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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.
| 201. |
when a piece of red glass heated in a furnace is taken out, it glows with green light. Why? |
| Answer» Red and green are complement colours. When a piece of red glass is heated in a furnace , it absorbs green light but transmits red light. When this piece of red glass is taken out, it emits green light, i.e., glows with green light. It is so because a good absorber when heated is a good radiator when cooled. | |
| 202. |
A hot liquid kept in a beaker cools from `80^(@)C` to `70^(@)C` in two minutes. If the surrounding temperature is `30^(@)C`, find the time of coolilng of the same liquid from `60^(@)C` to `50^(@)C`. |
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Answer» Correct Answer - `[216 s]` As, `(dT)/(dt)= -K[(T_(1)+T_(2))/(2)-T_(0)]` `:. (80-70)/(2xx60)= -K[(80+70)/(2)-30]` or `(10)/(2xx60)= -Kxx45` And ` (60-50)/(t)= -K[(60+50)/(2)-30]` or `(10)/(t)= -Kxx25` …(ii) Dividing (i) by (ii), we get `(t)/(2xx60)=(45)/(25)=(9)/(5)` or `t=216 s` |
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| 203. |
A liquid initally at `70^(@)C` cools to `55^(@)C` in 5 minutes and `45^(@)C` in 10 minutes. What is the temperature of the surrounding ? |
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Answer» Correct Answer - `[25^(@)C]` 1st case, change in temperature, `dT=70-55=15^(@)C` time interval, dt=5 min Average temperature, `T=(70+55)/(2)=62.5^(@)C` Temperature of surrounding, `T_(0)=?` As, `(dT)/(dt)=-K(T-T_(0))` `:. (15)/(5) = - K(62.5-T_(0))` .....(i) 2nd case, `dT=55-45=10^(@)C`, `dt=10-5=5 min` `T=(55+45)/(2)=50^(@)C` Then, `(10)/(5)= -K(50 - T_(0))` ....(ii) Dividing (i) by (ii) and solving we get, `T_(0)=25^(@)C` |
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| 204. |
A body cools from `80^(@)C` to `70^(@)C` in 5 minutes and further to `60^(@)C` in 11 minutes. What will be its temperature after 15 minutes from the start ? Also determine the temperature of the surroundings. |
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Answer» Correct Answer - `[54.4^(@)C; 15^(@)C]` In first case Change in temperature, `dT=80-70=10^(@)C` Time interval, `dt= 5` minutes Average temperature, `T=(80+70)/(2)=75^(@)C` Temperature of the surrounds `=T_(0)` (say) As `(dT)/(dt)= -K(T-T_(0))` `:. (10)/(5)= -K(75-T_(0))` ....(ii) In second case Change in temperature, `dT=70-60=10^(@)C` Time interval, `dt=11-5=6 min`. Average temperature, `T=(70+60)/(2)=65^(@)C` Using, `(dT)/(dt)= -K[T-T_(0)]` `(10)/(6)= -K[65-T_(0)]` ...(ii) Dividing (i) by (ii), we have `(6)/(5)=(75-T_(0))/(65-T_(0))` On solving, `T_(0)=15^(@)C` From (i), `(10)/(5)= -K[75-15]= -Kxx60` .....(iii) In third case Let `T_(1)` be tghe temperature of the body after 15 minutes from start. Change in temperature, `dT=60-T_(1)^(@)C` Time interval, `dt=15-11 =4 min` Average temperature, `T=((60+T_(1))/(2)) .^(@)C` As, ` (dT)/(dt)= -K[T-T_(0)]` `:. (60-T_(1))/(4)= -K[(60+T_(1))/(2)-15]= -K[(T_(1)+30)/(2)]` ...(iv) Dividing (iv) by (iii), we have `[(60-T_(1))/(4)xx(5)/(10)]=((T_(1)+30)/(2))xx(1)/(60)` `60-T_(1)=(T_(1)+30)/(15)` or `60xx15-15 T_(1)=T_(1)+30` `900-30= 16 T_(1)` or `T_(1)= 870//16=54.4^(@)C` |
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| 205. |
A body initially at `80^(@)C` cools to `64^(@)C` in 5 minutes and to `52^(@)C` in 10 minutes. What is the temperature of the surroundings? |
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Answer» First method : In first case. `T_(1) = 80^(@)C, T_(2) = 64^(@)C, t = 5 min` Let `T_(0)` be the temperature of the surroundings. Using the relation `2.3026 log_(10) ((T_(1)-T_(0))/(T_(2)-T_(0))) = Kt` `2.3026 log_(10) = (80-T_(0))/(64-T_(0)) = K xx 5` ..(i) 2nd case, `T_(1) = 64^(@)C, T_(2)= 52^(@)C`, `t=10- 5 = 5 mi n` the temperature of surrounding remains unchanged, i,e`T_(0)` `2.3026log_(10)((64-T_(0))/(52-T_(0))) = Kxx5` ..(ii) From (i) and (ii) , we have `2.3026 log_(10)((64-T_(0))/(52-T_(0))) = 2.3026 log (( 80-T_(0))/(64-T_(0)))` or `log_(10)((64-T_(0))/(52-T_(0)))=log_(10) ((80-T_(0))/(64-T_(0)))` Taking antilogs, we have `(64-T_(0))/(52-T_(0))=(80-T_(0))/(64-T_(0))` On solving , `T_(0) = 16^(@)C` Second method : In first case, Change in temperature, `dT=80 - 64 = 16^(@)C` time interval , `dt = 5 min` Average temperature, `T = (80+64)/(2) = 72^(@)C` Temperature of surronding , `T_(0)=?` As, `(dT)/(dt) = -K(T-T_(0))` `:. 16/5 = -K(72-T_(0))` ...(i) In second case, change in temperature, `dT = 64-52 = 12^(@)C` time temperature, `T=(64+52)/(2)=58^(@)C` Using ,`(dT)/(dt) = -K(T-T_(0))`, we have `12/5 = -K(58-T_(0))` Dividing (i) by (ii), we have `16/5 xx 5/12 = (72-T_(0))/(58-T_(0))` on solving, `T_(0) = 16^(@)C`. |
