Saved Bookmarks
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.
| 301. |
A small sphere of mass m is dropped from a height After it has fallen 100 m it has attained its terminal velocity and continues to fall at that speed. The work done by air friction against the sphere during the first 100 m of fall is-A. Greater than the work done by air friction in the second `100 m`B. Less than the work done by air friction in the second `100 m`C. Equal to `100 mg`D. Greater than `100 mg` |
|
Answer» Correct Answer - B In the first `100m` body starts from rest and its velocity goes on increasing and after `100m` it acquire maximum velocity (terminal velocity). Further, air friction i.e viscous force which is proportional to velocity is low in the beginning and maximum at `v=v_(T)`. |
|
| 302. |
An expansible balloon filled with air floats on the surface of a lake with `2//3` of its volume submerged. How deep must it be sunk in the water so that it is just in equilibrium neither sinking further nor rising? Is is assumed that the temperature of the water is constant `&` that the height of the water barometer is `9` meters. |
|
Answer» Correct Answer - 4.5 m Let mass of air in ballon is m and its volume at surface is V. `mg=B" "=mg=(2)/(3)Vrho_(w)g" "impliesV=(3)/(2)mcm^(3)` at depth d below let its volume is `V_(2)` `W=B" "impliesmg=V_(2)rho_(w)g" "impliesimpliesV_(2) = m cm^(3)` Since `P_(1)V_(1)=P_(2)V_(2)` `9xx(3)/(2)m=(9+d)xxm" "impliesd=4.5m` |
|
| 303. |
A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of the U-shaped tube until the vertical height of the water column is `15.0 cm`. (a) What is the gauge pressure at the water mercury interface ? (b) Calculate the vertical distance h from the top of the mercury in the right hand arm of the tube to the top of the water in the left-hand arm. . |
|
Answer» Correct Answer - A::C (a) Gauge pressure=`rho_(w)gh_(w)` `=(10^(3))(9.8)(0.15)` `=1470 N//m^(2)` (b) Let us calculate pressure of two sides at the level of water and mercury interface. `p_(0)+rho_(w)gh_(w)=p_(0)+rho_(Hg)gh_(Hg)` `:. (1) (15) = (13.6) (15-h)` `h =13/9 cm` |
|
| 304. |
Some questions (Assertion-Reason Type) are given below. Each question contains Statement I (Assertion) and statement II(reason). Each question has 4 choices (a),(b),(c ) and (d) out of which only one is correct. So select the correct choise. a. Statement I is True, Statement II is True,Statement II is a correct explanation for Statement I b. Statement I is True, Statement II is True, Statement II is NOT a correct ecplanation for Statement I c. Statement I is True, Statement II is False . d. Statement I is false, Statement II is True. 2. Statement:I Though light of a single frequency (monochromatic light) is incident on a metal, the energies of emitted photoelectrons are different. Statement II: The energy of electrons just after they absorb photons incident on the metal surface may be lost in collision with other atoms in the metal before the electron is ejected out of the metal.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
|
Answer» Correct Answer - D The buoyant force acts at the centre of buoyancy of the body i.e., at the centre of mass of the fluid displaced. |
|
| 305. |
