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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. |
Derive the expression for Carnot engine efficiency. |
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Answer» Solution :Efficiency of a Carnotenginr: Efficiency is DEFINED as the ratio of WORK done by the working substance in one cycle to the amount of heat extrcted from the source. `eta=("work done" )/("heat extracted")=(W)/(Q_(H))` From the FIRST law of thermodynamics, `W=Q_(h)-Q_(L)` `eta=(Q_(H)-Q_(L))/(Q_(H))=(Q_(L))/(Q_(H))` Applying isothermal conditions, we GET `Q_(H)=muRT_(H) In((V_(2))/(V_(1)))` `Q_(L)=muRT_(L) In((V_(3))/(V_(4)))` `therefore (T_(L))/(T_(H))=(Q_(L) In(V_(3))/(V_(4)))/(Q_(H)In(V_(2))/(V_(1)))` By applying adiabatic conditions we get , `T_(H)V_(2)^(y-1) = T_(L) V_(3)^(y-1)` `T_(H)V_(1)^(y-1) = T_(L) V_(4)^(y-1)` By dividing the above two equations we get, `(V_(2)/(V_(1)))^(y-1)=(V_(3)/(V_(4)))^(y-1)` Subsitituting EQUATION (5) in (4), we get `(Q_(L))/(Q_(H)) =(T_(L))/(T_(H))` `therefore` The effiency `mu=1-(T_(L))/(T_(H))` |
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| 2. |
If a bulbof 60 Wis keptON for200 hours then ……… unit of electrical energy has tobe spend . |
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Answer» `1.2` |
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| 3. |
Using constraint equations find the relation between a_1 and a_2 |
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Answer» Solution :Point 1,2,3 and 4 are movable. LET their dis-placement from a fixed line be ` x_1 , x_2 , x_3 and x_4` `x_1 + x_3 =l_1` `(x_1 - x_3 )+ ( x_4- x_3) =l_2` ` (x_1 - x_4 ) + (x_2 - x_4) = l_3` On double differentiating with respect to TIME , we will get following three constraint RELATIONS : ` a_1 + a_3 =0` ....... (i) `a_1 + a_4 - 2a_3 =0`........ (ii) `a_1 + a_2 - 2a_4 =0`............ (III) SolvingEqs. (i) ,(ii) and (iii) , we get ` a_2=-7a_1 ` Which is the desired relation between ` a_1 and a_2` 
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| 4. |
A block is kept on a rough horizontal plank. The coefficient of friction between the block and the plank is 1//2. The plank is undergoing SHM of angular frequency 10 rad//s. Find the maximum umplitude of plank in which the block does not slip over the plank (g = 10 m//s^(2)). |
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Answer» SOLUTION :Maximum acceleration in SHM is `a_("max")= omega^(2)A` this will be provided to the block by friction. HENCE, `a_("max")= mug` or `omega^(2)A=mug` or `a= (mug)/(omega^(2))= (((1)/(2))(10))/((10)^(2))= 0.05 m= 5CM` |
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| 5. |
A particle is moving along a circle and its linear velocity is 0.5ms when its angular velocity is 2.5rads^(-1). Its tangential acceleration is 0.2ms^(-2) then its angular acceleration is |
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Answer» `1rads^(-2)` |
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| 6. |
How do you deduce that two vectors are perpendicular? |
| Answer» Solution :If TWO vectors `vecA` and `vecB` are perpendicular to each other than their scalar product `vecA.vecB=0` because `cos90^(@)=0`. Then the vectors `VECAANDVECB` are said to be mutually ORTHOGONAL. | |
| 7. |
Find the temperature of an oven if it radiates 8.28 cal per second through an opening, whose area is6.1cm^(2). Assume that the radiation is close to that of a black body. |
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Answer» SOLUTION :The emittance of the oven (the ENERGY radiated in ONE second by unit SURFACE area) is `E=(8.28cal//s)/(6.1cm^(2))` = `(8.28xx4.2J//s)/(6.1xx10^(-4)m^(2))=5.7xx10^(4)"watt"//m^(2)` From stefan - BOLTZMANN formula `E=sigma-T^(4)`, where `sigma=5.67xx10^(-8)W//m^(2).K^(4)` `T^(4)=E/sigma=(5.7xx10^(4))/(5.67xx10^(-8))=1xx10^(12)K^(4)` `thereforeT=10^(3)K=1000K`. |
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| 8. |
If C_(P) and C_(V) denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively then |
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Answer» `C_(P) - C_(V) = R` |
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| 9. |
A ball of radius r fits tightly inside a tube attached to a container. There is no friction between the tube wall and the ball. Volume of air inside the container is V_(0) when the ball is in equilibrium. Density of the material of the ball is d and atmospheric pressure is P_(0). If the ball is displaced a little from its equilibrium position and released, find time period of its oscillation. Assume that temperature in the container remains constant and that air is an ideal gas. |
