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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 gas expands in a piston cylinder device from volume V_1 to V_2,the process being p = a/V + b where p = pressure, V = volume. The work done in the process is |
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Answer» `-a LN (V_1/V_2) + B(V_2 -V_1)` |
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| 2. |
A spring balance shows wrong readings after using for a long time. Why? |
| Answer» Solution :When a spring balance is USED for long TIME, it starts loosing its elastic STRNGTH. i.e., such a spring starts LONGER time to regain its original position. So its shows WRONG readings. | |
| 3. |
Which is conserved in inelastic collision ? Total energy (or) Kinetic energy ? |
| Answer» SOLUTION :TOTAL ENERGY is CONSERVED. But KINETIC energy is not conserved. | |
| 4. |
A microscope consistsof two convex lenses of focal lengths 2cm and 5cm placed 20cm apart. Where must the object be placed so that the final virtual image is at a distance of 25cm from the eye? |
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Answer» Solution :For the eyepiece, FOCAL LENGTH `f=f_e=+5cm` `v=v_e=-25cm, U=u_e` substituting in `1/f=1/v-1/u implies 1/5=1/(-25)-1/u_e,1/u_e=-1/25 -1/5 =-6/25` `u_e=25/6 cm` object for the eyepiece is to be at a DISTANCE of `25/6cm` to its left But `V_0+u_e=20cm` where `u_c=25/6cm` , `v_0=20-u_e=20-25/6=95/6 cm` For the objective `v=v_0=+ 95/6 cm` `f=f_0=+2cm, u=? 1/f=1/v-1/u,1/2=6/95-1/u` `1/u= 6/95 -1/2= -83/190, u= -190/83=-2.29cm` The object is to be PLACED at a distance of 2.29cm to the left side of the objective. |
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| 5. |
Derive Poiseuille's formula for the volume of a liquid flowing per second through a pipe under streamlined flow. |
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Answer» Solution :Consider a liquid flowing steadily through a horizontal capillary tube. Let V = `(V/t)` be the volume of the liquid flowing out per second through a capillary tube. It DEPENDS on (l) coefficient of viscosity `(eta)` of the liquid, (2) radius of the tube (r), and (3) the pressure gradient `(P/l)` Then, v = `keta^(a)r^(b)(P/l)^(c)`...(1) where, k is a DIMENSIONLESS constant. Therefore, `[v] = ("Volume")/("Time") = [L^(3)T^(-1)] , [(dp)/(dx)] = ("Pressure")/("DISTANCE") = [ML^(-2)T^(-2)]` `[eta]=[ML^(-1)T^(-1)]` and [r]=[L] Substituting in equation (1) `[L^(3)T^(-1)] = [ML^(-1)T^(-1)]^(a) [L]^(b) [ML^(-2)T^(-2)]^(c)` `M^(0)L^(3)T^(-1) = M^(a+b)L^(-a+b-2c)T^(-a-2c)` So, equating the powers of M, L and T on both sides, we get a + c = 0, - a + b - 2c = 3, and - a - 2c = - 1 We have three unknowns a, b and c. We have three equations, on solving, we get a = -1, b = 4 and c = 1 Therefore, equation (1) becomes, v = `keta^(-1) r^(4) ((P)/(l))^(1)` Experimentally, the value of k is shown to be `pi/8` we have `v = (pir^(4)P)/(8etal)` The above equation is known as Poiseuille.s equation for the flow of liquid through a narrow tube or a capillary tube. This relation holds good for the fluids whose velocities are lesser than the critical velocity `(v_(c))`. |
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| 6. |
The method of dimensional analysis can be used to derive which of relation? |
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Answer» `A SIN (OMEGA t- kx)` |
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| 7. |
For the following situation select the correct statement. Abody is pojected from ground with some angle to the horizontal. The angualr momentum about the initial position will: |
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Answer» decreases |
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| 8. |
In the following question, statement 1 refers assertion and statement 2 refers reason. Statement 1: During perfect rolling, thefrictional force is zero. Statement 2: The velocity at the point of contact becomes zero. Select the correct option. |
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Answer» Statement 1 and 2 are true and statement 2 is a CORRECT explanation for statement 1. When velocity becomes ZERO and force of friction becomes zero. So, no work is done. |
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| 9. |
A particle of mass 50 gm moving in a circle at 5 cm radius with a constant speed of 10 cm/s. What is its angular momentum about (i) centre of the circle, (ii) a point on the axis of the circle at 4 cm distance from its centre ? Which of these will be always in the same direction ? [Fig.] |
