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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.
| 151. |
The force of buoyncy is equal toA. weight of the bodyB. weight of the liquid displaced by the bodyC. apparent weight of the bodyD. viscous force |
| Answer» Correct Answer - B | |
| 152. |
A fully loaded Boeing aircraft has a mass of `3.3xx10^(5) kg`. Its total wing area is `500 m^(2)` . It is in level flight with a speed of `960 km//h`. (a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increases in the speed of the air on the upper surfaces of the wing relative to the lower surface. The density of air is `1.2 kg//m^(3)`. |
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Answer» (a). The weighht of the boeing aircraft is balanced by the upward force due to the pressure difference `DeltaPxxA=3.3xx10^(5)kgxx9.8` `DeltaP=(3.3xx10^(5)kgxx9.8ms^(-2))//500m^(2)` `6.5xx10^(3)Nm^(-2)` (b). We ignore the small height difference between the top and bottom sides in eQ. The pressure difference between them is then `DeltaP=(rho)/(2)(v_(2)^(2)-v_(1)^(2))` When `v_(2)` is the speed of air over the upper surface and `v_(1)` is the speed under the bottom surface. `(v_(2)-v_(1))=(2DeltaP)/(rho(v_(2)+v_(1)))` Taking the average speed `v_(av)=(v_(2)+v_(1))//2=960km//h=267ms^(-1)` We have `(v_(2)-v_(1))/v_(av)=(DeltaP)/(rhov_(av)^(2))~~0.08` the speed above the wing needs to be only 8% higher than that below. |
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| 153. |
A large block of ice floats in a liquid. Whe ice melts the liquid level rises. The density of liquid isA. greater than that of waterB. less than that of waterC. equal to that of waterD. half of that of water |
| Answer» Correct Answer - A | |
| 154. |
The weight of the body is maximum inA. airB. hydrogenC. waterD. vacuum |
| Answer» Correct Answer - D | |
| 155. |
When a body is fully immersed in a liquid the loss of weight of the body is equal toA. apparent weight of the bodyB. force of buoyancyC. half the force of buoyancyD. twice the force of buoyancy |
| Answer» Correct Answer - B | |
| 156. |
A boat having length 2 m and width 1 m is floating in a lake. When a man stands on the boat, it is depressed by 3 cm. The mass of the man isA. 50 kgB. 55 kgC. 60 kgD. 70 kg |
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Answer» Correct Answer - C Total weight `=` Force of buoyancy, `Mg=Vdg` `(M+m)g=DeltaVrho g rArr (M +m) = Delta x A rho` |
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| 157. |
A boat carrying steel balls is floating on the surface of water in a tank. If the balls are thrown into the tank one by one, how will it affect the level of water ?A. go upB. for downC. remain the sameD. can not be decided |
| Answer» Correct Answer - B | |
| 158. |
The length of a rubber cord floating on water is 5 cm. The force needed to pull the cord out of water is ….N (surface tension of water is `7.2xx10^(-4)Nm^(-1)`).A. `7.2xx10^(-3)`B. `7.2xx10^(-4)`C. `7.2xx10^(-5)`D. `7.2xx10^(-2)` |
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Answer» Correct Answer - C `T=(F)/(2l)` |
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| 159. |
A solid rubber ball orf density `d` and radius `R` falls vertically through air. Assume that the air resistance acting on the ball is `F=KRV` where K is constant and V is its velocity. Because of this air resistance the ball attains a constant velocity called terminal velocity `v_(T)` after some time. Then `V_(T)`A. `(4piR^(2)dg)/(3K)`B. `(3K)/(4piR^(2)dg)`C. `(4)/(3)(pir^(3)dg)/(K)`D. `pirdgk` |
| Answer» Correct Answer - A | |
| 160. |
Two needles are floating on the surface of water. A hot needle when touches wate `3r` surface between the needles then they moveA. closerB. awayC. out of the liquidD. into the liquid. |
| Answer» Correct Answer - B | |
| 161. |
Two equal drops of water are falling through air with a steady velocity v. If the drops coalesced, what will be the new velocity? |
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Answer» `(4)/(3)piR^(3)xxrho=(4)/(3)pir^(3)xxrho+(4)/(3)pir^(3)xxrho` or `R=(2^((1)/(3)))r` and `v_(t)alphar^(2)` (stokes now) `(v^(1))/(v)=(R^(2))/(r^(2))=2^(2//3)` `:. v^(1)=2^(2//3)` |
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| 162. |