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Answer» |
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| 10. |
The internal radius of a 1m long resonance tube is measured as 3.0 cm. A tuning fork of frequency 2000 Hz is used. The first resonating length is measured as 4.5 cm and the second resonating length is measured as 14.0 cm. We shall calculate the following. |
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Answer» SOLUTION :(i) Maximum percentage error in measurement of as, given by Reyleigh'sformula (Given error in measurement of radius is 0.1 m) `Deltae= 0.6 Delta R = 0.6 xx 0.1 = 0.06 cm` Percentage error is `(Delta e)/(e) xx 100 = (0.06)/(0.6 xx 3) xx 100 = 3.33 %` (II) Speed of sound at the room temperature `l_(1)=4.6m,l_(2)=14.0cm,lamda = 2(l_(2)-l_(1))(14.0-4.6)=18.8cm,V =flamda = 2000 xx (18.8)/(100) = 376 m//s` (ii) End correction obtained in the experiment, `e = (l_(2)-3l_(1))/(2)=(14.0-3xx4.6)/(2)=0.1cm` (IV) Percentage error in the calculation of a with respect to theoretical valuess Percentage error `= (0.6xx3-0.1)/(0.6xx3)xx100=94.44%` |
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| 11. |
Streamline flow is more likely for liquids with |
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Answer» HIGH density |
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| 12. |
A body of mass 1kg is suspended from a weightless spring having force constant 600 N/m Another ody of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of 3m/s and gets embedded in it. Find amplitude of oscillation. |
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Answer» Solution :By CONSERVATION of linear momentum in the collision `mv=(m+M)vimpliesV=(mv)/(m+M)=(0.5xx3)/((1+0.5))=1m//s` Now just after collisions the SYSTEM will have `KE=1/2(m+M)V^(2)` at equilibrium position. So after collision by conservation of mechanical ENERGY `KE_("MAX")=PE_("max")` `1/2(m+M)V^(2)=1/2KA^(2)impliesA=Vsqrt(((m+M)/K))=1sqrt(1.5/600)=1/20m=5cm` |
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| 13. |
A stone of mass 2kg is tied to one end of a string of length 0.5m. It is whirled in a vertical circle. If the velocity of stone at the top of circle is 5ms^(-1), the tension in the string at the bottom of circle is |
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Answer» 125N |
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| 14. |
A lift is going up. The total mass of the lift and the passenger is 1500 kg. The variation in the speed of the lift is as shown in the figure. Find the tension in the rope pulling the lift at t = 11 th second. |
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Answer» Solution :Acceleration of LIFT at 11TH second = slope of the line CD `a_(11)=(0-3.6)/(12.10)` `=-1.8 m//s^(2)` `T=m(g+a)=1500(9.8-1.8)` `= 1500xx8=12000 N` 
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| 15. |
Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken. |
Answer» SOLUTION : Let `P,Q` be the POSITIONS at TIME .t. `barL` of .P. about .O. is `barL_(P)=(xbari+ybarj)XX(-mvbari)` `barL_(P) =mvybark` , `barL` of .Q. about O is `barL_(Q)=[xbari+(y+d)barj)]xxmvbari=mv(y+d)(bar-k)` `sumbarL=bar_(P)+barL_(Q)=mvd(bar-k)`. |
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| 16. |
A hemispherical bowl of radius R is set rotating about its axis of symmetry which is kept vertical. A small block kept in the bowl rotates with the bowl without slipping on its surface. If the surface of bowl is smooth, and the angle made by the radius through the block with the vertical is q, the angular speed at which the bowl is rotating is |
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Answer» `SQRT((g)/(R cos theta))` |
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| 17. |
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s^(-1). The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis? |
| Answer» Solution :KINETIC ENERGY = 3125 J, ANGULAR Momentum = 62.5 J s | |
| 18. |
A mass m moves with a speed v on a horizontal smooth surface and collides with a nearly massless spring whose spring constant is k. If the mass stops after collision, compute the maximum compression of the spring. |
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Answer» Solution :When the MASS collides with the spring, from the law of conservation of energy ..the loss in kinetic energy of mass is gain in elastic potential energy by spring... Let x be the DISTANCE of COMPRESSION of spring, then the law of conservation of energy `(1)/(2)mv^(2)=(1)/(2)KX^(2)impliesx=vsqrt((m)/(k))` |
