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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A spherical ball of radius 3.0xx10^(-4) m and density 10^(4) kg//m^(3) falls freely under gravity through a distance h before entering a tank of water. If after entering the water the velocity of the ball does not change, find h. Viscosity of water is 9.8xx10^(6) N-s//m^(2). |
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Answer» `sqrt(2gh)=2/9 (R^(2)(rho-sigma)g)/(eta)` `:. h={(2)/(9) (r^(2)(rho-sigma)g)/(eta)}^(2)` `=(2)/(81) XX(r^(4)(rho-sigma)^(2)g)/(eta^(2))` `=(2)/(81)xx((3xx10^(-4))^(4)(10^(4)-10^(3))^(2)xx9.8)/(9.8xx10^(-6))^(2)` `=1.65xx10^(3)m`. |
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| 2. |
The momentum of two unequal bodies is same , then which onehas larger kinetic energy ? |
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Answer» <P> Solution :` :. ` The kinetic energy of LIGHT BODY is larger .` K = 1/2 mv^(2) = (p^(2))/(2m) ,p/2 ` is CONSTANT , `K prop 1/m ` |
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| 3. |
How much below the Earth's surface the value of g reduces to 30% of its value at the surface of earth? Radius of earth = 6,400 km. |
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Answer» |
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| 4. |
A wire of length 'll meters, made of a material of specific gravity 8 is floating horizontally on the surface of water. If it is not wet by water, the maximum diameter of the wire (in millimeters) upto which it can continue to float is (surface tension of water is T= 70 xx 10^(-3) "Nm"^(-1) ) |
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Answer» 1.5 |
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| 5. |
Derive the equation of continuity of steady flow of incompressible fluid. |
Answer» Solution :If in steady flow , we draw streamline of every particle ,then creation form like tube.This tube is known as tube of slow .No any particle goes OUTSIDE from this tube and no anyy other particle enter in this tube. In above figure tube of flow is shown and SHOWS points , P,R and Q on the tube. Suppose `v_(p),v_(Q)andv_(R)` is the velocity of fluid particle at point P,Q and R respectively and cross sectional areas are `A_(P),A_(Q)andA_(R)`. Since flow is steady so number of particles passing point P,Q and R are same., Mass of fluid at cross section area `A_(P)` with velocity `v_(P)` in `Deltat` time. `m_(P)=A_(P)v_(P)Deltat_(rho)`, where `rho=` density of fluid The mass of fluidpassing through Qand R are respectively , `m_(Q)=A_(Q)v_(Q)Deltatrhoandm_(R)=A_(R)v_(R)Deltatrho` now in `Deltat` time mass entering is same as leaving from flow of tube. `thereforem_(p)=m_(R)=m_(Q)` `thereforeA_(P)v_(P)Deltat_(rho)=A_(R)v_(R)Deltat_(rho)=A_(Q)v_(Q)Deltat_(rho)` `thereforeA_(P)v_(P)=A_(R)v_(R)=A_(Q)v_(Q)` is the equation of continuity for steady flow. It is a statement for conservation of mass in incompressible liquid. In general Av = constant `vprop(1)/(A)` Av gives the volume fluxor flow rate and remains constant throughout the pipe of flow . At narrower portions where the STREAMLINES are closely spaced , velocity increases and its vice versa . Widely spaced streamlines indicate regions of low SPEED and closely spaced streamlines indicates regions of high speed. |
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| 6. |
A particle of mass 0.2kg is moving with linear velocity (i-j+2k). If the radius vector T = 4i+j-k, the angular momentum of the particle is |
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Answer» `2.14` UNITS |