An air bubble of radius 1 mm is allowed to rise through a long cylindrical column of a viscous liquid of radius 5 cm and travels at a steady rate of 2.1 cm per sec. if the density of the liquid is `1.47(g)/(c c)`. Its viscosity is nearly `(n)/(2)` poise. Then find the value of n. Assume `g=980(cm)/(sec^(2))` and neglect the density of air. |
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Answer» Correct Answer - 3 `v_(t)=(2r^(2)rhog)/(9eta)`. |
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| 163. |
A water filled cylinder of height 50 cm and base area `20cm^(2)` is placed on a table with the base on the table. The thrust offered by water on the table isA. 98 NB. 49 NC. 9.8 ND. 4.9 N |
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Answer» Correct Answer - 3 `F=PA,P=hrhog` |
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| 164. |
When a pressure measuring device is immersed in a liquid, a restoring force of 2 N is exerted by the spring. If the area of cross-section of the pistion is 10 `cm^(2)`, what pressure does the fluid exert on the piston? |
| Answer» `2 xx 10^(3) N m^(-2)` | |
| 165. |
As Fig. shows, `S_(1)` and `S_(2)`, are spring balances. Block `A` is hanging from spring balance `S_(1)` and immersed in liquid `L` which is contained in beaker `B`. The mass of beaker `B` is `1 kg` and mass of liquid `L` is `1.5 kg`. Balances `S_(1)` and `S_(2)` reads `2.5 kg` and `7.5 kg`, respectively. What will be the readings of `S_(1)` and `S_(2)` when block `A` is pulled up out of the liquid. Find the reading of `S_(1)` and `S_(2)`? A. 2.5 kgB. 2 kgC. 1.5 kgD. 3 kg |
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Answer» Correct Answer - A Balance D reads weight minus buoyancy force whereas E reads weight and reaction forces. |
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| 166. |
The pressure at the bottom of a lake due to water is `4.9xx 10^(6)N//m^(2)`. What is the depth of the lake? |
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Answer» Pressure `P=hrhog=4.9xx10^(6)N//m^(2)` `rho` density of water `=1000km//m^(3), g=9.8m//s^(2)` Hence, `h=(P)/(rhog)=(4.9xx10^(6))/(1000xx9.8)=500m` |
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| 167. |
A solid ball is immersed in a liquid. The coefficient of volume expansion of ball and liquid are `3 xx 10^(-6)` per/C and `9xx 10^(-6)` per `,^(@)`C respectively. Find the percentage change in upthrust when the temperature is increased by `25^(@)C` |
| Answer» Correct Answer - `0.015%` | |
| 168. |
A sphere of radius r is dropped in a liquid of viscosity x. Match the Column-I showing some graph with Column-II giving the names of graphs. |
| Answer» Correct Answer - A(s), B(q), C(p), D(r) | |
| 169. |
Statement-1 : When an air bubble moves up from the bottom of a lake, its acceleration decreases and becomes zero. And Statement-2 : When an air bubble moves up from the bottom of a lake, its velocity increases and become constant.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - B | |
| 170. |
If a body floats with `(m//n)^(th)` of its volume above the surface of water, then the relative density of the material of the body isA. `((n-m))/(n)`B. `(m)/(n)`C. `(n)/(m)`D. `((n-m))/(n)` |
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Answer» Correct Answer - A Specific `=("weight of the body")/("force of buoancy")=(W)/(F_(B))=(d_(B))/(d_(W))` `V_("in")=V-V_(out)=V-(m)/(n)V=V((n-m)/(n))` `implies(V_(I n))/(V)=(d_(B))/(d_(W))=S.G` |
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| 171. |
A solid body floating in water has `(1)/(5^(th))` of its volume immersed in it. What fraction of its volume will be immersed, if it floats in a liquid of specific gravity 1.2 ? |
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Answer» Let volume of the solid body = `V_(s)` Density of the solid body = `rho_(s)` Density of water = `rho_(w)` Given that the body floats in water with one-fifth of its volume immersed i.e., immersed volume = `(V_(s))/(5)` `because (rho_(s))/(rho_(w)) = ("Immersed volume of the body")/("Total volume of the body")` `rArr (rho_(s))/(rho_(w))=(V_(s)//5)/(V_(s))` `rArr (rho_(s))/(rho_(w))=(1)/(5)" "...(i)` Let `(1)/(n^(th))` volume of the solid body is immersed when it floats in a liquid of relative density 1.2. Density of the liquid `rho_(l)=(1.2) rho_(w)` So, `(rho_(s))/(rho_(l))=(V_(s)//n)/(V_(s))` `rArr (rho_(s))/(1.2 rho_(w))=(1)/(eta)` `rArr (1)/(1.2 xx 5)=(1)/(eta)" "("from equation (i)")` `rArr (1)/(eta)=(1)/(6)` Thus, one-sixth of the volume remains immersed. |
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| 172. |
