) N_(R)gt 2000` but less than `3000`, hence the flow is uncertain. |
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| 46. |
Which mode of transfer of heat is quickest? |
| Answer» Radiation is the quickest mode of transfer of heat. | |
| 47. |
A mode of transfer of heat from one part of the body to another part, from paticle to particle in the direction of fall of temperature without any actual movement of the heated particle is called…. |
| Answer» Correct Answer - conduction | |
| 48. |
Sea breeze is caused by …….. |
| Answer» natural convection process | |
| 49. |
A cylindrical metallic rod in thermal contact with two reservation of heat at its two ends conducts an amount of heat `Q` in time `t`. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod when placed in thermal contact with the two reservation in time t?A. `Q//4`B. `Q//16`C. 2 QD. `Q//2` |
| Answer» Correct Answer - B | |
| 50. |
Three identical rods A, B and C are placed end to end. A temperature difference is maintained between the free ends of A and C. The thermal conductitvity of B is thrice that of C and half that of A. The effective thermal conductivity of rod A)A. `(1)/(3)K_(A)`B. `3 K_(A)`C. `2 K_(A)`D. `(2)/(3)K_(A)` |
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Answer» Correct Answer - A Let L be the length and A be the area of cross-section of each rod. Given `K_(B) =3 K_(C), K_(B) =(1)/(2)K_(A)` `:. K_(C)=(1)/(3)K_(B) =(1)/(3)xx(1)/(2)K_(A) =(1)/(6)K_(A)` Thermal resistance of rod A, `R_(A) =(L)/(K_(A)A)` Thermal resistance of rod B, `R_(B) =(L)/(K_(B)A) =(2L)/(K_(A)A)` Thermal resistance of rod C, `R_(C) =(L)/(K_(C)A) =(6L)/(K_(A)A)` Here, the rods connected in series. If K is effective thermal conductivity of the system, and `R_(s)` is the total thermal resistance of the system then `R_(S) =R_(A) + R_(B) + R_(C)` `:. (3L)/(KA) =(L)/(K_(A)A) + (2L)/(K_(A)A) + (6L)/(K_(A)A) = (9L)/(K_(A)A)` or `(3)/(K) = (9)/(K_(A))` or `K=(K_(A))/(3)` |
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