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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.
| 13751. |
An object is moving in a straight line with uniform acceleration, the displacement-time relation is |
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Answer» (a) `S=ut^(2)+(1)/(2)at^(2)` |
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| 13752. |
Whichone of the following force is non- conservative ? |
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Answer» ELECTROSTATIC force |
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| 13753. |
A thin uniform cylindrical shell, closed at both ends is partially filled with water. It is floating vertically in water in half-submerged state. If rho_(c) is the relative density of the material of the shell with respect to water, then the correct statement is that the shell is |
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Answer» more than HALF FILLED is `rho_(C)` is LESS than 0.5 |
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| 13754. |
The angle of contact at the interface of water-glass is 0^(@), Ethylalcohol-glass is 0^(@), mercury-glass is 140^(@) and Methyliodide-glass is 30^(@). A glass capillary is put in a trough containing one of these four liquids. It is observed that the menisucs is convex. The liquid in the through is |
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Answer» water |
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| 13755. |
What is bending ? How bending problems prevents and what is buckling ? |
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Answer» Solution :A bridge has to be DESIGNED such that it can withstand the LOAD of the flowing traffic, the force of winds and its own weight. Means it should not bend too much or BREAK. Similarly, in the design of building use of beams and column is very common. In both the cases, the overcoming of the problem of bending of beam under a load is important.  Let us consider the case of a beam loaded at the centre and supported near its ends as SHOWN in figure (a). A bar of length l, breadth b and depth d when loaded at the centre by a load W sags by an amount given by, `delta = (Wl ^(3))/( 4b d ^(3) Y)` where Y = Young modulus This equation shows that , Bending `delta prop (l ^(3))/(bd ^(3) Y)` Means for reduce the bending for the given load, the distance between two support must be small or one should use a material of beam with large Youung.s modulus. Moreover for a given material increasing in the thickness (depth) d rather than the breadth b is more effective in reducing the bonding. Bun on increasing the depth the DEEP bar may bend as shown in figure. (b). This is called bucking.  To avoid bucking a common comporomise is the cross sectional shape (I shape) as shown in below figurre (c) .  Such a section provides a large load bearing surface and enough depth to prevent bending. This shape reduces the weight of the beam and hence reduces the cost. |
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| 13756. |
If R is the radius of earth , the height ,at which the weight of a body becomes 1/4 of its weight on the surface of earth , is |
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Answer» 2R |
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| 13757. |
A body of mass 5 kg has momentum of 10 kg ms^(-1) when a force of 0.2 N is applied on it for 10 seconds , what is the change in the kinetic energy ? |
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Answer» `1.1` J ACCELERATION `a = F/m = (0.2)/5 = 0.04 ms^(-2)` ` v = v_(0)+at` ` = 2+ 0.04 xx10` ` = 2+0.4` ` = 2.4 ms^(-1)` ` :. ` CHANGE in kinetic energy `DeltaK = 1/2 mv^(2) - 1/2 mv_(0)^(2)` ` = 1/2 m(v^(2) -v_(0)^(2))` `= 1/2 xx5(2.4^(2)-2^(2))` ` = 1/2 XX(5.76 -4.00)` `= 1/2 xx5xx1.76 ` = 4.4 J |
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| 13758. |
Angle of prism is A and its one surface is silvered. Light rays falling at an angle of incidence 2A on first surface return back through the same path after suffering reflection at second silvered surface. Refracting index of the material of the prism is |
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Answer» 2 SinA |
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| 13759. |
If the sun and the planets carried huge amounts of opposite charges, |
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Answer» all three of KEPLER's LAWS would STILL be valid |