Some question (Assertion-Reason type) are given below. Each question contains STATEMENT-1(Assertion) and STATEMENT-2(Reason) . Each question has 4 choices (A),(B),(C ) and (D) out of which ONLY ONE is correct. So select the correct choice : STATEMENT-1 Viscosity of liquid increases rapidly with the rise of temperature. STATEMENT-2 Viscosity of liquid is the property of liquid by virtue of which it opposes the relative motion amongst its different layers.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
|
Answer» Correct Answer - D Statement 1 is incorrect as viscosity decreases in the rise of temperature and statement 2 is true. |
|
| 306. |
A tube of insufficient length is immersed in water (surface tension`=0.7N//m`) with 1 cm of it Given, radius of tube`=1mm`. |
|
Answer» Correct Answer - A::D `h=(2T)/(Rrhog) (R=`radius of meniscus at the top) `:. R=(2T)/(h rho g)=(2xx0.07)/((10^(-2))(10^(3))(10))` `=1.4 xx 10^(-3) m` `= 1.4 mm` |
|
| 307. |
A water drop of radius `10^-2` m is brokenn into 1000 equal droplets. Calculate the gain in surface energy. Surface tension of water ils `0.075Nm^-1`A. `16.96xx10^(-4) J`B. `8.48 xx 10^(-4) J`C. `4.24 xx 10^(-4) J`D. `2.12 xx 10^(-4) J` |
|
Answer» Correct Answer - B Gain in surface energy `E=4piR^(2)(n^(1/3)-1)T` `=4 xx 3.14 xx (10^(-2))(1000)^(1//3)-1 xx 0.075` `=8.48 xx 10^(-4)` J |
|
| 308. |
Assertion : At same level of same liquid pressure is always same. Reason : When any fluid travels from a region of higher pressure to lower pressure (At same levels) it gains same speed.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
| Answer» Correct Answer - D | |
| 309. |
The surface tension of soap solution is `0.03 N//m` . The work done in blowing to from a soap bublle of surface area `40 cm^(2)`, (in J) , isA. `1.2 XX10^(-4)`B. `2.4 XX10^(-4)`C. `12 XX10^(-4)`D. `24 XX10^(-4)` |
|
Answer» (b) in case of soap buuble `W=Txx2xxDeltaA` `=0.03xx2xx40xx10^(-4)` `=2.4xx10^(-4)J` |
|
| 310. |
Two rain drops reach the earth with different terminal velocities having ratio `9:4` . Then , the ratio of their volumes isA. `3:4`B. `4:9`C. `9:4`D. `27:8` |
|
Answer» (d) Terminal velcoity,`V_(T)oor^(2)` or `(V_(T1))/(v_(T2))=(r_(1)^(2))/(r_(2)^(2))` `thereforesqrt((9)/(4))=(r_(2))/(r_(2))or(r_(1))/(r_(2))=(3)/(2)` `thereforeV=(4)/(3)pir^(3)or (V_(1))/(V_(2))=(r_(1))/(r_(2)^(3))=(27)/(8)` |
|
| 311. |
Assertion : In the given figure shown, `p_(1)=p_(2)`. Reason : Pressure at 1 is less than the atmospheric pressure.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
| Answer» Correct Answer - D | |
| 312. |
Assertion : A solid object of iron is dipped in water, both are at same temperature of `2^(@)C`. If the temperature of water is increased by `2^(@)C`, then the buoyancy force action of the object will increase. Reason : If we increase the temperature of water from `2^(@)C " to " 4^(@)C`, then density of water will increase. Ignore expansion of solid sphere.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
| Answer» Correct Answer - A | |
| 313. |
An engine pumps water continously through a hose. Water leave the hose with a velocity `v` and `m` is the mass per unit length of the Water jet. What is the rate at Which kinetic energy is imparted to water?A. `(1)/(2) mv^3`B. `mv^3`C. `(1)/(2) mv^2`D. `(1)/(2) m^2 v^3` |
|
Answer» Correct Answer - A Rate at which kinetic energy is imparted `(d K)/(dt) = (d)/(dt) ((1)/(2) M_0 v^2) = (v^2)/(2) (d M_0)/(dt)` =`(v^2)/(2) (d M_0)/( dl) xx (dl)/(dt) = (v^2)/(2) mv = (1)/(2) m v^3`. |
