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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. |
Calculate rate of flow of glycerin of density 1.25xx10^(3) kg//m^(3)? through the conical section of a horizontal pipe, if the radii of its ends are 0.1m and 0.04m and pressure drop across its length is 10 N//m^(2) |
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Answer» <P> SOLUTION :ACCORDING to continuity equation`(V_(2))/(V_(1))=(A_(1))/(A_(2))=(r_(1)^(2))/(r_(2)^(2))=((0.1)^(2))/((0.04)^(2))=(25)/(4)` and, according to Bernoulli.s equation for a horizontal tube, `P_(1)+(1)/(2)rhoV_(1)^(2)=P_(2)+(1)/(2)rhoV_(2)^(2)` `V_(2)^(2)-V_(1)^(2)=2((P_(1)-P_(2)))/(2)=2XX(10N//m^(2))/((1.25xx10^(3)kg//m^(3)))=16xx10^(-3)m^(2)//s^(2)` but `V_(2)=(25)/(4)V(1)=6.25V+_(1)therefore [(6.25)^(2)-1^(2)]V_(1)^(2)=16xx10^(-3)m^(2)//s^(2)` or `V_(1)~~0.0205 m//s` the RATE of volume flow=`A_(1)V_(1)=pi(0.1)^(2)xx(0.02)=6.28xx10^(-4)m^(3)//s` And the rate of mass flow is `(dm)/(dt)=rhoA.V.=(1.25xx10^(3) kg//m^(3))xx(6.28xx10^(-4)m^(3)//s)=0.785 kg//s` |
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| 2. |
One cubic metre of an ideal gas is at a pressure of 10^5 N//m^2and temperature 300K. The gas isallowed to expand at constant pressure to twice its volume by supplying heat. If the change in internal energy in this process is 10^4 J, then the heat supplied is |
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Answer» `10^5J` |
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| 3. |
Two non-mixing liquids of densities rho and n rho ( n gt 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length PL ( P lt 1) in the denser liquid. The density dis equal to |
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Answer» <P>`{1 + ( n+1) p} RHO` |
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| 4. |
If f is the number of degrees of freedon of a gas, show that gamma =1 + (2)/(f). |
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Answer» SOLUTION :We know `C _(p) - C _(v) = R and (C _(p))/( C_(v)) = gamma ` ALSO `C _(v) = (f R)/( 2) i.e., (R)/(C _(v)) = (2)/(f)` `therefore C _(p) = R + C _(v)` `(C _(p))/( C _(v)) = (R)/( C _(v) + 1 i.e., gamma = (2)/(f) + 1` `gamma = 1 + (2)/(f)` |
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| 5. |
Show that Reynold's number R=(rhov_(0)d)/(eta)is dimensionless. (rho is density of the fluid, v_(0) is the critical velocity, d is diameter through which the fluid is flowing and eta is the coefficient of viscosity of the fluid). |
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Answer» Solution :`R=([ML^(-3)][LT^(-1)][L])/([ML^(-1)T^(-1)])=[(ML^(-1)T^(-1))/(ML^(-1)T^(-1))]=[M^(0)L^(0)T^(0)]` `therefore R` is dimensionless. |
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| 6. |
Define angular acceleration. |
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Answer» Solution :The RATE of change of angular VELOCITY is called angular ACCELERATION. The angular acceleration is also a vector quantity . ` VEC(alpha) = (d vec(omega))/(dt)` |
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| 7. |
In the Ingen hauz's method to compare the thermal conductivities of different substances, the length upto which wax melted in copper and zinc rods are 9.3 cm and 5 cm respectively. Compare their thermal conductivities. |
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Answer» Solution :Given `l_(C)=9.3cmandl_(Z)=5cm` `becauseK_(C)/K_(Z)=l_(C)^(2)/l_(Z)^(2)=((9.3cm)/(5cm))^(2)=3.46` THUS, the THERMAL conductivity of COPPER is 3.46 TIMES that of zinc. |
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| 8. |
Find the emissive power from "0.3 cm"^(2) surface of filament of bulb of tungsten at 3000 K temperature. Take e=0.4 for tungsten bulb. |
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Answer» Solution :From Stefan - Boltzmaan law, `H=AesigmaT^(4)` `=0.3xx10^(-4)xx0.4xx5.67xx10^(-8)xx(3000)^(4)` `=55.1124` `:.` H `~~` 60 W (Acc. to textbook VALUE) |
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| 9. |
A rocket is fired vertically with a speed of 5 km s^(-1) from the earth’s surface. How far from the earth does the rocket go before returning to the earth ?Massof the earth = 6.0 × 10^(24) kg, mean radius of the earth = 6.4 xx 10^(6) m , G = 6.67 xx 10^(-11) " N "m^(2) kg^(-2). |