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| 206. |
Two differenent liqiud of masses 10g and 20 g and density `0.84 g//c c` and `0.95 g//c c` are mixed toghther. Find the density of the mixture. |
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Answer» Here, `m_(1)=10g, m_(2)=20g, rho_(1)=0.84g//c c , rho_(2)=0.95 g//c c`. Volume of first liquid, `V_(1)=(m_1)/(rho_1) = (10)/(0.84)c c`. Volume of second liquid, `V_(2)=(m_2)/(rho_2) = (20)/(0.95) c c`. Total volume `= V_(1)+V_(2) = (10)/(0.84) +(20)/(0.95)` `:.` Density of mixture` = (m_(1)+m_(2))/(V_(1)+V_(2))=(10+20)/((10//0.84)+(20//0.95)) = (30)/(11.9+21.1) = (30)/(33.0) = 0.90 g//c c`. |
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| 207. |
The radious of a metal sphere at room temperature T is R, and the coefficient of linar expansion of the metal is `alpha`. The sphere is heated a little by a temperature `Delta T` so that its new temperature is `T+ Delta T`. The increase in the volume of the sphere is approximatelyA. `2pi R alpha Delta T`B. `pi R^(2) alpha Delta T`C. `4 pi R^(3) alpha Delta t//3`D. `4 pi R^(3) alpha Delta T` |
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Answer» Correct Answer - D Let V be the volume of sphere of radiuas R at temp. T, then `V=(4)/(3) pi R^(3)`. Increase in volume of sphere with rise in temperature `Delta T` is `Delta V= gamma V Delta T =3alphaxx(4)/(3) pi R^(3) Delta T =4pi R^(3) alpha Delta T` |
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| 208. |
The volume of a metal sphere is increased by `2%` of its original volume when it is heated from `300 K` to `604 K`. Calculate the coefficient of linear, superficial and cubical expansion of the metal. |
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Answer» Correct Answer - `[2.19xx10^(-5) .^(@)C^(-1); 4.38xx10^(-5) .^(@)C^(-1); 6.58xx10^(-5) .^(@)C^(-1)]` Here, `(Delta V)/(V)=(2)/(100)` or `DeltaV=(2V)/(100)=(V)/(50)` `DeltaT=604-300=304 K` `Delta T=V gamma Delta T` or `(V)/(50)=Vxxgammaxx304` or `gamma=(1)/(50xx304)=6.58xx10^(-5) .^(@)C^(-1)` `alpha=(gamma)/(3)=(6.58)/(3)xx10^(-5)=2.19xx10^(-5) .^(@)C^(-1)` `alpha=2 alpha=2xx2.19xx10^(-5)=4.38xx10^(-5) .^(@)C^(-1)` |
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| 209. |
If n equal rain droplets falling through air with equal steady velocity of `10 cm s^(-1)` coalesce, find the terminal velocity of big drop formed. |
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Answer» Let, R be the radius of each droplet and big drop respectively. Volume of big drop = n x volume of a small droplet `(4)/(3) pi R^(3) =nxx(4)/(3)pi r^(3)` or `R=n^(1//3)r` Terminal velocity of small droplet, `v=(2r^(2)(rho-sigma)g)/(9eta)` Terminal velocity of big drop, `V=(2R^(2)(rho-sigma)g)/(9eta)` Dividing (ii) by (i), we get `(V)/(v)=(R^(2))/(r^(2))=(n^(1//3)r^(2))/(r^2)=n^(2//3)` or `V=n^(2//3) v cm//s` |
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| 210. |
A wire ring of 4 cm radius on the surface of a liquid and then raised. If surface tension of the tension of the liquid is `78.8 dyn e//cm`, find the pull (in g wt) required to raise the ring more before the film breaks than it is afterwards. |
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Answer» Here, `S=78.8 dyn e//cm , r=4cm` As the liquid touches the ring both along the inner and outer circumference, so force on the ring due to surface tension is `F= 2 xx 2 pi r xx S = 4 pi r S` `=4 xx 22/7 xx 4 xx 78.8 dyn e` `= 4 xx 22/7 xx (4 xx 78.8)/(980) g wt = 4.04 g wt`. |
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| 211. |
Calculate the work done in blowing out a soap bubble of diameter 1 cm. given that the surface tension of soap solution is `28 xx 10^(-3) Nm^(-1)` |
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Answer» Correct Answer - `1.76 xx 10^(-5) J` work done, W = surface tension xx increase in surface area ` = [Sxx4 pi (D//2)^2xx2]` |
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| 212. |
A glass tube of radius r is dipped vertically into a container of mercury of density `rho` with its lower end at a depth h below the mercury surface . If S is the surface tension of mercury, what must be the gauge pressure of air in the tube to blow a hemispherical bubble at the lower end? |
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Answer» Gauge pressure in te hemispherical eair bubble formed at the end of the glass tube in mercury is P = excess of pressure + Pressure due to h column of mercury `=(2S)/( r ) + h rho g`. |