|
| 314. |
Density of sea water is `1.03 ghc c^(-1)` . A ship passes from fresh water into sea water. It willA. riseB. sinkC. reamin at the same depthD. rise or sink depending on its shape and size |
| Answer» Fresh water has lesser density as compared ot sea water. So, upthrust in sea water will be more as compound to fresh water. | |
| 315. |
water is flowing through a tube of non-uniform cross-section. If the radii of the tube at the ebtrance and the exit are in the ratio `3:2` then the ratio of the velocites of flow of watern at the entrance and the exit isA. `9:4`B. `4:9`C. `8:27`D. `27 :8` |
|
Answer» (b) From equation of conityuity `A_(1)V_(1)=A_(2)V_(2)` `impliespir_(1)^(2)=pir_(2)^(2)V_(2)` `implies(V_(1))/(V_(2))=(r_(2)^(2))/(r_(1)^(2))=(4)/(9)` |
|
| 316. |
A small ball (mass m) falling under gravity in a viscous medium experience a drag force proportional to the instantaneous speed u such that `F_(drag)=ku`. Then the terminal speed of ball within viscous medium isA. `(k)/(mg)`B. `(mg)/(k)`C. `sqrt((mg)/(k))`D. None of these |
|
Answer» Correct Answer - D Weight-upthrust=dragforce. But no information is given reagarding the upthrust. |
|
| 317. |
Statement -1 : A rain drop after fallling through some height attains a constant velocity. Statement -2 : At constant velocity, the viscous drag is just to its weight.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. If assertion ture but reason is false.D. If the assertion and reason both are false. |
|
Answer» Correct Answer - A |
|
| 318. |
A glass full of water upto a height of 10 cm has a bottom of are `10 cm^(2)`, top of area `30cm^(2)` and volume 1 litre. (a) Find the force exerted by the water on the bottom. (b) Find the resultant force exerted by the sides of the glass on the water. (c ) If the glass is convered by a jar and the air inside the jar is completely pumped out, what will be the answer to parts (a) and (b). (d) If a glass of different shape is used, provided the height, the bottom area, the top area and the volume are unchanged, will the answer to parts (a) and (b) change. Take `g=10m//s^(3)`, density of water`=10^(3) kg//m^(3)` and atmospheric pressure `=1.01xx10^(5)N//m^(2)`. |
|
Answer» Correct Answer - A::B::C::D (a) Force exerted by the water on the bottom `F_(1)=(p_(0)+rho gh)A_(1)` …(i) Here, `p_(0)= `atmospheric pressure `=1.01xx10^(5)N//m^(2)` `rho=`density of water `=10^(3)k//gm^(3)` `g=10m//s^(2), h=10cm=0.1 m` and `A_(1) =` area of base `=10 cm^(2)=10^(-3)m^(2)` Subsituting in Eq. (i) and we get, `F_(1)=(1.01xx10^(5)+10^(3)xx10xx0.1)xx10^(-3)` or, `F_(1)=102 N` (downwards) (b) Force exerted by atmosphere on water `F_(2)=(p_(0))A_(2)` Here,` A_(2)=` area of top `=30cm^(2)=3xx10^(-3)m^(2)` `=303N` (downwards). Force exerted by bottom on the water `F_(3)=-F_(1)` or `F_(3)=102 N`(upwards) weight of water `W=("volume")("density")(g)=10^(3)10^(3)(10)` `=10N` (downwards) Let` F` be the force= net downward froce or, `F+F_(3)=F_(2)+W` `:. F=F_(2)+W-F_(3)=303+10-102` or,` F=211N` (upwards) (c ) If the air inside the jar is completely pumped out, ` F_(1)=(rho gh)A_(1) (as p_(0)=0)` `=10^(3)(10)(0.1)10^(3)=1N` (downwards) In this case, `F_(2)=0` and `F_(3)=1N` (upwards) `:. F=F_(2)+W-F_(3)=0+10-1=9 N` (upwards) (d) No, the answer will remain the same. Because the answer depend upon `p_(0) rho,g,h,A_(1)` and `A_(2)`. |