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| 7. |
Volume of a room is 15 m times 12 m times 8m. The room was at 22^@C in the morning.What is the percentage of initial volume of air of the room that is expelled when the room temperature reaches 30^@C at noon? The pressure remains constant during the change of temperature. |
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Answer» Solution :From Charle.s LAW , `V_1/T_1=V_2/T_2` or `V_1/V_2=T_1/T_2` or,`V_1/(V_2-V_1)=T_1/(T_2-T_1)or,(V_2-V_1)/V_1times100=(T_2-T_1)/T_1times100` Given, `T_1=22^@C=295 K,T_2=30^@C=303 K,V_2-V_1`=volume of EXPELLED air `THEREFORE(V_2-V_1)/V_1times100=(T_2-T_1)/T_1times100=(303-295)/295times100=2.7` `therefore 2.7%` of air will be expelled from the room. |
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| 8. |
Two particlesare projected horizontally from the same elevated point in opposite directions with velocities 4ms^(-1)and 9ms^(-1)respectively. At the moment when their velocity vectors vectors are mutually perpendicular, the separation between them is(g=10ms^(-2)) |
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Answer» 5 m at angle `TAN^(-1). 9/4` |
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| 9. |
Obtain the equatin of frequency observed by observer for moving source and moving observer at different velocities. |
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Answer» Solution :Let us take the observer to the source as the positive direction. Let the source and the observer be moving with velocities `v _(s) and v _(0)` respectively as shown in figure.  Suppose at time `t =0,` the observer is at `O_(1)` and the source is at `S_(1), O _(1)` being to the left of `S_(1).` The source EMITS a wave of velocity v of frequency v and period `T_(0)` all measureed by an observer at rest with respect to the medium. Let L be the distance between `O_(1) and S_(1)` at `t=0,` when the source emits the first crest. Now, since the observer is moving, the velocity of the wave relative to the observer is `v+v_(0).` Therefore, the first crest reaches the observer at time `t _(1) =(L)/(v + v _(0)).` At time `t = T_(0),` both the ovserver and the source have MOVED to their new positions `O_(2) and S_(2)` respecitvely. THe new DISTNACE between the observer and the source `O _(2) S _(2)` would be `L + (v _(s) -v _(0)) T _(0).` At `S_(2)` the source emits a second crest. THis reaches the observer at time . `t _(2) =T_(0) +([ L + (v _(s) - v _(0)) T _(0)])/((v + v _(0)))` At time `n T _(0)` the source emits its `(n +1) ^(th)` crest and this reaches the observer at time, `t _(n +1) = nT _(0) + (L + n (v _(s) -v _(0)) T _(0))/(v + v _(0))` Hence, in a time intervel, `t _(n +1) -t _(1) = n T _(0) + (L + n (v _(s - v _(0)) T _(0)))/(v +v_(0)) - (L)/(v + v _(0)) ,` the observer counts n crests. The observer records the period of the wave as equal to T given by, ` T = (t _(n)+1_ -t _(1))/(n)` `therefore = n T _(0) + (n (v _(s -v _(0)) T _(0))/(v + v _(0)))/(n )` ` therefore T = T _(0) + ((v _(s) -v _(0)) T _(0))/(v + v _(0))` `therefore T = T _(0) [1 + (v _(s) -v _(0))/(v + v _(0))]` `thereore T= T_(0) [1 + (v _(s) -v _(0))/(v + v _(0)) ]= T _(0) ((v + v _(s))/( v+ v _(0)))` The frequency v observed by the observer is given by, `V = 1/T` `therefore v =v _(0) ((v + v _(s))/( v + v _(0))) ^(-1)` ` therefore v =v _(0) ((v + v _(0))/(v +v _(s)))` This is general equation of DOPPLER effect. |
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| 10. |
Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in figure. (a) Find the work done when the gas is taken from state 1 to state 2. (b) What is the ratio of temperature T_1/T_2 if V_2=2V_1 ? (c) Given the internal energy for one mole of gas at temperature T is 3/2RT, find the heat 2 supplied to the gas when it is taken from state 1 to 2, with V_2 = 2V_1 |
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Answer» Solution :From graph, `PV^(1/2)`=constant K `therefore P=K/sqrtV` (a)WORK DONE for the process 1 to 2, `DeltaW=int_(V_1)^(V_2)P dV = int_(V_1)^(V_2)K/sqrtV xx dV` `DeltaW=K[V^(1/2)/(1/2)]_(V_1)^(V_2) =2K[V_2^(1/2)-V_1^(1/2)]`…(1) but `P_1V_1^(1/2)=K` and `P_2V_2^(1/2)=K` `therefore W=2P_1V_1^(1/2)[V_2^(1/2)-V_1^(1/2)]` ....(2) (b) From ideal gas equation of state , PV=nRT `therfore T=(PV)/(nR)` `therefore T=(PsqrtV.sqrtV)/(nR)` `therefore T=(KsqrtV)/(nR)[ because PsqrtV=K]` `therefore T_1=(KsqrtV_1)/(nR)` and `T_2=(KsqrtV_2)/(nR)` `therefore T_1/T_2=sqrt(V_1/V_2)=sqrt((V_1)/(2V_1))[ because V_2=2V_1]`....(3) `therefore T_1/T_2=1/sqrt2 therefore T_2=sqrt2T_1`...(4) (c) Internal energy of gas , `U=3/2 RT` `therefore DeltaU=U_2-U_1=3/2R(T_2-T_1)` `DeltaU=3/2R(sqrt2T_1-T_1)=3/2RT_1(sqrt2-1)`....(5) `rArr DeltaW=2P_1V_1^(1/2)[sqrtV_2- sqrtV_1]` (From equation 2) `=2P_1V_1^(1/2)[sqrt(2V_1)-sqrtV_1]` `=2P_1V_1[sqrt2-1]` `=2RT_1[sqrt2-1][ because P_1V_1=RT_1]` ....(6) From first rule of thermodynamics , `DeltaQ=DeltaU+DeltaW` `=3/2RT_1(sqrt2-1)+2RT_1(sqrt2-1)` `therefore DeltaQ=(7RT_1)/2 (sqrt2-1)` |
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| 11. |
A rocket is fired with a speed v = 2sqrt(gR) near the earth's surface and directed upwards. (a) Show that it will escape from the earth. (b) Show that in interstellar space it speed is v = sqrt(2gR). |
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Answer» Solution :(a) As PE of the rocket at the surface of the earth is (-GMm/R) and at `oo`, zero, so ENERGY required for escaping from earth `= 0-((GMm)/(R)) = mgR ["as g" = (GM)/(R^(2))]` And as initial KE of the rocket `(1)/(2)mv^(2) = 2mgR` isgreater than the energy required for escaping (=MG R), the rocket will escape. (b) If `v` is the VELOCITY of the rocket in interstellar space (free from gravitational effects) then by conservation of energy. `(1)/(2)mn(2sqrt(gR))^(2) - (1)/(2)m(sqrt(2gR)) = (1)/(2) m v^(2)` `v^(2) = 4gR - 2gR` or `v = sqrt(2gR)` |
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| 12. |
A motorcycle moving with a velocity of 72 km h^(-1) on a flat road takes a turn on the road at a point where the radius of curvature of the road is 20 m. The acceleration due to gravity is 10 ms^(-2). In order to avoid skidding, he must not bent with respect to the vertical plane by an angle greater than |
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Answer» `theta = tan^(-1) (2) ` ` theta= tan^(-1) ""( v^2)/(RG)` here` r= 20m , v = 72kmh^(-1)= 72 XX (5)/(18) = 20ms^(-1)` `thereforetheta = tan^(-1)(( 20 xx 20)/( 20 xx 10)) or theta= tan^(-1)(2)` |
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| 13. |
Two particles starting from a point on a circle of radius 4m in horizontal plane move along the respectively in opposite directions. The particle will collide with each other after a time of |
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Answer» `3.0s` |
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| 14. |
A rod of mass m and length / is made to stand at an angle of 60^@ with the vertical. Potential energy of the rod in this position is |
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Answer» `MGL` Height of the mass from ground is `h = (l//2) sin 30^@` POTENTIAL energy of the rod `= MGH xx m xx g xx l/2 sin 30^@` `= m xx g xx l/2 xx 1/2 = (mgl)/(4)`. 