Statement I: A needle placed carefully on the surface of water may float, whereas the ball of the same material will always sink.Statement II: The buoyancy of an object depends both on the material and shape of the object.A. Both statement 1 and statemet 2 are true and statement 2 is the correct explanation of statement 1.B. Both statement 1 and statement 2 are true but statement 2 is not the correct explanantion of statement 6C. Statement 1 is true but statement 2 is false.D. Statement 1 is false but statement 2 is false. |
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Answer» Correct Answer - C When a raindrop falls in air (viscous medium) after falling through the same height, the viscous dragbalances the weight of the drop. Through the rest od its height, velocity is constant or it attains a terminal velocity. |
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| 173. |
The pressure at the bottom of a lake due to water is `4.9xx 10^(6)N//m^(2)`. What is the depth of the lake?A. 500 mB. 400 mC. 300 mD. 200 m |
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Answer» Correct Answer - 1 `P=hrhog` |
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| 174. |
statement 1 is false, statement 2 is true. Statement-1: When a soda water bottle fals freely from a height `h`, the gas bubble rises in water from the bottom. Statement-2:Air lighter than liquid.A. Statement-I is true, statement-2 true and statements-2 is a correct explanation for statements-1B. Statement 1 is true, statement 2 is true, statement-2 is not a correct explanation for statement 1C. Statement 1 is true, statement 2 is falseD. Statement 1 is false, statement 2 is true |
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Answer» Correct Answer - D Uptrust will be zero. |
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| 175. |
A spherical soap bubble of radius 1 cm is formed inside another of radius 3 cm the radius of single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is ____cmA. `(4)/(3)`B. `(3)/(4)`C. `(1)/(2)`D. `2` |
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Answer» Correct Answer - 2 `r=(r_(1)r_(2))/(r_(1)+r_(2))` |
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| 176. |
To what height should a cyclindrical vessel be filled with a homogeneous liquid to make the force with which the liquid pressure on the sides of the vessel equal to the force exerted by the liquid on the bottom of the vessel ?A. equal to the radiusB. less than radiusC. more than radiusD. four times of radius |
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Answer» Correct Answer - A Let `h` be the desired height of liquid in cylinder for which the force on the bottom and sides of the vessel is equal. Force on bottom`=rhoghxxpiR^(2)` Force on walls of vessel `=rhog(h//2)xx2piRh` According to question, `rhoghpiR^(2)=rhogpiRh^(2)` or `R=h` |
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| 177. |
When liquid medicine of density `rho` is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of te drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surfacetension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper. If the radius of the opening of the dropper is `r`, the vertical force due to the surface tension on the top of radius `R` (assuming `rltltR`) isA. `2pirT`B. `2pirRT`C. `(2pir^(2)T)/(R)`D. `(2piR^(2)T)/(r)` |
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Answer» Correct Answer - C `2pirT(r)/(R)=F` |
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| 178. |
The high domes of ancient buildings have structural value (besides beauty). It arises from pressure difference on the 2 faces due to curvature (as in soap bubbles). There is a dome of radius 5 m and uniform (but small ) thickness. The surface tension of its masonry structure is about 500 N/m. Treated as hemispherical, the maximum load that the dome can support is nearest toA. `1500kg-Wt`B. `3000kg-Wt`C. `6000kg-Wt`D. `12000kg-Wt` |
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Answer» Correct Answer - B `DeltaP=(4T)/(r)` Maximum load that can be supported by the dome `=(DeltaP)pir^(2)=(4T)/(R)xxpiR^(2)=4piTR` `=4xx(22)/(7)xx500xx5=31400N` `approx3000kgWt` (nearly) |
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| 179. |