|
| 319. |
A glass full of water has a bottom of area `20 cm^(2)`, top of area `20 cm`, height `20 cm` and volume half a litre. a. Find the force exerted by the water on the bottom. b. Considering the equilibrium of the `v`, after. find the resultant force exerted by the side, of the glass on the water. Atmospheric pressure `= 1.0 xx 10^(5) N//m^(2)`. Density of water `= 1000 kg//m^(-3)` and `g = 10 m//s^(2)` |
|
Answer» Correct Answer - A::B::D `Pa=1.0xx10^5N/m^2`, `=Pw=10^3mg/m^3` g=10m/s^2` `v=500m/=500g=0.5kg` a. Force exerted at the bottom = Force due to cylindrical water column + atm. Force `=AxxhxxP_wxxg+P_axxA` `=A(hrho_wg+P_a)` `=204N` b. To find out the resultant force exerted by the sides of the glass, from the free body diagram of water inside the glass, `P_axxA+mg=Axxhxxrho_2xxg+F_s+P_axxA` `rarr mg=Axxhxxp_wxg+F_s` `rarr 0.5xx1=20xx10^-4xx20x10^-2xx10^-3xx10+F_s` `rarr F_s=5-4=1N` (upward) This force 1N is provided by the sides of the glass. |
|
| 320. |
In a hydraulic lift, used at a service station the radius of the large and small piston are in the ration of `20 : 1`. What weight placed on the small piston will be sufficient to lift a car of mass `1500 kg` ?A. `3.75` kgB. `37.5` kgC. `7.5` kgD. `75` kg |
|
Answer» Correct Answer - A `(m_(1)g)/(A_(1)) = (m_(2)g)/(A_(2))` Solving. `m_(2) = 3.75 kg`. |
|
| 321. |
Assertion: To floats, a body must displace liquid whose weight is greater than actual weight of the body. Reason: The body will experience no net downward force in that case.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. If assertion ture but reason is false.D. If the assertion and reason both are false. |
|
Answer» Correct Answer - C |
|
| 322. |
Assertion: To floats, a body must displace liquid whose weight is greater than actual weight of the body. Reason: The body will experience no net downward force in that case.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertionC. if assertion is true but reason is false.D. if both assertion and reason are false. |
|
Answer» Correct Answer - B A body much displaces as much amount of liquid greater than the actual weight of the body to float in it. In case of floating, no net downward force acts on the body. |
|
| 323. |
Water flows ina horizontla tube as shown in figure. The pressure of water changes by 600 N m^-2 between A and B whre the areas of cross section asre `30cm^2 and 15cm^2 respectively. Find the rate of flow of water through the tube. , |
|
Answer» Let the velocity at `A=v_A and that at B=v_B` By the equation of continuity `v_B/V_A=(30cm^2)/(15cm^2)=2` By Bernoulli equation `P_A+1/2rhov_A^2=P_B+1/2rhov_B^2` or `P_A-PB=1/2rho(2v_A)^2-1/2rho v_A^2=3/2rhov_A^2` or v_A=sqrt(0.4ms^2s^-2)=0.63ms^-1` the rate of flow =`(30cm^2)(0.63ms^-1)=1890cm^3s^-1` |
|
| 324. |
A cylinderical tank of radiuus `20 cm` and height `50 cm` has water up to `30 cm` of height. What will be the rise in level of liquid at the periphery if the cylinder be givenan angular velocity of `10 rad s^(-1)`? Also determine the frequency of rotation when water just starts spilling over the sides of the vessel. |
|