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Answer» Solution :`m=0.050 kg, r = 5cm = 0.05 m, v= 10 cm//s` (i) Angular momentum about the centre of the circle `L_(1) = mvr = 0.050 XX 0.1 xx 0.05` `=2.5 xx 10^(-4) J.s` The direction of `L_(1)`is PERPENDICULAR to the plane ofthe circle (ii) Angular momentum about a point on the axis 4 cm distance from its centre `=L_(2)` = ? OB = 4 cm, `BA = sqrt(4^(2) + 5^(2)) = 6.4 cm = 0.064 m` `L_(2)=0.05 xx 0.1 xx 0.064` `=3.2 xx 10^(-4) J.s` `L_2` is perpendicular to the plane of its position vector `vec(OA)`and the instantaneous velocity of the particle. Direction of `L_2` changes with time. The direction of `L_(1)`REMAINS constant. |
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| 10. |
In above diagram, a combined rod is formed by connecting two rods of length l_(1) and l_(2) respectively and thermal conductivity k_(1) and k_(2) respectively. Temperature of ends of rods are T_(1) and T_(2) constant, then find the temperature of their contact surface ? |
| Answer» SOLUTION :`T_(X)=(k_(1)l_(2)T_(1)+k_(2)l_(1)T_(2))/(k_(1)l_(2)+k_(2)l_(1))` | |
| 11. |
At a certain heigh, a body ar rest expoldes into two equal fragments with one fragment receving a horizontal velocity of 10sqrt(3)ms^(-1). The horizontal distance between the two fragments, when their displacement vectors are inclined at 60^(@) relative to each other is (g = 10 ms^(-2)) |
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Answer» `40sqrt(3)m` |
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| 12. |
When ‘n’ identical droplets are combined to form a big drop, then the energy will be released If n identical small droplets each of radius r merge to form a big drop of radius R and the energyreleased in this process is absorbed by the big drop then rise in temperature (T be the surface tension, d be the density and S be the specific heat of the liquid droplets) |
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Answer» `(3Td)/( s)((1)/(R ) -(1)/(r ))` |
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| 13. |
Two waves are propagating with same amplitude and nearly same ferquency in opposite direction, they result in |
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Answer» beats |
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| 14. |
Identify the pair whose dimensions are equal |
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Answer» torque and work |
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| 15. |
A particle starts from the origin, goes along the x-axis to the point (20m, 0) and then returns along the same line to the point (-20m, 0). The distance and displacement of the particle during the trip are |
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Answer» 60M, -20m |
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| 16. |
A ring of mass M and radius R is held at rest on a rough horizontal surface. A rod of mass M and length L = 2 sqrt3 R is pivoted at its end A on the horizontal surface and is supported by the ring. There is no friction between the ring and the rod. The ring is released from this position. Find the acceleration of the ring immediately after the release if theta = 60^(@). Assume that friction between the ring on the horizontal surface is large enough to prevent slipping of the ring. |
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Answer» |
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| 17. |
A spring of force constant 1200 Nm^(-1) is mounted on a horizontal table as shown in Fig. A mass of 3.0 kg is attached to the free end of the spring, pulled side ways to a distance 2.0 cm and released. Determinethe frequency of oscillation of the mass. |
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Answer» Solution :Here, `k=1200 NM^(-1) , m= 3.0 kg , A = 2.0 cm = 0.02 m` (a) Frequency , `f= 1/T = 1/(2A) SQRT(k/m) = 1/(2xx3.14) sqrt((1200)/(3)) = 3.2 s^(-s)` |
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| 18. |
A uniform rod of mass m and length is acted upon by the forces F_(1) and F_(2) Find that: a. linear-and angular acceleration of the rod. b. value of x for which the point P does not accelerate. |
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Answer» Solution :a. Equation of motion `F_(1)+F_(2)=ma` or `2F-F=maimpliesa=(3F)/m`…….i TORQUE equation about centre of mass. `2F.l/2-F.x=((ML^(2))/12)alpha` `implies alpha(12F(l-x))/(ml^(2))` ………ii b. If point `P` does not ACCELERATION `veca_(P)=veca_(P,C)+veca_(C)` As `a_(P)=0=-ax+aimpliesax=a` `(12F(l-x))/(ml^(2))=(3F)/m` which given `x=l/2` |
