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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 37451. |
A satellite is revolving very close to a planet of density D.The time period of revolution of that planetis |
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Answer» `SQRT((3PI)/(DG))` |
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| 37452. |
The ratio of distances travelled by a body starting from rest with constant acceleration in g^(th) and 8^(th) second is ......... |
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Answer» `17/15` `X _(n) = v _(0)+ a/2 (2n -1)` `therefore` Distance coverd in `9^(th)` second, `x _(9) =(a)/(2) (2 (9) -1) = (17a)/(2) (because v _(0) =0)` Distance covered in `8^(th)` second, `x _(8) = a/2(2 (8) -1) = (15a )/(2)` `therefore (x _(9))/(x _(8)) = (17a)/(2) xx (2)/(15 a)= (17)/(15)` |
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| 37453. |
A mercury thermometer contains 2c.c. of Hg. at 0°C. Distance between 0°C and 100°C marks on the stem is 35cm and diameter of the bore is 0.02cm. gamma_(A) of liquid is |
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Answer» 0.000055/°C |
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| 37454. |
A man is standing in the middle of a perfectly smooth ‘island of ice’ where there is no friction between the ground and his feet. Under these circumstances |
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Answer» he can reach the DESIRED corner by throwing any object in the same DIRECTION |
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| 37455. |
One end of a uniform glass capillary tube of radius r = 0.025 cm is immersed vertically in water to a depth h = 1 cm. The excess pressure in N/m^2 required to blow an air bubble out of the tube is : [Surface tension of water = 7 xx 10^(-2) Nm, Density of water = 10^3 kg m^(-3) , Acceleration due to gravity = 10ms^(-2)] |
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Answer» `0.0048 XX 10^5` |
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| 37456. |
Find the acceleration of a particle placed on the surface of the earth at the equator, due to the earth rotation. The radius of earth is 6400 km and time period of revolution of the earth about its axis is 24 h. |
| Answer» SOLUTION :CENTRIPETAL FORCE is MAXIMUM at the EQUATOR | |
| 37457. |
Which of the following functions of time represent (a) periodic and (b) non-periodic motion? Give the period for each case of periodic motion [omega is any positive constant]. (i) sinomegat+cosomegat (ii) sinomegat+cos2omegat+sin4omegat (iii) e^(-omegat) (iv) log(omegat) |
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Answer» Solution :(i) sin `omegat + COS omegat` is a periodic function, it can ALSO be written as `sqrt2 sin (omegat + pi/4)`. Now `sqrt2 sin (omegat + pi//4)= 2 sin (omegat + pi//4+2pi)` `=sqrt2sin[omega(t+2pi//omega)+pi//4]` The periodic time of the function is `2pi//omega`. (ii) This is an EXAMPLE of a periodic motion. It can be noted that each term represents a periodic function with a different angular FREQUENCY. Since period is the least interval of time after which a function repeats its value, `sinomegat` has a period `T_(0)= 2pi//omega , cos 2 omegat` has a period `pi//omega =T_(0)//2`, and `sin 4omegat` has a period `2pi//4omega = T_(0)//4`. The period of the first term is a MULTIPLE of the periods of the last two terms. Therefore, the smallest interval of time after which the sum of the three terms repeats is `T_(0)`, and thus, the sum is a periodic function with a period `2pi//omega`. (iii) The function e–wt is not periodic, it decreases monotonically with increasing time and tends to zero as `t to oo` and thus, never repeats its value. (iv) The function `log(omegat)` increases monotonically with time t. It, therefore, never repeats its value and is a nonperiodic function. It may be noted that as `t to oo, log(omegat)` diverges to `oo`. It, therefore, cannot represent any kind of physical displacement. |
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| 37458. |
Assertion : Vibratory motion is sometimes called oscillatory motion.Reason : Spinning of the Earth about its axis .Choose the correct statement from the following : |
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Answer» Assertion is FALSE and reason is true. |
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| 37459. |
(A) : When range of a projectile is maximum, its angle of projection may be 45^(@) or 135^(@). ( R) : Whether theta is 45^(@) or 135^(@) value of range remains the same, only the sign changes. |
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Answer» Both (A) and ( R) are TURE and ( R) is the correct EXPLANATION of (A) |
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| 37460. |