Answer» At an angular sped of `10 rads^(-1)`, let us assume that the water does not spill. `h_(2)=H+(omega^(2)r^(2))/(4g)impliesh_(2)-H=(omega^(2)r^(2))/(4g)` Therefore, rise in the liquid level at the periphery is `10 cm`. For water to just spill over the sides ,the maximum height `(h_("max"))` is `50 cm` Again `h_("max")=y_(0)+(omega^(2)R^(2))/(4g)` Here `h_("max")=50 cm, y_(0)=30 cm, R=20 cm, g=10ms^(-2)` `:. 50xx10^(-2)=30xx10^(-2)+(omega^(2)(20xx10^(-2))^(2))/(4xx10)` `implies omega^(2)xx10^(-3)=20xx10^(-2)` `omega^(2)=200implies omega=10sqrt(2)rads^(-1)` Hence, freqency of rotation is `f=omega/(2pi)=(5sqrt(2))/pis^(-1)` |
|
| 325. |
There are two holes `O_(1) " and " O_(2)` in a tank of height H. The water emerging from `O_(1) " and " O_(2)` strikes the ground at the same points, as shown in figure. Then A. `H=h_(1)+h_(2)`B. `H=h_(2)-h_(1)`C. `H=sqrt(h_(1)h_(2))`D. None of these |
|
Answer» Correct Answer - A `R_(h)=R_(H+h)" or "h_(1)=H" or "h_(1)+h_(2)=H` |
|
| 326. |
figure shows two holes in a wide tank containing a liquid common. The water streams coming out of these holes strike the ground at the same point. The heigth of liquid column in the tank is A. `10cm`B. `8cm`C. `9.8 cm`D. `980 cm` |
|
Answer» Correct Answer - A Range `R = 2 sqrt(h(H-h))` `R_(1) = R_(2) implies 2sqrt(h_(1)(H-h_(1))) = 2sqrt(h_(2)(H-h_(2)))` `4(H-4)=6(H-6)` or `2H = 36-16 = 20or H = 10cm`. |
|
| 327. |
A liquid of density `rho` is rotated by an angular speed `omega` as shown in figure,Using the concept of pressure equation, find a relation between `h_(1),h_(2),x_(1)` and `x_(2)`. |
|
Answer» Writing pressure equation between points `A` and `B`, we have `p_(A)+rho gh_(1)-(rho omega^(2)x_(1)^(2))/(2)+(rho omega^(2)x_(2)^(2))/(2)-rho gh_(2)=p_(B)` Points `A` and `B` are open to atmosphere. Therefore, `p_(A) = p_(B) = p_(0) =` atmospheric pressure. Substituting the values in above equation we have, `rho gh_(1) - (rho omega^(2)x_(1)^(2))/(2) +(rho omega^(2) x_(2)^(2))/(2) -rho g h_(2) = 0` On simplifying we get, `(h_(2)-h_(1))=(omega^(2))/(2g)(x_(2)^(2)-x_(1)^(2))` This is the desired relation between `h_(1),h_(2),x_(1)` and `x_(2)`. |
|
| 328. |
A sphere of solid material of specific gravity `8` has a concentric spherical cavity and just sinks in water. The ratio of radius of cavity to that of outer radius of the sphere must beA. `7^(1//3)/2`B. `5^(1//3)/2`C. `9^(1//3)/2`D. `3^(1//3)/2` |
|
Answer» Correct Answer - A Let `rho` be the density of the material `rho_(0)` be the density of water.When the sphere has just started sinking, the weight of the sphere `=` weight of water displaced (approx). `implies 4/3pi(R^(3)-r^(3))rhog=4/3piR^(3)rho_(0)g` `implies (R^(3)-r^(3))rho=R^(3)rho0implies((R^(3)-r^(3)))/R^(3)=(rho_(0))/(rho)` `rarr r/R=(7^(1/3))/2` |
|
| 329. |
Assertion: The velocity increases, when water flowing in broader pipe enter a narrow pipe. Reason: According to equation of continuity, product of area and velocity is constant.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. If assertion ture but reason is false.D. If the assertion and reason both are false. |
|
Answer» Correct Answer - A |
|
| 330. |
Assertion: Sudden fall of pressure of at a place indicates storm. Reason: air flows from higher pressure to lower pressure.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertionC. if assertion is true but reason is false.D. if both assertion and reason are false. |
|