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| 13760. |
Read the following statements: S_(1) : An object shall weight the same at pole and equator when weighed by using a physical balance. S_(2) : If shall weight the same at pole and equator when weighed by using a physical balance. S_(3) : If shall weight the same at pole and equator when weighed by using a spring balance. S_(4) :If shall weight the same at pole and equator when weighed by using a spring balance. Which of the above statements is/are correct ? |
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Answer» `S_(1)` and `S_(2)` |
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| 13761. |
A block of mass M slides on a frictionless surface with an initial speed of v_(0) . On top of block is small box of mass m . The coefficient of friction between box and block are mu_(s) and mu_(k) . The sliding block encounters an ideal spring with forceconstant k . Answer the following questions : What is maximum value of k for which it remains true that box does not slide ? |
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Answer» `((mu_(s) g )/(v_(0)))^(2) ( M)/(( M + m))` `mu_(s) "MG" = "ma""" …(ii)` |
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| 13762. |
A block of mass M slides on a frictionless surface with an initial speed of v_(0) . On top of block is small box of mass m . The coefficient of friction between box and block are mu_(s) and mu_(k) . The sliding block encounters an ideal spring with forceconstant k . Answer the following questions : Assuming no relative motion between box and block what is the maximum possible acceleration of block and box at the instant of maximum compression ? |
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Answer» `mu_(s) G` 
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| 13763. |
A block of mass M slides on a frictionless surface with an initial speed of v_(0) . On top of block is small box of mass m . The coefficient of friction between box and block are mu_(s) and mu_(k) . The sliding block encounters an ideal spring with forceconstant k . Answer the following questions : Suppose the value of k is just slightly greater than the value found in Q.2 , so that the box beginsto slide just as the spring reaches maximum compression . The accelerations of box and block are respectively : |
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Answer» `a_("BOX") = mu_(K) G , a_("Block")= (m(mu_(s) - mu_(k)) + mu_(s) M)/(M) g ` |
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| 13764. |
Two identical wires carry two spheres of radii 2cm and 3 cm. Their densities are 6g//cm^(3)and 4g//cm^(3). Find the ratio of the elongations produced in them. |
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| 13765. |
Given two vectors vecA=2hati+4hatj+5hatkandvecB=hati+3hatj+6veck. Find the product vecA.vecB and the magnitudes of vecAandvecB. What is the angle between them? |
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Answer» SOLUTION :`vecA.vecB=2+12+30=44` Magnitude `A=sqrt(4+16+25)=sqrt45` UNITS Magnitude `B=sqrt(1+9+36)=sqrt46` units The ANGLE between the two VECTORS is given by `theta=cos^(-1)((vecA.vecB)/(AB))=cos^(-1)((44)/(sqrt(45)xxsqrt46))=cos^(-1)((4)/(45.49))=cos^(-1)(0.967)` `:.theta~=15^(@)`. |
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| 13766. |
Water rises in a capillary tube of diameter 0.2xx10^(-2)m, upto a height of 1.5 cm. The surface tesion of water is |
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Answer» `73.5xx10^(-3)NM^(-1)` |
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| 13767. |
A ring of mass M is rolling down an inclined plane . The length of this plane is L . It makes an angle theta with the horizontal . Find the velocity of the centre of mass of the ring at the centre of the plane. |
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Answer» SOLUTION :The velocity of a body rolling down an inclined plane , `v = sqrt((2 g S SIN theta)/(1 + (K^(2))/(R^2)))` For ring `K= R , S = (L)/(2)` . `therefore v = sqrt((2 g xx (L)/(2) xx sin theta)/(1 + l)) = sqrt((gL sin theta)/(2))` |
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| 13768. |