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Answer» Solution :`implies` Suppose, initial speed of rocket is v and at maximum height its final speed will be ZERO. Suppose, mass of rocket is m. Radius of earth `R_E`and from SURFACE its height is h. From conservation of energy, on surface, `K_(1)+U_(1) = K_(2)+U_(2)`at maximum height `:. 1/2 mv^2 +(-(GM_Em)/R_E)=0 + (GM_Em)/((R_E+h))` `:. 1/2 mv^2= GM_(E)m[1/R_E-1/((R_(E)+h))]` `:. v^2=2GM_E[h/(R_E(R_E+h))]` `:. v^2 =2gR_E^2[(h)/(R_E(R_E+h))] ""[:. GM_E = gR_E^2]` `:. v^2 = 2gR_E [h/(h(R_E/h+1))]` `:. v^2=2 g R_E[1/((R_E)/h+1)]` `:. R_E/h+1 = (2g R_E)/(v^2)` `:. R_E/h = (2gR_E)/(v^2) -1` `:. R_E/h=(2 xx 9.8 xx 6.4 xx 10^6)/(25xx10^6)-1` `= 5.0176 - 1.0000` = 4.0176 `:. h= R_E/(4.0176)` `:. h = (6.4 xx10^6)/(4.0176)` `:. h = 1.59 xx10^(6)` `:. h = 1.6 xx10^(6) m` `:. h = 1.600 xx10^3 m` `:. h = 1600 km` `implies` The distance of rocket from the centre of earth, `= 6.4 xx10^(6) m + 1.6 xx10^(6) m` `= 8.0 xx 10^(6) m` |
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| 10. |
Explain Hooke's Law. |
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Answer» Solution :"For small DEFORMATIONS the stress and strain are PROPORTIONAL to each other". This is known as Hooke.s law. stress `prop` `therefore` strain stress `= kxx` strain where k is the proportionality constant and is known as modulus of elasticity. Hooke.s law is an empirical law and it is VALID for most materials. However, there are some materials which do not exhibit this linear relationship. SI unit of k is `Nm^(-2) or Pa` Dimensional formula is `[M^(1) L^(-1) T ^(-2)]` |
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| 11. |
If an object is dropped from the top of a building and it reaches the ground at t = 4s, then the height of the building is (ignoring air resistance)(g=9.8ms^(-2)) |
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Answer» 77.3m |
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| 12. |
Which of the following combinations of properties would be most desirable for a cooking pot ? |
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Answer» High specific HEAT and LOW thermal conductivity |
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| 13. |
What is the pressure on a swimmer 10 m below, the surface of a lake ? |
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Answer» SOLUTION :Here, `h = 10 m, G = 9.8 ms^(-2) , rho_w = 1000 KG m^(-3)` PRESSURE on a swimmer10 m below the surface of the lake. `P = P_(a) + h rho g = (1 xx 10^5) + (10 xx 9.8 xx 10^3)` `=10 xx 10^4 + 9.8 xx 10^4 = 19.8 xx 10^4` `P = 1.98 xx 10^5 ~~ 2 atm`. |
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| 14. |
A pendulum clock of an iron rod is connected to a small, heavy bob. If it shows correct time at 20^(0)C, how fast or slow will itgo in one day at 40^(0)C ? (Coefficient of linear expansion for steel is 12 xx 10^(-6)//^(0)C . ) |
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Answer» Solution :We known `(DELTA T)/(T) = (1)/(2) alpha Delta t` `rArr (T_(2) - T_(1))/(T_(1)) = (1)/(2) alpha (t_(2) - t_(1))` Which is the fractional LOSS of time . As the temperature period increases and the clock loses time or goes slow. The time lost in one day (or in 24 HOURS) `((1)/(2) alpha (t_(2) - t_(1)) ) = (86400 s) (1)/(2) alpha (t_(2) - t_(1))` = 43200 `alpha (t_(2) - t_(1)) s ` In this problem, `alpha = 12 xx 10^(-6)//^(0)C`, `t_(1) = 20^(0)C , t_(2) = 40^(0)C ` The time lost in one day = 43200 `xx 12 xx 10^(-6) (40 - 20)` = 10.368 s. |
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| 15. |
Define angular frequency. Giveits S.I. unit. |
| Answer» Solution :It is the angel COVERED per unit time or it is the QUANTITY OBTAINED by multiplying frequency by a factor of `2pi`. `omega = 2pin`, S.I. Unit is RAD`s-^1` | |
| 16. |
The escape velocity for a body projected vertically upwards from the surface of Earth is 11"kms"^(-1)If the body is projected at an angle of45^(@) with verical, the escape velocity will be ........ |
| Answer» Solution :Escape velocity does not depends upon the ANGLE of projection | |
| 17. |
The displacement of particle is represented by the equation y=3cos((pi)/4-2omegat). The motion of the particle is |
