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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. |
If the angular momentum of a planet is given by .vec(L) = 5t^(2)hat(i) - 6that(j) + 3hat(k). What is the torque experienced by the planet? Will the torque be in the same direction as that of the angular momentum? |
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Answer» Solution :The torqueexperiencedby the PLANET `vec (tau)= (dvec(L))/(DT) "" [ d/(dt ) (t^(n)) = n^(n-1)] ` `= d/(dt)(5t^(2)hat(i) - 6hat(tj) +3hat(k))` ` = 5. d/(dt) (t^(2)) hat(i) - 6 d/(dt)(t) hat(j)+ 3 d/(dt) (hat(k)) = 10 hat(i) - 6hat(j) + 0 ` `vec(tau) = 10 t hat(i) - 6hat(j)` |
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| 2. |
Calculate the rate at which water flows through a capillary tube of length 0.5m with an internal radius of 1mm. Coefficient of viscosity is 1.3xx10^(-3)kg//m-s. The pressure head is 20 cm of water. |
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Answer» Solution :From Poiseuille.s formula `V=(piPr^(4))/(8etal)` `=(3.14xx20xx10^(-2)xx1000xx9.8xx(10^(-3))^(4))/(8xx1.3xx10^(-3)xx0.5)=7.4xx10^(-8)m^(3)s^(-1)`. |
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| 3. |
Find the value of pi//53.2 with due regard to significant figures. |
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Answer» Solution :Out of the TWO NUMBERS 53.2 has three significant DIGITS. So `pi` should be written with `3 + 1= 4` significant figures. `pi= 3.1415= 3.142` (as 1 is odd it is RAISED by one). Now `(3.142)/(53.2)= 0.0590601` This is to be rounded off to three significant figures 0.059060= 0.0591 |
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| 4. |
The spin drier of a washing machine revolving at 15 r.p.s slows down to 5 r.p.s., while making 50 revolutions. Find (a) angular acceleration and (b) time taken. |
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Answer» SOLUTION :Here `omega_(0)=15` r.p.s `=30rads^(-1)` `omega_(0)=5`r.p.s `=10pirads^(-1)`, `theta=50xx2pi=100pi` Rad (a) Now, `OMEGA^(2)-omega_(0)^(2)=2alphatheta` `:.(10pi)^(2)-(30pi)^(2)=2alphaxx100pi` or `alpha=-4pirads^(-2)` (b) Also `omega=omega_(0)+alphat` `:.10pi=30pi+(-4PI)xxt` or `t=5s` |
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| 5. |
The ratio of escape velocity at earth to the escape velocity at a planet whose radius and mean density are twice as that of earth is |
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Answer» `1:2` |
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| 6. |
Fig. depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion. What is the period of motion (in case of periodic motion)? |
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Answer» Solution :a) is not PERIODIC B) It is a periodic motion with a period of 2 seconds. c) It is not a periodic motion. d) It is a periodic motion with a period of 2 seconds. |
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| 7. |
Explain work doneby a constant force . |
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Answer» Solution :Constant force : The magnitude and direction offorce actingon the body remains constant during the motion of body , then such type of force is known as constant force . DEFINITION of work on a trial basis is as under : Work : " The product of magnitude of force and dispalcement during which te force is acts is called work " . ` :. `Work = Force F `xx` Displacementd ...(1) where F is the forceactingon the bodyand d isthe displacementin the direction of force . Work is denotedby thedot product of force `vec(F)`and displacement in the direction of force . It is not necessary that the displacementof a body is always in the direction of force of force applied . If the displacement`vec(d)` It is not necessary that thedisplacement of a body is always in the dirction of force applied . If the displacement of bodyis not in the direction of force applied then the definition of work is given as under by two ways . (1) Work= Force `xx` Component of displacement in the direction of force (2) Work = Component of force in the direction of displacement `xx` displacement...(3) Understanding these by following figure (a) a body UNDERGOES a displacement `vec(d)`under the influence of the force `vec(F)` on the bodywith `theta `ALONG the horizontal direction .  Force is not fully responsible for displacement `vec(d)` for this take the componentof force as shown in figure . (i) The componentof force in the horizontal ( displacement `vec(d)` is not ) direction is `F cos theta ` (II) Perpendicular component `f sin theta ` to the displacement `vec(d)`is not responsible for horizontaldispalcement .  ` :. ` From the equation (3) of definition of work `W = |vec(F)| cos theta xx |vec(d)|` ` :. W = F cos theta.d` ` [ :. |vec(F)| = F and |vec(d) |=d]` The perpendicular components of displacement `vec(d)`in the directionof forceas shown in figure (b) are , (1) Component `dcos theta `in the direction of force `vec(F)` (2) Component `dsin theta ` in the perpendiculardirection of force `vec(F)` ` :. ` From the equation(2) of definition of work , `W = |vec(F)| xx|vec(d)| cos theta` ` :. W = F d cos theta ""`...(5) ` [ :.|vec(F ) |= F and |vec(d) | = d.|` From equation (4) and (5) , the result obtained by two different definition of work is same . If `theta = 0^(0), cos 0^(0)=1 :. ` Work in general `W = d cos theta ""....(5)` ` [ :. |vec(F)|=F and |vec(d)| = d]` From equation (4) and (5) , the result obtained by two different definition of work is same . If `theta =0^(0) ,cos0^(0)=1 :. ` Work in general W = Fdwhich is as equation (1) . The work obtained by the force acting on the body for definite displacement is proportional to the magnitude of force . |
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| 8. |
Four identical particles each of mass 1 kg are arranged at the corners of a square of side length 2sqrt(2)m. If one of the particles is removed, the shift in the centre of mass is |
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Answer» `(8)/(3)m` |
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| 9. |
If impulse/ varies with time t as f (kg ms^(-1))=20t^(2)-40t.The change in momentum is minimum at |
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Answer» t=2s `(d)/(dt) ( 20t^(2)-400t)=0` 40t-40=0 t=1s |
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| 10. |
Statement A: An engine A can perform a given work in 1 hr and engine B can perform the same work in 1/2 hr. then B has greater power than A. Statement B: Power is the dot product of force and velocity |
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Answer» A & B is correct |
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| 11. |
In the question number 11, the percentage strain produced in the rod is |
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Answer» Â 0.04% `DeltaL = ((F//A)L)/(Y) = ((3.18xx10^(8) N m^(-2))(1 m))/(2XX10^(11)N m^(-2))` `=1.59xx10^(-3) m = 1.59 mm` STRAIN produced in the rod is Strain `= (DeltaL)/(L) = (1.59 xx 10^(-3)m)/(1m) = 1.59xx10^(-3) = 0.16%` |
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| 12. |
S= (1)/(2) a t^(2) where a is a constant. If relative error in measurement of t is y, relative error in measurement of S will be |
| Answer» Answer :C | |
| 13. |
Two beakers, identical in all respects, but made of different materials, are filled with equal amounts of ice at 0^(@) C. Toe ice in one beaker melts in one minute whereas that in the other beaker melts in five minutes. Find the ratio of the thermal conductivities of the material of the two beakers. |
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Answer» |
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| 14. |
If direction of velocity and acceleration are in mutually opposite direction then what will be the change in speed ? |
| Answer» SOLUTION :SPEED (VELOCITY) will be DECREASED. | |
| 15. |
Calculate the amount of energy dissipated when 8 drops of water of 0.3 mm radius coalesce from one big drop surface tension of water is 7.2xx10^(-2)Nm^(-1). |
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Answer» |
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| 16. |
The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1 % the maximum error in determining the density is |
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Answer» `2.5%` `=(M)/((4)/(3)pir^(3))` `:.` Percentage error in density `=(Delta rho)/(rho)xx100%` `=(DELTAM)/(M)xx100%+3(DELTAR)/(r)xx100%` `=1.5%+3xx1%` `=4.5%` |
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| 17. |
When a conservative force does positive work on a body, what is the change in its potential energy ? |
| Answer» SOLUTION :Its potential energy decreases SINCE dU = - Fdx (or) `U_(f) - U_(i)` = - dW where `U_(f) and U_(i)` are the FINAL and INITIAL potential energies RESPECTIVELY. | |
| 18. |
To what height should a cylindrical vessel be filled with a homogeneous liquid such that the force exerted by the liquid on the wall of the vessel be equal to the force exerted by the liquid on the bottom of the vessel? |
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Answer» Solution :Let h be the height of the liquid of density `rho` in the cylinder of radius r. The pressure of the liquid COLUMN at the bottom of the cylinder is `hrhog`. THEREFORE, the FORCE exerted by the liquid column on the bottom of the cylinder is `F_(1)="pressure"xx"area of the bottom"` `=hrhogxxpir^(2)` The pressure of the liquid column on the wall of the vessel VARIES from zero at the TOP of the column to `hrhog` at the bottom. Thus, the average pressure on the wall is `(0+hrhog)/2=1/2hrhog` Therefore, the force exerted by the liquid column on the wall of the cylinder is `F_(2)="pressure"xx"area of wall"=1/2hrhogxx2pirh` For `F_(2)=F_(1)` we should have `1/2hrhogxx2pirh=hrhogxxpir^(2)` or, h = r i.e., the height of the liquid column in the cylinder should be equal to the radius of the cylinder. |