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| 19. |
Water from a tap emerges vertically downwards with initial velocity 4ms^(-1). The cross sectional area of the tap is A. The flow is steady and pressure is constant throughout the stream of water. The distance h vertically below the tap, where the cross-sectional area of the stream becomes (2/3) A. is (g=10m//s^(2)): |
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Answer» `0.5m` |
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| 20. |
A force applied by the engine of a train of mass 2.05xx10^(6) kg changes its velocity from 5ms^(-1) to 25 ms^(-1) in 5 minutes. The power of the engine. |
| Answer» ANSWER :B | |
| 21. |
For solids with elastic modulus of rigidity , the shearing force is proportional to ……. |
| Answer» SOLUTION :SHEAR STRAIN , RATE of shearstrain , | |
| 22. |
A 50kg girl wearing high heel shoes balances on a single heel.the heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor? |
| Answer» SOLUTION :`6.2 TIMES 10^6 PA` | |
| 23. |
The density of a material in C.G.S. system is 8g/cm^3 . In a system of units in which unit of length is 5 cm and unit of mass is 20g. What is the density of the material? |
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Answer» |
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| 24. |
A wheel rotates with an angular acceleration given by alpha= 4at^(2)-3bt^(2), where t is the time and a and bare constants. If the wheel has an initial angular speed w_(0), write the equations for (a) the angular speed and (b) the angle tumed through as functions of time. |
| Answer» Solution :(a)`at^(4)-bt^(3)+omega_(0)`, (B)`a(t^(5))/(5)-b(t^(4))/(4)+omega_(0)t` | |
| 25. |
Solve with due regard to significant figures. 46.7-10.4(3.0xx10^(-8))+(4.5xx10^(-6))= |
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Answer» Solution :`46.7-10.04` Here `46.7`has one decimal PLACE , and `10.04`has TWO decimal places. `:. 46.7-10.04=36.66`. The RESULT should have only one decimal place . The result is `36.7`. `3.0xx10^(-8)+4.5xx10^(-6)=0.03xx10^(-6)=0.003xx10^(-6)+4.5xx10^(-6)=4.53xx10^(-6)` here `4.5xx10^(-6)`has only one decimal place and `0.03xx10(-6)`can have two decimal places. this result should be rounded off to one decimal place . `:.(3.0xx10^(-8))+(4.5xx10^(-6))=4.5xx10^(-6)` |
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| 26. |
The volume of the air bubble increases by 10%, when it rises from bottom to the surface of the water pond. If the temperature of the water is constant, find the depth of the pond. ( 1 atm = 10 m of water column) |
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Answer» SOLUTION :Volume of the air bubble at the bottom of the POND `(V_(1) ) = V` Volume of the air bubble at the surface of the pond`V_(2) = V xx (110)/(100)` Pressure at the bottom of the pond `(P_(1) )` = (H+ H) cm of water column Where .h. is the DEPTH of the pond and .H. is the atmospheric pressure pressure on surface of the pond `(P_(2)) = H ` cm of water column From Boyle.s law, `P_(1) V_(1) = P_(2) V_(2) ` or `(h + H) V = H (V xx (110)/(100) ) , "" h + H = H xx (110)/(100)` h = `H xx (110)/(100) - H = H xx (10)/(100) = (H)/(10)` `therefore` Depth of the pond = H/10 cm `= (1000)/(10) = 100 ` cm = 1m |
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| 27. |
The amplitude of a damped oscillator decreases to 0.9 times its original value in 5s. In another 10s it will decreases to alpha times its original magnitude, where alpha is |
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Answer» `0.81` |
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| 28. |
For which of the following doesthe centre of mass lie outside the body ? |
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Answer» A pencil |
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| 29. |
A cricket ball is thrown at a speed ofms^(-1) in a direction 30^(@) above the horizontal . Calculate (a) the maximum height , (b) the time taken by the ball to return to the same thrower to the point where the ball returns to the same level. |
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Answer» Solution :The maximum HEIGHT is GIVEN by `h_(m)=((v_(0)sintheta_(@))^(2))/(2g)=((28sin30^(@))^(2))/(2(9.8))m` `=(14xx14)/(2xx9.8)=10.0m` (b)The time taken to return to the same level is `T_(f)=(2v_(@)sintheta_(@))//G=(2xx28xxsin30^(@))//9.8=28//9.8s=2.9s` (c ) The distance from the thrower to the point where the ball returns to the same level is `R=((v_(@)^(2)sin2theta_(@)))/(g)=(28xx28xxsin60^(@))/(9.8)=69m` |