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| 213. |
A capillary tube of inner radius `0.5 mm` is dipped keeping it vertical in a mercury of soecific gravity 13.6, surface tension `545 dyn e//cm` and angle of contact `135^(@)` . Find the depression or elevation of liquid in the tube. |
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Answer» Here, `r=0.5//2 = 0.25 mm = 0.0025 cm`, Density of mercury , `rho =` specific . gravity `xx` density ofwater` = 13.6 xx 1 = 13.6 g//c c` Surface tension , `S=545 dyn e//cm`, angle of contact `theta= 135^(@)`. The height of liquid in tube is `h=(2S cos theta)/(r rho g) = (2 xx 545 xx cos 135^(@))/((0.025)xx13.6 xx 980) = (2xx545xx(-1//sqrt(2)))/(0.025xx13.6xx980)=-2.31 cm`. |
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| 214. |
A tube of 1mm bore is dipped into a vessel containing a liquid of density `rho = 800 kg m^(-3)` and of surface tension , `S=49 xx 10^(-3)Nm^(-1)` and angle of contact , `theta = 0^(@)`. The tube is held inclined to the verticle at an angle of 60^(@), find the height to which the liquid can rise and the length which the liquid will oc cupy in the tube. |
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Answer» Here, `r=1.2 mm = 1/2 xx 10^(-3)m, rho = 800 kg//m^(3), S=49 xx 10^(-3) N//m, theta = 0^(@) , h=?` Angle with the horizontal of inclined tube `alpha= 90^(@)-60^(@)=30^(@)` Alsom `h=(2S cos theta)/(r rho g) = 2 xx 49 xx 10^(-30 xx cos 0^(@))/((1//2 xx 10^(-))xx 800 xx 9.8) = 0.025 m = 2.5 cm` if l is the length oc cupied by the liquid in the tube sin `alpha = h//l` or `l=(h)/(sin alpha) = (2.5) /(sin 30^(@) ) = 2.5 xx 2 = 5 cm`. |
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| 215. |
Calculate the total inside a spherical air bubble of radius 0.1 mm at a depth of 10 cm below the surface of a liquid of density 1.1 g/c.c and surface tension 50 dynes/cm. (Height of Hg barometer = 76 cm). |
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Answer» Correct Answer - `1.0337 xx 10^6 dynes//sq. cm` Total pressure = atmospheric pressure + pressre due to h cm of liquid columm + execess pressure |
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| 216. |
There is an air bubble of radius 2.0 mm in a liquid of surface tension `0.070 Nm^(-1)` and density `10^3 kg m^(-3)` The bubble is at a depth of 12.0 cm below the free surface of liquid. By what amount is the pressure inside the bubble is greater then the atmospheric pressure ? Use `g =10 m//s^2.` |
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Answer» Correct Answer - `1270 Nm^(-2)` Pressure inside the bubble greater than atmospheric pressure ` =(2S)/(R ) + h rho g = (2xx0.070)/((2.0xx10^(-3)) + 0.12xx10^3 xx 10` `=1270 Nm^(-2)` |
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| 217. |
If the excess pressure inside a spherical soap bubble of radius 1 cm is balanced by that due to column of oil of specific gravity 0.9, height 1.36 mm. Calculate the value of surface tension of soap solution. |
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Answer» Here, `r=1 cm` density `rho = sp.gravity xx density` of water `= 0.9 xx 1 = 0.9 g//c c` `h=1.36 mm = 0.136 cm` Given excess pressure in a soap bubble = pressure due to column of oil of height h So, `(4S)/(r ) = h rho g` or `S= (h rho g r)/(4) = (0.136 xx 0. 9 xx 980 xx 1)/(4)` `= 29.988 dyn e//cm`. |
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| 218. |
The excess pressure inside a soap bubble of radius 5 mm in balanced by 3 mm column of oil of specific gravity 0.6 Find the surface tension of soap solution. |
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Answer» Correct Answer - `2.205 xx 10^(-2) Nm^(-1)` Here, `R = 5mm = 5xx10^(-3)m ,` `h = 3mm = 3xx10^(-3)m` `rho = 0.6 xx10^3 kg m^(-3)` Excess of pressure inside the soap bubble = pressure exerted by 3 mm of oil columm `(4S)/(R ) = h rho g or S = (h rho g R)/(4)` `:. S = (3xx10^(-3))xx(0.6xx10^3)xx9.8 xx(5xx10^(-3))/(4)` `=2.205 xx 10^(-2) Nm^(-1)` |
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| 219. |
A capillary tube is dipped in water with the lower and 10 cm below the surface. Water rises in the tube to a height of 5 cm. The pressure required to blow a bubble at the lower end of the tube will be (atmospheric pressure ` = 10^5 Nm^(-2) and g = 10 m//s^2`)A. `10^5 Nm^(-2)`B. `1.015 xx 10^5 Nm^(-2)`C. `2xx10^5 Nm^(-2)`D. `2.5 xx10^5 Nm^(-2)` |
| Answer» Correct Answer - B | |
| 220. |