Surface tension arises from the cohesive force between the surface molecules. Interplay between cohesion and adhesion force make the surface inclined at acute or obtuse angle with the contacting solid surfaces. This causes a capillary rise (or fail) given as `h=(2Tcostheta)/(rhogr)`, where `theta=` angle of contact `T=` surface tension `rho=` density of the liquid, `g=` acceleration due to gravity and `r=` radius of the capillary tube. Q. If the vessel accelerates up, capillary rise,A. increasesB. decreasesC. remains the sameD. becomes zero |
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Answer» Correct Answer - B `h=(2Tcostheta)/(rhogr)` and `g_(eff)` increases therefore h decreases. |
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| 180. |
When a wire o length `l(lltltr)` and ceoss sectional radius `r` is kept floating on surface of a liquid. Maximum radius of wire such that it may not sink. Is |
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Answer» Weight `u=` maximum force of surface tension `Mg=T[2l]impliesrho(pir^(2))lg=T(2l)` `r_(max)=sqrt((2T)/(pirhog)) (rho` is the density of the wire) Note: `D_(max)=2r_(max)=sqrt((8T)/(pirhog)),A_(max)=pir^(2)=(2T)/(rhog)` |
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| 181. |
If a soap bubble of radius 3 cm coalesce with another soap bubble of radius 4 cm under isothermal conditions the radius of the redultant bubble formed is in cmA. 7B. 1C. 5D. 12 |
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Answer» Correct Answer - C `r=sqrt(r_(1)^(2)+r_(2)^(2))` |
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| 182. |
one end of a U-tube of uniform bore (area A) containing mercury is connected to a suction pump. Because of it the level of liquid of density `rho` falls in one limb. When the pump is removed, the restoring force in the other limb is: A. `2xrhoAg`B. `xrhog`C. `Arhog`D. `xrhoAg` |
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Answer» Correct Answer - A Force due to excess pressure `=` Restoring force `=(rhogh)A=rhog(2x)A` |
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| 183. |
Excess pressure can be `(2T//R)` forA. spherical dropB. spherical meniscusC. clindrical bubble in airD. spherical bubble in water |
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Answer» Correct Answer - A::B::C::D i. Excess pressure inside a spherical liquid drop `(2T)/(R)` because there is only one free surface here (ii). In case of spherical meniscus of radius of curvature R, also excess pressure `=(2T)/(R)` because again there is only one free surface. (iii). Excess pressure inside a cylindrical drop of radius `R=(T)/(R)`, hence, for a culindrical bubble in air, excess pressure is `(2T)/(R)` because there exists two free surface `=(T)/(R)` in this case (iv). for a spherical bubble in water, excess pressure `=(T)/(R)`, as there is only one free surface. Hence, all the four optional are correct. |
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| 184. |
On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If `rlt ltR` and the surface tension of water is T, value of r just before bubbles detach is: (density of water is `rho_w`) A. `R^(2)sqrt((2rho_(w)g)/(2T))`B. `R^(2)sqrt((rho_(w)g)/(6T)`C. `R^(2)sqrt((rho_(w)g)/(T))`D. `R^(2)sqrt((3rho_(w)g)/(T))` |
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Answer» Correct Answer - A The bubble will detach if, `intsinthetaTxxd,=T(2pir)sintheta` Buoyant force `ge` surface tension force `(rho_(w))((4)/(3)piR^(3))ggt(T)(2pir)sintheta` `impliessintheta=(r)/(R)` |
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| 185. |
With increase in temperature the viscosity ofA. liquids increases and of gases decreasesB. liquids decreases and gases increasesC. both liquids and gases increasesD. both liquids and gases decreases |
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Answer» Correct Answer - B With the increase in temperature , the viscosity of liquids decreases and that of gases increases. |
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| 186. |
With increase in temperature the viscosity ofA. gases decreasesB. liquids increasesC. gases increasesD. liquids decreases |
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Answer» Correct Answer - C::D For liquids coefficient of viscosity, `eta prop (1)/(sqrt(T))` i.e., with increase in temperature `eta` decreases. For gases coefficient of viscosity, `eta prop sqrt(T)` i.e., with increase in temperature `eta` increases. |
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| 187. |