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| 19. |
A spring of force constant 1200 Nm^(-1) is mounted on a horizontal table as shown in Fig. A mass of 3.0 kg is attached to the free end of the spring, pulled side ways to a distance 2.0 cm and released. Determine the maximum speed of the mass. |
| Answer» SOLUTION :Max. speed of the MASS will be when it is PASSING through the mean position , which is given by `V_("max") = A omega = A sqrt(k/m) =0.02 xx sqrt((1200)/(3)) = 0.4 ms^(-1)` | |
| 20. |
You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will (i) deviate a pencil of white light without much dispersion. (ii) disperse and displace a pencil of white light without much deviation. |
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Answer» Solution :(i) For no dispersion, total ANGULAR dispersion produced by TWO prisms should be zero. i.e., `(mu_b - mu_r) A + (mu'_b - mu'_r) A' = 0` As `(mu'_b - mu'_r)` for flint GLASS than that for crown glass, therefore, `A' lt A` i.e., flint glass prism of SMALLER angle has to be suitably combined with crown glass prism of larger angle. (ii) For almost no deviation, `(mu_y - 1)A + (mu'_y - 1) A' = 0` Taking crown glass prism of certain angle, we go on increasing angle of flint glass prism till this condition is met. In the final combination HOWEVER, angle of flint glass prism will be smaller than the angle of crown glass as `mu'_y` for flint glass is more than `mu_y` for crown glass. |
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| 21. |
A spring of force constant 1200 Nm^(-1) is mounted on a horizontal table as shown in Fig. A mass of 3.0 kg is attached to the free end of the spring, pulled side ways to a distance 2.0 cm and released. Determine the maximum acceleration of the mass. |
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Answer» Solution :Here, `k=1200 NM^(-1) , m= 3.0 kg , A = 2.0 CM = 0.02 m` (a) Frequency , `f= 1/T = 1/(2A) sqrt(k/m) = 1/(2xx3.14) sqrt((1200)/(3)) = 3.2 s^(-s)` Acceleration `a= omega^2 y = k/my ` Accerlation will be maximum when y is maximum i.e. y = A `therefore ` Max. acceleration , `a_("max") = (kA)/(m) = (1200xx0.02)/(3) = 8 ms^(-2)` |
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| 22. |
The kilowatt-hour is the unit of |
| Answer» SOLUTION :The kilowatt-hour is the unit of energy. | |
| 23. |
Two aeroplanes are approaching each other and their velocities are 160 and 240 kilometres per hour. The frequency of the note emitted by the first as heard by the passengers in the other is 1000 Hz. Calculate the true frequency of the note as heard by its own passengers. (Velocity of sound = 1200 kmph) |
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Answer» |
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| 24. |
A particle travelling horizontally with speed u collides and sticks with a particle of equal mass hanging at rest at the end of a light enextensible string of length 2l. If thestring rotates through an angle of 60^(@) before its velocity becomes zero, then u is |
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Answer» `sqrt(4GL)` |
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| 25. |
An object is moving with constant acceleration. Its velocity is 48 ms^(-1)at the end of 10 second and becomes 68 ms^(-1)at the end of 15 second. What would be the distance travelled by the object in 15 second ? |
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Answer» Solution :In equation of motion `v =v _(0) + at` `t _(1) = 10 s IMPLIES v _(1) = 48 ms ^(-1)` `therefore 48 = v _(0) + 10a ""…(1) and t _(2) = 15 s implies v _(2)= 68 ms ^(-1)` `therefore 68 = v _(0) + 15 a ""…(2)` Solving equation (1) and (2), `({:(68=, v _(0)+, 15A), (48=, v_(0)+, 10 a),(-,-,-):})/({:(20=, 5A):})` `a4 ms ^(-2)` now from equation (1) `48= v _(0) + 10 a, `put `a = 4 ms ^(-2)` `therefore 48 = v _(0) + 40` `therefore v _(0) = 8 ms ^(-1)` Now displacement `d = v _(0)t + 1/2 at ^(2)` Put `v _(0) = 8 ms ^(-1), t = 15 s, a = 4 ms ^(-2)` ` therefore d = 8 xx 15 + 1/2 xx 4 xx 225` `therefore d = 120 + 450 therefore d = 570 m` |
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| 26. |
The time period of oscillation of a spring is 1.57s when a mass of 100 gm is suspended from its lower end. Calculate (i) the force constant of the spring (ii) the K.E. of the mass when its displacement'is equal to the amplitude. |
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Answer» Solution :`m = 100gm=0.100 kg , T = 1.57 s, K = ? ` `T = 2pi SQRT(m/k)` `T^2 = 4pi^2(m)/k` `k = (4pi^2m)/(T^2)` `=(4xx(3.14)^2xx0.100)/((1.57)^2)=1.6 NM^(-1)` When the displacement is equal to the amplitude the velocity of the body is zero and hence K.E. is zero. |