A bob tied to an ideal string of length l is released from the horizonatal position shown. A peg P whose height is adjustable, can arrest the free swing of the pendulum, as shown in figure {:(,"Column-1",,"Column-2"),("(A)",underset("remaining taut throughout the swing")"For what range of y will the string wind up on the peg",,(P) (2l)/5 lt y lt (2l)/3),("(B)",underset("pendulum become projectile")"For what range of y will the",,(Q)0 lt y lt (2l)/5),("(C)",underset("mechanical energy always remain conserved")"For what value of y will",,(R)(2l)/5 lt y lt l),(,,,(S) l/3 lt y lt (2l)/3):} |
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Answer» <P> |
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| 37461. |
A lift is ascending with a constant speed ''V''. A passenger in the lift drops a coin. The acceleration of the coin towards the floor will be |
| Answer» ANSWER :B | |
| 37462. |
Mention few illustrationfor damped oscillations. |
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Answer» SOLUTION :(i) The OSCILLATIONS of a PENDULUM. (ii) Electromagnetic DAMPING in galvanometer (oscillations of a coil in galvanometer). (iii) Electromagnetic oscillations in tank circuit. |
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| 37463. |
A man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weighs 1,000 N exerts a force of 450 N on the chair downwards while pulling the rope on the other side. If the chair weighs 250N, then the acceleration of the chair is |
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Answer» `0.45m//s^(2)` For CHAIR : `T-F-W_(C)=m_(c)a` |
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| 37464. |
In an open pipe when air column is 20cm it is in resonance with tuning fork A. When length is increased by 2cmthen the air column is tn resonance with fork B. When A and B are sounded together 4 beats/sec are heard. Frequencies of A and B are respectively (in Hz) |
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Answer» `40,44` |
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| 37465. |
The rectangular surface of area 8cm xx 4cm of a black body at a temperature of 127^(@)C emits energy at rate of E per second. If the length and breadth are reduced to half of its initial value and the temperature is raised to 327^(@)C, the rate of emission of energy will be |
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Answer» `((3E)/(8))` |
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| 37466. |
What is meant by Unit? What is the importance of a unit? |
| Answer» Solution :UNIT is a standard measure of a PHYSICAL QUANTITY . From unit, nature of the quantityis REVEALED. | |
| 37467. |
In the following figure which of the statements are correct? |
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Answer» the sign of x-component of `vecl_1` is positive and that of `vecl_2` is NEGATIVE |
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| 37468. |
Which one of the following statem ents is incorrect? |
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Answer» rolling fricition is smaller than slidingfriction |
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| 37469. |
A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross sectional area is 4.9 times 10^-7 m^2. IF the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad.s^-1. If the young's modulus of the material of the wire is n times 10^9 N.m^-2 Calculate the value of n. |
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Answer» Solution :YOUNG modules of the material of the WIRE, `Y=(F/A)/((DeltaA)/L)=(FL)/(A DeltaL) or,F=((YA)/L)DeltaL`......(1) IF mass m is pulled by a length `DeltaL` then restoring FORCE developed in the wire, `F=k DeltaL`…….(2) Comparing equation (1) and (2) we GET , `k=(YA)/L` Angular frequency , `omega=sqrt(K/m)=sqrt((YA)/(ML))` or,`140=sqrt((ntimes10^9times4.9times10^-7)/(0.1times1))=70sqrtn` `therefore sqrtn=2 or n=4` |
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| 37470. |
The mass of a box measured by a grocer's balance is 2.3 kg. Two gold pieces 20.15 g and 20.17 g are added to the box. (i) What is the total mass of the box? (ii) The difference in masses of the pieces to correct significant figures. |
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Answer» Solution :(i) MASS of box = 2.3 kg Mass of GOLD pieces = 20.15 + 20.17 = 40.32 g = 0.04032 kg TOTAL mass = 2.3 + 0.04032 = 2.34032 kg In correct significant figure mass = 2.3 kg (as least decimal) (II) Difference in mass of gold pieces = 0.02 g In correct significant figure (2 significant FIG, minimum decimal) will be 0.02 g |
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| 37471. |
A physics textbook of mass m rests flat on a horizontal table of mass M placed on the ground Let N_(a-b) be the contact Newton's 3 rd law, which of the following is an action reaction pair of forces? |
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Answer» mg and `N_("table" RARR "book")` |
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| 37472. |
A heavy brass sphere is hung from a weightless inelastic spring and as a simple pendulum its time period of oscillation is T. When the sphere is immersed in a non-viscous liquid of density 1/10 that of brass, find the time period of oscillation of the pendulum. |