The lower end of a capillary tube is dipped in water. Water rises to a height of 8 cm. The tube is then broken at a height of 6 cm. The height of water colume and angle of contact will beA. `6cm, sin^(-1)((3)/(4))`B. `6cm, cos^(-1)((3)/(4))`C. `4cm, sin^(-1)((1)/(2))`D. `4cm, cos^(-1)((1)/(2))` |
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Answer» Correct Answer - B When a capillary tube is broken at a height of 6 cm, the height of water column will be 6 cm. As `h=(2S cos theta)/(r rho g)` or `(h)/(cos theta) =(2S)/(r rho g)=a` constant `:. (8)/(cos theta^(@)) =(6)/(cos theta)` or `cos theta =(6 cos 0^(@))/(8) =(3)/(4)` or `theta=cos^(-1)(3//4)` |
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| 221. |
A glass tube of 1 mm bore is dipped vertically into a container of mercury, with its lower end 5 cm below the mercury surface. What must be the gauge pressure of air in the tube to a hemispherical bubble at its lower end? Given density of mercury ` = 13.6 xx 10^(3) kg//m^(3)`, surface tension of mercury = `440 xx 10^(-3) Nm^(-1)` and `g = 10m//s^(2)` |
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Answer» Here, `R=1.2 mm = 0.5 mm` `= 5 xx 10^(-4)m`, `h= 5 cm = 0.05 m , rho = 13.6 xx 10^(3) kg//m^(3)` `S= 440 xx 10^(-3) Nm^(-1)` Gauge pressure in hemispherical bubble `= h rho g + (2S)/(R )` `= 0.05 xx (13.6 xx 10^(3)) xx 10 + ( 2 xx 440 xx10^(3))/(5 xx 10^(-4))` `= 6800+1760 = 8560 Pa`. |
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| 222. |
What will be the total pressure inside a spherical air bubble of radius 0.2 mm at a depth of 2m below the surface of a liquid of density `1.1 g//cm^(3)` and surface tension `50 dyn e//cm`. Atmospheric pressure is `1.01 xx 10^(5) Nm^(-2)`. |
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Answer» Here, `r=0.2 mm = 0.2 xx 10^(-3) m` `= 2 xx 10^(-4) m` , `h= 2m`, `rho = 1.1 g cm^(-3) = 1.1 xx 10^(3) kg//m^(3)` , `S=50 dyn e//cm = 50 xx 10^(-3) N//m` `P_(0) = 1.01 xx 10^(5) N//m^(2)` Total pressure inside the bulb = atmospheric pressure + pressure due to column of liquid + excess of pressure `= P_(0)+h rho g + (2S)/(r) = 1.01 xx 10^(5) + 2 xx 1.1` `xx 10^(3) xx 9.8 + 2 z (50xx10^(-3))/(2 xx 10^(-4))` `= 1.01 xx 10^(5) +20.58 xx 10^(3) +500` `= 1.2158 xx 10^(5) = 1.22 xx 10^(5) N//m^(2)`. |
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| 223. |
A solid material is supplied heat at a constant rate. The temperature of material is changing with heat input as shown in the figure. What does the slope of DE represent ? A. AB and CD of the graph represent phase changesB. AB represents the change of state from solid to liquidC. Latent heat of fusion is twice the latent heat of vaporisationD. CD represents change of state from liquid to vapour |
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Answer» Correct Answer - C Latent heat of fusion is smaller than the latent heat of vaporisation. |
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| 224. |
A water cooler of storages capacity 120 liters can cool water at a constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3kW of heat (thermal load). The temperature of water fed into the device cannot exceed `30^@C` and the entire stored 120 liters of water is initially cooled to `10^@C.` The entire system is thermally insulated. The minimum value of P ( in watts) for which the device can be operated for 3hours is (Specific heat of water is `4.2kJkg^-1K^-1` and the density of water is `1000kgm^-3`)A. 1600B. 2067C. 2533D. 3933 |
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Answer» Correct Answer - B Heat generated in device in 3 hours = power x time = (3 kW) x 3 h `=3xx10^(3)xx(3xx60xx60)=324xx10^(5) J` Heat used to heat water `=ms Delta theta` `=(120xx1)(4.2xx10^(3))xx(30-10)` `=120xx4.2xx20xx10^(3) J =100.8xx10^(5) J` Heat absorbed by coolant, ltbRgt `Pt =324xx10^(5) - 100.8xx10^(5) J` `=(324 - 100.8)xx10^(5) =223.2xx10^(5) J` `P =(223.2xx10^(5))/(t) = (223.2xx10^(5))/((3xx60xx60)) =2067 wat t` |
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| 225. |
In a cylindrical water tank there are two small holes `Q` and `P` on the wall at a depth of `h_(1)` from the upper level of water and at a height of `h_(2)` from the lower end of the tank, respectively, as shown in the figure. Water coming out from both the holes strike the ground at the same point. The ratio of `h_(1)` and `h_(2)` is |