Answer» Correct Answer - A We know that the pressure of air decreases with hight because of it the air having lower pressure comes downwards near to surface of ocean. This cause air to flow inwards towards towards the centre of storm at higher pressure around to it. Now on account of rotational of earth, this inflow of wind is deflected in counter clockwise, spiral in northern hemisphere. |
|
| 331. |
A concrete sphere of radius `R` has cavity of radius `r` which is packed with sawdust. The specific gravities of concrete and sawdust are respectively `2.4 and 0.3` for this sphere to float with its entire volume submerged under water. Ratio of mass of concrete to mass of swadust will beA. 8B. 4C. 3D. Zero |
|
Answer» Correct Answer - B Let specific gravities of concrete and saw dust are `rho_(1)` and `rho_(2)` respectively According to pricnciple of floation weight of a whole sphere = upthrust on the sphere ` 4/3 pi(R^(3)-r^(3))rho_(1) g +4/3 pir^(3) rho_(2) g = 4/3piR^(3)xx1xxg` `implies R^(3)rgho_(1) -r^(3)rho_(1) +r^(3)rho_(2)=R^(3)` `implies R^(3) (rho_(1)-1) = r^(3)(rho_(1)-rho_(2)) implies (R^3)/(r^3) = (rho_(1)-rho_(2))/(rho_(1)-1)` `implies (R^(3)-r^(3))/(r^3)` = (rho_(1)-rho_(2)-rho_(1)+1)/(rho_(1)-1)` `implies ((R^(3)-r^(3))rho_(1))/(r^(3)rho_(2)) = ((1-rho_(2))/(rho_(1)-1))(rho_1)/(rho_2)` `implies (Mass of concrete)/(Mass of saw dust) = ((1.03)/(2.4-1) xx (2.4)/(0.3) = 4`. |
|
| 332. |
A concrete sphere of radius `R` has cavity of radius `r` which is packed with sawdust. The specific gravities of concrete and sawdust are respectively `2.4 and 0.3` for this sphere to float with its entire volume submerged under water. Ratio of mass of concrete to mass of swadust will beA. `8`B. `4`C. `3`D. Zero |
|
Answer» Correct Answer - B |
|
| 333. |
Assertion: If an object is submerged in fluid at rest, the fluid exerts a force on its surface. Reason: The force exerted by the fluid at rest has to be parallel to the surface in contact with it.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertionC. if assertion is true but reason is false.D. if both assertion and reason are false. |
|
Answer» Correct Answer - C When an object is placed in fluid at rest, the fluid always exerts a force normal to the objects surface. |
|
| 334. |
A beaker containing water with a total mass of `10kg` is placed on the pan of a balance `A` . A solid body of mass `5 kg` and density `50g//c c` suspended from a spring balance `B` is gently lowered in the water contained in the beaker. So that it gets fully immersed with out any contact with the beaker. Find the ratio of readings shown by the balance `A` and `B`.A. `11/4`B. `7/4`C. `9/4`D. None of these |
|
Answer» Correct Answer - A Volume of body = (5)/(5xx10^3) = 1xx10^(-3)m^(3)` `:. Force of upthrust = weight of water displaced ` `=1xx10^(-3)m^(3)xx1000kg//m^(3) = 1kg` `:. Reading of balance A = 10 kg + upthrust` `=10+1=11kg` and reading of balance `B=5kg-upthrust ` `=5-1 = 4kg`. |
|
| 335. |
Two water pipes `P` and `Q` having diameters `2 xx 10^(-2) m ` and `4 xx 10^(-2) m`, respectively, are joined in series with the main supply line of water. The velocity of water flowing in pipe `P` isA. `4` times that of `Q`B. `2` times that of `Q`C. `1//2` times that of `Q`D. `1//4` times that of `Q` |
|
Answer» Correct Answer - A by `A_(1)V_(1) = A_(2)V_(2)` `((piD_(1)^(2))/(4))V_(1) = ((piD_(2)^(2))/(4))V_(2)` `V_(2) = 4V_(1)` |