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Answer» |
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| 37473. |
The frequency of a pendulum whose normal period 2s, when it is in an elevator in free fall, then the frequency will be |
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Answer» zero |
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| 37474. |
How much radius of earth at equator is grater than the radius at poles of earth |
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Answer» |
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| 37475. |
The temperature of an elactric bulb changes from 2000K to 3000Kdue to a.c. voltage fluctuations. Calculate the percentage rise in electric power consumed ? |
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Answer» <P> |
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| 37476. |
A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15ms^(-1) Neglecting air resistance, find the time taken by the stone to reach the ground and the speed with which it hits the ground? (g=9.8ms^(-2)) |
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Answer» Solution :TIME =10 SECONDS `V=sqrt(V_(s)^(2)+V_(y)^(2))=sqrt(15^(2)+98^(2))=99.1ms^(-1)` |
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| 37477. |
The vertical extension in a light spring by a weight of 1 kg, in equilibrium is 9.8 cm. The period of oscillation of the spring, in seconds, will be……………. |
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Answer» `(2PI)/(10)` |
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| 37478. |
An ideal gas passesfrom one equllibriumstate (P_(1) ,V_(1), T_(1), N)to anoterequilibriumstate (2P_(1) , 3V_(1), T_(2),N) Then |
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Answer» `T_(1)=T_(2)` One equilibrium STATE `(P_(1),V_(1),T_(1),N)` Another equilibrium state `(P_(2), V_(2),T_(2),N)` `P_(2)=2P_(1) and V_(2)=3V_(1)` `(P_(1)V_(1))/(P_(2)V_(2))=(T_(1))/(T_(2))` `(P_(1)V_(1))/(P_(2)V_(2))=(T_(1))/(T_(2))` `(P_(1)V_(1))/((2P_(1))(3V_(1)))=(T_(1))/(T_(2)) RARR (1)/(6)=(T_(1))/(T_(2))` `THEREFORE""T_(1)=(T_(2))/(6)` |
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| 37479. |
Two bulbs have filaments of lengths, Emissi-vities and diameters in the ratio of 2 : 1. If ratio of their powers is 1 : 2 then, ratio of their temperatures is |
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Answer» `1:1` |
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| 37480. |
The Poission.s ratio (sigma) should satisfy the relation |
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Answer» `-1 LT SIGMA lt 0.5` |
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| 37481. |
A body can have |
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Answer» ZERO MOMENTUM and finite kinetic ENERGY |
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| 37482. |
Is centre of gravity of a body to be many? |
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Answer» |
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| 37483. |
The square of the resultant of two forces 4 N and 3 N exceeds the square of the resultant of the two forces by 12 when they are mutually perpendicular. The angle between the vectors is |
| Answer» ANSWER :B | |
| 37484. |
For any arbitrary motion in space , which of the following relations aretrue : (a) V_("average")=(1//2)(v(t_(1))+v(t_(2)))(b)V_("average")=[r(t_(2))-r(t_(1))]//(t_(2)-t_(1)) (c )V(t)=V(0)+at(d)r(t) =r(0)+v(0)T+(1//2)at^(2) (e )a_("average")=[v(t_(2))-v(t_(1))]//(t_(2)-t_(1))(The average stands for average of the quantity over the time interval t_(1)" to "t_(2)) |
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Answer» |
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| 37485. |
When a rigid body is rotating, every point in it describes a circular path |
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Answer» Only if the axis of ROTATION PASSES symmetrically through the centre of GRAVITY of the body |
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| 37486. |
A 10 kg wooden block is lying stationary on rough horizontal surface. To pull this block 49 N force is req u ired . Find value of co-efficient of friction and angle of friction. |
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Answer» SOLUTION :`MU = tantheta` `THETA =tan^(-1)(mu)` `=tan^(-1)(0.5)` `26^(@)34` |
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| 37487. |
A particle execute SHM from the mean positon. Its amplitude is A, its time period is T. At what displacement its speed is half of |
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Answer» `(sqrt(3)A)/2` |
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| 37488. |
The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of paesecs ? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun ? |