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| 15. |
The external and internal diameters of a hollow cylinder are determined with vernier callipers and the results are recorded as 4.23pm 0.01 cm and 3.89pm 0.01 cm. The thickness of the cylinder wall within the limits of error is |
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Answer» `0.34 PM 0.01cm` |
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| 16. |
Twenty identical cubical blocks each of mass 100 gm and of side 10 cm are lying on a horizontal surface. Work done in piling them one above the other is, (g= 10 ms^(-2)) |
| Answer» ANSWER :C | |
| 17. |
A capillary tube is attached horizontally to a constant head arrangement. If the radius capillary tube is increased by 10% then the rate of flow of liquid will change nearly by |
| Answer» ANSWER :B | |
| 18. |
The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it? (a) The forces may be acting radially from a point on the axis b) The forces may be acting on the axis of rotation (c) The force may be acting parallel to the axis of rotation d) The torque caused by some forces may be equal and opposite to that caused by other forces. |
| Answer» Answer :D | |
| 19. |
(A) : Graph between potential energy of a spring versus the extension or compression of the spring is a parabola.(R ) : Potential energy of a stretched or compressed spring is proportional to square of extension or comession. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 20. |
What is the specific heat of a gas during an isothermal process? |
| Answer» Solution :`C=(DELTAQ)/(mDeltaT)`. For an ISOTHERMAL PROCESS, `DeltaT=0` So C is INFINITY. | |
| 21. |
A small sphere of radius 'r' falls from rest in a viscous liquid Asa result heat is produced due to viscous force .the rate of production of heat when the sphere attains its terminal velocity is proportional to : |
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Answer» `r^(1)` where `F_(v)=` viscous FORCE `v_(tau)`= TERMINAL velocity `=6pietarv_(tau)xxv_(tau)` `proprv_(T)^(2)[because6pieta` is constant] But `V_(T)=(2)/(9)(r^(2)g)/(eta)(rho-sigma)` `thereforeV_(T)propr^(2) "" [because` all other terms are constant) `therefore(dQ)/(dt)propr(r^(2))^(2)` `propr^(5)` |
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| 22. |
When water at zero degree celsius is gradually heated, the volume of water |
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Answer» Remains constant |
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| 23. |
Conservation of momentum in a collision between particles can be understood from . |
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Answer» conservation of energy |
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| 24. |
A motor of power 2.45 kW draws water from a well of depth 20m. If it fills a tank of capacity 3000 lit in 7 minutes. The height of the tank above ground is |
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Answer» 5m |
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| 25. |
Draw a graph of force of friction v/s applied force. |
Answer» Solution : Note : We note that `F_(t)` INCREASES with the increase of applied FORCE upto the MAXIMUM force of friction. Hence STATIC friction is a self adjusting force. |
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| 26. |
In a horizontal pipe of non-uniform cross section, water flows with a velocity of 1 ms^(-1) at a point where the diameter of the pipe is 20 cm. The velocity of water (1.5 ms^(-1)) at a point where the diameter of the pipe is (in cm) |
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Answer» 8 `a_1v_1 = a_2v_2 ""v_1 = 1 ms^(-1)` `PI r_1^2 v_1 = pi r_2^2 v_2 "" v_2 = 1.5 ms^(-1)` `(10 xx 10^(-2))^(2) xx 1 = r_2^2 xx 1.5 "" d_1 = 20 CM` `r_2^2 = (100 xx 10^(-4))/(1.5) "" r_1 = (d_1)/2 =10 cm` `:. r^2 = 8CM` `:. d_2 = 16 cm`. |
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| 27. |
A bomb at rest explodes. The total momentum of all its fragments is |
| Answer» Solution :zero | |
| 28. |
What is .Chandrasekhar limit. ? |
| Answer» SOLUTION :The mass of the STAR equal to 1.4 times the SOLAR mass is CALLED chandrasekhar limit.Chandrasekhar limit indicates the mass of the star for which the DEGENERATE electron pressure will not be sufficient to prevent are collapse of the star. | |
| 29. |
(A): At room temperature water does not sublimate from ice to steam. (R ): The critical points of water is much above the room temperature.(1994) |
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Answer» Both (A) and(R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 30. |
Express incident average solar energy of5xx10^(24) J on earth in eV . |
| Answer» SOLUTION :`3.125 XX1^(43) EV` | |
| 31. |
The work done on an object does not depend upon the |
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Answer» displacement |
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| 32. |
An engine develops 10 kW of power. How much time will it take to lift a mass of 200 kg to a height of 40m. (g=10ms^(-2)) |
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Answer» 4s |
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| 33. |
An aluminium rod has a breaking strain 0.2%. The minimum cross-sectional area of the rod in m^2in order to support a load of 10^4N is if Youngs modulus is 7 xx 10^9 Nm^(-2) |
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Answer» `1.7xx10^(-4)` |
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| 34. |
In a region, magnetic field along X-axis changes with time according to the given garph. A charged particle of mass 1 kg and charge 1C located at origin starts moving with initial velocity ((2)/(pi)hati+2sqrt3hatj)m//s. Completes one revolution in T_(0) time. Find the displacement (in m) of particle at time (5T_(0))/(4). |