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| 19. |
The workdone in raising a stone of mass 4 kg, specific gravity 2 immersed in water, from a depth of 3 m to 1 m below the surface is |
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Answer» 98 J |
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| 20. |
Why a hot glass beaker when cold water is sprinkled on it? |
| Answer» SOLUTION :DUE to UNEVEN EXPANSION | |
| 21. |
Which of the following statements is/are true for a simple harmonic oscillator? |
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Answer» Force ACTING is DIRECTLY proportional to displacement from the mean POSITION and opposite to it |
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| 22. |
Hydrostatic pressure 'P' varies with displacement'x' as P = (A)/(B) log (Bx^(2) + C) whre A,B and C are constants. The dimensional formula for 'A' is |
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Answer» `[M^(1) L^(-1) T^(-2)]` |
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| 23. |
A particle is revolving along a circular path with decreasing speed. Which one of the following is true for the particle? |
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Answer» angular momentum of the particle is constant |
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| 24. |
If V_(H),V_(N) and V_(O) denote the root mean square velocities of molecules of hydrogen ,nitrogen and oxygen respectively at a given temperature ,then |
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Answer» `v_(N) gt V_(O) gt V_(H)` |
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| 25. |
Three identical uniform thin metal rods form the three sides of an equilateral triangle. If the moment of inertia of the system of these three rods about an axis passing through the centroid of the triangle and perpendicular to the plane of the triangle is .n. times the moment of inertia of one rod separately about an axis passing through the centre of the rod and perpendicular to its length, the value of .n. is |
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Answer» 3 |
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| 26. |
The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B . Then the angle between A and B is :(a)30^(@)(b)45^(@)(c) 150^(@)(d)120^(@) |
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Answer» `30^(@)` `C^(2)=A^(2)+B^(2)+2ABcostheta` `(B^(2))/(4)=A^(2)+B^(2)+2AB((A)/(B))` `-(3)/(4)B^(2)=3A^(2)impliesB^(2)=4A^(2)` i.e., B=2A (ii) Directionof resultant `tanalpha=(Bsintheta)/(A+Bcostheta)` `oo=(Bsintheta)/(A+Bcostheta)` `:.A+Bcostheta=0` `COSTHETA=-(A)/(B)=-(A)/(2A)=-(1)/(2)` `costheta=-(1)/(2)""theta=cos^(-1)(-(1)/(2))` `theta=120^(@)` |
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| 27. |
A car travelling at 108 km/hour has its speed reduced to 36 km/hour while travelling a distance of 200 m. Its acceleration and the time taken are |
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Answer» `2m//s^(2),10SEC` |
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| 28. |
The ratio (C_(p))/(C_(v))for a gas. Its molecularweight is M. Its specific heat capacity at constant pressure is |
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Answer» `(R )/(gamma-1)` |
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| 29. |
A uniform circular ring radius 'R' lies in the xy plane with its centre at origin Its moment of inertia about axis lying The XY-plane parallel to X-axis and passing through a point on the Y-axis at a distance Y=2R is I_(1). Its M.I. about axis lying in a plane perpendicular to XY-plane passing through a point on the - axis at a distance x=d is I_(2). I f I_(1)=I_(2), the value of 'd' is |
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Answer» `Rsqrt(7//2)` |
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| 30. |
If a spring of force constant 'K' is cut into three equal parts, then the force constant of each part will be |
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Answer» `K//3` |
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| 31. |
Dimensional method helps in converting the units from one system into another.Name the principle used for the above purpose. |
| Answer» SOLUTION :PRINCIPLE of HOMOGENEITY of DIMENSIONS | |
| 32. |
A body of mass 4kg is attached to another body of mass 2kg with a massless rod. If the 4kg mass is at (2i+5j)m and 2kg mass is at (4i+2j)m, the centre of mass of the system is at (in m) |
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Answer» `8i+12j` |
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| 33. |