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| 30. |
Match Column-I with Column-II : {:("Column-I","Column-II"),("(1) Conservative force","(a) Friction force"),("(2) Non-conservative force","(b) Gravitational force"),(,"(c) Internal force"):} |
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Answer» |
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| 31. |
A helicopter of mass 1000 kg rises with vertical acceleration of 15 ms^(-2). The crew and the passengers weight 300 kg. Give the magnitude and direction of the(a) force on the floor by the crew and passengers(b) action of the rotor of the helicopter on the surrounding air (c ) force on the helicopter due to the surrounding air |
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Answer» Solution :Here, mass of the helicopter, `m_(1)=1000 kg` Mass of the crew and passengers `m_(2)=300 kg` Upward acceleration `a = 15 ms^(-2)` and `g=10 ms^(-2)` (a) Force on the floor of helicopter by the crew and passenger = appeared weight of crew and passengers `= m_(2)(g+a)` `=300 (10+15)=7500 N` (b) Action of rotor of helicopter on SURROUNDING air is obviously vertically downwards because helicopter rises on account reaction to this force. Thus, force of action `F=(m_(1)+m_(2))(g+a)` `= (1000+300)(10+15)` `= 1300xx25=32500 N` (C ) Force on the helicopter due to surounding air is the reaction. As action and reaction are equal and OPPOSITE, therefore, force of reaction `F^(1)=32500 N`, vertically upwards. |
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| 32. |
One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). The ratio of molar volume to the atomic volume of a mole of hydrogen is (Take the size of hydrogen molecule to be about 1 Å) |
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Answer» `10^4` |
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| 33. |
The velocity of a projectile when at its greatest height is sqrt((2)/(5)) of its velocity when at half of its greatest height find the angle of projection |
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Answer» Solution :Step: we know that, VELOCITY of a projectile at half of maximum HEIGHT =`usqrt((1+cos^(2)theta)/(2))` Step 2 : given that `u cos theta=sqrt((2)/(5))xxusqrt((1+cos^(2)theta)/(2))` SQUARING on both SIDES `u^(2)cos^(2)theta = (2)/(5)u^(2)((1+cos^(2)theta)/(2))` `10costheta = 2 + cos theta` `RARR 8cos^(2)theta = 2 rArr cos^(2)theta=(1)/(4)rArr theta=60^(@)` |
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| 34. |
Which of the followinf units denotes the dimensions of ML^(2)//Q^(2), where Q denotes the electric charge, M denotes mass, L denotes length |
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Answer» Weber (Wb) |
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| 35. |
There are two pendulums with bobs having indencital size and mass. The pendulum A is released from rest in the position as shown in the figure. If the maximum angle formed by cord BO' with vertical in the subsequent motion of sphere B is equal to the angle theta_(0) If the coefficient of restitution between sphere A and sphere B is l. find a. the velocities of sphere A and sphere B just after collisions b. the ratio of lengths of pendulums l_B//l_A. |
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Answer» Solution :Velocity of `A` just before collision `v_(0)=sqrt(2gl_(A)(1-costheta_(A))`………..i Collision of `A` and `B` Using C.O.L.M. , `m.v_(0)+0=mv_(A)+mv_(B)` `v_(A)+v_(B)=v_(0)`……ii Using NEWTON's restitution law `v_(B)-v_(A)=v_(0)` `v_(B)-v_(A)=ev_(0)`………ii Solving EQN i and ii `v_(A)=(v_(0))/2(1-e)` and `v_(B)=(v_(0))/2(1+e)` As `B` swings angle `thetaB`, hence `v_(B)=sqrt(2gh_(B)(1-costheta_(B)))` ............iii `and (v_(0))/2(1+e)=sqrt(2gl_(B)(1-costheta_(B)))` `((2gl_(A)(1-costheta_(A)))/2(1+e) sqrt(2gl_(B)(1-costheta_(B)))` `implies (l_(B))/(l_(A))=((1+e)/2)^(2)` |
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| 36. |
A solid sphere is at a temperature T.K. The sphere is cut into two halves. The ratios of Intensities emitted by the half sphere to that by complete sphere is |
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Answer» `1:2` |
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| 37. |
If vec(P) and vec(q) are two unit vectors and the angle between them is 60^(@) then ((1+vec(p).vec(q))/(1-vec(p).vec(q))) is |