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Answer» |
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| 35. |
In a nuclear reactor a neutron of high speed("typically" 10^(7) ms^(-1)) must be slowed to 10^(3)ms^(-1) so that it canhave ahigh probality of interacting with isotope ""_(92)""^(235)Uand causing it to fission. Show that a neutron can lose most of its kinetic energy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutron mass. |
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Answer» Solution :The initial kinetic ENERGY of the NEUTRON is `K_(1i)=(1)/(2)m_(1)v_(1i)^(2)rArrK_(1f)(1)/(2)m_(1)v_(1f)^(2)=(1)/(2)m_(1)((m_(1)-m_(2))/(m_(1) + m_(2)))^(2)v_(1i)^(2)` The fractional kinetic energy lost is `f_(1)=(K_(1f))/(K_(1i))=((m_(1)-m_(2))/(m_(1)+m_(2)))` While the fractional kinetic energy gained by the moderating nuclei `K_(2f)//K_(1i)` is `f_(2) = 1 - f_(1)` (elastic collision) = `(4m_(1)m_(2))/((m_(1) + m_(2))^(2))` For deutrium `m_(2) = 2m_(1)` and we obtain `f_(1)=1//9=11% "while"f_(1)=8//9=89%.` Almost90% of the neutrons energy is transfered to the deuterium. For CARBON `f_(1) = 71.6% and f_(2) = 28.4 %` In practice, however, this number is smaller since head - on collisions are rare. |
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| 36. |
A satellite of mass m is orbiting the earth at a height h above its surface. The mass of the earth is M and its radius R. Match the physical quantities in column-I with the expression in column-II {:("Column-I",,"Column-II"),( (A)(GMm)/(2(R+H)),,"(P) Potential energy of hte satellite"),((B) sqrt((GM)/(R+H)),,"(Q) Kinetic energy of the satellite"),((C ) (-GMm)/(R+H),,"(R) Orbital velocity of the satellite"),((D) sqrt((2GM)/(R +h)),,"(S) Escape velocity"):} |
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Answer» |
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| 37. |
A balloonwith its contents weighing 160 N is moving down with an acceleration of g//2 ms^(-2). The mass to be removed from it so that the balloon moves up with an acceleration of g//3 ms^(-2) is |
| Answer» Answer :B | |
| 38. |
In the system shown in Figure 6.13 P and Q are in equilibrium . If P is doubled, calculate the acceleration of P. The pulley is light. |
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Answer» |
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| 39. |
A beaker containing a liquid is kept inside a big closed jar. If the air inside the jar is continuously pumped out, the pressure in the liquid near the bottom of the liquid will |
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Answer» Increase |
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| 40. |
An engine develops 33250 W power, consuming 10 kg fule per hour. If the clorific value of the fuel is 11450 cal g^(-1), find the efficiency of the engine. |
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Answer» |
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| 41. |
The initial pressure and volume of a gas are P_i and V_i . The gas after expansion attains final volume V_f. Let W_1, W_2 and W_3are the corresponding work done by the gas, under isothermal, adiabatic and isobaric processes, respectively. Then, |
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Answer» `W_1 = W_2 = W_3` |
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| 42. |
One mole of an ideal gas initially kept in a cylinder at pressure 1 MPa and temperature 27^(@)C is made to expand until its volume is doubled. (a) How much work is done if the expansion is (i) adiabatic (ii) isobaric (iii) isothermal? (b) Identify the processes in which workdone is least and is maximum. (c) Show each process on a PV diagram. (d) Name the processes in which the heat transfer is maximum and minimum. ("Take "gamma = 5//3 and R = 8.3 J mol^(-1) K^(-1)) |
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Answer» Solution :(a) (i) In an adiabatic process the work done by the system is `W_("adia")=(muR)/(gamma-1)[T_(i)-T_(f)]` To find the temperature `T_(f)`, we can use adiabatic equation of state. `T_(f)V_(f)^(gamma-1)=T_(i)V_(i)^(gamma-1)` `T_(f)=T_(i)((V_(i))/(V_(f)))^(gamma-1)=300xx((1)/(2))^((2)/(3))=0.63xx300K=189.8K` `W=1xx8..3xx(3)/(2)(300-189.8)=1.37kJ` (ii) In an isobaric process the work done by the system `W=PDeltaV=P(V_(f)-V_(i)) and V_(f) =2V_(i)` so `W=2PV_(i)` To find `V_(i)`, we can use the ideal gas for initial state, `P_(i)V_(i)=RT_(i)` `V_(i)=(RT_(i))/(P_(i))=8.3xx(300)/(1)xx10^(-6)=24.9xx10^(-4)m^(3)` The work doen during isobaric process, `W=2xx10^(6)xx24.9xx10^(-4)=4.9kJ` (iii) In an isothermal process the work doen by dthe system, `W=muRT LN ((V_(f))/(V_(i)))` In an isothermal process the initial room temperature is CONSTANT. `W=1xx8.3xx300xxln (2)=1.7 kJ` (b) Comparing all three processes, we see that the work done in the isobaric process is the greatest, and work done in the adiabatic process is the least least (C) The PV diagram is SHOWN in the Figure. The area under the curve AB=Work done during the isobaric process The area under the curve AC=Work done during the isothermal process The area under the curve AD= Work done during the adiabatic process From the PV diagram the area under the curve AB is more, implying that the work done in isobaric process is highest and work done in adiabatic process is least. (d) In an adiabatic process no HEAT enters into the system or leaves from the system. In an isobaric process the work done is more so heat supplied should be more compared to an isothermal process. 