|
| 336. |
Two water pipes of diameters 2 cm and 4 cm are connected with the main supply line. The velocity of flow of water in the pipe of 2 cmA. `4` times that in the other pipeB. `1/4` times that in the other pipeC. `2` times that in the other pipeD. `1/2` times that in the other pipe |
|
Answer» Correct Answer - A |
|
| 337. |
Assertion : A solid is floating in a liquid . If temperature is increased and expansion of solid is ignored, then fraction of volume immersed will increase. Reason : By increasing the temperature density of liquid will decrease.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
| Answer» Correct Answer - A | |
| 338. |
A uniform cube of mass `M` is floating on the surface of a liquid with three fourth of its volume immersed in the liquid `(density =rho)`. The length of the side of the cube is equal to A. `(4M//3rho)^(2//3)`B. `(M//3rho)^(2//3)`C. `(M//4rho)^(2//3)`D. None of these |
|
Answer» Correct Answer - D Weight = upthrust `:. Mg=((3)/(4) a^(3))rhog` `:.a=((4M)/(3 rho))^(1//3)`. |
|
| 339. |
A body of density `rho` is dropped from reat from a height h into a lake of density `sigma(sigmagtrho)`. The maximum depth the body sinks inside the liquid is (neglect viscous effect of liquid)A. `((hsigma)/(rho-sigma))`B. `hsigma/(rho)`C. `h((rho)/(rho-sigma))`D. `h((rho)/sigma-1)` |
|
Answer» Correct Answer - A If `v_(0)-sqrt(2gh)` is the velocity downwards just when the body enters the liquid the final velocity at the maximum depth it will sink is zero. Its acceleration is `a`. Hence `(-)y=` maximum depth `(v_(0)^(2))/(2a)=(2gh)/(2g(rho/sigma-1))` `:. Y=(hsigma)/(rho-sigma)` |
|
| 340. |
A body of density `rho` is dropped from reat from a height h into a lake of density `sigma(sigmagtrho)`. The maximum depth the body sinks inside the liquid is (neglect viscous effect of liquid)A. `(hrho)/(sigma-rho)`B. `(h sigma)/(sigma-rho)`C. `(h rho)/(sigma)`D. `(h sigma)/(rho)` |
|
Answer» Correct Answer - A Velocity of body just before touching the lake surface is, `v=sqrt(2gh)` Retardation in the lake , `a=("upthrust-weight")/("mass")` `=(V rhog-V rhog)/(Vrho) =((sigma-rho)/(rho))g` Maximum depth `d_(max)=(v_2)/(2a)=(h rho)/(sigma-rho)` . |
|
| 341. |
Let air at rest at the front edge of wing of an aeroplane and air passing over the surface of the wing at a fast speed `v`. If density of air is `rho`, then find out the highest value for `v` in stream line flow when atmospheric pressure is `P_(atm)`. |
| Answer» Correct Answer - `v_(max) = ((2p_(atm))/(P))^(1//2)` | |
| 342. |
Pressure gradient in the horizontal direction in a static fluid is represented by (z-direction is vertically upwards, and x-axis is along horizontal, d is density of fluid):A. `(delp)/(delz)=-dg`B. `(delp)/(delx)=dg`C. `(delp)/(delx)=0`D. `(delp)/(delz)=0` |
|
Answer» Correct Answer - C In a static fluid, pressure remains same at the same level, ie, pressure do not vary with x-coordinate |
|
| 343. |
In English the phrase tip of the iceberg is used to mean a small visible fraction of something that is mostly hidden . For a real iceberg. What is this fraction. If the density of sea water is `1.03 kg//c c` and that of ice is `0.92 g//c c`?mA. `0.106`B. `10.6`C. `0.901`D. `0.801` |
|