If the gauge pressure, at the bottom of a water tank is 2.7 kPa, what is the height of the water [Take g = 10 m `s^(-2)`] Hint : Gauge Pressure = absolute pressure - atmospheric pressure = pgh |
| Answer» Correct Answer - 27 cm | |
| 188. |
The density of the atmosphere at sea level is `1.29 kg//m^(3)`. Assume that it does not change with altitude. Then how high would the atmosphere extend ? `g=9.8 ms^(-2)`. Atmospheric pressure `=1.013 xx 10^(5) Pa`. |
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Answer» `rhogh=1.29kgm^(-3)xx9.8ms^(2)xxhm=1.01xx10^(5)Pa` `thereforeh=7989m~~8km` In reality the density of air decreases with height. So does the value of g. the atmospheric cover extends with decreasing pressure over 100 km. We should also note that the sea level atmospheric pressure is not always 760 mm of Hg. A drop in the Hg level by 100 or more is a sign of an approaching storm. |
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| 189. |
The density of the atmosphere at sea level is `1.29 kg//m^(3)`. Assume that it does not change with altitude, how high should the atmosphere extend? |
| Answer» Correct Answer - About 8 km | |
| 190. |
What is the terminal velocity of a copper ball of radius 2.0 mm, which falls through an oil tank? Given that density of oil = `1.5 xx 10^(3)` kg `m^(-3)`, density of copper `= 8.9 xx 10^(3) kg m^(-3)` and viscosity of oil is 0.99 kg `m^(-1)s^(-1)`. |
| Answer» Correct Answer - 6.5 cm `s^(-1)` | |
| 191. |
An iron ball of radius 0.3 cm falls through a column of oil of density 0.94 g `cm^(-3)`. If it attains a terminal velocity of 0.54 m `s^(-1)`, what is the viscosity of oil? Density of iron is 7.8 g `cm^(-3)` |
| Answer» Correct Answer - 248 mPI | |
| 192. |
Two rain drops reach the earth with different terminal velocities having ratio `94` then te ratio fo their volume isA. `3:2`B. `4:9`C. `9:4`D. `27:8` |
| Answer» `V_(t)=(2)/(9)(gr^(2)[rho-sigma])/(eta)impliesV_(t)propr^(2)prop(V^((1)/(3)))^(2)propV^((2)/(3))` | |
| 193. |
Assertion :The contact angle between water and glass is acute. Reason :The surface of water in the capillary is convex .A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true and reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C The surface of water in the capillary is concave. |
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| 194. |
If a big shop of liquid at `27^(@)` is broken into number of small drops then the termparature of the drplets isA. `=27^(@)C`B. `gt27^(@)C`C. `lt27^(@)C`D. `=54^(@)C` |
| Answer» Correct Answer - C | |
| 195. |
With the increase of temperatureA. the viscosity of a liquid increases.B. the viscosity of a gas decreasesC. the viscosity of a gas increasesD. the viscosity of a gas remains unchaged. |
| Answer» Correct Answer - B | |
| 196. |
With the increase in temperature the angle of contact glass and waterA. decreasesB. increasesC. remains contD. some times increases and some times decreases |
| Answer» Correct Answer - A | |
| 197. |
A small block of wood of relative density 0.5 is submerged in water at a depth of 5 m When the block is released it starts moving upwards, the acceleration of the block is `(g=10ms^(-2)`)A. `5ms^(-2)`B. `10ms^(-2)`C. `7.5ms^(-2)`D. `15ms^(-2)` |
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Answer» `ma_("apparent")="force of buoyancy-weight of body"` `=Vd_(w)g-Vd_(B)g` `impliesa_("apparent")=(Vd_(w)g-Vd_(B)g)/(Vd_(B))=g((d_(w))/(d_(B))-1)` |
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| 198. |
A block of wood floats in water with `((4)/(5))^(th)` of its volume submerged. In an oil, it floats with `((9)/(10))^(th)` volume submerged. The ratio of the density of oil and water isA. `(8)/(9)`B. `(9)/(8)`C. `(19)/(25)`D. `(25)/(18)` |
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Answer» `d_(w)Vg=(4)/(5)Vd_(w)g` ..(1) `d_(w)Vg=(9)/(10)Vd_("oil")g` ..(2) |
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| 199. |
A capillary tube, made of glass is dipped into mercury. ThenA. mercury rises in the capillary tubeB. mercury descends in capillary tubeC. mercury rises and flows out of capillary tubeD. mercury neither rises nor descends in the capillary tube. |
| Answer» Correct Answer - B | |
| 200. |
The height upto which water will rise in a capillary tube will be:A. maximum when water temperature is `4^(0)C`B. minimum when water temperature is `4^(0)C`C. minimum when water temperature is `0^(0)C`D. same at all temperature |
| Answer» Correct Answer - B | |