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| 43. |
Statement I : The sound gets amplified when a vibrating tuning fork is made to touch the surface of a table. Statement II : Dimension of the table is more than that of the tuning fork. |
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Answer» Statement I is TRUE, statement II is true , statement II is a correct EXPLANATION for statement I. |
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| 44. |
A horizontal force F is applied to applied to a homogeneous rectangular block of massm, width b and height H. The block moves with constant velocity and the coefficient of friction is mu_(k) What is the greatest height h at which the foce F can be applied so that the block will slide without toppling ? |
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Answer» `(B)/(mu_(k))` |
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| 45. |
A horizontal force F is applied to applied to a homogeneous rectangular block of massm, width b and height H. The block moves with constant velocity and the coefficient of friction is mu_(k) |
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Answer» `(B)/(2mu_(s))` |
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| 46. |
A horizontal force F is applied to applied to a homogeneous rectangular block of massm, width b and height H. The block moves with constant velocity and the coefficient of friction is mu_(k) At what distance (from the left end) on the bottom face of the block will the normal reaction act when F acts at the height (H)/(2) ? (here, assume no toppling happens) |
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Answer» `(mu_(K)H)/(2)` |
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| 47. |
A brass wire 1.8 m long at 27^(@)C is held taut with little tension between two rigid supports. If the wire cooled to a temperature of -39^(@)C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass = 2.0 xx 10^(-5)//^(@)C, Young's modulus of brass = 0.91 xx 10^(11) Pa. |
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Answer» Solution :Here`I=1.8m,t_(1)=27^(@)C,t_(2)=-39^(@)C` `r=(2.0)/(2)=1.0 mm =1.0xx10^(-3)m` `r=2.0xx10^(-5)""^(@)C^(-1),Y=0.91xx10^(11)Pa` As `"" DeltaI=Ialpha(t_(2)-t_(1))` `:.` Strain `(DeltaI)/(I)=alpha(t_(2)-t_(1))` `"Stress" = "Strain" xx "Young’s modulus" =alpha(t_(2)-t_(1))xx`Y `=2.0xx10^(-5)xx(-39-27)xx0.91xx10^(11)=1.2xx10^(8)Nm^(-2)` [Numerically] Tension developed in the wire = Stress xx AREA of cross-section `= Strees xxPr^(2)=1.2xx10^(8)xx3.14xx(1.0xx10^(-3))^(2)=3.77xx10^(2)N`. |
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| 48. |
Containers A and B contain a liquid up to same height. They are connected by a tube (see figure).(a) If the liquid in container A is heated, in which direction will the liquid flow through the tube. (b) If the liquid in the container B is heated in which direction will the liquid flow through the tube?Assume that the containers do not expand on heating. |
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Answer» |
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| 49. |
A train starting from rest is accelerated and the instantaneous acceleration is given by (10)/(v+1)m//s^(2) . Where v is the velocity in m/s. Find the distance in which the train attains a velocity of 54 km/hr. |
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Answer» 10.75 sec |
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| 50. |
A train starting from rest is accelerated and the instantaneous acceleration is given by (10)/(v+1)m//s^(2). Where v is the velocity in m/s. Find the distance in which the train attains a velocity of 54 km/hr. |
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Answer» 123.75 m |
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