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Answer» Let H be the total height of the water column in cylindrical tank. Let `upsilon_1, upsilon_2` = velocity of efflux at P and Q respectively. Then `upsilon_1 = sqrt(2 gh_1) and upsilon_2 =sqrt(2g(H - h_2))` Time taken by water stream to fall from P to S on the ground is ,`t_1 = (sqrt(2(H - h_1))/(g))` Time taken by water stream to fall from Q to S on the ground is, `t_2 = (sqrt(2h_2))/(g)` Since, the two water streams from P and Q reach the same point S on the ground, so `R_1 = R_2 or upsilon_1 t_2 or sqrt(2gh_1) xx sqrt((2(H -h_1))/(g)) = sqrt(2g (H - h_2)) xx sqrt((2h_2)/(g))` ro `h_1 (H - h_1) = (H - h_2) h _2 or H(h_(1) - h_(2) - h_1^2 + h_2^2 = 0` or `[H - (h_1 + h_2] [h_1 - h_2] = 0` Since `h_1+ h_(2) =H` is not possible as the holes are at different height so `h_1 = h_2.` Hence `h_1//h_2 =1` |
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| 226. |
The approximate depth of an ocean is `2700m`. The compressibility of water is `45.4xx10^(-11)Pa^-1` and density of water is `10^3(kg)/(m^3)`. What fractional compression of water will be obtained at the bottom of the ocean?A. `1.0xx10^(-2)`B. `1.2xx10^(-2)`C. `1.4xx10^(-2)`D. `0.8xx10^(-2)` |
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Answer» Correct Answer - B Here, `h=2700m,k=45.4xx10^(-11) Pa^(-1)`, `rho=10^(3) kg//m^(3)` As `B=(P)/(Delta V//V)` or `(Delta V)/(V)=(P)/(B)` As `P=(F)/(A)=(mg)/(A)=(Vxxrhoxxg)/(A)=h rho g` Also `B=(1)/(k)` or `(Delta V)/(V)=Pk=h rho g k` `=2700xx10^(3)xx9.8xx45.4xx10^(-11)` `=1201284xx10^(3)xx10^(-11)` `=1201284xx10^(-8) =1.2xx10^(-2)` |
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| 227. |
An iceberg is floating partly immersed in sea water, the density of sea water is `1.03 g cm^(-3)` and that of ice is `0.92 g cm^(-3)`. The fraction of the total volume of the iceberg above the level of sea water isA. `8.1 %`B. `11%`C. `34%`D. `0.8%` |
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Answer» Correct Answer - B Let V be the volume of the iceberg and x be the volume of iceberge out of sea water. The iceberge is floating in sea water, then `Vxx0.92xxg =(V-x)xx1.03xxg` or `0.92 V =1.03 V -1.03 x` or `(x)/(V)=(0.11)/(1.03)` `:. %` fraction of the volume of ice berg above the level of sea water `=(x)/(V)xx100=(0.11xx100)/(1.03) =11%` |
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| 228. |
A venturimeter is connected to two points in the mains where its radii are `20 cm`. And `10 cm`. Respectively and the levels of water column in the tube differ by `10 cms`. How much water flows through the pipe per minute? |
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Answer» Correct Answer - [`2726.6 litres//minute`] Volume of water flowing per sec, `V=a_(1)a_(2)sqrt((2gh)/(a_(1)^(2)-a_(2)^(2)))` where, `1a_(1)=pi r_(2)^(2)=(22)/(7)xx20^(2) cm^(2)`, `a_(2)=pi r_(1)^(2)=(22)/(7)xx10^(2) cm^(2)`, `g=980 cm//s^(2), h=10 cm`, Volume of water flowing per minute `=vxx60=a_(1) a_(2)sqrt((2gh)/(a_(1)^(2)-a_(2)^(2)))xx60 c.c//m` minute |
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| 229. |
The flow of blood in a large artery of an anaeshetized dog is diverted throg]ugh a venturimeter. The wider part of the meter has a cross sectional area equal to that of the artery i.e. `8 mm^(2)`. The narrower parts has an are `4 mm^(2)`. The pressure dorp in the artery is `24Pa`. what is the speed of the blood in the artery ? Given that density of the blood = `1.06 xx 10^(3) kg//m^(3)` |
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Answer» Here, `a_(1) = 8 mm^(2) = 8 xx 10^(-6)m^(2)`, `a_(2) = 4mm^(2) = 4 xx 10^(-6) m^(2)` `P_(1) - P_(2) = 24 Pa, rho = 1.06 xx 10^(3) kg//m^(3)` `upsilon = V/(a_(1)) = a_(2) sqrt((2(P_(1)-P_(2)))/(rho(a_(1)^(2)-a_(2)^(2))))` `= 4 xx 10^(-6)` ` xx [(2 xx 24)/((1.06 xx 10^(3))[(8xx10^(-6))^(2)-(4 xx 10^(-6))^(2)])]^(1//2)` `= 0.123 ms^(-1)` . |
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| 230. |
Two narrow bores of diameters 3.0mm and 6.0 mm are joined together to form a U-shaped tube open at both ends. If th U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is `7.3xx10^(-2)Nm^(-1)`. Take the angle of contact to be zero. and density of water to be `1.0xx10^(3)kg//m^(3)`. `(g=9.8 ms^(-2))` |
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Answer» Here , `S = 7.3 xx 10^(-2) Nm^(-1),rho = 1.0 xx 10^(3), theta = 0^@)` For narrow tube, `2r_(1) = 3.00mm = 3xx10^(-3)m or r_(1)=1.5xx10^(-3)m` For wider tube ,`2r_(2) = 6.00 mm = 6xx10^(-3)m` or `r_(2) = 3 xx10^(-3)m` Let `h_(1),h_(2)` be the heights to which water rises in narrow tube and wider tube respectively. Then, `h_(1) = (2S cos theta)/(r_(1)rho g)` and `h_(2) = (2S cos theta)/(r_(2)rho g)` `:.` Differnece in levels of water in two limbs of U tube is , `h_(1)-h_(2) = (2S cos theta)/(rho g) [(1)/(r_1)-(1)/(r_2)]` `=(2 xx 7.3 xx 10^(-2)xxcos 0^(@))/(10^(3)xx9.8) xx [(1)/(1.5xx10^(-3))-(1)/(3xx10^(-3)) = 4.97 xx 10^(-3)m]`. |