The workdone in blowing a spherical soap bubble of diameter 2 cm will be (the surface tension of soap solution is 2xx10^(-2)Nm^(-1)) |
| Answer» Answer :A | |
| 34. |
A helicopter of mass M = 15000 kg is lifting a cubical box of mass m = 2000 kg. The helicopter is going up with an acceleration of a = 1.2 m//s^(2). The four strings are tied at mid points of the sides of the square face PQRS of the box. The strings are identical and form a knot at K. Another string KH connects the knot to the helicopter. Neglect mass of all strings and take g = 10 m//s^(2). Length of each string AK, BK, CK and DK is equal to side length of the cube.(a) Find tension T in string AK.(b) Find tension T_(0) in string KH. (c) Find the force (F) applied by the atmosphere on the helicopter. Assume that theatmosphere exerts a negligible force on the box. (d) If the four strings are tied at P,Q,R and S instead of A, B, C & D, how will the quantities T, T_(0) and F change? Will they increase or decrease? Assume that length of the four identical strings remains same. |
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Answer» (B) 22400 N (C) 190400 N (d) `T_(0)` and F do not CHANGE . T will INCREASE . |
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| 35. |
Four particles of masses m, 2m, 3m and 4m are arranged at the corners of a parallelogram with each side equal to a and one of the angle between two adjacent sides as 60^(@) The parallelogram lies in the x-y plane with mass m at the origin and 4m on the r-axis. The centre of mass of the arrangement will be located at |
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Answer» `((SQRT3)/(2) a , 0.95 a)`  `X_(CM) = (m_(1) x_(1) + m_(2) x_(2) + m_(3) x_(3) + m_(4) x_(4))/(m_(1) + m_(2) + m_(3) + m_(4))` `= ((m xx 0) + (2m xx (a)/(2)) + (3m xx (3a)/(2)) + (4M xx a))/(m+ 2m + 3m + 4m)` `= (ma + 4.5 ma + 4MA)/(10 m) = (9.5 ma)/(10m) = 0.95` a `Y_(CM) = (m_(1) y_(1) + m_(2) y_(2) + m_(3) y_(3) + m_(4) y_(4))/(m_(1) + m_(2) + m_(3) + m_(4))` `(( m xx 0) + (2m xx (a sqrt3)/(2)) + (3m xx (a sqrt3)/(2)) 4 (4m xx0))/(m + 2m + 3m + 4m)` `= (sqrt3 am + sqrt3 xx 1.5 ma)/(10 m) = (2.5 sqrt3 am)/(10 m) = (sqrt3a)/(4)` `therefore` Centre of MASS is at `(0.95a , (sqrt3a)/(4))` |
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| 36. |
A stone of mass 0.5 kg is attached to a string of length 2m and is whirled in a horizontal circle. If the string can withstand a tension of 9N, the maximum velocity with which the stone can be whirled is |
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Answer» `6 MS^(-1)` |
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| 37. |
Can transverse waves be produced in air ? |
| Answer» SOLUTION :No transverse waves travel in the from of CRESTS and troughs and so involve CHANGE in shape. so the transverse waves can be produced in a medium which has ELASTICITY of shape. as AIR has no elasticity of shape, hence transverse waves cannot be produced in it. | |
| 38. |
What will be the time period of seconds pendulum if its length is doubled? |
| Answer» SOLUTION :`L= 2.828 s` | |
| 39. |
Under which condition will the magnitude of the resultant of two vectorsbe equal to that of any one of theconstituent vectors ? |
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Answer» Solution :Suppose, a and b are the magnitudes of two vectors and the angle between them is `theta` . ACCORDING to the problem let the MAGNITUDE of theresultant of `veca and vecb` be EQUAL to a . Therefore, `a^2=a^2+b^2+2abcos theta or, b(b+2a cos theta)=0` `therefore b+2acos theta=0[because b ne0]` or, `theta = cos^(-1)(-b/(2a))` that is , if the angle between `veca and vecb` is =`cos^(-1)(-b/(2a)) or cos^(-1)(-a/(2b))`, the magnitude of the resultant vector is equal to a or b respectively. |
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| 40. |
Name the device that can be used to measure the number of wavelengths of light in a given distance. |
| Answer» SOLUTION :OPTICAL INTERFEROMETER. | |
| 41. |
Figure shows the position - time graph of a particle of mass 4 kg, then (a) force on the particle for r lt 0, t gt 4 s, 0 lt t lt 4 s ?(b)impulse at t = 0 and t = 4 s. (consider one - dimensional motion only) |
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Answer» Solution :(a) (i) For `t lt 0`, in the position time GRAPH is at O which means displacement of the particle is zero. i.e., particle is at rest at the origin. Hence force on the particle MUST be zero. (ii) For `0 lt lt 4` s, the position time graph OA has a constant slope. THEREFORE velocity of the particle is constant in this interval i.e., particle has zero acceleration. Hence force on he particle must be zero. (iii) Fro `t GT 4` s, the position time graph AB is parallel to time axis. Therefore, the particle remains at a distance 3 m from the origin, i.e., it is at rest. Hence force on the particle is zero. (b) (i) impulse at t = 0 We know, Impulse = change in linear momentum, Before t = 0 particle is at rest i.e., U = 0. After t = 0, particle has a constant velocity `v=(3)/(4)=0.75 m//s` `therefore` Impulse `= m(v-u)=m(0.75-0)=3kgm//s` (ii) `therefore` Impulse at t = 4 s. Before t = 4s, particle has a constant velocity u = 0.75 m/s After t = 4s, particle is at rest i.e., v = 0 Impulse `= m(v-u)=4(0-0.75)=-3 kg m//s`. |