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Answer» 2 |
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| 38. |
Why canals are constructed in 'V' shape? Explain. |
| Answer» SOLUTION :The water in canals flow in the form of layers. When the canals are constructed in "V" shape, the AREA of contact between the layers decreases from TOP to bottom of the canal. The viscous force is directly proportional to the area of contact between the layers. Because of this REASON the viscous force between the layers decreases and the water FLOWS freely along the canal. | |
| 39. |
Two identical cyclindrical vessels (area of cross section 3.5 xx 10^(-3) m^(2)) with their bases at the same level contain liquid of density 800 kg//m^(3). The height of liquid in one vessel is 0.3 m and in the other it is 0.1 m. Assuming g = 10 m//s^(2),find the workdone by gravity in equalising levels when the vessels are interconnected at bottom. |
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| 40. |
Give the vector form of the relation between L and p |
| Answer» SOLUTION :VECTOR relation between angular MOMENTUM and LINEAR momentum `BARP` is `barL` =`barrxxbarp` | |
| 41. |
The compressibility of water is 4 xx 10 ^(-5)per unit atmospheric pressure. The decrease in volume of 100 cm^(3) of water under a pressure of 100 atmosphere will be ......... |
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Answer» `0.4 CM ^(3)` `K = 1/B` `therefore K = (1)/(- (Delta P )/( Delta V //V )) = - (Delta V//V)/( Delta P ) =- (Delta V )/( V Delta P )` `therefore Delta V =- KV Delta P` `=-4 xx 10 ^(-5) (m ^(2))/( N ) xx 100 cm ^(3) xx 100 (N)/(m ^(2))=-0.4 cm ^(3)` `therefore` Decrease in volume `=0.4 cm ^(3)` |
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| 42. |
A simple pendulum has a bob of mass m and it is oscillating with a small angular ampitude of theta_(0). Calculate the average tension in the stringaveraged over one time period. [For small theta take cos theta =1-(theta^(2))/(2)] |
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Answer» |
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| 43. |
A small bead of mass M is launched from the end of a horizontal rod at a speed u along rod having same mass M. The rod is free to rotate in a vertical plane about a end. Simultaneously the rod is given a angular velocity omega as shown in the figure. Find the angular acceleration of the rod. [The rod is hinged at an end] |
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Answer» `(9g)/(8L)` |
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| 44. |
Consider the equations P=Lim_(Deltasto0)F/(DeltaS) and P_(1)-P_(2)=rhogz. In an elector acceleration upward |
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Answer» Both the EQUATIONS are VALID |
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| 45. |
If a body is projected with an angle theta to the horizontal, then |
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Answer» its velocity is ALWAYS PERPENDICULAR to its acceleration. |
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| 46. |
A particle moves along the X-axis as x=u(t-2)+a(t-2)^(2) a) the initial velocity of the particle is u b) the acceleration of the particle is 2a c) at t=2s particle is at the origin |
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Answer» a & b are correct |
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| 47. |
About which axis, the moment of inertia of a body is minimum? |
| Answer» Solution :The MOMENT of INERTIA of a body is MINIMUM about an AXIS passsing through its centre of mass. | |
| 48. |
What does speedometer of vehicle measure ? Instantaneous velocity or instantaneous speed ? |
| Answer» SOLUTION :INSTANTANEOUS SPEED. | |
| 49. |
Let the speed of the planet at the perihelion P be V_(P) and the sun-planet distance SP be r_(p). Relates (r_(p), V_(P)) to the corresponding quantities at the aphelion (r_(A), V_(A)). Will the planet take equal times to tranverse BAC and CPB ? |
Answer» Solution : According to law of conservation of angular momentum. Angular of the planet at P = Angular momentum of the planet at A `RARR mV_(P)r_(P) = mV_(A)r_(A)` (or) `(V_(P))/(V_(A)) = (r_(A))/(r_(P))` Since `r_(A) GT r_(P)` so `V_(P) gt V_(A)` Here area SBAC is greater than the area SCPB. According to Kepler.s second law, as the arcal velocity of the planet is contant AROUND the sun , i.e. equal areas are swept on equal times, hence the planet will take longer TIME to traverse BAC than CPB. |
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