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| 231. |
Mercury has an angle of contact equal to `140^(@)` with soda lime galss. A narrow tube of radius `1.00mm` made of this glass is dipped in a through containing mercury. By what amount does the mercury dip down in the tube relative to the mercury surface outside? Surface tension of mercury at the temperature of the experiment is `0.465 Nm^(-1)`. Density of mercury = `13.6xx10^(3) kg m^(-3)`. |
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Answer» Here, `theta = 140^(@), r =10^(-3)m, S=0465 Nm^(-1), rho = 13.6 xx10^(3) kg, h=?` `cos 140^(@) = -cos 40^(@) = -0.7660` Now, `h=(2S cos theta)/(r rho g) = (2xx0.465xxcos 140^(@))/(10^(-3)xx13.6xx10^(3)xx9.8) = 2xx0.465xx(-0.7660)0/(10^(-3)xx13.6 xx 10^(3)xx9.8) = -5.34 xx 10^(-3)m = -5.34mm`. Negative value of h shows that the mercury level is depressed in the tube. |
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| 232. |
Iceberg floats in water with part of it submerged. What is the fraction of the volume of iceberg submerged if the density of ice is `rho_i = 0.917 g cm^(-3) ?` |
| Answer» Let V be the total volume of the ice berg and `upsilon` be the volume of ice berg submerged in water. As ice berg is floating, the weight of ice berge is equal to weight of water dieplaced by immersed part of ice berg. So `V rho_(ice) g= upsilon rho_w g or (upsilon)/(V) = (rho_(ice))/(rho_w) = (0.917 g//cm^3)/(1g//cm^3) = 0.917` | |
| 233. |
The area of cross-section of the wider tube shown in fig., is `800cm^(2)` . If a mass of 12 kg is placed on the massless piston, what is the difference in the level of water in two tubes. |
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Answer» Here, `A= 800 cm^(2) ` , `F= mg = (12 xx 1000 xx 980) dyn es`. Thus, pressure on the liquid, `p= F/A = (12 xx 1000 xx 980)/(800) dyn e//cm^(2)` If h is the difference in level of liquid in the two tubes then, `p = h rho g = hxx 1 xx 980` `:. h xx 1 xx 980 = (12 xx10 xx 980)/(800)` or, ` h=15.0 cm`. |
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| 234. |
To lift an automobile of 2000 kg, a hydraulic pump with a large piston 30 cm square in area is employed. Calculate the force that must be applied to pump a small piston of area 10 square cm to achieve it. |
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Answer» Here, `F_(2) = Mg = 2000 xx 9.8 = 19600 N` `A_(2) = 30 x` cm square `= 30 xx 30 sq cm` `= 900 xx 10^(-4)m^(2)` As `(F_1)/(A_1) = (F_2)/(A_2)` so `F_(1) =(F_(2) xx A_(1))/(A_2)` ` = (19600 xx 10 xx 19^(-4))/(900 xx 10^(-4))` `= 217.8 N`. |
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| 235. |
The average mass that can be lifted by a hydraulic press is 100kg. If the radius of the larger piston is six times that of a smaller poston. What is the minimum force (in kg wt) that must be applieed? Use `g=10ms^(-2)`. |
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Answer» Here `F_(2) = 100 kgf = 100 xx 10 N` `A_(2) = 6 A_(1), F_(1)=?` As `(F_1)/(A_1) = (F_1)/(A_2) or F_(1) = (A_1)/(A_2) xx f_(2) = (A_(1) xx 100 xx 10)/(6 A_(1))` `=(100 xx 10 )/(6) N = (100 xx 10)/(6xx10) kgf` `=16.67 kgf`. |
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| 236. |
A hydraulic press with the larger piston of diameter 35 cm at a heigth of 1.5 m relative to the smaller piston of diameter 10 cm. The mass on the smaller piston is 20 kg. What is the force exerted on the load placed. On the larger piston ? The density of oil in the press is `750 kg m^(-3)` |
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Answer» Correct Answer - `1.34 xx 10^3 N` Pressure on the larger pistion =Pressure on smaller piston - pressure due to column 1.5 m |
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| 237. |
Two pistons of hydraulic press have diameter of `30.0 cm` and `2.5 cm`, find the force exerted on the longer piston when 50.0 kg wt. is placed on smaller piston. |
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Answer» Correct Answer - 7200 kg wt ; 0.28 cm Here, `A_1 = (piD_1^2)/(4) = (22)/(7)xx(2.5)^2cm^2` `A_2 = (pi D_2^2)/(4) = (22)/(7) xx(30)^2cm^2,` `F_1 = 50 kg wt , l_1 = 4.0cm` Force on longer piston, `F_2 = (F_1 xxA_2)/(A_1) = 50 xx ((30)^2)/(2.5)^2` =` 50 xx 144 = 7200 kg wt` As, `F_1 l_1 =F_2 l_2, so ` `l_2 = (F_1l_1)/(F_2) = (50xx4)/(7200) = 0.028cm` `:.` Distance covered in 10 strokes `= 10 xx 0.028 = 0.28 cm` |