| Answer» SOLUTION :1.32 PARSEC, 2.64" (SECOND of ARC) | |
| 37489. |
(A) Clouds appear white due to scattering of light ( R) When size of objects are large compared to wavelength, all wavelengths scatter equally. |
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Answer» Both A and R are TRUE and R is the CORRECT explanation of A |
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| 37490. |
Radio and television are based on………….. |
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Answer» |
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| 37491. |
A body travels uniformly a distance of (13.8 pm 0.2)m in a time (4.0 pm 0.3)s. Determine velocity of the body within error limits. |
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Answer» Solution :Here `s=(13.8pm0.2)m` and `t=(4.0 pm 0.3)SEC` Expressing it in percentage error, we have, `s=13.8 pm (0.2)/(13.8)xx100%` `=13.8pm 1.4%` and `t=4.0pm (0.3)/(4)xx100%` `=4PM 7.5%` `THEREFORE V=(s)/(t)=(13.8 pm 1.4)/(4pm 7.5)` `=(3.45pm0.18)m//s` |
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| 37492. |
One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1Å). Why is this ratio so large ? |
| Answer» Solution :`~= 10^(4)` , intermolecular SEPARATION in a GAS is much larger than the SIZE of a molecule. | |
| 37493. |
On loading a metal wire of cross section 10^(-6) m^2 and length 2m by a mass of 210 kg it extends by 16mm and suddenly broke from the point of support. If density of that metal is 8000 kgm^(-3)and its specific heat is 420 Jkg^(-1)K^(-1) the rise in temperature of wire is |
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Answer» `2.5^@C` |
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| 37494. |
A body weighs 500 N on the surface of the earth. How much would it weight half way below the surface of the earth ? |
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Answer» 100 N |
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| 37495. |
A log of wood of length l and mass M is floating on the surface of a river perpendicular to the banks. One end of the log touches the banks. A man of mass m standing at the other end walks towards the bank. Calculate the displacement of the log when he reaches nearer end of the log |
Answer» Solution :Let `PQ` be the log of wood. As there is no external FORCE, the CENTRE of MASS of man and the log system remains at rest. Let the bank of the river be the origin `A`. Initially, the man is at point `Q`.  Let `m=`mass of man `M=`mass of log `x=`displacement of log w.r.t GROUND (here water) `X_(CM)` (initial) `=(m(L)+m(l/2))/(m+M)` `X_(CM)`(final)`=(m(x)+M(l//2+x))/(m+M)` Now `X_(CM)`(initial) `=X_(CM)` (final) `implies ml+(Ml)/2=mx+(Ml)/2+Mx` `impliesx=(ml)/(m+M)` Hence the log moves away from the bank through a distance of `(ml)/((m+M))` Alternative method: Displacement of the log `=/_\x_(1)=x` Displacement of the man `=/_\x_(2)=l-x` Apply `m_(1)x_(1)=m_(2)x_(2)implies`(if the centre of mass remains at the same place) `impliesMx=m(l-x)impliesx(ml)/((m+M))` |
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| 37496. |
About floaring in a water tank is carrying a number of large stones. If the stories were unloaded into water what will happen to water level? |
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Answer» SOLUTION :The water level will come down. This is because when the stones were INSIDE the boat, the volume of water displaced by the boat and the stones. `=("TOTAL mass of stones and boat")/ ("density of water")` But when UNLOADED, the volume of water displaced by stones `("mass of stones")/(" density of stones")` which is less than that in the previous CASE, as density of stone is greater than that of water. |
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| 37497. |
A satellite is revolving in a circular orbit of roundthe earth at a height of h. If the acceleration due to gravity at that height is g. The orbital speed of the satellite . |
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Answer» `SQRT(GH)` |
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| 37498. |
A liquid of density rho flows along a horizontal pipe of uniform cross - section A with a velocity v through a right angled bend as shown in Fig. What force has to be exerted at the bend to hold the pipe in equilibrium ? |
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Answer» <P> Solution :Change in MOMENTUM of mass `Deltam` of LIQUID as it passes through the bend `dP=P_(f)-P_(i)=sqrt2d Delta mv` `F=(dP)/(dt)=(sqrt2)v(dm)/(dt),["as dm "=rho A DL]` `F=sqrt2v((rho. AdL))/(dt),["as "dL//dt=v]` `F=sqrt2rhoA v^(2)` |
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| 37500. |
1 g of steam at 100^(@)C is passed into a calorimeter of water equivalent 10g containing water of mass 90 g. Find the resultant temperature of the mixture if the initial temperature of water is 27^(@)C. |
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Answer» `36.2^(@) C` |
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