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Answer» SIMPLE harmonic with particle is `(2PI)/(OMEGA)` |
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| 18. |
A screen is placed at a distance of 90 cm from an object. The image of this object is formed on the screen by a convex lens at two different location separated by 20cm. The focal length of the lens is |
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Answer» SOLUTION :By the displacement method `f=(d^2-x^2)/(4D)= f=(90^2-20^2)/( 4 TIMES 90)= 21.4 CM` |
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| 19. |
A metal sphere is electrically charged to Q_(0) and hangs on an insulting cord. The sphere slowly loses it charge because the air has a non zero resistivity rho. The potential of sphere of radius r_(0) is to be calculated with reference to infinity. Assume that resistivity of air is every where the same. Mass the correct statements |
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Answer» Capacitance of sphere is `4 pi epsilon_(0)r_(0)` |
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| 20. |
Two soap bubbles of radii in the ratio of 2 : 1. What is the ratio of excess pressure inside them? |
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Answer» `1:2` |
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| 21. |
The linear momentum of a particle varies with time as p=a_(0)+at+bt^(2). Which of the following graph represents force and time relation ? |
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Answer»
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| 22. |
A particle falls towards the earth from infinity. The velocity with which it reaches earth's surface (g rarr acceleration due to gravity on earth's surface and R is the radius of the earth) |
| Answer» Answer :B | |
| 23. |
A fully loaded boeing aircraft has a mass of 3.3xx10^(5)kg . Its total wing area is 500m^(2) .It is in level flight with a speed of 96.km//h. (a)Estimate the pressure difference between the lower and upper surfaces of the wings . (b) Estimate the fractional increase in thespeed of the air on the upper surface of the wing relative to the lower surface. (The density of air is rho=1.2kgm^(-3))(g=9.8ms^(-2)) |
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Answer» <P> Solution :The weight of the boeing aircraft is balanced by the upward force DUE to the pressure difference `=DeltaPxxA``mg=DeltaPxxA` `thereforeDeltaP=(mg)/(A)` `=(3.3xx10^(5)xx9.8)/(500)` `0.06468xx10^(5)` `therefore` Pressure difference `=6468Nm^(-2)` (b)From Bernoulli.s equation, `P_(1)-P_(2)=(1)/(2)rho(v_(2)^(2)-v_(1)^(2))+rhog(h_(2)-h_(1))` Neglecting `h_(2)-h_(1)` for its small value, `thereforeP_(1)-P_(2)=(1)/(2)(rho(v_(2)^(2)-v_(1)^(2))` `therefore(2DeltaP)/(rho)=(v_(2)-v_(1))(v_(2)+v_(1))` `thereforev_(2)-v_(1)=(2DeltaP)/(rho(v_(2)+v_(1))=(DeltaP)/(rho((v_(2)+v_(1))/(2)))`...(1) Now average speed`ltvgt=(v_(1)+v_(2))/(2)` `=(960xx1000)/(3600)` `=266.6ms^(-1)` `thereforev_(2)-v_(1)=(DeltaP)/(rholtvgt)=(6468)/(1.2xx266.6)` `=20.21ms^(-1)` `therefore(v_(2)-v_(1))/(ltvgt)=(20.21)/(267)"" (because` From equation (1)) `=0.0756` `=0.8` Hence , speed above the wing needs to be only `8%` higher than that below. |
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| 24. |
In which process (isothermal or adiabatic) there is a greater change in internal energy? |
| Answer» SOLUTION :ADIABATIC | |
| 25. |
If 300 ml of a gas at 27^(@)C is cooled to 7^(@)Catconstant pressure, then its final volume will be |
| Answer» ANSWER :C | |
| 26. |
Two plane mirrors are arranged at right angles to each other as shown in figure. A ray of light is incident on the horizontal mirror at an angle theta For what value of theta the ray emerges parallel to the incoming ray after reflection from the vertical mirror? |
| Answer» Answer :D | |
| 27. |
Find the minimum radius of a planet of mean density 5500kgm^(-3) and temperature 400^(@)C which has retained oxygen in its atmosphere. Density of oxygen at STP =1.424kgm^(-3). G=6.6xx10^(-11)Nm^(2)kg^(-2) |
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Answer» |
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| 28. |
In the question number 55, the ratio of elongation of steel to the copper wires is |