Answer» (a) For equilibrium `:.` weight =force of buoyancy or `mg=V_("in")sigmag` `rhoV=V_("in")sigma :. V_("in") =(rhoV)/(sigma)` `:. V_("out")=V-V_("in")=V(1-(rho)/(sigma))` Fraction `V_("out)/(V)=1-(rho)/(sigma)=(1-(0.92)/(1.03))=0.106` |
|
| 344. |
Mark the correct options (s).A. two stram lines may cross each otherB. two stram lines must cross each otherC. two stram lines never cross each otherD. None of the above |
|
Answer» (c) If two strem lines cross each other , then at a given point, there are two directions of motion of fluid particles. This is physically not possible. Hence ,(c) is correct |
|
| 345. |
Water is flowing continuously from a tap having an internal diameter `8xx10^(-3)`m. The water velocity as it leves the tap is `0.4 ms^(-1)`. The diameter of the water stream at a distance `2xx10^(-1)`m below the tap is close to `(g=10m//s^(2))`A. `5.0 xx 10^(-3)m`B. `7.5 xx 10^(-3)m`C. `9.6 xx 10^(-3)m`D. `3.6 xx 10^(-3)m` |
|
Answer» Correct Answer - D Diameter `= 8 xx 10^(-3)m` `v = 0.4 m//s` `v = sqrt(u^(2) +2gh)` `= sqrt((0.4)^(2) + 2 xx 10 xx 0.2) = 2 m//s` `A_(1)v_(1) = A_(2)v_(2)` `pi((8 xx 10^(-3))/(2))^(2) xx 0.4 = pi xx d^(2)/(4) xx 2` `d approx 3.6 xx 10^(-3)m`. |
|
| 346. |
A small metal sphere of radius a is falling with a velocity `upsilon` through a vertical column of a viscous liquid. If the coefficient of viscosity of the liquid is `eta`, then the sphere encounters an opposing force ofA. `6 pi eta a^(2)upsilon`B. `(6eta upsilon)/(pia)`C. `6pi eta a upsilon`D. `(pi eta upsilon)/(6a^(3))` |
|
Answer» Correct Answer - C Stock estbalished that if a sphere of radius a moves with velocity v through a fluid of viscosity n, the the viscous force opposing the motion of the sphere if `F=6pieta av` |
|
| 347. |
A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal toA. `2 pi L`B. `(L)/(2pi)`C. LD. `L/(2pi)` |
|
Answer» Correct Answer - B Velocity of efflux when the hole is a t depth `h`, `v = sqrt(2gh)` Rate of flow of water from square hole `Q_(1) = a_(1)v_(1)= L^(2)sqrt(2gy)` Rate of flow of water from circular hole `Q_(2) = a_(2)v_(2) = piR^(2)sqrt(2g(4y))` According to problem, `Q_(1) = Q_(2)` `implies L^(2) sqrt(2gy) = pi R^(2)sqrt(2g(4y)) implies R = (L)/(sqrt(2pi)`. |
|
| 348. |
A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance l and h are shown here. After some time the coin falls into water. Then A. `l` decreases and `h` increases.B. `l` increases and `h` decreases .C. Both `l` and `h` increase.D. Both `l` and `h` increase. |
|
Answer» Correct Answer - D As the block moves up with the fall of coin, `l` decreases, similarly `h` will also decrease because when the coin is in water, it displaces water equal to its own volume only. |
|
| 349. |
A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance l and h are shown here. After some time the coin falls into water. Then A. `l` decreases and `h` increasesB. `l` increases and `h` decreasesC. both `l` and `h` increaseD. both `l` and `h` decrease |
|
Answer» Correct Answer - D When coin sinks in liquid, it will displace less volume of water, level will fall i.e., `h` decreases. When no coin, less weight and hence upthrust decreases, so `l` decreases. |
|
| 350. |
Figure shows for containers of olive oil. The pressure at depth h is A. least in B and C bothB. greatest in AC. greatest in DD. equal in all the containers |
|
Answer» Correct Answer - D (d) `P = h rho g` |
|