Two samples of a liquid have volumes 400 cc and 220 cc and their temperature are 10^(@)C and 110^@C respectively. Find the final temperature and volume of the mixture if the two samples are mixed. Assume no heat exchange with the surroundings. Coefficient of volume expansion of the liquid is g = 10^(-3).^(@)C^(-1) and its specific heat capacity is a constant for the entire range of temperature. |
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| 13769. |
Determine the equation for the volume of body's partially part immersed in a fluid for the floating body. |
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Answer» Solution :When any body float on the SURFACE of liquid then weight of body =weight of displaced liquid by the body. `Vrhog=V.rho_(l)g` (where V= volume PF body) V. =volume of partially PART of immerged in liquid = volume of displaced liquid `rho=` Density of body `rho_(l)=` Density of liquid `therefore(V.)/(V)=(rho)/(rho.)=("Volume of immerged part of body")/("Total volume of body")` `=("Density of body")/("Density of liquid")` |
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| 13770. |
A solid shere is rotating with an angular velocity omega. It is of moment is inertia l about its diametrical axis and is of linear expansion coefficient alpha. When temperature is increased by DeltaT, the fractional change in its angular velocity is |
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Answer» `2alphaDeltaT` |
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| 13771. |
Good quality glass tumblers are made of special material whose linear coefficient of expansion is |
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Answer» LARGE |
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| 13772. |
When a ball hits a floor and rebounds after an inelastic collision, |
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Answer» the momentum of the ball just before collision is EQUAL to its momentum just after collision |
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| 13773. |
Falling rain drops acquire terminal velocity due to |
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Answer» UPTHRUST of air |
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| 13774. |
Energy of 1000 J is spent in increasing the speed of a flywheel from 30 rpm to 720ppm, find the moment of inertia of the wheel |
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Answer» Solution :`omega_(1)=30 rpm= 2pi xx(30)/(60)"rad" s^(-1)= pi "rad" s^(-1)` `omega=720 rpm= 2pi xx(720)/(60) "rad" s^(-1)=24 pi "rad"` Change in kinetic energy `DeltaKE=(1)/(2)(omega_(2)^(2)-omega_(1)^(2))` `I=(2xxDeltaKE)/((omega_(1)^(2)-omega_(1)^(2)))=(2xx1000)/((24pi)^(2)-(pi)^(2))` `I=(2000)/(25 pixx23 pi)` REMEMBER : `a^(2)-b^(2)=(a+b)(a-b)` `I~~0.35 kg m^(2) and pi^(2)=10` |
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| 13775. |
An artificial satellite is orbiting around the earth with the speed of 4 km s^(-1) at a distance of 10^4 km from the earth. Calculate its centripetal acceleration. |
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| 13776. |
A pendulum clock is taken to moon. Will it gain or lose time? |
| Answer» SOLUTION :The period of OSCILLATION of a SIMPLE pendulum is `T = 2pisqrt(l/g)` The value of g is less in moon than thaton the surface of EARTH. So T increases. The pendulum TAKES more time to complete one oscillation. Hence it will lose time. | |
| 13777. |
The smallest mass is that of ………….. Of the order of …………… kg. |
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| 13778. |
A faulty thermometer reads -0.5^(@)C in boiling water at the pressure of 747 mm of Hg. What is the correct temperature when the faulty thermometer reads 45^(@)C ? Actual boiling point of water is 99^(@)C at 734 mm of Hg. |
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Answer» Solution :Actual boiling point of water at the PRESSURE of 760 mm of Hg is `100^(@)C`. Now, a decrease in pressure by (760-734) or 26 mm of Hg DECREASES the boiling point of water by (100-99) = `1^(@)C`. `therefore` The decrease in pressure by (760-747) or 13 mm of Hg decreases the boiling point of water by `1/26 times 13 = 0.5^(@)C`. Hence, the boiling point of water at 747 mm of Hg should be (100-0.5) = `99.5^(@)C.` Let the correct TEMPERATURE be `X^(@)C`, when the faulty thermometer reads `45^(@)C`. Hence, `x/(99.5)=(45-(-0.5))/(99-(-0.5))=(45.5)/(99.5) " " therefore x=45.5^(@)C.` |
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| 13779. |