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| 27. |
A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weight 12 kg. It is hinges at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it. |
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Answer» 0.25 rad/s |
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| 28. |
A force of 100 dyne acts on a mass of 5 grams for 10 sec . Find the velocity produced?. |
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Answer» SOLUTION :`a=F/m=(100)/5=20 cm//s^(2)` `v=u+atrArr v=0+20xx10=200 cm//sec` |
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| 29. |
Express newton secondlaw of motion in component form given its significance |
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Answer» Solution :Intermsof cartesiancoordinates , thisequationcan bewritten as Bycomparingbothsidesthe three scalar EQUATIONARE Theaccelerationalongthe ydirectiondepends only on thecomponentof forceactingalong the y -direction `f_(y) = ma_(y)` SIGNIFICANCE: From the aboveequationsit isinferredthat theforceactingalong ydirectioncannotalterthe accelerationalong X direction. In the SAMEWAY `F_(z)` cannotaffect`a_(y)` and `a_(x)` |
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| 30. |
A man jumping out of a slow bus falls forward. Ths is due to ………….. |
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Answer» INERTIA of MOTION |
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| 31. |
Find|vec a xx vecb| where veca = 3hati + 4 hatj and vecb= hati + hatj + hatk |
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Answer» `45^(@)` |
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| 32. |
Calculate the resultant of the following forces at a point , making use of resolution process. (ii) 980sqrt(2) dyn along north -west |
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Answer» |
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| 33. |
A circular wheel of MI 0.08kgm^2 is rotating 30 rotations per second. When a retarding troque is applied the rate of rtation is reduced to 10 rotations per second in 2 minutes. Find the torque. |
| Answer» SOLUTION :8.376 `TIMES 10^(-2)`NM | |
| 34. |
When two tuning forks when sounded together,20 beats were produced in 8 seconds .After loading one of the forks with a littile wax,they produce 32 beats in 8 seconds.If the unloaded fork had a frequency of512 Hz,calculate the frequency of the other. |
| Answer» SOLUTION :509.5 HZ | |
| 35. |
The radius of hydrogen atom is ground state is 5xx10^(-11)m. Find the radius of hydrogen atom is fermimetre.(1fm=10^(-15)m) |
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Answer» `5XX10^(4)fm` `:.1m=10^(15)fm` `:. 5xx10^(-11)m=5xx10^(-11)xx10^(15)fm=5xx10^(4)fm` |
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| 36. |
A heavy particle is attached to one end of a light string of length l whose other end is fixed at O. The oarticle is projectyed horizontally with a velocity v_0 from its lowest position A. When the angular displacement of the string is more than90^(@), the particle leaves the circular path at B. The string again becomes taut at C such that B,O,C are collinear. Find v_0 in terms of l and g |
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Answer» |
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| 37. |
There is an infinite thin flat sheet with mass density eta per unit area. Find the gravitational force, due to sheet, on a point mass m located at a distance x from the sheet.Consider a large flat horizontal sheet of material density r and thickness t, placed on the surface of the earth. The densityphiof the earth is phi_(0) If it is found thatravitational field intensity just between the sheet is larger than field just above it, prove that rho_(@)lt(3)/(2)rho. |
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Answer» |
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| 38. |
If a,b are the magnitudes of vectors veca & vec b and veca vecb=0 the value of |vecx xx vecb| is |
| Answer» ANSWER :B | |
| 39. |
Two billiard balls of the same size (radius r) and same mass are in contact on a billiard table. A third ball also of the same size and mass strikes them symmetrically and remains at rest after the impact. The coefficient of restitution between the balls is |