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Answer» `(5)/(2)` Substitution the GIVEN values,we get `(DeltaL_(S))/(DeltaL_(C)) = (1.6 m)/(2.4 m)xx (1.2 xx 10^(11) N m^(-2))/(2 xx 10^(11) N m^(-2)) = (2)/(5)` |
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| 29. |
The radius of the Earth is 6400 km and g = 10 in order that a body of 5 kg weights zero at the equator, the angular speed of the Earth is ............ rad |
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Answer» `1/80` |
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| 30. |
A particle is executing a simple harmonic motion. Its maximum acceleration is alpha and maximum velocity is beta. Then, its time period of vibration will be……… |
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Answer» `(2pi BETA)/(ALPHA)` `a_("max")= A omega^(2)` `therefore alpha = A omega^(2)` and Maximum velocity `v_("max")= A omega` `therefore beta = A omega` `therefore (a_("max"))/(v_("max"))= (A omega^(2))/(A omega)` `(alpha)/(beta)= omega` `therefore (alpha)/(beta)= (2pi)/(T)` `therefore T= 2pi (beta)/(alpha)`. |
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| 31. |
What are the different types of longtitudinal strain? |
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Answer» SOLUTION :Longitudinal strain can be classified into two types:(i) Tensile strain:If the length is incrases from its NATURAL length then it is knownas tensile strain. (ii) Compressive strain:If the length is decreased from its natural length then it is KNOWN as compressive strain. |
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| 32. |
During damped vibration, the quantity which gradually decrease is |
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Answer» velocity |
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| 33. |
Four moels of a perfect gas is heated to increase its temperature by 2^(0)C absorbsheat of 40 cal at constant volume. If the same gas is heated at constant pressure find the amount of heat supplied. (R = 2 cal/molK). |
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Answer» Solution :Number of moles n = 4 Change in TEMPERATURE`Deltat=2` Heat ENERGY absorbed `dQ=40` cal. GAS constant `R = 2` cal/mole K Heat energy absorbed at constant pressure `dQ = dU +dW` per mole per kelvin. But `dU = C_(v)DT and dW = P dV` `C_(p)=C_(v) +R`Multiply with n.dT `nC_(p)dT=nC_(v)dT +nRdT` `nC_(p)dT =` Heat energyabsorbed at constant pressure by n moles of gas. `nC_(v)dT=dU =`Heat energy absorbed at constant volume. `:.dQ=dU+nRdT =40+4xx2xx2=56`cal Heat energy absorbed at constant presure when 4 moles of air is heated through `2^(0)=56`cal. |
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| 34. |
A particle revolves round a circular path. The acceleration of the particle is inversely proportional to |
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Answer» MASS of particle |
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| 35. |
Arrange the following colours is increasing order of their critical angles with respect to glass (A) red (B) yellow ( C) blue (D) violet |
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Answer» a,b,C,d |
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| 37. |
Two blocks A and B of mass m and 2m respectively are moving towards a massive ("mass" gt gt 2m) cliff with velocities 2v and v respectively. The cliff moves with a velocity v. If the coefficient of restitution of collision at the surface of the cliff is e = 1//2, find the: a. velocity of the block B just after colliding with the cliff. b. work done by the cliff during collision. c. maximum compression of the spring of stiffness k which is fitted with the block B. |
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Answer» Let velocilty of BLOCK `B` just after Collision be `v_(1)` `v_(2)-v_(1)=E[u_(1)u_(2)]` `impliesv-v_(1)=1/2[2-v]impliesv_(1)=v/2` b. work DONE by clif =change in `KE` of `m` `1/2m[(v/2)^(2)-(2)^(2)]a=-15/8mv^(2)` c. Just after collision with cliff  At maximum compression `COLM: 2mv_(0)+mv_(0)=2mv+(mv)/2impliesv_(0)=5/6v` COME:` 1/2kx_(m)^(2)+1/2(3m)v_(0)^(2)=1/2(2m)v^(2)+1/2m(v/2)^(2)` `implies x_(m)=(sqrt(m/(6k)))v` |
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| 38. |