Suppose unknowingly you wrote the universal gravitational constant value as G=6.67xx10^(11) instead of the correct value G=6.67xx10^(-11), what is the acceleration due to gravity g' for this incorrect G? According to this new acceleration due to gravity, what will be your weight W? |
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Answer» Solution :INCORRECT value of G`=6.67xx10^(11)` `g=(GM)/(R^2)` `M_(E)=6.02xx10^(24)KG` `R_(E)=6.4xx10^(6)` `g'=(6.67xx10^(11)xx6.02xx10^(24))/((6.4xx10^6)^(2))` `=(40.1534)/(40.96)xx10^(11+24-12)` `=0.9798xx10^(35-12)` `=0.9798xx10^(23)` `=9.798xx10^(22)` `g'=9.798xx10^(22)` `=9.8xx10^(22)` If mass m = 1 then WEIGHT `W'=mg'=9.8xx10^(22)` `W'=9.8xx10^(22)W` |
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| 13780. |
Two friends A and B are running on a circular track of perimeter equal to 40 m. At time t = 0they are at same location running in the same direction. A is running slowly at a uniform speed of 4.5 km//hr whereas B is running swiftly at a speed of18 km//hr. (a) At what time t_(0) the two friends will meet again? (b) What is average velocity of A and B for the interval t = 0 to t = t_(0)? |
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| 13781. |
Earth's orbit is an ellipse with eccentricity 0.0167. Thus, the earth's distance from the sun and speed as it moves around the sun varies from day-to-day. This means that the length of the solar day is not constant through the year. Assume that the earth's spin axis is normal to its day. A day should be taken from noon to noon. Does this explain variation of length of the during the year ? |
Answer» Solution :Consider the DIAGRAM. Let `m` be the mass of the earth, `v_(p), v(a)` be the velocity of the earth at perigee and apogee respectively. Similarly, `omega_(p)` and `omega_(a)` are corresponding angular velocities.  Angular momentum and areal velocity are constant as the earth orbits the sun At perigee, `r_(p)^(2) omega_(p) = r_(a)^(2) omega_(a)` at apogee ....(i) If a is the semi-major axis of the earth's orbit, then `r_(p) = a(1 - e) " and" r_(a) = a (1 + e)`....(ii) `:. (omega_(p))/(omega_(a)) = ((1 + e)/(1 - e))^(2), e = 0.0167` [from Eqs. (i) and (ii)] `:. (omega_(p))/(omega_(a)) = 1.0691` Let `omega` be angular speed which is geometric mean of `omega_(p)` and `omega_(a)` and corresponds to mean solar day, `:. ((omega_(p))/(omega)) ((omega)/(omega_(a))) = 10691` `:. (omega_(p))/(omega) = (omega)/(omega_(a)) = 1034` If `omega` corresponds to `1^(@)` per day (mean angular speed), then `omega_(p) = 1034^(@)` per day and `omega_(a) = 0.967^(@)` per day. Since, `361^(@) = 24`, mean solar day, we get 361.034 which corresponds to `24h, 8.14''` (8.1'' longer) and `360.967^(@)` corresponds to `23h 59 min 52''` (7.9'' smaller). This does not explain the actual VARIATION of the length of the day during the year. |
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| 13782. |
Which one of the following statements is a correct statement? |
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Answer» STICKTO each other |
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| 13783. |
Find the pressure to be applied to increase the density of water by 1% if the bulk modulus of water is 2.2xx10^(9)Nm^(-2). |
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Answer» `2.2xx10^(7) N//m^(2)` |
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| 13784. |
Show that the velocity of an object released from rest at infinity reaches the earth with a velocity equal to the escape velocity . |
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Answer» Solution :By the principle of conservation of energy for the OBJECT , TE at infinity = (KE +PE) at the surface of the earth. As the body is released from rest at infinity , its KE at infinity is zero and its PE at infinity isalso zero. If v is its VELOCITY on reaching the earth its KE BECOMES equal to `1/(2)mv^2 and PE = (GMM)/R` `:. 0 +0 =1/(2)mv^(2) -(GMm)/R` `1/(2)v^(2)=(GM)/R` The velocity of the object, `V = sqrt((2GM)/R) =V_(e)` |
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| 13785. |
A hotcup ofcoffee iskepton thetable , Aftersome timeis attains a thernalequilibriumwith thesurroundingByconsideringthe airmoleculesin the roomas a thermodynamicsystemwhichof the followingis true |
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Answer» `DeltaU GT 0, Q=0` |