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Answer» Solution :`sin theta =(r )/(2R) = (1)/(2), THEREFORE 3 = 30^(@)` From conservation of linear momentum `m u = 2 mv = cos 30^(@) or v = (u)/(sqrt3)` Now ` E = ("relative velocity of separation")/("relative velocity of approach")` in common normal direction HENCE, `e = (v )/( u cos 30 ^(@)) = (u // sqrt3)/( u sqrt3 //2) = 2/3` |
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| 40. |
(A): Silver is a ductile material(R): For a ductile material yield point and Breaking point are seperated by larger distance than for Brittle materials on the stress-strain curve. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 41. |
Frequency of a particle executing SHM is 10 Hz. The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is unstretched. Find the maximum speed of the particle : (g= 10m//s^(2)) |
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Answer» Solution :Mean position of the particle is `(mg)/(k)` distance below unstretched position of SPRING. THEREFORE, amplitude of oscillation is `A= (mg)/(k) OMEGA= sqrt((k)/(m))= 2pif= 20pi` `:. (m)/(k)= (1)/(400pi^(2))` Therefore, the maximum speed of particle will be: `v_("max")= Aomega= ((g)/(400pi^(2)))(20pi)= (1)/(2pi)m//s` |
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| 42. |
A lightmetrescaleis placedhorizontallyon twosupportsat 10cm markand 80 cmmark. If70 gmassIshangedfrom50 cmmark, findforcesupliedby thesupportat 10cmmark, on thescale. |
Answer» Solution : fromthefigure `OA = 10 cm, OB= 50 cm, OC = 80 cm ` ` thereforeAC = OC- OA = 80 -10 = 70cm ` ` BC = OC - OB = 80 - 50 = 30 cm ` `F_(3)= MG= 70xx 980 dyn ` for abodyequilibriumthealgebraicsum of themoments oftheforcesaboutany pointis zero ` therefore `Sumof themomentsof thethreeforcesaboutpointC `G = 0-F_(1 )* AC + F_(3)* BC=0 ` Hence`F_(1)= F_(3) *(BC )/(AC )=(70 xx(80 )xx (3) /(70)=29400dyn ` |
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| 43. |
A shell fired such that it hits a target at a distance 'b' and having height 'h' Minimum velocity of projection is |
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Answer» `SQRT(2gh)` |
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| 44. |
A vessel contains a liquid filled with 1/10th of its volume. Another vessel contains same liquid upto 1/8th of its volume. In both case the volume of empty space remains constant at all temperatures. Then the space remains constant at all temperatures. Then the vessels is |
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Answer» `2:5` |
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| 45. |
Define strain energy. Write formula for the same |
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Answer» Solution :Strain energy is DEFINED as the energy stored in a body due to DEFORMATION. Strain energy = Work done `=(1)/(2)xx` APPLIED FORCE `xx` extension `=(1)/(2)`Fe. |
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| 46. |
A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s. (a) How far will the cylinder go up the plane? (b) How long will it take to return to the bottom? |
| Answer» SOLUTION :(a) 3.8 m (B) 3.0 s | |
| 47. |
Three masses are placed on the x-axis : 300 g at origin , 500 g at x = 40 cm and 400 g at x = 70 cm . The distance of the centre of mass from the origin is |
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Answer» 40 cm  The distance of the centre of mass of the system of three masses from the origin O is `X_(CM) = (m_(1) x_(1) + m_(2) x_(2) + m_(3) x_(3))/(m_(1) + m_(2) + m_(3)) = (300 xx 0 + 500 xx 40 + 400 xx 70)/(300 + 500 + 400)` `= (500 xx 40 + 400 xx 70)/(1200) = (400 [50 + 70])/(1200)` ` = (50 + 70)/(3) = (120)/(3) = 40` cm |
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| 48. |
A mass.m. is released, with a horizontal speed v from the top of a smooth and fixed, hemispherical bowl, of radius r. The angle theta w.r.t. the vertical where it leaves contact with the bowl is |
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Answer» `SIN^(-1)(v^(2)/(3rg) + 2/3)` |
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| 49. |
Does a scalar quantity depend upon the frame of reference chosen . |
| Answer» SOLUTION :No. A SCALAR does not DEPEND on the FRAME of REFERENCE. | |
| 50. |
What do you understand by the term measurement ? Explain with example. |
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Answer» Solution :(i) The process of measurement is BASICALLY a process of comparison of any physical quantity with its standard unit. (ii) For example, when we state that a rope is 10 METER LONG, it is to say that it is 10 times as long as an object whose length is defined as 1 meter. |
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