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| 238. |
The area of smaller piston of a hydraulic pressure is `2cm` square and that of larger piston is 20 cm square. How much weight can be raised on the larger piston when a force 200kg f is exerted on the smaller piston? `g=10m//s^(2)` |
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Answer» Here` A_(1)=2 cm sq = 2 xx 2 sq cm` `A_(2) = 20 xx 20 sq cm = 400 xx 10^(-4)m^(2)` `F_(1) = 200 kg f=200 xx 10N , F_(2)=?` `F_(2) =(A_2)/(A_1) F_(1) = ((400 xx 10^(-4)) xx (200 xx 10))/(4 xx 10^(-4))` `= 2 xx 10^(5) N = 2 xx 10^(4) kg f`. |
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| 239. |
The meterial of a wire has density of 1.4 g/c c. If it is not wetted by a liquid of surface tension 44 dyne / cm., find the maximum radius fo the wire which can float on the surface of the liquid. |
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Answer» Correct Answer - 1.43 mm Here `rho = 1.4 g//c c , S = 44 dyn e// cm ,r = ?` Let l be the length of the wire. As the wire floats, So, `2Sl = pi r^2 l rho g` or `r = sqrt((2S)/(pi rho g)) = sqrt((2xx44)/((22//7)xx1.4 xx980))` `=(1.0)/(7)cm = (10)/(7)mm = 1.43 mm` |
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| 240. |
A rain drop of radius 0.5 mm has a terminal velocity in air `2 ms^(-1)` . The viscosity of air is `18 xx 10^(-5)" dyne "cm^(-2)s` . Find the viscous force on the rain drop. |
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Answer» Here, `r = 0.5 mm = 0.05 cm, upsilon = 2 ms^(-1) = 200 cm//s , eta = 18 xx 10^(5)" dyne "cm^(-2) s, F=?` Now, `F= 6 pi eta r upsion = 6 xx 22/7 xx 18 xx 10^(-5) xx 0.05 xx 200 = 0.034 dyn es`. |
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| 241. |
Let n number of little droplets of water of surface tension S (dyne/cm), all of the same radius r cm combine to from a single drop of radius R heat, while using cgs system of units answer the following questions. If the radius of the big drop formed is made two times without any change in temperature then the work done isA. `4 pi R^(2) S`B. `8 pi R^(2) S`C. `12 pi R^(2) S`D. `16 pi R^(2) S` |
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Answer» Correct Answer - C Work done = S.T. x change in surface area `=S[4 pi(2 R)^(2) - 4pi R^(2)]` `=Sxx12 pi R^(2)` |
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| 242. |
Water flows through a horizontal pipe of varying cross-section at the rate of 20 litres per minuts , determine the velocity of water at a point where diameter is `4 cm` |
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Answer» Volume of the water floeing per second `V = 20 liter // min = (20 xx 1000)/(60 xx (100)^(3)) m^(3) s^(-1)` ` = (1)/(3) xx 10^(-3) m^(3) s^(-1)` Ratio of the pipe `r = (4)/(2) = 2 cm = 0.02 m` Area of cross section ` a = pi r^(2) = (22)/(7) xx (0.02)^(2) m^(2)` Let `upsilon` be the velocity of flow of water at the given ,Clearly `V = a upsilon` or ` (1)/(3) xx 10^(-3) = (22)/(7) xx (0.02)^(2) xx upsilon ` or `upsilon = (7 xx 10^(-3))/(3 xx 22 xx (0.02)^(2)) = 0.2639 ms^(-1)` |
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| 243. |
A liquid Flows through a pipe 2.0 mm radius and 20 cm length under a pressure `10^(3)" dyne "cm^(-2)`. Determine the rate of flow and the speed of the liquid coming out of the tube. The coefficient of viscosity of the liquid is 1.25 centipoise. |
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Answer» Here, `r = 2.0 mm = 0.2 cm , l=20 cm , p=10^(3)" dyne "cm^(-2) , eta = 1.25 " centipoise "= 0.0125" poise"` Rate of flow of liquid through the tube is `V= (pi p r^(4))/(8 eta l) = ((3.142)xx10^(3)xx (0.2)^(4))/(8 xx 0.0125 xx 20) = 2.51 cm^(3) s^(-1)` Speed of liquid, `upsilon = V/(pi r^(2)) = (2.51)/(3.142 xx (0.2)^(2))= 20 cm s^(-1)`. |
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| 244. |
Water is flowing with a speed of `2m//s` in a horizontal pipe with cross-sectional area decreasing from `2xx10^(-2) m^(2)` to `0.01 m^(2)` at pressure `4xx10^(4)` pascal. What will be the pressure at smaller cross-section? |
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Answer» Here, `upsilon_(1) = 2 ms^(-1), a_(1)=2xx10^(-2)m^(2), P_(1) = 4xx10^(4) pascal , a_(2) = 0.01 m^(2) , P_(2)=?` As, `a_(1)upsilon_(1) = a_(2) upsilon_(2) or upsilon_(2) =(a_(1)upsilon_(1))/(a_(2)) = ( 2xx 10^(-2) xx 2)/(0.01) = 4ms^(-1)` Now, `P_(1) + 1/2 rho upsilon_(1)^(2) = P_(2) + 1/2 rho upsilon_(2)^(2) or P_(2)= 1P_(1)+1/2 rho (upsilon_(1)^(2)-upsilon_(2)^(2))` `:. P_(2) = 4xx10^(4) +1/2 xx 10^(3) (2^(2)-4^(2)) = 4 xx 10^(4) -6 xx 10^(3) = 3.4 xx 10^(4)` pascal. |