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| 42. |
A stone is dropped into a well and the sound of splash is heard after 5.3sec. If the water is at a depth of 122.5m from the ground, the velocity of sound in air is |
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Answer» Solution :If `t_(1)` is the time taken by stone to REACH the GROUND and `t_(2)` the time taken by sound to go up, then `t_(1)+t_(2)=5.33` SINCE `s=ut+(1)/(2)at^(2) 122.5=0t+(1)/(2)xx9.8xxt_(1)^(2)` `THEREFORE t_(1)^(2)=(245)/(9.8)=(2450)/(98)=25 therefore t_(1)=5s therefore t_(2)=0.33s` Velocity of sound = `therefore t_(1)^(2)=(245)/(9.8)=(2450)/(98)=25 therefore t_(1)=5s therefore t_(2)=0.33s` |
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| 43. |
A ball of mass 'm', moving with uniform speed, collides elastically with another stationary ball. The incident ball will lose maximum kinetic energy when the mass of the stationary ball is |
| Answer» Answer :A | |
| 44. |
Polar satellites go round the poles of earth in |
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Answer» South-east DIRECTION |
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| 45. |
A flat horizontal belt is running at a constant speed V. There is a uniform solid cylinder of mass M which can rotate freely about an axle passing through its centre and parallel to its length. Holding the axle parallel to the width of the belt, the cylinder is lowered on to the belt. The cylinder begins to rotate about its axle and eventually stops slipping. The cylinder is, however, not allowed to move forward by keeping its axle fixed. Assume that the moment of inertia of the cylinder about itsaxle is (1)/(2) MR^(2) where M is its mass and R itsradius and also assume that the belt continues to move at constant speed. No vertical force is applied on the axle of the cylinder while holding it. (a) Calculate the extra power that the motor driving the belt has to spend while the cylinder gains rotational speed. Assume coefficient of friction = mu. (b) Prove that 50% of the extra work done by the motor after the cylinder is placed over it, is dissipated as heat due to friction between the belt and the cylinder |
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Answer» <P> |
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| 46. |
A longsighted person has a minimum distance of distinct vision of 50 cm. He wants to reduce it to 25 cm. He should use a |
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Answer» CONCAVE LENS of FOCAL LENGTH 50cm |
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| 47. |
Politics is the art of possible'. Similarly, Science is the artof the soluble'. Explain this beautiful aphorism on the nature and parcitce of science. |
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Answer» SOLUTION :It is well that to win over votes, politicians would make anything and everything possible even when they are least sure of the same. That is why we say that politics is the ART of possible. The statement that science is the art of the soluble implies that a wide veriety of physical phenomena are understood in terms of only a few BASIC concepts, i.e., there appears to be unity in diversity as if widley different phenomena are soluble and can be EXPLAINED in terms of only a few fundamental LAWS. |
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| 48. |
In Column-I devices and in Column-II efficiency are given. Match them appropriately: |
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Answer» |
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| 49. |
An orbiting satellite will escape if (a) its speed increases by 41.4% (b) its speed in the orbit is made sqrt(1.414) times that of its initial value (c) its kinetic energy increases by 100% (d) its kinetic energy is made sqrt(2) times that of its initial value |
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Answer» only a & B are correct |
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| 50. |
A wave travelling in the +ve X-direction having maximum displacement along Y-direction as 1 m, wavelength 2pim and frequency 1/pi and initial phase zero is represented by ..... |
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Answer» `y = sin (2pi x 0 2pit )` `therefore y = a sin (2pi F t -(2pi)/( LAMDA) x + phi)` `therefore y = 1 XX sin (2pi xx (1)/(pi) xx t - (2pi)/(2pi) x + 0)` `therefore y = sin (2t -x)` |
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