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| 13786. |
An engine of power 3 MW applies a force for 6 minutes on a train moving with a velocity of 10 ms^(-1). If there is no friction and the velocity attained is 25 ms^(-1), find the mass of the train. |
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| 13787. |
What is reversible and irreversible processes ? |
Answer» SOLUTION :A REVERSIBLE PROCESS is one which can be RETRACED in opposite direction so that it passes through exactly the same states as in the direct process.  and anirreverssible process is one which can notbe revesed that is it does not pass through the siilar states as in the direct process when retraced . |
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| 13788. |
When you singin the shower why is the sound heard |
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| 13789. |
If the length of the simple pendulum is equal to the radius of the earth, anf it oscillates just above the surface of the earth then its time period is a. 2pisqrt(R/(2g)) b. 2pisqrt(R/g) c. nearly 59.5 minutes d. 84.6 minutes |
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Answer» a is only CORRECT |
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| 13790. |
Can an object accelerate if its speed is constant ? |
| Answer» Solution :Yes, an object MOVING along a curved path with constant SPEED has verying velocity because its DIRECTION of velocity CHANGES from point to point along the trajectory. | |
| 13791. |
Two balls with masses m_(1) = 3kg and m_2 = 5kg have intial velocities V_1 = 5ms^(-1)towards east and V_2 = 5ms^(-1)towards North. They collide at the origin. The ratio of velocity of CM. 3s before collision to that 5s after collision is......... |
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Answer» `1:1` |
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| 13792. |
A liquid flows through two capillary tube under the same pressure head. The lengths of the tube are in the ratio 2:1 and the ratio of their diameters is 2:3. Compare the rates of flow of liquid through the tubes ? |
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| 13793. |
A Merr-go-raound, made of a ring - like platform of radius R and mass M, is revolving with angular speed omega. A person of mass M is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is |
| Answer» Answer :B | |
| 13794. |
Two particles of mass 1 kg and 3 kg have position vectors 2hati + 3hatj+4hatk and -2hati+ 3hatj - 4hatk respectively. The centre of mass, has a position vector |
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Answer» `hati+3hatj-2hatk` |
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| 13795. |
One mole of an ideal gas undergoes a process P= ( P_0)/(1+ ((V_0)/( V) )_(1)^(2) ) . Here, P_(0) and V_(0) are constants. Change in temperature of the gas when volume is changed from V=V_0 to V=2V_0 is : |
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Answer» `(2 P_(0) V_(0) )/( 5R)` |
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| 13796. |
If the coefficient of linear expansion of a solid is (1.5xx10^-5)^@C^-1, its coefficient of volume expension is nearly |
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Answer» `(3xx10^-5)^@C^-1` |
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| 13797. |
Fundamental frequency of sonometer wire is n. If the length, tension and diameter of wire are tripled the new fundamental frequency is |
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Answer» `(N)/SQRT3` |
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| 13798. |
Water flows out of a big tank along a horizontal tube of length l and radius r and bent at right angles at the other end. The rate of flow is Q. Calculate the moment of the force exerted by the water on the tube about the other end. Neglect viscosity of water. |
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| 13799. |
What is viscosity ? Explain its causes. |
| Answer» SOLUTION :Viscosity is DEFINED as 'the property of a FLUID to oppose the RELATIVE motion between its layers. | |
| 13800. |
A body of mass 2kg is moving along positive X-direction with a velocity of 5 ms^(-1). Now a force of 10sqrt(2) N is applied at an angle 45^(@) with X-axis. Its velocity after 3s is, |
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Answer» `20 MS^(-1)` |
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