When a satellite moves around the earth in a certain orbit, the quantity which remains constant is : |
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Answer» ANGULAR VELOCITY |
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| 39. |
Velocity of a wave y = 10 ^(-4) sin (100 t - (x)/(10)) m is …...m/s |
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Answer» 100 `y = a sin (OMEGA t - KX) ` we get, `v = (omega )/(K) = (100)/(1//10) = 1000 (m)/(s)` |
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| 40. |
(A) : Collision between two particles is not necessarily associated with physical contact between them (R): Only in physical contact, momentum transfer takes place |
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Answer» Both .A. and .R. are TRUE and .R. is the CORRECT explanation of .A. |
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| 41. |
velocity of ground with respect to B |
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Answer» SOLUTION :Velocity of ground with respect to B `=0-v_B-v_A=0-(-25)=25m//s` TOWARDS north |
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| 42. |
Two particles of masses in the ratio 1: 2 you're moving in circles of radii in the ratio 2:3 with time periods in the ratio 3: 4. The ratio of their centripetal forces is |
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Answer» 9:4 |
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| 43. |
10 grams of ice at -20^(@)C is added to 10 grams of water at 50^(@)C. The amount of ice and water that are present at equilibrium respectively |
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Answer» 0,20g |
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| 44. |
The average translational kinetic energy of O_2, molecules at a particular temperature is 0.048 eV. The average translational kinetic energy of equal number of N, molecules at the same temperature is (in eV) |
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Answer» 0.016 |
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| 45. |
For which of the following does the centre of mass lie outside the body? |
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Answer» A PENCIL |
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| 46. |
What do you understand by the word speculaar reflection ? |
| Answer» Solution :Specular REFLECTION is OBSERVED only when the WAVELENGTH of the source is smaller than DIMENSIONS of the reflecting surface, as well as smaller than surface irregularities. | |
| 47. |
A circular disc rolls on a horizontal floor without slipping and the centre of the disc moves with a uniform velocity v. Which of the following values the velocity at a point on the rim of the disc can have? |
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Answer» v Velocity of the farthest point = v + v = 2v Horizontal velocity of the left or right end point = velocity of the CENTRE of the disc = v The option A, C and D is correct. |
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| 48. |
A 5 kg collar is attached to a spring of spring constant 500 N m^(-1). It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10.0 cm and released. Calculate (a) the period of oscillation, (b) the maximum speed and (c) maximum acceleration of the collar. |
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Answer» Solution :(a) The period of OSCILLATION as given by Eq. (14.21) is, `T=2pisqrt(m/k)=2pisqrt((5.0kg)/(500Nm^(-1)))` `=(2pi//10)s` `=0.63s` (b) The VELOCITY of the collar executing SHM is given by, `v(t)=Aomega` `=0.1xxsqrt(k/m)` `=0.1xxsqrt((500Nm^(-1))/(5KG))` `=1MS^(-1)` and it occurs at x = 0 (c) The acceleration of the collar at thedisplacement x (t ) from the equilibrium is given by, `a(t)=-omega^(2)xt` `=-k/mx(t)` Therefore, the maximum acceleration is, `a_("max")=omega^(2)A` `(500Nm^(-1))/(5kg)xx0.1m` `=10MS^(-2)` and it occurs at the extremities. |
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| 49. |
Two balls of equal masses movin with velocities 10m/s and -7m/s respectively collide elatically. There velocities after collision with be |
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Answer» `7 MS^(-1) and 17ms^(-1)` |
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