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| 245. |
Water flowing steadily through a horizontal pipe of non-unifrom cross-section. If the pressure of water is `4xx10^(4)N//m^(2)` at a point where cross-section is `0.02 m^(2)` and velocity of flow is `2 ms^(-1)` . The pressure at a point where cross-section reduces to `0.01 m^(2)` is `3.4xx10^(n)` Pa. What is the value of n ? |
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Answer» Correct Answer - 4 Here, `a_(1) =0.02 m^(2), a_(2) =0.01 m^(2)`, `v_(1) =2 ms^(-1), v_(2)= ?` From Bernoullis theorem for horizontal flow `P_(1) +(1)/(2)rho v_(1)^(2) = P_(2) +(1)/(2)rho v_(2)^(2)` or `P_(2) = P_(1) + (1)/(2)rho 9v_(1)^(2) - v_(2)^(2)` `=4xx10^(4) + (1)/(2)xx10^(3) (2^(2) - 4^(2))` `=3.4xx10^(4)` Pa `=3.4xx10^(n)` Pa (Given) `:. n=4` |
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| 246. |
A liquid is flowing through a horizontal pipe line of varying cross - section .At a certin point the diameter of the pipe is `6cm` and the velocityof flow of liquid is`2.0cm s^(-1)` Calculate the velocity of flow at another point where the diameter is `1.5 cm. |
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Answer» Here, `D_(1) = 6 cm , upsilon_(1) = 2.0 cm s^(-1)`, `upsilon_(2) = ?, D_(2) = 1.5 cm` From equation of continuty `(pi D_(1)^(2))/(4) upsilon_(1) = (pi D_(2)^(2))/(4) upsilon_(2) or upsilon_(2) = upsilon_(2) (D_(1)^(2))/(D_(2)^(2))` `upsilon_(2) = 2.0 xx (6/1.5)^(2) = 32 cm//s`. |
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| 247. |
Water flow through horizontal pipe of varying cross-section at the rate of `1//2` liter per second. Determine the velocity of flow of water at a point where the diameter is (a) `4 cm` (b) `2 cm`. |
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Answer» `V=(1)/(2) litre//s =500 c.c.//s`, (a)`2 r=4 cm` or `r=2 cm`. `v=(V)/(pir^(2))=(500)/((22//7)xx2^(2))=(500xx7)/(22xx4)` `=39..8 cm s^(-1)` |
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| 248. |
The liquid is flowing steadily through a tube of varying diameter. How are the velocity of liquid flow (V) in any portion and the diameter (D) of the tube in that portion are related? |
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Answer» For a steady flow, `((piD^(2))/(4)) xx V="a constant"` or `V prop 1/(D^2)` . |
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| 249. |
A mild steel wire of length 1.0 m and cross-sectional are `0.5 xx 10^(-20)cm^(2)` is streached, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid point of the wire, calculate the depression at the mid point. `g = 10ms^(-2), Y=2 xx 10^(11)Nm^(-2`. |
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Answer» Let x be the depression at the mid point ie, CD = x In fig., `AC = CB = l = 0.5m , m=100 g = 0.100 kg` `AD = BD = (l^(2) + x^(2))^(1//2)` Increase in length, `Delta l = AD + DB - AB = 2 AD - AB` `=2 (l^(2)+x^(2))^(1//2) -2l` `=2l(1+(x^2)/(l^2))^(1//2) - 2l = 2l[1+(x^2)/(2l^2)] - 2l = (x^2)/(l)` `:. Strai n = (Delta l)/(2l) = (x^2)/(2l^(2))` If T is the tension in the wire, then `2 T cos theta = mg or T = (mg)/(2 cos theta)` Here, `cos theta = (x)/((l^(2)+x^(2))^(1//2)) = (x)/(l(1+(x^2)/(l^2))^(1//2)) = (x)/(l(1+1/2(x^2)/(l^2)))` As, `x lt lt l, so l=1/2 (x^2)(l^2) and 1/2 (x^2)(l^2)~~1 :. cos theta = x/l` Hence, `T = (mg)/(2(x//l)) = (mgl)/(2x)` Stress `= T/A = (mgl)/(2Ax)` `Y= (stress)/(strai n) = (mgl)/(2Ax) xx (2l^2)/(x^2) = (mgl^3)/(Ax^3)` `:. x = l[(mg)/(YA)]^(1//3) = 0.5[(0.1 xx 10)/(20 xx 10^(11) xx 0.5 xx 10^(-6))]^(1//3) = 1.074 xx 10^(-2)m = 1.074 cm`. |
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| 250. |
An external pressure P is applied on a cube at `0^(@)C` so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and `alpha` is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised byA. `(3alpha)/(PK)`B. `3 PK alpha`C. `(P)/(3alpha K)`D. `(P)/(alpha K)` |
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Answer» Correct Answer - C Bulk modulus of material, `K= (Delta p)/((-Delta V//V_(0)))` or `(Delta V)/(V_(0)) =(Delta p)/(K)` (in magnitude) Due to thermal expansion for rise in temperature `Delta t`, `V=V_(0)(1+gamma Delta t)=V_(0) + V_(0)gamma Delta t` or `(V-V_(0))/(V_(0)) =gamma Delta t=3 alpha Delta t` `[.: gamma =3 alpha]` or `(Delta V)/(V_(0)) =3alpha Delta t` `:. (Delta p)/(K) =3alpha Delta t` or `Delta t =(Delta p)/(3 alpha K) =(P)/(3 alpha K)` `[.: Delta p =p` (Given)] |
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